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# Solve for $x$ implicitly when $x(0)=3$.
Let $$\frac{dx}{dt}=\frac{1}{16}x(4-x)^2.$$ Solve for $$x$$ implicitly when $$x(0)=3$$.
Can anyone explain what this means or how to solve it. Thank you.
Implicitly, you have already described what $$x$$ is. I think you seek an explicit solution. If you separate your variables, you get $$\frac{16\,dx}{x(4-x)^2}=dt$$ Partial fraction decomposition gives (steps left as a learning exercise) $$\frac{16}{x(4-x)^2}=- \frac{1}{x - 4} + \frac{4}{\left(x - 4\right)^{2}} + \frac{1}{x}$$ Then integrating these terms with respect to $$x$$ is simple (just $$u$$-sub with power rule and $$\ln$$), and integrating $$dt$$ is trivial. Solve for the constant of integration by subbing 0 for $$t$$ and 3 for $$x$$ since $$x(0)=3$$. I leave the details for you to do as a learning exercise.
The complete solution for $$\frac{dx}{dt}=\frac{1}{16}x(4-x)^2$$ with $$x(0) = 3$$ is seen by the following.
Using $$\frac{16}{x (x-4)^2} = \frac{4}{(x-4)^2} - \frac{1}{x-4} + \frac{1}{x} = \frac{d}{dx} \left( 1 + \ln\left(\frac{x}{x-4} \, e^{-x/(x-4)} \right) \right)$$ and $$y \, e^{y} = q$$ has the Lambert W-function solution $$y = W(q)$$, then:
\begin{align} \frac{dx}{dt} &= \frac{x(x-4)^2}{16} \\ \frac{16 \, dx}{x(x-4)^2} &= dt \\ \int \frac{16 \, dx}{x(x-4)^2} &= t + c_{0} \\ \ln\left(\frac{x}{x-4} \, e^{-x/(x-4)} \right) &= t - 1 + c_{0} \\ \frac{x}{x-4} \, e^{-x/(x-4)} &= c_{1} \, e^{t-1}. \end{align} Now use $$x(0) = 3$$ to obtain $$c_{1} = - 3 \, e^{4}$$ and \begin{align} - \frac{x}{x-4} \, e^{-x/(x-4)} &= 3 \, e^{t + 3} \\ - \frac{x}{x-4} &= W(3 \, e^{t+3}) \end{align} or $$x(t) = \frac{4 \, W(3 \, e^{t+3})}{1 + W(3 \, e^{t+3})}.$$
To check the boundary condition use $$W(x \, e^{x}) = x$$. This is used in the following way: $$x(0) = \frac{4 \, W(3 \, e^{3})}{1 + W(3 \, e^{3})} = \frac{4 \cdot 3}{1 + 3} = 3$$ as required. |
# 7 8 9 Multiplication Chart
Understanding multiplication right after counting, addition, as well as subtraction is ideal. Kids understand arithmetic using a natural progression. This progression of discovering arithmetic is truly the adhering to: counting, addition, subtraction, multiplication, and lastly department. This statement leads to the question why learn arithmetic with this pattern? More importantly, why understand multiplication right after counting, addition, and subtraction just before section?
## These information respond to these queries:
1. Youngsters understand counting initial by associating graphic objects with their fingers. A real instance: How many apples exist in the basket? More abstract instance is just how older are you?
2. From counting amounts, another plausible stage is addition combined with subtraction. Addition and subtraction tables are often very useful instructing helps for kids because they are graphic instruments making the changeover from counting easier.
3. Which should be learned after that, multiplication or division? Multiplication is shorthand for addition. At this moment, young children possess a firm understanding of addition. As a result, multiplication is definitely the up coming logical kind of arithmetic to discover.
## Evaluate basics of multiplication. Also, evaluate the basic principles utilizing a multiplication table.
Allow us to review a multiplication example. Employing a Multiplication Table, flourish several times 3 and obtain an answer twelve: 4 x 3 = 12. The intersection of row three and column four of your Multiplication Table is 12; 12 is definitely the respond to. For kids starting to find out multiplication, this really is effortless. They can use addition to resolve the problem thus affirming that multiplication is shorthand for addition. Illustration: 4 x 3 = 4 4 4 = 12. It is really an superb guide to the Multiplication Table. A further advantage, the Multiplication Table is graphic and displays returning to discovering addition.
## Exactly where can we begin studying multiplication while using Multiplication Table?
1. First, get knowledgeable about the table.
2. Start with multiplying by one particular. Start off at row number 1. Go on to line primary. The intersection of row one particular and line one is the solution: a single.
3. Perform repeatedly these techniques for multiplying by one. Flourish row one particular by posts one particular by means of twelve. The replies are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Perform repeatedly these steps for multiplying by two. Increase row two by posts one particular by means of several. The answers are 2, 4, 6, 8, and 10 respectively.
5. Allow us to hop in advance. Replicate these techniques for multiplying by 5. Flourish row several by posts one via twelve. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now allow us to boost the amount of issues. Repeat these actions for multiplying by 3. Flourish row a few by posts one particular via twelve. The responses are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. In case you are at ease with multiplication to date, try out a analyze. Fix the subsequent multiplication issues in your mind and after that compare your answers to the Multiplication Table: increase half a dozen as well as two, multiply nine and 3, multiply one and 11, flourish 4 and four, and grow seven as well as two. The issue answers are 12, 27, 11, 16, and 14 correspondingly.
When you got a number of out of 5 issues right, build your individual multiplication exams. Estimate the answers in your thoughts, and look them making use of the Multiplication Table. |
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# 5.5: Averages and Probability (Part 2)
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### Use the Basic Definition of Probability
The probability of an event tells us how likely that event is to occur. We usually write probabilities as fractions or decimals. For example, picture a fruit bowl that contains five pieces of fruit - three bananas and two apples.
If you want to choose one piece of fruit to eat for a snack and don’t care what it is, there is a $$\frac{3}{5}$$ probability you will choose a banana, because there are three bananas out of the total of five pieces of fruit. The probability of an event is the number of favorable outcomes divided by the total number of outcomes.
Definition: Probability
The probability of an event is the number of favorable outcomes divided by the total number of outcomes possible.
$$Probability = \frac{number\; of\; favorable\; outcomes}{total\; number\; of\; outcomes}$$
Converting the fraction $$\frac{3}{5}$$ to a decimal, we would say there is a 0.6 probability of choosing a banana.
Probability of choosing a banana = $$\frac{3}{5}$$
Probability of choosing a banana = 0.6
This basic definition of probability assumes that all the outcomes are equally likely to occur. If you study probabilities in a later math class, you’ll learn about several other ways to calculate probabilities.
Example 5.56:
The ski club is holding a raffle to raise money. They sold 100 tickets. All of the tickets are placed in a jar. One ticket will be pulled out of the jar at random, and the winner will receive a prize. Cherie bought one raffle ticket. (a) Find the probability she will win the prize. (b) Convert the fraction to a decimal.
##### Solution
(a)
What are you asked to find? The probability Cherie wins the prize. What is the number of favorable outcomes? 1, because Cherie has 1 ticket. Use the definition of probability. $$Probability\; of\; an\; event = \frac{number\; of\; favorable\; outcomes}{total\; number\; of\; outcomes}$$ Substitute into the numerator and denominator. Probability Cherie wins = $$\frac{1}{100}$$
(b)
Write the probability as a fraction. Probability = $$\frac{1}{100}$$ Convert the fraction to a decimal. Probability = 0.01
Exercise 5.111:
Ignaly is attending a fashion show where the guests are seated at tables of ten. One guest from each table will be selected at random to receive a door prize. (a) Find the probability Ignaly will win the door prize for her table. (b) Convert the fraction to a decimal.
Exercise 5.112:
Hoang is among 20 people available to sit on a jury. One person will be chosen at random from the 20. (a) Find the probability Hoang will be chosen. (b) Convert the fraction to a decimal.
Example 5.57:
Three women and five men interviewed for a job. One of the candidates will be offered the job. (a) Find the probability the job is offered to a woman. (b) Convert the fraction to a decimal.
##### Solution
What are you asked to find? The probability the job is offered to a woman. What is the number of favorable outcomes? 3, because there are three women. What are the total number of outcomes? 8, because 8 people interviewed. Use the definition of probability. $$Probability\; of\; an\; event = \frac{number\; of\; favorable\; outcomes}{total\; number\; of\; outcomes}$$ Substitute into the numerator and denominator. Probability = $$\frac{3}{8}$$
(b)
Write the probability as a fraction. Probability = $$\frac{3}{8}$$ Convert the fraction to a decimal. Probability = 0.375
Exercise 5.113:
A bowl of Halloween candy contains 5 chocolate candies and 3 lemon candies. Tanya will choose one piece of candy at random.(a) Find the probability Tanya will choose a chocolate candy.(b) Convert the fraction to a decimal.
Exercise 5.114:
Dan has 2 pairs of black socks and 6 pairs of blue socks. He will choose one pair at random to wear tomorrow. (a) Find the probability Dan will choose a pair of black socks (b) Convert the fraction to a decimal.
### Practice Makes Perfect
#### Calculate the Mean of a Set of Numbers
In the following exercises, find the mean.
1. 3, 8, 2, 2, 5
2. 6, 1, 9, 3, 4, 7
3. 65, 13, 48, 32, 19, 33
4. 34, 45, 29, 61, and 41
5. 202, 241, 265, 274
6. 525, 532, 558, 574
7. 12.45, 12.99, 10.50, 11.25, 9.99, 12.72
8. 28.8, 32.9, 32.5, 27.9, 30.4, 32.5, 31.6, 32.7
9. Four girls leaving a mall were asked how much money they had just spent. The amounts were $0,$14.95, $35.25, and$25.16. Find the mean amount of money spent.
10. Juan bought 5 shirts to wear to his new job. The costs of the shirts were $32.95,$38.50, $30.00,$17.45, and $24.25. Find the mean cost. 11. The number of minutes it took Jim to ride his bike to school for each of the past six days was 21, 18, 16, 19, 24, and 19. Find the mean number of minutes. 12. Norris bought six books for his classes this semester. The costs of the books were$74.28, $120.95,$52.40, $10.59,$35.89, and $59.24. Find the mean cost. 13. The top eight hitters in a softball league have batting averages of .373, .360, .321, .321, .320, .312, .311, and .311. Find the mean of the batting averages. Round your answer to the nearest thousandth. 14. The monthly snowfall at a ski resort over a six-month period was 60.3, 79.7, 50.9, 28.0, 47.4, and 46.1 inches. Find the mean snowfall. #### Find the Median of a Set of Numbers In the following exercises, find the median. 1. 24, 19, 18, 29, 21 2. 48, 51, 46, 42, 50 3. 65, 56, 35, 34, 44, 39, 55, 52, 45 4. 121, 115, 135, 109, 136, 147, 127, 119, 110 5. 4, 8, 1, 5, 14, 3, 1, 12 6. 3, 9, 2, 6, 20, 3, 3, 10 7. 99.2, 101.9, 98.6, 99.5, 100.8, 99.8 8. 28.8, 32.9, 32.5, 27.9, 30.4, 32.5, 31.6, 32.7 9. Last week Ray recorded how much he spent for lunch each workday. He spent$6.50, $7.25,$4.90, $5.30, and$12.00 . Find the median.
10. Michaela is in charge of 6 two-year olds at a daycare center. Their ages, in months, are 25, 24, 28, 32, 29, and 31. Find the median age.
11. Brian is teaching a swim class for 6 three-year olds. Their ages, in months, are 38, 41, 45, 36, 40, and 42. Find the median age.
12. Sal recorded the amount he spent for gas each week for the past 8 weeks. The amounts were $38.65,$32.18, $40.23,$51.50, $43.68,$30.96, $41.37, and$44.72. Find the median amount.
#### Identify the Mode of a Set of Numbers
In the following exercises, identify the mode.
1. 2, 5, 1, 5, 2, 1, 2, 3, 2, 3, 1
2. 8, 5, 1, 3, 7, 1, 1, 7, 1, 8, 7
3. 18, 22, 17, 20, 19, 20, 22, 19, 29, 18, 23, 25, 22, 24, 23, 22, 18, 20, 22, 20
4. 42, 28, 32, 35, 24, 32, 48, 32, 32, 24, 35, 28, 30, 35, 45, 32, 28, 32, 42, 42, 30
5. The number of children per house on one block: 1, 4, 2, 3, 3, 2, 6, 2, 4, 2, 0, 3, 0.
6. The number of movies watched each month last year: 2, 0, 3, 0, 0, 8, 6, 5, 0, 1, 2, 3.
7. The number of units being taken by students in one class: 12, 5, 11, 10, 10, 11, 5, 11, 11, 11, 10, 12.
8. The number of hours of sleep per night for the past two weeks: 8, 5, 7, 8, 8, 6, 6, 6, 6, 9, 7, 8, 8, 8.
#### Use the Basic Definition of Probability
In the following exercises, express the probability as both a fraction and a decimal. (Round to three decimal places, if necessary.)
1. Josue is in a book club with 20 members. One member is chosen at random each month to select the next month’s book. Find the probability that Josue will be chosen next month.
2. Jessica is one of eight kindergarten teachers at Mandela Elementary School. One of the kindergarten teachers will be selected at random to attend a summer workshop. Find the probability that Jessica will be selected.
3. There are 24 people who work in Dane’s department. Next week, one person will be selected at random to bring in doughnuts. Find the probability that Dane will be selected. Round your answer to the nearest thousandth.
4. Monica has two strawberry yogurts and six banana yogurts in her refrigerator. She will choose one yogurt at random to take to work. Find the probability Monica will choose a strawberry yogurt.
5. Michel has four rock CDs and six country CDs in his car. He will pick one CD to play on his way to work. Find the probability Michel will pick a rock CD.
6. Noah is planning his summer camping trip. He can’t decide among six campgrounds at the beach and twelve campgrounds in the mountains, so he will choose one campground at random. Find the probability that Noah will choose a campground at the beach.
7. Donovan is considering transferring to a 4-year college. He is considering 10 out-of state colleges and 4 colleges in his state. He will choose one college at random to visit during spring break. Find the probability that Donovan will choose an out-of-state college.
8. There are 258,890,850 number combinations possible in the Mega Millions lottery. One winning jackpot ticket will be chosen at random. Brent chooses his favorite number combination and buys one ticket. Find the probability Brent will win the jackpot. Round the decimal to the first digit that is not zero, then write the name of the decimal.
### Everyday Math
1. Joaquin gets paid every Friday. His paychecks for the past 8 Fridays were $315,$236.25, $236.25,$236.25 $315,$315, $236.25,$393.75. Find the (a) mean (b) median, and (c) mode.
2. The cash register receipts each day last week at a coffee shop were $1,845,$1,520, $1,438,$1,682, $1,850,$2,721, \$2,539. Find the (a) mean, (b) median, and (c) mode.
### Writing Exercises
1. Explain in your own words the difference between the mean, median, and mode of a set of numbers.
2. Make an example of probability that relates to your life. Write your answer as a fraction and explain what the numerator and denominator represent.
### Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) After looking at the checklist, do you think you are well prepared for the next section? Why or why not? |
I started reviewing a book draft that mentions the adjoint in passing, but I’ve forgotten what I knew about the adjoint (not counting self-adjoint operators, which is different.) I do recall that adjoint matrices were covered in high school linear algebra (now 30+ years ago!), but never really used after that.
It appears that the basic property of the adjoint $$A$$ of a matrix $$M$$, when it exists, is
M A = \Abs{M} I,
so it’s proportional to the inverse, where the numerical factor is the determinant of that matrix. Let’s try to compute this beastie for 1D, 2D, and 3D cases.
## Simplest case: $$1 \times 1$$ matrix.
For a one by one matrix, say
M =
\begin{bmatrix}
m_{11}
\end{bmatrix},
the determinant is just $$\Abs{M} = m_11$$, so our adjoint is the identity matrix
A =
\begin{bmatrix}
1
\end{bmatrix}.
Not too interesting. Let’s try the 2D case.
## Less trivial case: $$2 \times 2$$ matrix.
For the 2D case, let’s define our matrix as a pair of column vectors
M =
\begin{bmatrix}
\Bm_1 & \Bm_2
\end{bmatrix},
and let’s write the adjoint out in full in coordinates as
A =
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}.
We seek solutions to a pair of vector equations
\begin{aligned}
\Bm_1 a_{11} + \Bm_2 a_{21} &= \Abs{M} \Be_1 \\
\Bm_1 a_{12} + \Bm_2 a_{22} &= \Abs{M} \Be_2.
\end{aligned}
We can immediately solve either of these, by taking wedge products, yielding
\begin{aligned}
\lr{ \Bm_1 \wedge \Bm_2 } a_{11} + \lr{ \Bm_2 \wedge \Bm_2 } a_{21} &= \Abs{M} \lr{ \Be_1 \wedge \Bm_2 } \\
\lr{ \Bm_1 \wedge \Bm_1 } a_{11} + \lr{ \Bm_1 \wedge \Bm_2 } a_{21} &= \Abs{M} \lr{ \Bm_1 \wedge \Be_1 } \\
\lr{ \Bm_1 \wedge \Bm_2 } a_{12} + \lr{ \Bm_2 \wedge \Bm_2 } a_{22} &= \Abs{M} \lr{ \Be_2 \wedge \Bm_2 } \\
\lr{ \Bm_1 \wedge \Bm_1 } a_{12} + \lr{ \Bm_1 \wedge \Bm_2 } a_{22} &= \Abs{M} \lr{ \Bm_1 \wedge \Be_2}.
\end{aligned}
Any wedge with a repeated vector is zero.
Provided the determinant is non-zero, we can divide both sides by $$\Bm_1 \wedge \Bm_2 = \Abs{M} \Be_{12}$$ to find a single determinant for each element in the adjoint
\begin{aligned}
a_{11} &= \begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} \\
a_{21} &= \begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} \\
a_{12} &= \begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} \\
a_{22} &= \begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix}
\end{aligned}
or
A =
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} \\
& \\
\begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix}
\end{bmatrix},
or
A_{ij} =
\epsilon_{ir}
\begin{vmatrix}
\Be_j & \Bm_r
\end{vmatrix},
where $$\epsilon_{ir}$$ is the completely antisymmetric tensor, and the Einstein summation convention is in effect (summation implied over any repeated indexes.)
### Check:
We should verify that expanding these determinants explicitly reproduces the usual representation of the 2D adjoint:
\begin{aligned}
\begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} &= \begin{vmatrix} 1 & m_{12} \\ 0 & m_{22} \end{vmatrix} = m_{22} \\
\begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} &= \begin{vmatrix} m_{11} & 1 \\ m_{21} & 0 \end{vmatrix} = -m_{21} \\
\begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} &= \begin{vmatrix} 0 & m_{12} \\ 1 & m_{22} \end{vmatrix} = -m_{12} \\
\begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix} &= \begin{vmatrix} m_{11} & 0 \\ m_{21} & 1 \end{vmatrix} = m_{11},
\end{aligned}
or
A =
\begin{bmatrix}
m_{22} & -m_{12} \\
-m_{21} & m_{11}
\end{bmatrix}.
Multiplying everything out should give us determinant weighted identity
\begin{aligned}
M A
&=
\begin{bmatrix}
m_{11} & m_{12} \\
m_{21} & m_{22}
\end{bmatrix}
\begin{bmatrix}
m_{22} & -m_{12} \\
-m_{21} & m_{11}
\end{bmatrix} \\
&=
\lr{ m_{11} m_{22} – m_{12} m_{21} }
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\
&= \Abs{M} I,
\end{aligned}
as expected.
## 3D case: $$3 \times 3$$ matrix.
For the 3D case, let’s also define our matrix as column vectors
M =
\begin{bmatrix}
\Bm_1 & \Bm_2 & \Bm_3
\end{bmatrix},
and let’s write the adjoint out in full in coordinates as
A =
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}.
This time, we seek solutions to three vector equations
\begin{aligned}
\Bm_1 a_{11} + \Bm_2 a_{21} + \Bm_3 a_{31} &= \Abs{M} \Be_1 \\
\Bm_1 a_{12} + \Bm_2 a_{22} + \Bm_3 a_{32} &= \Abs{M} \Be_2 \\
\Bm_1 a_{13} + \Bm_2 a_{23} + \Bm_3 a_{33} &= \Abs{M} \Be_3,
\end{aligned}
and can immediately solve, once again, by taking wedge products, yielding
\begin{aligned}
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{11} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{21} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{31} &= \Abs{M} \Be_1 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{11} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{21} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{31} &= \Abs{M} \Bm_1 \wedge \Be_1 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{11} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{21} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{31} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_1 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{12} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{22} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{32} &= \Abs{M} \Be_2 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{12} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{22} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{32} &= \Abs{M} \Bm_1 \wedge \Be_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{12} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{22} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{32} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_2 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{13} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{23} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{33} &= \Abs{M} \Be_3 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{13} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{23} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{33} &= \Abs{M} \Bm_1 \wedge \Be_3 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{13} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{23} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{33} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_3,
\end{aligned}
Any wedge with a repeated vector is zero.
Like before, provided the determinant is non-zero, we can divide both sides by $$\Bm_1 \wedge \Bm_2 \wedge \Bm_3 = \Abs{M} \Be_{123}$$ to find a single determinant for each element in the adjoint
\begin{aligned}
A &=
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_2 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Bm_1 & \Be_1 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_3 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Bm_1 & \Bm_2 & \Be_1 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Bm_2 & \Be_2 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Bm_2 & \Be_3 \end{vmatrix}
\end{bmatrix} \\
&=
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_2 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_3 & \Bm_1 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_1 & \Bm_2 \end{vmatrix}
\end{bmatrix},
\end{aligned}
or
A_{ij} = \frac{\epsilon_{irs}}{2!} \begin{vmatrix} \Be_j & \Bm_r & \Bm_s \end{vmatrix}.
Observe that the inclusion of the $$\Be_j$$ column vector in this determinant, means that we really need only compute a $$2 \times 2$$ determinant for each adjoint matrix element. That is
A_{ij} = \frac{(-1)^j \epsilon_{irs}\epsilon_{jab}}{(2!)^2}
\begin{vmatrix}
m_{ar} & m_{as} \\
m_{br} & m_{bs}
\end{vmatrix}
.
This looks a lot like the usual minor/cofactor recipe, but written out explicitly for each element, using the antisymmetric tensor to encode the index alternation. It’s worth noting that there may be an error or subtle difference from the usual in my formulation, since wikipedia defines the adjoint as the transpose of the cofactor matrix, see: [1].
## General case: $$n \times n$$ matrix.
It appears that if we wanted an induction hypotheses for the general $$n > 1$$ case, the $$ij$$ element of the adjoint matrix is likely
\begin{aligned}
A_{ij} &= \frac{\epsilon_{i s_1 s_2 \cdots s_{n-1}}}{(n-1)!} \begin{vmatrix} \Be_j & \Bm_{s_1} & \Bm_{s_2} & \cdots & \Bm_{s_{n-1}} \end{vmatrix} \\
&= \frac{(-1)^j \epsilon_{i r_1 r_2 \cdots r_{n-1}} \epsilon_{j s_1 s_2 \cdots s_{n-1}} }{\lr{(n-1)!}^2}
\begin{vmatrix}
m_{r_1 s_1} & \cdots & m_{r_1 s_{n-1}} \\
\vdots & & \vdots \\
m_{r_{n-1} s_{1}} & \cdots & m_{r_{n-1} s_{n-1}}
\end{vmatrix}.
\end{aligned}
I’m not going to try to prove this, inductively or otherwise.
# References
[1] Wikipedia contributors. Minor (linear algebra) — Wikipedia, the free encyclopedia, 2023. URL https://en.wikipedia.org/w/index.php?title=Minor_(linear_algebra)&oldid=1182988311. [Online; accessed 16-January-2024]. |
Parallel and Transversal Lines
Here we discuss how the angles formed between parallel and transversal lines.
When the transversal intersects two parallel lines:
• Pairs of corresponding angles are equal.
• Pairs of alternate angles are equal
• Interior angles on the same side of transversal are supplementary.
Worked-out problems for solving parallel and transversal lines:
1. In adjoining figure l ∥ m is cut by the transversal t. If ∠1 = 70, find the measure of ∠3, ∠5, ∠6.
Solution:
We have ∠1 = 70°
∠1 = ∠3 (Vertically opposite angles)
Therefore, ∠3 = 70°
Now, ∠1 = ∠5 (Corresponding angles)
Therefore, ∠5 = 70°
Also, ∠3 + ∠6 = 180° (Co-interior angles)
70° + ∠6 = 180°
Therefore, ∠6 = 180° - 70° = 110°
2. In the given figure AB ∥ CD, ∠BEO = 125°, ∠CFO = 40°. Find the measure of ∠EOF.
Solution:
Draw a line XY parallel to AB and CD passing through O such that AB ∥ XY and CD ∥ XY
∠BEO + ∠YOE = 180° (Co-interior angles)
Therefore, 125° + ∠YOE = 180°
Therefore, ∠YOE = 180° - 125° = 55°
Also, ∠CFO = ∠YOF (Alternate angles)
Given ∠CFO = 40°
Therefore, ∠YOF = 40°
Then ∠EOF = ∠EOY + ∠FOY
= 55° + 40° = 95°
3. In the given figure AB ∥ CD ∥ EF and AE ⊥ AB.
Also, ∠BAE = 90°. Find the values of ∠x, ∠y and ∠z.
Solution:
y + 45° = 1800
Therefore, ∠y = 180° - 45° (Co-interior angles)
= 135°
∠y =∠x (Corresponding angles)
Therefore, ∠x = 135°
Also, 90° + ∠z + 45° = 180°
Therefore, 135° + ∠z = 180°
Therefore, ∠z = 180° - 135° = 45°
4. In the given figure, AB ∥ ED, ED ∥ FG, EF ∥ CD
Also, ∠1 = 60°, ∠3 = 55°, then find ∠2, ∠4, ∠5.
Solution:
Since, EF ∥ CD cut by transversal ED
Therefore, ∠3 = ∠5 we know, ∠3 = 55°
Therefore, ∠5 = 55°
Also, ED ∥ XY cut by transversal CD
Therefore, ∠5 = ∠x we know ∠5 = 55°
Therefore,∠x = 55°
Also, ∠x + ∠1 + ∠y = 180°
55° + 60° + ∠y = 180°
115° + ∠y = 180°
∠y = 180° - 115°
Therefore, ∠y = 65°
Now, ∠y + ∠2 = 1800 (Co-interior angles)
65° + ∠2 = 180°
∠2 = 180° - 65°
∠2 = 115°
Since, ED ∥ FG cut by transversal EF
Therefore, ∠3 + ∠4 = 180°
55° + ∠4 = 180°
Therefore, ∠4 = 180° - 55° = 125°
5. In the given figure PQ ∥ XY. Also, y : z = 4 : 5 find.
Solution:
Let the common ratio be a
Then y = 4a and z = 5a
Also, ∠z = ∠m (Alternate interior angles)
Since, z = 5a
Therefore, ∠m = 5a [RS ∥ XY cut by transversal t]
Now, ∠m = ∠x (Corresponding angles)
Since, ∠m = 5a
Therefore, ∠x = 5a [PQ ∥ RS cut by transversal t]
∠x + ∠y = 180° (Co-interior angles)
5a + 4a = 1800
9a = 180°
a = 180/9
a = 20
Since, y = 4a
Therefore, y = 4 × 20
y = 80°
z = 5a
Therefore, z = 5 × 20
z = 100°
x = 5a
Therefore, x = 5 × 20
x = 100°
Therefore, ∠x = 100°, ∠y = 80°, ∠z = 100°
Lines and Angles
Fundamental Geometrical Concepts
Angles
Classification of Angles
Related Angles
Some Geometric Terms and Results
Complementary Angles
Supplementary Angles
Complementary and Supplementary Angles
Linear Pair of Angles
Vertically Opposite Angles
Parallel Lines
Transversal Line
Parallel and Transversal Lines
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# Tangents and Normals
Not what you're looking for?
Find the tangents and the normals at any point of the following curves:
1) y=1/x^2+2 (1,1/2)
2) y=6/(x^2+1)^2 (1,3/2)
##### Solution Summary
Finding the tangent and normal at a point to a curve defined by a Cartesian equation is one of the basic problems in elementary calculus. The derivative of y with respect to x gives the slope of the tangent and this can be used to find the equations of the tangent and the normal. This method is explained here using two examples. First, a brief account of the theory involved is given, then the problems are solved, showing each step accompanied by explanations. Links are also included for further reference, at each point in the solution.
##### Solution Preview
Theory:
Given the slope of a straight line, m, and any point it passes through, (x1, y1), the equation of the line is given by the point-slope formula:
(y - y1) = m(x- x1)
[
Refer: http://en.wikipedia.org/wiki/Linear_equation#Point.E2.80.93slope_form
http://www.purplemath.com/modules/strtlneq2.htm
]
The normal to a curve at a point is the line through the point that is perpendicular (or orthogonal) to the tangent to the curve at the same point. When two lines are perpendicular to each other, the product of their slopes is -1. That is, if the tangent has slope m, and the normal has slope n, then:
mn = -1
Or, given the slope of the tangent, m, the slope of the normal (at the same point) is:
n = -1/m
[ Refer: http://www.tpub.com/math2/28.htm ]
Using the above, we can solve the given problems. First, we find the slopes of the tangents, and use ...
##### Free BrainMass Quizzes
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This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
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Some questions on probability
##### Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.
##### Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them. |
# Combinatorics – combinations, arrangements and permutations
This calculator calculates the number of combinations, arrangements and permutations for given n and m
Below is a calculator that computes the number of combinations, arrangements and permutations for given n and m. A little reminder on those is below the calculator.
#### Combinatorics. Combinations, arrangements and permutations
Number of permutations from n
Number of arrangements of m from n
Number of combinations of m from n with repetitions
Number of combinations of m from n
So, assume we have a set of n elements.
Each ordered set of n is called a permutation.
For example, we have a set of three elements – А, В and С.
An example of an ordered set (one permutation) is СВА.
The number of permutations from n is
$P_n = n!$
Example: For the set of А, В and С, the number of permutations is 3! = 6. Permutations: АВС, АСВ, ВАС, ВСА, САВ, СВА
If we choose m elements from n in a certain order, it is an arrangement.
For example, the arrangement of 2 from 3 is АВ, and ВА is the other arrangement. The number of arrangements of m from n is
$A_{n}^m=\frac{n!}{(n-m)!}$
Example: For the set of А, В and С, the number of arrangements of 2 from 3 is 3!/1! = 6.
Arrangements: АВ, ВА, АС, СА, ВС, СВ
If we choose m elements from n without any order, it is a combination.
For example, the combination of 2 from 3 is АВ. The number of combinations of m from n is
$C_{n}^m=\frac{n!}{m!(n-m)!}$
Example: For the set of А, В and С, the number of combinations of 2 from 3 is 3!/(2!*1!) = 3.
Combinations: АВ, АС, СВ
Here is the dependency between permutations, combinations and arrangements
$C_{n}^m=\frac{A_{n}^m}{P_m}$
Note $P_m$ – the number of permutations from m
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Question Video: Finding the Total Amount of Money by Counting Different Coins | Nagwa Question Video: Finding the Total Amount of Money by Counting Different Coins | Nagwa
# Question Video: Finding the Total Amount of Money by Counting Different Coins Mathematics
Daniel is trying to count the money he has. I have £10 and 52p. Is Daniel correct?
02:40
### Video Transcript
Daniel is trying to count the money he has. I have 10 pounds and 52 pence. Is Daniel correct?
In the picture, we can see the group of coins that Daniel’s got. And did you notice the way that Daniel describes his coins is by using two units of measurement? This symbol represents the number of pounds that there are. And Daniel says that he has 10 pounds. And we can also see the letter p, which stands for pence. So Daniel has 10 pounds and 52 pence, or does he? Because we’re asked, is he correct?
The only way to find out the answer is for us to count the coins for ourselves. Let’s start by adding up the number of pounds that he has. Which coins can we find that have a value in pounds? Well, it’s quite small on this video, but if we look really closely we can see the words “two pounds” on this particular coin. We know that it’s a two-pound coin because it’s circular. And the color of the coin is silver in the middle with a sort of gold color around the outside.
And there’s another type of coin that has these colors. But this one’s a bit smaller. And it’s not a circle. It has sort of rounded edges to it. And this is a one-pound coin. And again, if we look really closely, we can see the words “one pound” on this coin. So in Daniel’s pile of coins, we can see some are worth two pounds and some are worth one pound.
Let’s begin by counting the pounds then. To begin with, there are four two-pound coins. So these have a value of two, four, six, eight pounds. And then we need to add our three one-pound coins. So if we start from eight, that’s nine, 10, 11. The number of pounds that Daniel has is 11 pounds. And we can see that he tells us he thinks that he has 10 pounds and 52 pence. So it looks like Daniel’s wrong, doesn’t it? Let’s count the number of pence just to finish off.
The remaining coins are a 50-pence coin and then these two coins here, which we know are worth two pence each. So if you start with the coin with the largest value, that’s 50, and then add two twos, that takes us to 52 and then 54. The number of pence that Daniel has is 54 pence. Instead of 10 pounds and 52 pence, Daniel has 11 pounds and 54 pence. We counted the pounds and the pence separately to find our answer. Is Daniel correct? No, he’s not.
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# MAT385 Recursion Examples
## Fibonacci Numbers
The Fibonacci numbers are generally defined via the recursive algorithm
F(1)=1
F(2)=1
F(n)=F(n-1)+F(n-2), ${\displaystyle n\geq 3}$
These numbers appear all over in nature: in pinecones, pineapples, artichokes, pussywillows, daisys, etc.
But they started out as pairs of immortal bunnies, which mature in a month, and then produce a single new pair every month thereafter:
This diagram illustrates why the recursive algorithm works:
• we assume that the number of bunny pairs arising from a single immature pair is given by the sequence F(n), with generations 1, 2, 3, ....
• We notice that at the third generation, the total population of pairs is given by the sum of two new populations: a perfect copy of the original tree, starting from the first generation, and a perfect copy of the original tree starting from the second generation.
• Hence ${\displaystyle F(3)=F(2)+F(1)}$
I class I presented the following algorithm in lisp for calculating the ${\displaystyle n^{th}}$ term in this sequence:
(defun fib(n)
(case n
;; the following two cases are the basis cases:
(1 1)
(2 1)
;; and, if we're not in a basis case, then we should use recursion:
(t (+ (fib (- n 1)) (fib (- n 2))))
)
)
If you calculate the ratio of successive Fibonacci numbers, ${\displaystyle {\frac {F(n)}{F(n-1)}}}$, you notice something interesting happening:
n ratio
2 1
3 2
4 1.50000
5 1.66667
6 1.60000
7 1.62500
8 1.61538
9 1.61905
10 1.61765
11 1.61818
12 1.61798
13 1.61806
14 1.61803
15 1.61804
16 1.61803
17 1.61803
18 1.61803
19 1.61803
20 1.61803
It appears that the ratio ${\displaystyle {\frac {F(n)}{F(n-1)}}\longrightarrow \approx 1.61803}$ as ${\displaystyle n\longrightarrow \infty }$.
We can use this idea to find the closed form solution of F(n):
1. assume that
${\displaystyle {\frac {F(n+1)}{F(n)}}={\frac {F(n)}{F(n-1)}}}$
in the limit. Use the recursive step (in the form ${\displaystyle F(n+1)=F(n)+F(n-1)}$) and solve for the ratio ${\displaystyle {\frac {F(n)}{F(n-1)}}}$ to get two solutions for the ratio.
2. You can write ${\displaystyle F(n)}$ as a linear combination of the two solutions:
${\displaystyle F(n)=ar_{1}^{n-1}+br_{2}^{n-1}}$
Find the closed form solution of F.
3. Check the formula for the first four Fibonacci numbers.
## How many invocations?
As we were talking about computing the Fibonacci numbers via the recursive algorithm
F(1)=1
F(2)=1
F(n)=F(n-1)+F(n-2), ${\displaystyle n\geq 3}$
we realized that the computations would grow excessively as n gets large. Consequently the time it takes to compute a Fibonacci number this way grows excessively long as n grows:
> (time (fib 20))
The evaluation took 0.02 seconds; 0.00 seconds in gc.
6765
> (time (fib 30))
The evaluation took 2.85 seconds; 0.05 seconds in gc.
832040
> (time (fib 35))
The evaluation took 31.61 seconds; 0.70 seconds in gc.
9227465
We argued this way: as soon as I go to calculate F(n), I need to invoke F two more times -- to compute F(n-1) and F(n-2). So it looks like the number of invocations is going to double each time. In addition, the invocation of F(n-1) is going to be wasteful, because it's going to invoke F(n-2) again (already invoked by F(n)). So we're redoing calculations -- very bad.
In class I asserted that the growth in the number of calculations is not really purely exponential. In this exercise, I want you to find out how bad it is. In order to do so, you need to find a recurrence relation for the number of invocations of F. Use a modified "expand, guess, check" strategy:
1. How many times will you invoke F if n=1?
2. How many times will you invoke F if n=2?
3. How many times will you invoke F if n=3?
4. How many times will you invoke F if n=4?
5. How are the invocations of F at n related to the invocations of F at n-1 and n-2?
6. Now, can you quickly write the first 10 terms in the sequence?
So why is it so easy to "recursively" write the first 10 terms? Because you're storing pairs. |
# Multicolor Squares Wall
1,455
49
6
This is easier than it looks. It's basically math and plotting. I have provided the equation and plotting technique I used to get my results. The size of the square varies on the number of squares wanted on the wall. The colors and styles are limitless. I did it using small cans of oops paints from the local hardware store, that ran me about 2 dollars a quart.
### 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: Materials
Paint - in a variety of colors of your liking.
Painter's tape- width of your choice
Paint brush
Tape measure
Paper and pencil/pen (not pictured)
Calculator (not pictured)
## Step 2: Doing the Math!
Take the length of the wall and subtract the width of the tape times the number of squares you want across the length of the wall plus one. Divide the answer by the number of squares and this will give you width of each square.
For example, You have a wall that is 100 inches long, you have 2 inch tape and want 10 squares across.
100 inches (Wall length) - (2 in (tape width)x 11 (number of squares +1) )
100 - 22= 78
Divide the result, in this case, 78, by the number of squares (10) and this will give you the width of each square.
78/10 = 7.8
Each square will be 7.8 inches wide for this wall.
So.... (Wall length) - ( Tape width x (number of squares +1) ) . Then divide and answer by number of squares.
Use a calculator if needed.
## Step 3: Marking the Wall and Taping
Continuing the example...
You will mark the wall using a tape measure and pencil, accounting for the tape. So you will mark 2 inches, 7.8 inches, 2 inches, 7.8 inches,2 inches, 7.8 inches, until the end of the the wall.
You will mark the height from top to bottom the same way. Once marked, apply your tape horizontally and then vertically.
## Step 4: Plotting the Paint
Make a legend of your colors on a separate page.
Using the pencil and paper, draw lines to mimic your wall. Label the lined paper with the colors making sure they do not end up repeating next to each other.
See pictures.
You can also lightly pencil the color code onto each square in the wall, so you don't need to keep referring to the main sheet.
Pretty simple.
## Recommendations
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## 6 Discussions
I love this idea. Where did you find that bed comforter? |
Hong Kong
Stage 4 - Stage 5
# Applications of Geometric Series
Lesson
There are many real life applications of geometric series and we discuss a few of these here.
Formulae are often developed for many of these applications, particularly when they occur regularly in industry.
For our purposes, it is often best to find solutions to various problems starting from first principles.
##### Example 1
A bank client deposits $\$1000$$1000 at the beginning of each year, and is given 7%7% interest per year for 5050 years. How much will accrue in the account over that time? To answer this, we might begin by searching for a pattern by examining what happens in the first few years. If we set A_nAn as the amount of money accrued after nn years have elapsed, then we have: A_0=1000A0=1000 Then A_1=1000+1000\times\frac{7}{100}=1000\left(1+\frac{7}{100}\right)=1000\times\left(1.07\right)^1A1=1000+1000×7100=1000(1+7100)=1000×(1.07)1 This means that the amount accrued after 11 year becomes \1070$$1070
By the end of the second year, another $\$1000$$1000 has been added, with interest, but the original \1000$$1000 has been boosted by two interest payments.
The total amount is determined as:
$A_2=1000\left(1.07\right)^1+\left[1000\left(1.07\right)\right]\times1.07=1000\left[1.07+1.07^2\right]$A2=1000(1.07)1+[1000(1.07)]×1.07=1000[1.07+1.072]
By the end of the third year, the total accrual becomes:
$A_3=1000\left[1.07+1.07^2+1.07^3\right]$A3=1000[1.07+1.072+1.073]
A pattern is emerging, and so by the end of $50$50 years, the total accrued becomes:
$A_{50}=1000\left[1.07+1.07^2+1.07^3+...+1.07^{50}\right]$A50=1000[1.07+1.072+1.073+...+1.0750].
Inside the square brackets is a geometric series with first term and common ratio both equal to $1.07$1.07
Recalling the formula for the sum of a geometric sequence as $S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn=a(rn1)r1 we have for this series:
$A_n=1000\times\frac{1.07\left(1.07^{50}-1\right)}{1.07-1}=434985.96$An=1000×1.07(1.07501)1.071=434985.96
Hence the amount accrued in the account will be approximately $\$434986$$434986 ##### Example 2 For example 11 above, devise a formula that a banker might use for any client wishing to deposit an amount PP at the beginning of ever year for nn years where an interest rate of r%r% p.a. is applied. Use the formula to find the accrued amount of the regular annual payment of \1000$$1000 after $50$50 years where $8%$8% is applied each year.
From our solution to example $1$1, and calling $R=1+\frac{r}{100}$R=1+r100, we have the generalised formula given by:
$A_n$An $=$= $P\times\frac{R\left(R^n-1\right)}{R-1}$P×R(Rn−1)R−1
Hence, at $8%$8%, we have:
$A_n=1000\times\frac{1.08\left(1.08^{50}-1\right)}{1.08-1}=573770.16$An=1000×1.08(1.08501)1.081=573770.16
This means that one extra percentage in interest each year makes a difference in total accrual of approximately $\$138784$$138784. ##### Example 3 Show that the repeating decimal N=0.2323232323...N=0.2323232323... is a rational number. That is to say, the number can be put in the form \frac{p}{q}pq where pp and qq are integers and q\ne0q0 The repeating decimal can be written: N=0.23+0.0023+0.000023+...N=0.23+0.0023+0.000023+... which is an infinite geometric series whose first term is given by a=0.23a=0.23 and whose common ratio is given by r=0.001r=0.001. The limiting sum becomes: NN == \frac{a}{1-r}a1−r == \frac{0.23}{1-0.001}0.231−0.001 == \frac{\frac{23}{100}}{1-\frac{1}{100}}231001−1100 == \frac{\frac{23}{100}}{\frac{100-1}{100}}23100100−1100 == \frac{23}{99}2399 The same strategy can be applied to any repeating decimal. ##### Example 4 The recipe for making a Koch snowflake is as follows; 1. Draw an equilateral triangle with area AA as shown in figure (a)(a) below. 2. To each of the three sides of figure (a)(a) add an equilateral triangle with sides of length \frac{1}{3}13 that of the original triangle, as shown in figure (b)(b). Note that the area of each of the new triangles is \frac{A}{9}A9. 3. To each of the 1212 sides of figure (b)(b) add an equilateral triangle with sides of length \frac{1}{3}13 that of the triangles formed in step 22. The area of each of these 1212 new triangles becomes \frac{A}{9^2}A92. 4. To each of the 4848 sides of figure (c)(c) add an equilateral triangle with sides of length \frac{1}{3}13 of the 1212 triangles formed in step 33. 5. Continue in this manner indefinitely to form what becomes the Koch snowflake. The emerging figure has an infinite perimeter but we can show that is has a finite area as follows. The total area A_TAT of the snow flake becomes: A_TAT == A+3\left(\frac{A}{9}\right)+12\left(\frac{A}{9^2}\right)+48\left(\frac{A}{9^3}\right)+192\left(\frac{A}{9^4}\right)+...A+3(A9)+12(A92)+48(A93)+192(A94)+... == A\left[1+\left(\frac{3}{9}\right)+\left(\frac{12}{9^2}\right)+\left(\frac{48}{9^3}\right)+\left(\frac{192}{9^4}\right)+...\right]A[1+(39)+(1292)+(4893)+(19294)+...] The expression on the right and within the main bracket begins with the number 11, but all the other terms form a geometric series with first term \frac{1}{3}13 and common ratio \frac{4}{9}49. We can see this because the numerator of each fraction is increasing by a factor of 44 and the denominator is increasing by a factor of 99 Since |\frac{4}{9}||49| is less than 11, the series sums to \frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{4}{9}}=\frac{3}{5}a1r=13149=35. Adding the extra 11 at the beginning, we see that the total sum within the main brackets is \frac{8}{5}85. This means that Koch snowflake has a total area given by A_T=\frac{8A}{5}AT=8A5. #### Worked Examples ##### QUESTION 1 The recurring decimal 0.8888\dots0.8888 can be expressed as a fraction when viewed as an infinite geometric series. 1. Express the first decimal place, 0.80.8 as an unsimplified fraction. 2. Express the second decimal place, 0.080.08 as an unsimplified fraction. 3. Hence write, using fractions, the first five terms of the geometric sequence representing 0.8888\dots0.8888 4. State the values of aa, the first term, and rr, the common ratio, of this sequence. aa==\editable{} rr==\editable{} 5. If we add up infinitely many terms of this sequence, we will have the fraction equivalent of our recurring decimal. Calculate the infinite sum of the sequence as a fraction. ##### QUESTION 2 At the start of 2014 Pauline deposits \5000$$5000 into an investment account. At the end of each quarter she makes an extra deposit of $\$700$$700. By looking at the pattern investment, Pauline realises she can use her knowledge of geometric series to find the balance in the account at some point in the future. The table below shows the first few quarters of 2014. All values in the table are in dollars. Quarter Opening Balance Interest Deposit Closing Balance Jan-Mar 50005000 200200 700700 59005900 Apr-Jun 59005900 236.00236.00 700700 6836.006836.00 Jul-Sep 6836.006836.00 273.44273.44 700700 7809.447809.44 1. Use the numbers for the January quarter to calculate the quarterly interest rate. 2. Write an expression for the amount in the account at the end of the first quarter. Do not evaluate the expression. \editable{}\times\editable{}+\editable{}×+ 3. Using 5000\times1.04+7005000×1.04+700 as the starting balance, write an expression for the amount in the account at the end of the second quarter. \editable{}\times\left(\editable{}\right)^2+\editable{}\times\editable{}+\editable{}×()2+×+ 4. Given that the amount in the account at the end of the second quarter can be expressed as 5000\times\left(1.04\right)^2+700\times1.04+7005000×(1.04)2+700×1.04+700, write a similar expression for the amount in the account at the end of the third quarter. \editable{}\times\left(\editable{}\right)^3+\editable{}\times\left(\editable{}\right)^2+\editable{}\times\editable{}+\editable{}×()3+×()2+×+ 5. The amount in the account after nn quarters can be expressed as a term of a geometric sequence plus the sum of a geometric sequence. Write an expression for the amount in the investment account after nn quarters. 6. Hence determine the total amount in Pauline’s account at the beginning of 2016 to the nearest dollar. ##### QUESTION 3 Rochelle invests \190000$$190000 at a rate of $7%$7% per annum compounded annually, and wants to work out how much she can withdraw each year to ensure the investment lasts $20$20 years.
We will use geometric sequences and series to determine what Rochelle's annual withdrawal amount should be if she wants the investment to last $20$20 years.
1. The amount in the account after $n$n years can be expressed as the $n$nth term of a geometric sequence minus the sum of a different geometric sequence.
Write an expression for the amount in the investment account after $n$n years.
Use $x$x to represent the amount to be withdrawn each year.
2. Hence determine Rochelle's annual withdrawal amount, correct to the nearest cent. |
SOLUTION 9: Compute the area of the region enclosed by the graphs of the equations $y=\ln x$ and $y= (\ln x)^2$ . Begin by finding the points of intersection of the two graphs. From $y=\ln x$ and $y= (\ln x)^2$ we get that $$\ln x = (\ln x)^{2} \ \ \longrightarrow$$ $$\ln x - (\ln x)^{2} = 0 \ \ \longrightarrow$$ $$\ln x (1 - \ln x) = 0 \ \ \longrightarrow$$ $$\ln x = 0 \ \ or \ \ \ln x = 1 \ \ \longrightarrow \ \ x = 1 \ \ or \ \ x = e$$ $$\displaystyle 2x = 2 \ \ \longrightarrow \ \ x = 1$$ Now see the given graph of the enclosed region.
Using vertical cross-sections to describe this region, we get that $$1 \le x \le e \ \ and \ \ (\ln x)^{2} \le y \le \ln x ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{1}^{e} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{1}^{e} (\ln x - (\ln x)^{2}) \ dx }$$ $$= \displaystyle{ \int_{1}^{e} \ln x \ dx} - \displaystyle{ \int_{1}^{e} (\ln x)^{2} \ dx }$$ $\Big($ Use Integration by Parts for $\ \displaystyle \int (\ln x)^{2} \ dx$ . Recall that the Integration by Parts Formula is $\ \int u \ dv = uv - \int v \ du$. Let $\ u = (\ln x)^{2} \$ and $\ dv = dx \$, so that $\ du = 2 \ln x \cdot \displaystyle \frac{1}{x} \ dx \$ and $\ v = x$. Then $$\displaystyle \int (\ln x)^{2} \ dx = x (\ln x)^{2} - \displaystyle \int 2 x \ln x \cdot \frac{1}{x} \ dx = x (\ln x)^{2} - 2 \displaystyle \int \ln x \ dx$$ Use Integration by Parts again. Let $\ u = \ln x \$ and $\ dv = dx \$, so that $\ du = \displaystyle \frac{1}{x} \ dx \$ and $\ v = x$. Then $$x (\ln x)^{2} - 2 \displaystyle \int \ln x \ dx = x (\ln x)^{2} - 2 \Big[ x \ln x - \int x \cdot \frac{1}{x} \ dx \Big] = x (\ln x)^{2} - 2 \Big[ x \ln x - \int 1 \ dx \Big] = x (\ln x)^{2} - 2 (x \ln x - x ) + C = x (\ln x)^{2} - 2 x \ln x + 2x + C \ \Big)$$ Continuing with the definite integral, we get that $$\displaystyle{ \int_{1}^{e} \ln x \ dx} - \displaystyle{ \int_{1}^{e} (\ln x)^{2} \ dx } = \displaystyle{ \Big( \Big(x \ln x - x \Big) - \Big( x (\ln x)^{2} - 2 x \ln x + 2x \Big) \Big) \Big\vert_{1}^{e} }$$ $$= \displaystyle{ \Big( - x (\ln x)^{2} + 3 x \ln x - 3x ) \Big) \Big\vert_{1}^{e} }$$ $$= \displaystyle{ \Big( - e (\ln e)^{2} + 3 e \ln e - 3e \Big) - \Big( - (\ln 1)^{2} + 3 \ln 1 - 3 \Big) }$$ $$= \displaystyle{ \Big( - e (1)^{2} + 3 e (1) - 3e \Big) - \Big( - (0)^{2} + 3(0) - 3 \Big) }$$ $$= \displaystyle{ ( - e ) - ( - 3 ) }$$ $$= \displaystyle{ 3-e }$$ |
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# How to add fractions with unlike denominators
Following on from our last blog on adding fractions with like denominators, adding fractions with unlike denominators is a little more complicated.
Improper fractions, or fractions with unlike denominators, may look a bit difficult. However, once you make the denominators the same, the addition is easy.
Let’s use an example:
## Find the least common multiple
First, we need to find the Least Common Multiple (LCM) for the denominators. Basically, we need to find a common multiple that they share.
The common multiples of 4 are 4, 8, 12, etc.
The common multiples of 12 are 12, 24, 36, etc.
Now we need to find the first common value. For most students new to this, it’s easiest to sketch it out on a number line:
The least common multiple is (1, 3) = 12.
## Multiply the numerator and denominator to get like denominators
Now, you’ll need to multiply the entire fraction to make the denominator become the least common multiple. In our example, this is what that looks like:
Important: you multiply both the top and the bottom by the same amount to keep the value of the fraction the same.
Now the denominators (the bottom numbers) are the same. This means we can add the numerators (the top numbers):
## Simplifying fractions
Finally, let’s simplify the fraction.
Think about the Greatest Common Factor: the highest number that divides evenly into both the numerator and the denominator.
8 divides into 1, 2, 4, 8
12 divides into 1, 2, 4, 6, 12
The Greatest Common Factor is 4. 8/4 = 2, and 12/4 = 3: |
# Exploring domain and range in word problems
How to Find Domain and Range in a Word Problem
Welcome to Warren Institute, where we explore the fascinating world of Mathematics education! In this article, we will delve into the important concept of finding the domain and range in word problems. Understanding how to identify the domain, which represents the possible input values, and the range, which represents the possible output values, is crucial for solving real-life mathematical scenarios. Join us as we unravel the complexities of domain and range through practical examples and step-by-step explanations. Get ready to enhance your problem-solving skills and gain a deeper understanding of Mathematics education. Let's dive in!
## Subtitulo 1: Understanding the Concept of Domain and Range
The first step in finding the domain and range in a word problem is to have a clear understanding of what these terms mean in the context of mathematics education. The domain refers to the set of all possible values that the independent variable can take on, while the range represents the set of all possible values that the dependent variable can take on. By defining these terms, we can accurately analyze a word problem and determine the appropriate domain and range.
## Subtitulo 2: Identifying the Independent and Dependent Variables
Once we have a grasp of the concept of domain and range, the next step is to identify the independent and dependent variables in the word problem. The independent variable is the variable that is being manipulated or controlled, while the dependent variable is the variable that is being observed or measured. By identifying these variables, we can determine which variable's values correspond to the domain and which correspond to the range.
## Subtitulo 3: Analyzing Constraints and Relationships
After identifying the independent and dependent variables, it is important to analyze any constraints or relationships mentioned in the word problem. Constraints are conditions that limit the possible values of the variables, while relationships describe how the variables interact with each other. By carefully examining the word problem for any constraints or relationships, we can refine our understanding of the domain and range and take them into account when determining their values.
## Subtitulo 4: Determining the Domain and Range
With a clear understanding of the concept of domain and range, identification of the independent and dependent variables, and analysis of any constraints or relationships, we are now ready to determine the domain and range in the word problem. This involves considering all the possible values that the independent variable can take on and determining the corresponding values of the dependent variable. By carefully applying the concepts and information gathered from the previous steps, we can accurately find the domain and range in a word problem.
### What strategies can be used to determine the domain and range in a word problem?
One strategy that can be used to determine the domain and range in a word problem is to carefully analyze the given information and constraints. By identifying the variables involved and understanding their limitations, we can determine the possible values for the input (domain) and the corresponding output (range). Additionally, considering any restrictions or conditions mentioned in the problem can help refine the domain and range further.
### How can I identify the domain and range of a function given a real-life scenario?
The domain of a function in a real-life scenario represents all the possible input values that are valid for the given situation. To identify the domain, you need to consider any restrictions or limitations within the context. For example, if you are measuring the height of students in a classroom, the domain would be limited to positive real numbers since negative heights don't make sense.
The range of a function in a real-life scenario represents all the possible output values that can be obtained from the given situation. To identify the range, you need to analyze the possible outcomes or results. Using the previous example, if the height is measured in centimeters, the range would typically be a set of positive real numbers representing the heights of the students in the class.
Remember to carefully analyze the context and any relevant constraints to determine the appropriate domain and range of a function in a real-life scenario.
### Are there any specific steps or guidelines to follow when finding the domain and range in word problems?
Yes, there are specific steps to follow when finding the domain and range in word problems.
1. Identify the variables involved in the problem and determine what they represent.
2. Determine any restrictions or conditions given in the problem that limit the possible values for the variables.
3. Consider the nature of the problem and think about what values would make sense in the given context.
4. Write down the domain as a set of all possible input values for the variables.
5. Write down the range as a set of all possible output values for the variables.
Remember to always check your solution against the problem's context to ensure it makes sense.
### Can you provide examples of word problems where determining the domain and range is crucial?
Yes, I can provide examples of word problems where determining the domain and range is crucial. One example could be a problem involving a function that represents the distance traveled by a car over time. In this case, determining the domain would be crucial to identify the valid time intervals for the problem, such as when the car started or stopped. The range, on the other hand, would help determine the maximum distance the car can travel within a given time frame. Another example could be a problem involving a function representing the height of a projectile as a function of time. Here, determining the domain would be crucial to identify the time interval during which the projectile is in motion, while the range would help determine the maximum height the projectile can reach.
### What are some common misconceptions or challenges students face when trying to find the domain and range in word problems?
Some common misconceptions or challenges students face when trying to find the domain and range in word problems include misunderstanding the context of the problem, not identifying the relevant variables, and overlooking restrictions or limitations. Students may struggle with translating the given information into mathematical language or determining the appropriate intervals for the domain and range. Additionally, they might forget to consider any constraints that could restrict the possible values.
In conclusion, understanding how to find the domain and range in a word problem is an essential skill in mathematics education. By carefully analyzing the given information and identifying the variables involved, one can determine the set of input values (domain) and output values (range) that are valid for the problem. Remember to pay attention to any restrictions or limitations stated in the problem, as they may impact the domain. By utilizing logical reasoning and mathematical understanding, students can confidently solve word problems and apply their knowledge to real-world situations. Practice and familiarity with different types of word problems will further enhance their proficiency in this area. So, continue honing your skills, and embrace the challenge of finding the domain and range in word problems to become a proficient problem solver in mathematics education.
See also Understanding Impulse and Momentum Conservation in Inelastic & Elastic Collisions
If you want to know other articles similar to Exploring domain and range in word problems you can visit the category General Education.
Michaell Miller
Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.
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Found in: Page 315
### Precalculus Enhanced with Graphing Utilities
Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465
# Solve $\text{}{x}^{2}-7x-30=0\text{.}$
The solution of the given equation is and $x=-3andx=10.$
See the step by step solution
## Step 1. Given information
The given equation is $\text{}{x}^{2}-7x-30=0\text{}$
We have to find the solution of the equation i.e the value of $x$ that satisfies the given equation.
## Step 2. Form the factors
$\begin{array}{l}\text{The given equation is}{x}^{2}-7x-30=0.\\ ⇒{x}^{2}-7x-30=0\\ ⇒{x}^{2}+3x-10x-30=0\text{}\left[\text{Split the middle term}\right]\\ ⇒x\left(x+3\right)-10\left(x+3\right)=0\text{}\left[\text{Take out the common terms}\right]\\ ⇒\left(x-10\right)\left(x+3\right)=0\text{}\left[\text{Form the factors}\right]\end{array}$
## Step 3. Solve for x
$\begin{array}{l}\text{By zero product rule,}\\ ⇒x-10=0\\ ⇒x=10\\ \text{and,}\\ ⇒x+3=0\\ ⇒x=-3\\ \text{The solution set is}\left\{-3,10\right\}\end{array}$ |
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# Proportions
## Multiplication to solve for an unknown given two equal ratios.
%
Progress
Practice Proportions
Progress
%
Write and Solve Proportions by Using Equivalent Rates
Have you ever faced a reading challenge? Take a look at this dilemma.
Jamie looked up at the clock when the bell rang. He had read 15 pages in 20 minutes. If she had read 15 pages in 20 minutes, how many pages would she read in 40 minutes?
Do you know how to figure this out? You can use equivalent rates to help you solve this problem. Pay attention to this Concept and you will understand the solution by the end of it.
### Guidance
A ratio is a comparison between two quantities or numbers. Ratios can be written in fraction form, with a colon or by using the word “to”.
Sometimes, you will compare ratios. Sometimes one ratio will be greater than another, and other times they can be equal or equivalent . When you have two equal ratios, you have a proportion.
A proportion is created when two ratios are equal, or we can say that two equal ratios form a proportion.
We can write a proportion when we know that two ratios are equivalent.
$1 : 2 = 2 : 4$
These two ratios are equivalent. We can say that the two ratios form a proportion.
Do these two ratios form a proportion?
$\frac{3}{4}$ and 4 : 24
To figure this out, we have to figure out if the two ratios are equivalent. If they are, then we know that they form a proportion. If not, then they don’t. To figure this out, we can simplify the ratios.
$& \qquad \ \frac{3}{4} \ is \ in \ simplest \ form.\\& 4 : 24 \ can \ be \ written \ as \ \frac{4}{24} = \frac{1}{6}\\& \qquad \qquad \quad \ \ \frac{3}{4} \ne \frac{1}{6}$
These ratios do not form a proportion.
These proportions were given to you. You can also write your own proportions.
To write a proportion, set two equivalent fractions equal to each other, using the information in the problem.
If you know the ratio of girls to boys in a class is 2 : 3, and you know there are 24 boys in the class, you can write a proportion in order to find the number of girls in the class.
The most important thing to remember when writing a proportion is to keep the units the same in both ratios.
$\frac{girls}{boys}: \frac{2}{3} = \frac{x}{24}$
You know the fractions are equivalent because each shows the ratio of girls to boys in the class. The first fraction shows the known ratio of girls to boys. The second ratio shows the known number of boys in the class, 24, and uses a variable to stand for the unknown number of girls.
Now let's use equivalent rates to solve a proportion.
The ratio of teachers to students in a certain school is 2 : 25. If there are 400 students in the eighth-grade class, how many teachers are there?
First set up a proportion. The problem gives a ratio of teachers to students, so set up two equivalent ratios comparing teachers to students.
$\frac{teachers}{students} = \frac{8^{th} \ grade \ teachers}{8^{th} \ grade \ students}$
You can see that we are comparing teachers to students in both ratios. The first one shows the ration in the whole school and the second ratio represents the eighth grade ratios. Next, we fill in the given information.
$\frac{2}{25} = \frac{x}{400}$
Now use what you know about equivalent ratios to solve the proportion.
Look at the denominators. You know that the first fraction, when the numerator and denominator are multiplied by some number, will equal the second fraction. What number, when multiplied by 25, will equal 400? Since $25 \times 16 = 400$ , the denominator was multiplied by 16. That means you can multiply the numerator by the same number to find the value of $x$ .
$2 \times 16 = 32$ , so $x = 32$
There are 32 teachers in the eighth-grade class.
Note: You can check that your answer is correct by making sure that the two ratios are equivalent.
$\frac{32}{400} = \frac{8}{100} = \frac{2}{25}$
Since the second ratio simplifies to the first, the ratios are equivalent.
Yes it is. Just remember that what you do to the numerator you have to do the denominator. If you can remember to always apply this rule, then you will create equal ratios.
Solve each proportion by using equal ratios.
#### Example A
$\frac{3}{4} = \frac{6}{x}$
Solution: $x = 8$
#### Example B
$\frac{9}{50} = \frac{x}{100}$
Solution: $x = 18$
#### Example C
$\frac{3.5}{7} = \frac{x}{35}$
Solution: $x = 17.5$
Now let's go back to the dilemma from the beginning of the Concept.
Jamie read 15 pages in 20 minutes. She wonders how many pages she will read in 40 minutes.
We can set up a proportion and look for an equivalent rate.
$\frac{pages}{minutes} = \frac{15}{20} = \frac{x}{40}$
Here is our proportion.
Next, we can look at the relationship between the denominators.
$20 \times 2 = 40$
What we do to the bottom, we can do to the top. This will give us the equivalent rate.
$15 \times 2 = 30$
At this rate, Jamie will read 30 pages in 40 minutes.
### Guided Practice
Here is one for you to try on your own.
Write a proportion to describe this situation.
The proportion of red paper to white paper in a stack is 2 to 7. If there are 32 red pieces of paper, what proportion could be used to find the number of pieces of white paper?
Solution
Write the known ratio of red paper to white paper as the first fraction: $\frac{2}{7}$ .
Now write the second ratio, using $x$ to stand for the unknown amount. Make sure to keep the units the same as in the first fraction. In this case, the unknown is the amount of white paper, which is in the denominator of the fraction.
$\frac{\text{red paper}}{\text{white paper}} : \frac{2}{7} = \frac{32}{x}$
The proportion $\frac{2}{7} = \frac{32}{x}$ could be used to find the number of pieces of white paper in the stack.
### Explore More
Directions: Solve each proportion using equal ratios.
1. $\frac{3}{4} = \frac{x}{12}$
2. $\frac{5}{6} = \frac{x}{12}$
3. $\frac{4}{7} = \frac{8}{y}$
4. $\frac{2}{3} = \frac{12}{y}$
5. $\frac{4}{5} = \frac{44}{y}$
6. $\frac{12}{13} = \frac{x}{26}$
7. $\frac{9}{10} = \frac{81}{y}$
8. $\frac{6}{7} = \frac{18}{y}$
9. $\frac{7}{8} = \frac{x}{56}$
10. $\frac{12}{14} = \frac{36}{x}$
11. $\frac{6}{4} = \frac{x}{12}$
12. $\frac{12}{14} = \frac{24}{x}$
13. $\frac{13}{14} = \frac{x}{42}$
14. $\frac{1.5}{4} = \frac{x}{8}$
15. $\frac{3.5}{4.5} = \frac{x}{9}$
16. $\frac{9}{14} = \frac{108}{x}$
### Vocabulary Language: English
Equivalent
Equivalent
Equivalent means equal in value or meaning.
Proportion
Proportion
A proportion is an equation that shows two equivalent ratios. |
# From cosh2A = 1+2(sinh^2)A, how do you prove that (sinh^4)A + (cosh^4)A=(cosh4A + 3)/4 and also (cosh^4)A - (sinh^4)A = cosh2A?
Feb 7, 2017
see below
#### Explanation:
Part I .
${\sinh}^{4} A + {\cosh}^{4} A = \frac{\cosh 4 A + 3}{4}$
Use the formulas:
$\cosh 2 A = 2 {\cosh}^{2} A - 1$ -->Solve for ${\cosh}^{2} A$
$\cosh 2 A = 1 + 2 {\sinh}^{2} A$ --->Solve for ${\sinh}^{2} A$
Left Hand Side:
${\sinh}^{4} A + {\cosh}^{4} A = {\left({\sinh}^{2} A\right)}^{2} + {\left({\cosh}^{2} A\right)}^{2}$
$= {\left(\frac{\cosh 2 A - 1}{2}\right)}^{2} + {\left(\frac{\cosh 2 A + 1}{2}\right)}^{2}$-->FOIL
$= \frac{{\cosh}^{2} 2 A - 2 \cosh 2 A + 1}{4} + \frac{{\cosh}^{2} 2 A + 2 \cosh 2 A + 1}{4}$
$= \frac{{\cosh}^{2} 2 A - 2 \cosh 2 A + 1 + {\cosh}^{2} 2 A + 2 \cosh 2 A + 1}{4}$
$= \frac{{\cosh}^{2} 2 A - \cancel{2 \cosh 2 A} + 1 + {\cosh}^{2} 2 A + \cancel{2 \cosh 2 A} + 1}{4}$
$= \frac{{\cosh}^{2} 2 A + 1 + {\cosh}^{2} 2 A + 1}{4}$
$= \frac{2 {\cosh}^{2} 2 A + 2}{4}$
Note: $\cosh 4 A = 2 {\cosh}^{2} 2 A - 1$ $\therefore$ $\cosh 4 A + 1 = 2 {\cosh}^{2} 2 A$
$= \frac{\cosh 4 A + 1 + 2}{4}$
$= \frac{\cosh 4 A + 3}{4}$
$\therefore =$ Right Hand Side
Part II
${\cosh}^{4} A - {\sinh}^{4} A = \cosh 2 A$
Use the properties :
${\cosh}^{2} A - {\sinh}^{2} A = 1$ and $\cosh 2 A = 1 + 2 {\sinh}^{2} A$
Left Hand Side :
${\cosh}^{4} A - {\sinh}^{4} A = \left({\cosh}^{2} A + {\sinh}^{2} A\right) \left({\cosh}^{2} A - {\sinh}^{2} A\right)$
$= \left({\cosh}^{2} A + {\sinh}^{2} A\right) \cdot 1$
$= {\cosh}^{2} A + {\sinh}^{2} A$
$= 1 + {\sinh}^{2} A + {\sinh}^{2} A$
$= 1 + 2 {\sinh}^{2} A$
$= \cosh 2 A$
$\therefore =$ Right Hand Side |
# Inclined Planes: Practice Problems with Solutions
In this article, we explore the fascinating world of inclined planes in physics with a number of solved practice problems designed for high school and college students.
Get ready to sharpen your problem-solving skills and deepen your knowledge of this fundamental concept in physics!
By the way, if you’re preparing for the AP Physics 1 exam, feel free to download this comprehensive AP formula sheet. It’s the ultimate resource you need!
## Inclined Plane Problems without Friction:
Problem (1): A 45 kg object rests on a 30° inclined plane.
(a) What force is exerted down the inclined plane?
(b) What force acts perpendicular to the plane?
Solution: In all inclined plane problems, the first step is to choose a suitable coordinate system. It is common to choose the positive direction along and down the plane, with the up direction perpendicular to the plane.
The weight force and normal force always act on an object resting on a smooth inclined plane (without friction). The normal force acts along the upward direction of our chosen coordinate system, while the weight force should be decomposed into components in the tilted coordinate system.
As shown in the figure below, one component of the weight force is parallel to the plane and acts downward, while the other is perpendicular and acts in the $-y$ direction.
(a) With these brief explanations, the downward weight component parallel to the slope always pulls an object down on an inclined plane.
Using trigonometry, we can find that this component is $mg\sin\theta$, where $m$ is the mass of the object and $\theta$ is the angle of the inclined plane. \begin{align*} F_x&=mg\sin\theta \\ &=45\times 9.8\sin 30^\circ \\ &=\boxed{220.5\,\rm N} \end{align*}
(b) As you noticed along the perpendicular direction to the plane, two forces act on the object while it has no motion (i.e., $a_y=0$). To determine the force acting perpendicular to the plane, we need to find the normal force. Applying Newton's second law to this direction gives us the normal force on the object \begin{gather*} F_{net-y}=ma_y \\\\ F_N-w_{\bot}=0 \\\\ F_N-mg\cos\theta=0 \\\\ \Rightarrow F_N=\underbrace{mg\cos\theta}_{w_{\bot}} \end{gather*} Substituting the given numerical values into this, we have $\boxed{F_N=45\times 9.8 \cos 30^\circ=382\,\rm N}$
Problem (2): A car of mass $m$ is at the top of an icy 15° slope whose length is 30 m.
(a) Find the acceleration of the car.
(b) How long does it take for the car to reach the bottom?
(c) How fast is the car at the bottom?
Solution: The slope is icy, so it has no friction. Recall from the previous problem that the downward weight component parallel to the inclines is $mg\sin\theta$. This force is the cause of the car going down from the top of the inclined plane.
(a) Because the incline is frictionless, the only force that acts on the car is $w_{\parallel}=mg\sin\theta$. Applying Newton's second law and substituting the numerical value yields \begin{gather*} F_{net}=ma_x \\\\ mg\sin\theta=ma_x \\\\ \rightarrow a_x=g\sin\theta \\\\ a_x=(9.8) \sin 15^\circ \\\\ \Rightarrow \boxed{a_x=2.6\,\rm m/s^2} \end{gather*} As this equation shows, the acceleration along the inclined plane does not depend on the object's mass.
(b) The car starts its motion from rest, i.e., $v_0=0$. The car's acceleration is also found in the previous part. The only kinematic equation that relates these known values together is $\Delta x=\frac 12 at^2+v_0t$, where $\Delta x$ is the total displacement traveled by the object along the slope, which here is $\Delta x=30\,\rm m$. \begin{gather*} \Delta x=\frac 12 at^2+v_0t \\\\ 30=\frac 12 (2.6)t^2+0 \\\\ \rightarrow t=\sqrt{\frac{2\times 30}{2.6}} \\\\ \Rightarrow \boxed{t=4.8\,\rm s} \end{gather*}
(c) To find the velocity at the bottom, we can use either of two kinematic equations $v^2-v_0^2=2a\Delta x$ or $v=v_0+at$. We choose the first equation and put the numbers into it \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ v^2-0=2(2.6)(30) \\\\ \Rightarrow \boxed{v=12.48\,\rm m/s} \end{gather*}
Problem (3): A hockey puck with an acceleration of $6\,\rm m/s^2$ is sliding down a $12-\rm m$-long frictionless ramp.
(a) What is the angle of the ramp?
(b) How long does it take the puck to reach the bottom?
(c) If the mass of the puck is doubled, how does its acceleration change down the incline?
Solution: The ramp or inclined plane has an unknown angle $\theta$ that we need to determine. We assume the positive $x$-axis to be down the ramp.
(a) The only force that pulls the puck down the slope and accelerates it is the component of weight force parallel to the inclined plane, $w_{\parallel}=mg\sin\theta$. Therefore, by using the second law, we get \begin{gather*} F_{net}=ma_x \\\\ mg\sin\theta=ma_x \\\\ \Rightarrow a_x=g\sin\theta \end{gather*} Substituting the numerical values into this, solving for angle $\theta$, and taking the inverse sine of both sides, we get \begin{align*} \theta &=\sin^{-1} \left(\frac{a_x}{g}\right) \\\\ &=\sin^{-1} \left(\frac{6}{9.8}\right) \\\\ &=37.7^\circ \end{align*}
(b) The puck is initially held at rest, $v_0$, and the ramp's length $\Delta x=12\,\rm m$ is also known, so use the following kinematic equation to find the required time. \begin{gather*} \Delta x=\frac 12 at^2+v_0t \\\\ 12=\frac 12 (6)t^2+0 \\\\ \Rightarrow \boxed{t=2\,\rm s} \end{gather*}
(c) In part (a), we saw that the acceleration of an object sliding down a frictionless ramp depends on $g$ and the ramp's angle $\theta$, but not on the mass $m$. Therefore, by doubling the puck's mass, its acceleration does not change along the inclined plane. $a_{new}=a_{old}=6\,\rm m/s^2$
Problem (4): A $1500-\rm kg$ car is at rest on a frictionless hill inclined at $17^\circ$.
(a) Sketch a free-body diagram and calculate the components $w_{\parallel}$, $w_{\bot}$, and the normal force acting on the car.
(b) Find the car's acceleration when it is released.
(c) If the hill is $55-\rm m$-long, what would be the car's final velocity at the bottom of the hill?
(d) How long will it take for the car to reach the end of the hill?
Solution: (a) Two forces act on the car. The normal force perpendicular to the hill, $F_N$, and the weight force, $w=mg$. The weight force has two components that are parallel $w_{\parallel}$ and perpendicular $w_{\bot}$ to the inclined plane.
From the above free-body diagram, we can calculate these components as \begin{align*} w_{\parallel} &=mg\sin\theta \\ &=(1500\times 9.8) \sin 17^\circ \\ &=\boxed{4297.86\,\rm N} \\\\ w_{\bot}&=mg\cos\theta \\&=1500\times 9.8\cos 17^\circ \\&=\boxed{14057.68\,\rm N} \end{align*}
(b) The car perpendicular to the hill does not have acceleration, so $a_y=0$. However, the car parallel to the hill is accelerating $a_x=?$, and we need to determine $a_x$.
By applying Newton's second law along the $x$- and $y$-axes on the hill, we get \begin{gather*} F_{net-x}=ma_x \\ mg\sin\theta=ma_x \\ \Rightarrow a_x=g\sin\theta \\\\ F_{net-y}=ma_y \\ F_N-w_{\bot}=0 \\\Rightarrow F_N=w_{\bot}=\boxed{14057.68\,\rm N} \end{gather*} Therefore, the car's acceleration is $a_x=9.8\times \sin 17^\circ =2.86\,\rm m/s^2$
(c) This part is related to a kinematics problem in physics. The car starts from rest ($v_0=0$) and is displaced along the hill by $\Delta x=55\,\rm m$. Its acceleration is also found in the previous part, $a_x=2.86\,\rm m/s^2$. Using the following kinematics equation and solving for $v$, we get \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ v^2-0=2(2.86)(55) \\\\ \Rightarrow \boxed{v=17.73\,\rm m/s} \end{gather*}
(d) To find the unknown time $t$, we apply the equation $v=v_0+at$ and solve as follows: \begin{gather*} v=v_0+at \\ 17.73=0+(2.86)t \\ \Rightarrow \boxed{6.2\,\rm s} \end{gather*}
Problem (5): A box with a mass of $45\,\rm kg$ is held at rest by a rope on a frictionless ramp inclined at $30^\circ$. Draw a free-body diagram and find the value of the tension force exerted by the rope.
Solution: The box is at rest, so there is no motion or acceleration along the $x$- and $y$- axes parallel and perpendicular to the inclined plane, respectively.
As the free-body diagram shows, the tension force is upward on the ramp, and the weight component $mg\sin\theta$, as always, is down the incline. Applying Newton's second law gives us \begin{gather*} F_{net-x}=ma_x \\\\ T-mg\sin\theta=0 \\\\ \Rightarrow T=mg\sin\theta \end{gather*} By substituting the given values into this, the value of tension force is obtained as follows $T=45\times 9.8\times \sin 30^\circ=\boxed{220.5\,\rm N}$
## Inclined Plane Problems with Friction:
Problem (6): A slippery object slides down a ramp of $9\,\rm m$ long inclined at $10^\circ$. The kinetic friction between the object and the ramp is $\mu_k=0.06$. How long does it take for the object to reach the bottom?
Solution: The only force exerted on the object and forcing it to go down is the component of weight parallel to the inclined plane. The kinetic friction force up the incline opposes the weight force down the incline. Balancing these two exerting forces gives acceleration to the slippery object.
Consider the downward direction of the ramp as the positive $x$ direction. Applying Newton's second law to forces along the inclined plane yields \begin{gather*} F_{net-x}=ma_x\\\\ mg\sin\theta-f_k=ma_x \\\\ mg\sin\theta -\mu_k N=mg \end{gather*} where in the last line, we used the definition of friction force. Applying the above law along the direction normal to the slope, given that in this direction the object does not move, we can find the normal force on the object \begin{gather*} F_N-mg\cos\theta=0 \\ \Rightarrow F_N=mg\cos\theta \end{gather*} Substituting this into the first expression and removing the common factor $m$, we get the object's acceleration along the rough inclined plane \begin{align*} a_x &=g(\sin\theta-\mu_k \cos\theta) \\\\ &=(9.8) \left(\sin 10^\circ-(00.6) \cos 10^\circ \right) \\\\ &=1.12\,\rm m/s^2 \end{align*} Assume the object was initially at rest, $v_0=0$. On the other hand, the total displacement or distance that the object covers is the length of the incline, $\Delta x=9\,\rm m$.
By knowing that information, we can use the displacement kinematic equation $\Delta x=\frac 12 at^2+v_0t$ to find the required time as below \begin{gather*} \Delta x=\frac 12 at^2+v_0t \\\\ 9=\frac 12 (1.12)t^2 + 0 \\\\ \rightarrow t=\sqrt{\frac{2\times 9}{1.12}} \\\\ \Rightarrow \quad \boxed{t=4\,\rm s} \end{gather*} Thus, our slippery object takes about $4$ seconds to reach the bottom of the inclined plane.
Problem (7): A $54.6\,\rm kg$ block is resting on a $7^\circ$ slope.
(a) What is the normal force acting on the block?
(b) What is the friction force holding the block on the slope?
Solution: except for the weight and friction forces, no other external forces at angles, act on the block resting on the inclined plane.
(a) Recall that, as always in all inclined plane problems, in the presence of these two forces, the normal force on an object on a slope is $F_N=mg\cos\theta$. Substituting the given values gives us $F_N=54.6\times 9.8\times \cos 7^\circ=531\,\rm N$
(b) The block is at rest on the slope, so there should be a force to oppose the weight component $mg\sin\theta$ to hold the block motionless on the inclined plane. This opposing force is the same friction force.
Consequently, in this special case, the friction force is found to be $f=mg\sin\theta$ whose magnitude is also as below $f=54.6\times 9.8\times \sin 7^\circ= 65.2\,\rm N$
Problem (8): A $46.5\,\rm kg$ crate is resting motionless on a ramp inclined at an angle of $20^\circ$. What is the coefficient of friction between the crate and the ramp?
Solution: The crate does not have acceleration along the ramp, so the downward weight component $mg\sin\theta$ must be balanced with the upward friction force on the incline. Therefore, we have $f=mg\sin\theta$. On the other hand, in these situations, the normal force acting on the object also has a magnitude of $F_N=mg\cos\theta$.
From the problems on the coefficient of static friction force, we remember that $f=\mu F_N$. Combining everything together, we get \begin{gather*} f=\mu F_N \\\\ mg\sin\theta=\mu (mg\cos \theta) \\\\ \rightarrow \mu=\frac{\sin\theta}{\cos\theta}=\tan\theta \end{gather*} Substituting the numerical values into this equation, we get $\boxed{\mu=\tan 20^\circ=0.36}$
This question has an important note for us. It tells us that when an object is either at rest (motionless) or moving with a constant velocity along a rough inclined plane, the coefficient of friction between the incline and the object is independent of the object's mass and only depends on the angle of the incline.
Problem (9): A box weighing $95\,\rm N$ is held at rest on a slope inclined at an angle of $25^\circ$.
(a) Find the magnitudes of the weight forces parallel and perpendicular to the inclined plane.
(b) Assuming the coefficient of static friction between the box and the incline is $0.55$, what is the maximum frictional force?
(c) Does the box move? If yes, at what rate does the box move down? If not, with what $\mu_s$ will the box be on the verge of sliding?
Solution: The weight of the box and the angle of incline $\theta=25^\circ$ are given.
(a) As you have learned so far, the weight force components parallel ($w_\parallel$) and perpendicular ($w_\bot$) to the incline are given by the following formulas: \begin{align*} w_\parallel&=mg\sin\theta \\&=95\times \sin 25^\circ \\ &=\boxed{40.14\,\rm N} \\\\ w_\bot&=mg\cos\theta \\&=95\cos 25^\circ \\&=\boxed{86\,\rm N} \end{align*}
(b) Static friction is the maximum frictional force that opposes motion and prevents an object from moving. It is also defined as $f_{s,max}=\mu_s F_N$, where $F_N$ is the normal force exerted on an object sitting on the incline.
Recall that in the case of an external force applied to the object at an angle, the magnitude of the normal force is calculated as follows: \begin{align*} f_{s,max}&=\mu_s (mg\cos\theta) \\ &=(0.55)(95) \cos 25^\circ \\ &=\boxed{47.35\,\rm N} \end{align*}
(c) To determine whether the box moves down the incline or remains at rest, we must compare the weight component $w_\parallel$ down the incline with the maximum static friction upward along the incline.
In the last two parts, we found that $\underbrace{w_\parallel}_{40.14} < \underbrace{f_{s,max}}_{47.35}$ The force opposing the motion is greater than the force pulling the box down the inclined plane, so the box remains motionless.
For the box to start sliding down from the top of the incline, the parallel weight component $w_\parallel$ must be equal to the frictional force. Equating these two forces and solving for $\mu_s$, we obtain the required coefficient of static friction to put the box on the verge of sliding. \begin{gather*} w_\parallel =f_{s,max} \\ mg\sin\theta =\mu_s (mg\cos\theta) \\\\ \rightarrow \mu_s=\tan\theta \\ \Rightarrow \boxed{\mu_s=\tan 25^\circ=0.466}\end{gather*}
Problem (10): A box of mass $35\,\rm kg$ sits on a $30^\circ$ incline. Consider these two cases and find out what is being said.
(a) If the box does not move down the incline, what can you say about the static frictional force acting on the box?
(b) If the box slides down the incline with an acceleration of $3\,\rm m/s^2$, find the kinetic frictional force and the coefficient of kinetic friction between the box and the incline.
Solution: Like any other inclined plane practice problem, first find the parallel and perpendicular components of the object's weight along the incline as below \begin{align*} w_\parallel&=mg\sin\theta \\&=35\times 9.8 \times \sin 30^\circ \\ &=\boxed{171.5\,\rm N} \\\\ w_\bot&=mg\cos\theta \\&=35 \times 9.8 \cos 30^\circ \\&=\boxed{297.0\,\rm N} \end{align*} Now we want to examine the behavior of the box along the slope with/without the frictional force.
(a) If the box is not going to slide down the incline, the net force along the slope must be zero. This means that the parallel component of the weight, $w_\parallel$, must be balanced by the static frictional force, $f_s$. Therefore we have: $f_s = w_\parallel \Rightarrow \boxed{f_s = 171.5\,\rm N}$ This is the actual static friction force required to keep the box at rest on the incline.
(b) If the box slides down the incline with an acceleration of $3\,\rm m/s^2$, it means that the net force along the slope is positive and equal to the product of the mass and the acceleration, $ma_x$. This implies that the parallel component of the weight, $w_\parallel$, is greater than the kinetic frictional force, $f_k$. Therefore, we can write: $w_\parallel-f_k=ma_x$ To find the value of $f_k$, we need to rearrange the equation and plug in the given values: \begin{align*} f_k &= w_\parallel - ma_x \\ &= 35\times 9.8 \times \sin 30^\circ - (35)(3) \\ &= 66.5\,\rm N \end{align*} Therefore, the kinetic frictional force is: $f_k = 66.5\,\rm N$ To find the coefficient of kinetic friction, $\mu_k$, we need to use the formula for $f_k$, which is given by: $f_k = \mu_k F_N$ where $F_N$ is the normal force, which is equal and opposite to the perpendicular component of the weight, $w_\bot$. The formula for $w_\bot$ is given by: $w_\bot = mg\cos\theta$ Plugging in the given values, we get: \begin{align*} w_\bot &= 35\times 9.8 \times \cos 30^\circ \\ &= 297.0\,\rm N \end{align*} Therefore, the normal force is: $F_N = 297.0\,\rm N$ To find the value of $\mu_k$, we need to divide both sides of the equation by $F_N$: $\mu_k = \frac{f_k}{F_N} = \frac{66.5}{297.0} = 0.22$ Therefore, the coefficient of kinetic friction is: $\mu_k = 0.22$
Problem (11): A $5-\rm kg$ box is placed on a surface with a kinetic coefficient of friction of $0.28$. The surface is slowly raised and reaches an angle of $48^\circ$. What would be the acceleration of the box down the incline?
Solution: In the earlier questions, we saw that the acceleration of an object on a rough inclined plane is written as $a=g(\sin\theta-\mu_k \cos\theta)$ Substituting the numerical values into it gives \begin{align*} a&=(9.8) \left(\sin 48^\circ-(0.28)\cos 48^\circ \right) \\&=\boxed{5.44\,\rm m/s^2} \end{align*}
Problem (12): A car weighing $2\times 10^4\,\rm N$ is at rest at the bottom of a $22^\circ$ hill. Several people are pushing the car uphill with a combined force of $8500\,\rm N$ parallel to the incline. (Assume the hill has negligible friction.)
(a) What acceleration does the car obtain?
(b) The people push on the car for $15\,\rm s$. How far does the car go up the incline?
Solution: Two forces are acting on the car in this direction: the parallel component of the weight,$w_{\parallel}$, and the pushing force, $F_p$. The net force is given by $F_{net} = F_p - w_\parallel$ As always the weight components are found as follows: \begin{align*} w_{\parallel}&=mg\sin\theta \\&= (2\times 10^4) \sin 22^\circ \\ &=\boxed{7492.13\,\rm N} \\\\ w_{\bot}&=mg\cos\theta \\ &=(2\times 10^4) \cos 22^\circ \\&=\boxed{18543.67\,\rm N} \end{align*} Take the direction of the movement as the positive $x$-direction. In this case, the weight component $w_{\parallel}$ is in the negative $x$ direction or downhill, but the pushing force, exerted by people, is uphill.
(a) Applying the second law along the $x$-direction (on the inclined plane) gives us \begin{gather*} F_{net-x}=ma_x \\ F_{ext}-w_{\parallel}=ma_x \\ 8500-7492.13=2040.81 a_x \\ \Rightarrow \boxed{a_x=0.49\,\rm m/s^2} \end{gather*}
(b) The car is initially at rest at the bottom of the hill, $v_0=0$. The only kinematic relationship that connects our known values, $a=0.49\,\rm m/s^2$ and $t=15\,\rm m$, is $\Delta x=\frac 12 at^2+v_0t$, where $\Delta x$ is the distance traveled on the incline. Plugging in the values, we get: \begin{align*} \Delta x&=\frac 12 at^2+v_0t \\\\ &=\frac 12 (0.49)(15)^2+0 \\\\ &=\boxed{55.125\,\rm m} \end{align*}
Problem (13): A box weighing $175\,\rm N$ rests on a horizontal surface with a static coefficient of friction of $\mu_s=0.85$.
(a) At what angle should the surface be raised so that the box would just start to move?
(b) Assume the kinetic coefficient of friction is $\mu_k=0.42$. At what angle should the ramp be lowered so that the box moves without acceleration?
Solution: On the horizontal surface, the only forces initially acting on the box are the normal $F_N$ and the weight $w=mg$ forces. As soon as it makes an angle with the level, no longer the weight pulls the box perpendicularly.
At this point, we must choose a coordinate system on the inclined plane, as you saw in the previous problems.
(a) Before reaching a certain angle, the box remains at rest because the upward static friction force on the incline is greater than the downward force of $mg\sin\theta$ along the incline. As the angle increases, the weight component parallel to the incline gradually increases until it equals the maximum static friction force, causing the box to be on the verge of slipping downward. This can be expressed as $f_{s,max}=mg\sin\theta$, which can be rearranged to solve for the static coefficient of friction $\mu_s$. \begin{gather*} f_{s,max}=mg\sin\theta \\ \mu_s F_N=mg\sin\theta \\ \mu_s (mg\cos\theta)=mg\sin\theta \\ \Rightarrow \boxed{\mu_s=\tan \theta} \end{gather*} The desired angle is then found by taking the arctangent (inverse of tangent) of $\mu_s$. Using this method, we find that the box will remain at rest until the surface reaches an angle of approximately $40.36^\circ$. \begin{align*} \theta&=\arctan{\mu_s} \\ &=\arctan{0.85} \\ &=\boxed{40.36^\circ} \end{align*}
(b) As you have seen in part (a), the box is at rest until the surface reaches an angle of about $40^\circ$. Beyond that angle, the box would accelerate downslope. Now, by reducing the angle we can make a situation in which the box would travel with constant velocity (no acceleration). This occurs when the weight force component $mg\sin\theta$ down slope becomes equal to the kinetic friction force $f_k=\mu_k F_N$ upslope. Equating these two forces will give us the desired angle \begin{gather*} f_k=mg\sin\theta \\ \mu_k (mg\cos\theta)=mg\sin\theta \\ \Rightarrow \boxed{\tan\theta=\mu_k} \end{gather*} Substituting the given value into this equation and taking the inverse tangent of both sides, we get \begin{gather*} \tan\theta=0.42 \\ \rightarrow \theta =\tan^{-1}(0.42)=\boxed{22.7^\circ} \end{gather*}
## Summary:
In this long article, we solved some problems on inclined planes with and without frictional force. In both cases, we learned that the normal force acting perpendicularly on an object on an inclined plane is written as $F_N=mg\cos\theta$.
On the other hand, the weight force component $w_{\parallel}$ parallel to the incline pulls the object downslope whose magnitude is $w_{\parallel}=mg\sin\theta$.
Author: Dr. Ali Nemati
Page Published: Sep 18, 2022 |
# Should a tennis player ever serve two first serves?
On a recent episode of The Tennis Podcast, presenter Catherine Whitaker was analysing Andy Murray’s shock exit from the 2017 Australian Open. A feature of the match was that Murray’s opponent, Mischa Zverev, had been eating Murray’s second serve for breakfast — Murray won only 36% of points behind his second serve in the match.
“When you’re winning that few points on second serve, why not just hit two first serves?” Catherine asked.
Good question. Would Andy Murray have actually been better off hitting two first serves in that match? Let’s do the maths.
Disclaimer: There’s a bit of algebra coming up. It’s not complicated but if it’s not your bag just scroll down to ‘TL;DR’.
Firstly let’s build the equation for proportion of service points won.
Let first serve percentage be ‘a’
Let percentage of points won behind first serve be ‘b’
Let percentage of points won behind second serve be ‘y’
(For simplicity all percentages here will be expressed as proportions in the equations, where 100% = 1, 50% = 0.5 and so on).
A player’s total proportion of points won on serve (call this W1) will be given by this equation:
W1 = ab + y(1-a)
In his match against Zverev, Murray got 66% of his first serves in, and won 65% of points behind his first serve. So we can plug these numbers in and get his overall % of points won on serve:
W1 = (0.66 x 0.65) + 0.36(1–0.66)
W1 = 0.429 + 0.1224
W1 = 0.5514
This tallies with what the match stats show: Murray won 55.14% of his service points in this match.
But what would his hypothetical % of service points won have been had he served two first serves?
We can build an equation in the same way. Let’s call his hypothetical win percentage for two first serves W2.
W2 = ab + ab(1-a)
Now plugging in Murray’s numbers:
W2 = (0.66 x 0.65) + [(0.66 x 0.65)(1–0.66)]
W2 = 0.429 + 0.14586
W2 = 0.57486
The maths shows that Murray would theoretically have increased his percentage of points won on serve from 55.1% to 57.5% by serving two first serves. In a sport with margins as fine as those in tennis, this is worth taking note of.
#### Generalising
We’ve plugged in the numbers for this specific match, but the equations can be simplified to give a simple rule to answer the question ‘when should a player serve two first serves?’.
We’re looking for instances where W2 would be greater than W1. So:
W1 < W2
ab + y(1-a) < ab + ab(1-a)
y(1-a) < ab(1-a)
y < ab
Basically, if a player’s win percentage on second serve (y) is dipping below his or her first serve percentage (a) multiplied by his or her win percentage behind first serve (b), then he or she would theoretically be better off serving two first serves.
Andy Murray is particularly prone to this phenomenon, as his first serve is strong and generally consistent while his second serve is still relatively weak, even though he upgraded it significantly last year. But he’s not unique. The stats reveal a significant number of matches where a player’s hypothetical W2 percentage would have exceeded their actual W1 percentage. In Novak Djokovic’s surprise loss to Kei Nishikori in the 2014 US Open, for example, Djokovic only won 37% of points behind his second serve. He could hypothetically have improved his overall win percentage on serve from 62% to 67% by serving two first serves.
Without going back through all the records, one suspects that a large number of Murray’s matches against top players pre-2016 would also have had better W2 scores than W1 scores.
TL;DR: Going purely by the stats, Andy Murray could theoretically have won more points on serve in his match against Zverev by serving two first serves. There are also many other matches where the stats show that a player could have increased his or her win percentage on serve by serving two first serves.
#### So why don’t players serve two first serves?
The clue is probably in the word ‘theoretically’. Tennis isn’t all about maths; there is complex psychology at play. Perhaps the prospect of double faulting on around 15% of your service points (as you would if you served two first serves) would be so alarming that your serve would go to pieces entirely. Or perhaps you would be handing a massive psychological advantage to your opponent, by essentially saying ‘I don’t think I can get the better of you in a rally’.
The truth is probably a combination of these and other factors. But we are in an era when the game’s best returners — Djokovic, Murray, Nadal, Nishikori — have turned the second serve into a significant disadvantage for their opponents (all these players win more than 55% of their return points when facing a second serve).
Maybe the psychology of ‘I’m serving therefore I have the upper hand’ is an outdated hangover from the days of faster courts and less athletic opponents, and players should be thinking differently about their serves. As Big Bill Tilden said, ‘Never change a winning game; always change a losing one’. Whatever the psychology of the situation, there is no getting around the fact that 36% on your second serve is emphatically not a winning game, and you would think that a player in that situation should be willing to try some pretty radical tactics.
In a sport where players sleep in oxygen tents and calibrate their food intake down to the last calorie to gain a few fractions of a percent advantage over their opponents, there may yet be some unrealised marginal gains out there for a player who is willing and able to overcome the psychological factors and cash in on a few extra service points by ditching their second serve.
You can listen to the post-Murray-match episode of The Tennis Podcast here:
And you can view a full breakdown of the match stats here:
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# Subtract Numbers with Decimal Places (2)
In this worksheet, students subtract numbers with decimal places.
Key stage: KS 2
Curriculum topic: Number: Fractions, Decimals and Percentages
Curriculum subtopic: Use up to Three Decimal Places
Difficulty level:
### QUESTION 1 of 10
In this worksheet, we will learn about the subtraction of numbers with decimals.
Example:
Question:
Find the difference between 17.5 - 2.75.
Finding the difference means subtracting, so we need to work out 17.5 - 2.75
Step 1. Line up the decimal points and add zeros when needed, next to the 17.50.
Step 2. Subtract, starting with the hundredths place. We need to regroup, so 5 in the tenths place becomes 4 and 0 in the hundredths place becomes 10. Subtract 10 - 5 = 5.
Step 3. Move to the tenths place and regroup with the ones place (to the left of the decimal), so the 7 becomes 6 and the 4 in the tenths place becomes 14. Subtract 14 - 7 = 7.
Step 4. Move to the ones place, to the left of the decimal. Subtract 6 - 2 = 4 and then in the tens place, 1 - 0 = 1.
The answer is 14.75. Subtract as normal, but remember to line up the decimals so that the place value in the answer is correct.
Find the difference between the two numbers
729.7 - 290.5 = ________
Find the difference between the two numbers
733.1 - 698.7 = ________
Find the difference between the two numbers
961.9 - 568.4 = ________
Find the difference between the two numbers
828.6 - 379.6 = ________
Find the difference between the two numbers
989.4 - 829.3 = ________
Find the difference between the two numbers
407.1 - 380.7 = ________
Find the difference between the two numbers
422.2 - 420.6 = ________
Find the difference between the two numbers
941.7 - 545.6 = ________
Find the difference between the two numbers
476.9 - 416.5 = ________
Find the difference between the two numbers
969.4 - 870.6 = ________
• Question 1
Find the difference between the two numbers
729.7 - 290.5 = ________
439.2
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place, 7 - 5 = 2. Step 3. Move to the ones place, to the left of the decimal, and subtract 9 - 0 = 9. Step 4. Move to the tens place. We need to regroup, so 7 in the hundreds place becomes 6 and 2 in the tens place becomes 12. Subtract 12 - 9 = 3. Step 5. Subtract in the hundred place, 6 - 2 = 4. Answer: 439.2
• Question 2
Find the difference between the two numbers
733.1 - 698.7 = ________
34.4
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place. We need to regroup, so 3 in the ones place becomes 2 and 1 in the tenths place becomes 11. Subtract 11 - 7 = 4. Step 3. Move to the ones place, to the left of the decimal. We need to regroup, so 3 in the tens place becomes 2 and 2 in the ones place becomes 12. Subtract 12 - 8 = 4. Step 4. Move to the tens place. We need to regroup, so 7 in the hundreds place becomes 6 and 2 in the tens place becomes 12. Subtract 12 - 9 = 3. Step 5. Subtract in the hundreds place, 6 - 6 = 0. Answer: 34.4
• Question 3
Find the difference between the two numbers
961.9 - 568.4 = ________
393.5
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place, 9 - 4 = 5. Step 3. Move to the ones place, to the left of the decimal. We need to regroup, so 6 in the tens place becomes 5 and 1 in the ones place becomes 11. Subtract 11 - 8 = 3. Step 4. Move to the tens place. We need to regroup, so 9 in the hundreds place becomes 8 and 5 in the tens place becomes 15. Subtract 15 - 6 = 9. Step 5. Subtract in the hundreds place, 8 - 5 = 3. Answer: 393.5
• Question 4
Find the difference between the two numbers
828.6 - 379.6 = ________
449
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place, 6 - 6 = 0. Step 3. Move to the ones place, to the left of the decimal. We need to regroup, so 2 in the tens place becomes 1 and 8 in the ones place becomes 18. Subtract 18 - 9 = 9. Step 4. Move to the tens place. We need to regroup, so 8 in the hundreds place becomes 7 and 1 in the tens place becomes 11. Subtract 11 - 7 = 4. Step 5. Subtract in the hundreds place, 7 - 3 = 4. Answer: 449 or 449.0
• Question 5
Find the difference between the two numbers
989.4 - 829.3 = ________
160.1
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place, 4 - 3 = 1. Step 3. Move to the ones place, to the left of the decimal. Subtract 9 - 9 = 0. Step 4. Move to the tens place. Subtract 8 - 2 = 6. Step 5. Subtract in the hundreds place, 9 - 8 = 1. Answer: 160.1
• Question 6
Find the difference between the two numbers
407.1 - 380.7 = ________
26.4
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place. We need to regroup, so 7 in the ones place becomes 6 and 1 in the tenths place becomes 11. Subtract 11 - 7 = 4. Step 3. Move to the ones place, to the left of the decimal. Subtract 6 - 0 = 6. Step 4. Move to the tens place. We need to regroup, so 4 in the hundreds place becomes 3 and 0 in the tens place becomes 10. Subtract 10 - 8 = 2. Step 5. Subtract in the hundreds place, 3 - 3 = 0. Answer: 26.4
• Question 7
Find the difference between the two numbers
422.2 - 420.6 = ________
1.6
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place. We need to regroup, so 2 in the ones place becomes 1 and 2 in the tenths place becomes 12. Subtract 12 - 6 = 6. Step 3. Move to the ones place, to the left of the decimal. Subtract 1 - 0 = 1. Step 4. Move to the tens place. Subtract 2 - 2 = 0. Step 5. Subtract in the hundreds place, 4 - 4 = 0. Answer: 1.6 or 001.6
• Question 8
Find the difference between the two numbers
941.7 - 545.6 = ________
396.1
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place, 7 - 6 = 1. Step 3. Move to the ones place, to the left of the decimal. We need to regroup, so 4 in the tens place becomes 3 and 1 in the ones place becomes 11. Subtract 11 - 5 = 6. Step 4. Move to the tens place. We need to regroup, so 9 in the hundreds place becomes 8 and 3 in the tens place becomes 13. Subtract 13 - 4 = 9. Step 5. Subtract in the hundreds place, 8 - 5 = 3. Answer: 396.1
• Question 9
Find the difference between the two numbers
476.9 - 416.5 = ________
60.4
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place, 9 - 5 = 4. Step 3. Move to the ones place, to the left of the decimal. Subtract 6 - 6 = 0. Step 4. Move to the tens place. Subtract 7 - 1 = 6. Step 5. Subtract in the hundreds place, 4 - 4 = 0. Answer: 60.4 or 060.4
• Question 10
Find the difference between the two numbers
969.4 - 870.6 = ________
98.8
EDDIE SAYS
Step 1. Line up the decimal points. Step 2. Subtract, starting with the tenths place. We need to regroup, so 9 in the ones place becomes 8 and 4 in the tenths place becomes 14. Subtract 14 - 6 = 8. Step 3. Move to the ones place, to the left of the decimal. Subtract 8 - 0 = 8. Step 4. Move to the tens place. We need to regroup, so 9 in the hundreds place becomes 8 and 6 in the tens place becomes 16. Subtract 16 - 7 = 9. Step 5. Subtract in the hundreds place, 8 - 8 = 0. Answer: 98.8 or 098.8
---- OR ----
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## Exploding Dots
### 4.4 (Optional) The Traditional Algorithm
Lesson materials located below the video overview.
How does this dots-and-boxes approach compare with the standard algorithm?
Consider again $$512-347$$.
The standard algorithm has you start at the right to first look at “$$2$$ take away $$7$$,” which you can’t do.
(Well you can do it, it is $$-5$$, but you are not to write that for this algorithm.)
So what do you do?
You “borrow one.” That is, you take a dot from the tens column and unexplode it to make ten ones. That leaves zero dots in the tens column. We should write ten ones to go with the two in the ones column.
But we are a bit clever here and just write $$12$$ rather than $$10+2$$. (That is, we put a $$1$$ in front of the $$2$$ to make it look like twelve.)
Then we say “twelve take seven is five” and write that answer.
The rightmost column is complete. Shift now to the middle column.
We see “zero take away four,” which can’t be done. So perform another unexplosion, that is, another “borrow,” to see $$10-4$$ in that column. We write the answer $$6$$.
We then move to the last remaining column where we have $$4-3$$, which is $$1$$.
Phew!
Here is a question for you to try, if you like. My answer to it appear in the final section of this chapter.
1. Compute each of the following two ways: the dots-and-boxes way (and fixing the answer for society to read) and then with the traditional algorithm. The answers should be the same.
Thinking question along the way: As you fix up your answers for society, does it seem easier to unexplode from left to right, or from right to left?
Additional question: Do you think you could become just as speedy the dots-and-boxes way as you currently are with the traditional approach?
Again. All correct approaches to mathematics are correct, and it is just a matter of style as to which approach you like best for subtraction. The traditional algorithm has you work from right to left and do all the unexplosions as you go along. The dots-and-boxes approach has you “just do it!” and conduct all the unexplosions at the end. Both methods are fine and correct.
## Books
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## Guides & Solutions
Dive deeper into key topics through detailed, easy to follow guides and solution sets.
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# Quadratic Simultaneous Equations. How to solve a pair of linear and quadratic equations by substitution.
Updated on January 2, 2013
There are many ways of solving a pair of equations when one is linear and the other is quadratic. But the easiest way is to substitute the linear equation into the quadratic equation. Before you attempt the example below, make sure you can solve a pair of linear equations using the substitution method. You can use these 5 steps below to help you solve a pair of quadratic and linear equations:
1) Make x or y the subject of the linear equation. If there is a 1x then make the x value the subject or if there is a 1y make y the subject – this will make the algebra easier to handle in later steps.
2) Next substitute your linear equation into the quadratic equation. Make sure you put a bracket around the algebra that you substituted into the equation. You should now have an equation just in terms of x or just in terms of y.
3) Next expand the brackets and simplify the equation so that is in the form ax2 + bx + c = 0
4) Factorise the equation and solve to find the value or values of x. If the equation cannot be factorised then apply the quadratic formula.
5) Finally substitute your x values back into the linear equation to work out the corresponding y values.
Let’s look at an example of solving simultaneous equations where one is linear and the other is quadratic.
Example 1
Solve this pair of simultaneous equations:
x2+ y2= 10 (A)
y = x – 2 (B)
1) Make x or y the subject of the linear equation. If there is a 1x then make the x value the subject or if there is a 1y make y the subject – this will make the algebra easier to handle in later steps.
y is already the subject of the linear equation so you need to put this into equation A.
2) Next substitute your linear equation into the quadratic equation. Make sure you put a bracket around the algebra that you substituted into the equation. You should now have an equation just in terms of x or just in terms of y.
Next put this linear equation into the quadratic equation.
x2 + (x-2)2 = 10
3) Next expand the brackets and simplify the equation so that is in the form ax2 + bx + c = 0
x2 + (x-2)(x-2) = 10
x2 + x2 -2x -2x + 4 = 10
2x2 -4x + 4 = 10
2x2 – 4x -6 = 0
Divide this equation by 2 as all the terms are multiples of 2:
x2 – 2x -3 = 0
4) Factorise the equation and solve to find the value or values of x. If the equation cannot be factorised then apply the quadratic formula.
(x+1)(x-3) = 0
So either x + 1 = 0 or x -3 = 0
Which gives x = -1 or x = 3
5) Finally substitute your x values back into the linear equation to work out the corresponding y values.
Start with x = -1 and substitute this into y = x – 2
y = -1 -2
y = -3
And when x = 3
y = 3 – 2
y = 1
So the solutions to this quadratic simultaneous equation are (-1,-3) and (3,1)
5
0
3
3
31
79
10 |
# NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3
NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3 – Linear Equation in one variable, has been designed by the NCERT to test the knowledge of the student on the topic – Solving Equations having the Variable on both Sides
### NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3
NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3
Solve the following and check your results.
1. 3x=2x+18
Solution:
3x-2x=18
x=18
To check: 3x=2x+18
3×(18)=2×(18)+18
54=36+18
54=54
LHS = RHS
Hence, x=18 is the correct solution
2. 5t-3=3t-5
Solution:
5t-3t=3-5
2t=-2
t=-(2/2)
t=-1
To check: 5t-3=3t-5
5×(-1)-3=3×(-1)-5
-5-3=-3-5
-8=-8
LHS = RHS
Hence, t=-1 is the correct solution
3. 5x+9=5+3x
Solution:
5x-3x=5-9
2x=-4
x=-(4/2)
x=-2
To check: 5x+9=5+3x
5×(-2)+9=5+3×(-2)
-10+9=5+(-6)
-1=-1
LHS = RHS
Hence, x=-2 is the correct solution
4. 4z+3=6+2z
Solution:
4z – 2z = 6 – 3
2z = 3
z = (3/2)
To check: 4z+3=6+2z
4 × (3/2) + 3 = 6 + 2x(3/2)
6 + 3 = 6 + 3
9 = 9
LHS = RHS
Hence, z = (3/2) is the correct solution
5. 2x-1=14-x
Solution:
2x+x=14+1
3x=15
x=(15/3)
x=5
To check: 2x-1=14-x
2×(5)-1=14-(5)
10-1=14-5
9=9
LHS = RHS
Hence, x=5 is the correct solution
6. 8x+4=3 (x-1)+7
Solution:
8x+4=3x-3+7
8x+4=3x+4
8x-3x=0
5x=0
x=0
To check: 8x+4=3 (x-1)+7
8×0+4=3 (0-1)+7
4=-3+7
4=4
LHS = RHS
Hence, x=0 is the correct solution
7. x=(4/5)(x+10)
Solution:
5x=4(x+10)
5x=4x+40
5x-4x=40
x=40
To check: x=(4/5)(x+10)
40=(4/5)(40+10)
40=(4/5) (50)
40=40
LHS = RHS
Hence, x=40 is the correct solution
8. (2x/3)+1 = (7x/15)+3
Solution:
(2x/3) – (7x/15) = 3-1
(3x/15) = 2
x=(2 × 15)/3
x=(30/3)
x=10
To check: (2x/3)+1=(7x/15)+3
(2 × 10)/3+1=(7 × 10)/15+3
(20/3)+1=(70/15)+3
(23/3)=(115)/15
(23/3) = (23)/3
LHS = RHS
Hence, x=10 is the correct solution
9. 2y + (5/3) = (26/3) – y
Solution:
2y + y = (26/3) – (5/3)
3y = (26-5)/3
3y = (21/3)
3y=7
y=(7/3)
To check: 2y+(5/3)=(26/3)-y
2×(7/3)+(5/3)=(26/3)-(7/3)
(14/3)+(5/3)=(19/3)
(19/3) = (19/3)
LHS = RHS
Hence, y=(7/3) is the correct solution
10. 3m=5m-(8/5)
Solution:
5m-3m=(8/5)
2m=(8/5)
m= (8/5×2)
m = 8/10 = 4/5
m = 4/5
To check: 3m = 5m – (8/5)
3×(4/5)=5×(4/5)-(8/5)
(12/5)=(20/5)-(8/5)
(12/5)=(12/5)
LHS = RHS
Hence, m=(4/5) is the correct solution
The next Exercise for NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.4 – Linear Equation in one variable can be accessed by clicking here |
# Video: AQA GCSE Mathematics Foundation Tier Pack 4 • Paper 2 • Question 15
There are 439 gallons of water in a pond. 1 gallon = 8 pints. 1 pint = 570 millilitres. Use the approximations above to show that the pond contains approximately 2000 litres of water.
03:00
### Video Transcript
There are 439 gallons of water in a pond. One gallon equals eight pints. One pint equals 570 millilitres. Use the approximations above to show that the pond contains approximately 2000 litres of water.
So therefore, the first thing we’re gonna do is use the fact that we know that one gallon is equal to eight pints. And if we have 439 gallons of water, then we want to work out how many pints this is going to be. Well, if we had one gallon and we want 439 gallons, then we’re gonna have to multiply this by 439. So therefore, if we want to work out how many pints this is going to be, we’re gonna multiply eight by 439. So this is gonna give us a total of 3512 pints. And that’s what we get when we multiply eight by 439. Okay, great. So we now know what our gallons are in pints.
So now, what we want to do is convert our 3512 pints into millilitres. And to do that, we’re gonna use the conversion that one pint is equal to 570 millilitres. So therefore, to work this out, what we’re gonna do is multiply each side of the conversion by 3512 because that’s what number of pints are. So when we do that, we’re gonna multiply 570 by 3512. So this is gonna give us that 3512 pints is equal to 2001840 millilitres. So have we finished here? No, because what we want to do is show that the pond contains approximately 2000 litres of water. And we’re currently in millilitres. So now, what we want to do is convert to litres.
Well, now what we’re gonna do is use an exact conversion. And that is that one litre is equal to 1000 millilitres. So therefore, we can say that 2001840 millilitres is gonna be equal to 2001840 divided by 1000 litres. And that’s because, as we said, one litre is 1000 millilitres. So if we want to convert back, we’re gonna divide by 1000. Well, this is gonna give us 2001.84 litres. So we know that 439 gallons of water is equal to 2001.84 litres. But we’re asked to show that it contains approximately 2000 of litres within the pond. Well, 2001.84 litres is approximately 2000 litres.
So therefore, we’ve used the approximations and shown that the pond contains approximately 2000 litres of water. |
# Solving Recurrence Relations (Part I)
## Introduction
In the previous post, we introduced the concept of recurrence relations. In this article and the following two articles, we will learn how to solve the recurrence relations to get the running time of recursive algorithms. Solving the recurrence relation means finding the closed form expression in terms of $n$. There are various techniques available to solve the recurrence relations. Some techniques can be used for all kind of recurrence relations and some are restricted to recurrence relations with a specific format.
## Forward substitution method
One of the simplest methods for solving simple recurrence relations is using forward substitution. In this method, we solve the recurrence relation for $n = 0, 1, 2, …$ until we see a pattern. Then we make a guesswork and predict the running time. The final and important step in this method is we need to verify that our guesswork is correct by using the induction. Consider a recurrence relation
$$T(n) = \begin{cases} 1 & \text{ if } n = 1 \\ T(n - 1) + 1 & \text{otherwise} \end{cases}$$
We can calculate the running time for $n = 0, 1, 2, ..$ as follows
$n$$T(n) 11 2T(2 - 1) + 1 = 1 + 1 = 2 3T(3 - 1) + 1 = 1 + 1 + 1 = 3 4T(4 - 1) + 1 = 1 + 1 + 1 + 1 = 4 We can easily see the pattern here. When the value of n = k, T(n) = k. So the running time is$$T(n) = n$$We need to verify that the running time we guessed is correct by induction (not mathematical induction ;)). Since T(n) = n, we can clearly see that T(1) = 1, T(2) = 2, T(3) = 3, …. Consider another example$$T(n) = \begin{cases} 1 & \text{ if } n = 1 \ 2T(n - 1) + 1 & \text{otherwise} \end{cases}$$As above, we can calculate the running time for n = 0, 1, 2, …, as follows n$$T(n)$
11
2$2.T(2 - 1) + 1 = 2.1 + 1 = 3$
3$2.T(3 - 1) + 1 = 2.2.1 + 2.1 + 1 = 7$
42.$T(4 - 1) + 1 = 2.2.2.1 + 2.2.1 + 2.1 + 1 = 15$
52.$T(5 - 1) + 1 = 2.2.2.2.1 + 2.2.2.1 + 2.2.1 + 2.1 + 1 = 31$
We already started to see the pattern here. When the value of $n$ is $k$ the patter would be
$$T(k) = 2^{k - 1} + 2^{k - 2} + … + 2^0$$
This is a geometric series whose sum can be easily calculated to be $2^k - 1$. Therefore, the running time of the algorithm is,
$$T(n) = 2^n - 1$$
The correctness of this running time can be easily proved by putting $n = 1, 2, 3, …$ in above equation.
## Back substitution method
In forward substitution method, we put $n = 0, 1, 2, …$ in the recurrence relation until we see a pattern. In backward substitution, we do the opposite i.e. we put $n = n, n - 1, n - 2, …$ or $n = n, n/2, n/4, …$ until we see the pattern. After we see the pattern, we make a guesswork for the running time and we verify the guesswork. Let us use this method in some examples.
Consider an example recurrence relation given below
$$T(n) = \begin{cases} 1 & \text{ if } n = 1 \\ 2T \left (\frac{n}{2} \right) + n & \text{otherwise} \end{cases}$$
Given $T(n)$, we can calculate the value of $T(n/2)$ from the above recurrence relation as
$$T(n/2) = 2T\left ( \frac{n}{4} \right ) + \frac{n}{2}$$
Now we back substitute the value of $T(n/2)$ in $T(n)$
$$T(n) = 2^2T\left (\frac{n}{2^2}\right ) + 2n$$
We proceed in a similar way
\begin{align}T(n) &= 2^3T \left (\frac{n}{2 ^3} \right ) + 3n \\ &= 2^4T \left (\frac{n}{2^4} \right ) + 4n\\ &= 2^kT \left (\frac{n}{2^k} \right ) + kn \end{align}
Now we should use the boundary (base) condition i.e. $T(1) = 1$. In order to use the boundary condition, the entity inside $T()$ must be 1 i.e.
$$\frac{n}{2^k} = 1$$
Taking $\log_2$ on both sides,
$$n = \log_2 n$$
The equation (6) becomes
\begin{align} T(n) &= 2^{\log_2 n}T \left (\frac{n}{2^{\log_2 n}} \right ) + \log_2 n.n \\ & = nT(1) + n\log_2 n \\ & = n\log_2 n + n \end{align}
The correctness of above running time can be proved using induction. Put $n = 2, 4, 8, 16, …$ and you can easily verify that the guessed running time is actually correct.
We rarely use forward and backward substitution method in the practical cases. There are much more sophisticated and fast methods. But these methods can be used as a last resort when other methods are powerless to solve some kinds of recurrences.
## Homogeneous recurrences
Some recurrences take the form
$$a_0T(n) + a_1T(n - 1) + … +a_kT(n - k) = 0$$
This recurrence is called Homogeneous linear recurrences with constant coefficients and can be solved easily using the techniques of characteristic equation. The steps to solve the homogeneous linear recurrences with constant coefficients is as follows.
1. Write the recurrence relation in characteristic equation form.
2. Change the characteristic equation into characteristic polynomial of degree $k$.
3. Find the roots of the characteristic polynomial. A polynomial $p(x)$ of degree $k$ has exactly $k$ roots i.e. $r_1, r_2, …, r_k$.
4. The solution of this recurrence relation, if the roots are distinct, is
$$T(n) = \sum_{i = 1}^k c_ir_i^n$$
Where $c_1, c_2, …, c_k$ are constants.
5. Find the value of constants $c_1, c_2, …, c_k$ by using the boundary conditions. If there are 3 constants then we need 3 equations.
6. If the roots are not distinct then the solution becomes
$$T(n) = \sum_{i = 1}^l \sum_{j = 0}^{m_i -1 } c_{ij}n^jr_i^n$$
Where $r_1, r_2,…, r_l$ are the $l$ distinct roots of the characteristics polynomial and $m_1, m-2, …, m_l$ are their multiplicities respectively.
We use these steps to solve few recurrence relations starting with the Fibonacci number. The Fibonacci recurrence relation is given below.
$$T(n) = \begin{cases} n & \text{ if } n = 1 \text{ or } n = 0\\ T(n - 1) + T(n - 2) & \text{otherwise} \end{cases}$$
First step is to write the above recurrence relation in a characteristic equation form. For this, we ignore the base case and move all the contents in the right of the recursive case to the left i.e.
$$T(n) - T(n - 1) - T(n - 2) = 0$$
Next we change the characteristic equation into a characteristic polynomial as
$$x^2 - x - 1 = 0$$
The roots of this characteristic polynomial are
$$r_1 = \frac{1 + \sqrt{5}}{2} \text{ and } r_2 = \frac{1 - \sqrt{5}}{2}$$
Since the roots are distinct, the solution is, therefore, of the form
$$T(n) = c_1r_1^n + c_2r_2^n = c_1\left (\frac{1 + \sqrt{5}}{2} \right )^n + c_2 \left (\frac{1 - \sqrt{5}}{2} \right )^n$$
The value of $c_1$ and $c_2$ can be calculated using the base conditions. We know that $T(0) = 0$ and $T(1) = 1$. If we put these two values we get,
\begin{align} & c_1 + c_2 = 0 \\ & r_1c_1 + r_2c_2 = 1\end{align}
Solving these equations, we get
$$c_1 = \frac{1}{\sqrt{5}} \text{ and } c_2 = -\frac{1}{\sqrt{5}}$$
Thus,
$$T(n) = \frac{1}{\sqrt{5}}\Bigg [ \left (\frac{1 + \sqrt{5}}{2} \right )^n - \left (\frac{1 - \sqrt{5}}{2} \right )^n \Bigg ]$$
Consider another recurrence,
$$T(n) = \begin{cases} n & \text{ if } n = 0, 1 \text{ or } 2\\ 5T(n - 1) - 8T(n - 2) + 4T(n - 3) & \text{otherwise} \end{cases}$$
First we write the characteristic equation
$$T(n) - 5T(n - 1) + 8T(n - 2) - 4T(n - 3) = 0$$
The characteristic polynomial is,
$$x^3 - 5x^2 + 8x - 4 = 0$$
The roots are $r_1 = 1$ of multiplicity $m_1 = 1$ and $r_2 = 2$ of multiplicity $m_2 = 2$. Since roots are repeated, the solution is
$$T(n) = c_11^n + c_22^n + c_3n2^n$$
The base conditions give,
\begin{align} & c_1 + c_2 = 0 \\ & c_1 + 2c_2 + 2c_3 = 1 \\ & c_1 + 4c_2 + 8c_3 = 2 \end{align}
Solving these equations, we get $c_1 = -2, c_2 = 2$ and $c_3 = -\frac{1}{2}$. Therefore
$$T(n) = 2^{n + 1} - n2^{n - 1} - 2$$
## Inhomogeneous recurrences
In homogeneous recurrences the linear combination of $T(n - i)$ terms is zero. But in inhomogeneous recurrences, the linear combination is not equal to zero and therefore the solution is more difficult than the homogeneous recurrences. Inhomogeneous recurrences take the following form
$$a_0T(n) + a_1T(n - 1) + … + a_kT(n - k) = b^np(n)$$
Where $b$ is a constant and $p(n)$ is a polynomial in $n$ of degree $d$.
To get the characteristic polynomial of an inhomogeneous recurrence, we follow the exact same procedure as the homogeneous for the left part of the equation and for the right part we multiply the characteristic polynomial by $(x - b) ^ {d + 1}$
Example 1: Consider the recurrence,
$$T(n) - 2T(n - 1) = 3^n$$
The right side must be in $b^np(n)$ format. We can easily change the right side to match the format as
$$T(n) - 2T(n - 1) = 3^nn^0$$
Here $b = 3$ and $d = 0$. The characteristic polynomial of the left part of the equation is (exactly same as the homogeneous one)
$$T(n) = (x - 2) = 0$$
Now we multiply this with $(x - b) ^ {d + 1}$ to get
$$(x - 2)(x - 3) = 0$$
This is the characteristic polynomial and we can easily solve this using the techniques discussed in the homogeneous section.
Example 2: Consider another recurrence,
$$T(n) = 2T(n - 1) + n$$
This can be written as
$$T(n) - 2T(n - 1) = 1^nn$$
Comparing the right hand side with $b^np(n)$, we get $b = 1$ and $d = 1$. Therefore the characteristic polynomial is
$$(x - 2)(x - 1)^2 = 0$$
This has repeated roots and we can solve this using the techniques discussed above.
## Change of variable
Sometimes changing the variable in a recurrence relation helps to solve the complicated recurrences. By changing the variable, we can convert some complicated recurrences into linear homogeneous or inhomogeneous form which we can easily be solved. At last, we put the original variable back to the recurrence to get the required solution. Let me explain this with the help of an example.
Example: Consider a recurrence
$$T(n) = 3T(n/2) + n$$
In above recurrence relation, the value of $n$ is reduced to half in every iteration. If we replace $n$ with $2^i$, we get
$$T(2^i) = 3T(2^{i - 1}) + 2^i$$
If we let $T(2^i) = t_i$, $T(2^{i - 1})$ becomes $t_{i - 1}$ and the recurrence relation converts to
$$t_i - 3t_{i - 1} = 2^i$$
This is linear inhomogeneous equation with $b = 1$ and $d = 0$. The characteristic polynomial is
$$(x - 3)(x - 2) = 0$$
It has roots $r_1 = 3$ and $r_2 = 2$. The solution will be in the form
$$t_i = c_13^i + c_22^i$$
Initially we changed variable $n$ to $2^i$. $n$ and $2^i$ will be equal when $i = \log_2 n$. Therefore,
$$t_{\log_2 n} = c_13^{\log_2 n} + c_22^{\log_2 n}$$
We let $T(2^i) = t_i$, and thus $T(n) = t_{\log_2 n}$
$$T(n) = c_13^{\log_2 n} + c_22^{\log_2 n}$$
or
$$T(n) = c_1n^{\log_2 3} + c_2n$$
Using the base conditions, value of $c_1$ and $c_2$ can be easily calculated.
## References
1. Brassard, G., & Bratley, P. (2008). Fundamentals of Algorithmics. New Delhi: PHI Learning Private Limited. |
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# Square Properties
A square is a quadrilateral with four equal sides and four right angles. Each diagonal of a square bisects the opposite angles and intersects at 90 degrees. Understanding the properties of squares is essential for mastering geometry concepts.
## Definition of Square Properties
A square is a special type of quadrilateral that has unique properties. Understanding these properties is crucial in various areas of mathematics, including geometry and algebra. Here, you will delve into the main characteristics and geometrical attributes of squares.
### Basic Characteristics
The most fundamental property of a square is that all four of its sides are of equal length. Additionally, the internal angles of a square are all right angles (90 degrees). These two characteristics distinguish squares from other types of quadrilaterals, such as rectangles and rhombuses.
Square: A quadrilateral with four equal sides and four equal angles (each 90 degrees).
Remember, a square is a special type of both rectangle and rhombus because it meets the criteria of both.
### Diagonals of a Square
Another important property of squares is related to their diagonals. In a square, the diagonals are equal in length and they bisect each other at right angles. This means that each diagonal divides the square into two congruent triangles. The length of a diagonal in a square can be calculated using the Pythagorean theorem.
If the side length of a square is denoted as $$a$$, the length of each diagonal $$d$$ is given by:
$d = a\sqrt{2}$
Why does the diagonal of a square equal $$a\sqrt{2}$$? This is derived from the Pythagorean theorem: $$a^2 + a^2 = d^2$$, which simplifies to $$d = a\sqrt{2}$$. Hence, knowing the side length of the square allows you to find the exact measurement of the diagonal.
### Perimeter and Area
The perimeter of a square is straightforward to calculate because it is the sum of all its sides. Since all four sides are equal, the formula for the perimeter $$P$$ is:
$P = 4a$
The area of a square is the amount of space enclosed within its sides. The formula for the area $$A$$ is:
$A = a^2$
If a square has a side length of 5 units, the perimeter is calculated as:
$P = 4 \times 5 = 20\text{ units}$
The area is:
$A = 5^2 = 25\text{ square units}$
Perimeter is a linear measurement, while area is a square measurement.
## Key Properties of a Square
A square is a type of quadrilateral that has several unique properties. These properties are fundamental in understanding various mathematical concepts and solving problems. In this section, you will learn about the key properties of squares.
### Equal Sides
One of the defining characteristics of a square is that all four sides are of equal length. This property not only distinguishes squares from other quadrilaterals but also provides a basis for many geometric calculations and proofs.
Equal Sides of a Square: In a square, each side is of equal length, often denoted as a.
If a square has a side length of 7 units, then each of its four sides is 7 units long.
This property of equal sides helps in calculating the perimeter and area of the square.
### Right Angles
Another key property of a square is that each of its internal angles is a right angle. This means that each angle measures 90 degrees, making the square a type of rectangle. This property is essential when studying the geometric properties and theorems related to squares.
Right Angles in a Square: All four internal angles of a square are right angles, measuring 90 degrees each.
If you draw the diagonals of a square, they meet at right angles, further illustrating this property.
Each right angle ensures that squares have the same angle properties as rectangles.
### Diagonals
Diagonals of a square have their own set of properties. In a square, the diagonals are equal in length and they bisect each other at right angles. Furthermore, they divide the square into two equal right triangles. The length of a diagonal can be calculated using the Pythagorean theorem.
If the side length of a square is denoted as $$a$$, the length of each diagonal $$d$$ is given by:
$d = a\sqrt{2}$
Why does the diagonal of a square equal $$a\sqrt{2}$$? This is derived from the Pythagorean theorem: $$a^2 + a^2 = d^2$$, which simplifies to $$d = a\sqrt{2}$$. Hence, knowing the side length of the square allows you to find the exact measurement of the diagonal.
The diagonals of a square not only help in dividing the square into equal parts but also intersect at right angles.
## Formula for Square Properties
Understanding the formulas related to the properties of squares is essential in various areas of mathematics. These formulas help in calculating the perimeter, area, and diagonal lengths of squares.
### Perimeter Formula
The perimeter of a square is the total length around the square. Since all four sides of a square are equal, the perimeter can be calculated by the following formula:
$P = 4a$
If a square has a side length of 6 units, the perimeter is:
$P = 4 \times 6 = 24\text{ units}$
The perimeter is simply the sum of all sides of the square multiplied by 4.
### Area Formula
The area of a square is the region enclosed within its four sides. The formula for calculating the area is:
$A = a^2$
If a square has a side length of 8 units, the area is:
$A = 8^2 = 64\text{ square units}$
Always remember that area is a measure of space, so it is expressed in square units.
### Diagonal Formula
The diagonals of a square have special properties. They are equal in length and bisect each other at right angles. The length of a diagonal can be calculated using the following formula:
$d = a\sqrt{2}$
If a square has a side length of 5 units, the length of each diagonal is:
$d = 5\sqrt{2} \approx 7.07\text{ units}$
The formula for the diagonal length $$d$$ originates from the Pythagorean theorem applied to the right triangles formed by the diagonals. If each side of the square is $$a$$, then:
$a^2 + a^2 = d^2$
Simplifying this, you get:
$2a^2 = d^2$
Hence, $$d = a\sqrt{2}$$. This means that knowing the side length allows for the exact calculation of the diagonal.
The diagonals intersect each other at right angles and form two equal right triangles within the square.
## Examples of Square Properties
Having a solid understanding of square properties is fundamental in geometry. Here, several examples of these properties are shown to illustrate their significance and application.
### Example 1: Calculating the Perimeter
Consider a square with a side length of 4 units. The formula to calculate the perimeter (P) of a square is:
$P = 4a$
Substituting the given side length:
$P = 4 \times 4 = 16\text{ units}$
The perimeter of a square is just four times the length of one of its sides.
### Example 2: Calculating the Area
If a square has a side length of 5 units, the area (A) can be calculated using the formula:
$A = a^2$
Substituting the side length:
$A = 5^2 = 25\text{ square units}$
The area of a square is the side length squared.
### Example 3: Determining the Diagonal Length
For a square with a side length of 3 units, the diagonal (d) length can be calculated using the formula:
$d = a\sqrt{2}$
Substituting the given side length:
$d = 3\sqrt{2} \approx 4.24\text{ units}$
This diagonal length formula is derived from the Pythagorean theorem. In a square, each diagonal divides it into two right-angled triangles. Thus, letting each side be denoted as a:
$a^2 + a^2 = d^2$
Simplifying, we get:
$2a^2 = d^2$
Thus, $$d$$ can be represented as:
$d = a\sqrt{2}$
Diagonal: A line segment connecting two non-adjacent vertices of a polygon.
### Example 4: Sum of Interior Angles
Each interior angle of a square is 90 degrees. Since a square has 4 interior angles, their sum is:
$90\times4 = 360 \text{ degrees}$
The sum of the interior angles of any quadrilateral is always 360 degrees.
## Square Properties - Key takeaways
• Definition of Square Properties: A square is a quadrilateral with four equal sides and four right angles (90 degrees each).
• Diagonal Properties: Diagonals of a square are equal in length, bisect each other at right angles, and divide the square into two congruent triangles. The length of each diagonal is given by the formula d = a√2.
• Perimeter Formula: The perimeter P of a square can be calculated using the formula: P = 4a.
• Area Formula: The area A of a square is calculated as: A = a².
• Examples of Calculations: The perimeter and area of a square, as well as the lengths of its diagonals, can be calculated using the respective formulas, such as P=4a, A=a², and d=a√2.
#### Flashcards in Square Properties 12
###### Learn with 12 Square Properties flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What are the defining properties of a square?
A square has four equal sides and four right angles. Its opposite sides are parallel, and its diagonals bisect each other at right angles and are of equal length. A square is a regular polygon and a type of rectangle and rhombus.
How do the properties of a square differ from those of a rectangle?
A square has all sides equal and all angles at 90 degrees. A rectangle has opposite sides equal and all angles at 90 degrees. Thus, a square is a specific type of rectangle with additional properties of equal side lengths.
How can the properties of a square be used to solve problems in geometry?
The properties of a square, such as equal side lengths, right angles, and symmetrical axes, can simplify calculations involving area, perimeter, and diagonal lengths. They aid in determining coordinates, analysing transformations, and solving complex geometric problems by reducing them to simpler, well-defined structures.
What are the symmetry properties of a square?
A square has four lines of symmetry (along its diagonals and midlines). It also exhibits rotational symmetry of order 4, meaning it can be rotated by 90°, 180°, 270°, and 360° and still look the same.
How do the angles of a square relate to its properties?
Each angle of a square is a right angle, measuring 90 degrees. These equal angles ensure the square has four sides of equal length and opposite sides parallel. Additionally, the angles contribute to the square's symmetrical properties and its classification as a type of rectangle.
## Test your knowledge with multiple choice flashcards
How can the length of a diagonal in a square be calculated?
What is a defining characteristic of a square?
What is the formula for the perimeter of a square with side length $$a$$?
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# Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 216: 27
$f{\,^,}\left( x \right) = \frac{{60{x^3} + 57{x^2} - 24x + 13}}{{{{\left( {5x + 4} \right)}^2}}}$
#### Work Step by Step
$\begin{gathered} f\,\left( x \right) = \frac{{\,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)}}{{5x + 4}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,the\,\,derivative \hfill \\ f{\,^,}\left( x \right) = \frac{{\,\left( {5x + 4} \right)\,\,{{\left[ {\,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)} \right]}^,} - \,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)\,{{\left( {5x + 4} \right)}^,}}}{{\,{{\left( {5x + 4} \right)}^2}}} \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{\left[ {\,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)} \right]^,} \hfill \\ f{\,^,}\left( x \right) = \frac{{\left( {5x + 4} \right)\left[ {\,\left( {3{x^2} + 1} \right)\,\,\left( 2 \right)\left( {2x - 1} \right)\,\left( {6x} \right)} \right] - \,\left( {3{x^2} + 1} \right)\,\left( {2x - 1} \right)\,\left( 5 \right)\,\,}}{{\left( {5x + 4} \right){\,^2}\,}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ f{\,^,}\left( x \right) = \frac{{\left( {5x + 4} \right)\,\left( {6{x^2} + 2 + 12{x^2} - 6x} \right) - \,\left( {3{x^2} + 1} \right)\,\left( {10x - 5} \right)}}{{{{\left( {5x + 4} \right)}^2}}} \hfill \\ f{\,^,}\left( x \right) = \frac{{\,\left( {5x + 4} \right)\,\left( {18{x^2} - 6x + 2} \right) - \,\left( {30{x^3} - 15{x^2} + 10x - 5} \right)}}{{{{\left( {5x + 4} \right)}^2}}} \hfill \\ f{\,^,}\left( x \right) = \frac{{90{x^3} - 30{x^2} + 10x + 72{x^2} - 24x + 8 - 30{x^3} + 15{x^2} - 10x + 5}}{{{{\left( {5x + 4} \right)}^2}}} \hfill \\ f{\,^,}\left( x \right) = \frac{{60{x^3} + 57{x^2} - 24x + 13}}{{{{\left( {5x + 4} \right)}^2}}} \hfill \\ \end{gathered}$
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This Rational root theorem solver supplies step-by-step instructions for solving all math troubles. Solve quadratic formula is the process of finding the value of the solution to a quadratic equation. This is one of the most common operations that students must learn in elementary and high school math classes. This problem can be solved in many different ways depending on what type of equation you are working with. The most common method is to use the quadratic formula, which involves solving for the root of the quadratic equation. There are also several other methods for solving quadratic equations. These include factoring, solving by completing the square, and trinomials. Factoring can be useful when dealing with non-square roots of numbers or when dealing with complex numbers. Solving by completing the square will only work if you know what values will give you a perfect square. Trinomials are an advanced method that requires knowledge of exponents and trigonometry, but they can also be used to solve quadratic equations as well as linear equations.
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Introduction In mathematics, an inequality is a statement that suggests that two things are not equal. Inequality equations are mathematical problems that involve finding the value of a variable that will make the two sides of the equation equal. In order to solve these equations, you must use algebraic methods to isolate the variable on one side of the equation. There are a few different types of inequality equations that you might encounter. The most common type is a linear inequality, which is an inequality that can be
Solve for x is the process of determining a value for a variable when given only one variable to work with. It is useful when solving for unknown values in different equations. For example, you could solve for the value of a variable in an equation by using the formula Solve for x = x + Solving for x involves solving an equation with one variable to determine the value of another variable. The solver chooses two variables—one to be solved and one to be used as a reference point. The solver then divides both sides of the equation by their respective reference points. The resulting numbers are added to find the solution. Solve for x can also be used when setting up an experiment or running a simulation. For example, if you want to establish an experiment where you measure the length of a sample, you can solve for x using the formula L = L 0 (1 + t)2 where L0 is the length of the sample at time t=0, and t is time in seconds.
Expression is a math word that means to write something as an equation. For example, 2 + 3 would be written as (2+3). There are many types of expressions in math. One type of expression is an equation. An equation is just a math word that means to write something as an equation. For example, 2 + 3 would be written as (2+3). Another type of expression is an equation with variables. In this type of expression, the variables replace the numbers in the equation. For example, x = 2 + 3 would be written as x = (2+3). A third type of expression is a variable in an equation. In this type of expression, the variable stands for one of the numbers in the equation. For example, x = 2 + 3 would be written as x = (2+3). A fourth type of expression is called a fraction in which you divide something by another thing or number. Fractions are written like regular numbers but with a '/' symbol before the number. For example, 4/5 would be written as 4/5 or 4 5/100. Anything that can be written as a number can also be used in an addition problem. This means that any number or group of numbers can be added together to solve an addition problem. For example: 1 + 1 = 2, 2 + 1 = 3, and 5 -
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Unformatted text preview: are 180° in a triangle. Since ∠FEG is the supplement of ∠HEG, ∠FEG = 70°. (C) The sum of the angles in a parallelogram is 360°. 12 x = 360° x = 30° Angle B = 5x = 5 · 30° = 150° 5. (A) The volume of a rectangular box is the product of its length, width, and height. Since 1 the height is 18 inches, or 1 feet, and the 2 length and width of the square base are the same, we have 1 x ⋅ x ⋅ 1 = 24 2 x 2 = 16 x=4 Angle O is a central angle equal to its arc, 100°. This leaves 80° for the other two angles. Since the triangle is isosceles (because the legs are both radii and therefore equal), angle ABO is 40°. 10. (A) d= = (5 − (3)) + (-5 - 1) 2 2 (8)2 + (-6)2 = 64 + 36 = 100 = 10 www.petersons.com 224 Chapter 13 Exercise 1 1. (C) Find the area in square feet and then convert to square yards by dividing by 9. Remember there are 9 square feet in one square yard. (18 · 20) ÷ 9 = 360 ÷ 9 = 40 square yards 2. (B) Area of parallelogram = b · h Exercise 2 1. (A) Area of equilateral triangle = s2 s2 3 4 Therefore, must equal 16 4 s 2 = 64 s=8 Perimeter is 8 + 8 + 8 = 24 2. (B) In 4 hours the hour hand moves through one-third of the circumference of the clock. C = 2π r = 2π ( 3) = 6π 1 ⋅ 6π = 2π 3 ( x + 7)( x − 7) = 15 x 2 − 49 = 15 x 2 = 64 x =8 Base = x + 7 = 15 3. 1 ⋅b⋅h 2 (D) Compare 2πr with 2π (r + 3). 3. (B) Area of triangle = 2π (r + 3) = 2πr + 6π Circumference was increased by 6π. Trying this with a numerical value for r will give the same result. 4. (E) In one revolution, the distance covered is equal to the circumference. C = 2πr = 2π (18) = 36π inches To change this to feet, divide by 12. 36π = 3π feet 12 Using one leg as base and the other as altitude, the area is 1 · 6 · 8 = 24. Using the hypotenuse 2 as base and the altitude to the hypotenuse will give the same area. 1 ⋅ 10 ⋅ h = 24 2 5h = 24 1 h = 4.8 ∴ ⋅ 10 ⋅ 4.8 = 24 2 4. (E) Area of rhombus = · product of 2 diagonals 1 1 Area = ( 4 x )( 6 x ) = 24 x 2 = 12 x 2 2 2 1 ( ) In 20 revolut...
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## This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.
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## University Calculus: Early Transcendentals (3rd Edition)
$$f^{-1}(x)=\frac{2x+b}{x-1}$$ - Domain: $(-\infty,1)\cup(1,\infty)$. - Range: $(-\infty,2)\cup(2,\infty)$.
$$y=f(x)=\frac{x+b}{x-2}\hspace{1cm}b\gt-2$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=\frac{x+b}{x-2}$$ $$y=\frac{x-2+(b+2)}{x-2}$$ $$y=1+\frac{b+2}{x-2}$$ $$\frac{b+2}{x-2}=y-1$$ $$x-2=\frac{b+2}{y-1}$$ $$x=\frac{b+2}{y-1}+2=\frac{b+2+2y-2}{y-1}=\frac{2y+b}{y-1}$$ 2) Interchange $x$ and $y$: $$y=\frac{2x+b}{x-1}$$ Therefore, $$f^{-1}(x)=\frac{2x+b}{x-1}$$ - Domain: $x$ is defined where $(x-1)\ne0$, or $x\ne1$. So the domain of $f^{-1}$ is $(-\infty,1)\cup(1,\infty)$. - Range: $$f^{-1}(x)=\frac{2x+b}{x-1}=\frac{2x-2+(b+2)}{x-1}=2+\frac{b+2}{x-1}$$ Since $b\gt-2$, which means $b+2\gt0$, in the domain for $x$ above, $\frac{b+2}{x-1}$ range from $-\infty$ to $\infty$, except that $\frac{b+2}{x-1}\ne0$ Thus, $\frac{b+2}{x-1}+2$ also ranges from $-\infty$ to $\infty$, except that $\frac{b+2}{x-1}+2\ne2$ In other words, the range of $f^{-1}$ is $(-\infty,2)\cup(2,\infty)$ *Check: $$f(f^{-1}(x))=\frac{\frac{2x+b}{x-1}+b}{\frac{2x+b}{x-1}-2}=\frac{\frac{2x+b+bx-b}{x-1}}{\frac{2x+b-2x+2}{x-1}}=\frac{2x+bx}{2+b}=x$$ $$f^{-1}(f(x))=\frac{2\times\frac{x+b}{x-2}+b}{\frac{x+b}{x-2}-1}=\frac{\frac{2x+2b+bx-2b}{x-2}}{\frac{x+b-x+2}{x-2}}=\frac{2x+bx}{2+b}=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$ |
# 3.7 Derivatives of inverse functions (Page 3/3)
Page 3 / 3
Find the derivative of $h\left(x\right)={\text{cos}}^{-1}\left(3x-1\right).$
${h}^{\prime }\left(x\right)=\frac{-3}{\sqrt{6x-9{x}^{2}}}$
## Applying the inverse tangent function
The position of a particle at time $t$ is given by $s\left(t\right)={\text{tan}}^{-1}\left(\frac{1}{t}\right)$ for $t\ge \frac{1}{2}.$ Find the velocity of the particle at time $t=1.$
Begin by differentiating $s\left(t\right)$ in order to find $v\left(t\right).$ Thus,
$v\left(t\right)={s}^{\prime }\left(t\right)=\frac{1}{1+{\left(\frac{1}{t}\right)}^{2}}·\frac{-1}{{t}^{2}}.$
Simplifying, we have
$v\left(t\right)=-\frac{1}{{t}^{2}+1}.$
Thus, $v\left(1\right)=-\frac{1}{2}.$
Find the equation of the line tangent to the graph of $f\left(x\right)={\text{sin}}^{-1}x$ at $x=0.$
$y=x$
## Key concepts
• The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative.
• We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.
## Key equations
• Inverse function theorem
${\left({f}^{-1}\right)}^{\prime }\left(x\right)=\frac{1}{{f}^{\prime }\left({f}^{-1}\left(x\right)\right)}$ whenever ${f}^{\prime }\left({f}^{-1}\left(x\right)\right)\ne 0$ and $f\left(x\right)$ is differentiable.
• Power rule with rational exponents
$\frac{d}{dx}\left({x}^{m\text{/}n}\right)=\frac{m}{n}{x}^{\left(m\text{/}n\right)-1}.$
• Derivative of inverse sine function
$\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{-1}x=\frac{1}{\sqrt{1-{\left(x\right)}^{2}}}$
• Derivative of inverse cosine function
$\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{-1}x=\frac{-1}{\sqrt{1-{\left(x\right)}^{2}}}$
• Derivative of inverse tangent function
$\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{tan}}^{-1}x=\frac{1}{1+{\left(x\right)}^{2}}$
• Derivative of inverse cotangent function
$\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{cot}}^{-1}x=\frac{-1}{1+{\left(x\right)}^{2}}$
• Derivative of inverse secant function
$\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{-1}x=\frac{1}{|x|\sqrt{{\left(x\right)}^{2}-1}}$
• Derivative of inverse cosecant function
$\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{csc}}^{-1}x=\frac{-1}{|x|\sqrt{{\left(x\right)}^{2}-1}}$
For the following exercises, use the graph of $y=f\left(x\right)$ to
1. sketch the graph of $y={f}^{-1}\left(x\right),$ and
2. use part a. to estimate ${\left({f}^{-1}\right)}^{\prime }\left(1\right).$
a.
b. ${\left({f}^{-1}\right)}^{\prime }\left(1\right)~2$
a.
b. ${\left({f}^{-1}\right)}^{\prime }\left(1\right)~-1\text{/}\sqrt{3}$
For the following exercises, use the functions $y=f\left(x\right)$ to find
1. $\frac{df}{dx}$ at $x=a$ and
2. $x={f}^{-1}\left(y\right).$
3. Then use part b. to find $\frac{d{f}^{-1}}{dy}$ at $y=f\left(a\right).$
$f\left(x\right)=6x-1,x=-2$
$f\left(x\right)=2{x}^{3}-3,x=1$
a. 6, b. $x={f}^{-1}\left(y\right)={\left(\frac{y+3}{2}\right)}^{1\text{/}3},$ c. $\frac{1}{6}$
$f\left(x\right)=9-{x}^{2},0\le x\le 3,x=2$
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x,x=0$
a. $1,$ b. $x={f}^{-1}\left(y\right)={\text{sin}}^{-1}y,$ c. $1$
For each of the following functions, find ${\left({f}^{-1}\right)}^{\prime }\left(a\right).$
$f\left(x\right)={x}^{2}+3x+2,x\ge -1,a=2$
$f\left(x\right)={x}^{3}+2x+3,a=0$
$\frac{1}{5}$
$f\left(x\right)=x+\sqrt{x},a=2$
$f\left(x\right)=x-\frac{2}{x},x<0,a=1$
$\frac{1}{3}$
$f\left(x\right)=x+\text{sin}\phantom{\rule{0.1em}{0ex}}x,a=0$
$f\left(x\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}x+3{x}^{2},a=0$
$1$
For each of the given functions $y=f\left(x\right),$
1. find the slope of the tangent line to its inverse function ${f}^{-1}$ at the indicated point $P,$ and
2. find the equation of the tangent line to the graph of ${f}^{-1}$ at the indicated point.
$f\left(x\right)=\frac{4}{1+{x}^{2}},P\left(2,1\right)$
$f\left(x\right)=\sqrt{x-4},P\left(2,8\right)$
a. $4,$ b. $y=4x$
$f\left(x\right)={\left({x}^{3}+1\right)}^{4},P\left(16,1\right)$
$f\left(x\right)=\text{−}{x}^{3}-x+2,P\left(-8,2\right)$
a. $-\frac{1}{96},$ b. $y=-\frac{1}{13}x+\frac{18}{13}$
$f\left(x\right)={x}^{5}+3{x}^{3}-4x-8,P\left(-8,1\right)$
For the following exercises, find $\frac{dy}{dx}$ for the given function.
$y={\text{sin}}^{-1}\left({x}^{2}\right)$
$\frac{2x}{\sqrt{1-{x}^{4}}}$
$y={\text{cos}}^{-1}\left(\sqrt{x}\right)$
$y={\text{sec}}^{-1}\left(\frac{1}{x}\right)$
$\frac{-1}{\sqrt{1-{x}^{2}}}$
$y=\sqrt{{\text{csc}}^{-1}x}$
$y={\left(1+{\text{tan}}^{-1}x\right)}^{3}$
$\frac{3{\left(1+{\text{tan}}^{-1}x\right)}^{2}}{1+{x}^{2}}$
$y={\text{cos}}^{-1}\left(2x\right)·{\text{sin}}^{-1}\left(2x\right)$
$y=\frac{1}{{\text{tan}}^{-1}\left(x\right)}$
$\frac{-1}{\left(1+{x}^{2}\right){\left({\text{tan}}^{-1}x\right)}^{2}}$
$y={\text{sec}}^{-1}\left(\text{−}x\right)$
$y={\text{cot}}^{-1}\sqrt{4-{x}^{2}}$
$\frac{x}{\left(5-{x}^{2}\right)\sqrt{4-{x}^{2}}}$
$y=x·{\text{csc}}^{-1}x$
For the following exercises, use the given values to find ${\left({f}^{-1}\right)}^{\prime }\left(a\right).$
$f\left(\pi \right)=0,f\prime \left(\pi \right)=-1,a=0$
$-1$
$f\left(6\right)=2,{f}^{\prime }\left(6\right)=\frac{1}{3},a=2$
$f\left(\frac{1}{3}\right)=-8,f{\prime }^{}\left(\frac{1}{3}\right)=2,a=-8$
$\frac{1}{2}$
$f\left(\sqrt{3}\right)=\frac{1}{2},f{\prime }^{}\left(\sqrt{3}\right)=\frac{2}{3},a=\frac{1}{2}$
$f\left(1\right)=-3,f\prime \left(1\right)=10,a=-3$
$\frac{1}{10}$
$f\left(1\right)=0,f{\prime }^{}\left(1\right)=-2,a=0$
[T] The position of a moving hockey puck after $t$ seconds is $s\left(t\right)={\text{tan}}^{-1}t$ where $s$ is in meters.
1. Find the velocity of the hockey puck at any time $t.$
2. Find the acceleration of the puck at any time $t.$
3. Evaluate a. and b. for $t=2,4,$ and $6$ seconds.
4. What conclusion can be drawn from the results in c.?
a. $v\left(t\right)=\frac{1}{1+{t}^{2}}$ b. $a\left(t\right)=\frac{-2t}{{\left(1+{t}^{2}\right)}^{2}}$ c. $\left(a\right)0.2,0.06,0.03;\left(b\right)-0.16,-0.028,-0.0088$ d. The hockey puck is decelerating/slowing down at 2, 4, and 6 seconds.
[T] A building that is 225 feet tall casts a shadow of various lengths $x$ as the day goes by. An angle of elevation $\theta$ is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the following figure. Find the rate of change of the angle of elevation $\frac{d\theta }{dx}$ when $x=272$ feet.
[T] A pole stands 75 feet tall. An angle $\theta$ is formed when wires of various lengths of $x$ feet are attached from the ground to the top of the pole, as shown in the following figure. Find the rate of change of the angle $\frac{d\theta }{dx}$ when a wire of length 90 feet is attached.
$-0.0168$ radians per foot
[T] A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by $\theta ={\text{tan}}^{-1}\left(\frac{x}{2000}\right),$ where $x$ is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5000 feet apart.
[T] A local movie theater with a 30-foot-high screen that is 10 feet above a person’s eye level when seated has a viewing angle $\theta$ (in radians) given by $\theta ={\text{cot}}^{-1}\frac{x}{40}-{\text{cot}}^{-1}\frac{x}{10},$
where $x$ is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure.
1. Find $\frac{d\theta }{dx}.$
2. Evaluate $\frac{d\theta }{dx}$ for $x=5,10,15,$ and 20.
3. Interpret the results in b..
4. Evaluate $\frac{d\theta }{dx}$ for $x=25,30,35,$ and 40
5. Interpret the results in d. At what distance $x$ should the person stand to maximize his or her viewing angle?
a. $\frac{d\theta }{dx}=\frac{10}{100+{x}^{2}}-\frac{40}{1600+{x}^{2}}$ b. $\frac{18}{325},\frac{9}{340},\frac{42}{4745},0$ c. As a person moves farther away from the screen, the viewing angle is increasing, which implies that as he or she moves farther away, his or her screen vision is widening. d. $-\frac{54}{12905},-\frac{3}{500},-\frac{198}{29945},-\frac{9}{1360}$ e. As the person moves beyond 20 feet from the screen, the viewing angle is decreasing. The optimal distance the person should stand for maximizing the viewing angle is 20 feet.
how does this work
Can calculus give the answers as same as other methods give in basic classes while solving the numericals?
log tan (x/4+x/2)
Rohan
Rohan
y=(x^2 + 3x).(eipix)
Claudia
Ismael
A Function F(X)=Sinx+cosx is odd or even?
neither
David
Neither
Lovuyiso
f(x)=1/1+x^2 |=[-3,1]
apa itu?
fauzi
determine the area of the region enclosed by x²+y=1,2x-y+4=0
Hi
MP
Hi too
Vic
hello please anyone with calculus PDF should share
Which kind of pdf do you want bro?
Aftab
hi
Abdul
can I get calculus in pdf
Abdul
How to use it to slove fraction
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
Jean
Im not in college but this will still help
nothing
how can we scatch a parabola graph
Ok
Endalkachew
how can I solve differentiation?
with the help of different formulas and Rules. we use formulas according to given condition or according to questions
CALCULUS
For example any questions...
CALCULUS
v=(x,y) وu=(x,y ) ∂u/∂x* ∂x/∂u +∂v/∂x*∂x/∂v=1
log tan (x/4+x/2)
Rohan
what is the procedures in solving number 1?
review of funtion role?
for the function f(x)={x^2-7x+104 x<=7 7x+55 x>7' does limx7 f(x) exist?
find dy÷dx (y^2+2 sec)^2=4(x+1)^2
Integral of e^x/(1+e^2x)tan^-1 (e^x)
why might we use the shell method instead of slicing
fg[[(45)]]²+45⅓x²=100 |
# Statistics » Range and quartiles
## Contents:
Range
Quartiles
Quartile range
Other methods of calculating the quartiles
## Range
The difference between the lowest value and the highest value is the range.
Example
3, 5, 5, 6, 7, 8, 9, 10
Range = 10 – 3 = 7
## Quartiles
The lower or first quartile (Q1) is the median of the first half of values.
The upper or third quartile (Q3) is the median of the second half of values.
The median itself is actually the second or middle quartile (Q2).
The quartiles divide the dataset in four groups with each 25% of the values.
### Example even number of values
1, 2, 5, 6, 7, 8, 9, 11
The median is (6 + 7) : 2 = 6.5.
The lower quartile is the median of the first half:
3, 5, 5, 6 so Q1 = (5 + 5) : 2 = 5.
The upper quartile is the median of the second half:
7, 8, 9, 10 so Q3 = (8 + 9) : 2 = 8.5.
### Example odd number of values
With an odd number of values the following problem arises. You can never make four equal groups with 25% of the values. Because of this, over the years, people came up with different methods to calculate the values between the quartiles (called hinges). Mathematicians do not agree which of these is the best method. In schools most of the time Tukey's method (including median) or the method of Moor & McCabe (excluding median) are taught.
Example odd number of values
1, 2, 4, 5, 7, 8, 9, 10, 12, 15, 16, 17, 20
The median is 9.
### Method of Tukey
The lower quartile is the median of the first half including the median:
1, 2, 4, 5, 7, 8, 9 so Q1 = 5.
The upper quartile is the median of the second half including the median:
9, 10, 12, 15, 16, 17, 20 so Q3 = 15.
### Method of Moore & McCabe
The lower quartile is the median of the first half excluding the median:
1, 2, 4, 5, 7, 8 so Q1 = (4 + 5) : 2 = 4.5.
The upper quartile is the median of the second half excluding the median:
10, 12, 15, 16, 17, 20 so Q3 = (15 + 16) : 2 = 15.5.
## Interquartile range
The interquartile range is the difference between the lower and upper quartile.
Example
In the examples above, the interquartile ranges are:
Example even number: 8.5 – 3.5 = 5
Example odd number Tukey: 15 – 5 = 10
Example odd number Moore & McCabe: 15.5 – 4.5 = 11
## Other methods of calculating the quartiles
Earlier on this page the methods of Tukey and Moore & McCabe were already mentioned. Tukey came up with the box plot, for this reason this method is widely used. Texas Instruments uses in their graphical calculators Moore & McCabe's method, so that's the reason it is also mentioned above.
As examples, I will use the same datasets as in the examples above.
### Method of Mendenhall & Sincich
With this method you will first calculate which values are the hinges.
You use the following formulas (where n is the number of values):
For Q1: 0.25(n + 1)
For Q3: 0.75(n + 1)
If the number is not a whole number you round off to the nearest integer, with one exception: if you get .5 at Q3 you round down.
The use of these formulas also results in different values for Q1 and Q3 with an even number of values.
Hint: If you know that with 0.5(n + 1) you can calculate which value is the median (Q2), you will probably understand better how they got these formulas.
Example even number of values
1, 2, 5, 6, 7, 8, 9, 11
There are 8 values therefore you will get:
For Q1: 0.25(8 + 1) = 2.25 thus the second value
For Q3: 0.75(8 + 1) = 6.75 thus the seventh value
With this method Q1 = 2 and Q3 = 9.
Note: this method makes sure by rounding off that the values of Q1 and Q3 are actual values from the dataset, while the median (Q2 = 6.5) is not ...
Example even number of values
1, 2, 4, 5, 7, 8, 9, 10, 12, 15, 16, 17, 20
There are 13 values therefore you will get:
For Q1: 0.25(13 + 1) = 3.5 thus the fourth value
For Q3: 0.75(13 + 1) = 10.5 thus the tenth value
With this method Q1 = 5 and Q3 = 15.
### Method of Minitab / =QUARTILE.EXC
Minitab is computer software with the functionality of Excel but with more possibilities for statistics. This method uses the same formulas as Mendenhall & Sincich, however you do not round off: For Q1: 0.25(n + 1)
For Q3: 0.75(n + 1)
Instead you use interpolation. This means that with 3.5 you use the value halfway between the third and fourth value and when you have 3.25 you will use the value a quarter of the way between the third and fourth value.
Excel uses this method with the formula =QUARTILE.EXC.
Example even number of values
1, 2, 5, 6, 7, 8, 9, 11
There are 8 values therefore you will get:
For Q1: 0.25(8 + 1) = 2.25
For Q3: 0.75(8 + 1) = 6.75
With this method Q1 and Q3 are:
Q1 = 0.75 × 2 + 0.25 × 5 = 2.75
Q3 = 0.25 × 8 + 0.75 × 9 = 8.75
Example odd number of values
1, 2, 4, 5, 7, 8, 9, 10, 12, 15, 16, 17, 20
There are 13 values therefore you will get:
For Q1: 0.25(13 + 1) = 3.5
For Q3: 0.75(13 + 1) = 10.5
With this method Q1 = 4.5 and Q3 = 15.5.
### Method van Freund & Perles / =QUARTILE.INC
This method is used by Excel, when you use the formula '=QUARTILE.INC' or the older '=QUARTILE'.
They use the following formulas.
For Q1: 0.25(n + 3)
For Q3: 0.25(3n + 1)
When the outcome is not a whole number, they also use interpolation.
Example even number of values
1, 2, 5, 6, 7, 8, 9, 11
There are 8 values therefore you will get:
For Q1: 0.25(8 + 3) = 2.75
For Q3: 0.25(3 × 8 + 1) = 6.25
With this method Q1 and Q3 are:
Q1 = 0.25 × 2 + 0.75 × 5 = 4.25
Q3 = 0.75 × 8 + 0.25 × 9 = 8.25
Example odd number of values
1, 2, 4, 5, 7, 8, 9, 10, 12, 15, 16, 17, 20
There are 13 values therefore you will get:
For Q1: 0.25(13 + 3) = 4
For Q3: 0.25(3 × 13 + 1) = 10
With this method Q1 = 5 and Q3 = 15.
### Comparison of methods using a table
Example even number of values
Example: 1, 2, 5, 6, 7, 8, 9, 11
Q1 Q3 Tukey 3.5 8.5 Moore & McCabe 3.5 8.5 Mendenhall and Sincich 2 9 Minitab / =QUARTILE.EXC 2.75 8.75 Freund & Perles / =QUARTILE.INC 4.25 8.25
Example odd number of values
Example: 1, 2, 4, 5, 7, 8, 9, 10, 12, 15, 16, 17, 20
Q1 Q3 Tukey 5 15 Moore & McCabe 4.5 15.5 Mendenhall and Sincich 5 15 Minitab / =QUARTILE.EXC 4.5 15.5 Freund & Perles / =QUARTILE.INC 4.5 15
### What is the best method?
Maybe you find it odd/weird/terrible/annoying there is more than one method to calculate the hinges between the quartiles. Especially when you are in secondary school and this is the first time you are confronted with the fact that there is not just one way to calculate something. In that case I want to you think about these two things:
Firstly: It shows beautifully that the definition of quartiles (four groups of 25%) directly gives an impossible task with an odd number of values. Because of this you will get different opinions and different methods. In the search for better formulas for calculating which value the hinge is, they actually also got different values for the hinges with an even number of values in the dataset. While Tukey's method at first sight looks so logical and neat, especially for an even number of values. Appearances are deceiving?
Secondly: Statisticians most often work with large datasets of >250 values. The values are therefore so close to each other that the different methods all give the same hinges. The whole question is more a problem for the perfectionists and curious sticklers. Statisticians themselves just choose one of the methods according to their preference. To illustrate, if I would have taken the dataset 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6 as an example, all methods give Q1 = 4 and Q3 = 6.
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# Chapter 5 Quiz: Exponential & Logarithmic Functions
Approved & Edited by ProProfs Editorial Team
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| By Sajones7668
S
Sajones7668
Community Contributor
Quizzes Created: 12 | Total Attempts: 5,115
Questions: 10 | Attempts: 139
Settings
• 1.
### Simplify
• A.
X^32
• B.
X^4
• C.
X^-4
• D.
X^12
D. X^12
Explanation
When multiplying variables with exponents, you add the exponents together. In this case, x^32 * x^4 * x^-4 * x^12 can be simplified by adding the exponents: 32 + 4 - 4 + 12 = 44. Therefore, the answer is x^44.
Rate this question:
• 2.
### Simplify
• A.
4x^3/y
• B.
4x/y
• C.
X^3y/4
• D.
4y/x^3
A. 4x^3/y
Explanation
The given expression can be simplified by dividing the numerator and denominator by 4. This gives us 4x^3/y. Therefore, the correct answer is 4x^3/y.
Rate this question:
• 3.
### Simplify
• A.
-9x^6y^3
• B.
9x^5y^2
• C.
9x^6y^2
• D.
-9x^5y^2
C. 9x^6y^2
Explanation
To simplify the given expression, we need to combine like terms. The negative sign in front of -9x^6y^3 indicates that the term is negative. When we divide -9x^6y^3 by 9x^5y^2, we get -x. Since the exponents of x and y are the same in both terms, we can subtract the exponents and keep the base. Therefore, the simplified expression is -x. However, the given answer is 9x^6y^2, which is incorrect.
Rate this question:
• 4.
### Consider the graph of the function , how will affect the graph?
• A.
Shift 3 left and 2 up
• B.
Shift 3 right and 2 up
• C.
Shift 3 up and 2 right
• D.
Shift 3 down and 2 right
B. Shift 3 right and 2 up
Explanation
When we shift the graph 3 units to the right, all the x-coordinates of the points on the graph will increase by 3. This means that the entire graph will move horizontally to the right. Additionally, when we shift the graph 2 units up, all the y-coordinates of the points on the graph will increase by 2. This means that the entire graph will move vertically upwards. Therefore, the correct answer is to shift the graph 3 units to the right and 2 units up.
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• 5.
### Sales of IPads The iPad has become the most quickly adopted nonphone consumer electronics product in history, topping the DVD player. Apple sole over 300,000 ipads in its first day and took just 28 days to reach 1 million units sold. Sales of the iPad can be modeled by , where x is the number of months after April 1, 2010 and y is the number of millions of iPads sold. According to this model, how many iPads would be sold by August 2011?
• A.
324 million
• B.
473 million
• C.
13 million
• D.
14.5 million
A. 324 million
• 6.
• A.
A
• B.
B
• C.
C
• D.
D
C. C
• 7.
• A.
3
• B.
1.442
• C.
0.815
• D.
2
D. 2
• 8.
• A.
20
• B.
100
• C.
1000
• D.
26.853
B. 100
• 9.
### Solve for x:
• A.
1/3
• B.
3
• C.
1/9
• D.
27
A. 1/3
Explanation
The given answer, 1/3, is the correct solution for the equation. By dividing 1 by 3, we get the value of x as 1/3.
Rate this question:
• 10.
### Life Span Based on data from 1920 and projected to 2020, the expected life span of people in the United States can be described by the function , where x is the number of years from 1900 to the person's birth year. Estimate the life span for someone born in 2008.
• A.
120 years
• B.
6 years
• C.
78 years
• D.
80 years
C. 78 years
Explanation
Based on the given function, the life span of people in the United States is described by the equation. Since the question asks for an estimate of the life span for someone born in 2008, we need to find the value of the function when x is equal to 108 (2008 - 1900). Therefore, the estimated life span for someone born in 2008 is 78 years.
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Related Topics |
# How do you write the slope-intercept equation of the line perpendicular to y = 5/2 x - 2, which passes through the point (0, 2)?
May 13, 2017
See a solution process below:
#### Explanation:
The equation in the problem is in slope intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.
$y = \textcolor{red}{\frac{5}{2}} x - \textcolor{b l u e}{2}$
Therefore the slope is: $\textcolor{red}{\frac{5}{2}}$
Next, let's call the slope perpendicular to this line ${m}_{p}$.
The slope of a perpendicular line is: ${m}_{p} = - \frac{1}{m}$
Substituting gives:
${m}_{p} = - \frac{1}{\frac{5}{2}} = - \frac{2}{5}$
Because the point given in the problem has $0$ for the $x$ value this is the y-intercept for the perpendicular line. Substituting for ${m}_{p} = - \frac{2}{5}$ and $2$ for $b$ into the slope-intercept formula gives:
$y = \textcolor{red}{- \frac{2}{5}} x + \textcolor{b l u e}{2}$ |
# Fibonacci polynomials
In mathematics, the Fibonacci polynomials are a polynomial sequence which can be considered as a generalization of the Fibonacci numbers. The polynomials generated in a similar way from the Lucas numbers are called Lucas polynomials.
## Definition
These Fibonacci polynomials are defined by a recurrence relation:[1]
$F_n(x)= \begin{cases} 0, & \mbox{if } n = 0\\ 1, & \mbox{if } n = 1\\ x F_{n - 1}(x) + F_{n - 2}(x),& \mbox{if } n \geq 2 \end{cases}$
The first few Fibonacci polynomials are:
$F_0(x)=0 \,$
$F_1(x)=1 \,$
$F_2(x)=x \,$
$F_3(x)=x^2+1 \,$
$F_4(x)=x^3+2x \,$
$F_5(x)=x^4+3x^2+1 \,$
$F_6(x)=x^5+4x^3+3x \,$
The Lucas polynomials use the same recurrence with different starting values:[2] $L_n(x) = \begin{cases} 2, & \mbox{if } n = 0 \\ x, & \mbox{if } n = 1 \\ x L_{n - 1}(x) + L_{n - 2}(x), & \mbox{if } n \geq 2. \end{cases}$
The first few Lucas polynomials are:
$L_0(x)=2 \,$
$L_1(x)=x \,$
$L_2(x)=x^2+2 \,$
$L_3(x)=x^3+3x \,$
$L_4(x)=x^4+4x^2+2 \,$
$L_5(x)=x^5+5x^3+5x \,$
$L_6(x)=x^6+6x^4+9x^2 + 2. \,$
The Fibonacci and Lucas numbers are recovered by evaluating the polynomials at x = 1; Pell numbers are recovered by evaluating Fn at x = 2. The degrees of Fn is n − 1 and the degree of Ln is n. The ordinary generating function for the sequences are:[3]
$\sum_{n=0}^\infty F_n(x) t^n = \frac{t}{1-xt-t^2}$
$\sum_{n=0}^\infty L_n(x) t^n = \frac{2-xt}{1-xt-t^2}.$
The polynomials can be expressed in terms of Lucas sequences as
$F_n(x) = U_n(x,-1),\,$
$L_n(x) = V_n(x,-1).\,$
## Identities
Main article: Lucas sequence
As particular cases of Lucas sequences, Fibonacci polynomials satisfy a number of identities.
First, they can be defined for negative indices by[4]
$F_{-n}(x)=(-1)^{n-1}F_{n}(x),\,L_{-n}(x)=(-1)^nL_{n}(x).$
Other identities include:[4]
$F_{m+n}(x)=F_{m+1}(x)F_n(x)+F_m(x)F_{n-1}(x)\,$
$L_{m+n}(x)=L_m(x)L_n(x)-(-1)^nL_{m-n}(x)\,$
$F_{n+1}(x)F_{n-1}(x)- F_n(x)^2=(-1)^n\,$
$F_{2n}(x)=F_n(x)L_n(x).\,$
Closed form expressions, similar to Binet's formula are:[4]
$F_n(x)=\frac{\alpha(x)^n-\beta(x)^n}{\alpha(x)-\beta(x)},\,L_n(x)=\alpha(x)^n+\beta(x)^n,$
where
$\alpha(x)=\frac{x+\sqrt{x^2+4}}{2},\,\beta(x)=\frac{x-\sqrt{x^2+4}}{2}$
are the solutions (in t) of
$t^2-xt-1=0.\,$
## Combinatorial interpretation
The coefficients of the Fibonacci polynomials can be read off from Pascal's triangle following the "shallow" diagonals (shown in red). The sums of the coefficients are the Fibonacci numbers.
If F(n,k) is the coefficient of xk in Fn(x), so
$F_n(x)=\sum_{k=0}^n F(n,k)x^k,\,$
then F(n,k) is the number of ways an n−1 by 1 rectangle can be tiled with 2 by 1 dominoes and 1 by 1 squares so that exactly k squares are used.[1] Equivalently, F(n,k) is the number of ways of writing n−1 as an ordered sum involving only 1 and 2, so that 1 is used exactly k times. For example F(6,3)=4 and 5 can be written in 4 ways, 1+1+1+2, 1+1+2+1, 1+2+1+1, 2+1+1+1, as a sum involving only 1 and 2 with 1 used 3 times. By counting the number of times 1 and 2 are both used in such a sum, it is evident that F(n,k) is equal to the binomial coefficient
$F(n,k)=\binom{\tfrac{n+k-1}{2}}{k}$
when n and k have opposite parity. This gives a way of reading the coefficients from Pascal's triangle as shown on the right.
## References
1. ^ a b Benjamin & Quinn p. 141
2. ^ Benjamin & Quinn p. 142
3. ^
4. ^ a b c Springer |
# 2021 AMC 10A Problems/Problem 11
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
For which of the following integers $b$ is the base-$b$ number $2021_b - 221_b$ not divisible by $3$?
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$
## Solution 1 (Factor)
We have \begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*} which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{\textbf{(E)} ~8}.$
~MRENTHUSIASM
## Solution 2 (Vertical Subtraction)
Vertically subtracting $2021_b - 221_b,$ we see that the ones place becomes $0,$ and so does the $b^1$ place. Then, we perform a carry (make sure the carry is in base $b$). Let $b-2 = A.$ Then, we have our final number as $$1A00_b.$$ Now, when expanding, we see that this number is simply $b^3 - (b - 2)^2.$
Now, notice that the final number will only be congruent to $$b^3-(b-2)^2\equiv0\pmod{3}.$$ If either $b\equiv0\pmod{3},$ or if $b\equiv1\pmod{3}$ (because note that $(b - 2)^2$ would become $\equiv1\pmod{3},$ and $b^3$ would become $\equiv1\pmod{3}$ as well, and therefore the final expression would become $1-1\equiv0\pmod{3}.$ Therefore, $b$ must be $\equiv2\pmod{3}.$ Among the answers, only $8$ is $\equiv2\pmod{3},$ and therefore our answer is $\boxed{\textbf{(E)} ~8}.$
~icecreamrolls8
By the definition of bases, we have $$2021_b - 221_b = \left(2b^3+2b+1\right) - \left(2b^2+2b+1\right).$$ For values $b_1$ and $b_2$ such that $b_1\equiv b_2\pmod{3},$ we get $$\left(2b_1^3+2b_1+1\right) - \left(2b_1^2+2b_1+1\right) \equiv \left(2b_2^3+2b_2+1\right) - \left(2b_2^2+2b_2+1\right) \pmod{3}.$$ Note that answer choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. So, $\textbf{(A)}$ and $\textbf{(C)}$ are either both correct or both incorrect. Since there is only one correct answer, $\textbf{(A)}$ and $\textbf{(C)}$ are both incorrect. Similarly, $\textbf{(B)}$ and $\textbf{(D)}$ are both incorrect. This leaves us with $\boxed{\textbf{(E)} ~8},$ the answer choice with a unique residue modulo $3.$
~emerald_block ~MRENTHUSIASM
~ pi_is_3.14
## Video Solution (Simple and Quick)
~ Education, the Study of Everything
## Video Solution
~North America Math Contest Go Go Go
~savannahsolver
~IceMatrix |
# 3/8 As a Decimal
Let us know today what is 3/8 as a decimal. A decimal number is defined as a number whose whole number part and fractional part can be separated by a decimal point.
let’s start from the beginning
Explanation:
To write 3/8 as a decimal using the division method:
To convert a fraction to a decimal form, we need to divide its numerator by the denominator.
Here, the fraction is 3/8, which means we need to do 3 8.
gives the answer as 0.375. So, 3/8 0.375 as a decimal. Is
#### The answer to the fraction 3/8 is 0.375.
##### How to convert numbers to decimal?
Converting a fraction to a decimal is a very simple process. First of all we take the fraction which we have to convert to decimal. Then we divide the numerator of that fraction by the denominator of that fraction. The quotient obtained by doing this is a decimal number.
##### What is the value of decimal?
The decimal system or decimal number system (decimal system, “base ten” or “denary”) is a number system in which a total of ten digits or ‘ten signs’ (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) is used. This is the most commonly used number system by humans.
##### How to solve fraction?
To solve for a fraction, simplify the first fraction. Begin by finding the greatest common factor (GCF) of both the numerator and the denominator of the fraction. The greatest common factor is a number by which both the numerator and the denominator can be divided. Then, simply divide the numerator and denominator by the greatest common denominator so that your fractions are simplified.
##### What will be the decimal expansion of 36 by 100?
The decimal form of 36 / 100 is 0.36, and this is the terminating decimal expansion. |
Center Home -> Content Areas Home -> Math Home -> Project Activities -> ExploreMath Activities ->
Exploring Period...Period.
Activity Description Activity Guide
## Part 1:Exploring Periodic Graphs
1.) Predict what the following graph would look like if the pattern was repeated indefinitely. Draw your prediction on the graph. (see Handout 1).
2.) Divide your graph using dashed lines into equal intervals so that the partitions are identical; that is, the portion of graph inside each partition looks exactly the same as the partitions adjacent to it. (An example is shown below.) What is the width of each of your partitions? For example, in the diagram below, the width of each partition is 2 units, or four tick marks.
3.) For the graph in #1, what is the minimum partition width you could use and still ensure that the partitions are identical?
4.) Put a piece of tracing paper over the graphs below and partition the graph into identical partitions using the minimum partition width. Compare your answer with two neighbors. Predict how many distinct ways you could construct identical partitions using the minimum partition width. Show at least two different ways to partition the graph into identical partitions using the minimum partition width. (see Handout 2)
5.) Put your piece of tracing paper with its partition lines over the graph so that you have identical partitions as in Task 4. Now, shift the tracing paper to the left one-half unit. (Do not redraw or shift the graph. Only shift the tracing paper on top of the graph.) Compare the portions of graph inside each shifted partition to the partition adjacent to it.
6.) Shift the tracing paper to the right one and one-fourth unit. Compare the portions of graph inside each shifted partition to the partition adjacent to it and discuss your findings.
7.) Now how many ways do you think you could construct identical partitions using the minimum partition width? Describe the ways you have discovered.
8.) The minimum partition width used to construct identical partitions is referred to as the length of one period or one cycle. In fact, we say that graphs like this one are periodic or cyclic. Why?
Part 2: Exploring the Graph of y = sin(x) and y = cos(x).
Using your web browser go to the “Shifting and scaling sine and cosine curves” Activity located at http://www.exploremath.com/activities/Activity_page.cfm?ActivityID=23. When the activity loads it will look like the following.
1.) Notice that the graph of y = sin(x) is shown. By looking at its graph, determine if the sine function is periodic. If it is, determine the length of one period.
To help you with your decision, you might to “zoom out” and “zoom in” by clicking on the magnifying glasses containing the minus and plus signs. You can also pan left or pan right by clicking on the red arrows pointing to the left and the right.
2.) Manipulate the red arrows and the magnifying glasses until the first x-intercept you see on the graph is at 0o and the last x-intercept you see is at 360o.
3.) Over the interval [0°, 360°], what are the minimum and maximum y-values of y = sin(x)? Find the global minimum and maximum y-values over the entire domain of real numbers. Explain and compare your answers with your neighbors.
4.) How far apart are the x-intercepts of y = sin(x) on the interval [0°, 360°]? Generalize your findings for the entire domain of y = sin(x).
Check your answers and predictions by using the red arrow buttons or zooming out and viewing as much of the full graph of y = sin(x) as you wish. Alternatively, you may verify your answers numerically by utilizing the “Calculate data values” clipboard tool to see a table of values.
5.) Repeat the tasks in Part 2 for y = cos(x).
# Part 3: The Variable b in the Equation y = a sin[b(x-c)] + d
The general form of the sine equation is:
y = a sin[b(x-c)] + d where a, b, c, d € R.
By substituting different numbers for a, b, c, and d, we can shift and scale the graph of the sine function.
1.) Write the sine equation for a = 1, c = 0o, and d = 0. What does this graph look like when b = 1?
Teacher Note: The next activities have students explore the graph of the sine function when b =/ 1. Activities #2 and #3 show how the value chosen for the variable b affects the graph at x = 0.
2.) Algebraically determine the values of the variable b for which (0,0) is a solution to the equation
y = sin(bx).
3.) In the ExploreMath activity, manipulate the b-slider and watch what happens to the graph of
y = sin(bx) for different values of b at x = 0. How do the graphical representations relate to the algebraic solutions?
In the ExploreMath activity, to center the graph about the origin, click on the red circle that is located between the red arrows.
4.) Click on the red circle if necessary and the magnifying glasses to make -360o the first labeled tic mark on the x-axis and 360o the last. Set b to 0.2 by clicking on the number next to the b-slider, typing in the number 0.2, and hitting enter. Now, slowly move the b-slider to the right. What happens to the graph as b gets larger?
5.) From what you have seen, make a conjecture about the relationship between b and the length of one period.
6.) To test your conjecture, repeat the instructions in #4, but this time click on the box marked “Show amplitude, period, frequency” in the ExploreMath activity. As you move the b-slider slowly to the right from 0.2, watch the value of the period change. Was your conjecture correct?
7.) From what you have seen, make conjectures about:
a.) the length of one period as b approaches 0, and
b.) the appearance of the full graph when b equals 0.
8.) Test your conjectures in #7 by setting b equal to 0.2, and then watching the changes in the graph and the changes in the period as you enter decreasing values of b, e.g., 0.1, 0.05, 0.03, 0.01, and 0.
To get a good look at what is happening in #8, center the graph using the red circle and zoom out as far as possible using the ‘-’ magnifying glass. Then, click on the right red arrow until 2880o is the last labeled tic mark you see on the x-axis. The graph should look like the one below.
9.) Algebraically test your conjecture in #7b by noticing the type of equation that results when you substitute 0 for b into the equation y = sin(bx).
# Part 4: Taking a Closer Look at the Variable b
1.) Fill in the table below for y = sin(bx) when b = 1. Then, use the table and the graph below to sketch one cycle of sine starting at the origin. (see Handout 3.)
x y=sin(x) 0o 90o 180o 270o 360o
2.) Fill in the tables for y = sin(bx) when b = 2 and for y = sin(bx) when b = 4. As you do, predict how the graphs of y = sin(2x) and y = sin(4x) will be different from the graph of y = sin(x).
3.) Check your conjectures by drawing the graphs of one cycle of y = sin(2x) and one cycle of
y = sin(4x) on the same axes as y = sin(x). What is the relationship between one period of
y = sin(2x), one period of y = sin(4x), and one period of y = sin(x)?
x 2x y = sin(2x) 0o 90o 180o 270o 360o
x 4x y = sin(4x) 0o 90o 180o 270o 360o
4.) Predict the length of one cycle of y = sin(0.5x). How does this length compare to the length of one cycle of y = sin(x)? Test your prediction using the ExploreMath activity by sliding b to 0.5 and looking at the graph and its period.
5.) Derive a formula for the period of y = sin(bx) when b > 0. Using a variety of b values, test the validity of your formula algebraically and using ExploreMath.
# Part 5: The Effect of Negative Values of b
1.) Zoom out so that -720 is the last labeled tic mark on the x-axis. Set b to 0 by clicking on the number next to the b-slider, typing in the number 0, and hitting enter. Now, slowly move the b-slider to the left. What happens to the period as b gets larger in the negative direction? How does this compare/contrast to what happens to the period as b gets larger in the positive direction?
Note: Be careful not to confuse period with frequency. Frequency refers to the number of cycles that occur in a certain interval. What happens to the frequency as b gets larger in the positive direction? In the negative direction?
2.) What kind of relationship exists between the period and the magnitude of b; that is, direct, inverse, or no relationship? Hint: magnitude of b = |b|).
Remember that as b gets larger in the positive direction, the period and b have an inverse relationship. Since the magnitude of any b-value always equals a b-value that is positive (or equal to zero), we know that the period and the magnitude of b have an inverse relationship as well.
Next, we’d like to know if the graph of the sine function for a particular positive value of b corresponds to the graph of the sine function for the opposite of that particular value of b. For example, we’d like to know how the graph of y = sin(2x) compares to the graph of y = sin(-2x).
3.) Complete the tables below for y = sin(-2x) and y = sin(-4x) without using a calculator or graph. (Hint: Use the fact that the sine function is an odd function; that is, sin(-x) = -sin(x).) (see Handout 4.)
x 2x y = sin(2x) y = sin(-2x) 0o 90o 180o 270o 360o
x 2x y = sin(4x) y = sin(-4x) 0o 90o 180o 270o 360o
4.) Using the tables, make a conjecture about how the graphs of y = sin(-2x) and y = sin(-4x) compare to the graphs of y = sin(2x) and y = sin(4x) respectively. How will the period be affected?
5.) Using the ExploreMath activity, check your conjectures in #4 by comparing the graphs of y = sin(2x) and y = sin(-2x) and by comparing the graphs of y = sin(4x) and y = sin(-4x).
To make it easier to see the relationship between the graphs in #5, click on the red arrows and the magnifying glasses until the first x-intercept you see on the graph is at 0o and the last x-intercept you see is at 180o.
Teacher Note: Some students may incorrectly assume that sin(-2x) = -2sin(x).
6.) Take the formula you developed earlier for finding the period when b > 0 and adapt it so it can be used to find the period for R - {0}. Using a variety of b values, test the validity of your formula algebraically and using ExploreMath. |
# Analyzing structure with linear inequalities: fruits
## Video udskrift
- [Instructor] Shantanu bought more apples than bananas and he bought more bananas than cantaloupes. Let A represent the number of apples Shantanu bought, let B represent the number of bananas, and let C represent the number of cantaloupes. Let's compare the expressions B plus C and A. Which statement is correct? So is B plus C greater than A? Is it less than A? Or are these two quantities equal? Or is there not enough information to tell? So like always, pause this video and see if you can work through it on your own and now I will work through it with you. All right, so let's just write down the information that they gave us. They say let A represent the number, well that's more straightforward, A for apple, B for banana, C for cantaloupe. Here we have more apples than bananas, so A is greater than B, and then they also tell us he bought more bananas than cantaloupes, so B is greater than C, or we could rewrite that as A is greater than B is greater than C or we could write that as C is less than B is less than A. This is essentially the information that they give us. So let's see, which of these is going to be true? B plus C greater than A, B plus C greater than A, B plus C less than A. So one thing that we can try is let's try to plug in some values, some numbers to see if we can get combinations that are consistently in one of these buckets or if they fall into multiple of these choices, then we say hey there's not enough information to tell. In general, this is a good strategy for things like this where we're dealing with very abstract quantities. So let's make a little table here. So A, B, C and then I can also figure out what B plus C is. So this is going to be A, B, C, and this is B plus C and we can compare that to A. So let's see a situation where let's see if we can make B plus C greater than A. So they both have to be less than A. So let's see, if C is five and B is six and let's make A seven. So in this situation, B plus C is going to be equal to 11. So we're able to find a situation where if B plus C are close enough to A, that B plus C is going to be greater than A. So we're able to find this scenario. Let's see if we can figure out a scenario where B plus C is less than A. Well if, well we could do the same B plus C, six, five. We could make A bigger than six plus five. We can make A 12. And now this is a situation, so this first situation we have B plus C is greater than A. The second situation right over here, you have B plus C is less than A and so depending on what your A, B, and C's are that meet these constraints and notice, both of these situations I meet all the constraints where A is greater than B is greater than C, but it could be either one of these. So that immediately tells us that there is not enough information to tell. Now, one thing that we, yeah there's just not enough information to tell. I can even come up with a scenario where B plus C is equal to A. If it's six, five, and 11. Then B plus C is equal to A. So based on the information they gave us, any of these are actually possible. So there's not enough information to tell. |
# Spectrum Math Grade 6 Chapter 3 Lesson 10 Answer Key Problem Solving
Go through the Spectrum Math Grade 6 Answer Key Chapter 3 Lesson 3.10 Problem Solving and get the proper assistance needed during your homework.
## Spectrum Math Grade 6 Chapter 3 Lesson 3.10 Problem Solving Answers Key
Solve each problem.
Question 1.
The sales tax on the purchase of a refrigerator that costs $695 is 7 percent. What is the amount of sales tax? The sales tax is ____. Answer: Amount of sales tax =$486.50.
The sales tax is $486.50. Explanation: Cost price of the refrigerator =$695.
Percentage of sales tax = 7%.
Amount of sales tax = Cost price of the refrigerator × Percentage of sales tax
= $695 × 7% = ($695 × 7) ÷ 100
= $4865 ÷ 100 =$486.50.
Question 2.
A stove that costs $695 will be on sale next week for 28 percent off its regular price. What is the amount of savings? The savings will be ____. Answer: Sales tax amount on the store =$194.60.
Final price of the store = $500.40. The savings will be$194.60.
Explanation:
Cost price of a stove = $695. Sales percent of the store next week = 28%. Sales tax amount on the store = Cost price of a stove × Sales percent of the store next week =$695 × 28 %
= ($695 × 28) ÷ 100 =$19,460 ÷ 100
= $194.60. Amount of savings = Regular price of a stove – Sales tax amount on the store =$695 – $194.60 =$500.40.
Question 3.
In math class, 60 percent of the students are males. There are 30 students in the class. How many students are males?
There are ____ males.
Number of students are males = 18.
There are 18 males.
Explanation:
Percentage of students are males in Math class = 60%.
Total students in the class = 30.
Number of students are males = Total students in the class × Percentage of students are males in Math class
= 30 × 60%
= (30 × 60) ÷ 100
= 3 × 6
= 18.
Question 4.
East Side Middle School has 1,500 students. Thirty-two percent of them are in sixth grade. How many sixth-grade students are there?
Number of students are sixth-grade = 480.
Explanation:
Number of students East Side Middle School has = 1,500.
Percentage of students are in sixth grade = 32%.
Number of students are sixth-grade = Number of students East Side Middle School has × Percentage of students are in sixth grade
= 1,500 × 32%
= (1,500 × 32) ÷ 100
= 15 × 32
= 480.
Question 5.
Lauren is saving for gymnastics camp. Camp costs $225 to attend. She has 40 percent of the money saved. How much money has she saved? Lauren has saved ______ Answer: Amount of money she saved =$90.00.
Lauren has saved = $90.00. Explanation: Cost to attend Camp =$225.
Percentage of money she saved = 40%.
Amount of money she saved = Cost to attend Camp × Percentage of money she saved
= $225 × 40% = ($225 × 40) ÷ 100
= $9000 ÷ 100 =$90.00.
Question 6.
Of the 1,500 students attending East Side Middle School, twenty-five percent are running for student council. How many students are running for student council?
____________ students are running for student council.
Number of students are running for student council = 375.
375 students are running for student council.
Explanation:
Total number of students attending East Side Middle School = 1,500.
Percentage of students are running for student council = 25%.
Number of students are running for student council = Total number of students attending East Side Middle School × Percentage of students are running for student council
= 1,500 × 25%
= (1500 × 25) ÷ 100
= 15 × 25
= 375.
Solve each problem.
Question 1.
The Jacksons’ dinner cost $125. They left$21.25 for a tip. What percent did they tip?
The Jacksons tipped ____.
Percentage they tipped = 17%.
The Jacksons tipped 17%.
Explanation:
Cost of the Jacksons’ dinner cost = $125. Cost of amount they left for a tip =$21.25.
Let the Percentage they tipped be X.
=> Cost of the Jacksons’ dinner cost × Percentage they tipped be = Cost of amount they left for a tip
= $125 × X% =$21.25
= X% = $21.25 ÷$125
= X% = 0.17
= X ÷ 100 = 0.17
= X = 0.17 × 100
= X = 17.
Question 2.
A sweater was originally $55. It is now marked down to 65% of its original price. How much is the sweater now? The sweater now costs ____. Answer: Cost of the sweater now =$19.25.
The sweater now costs = $19.25. Explanation: Cost price of a sweater =$55.
Percentage of a sweater marked down of its original price = 65%.
Cost of marked down on a sweater = Cost price of a sweater × Percentage of a sweater marked down of its original price
= $55 × 65% = ($55 × 65) ÷ 100
= $3575 ÷ 100 =$35.75.
Cost of the sweater now = Cost price of a sweater ÷ Cost of marked down on a sweater
= $55 –$35.75
= $19.25. Question 3. Ms. Martino’s new home cost$260,000. She paid $39,000 in a down payment. What percent of the home cost did she pay in the down payment? Ms. Martino paid __________ of the total price. Answer: Percentage of the home cost she paid in the down payment = 15%. Ms. Martino paid 15% of the total price. Explanation: Cost of the Ms. Martino’s new home =$260,000.
Cost of amount she paid in a down payment = $39,000. Let the percentage of the home cost she paid in the down payment be X. => Cost of the Ms. Martino’s new home × Percentage of the home cost she paid in the down payment = Cost of amount she paid in a down payment =$260,000 × X% = $39,000 = X% =$39,000 ÷ \$260,000
= X% = 0.15.
= X ÷ 100 = 0.15
= X = 0.15 × 100
= X = 15.
Question 4.
Workers have painted 920 square feet of an office. They have completed 80% of their job. How many square feet do they need to paint in all?
They need to paint ____ square feet.
Number of square feet they need to paint in all = 184.
They need to paint 184 square feet.
Explanation:
Number of square feet of an office workers have painted = 920.
Percentage of their job they completed = 80%.
Number of square feet they painted = Number of square feet of an office workers have painted × Percentage of their job they completed
= 920 × 80%
= (920 × 80) ÷ 100
= 92 × 8
= 736.
Number of square feet they need to paint in all = Number of square feet of an office workers have painted – Number of square feet they painted
= 920 – 736
= 184.
Question 5.
The Franklins are taking a cross-country trip. They will drive 3,150 miles in all. On the first day, they drove 567 miles. What percent of their trip did they drive?
The Franklins drove ___________ of their trip.
Percent of their trip did they drive = 18%.
The Franklins drove 18% of their trip.
Explanation:
Number of miles in all they will drive = 3,150.
Number of miles on first day, they drove = 567.
Let the Percent of their trip did they drive be X.
Number of miles in all they will drive × Percent of their trip did they drive = Number of miles on first day, they drove
= 3150 × X% = 567
= X % = 567 ÷ 3150
= X ÷ 100 = 0.18
= X = 0.18 × 100
= X = 18.
Question 6.
Jen is reading a 276-page book. She is 25% finished. How many pages has she read?
Number of pages she read = 69.
Explanation:
Number of page book Jen is reading = 276.
Percentage of she finished = 25%.
Number of pages she read = Number of page book Jen is reading × Percentage of she finished
= 276 × 25%
= (276 × 25) ÷ 100
= 6,900 ÷ 100
= 69.
Question 7.
Pete’s dog weighed 30 pounds. It then lost 16% of its weight. How much did Pete’s dog lose?
The dog lost __________ pounds.
Number of pounds Pete’s dog lose = 4.80.
The dog lost 4.80 pounds.
Explanation:
Number of pounds Pete’s dog weighed = 30.
Percentage lost in its weight = 16%.
Number of pounds Pete’s dog lose = Number of pounds Pete’s dog weighed × Percentage lost in its weight
= 30 × 16%
= (30 × 16) ÷ 100
= 480 ÷ 100
= 4.80.
Question 8.
Karla has read 85% of her book, which amounts to 238 pages. How long is the book?
The book is __________ pages long. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 12.1: Inverse Variation Models
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Distinguish direct and inverse variation.
• Graph inverse variation equations.
• Write inverse variation equations.
• Solve real-world problems using inverse variation equations.
## Introduction
Many variables in real-world problems are related to each other by variations. A variation is an equation that relates a variable to one or more variables by the operations of multiplication and division. There are three different kinds of variation problems: direct variation, inverse variation and joint variation.
## Distinguish Direct and Inverse Variation
In direct variation relationships, the related variables will either increase together or decrease together at a steady rate. For instance, consider a person walking at three miles per hour. As time increases, the distance covered by the person walking also increases at the rate of three miles each hour. The distance and time are related to each other by a direct variation.
\begin{align*}\text{distance}= \text{rate} \times \text{time}\end{align*}
Since the speed is a constant 3 miles per hour, we can write: \begin{align*}d=3t\end{align*}.
Direct Variation
The general equation for a direct variation is
\begin{align*}y = kx.\end{align*}
\begin{align*}k\end{align*} is called the constant of proportionality
You can see from the equation that a direct variation is a linear equation with a \begin{align*}y-\end{align*}intercept of zero. The graph of a direct variation relationship is a straight line passing through the origin whose slope is \begin{align*}k\end{align*} the constant of proportionality.
A second type of variation is inverse variation. When two quantities are related to each other inversely, as one quantitiy increases, the other one decreases and vice-versa.
For instance, if we look at the formula \begin{align*}\text{distance} = \text{speed} \times \text{time}\end{align*} again and solve for time, we obtain:
\begin{align*}\text{time}= \frac{\text{distance}}{\text{rate}}\end{align*}
If we keep the distance constant, we see that as the speed of an object increases, then the time it takes to cover that distance decreases. Consider a car traveling a distance of 90 miles, then the formula relating time and speed is \begin{align*}t = \frac{90}{s}\end{align*}.
Inverse Variation
The general equation for inverse variation is
\begin{align*}y=\frac{k}{x}\end{align*}
where \begin{align*}k\end{align*} is called the constant of proportionality.
In this chapter, we will investigate how the graph of these relationships behave.
Another type variation is a joint variation. In this type of relationship, one variable may vary as a product of two or more variables.
For example, the volume of a cylinder is given by:
\begin{align*} V=\pi r^2 \cdot h\end{align*}
In this formula, the volume varies directly as the product of the square of the radius of the base and the height of the cylinder. The constant of proportionality here is the number \begin{align*}\pi\end{align*}.
In many application problems, the relationship between the variables is a combination of variations. For instance Newton’s Law of Gravitation states that the force of attraction between two spherical bodies varies jointly as the masses of the objects and inversely as the square of the distance between them
\begin{align*} F=G\frac {m_1m_2}{d^2}\end{align*}
In this example the constant of proportionality, \begin{align*}G\end{align*}, is called the gravitational constant and its value is given by \begin{align*}G = 6.673 \times 10^{-11} N \cdot m^2/kg^2\end{align*}.
## Graph Inverse Variation Equations
We saw that the general equation for inverse variation is given by the formula \begin{align*}y=\left( \frac{k}{x} \right )\end{align*}, where \begin{align*}k\end{align*} is a constant of proportionality. We will now show how the graphs of such relationships behave. We start by making a table of values. In most applications, \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are positive. So in our table, we will choose only positive values of \begin{align*}x\end{align*}.
Example 1
Graph an inverse variation relationship with the proportionality constant \begin{align*}k = 1\end{align*}.
Solution
\begin{align*}x\end{align*} \begin{align*}y =\frac {1}{x}\end{align*}
0 \begin{align*}y=\frac {1}{0} = \text{undefined}\end{align*}
\begin{align*}\frac {1}{4}\end{align*} \begin{align*}y =\frac {1}{\frac{1}{4}}=4\end{align*}
\begin{align*}\frac {1}{2}\end{align*} \begin{align*}y=\frac {1}{\frac{1}{2}}=2\end{align*}
\begin{align*}\frac {3}{4}\end{align*} \begin{align*}y =\frac {1}{\frac{3}{4}}=1.33\end{align*}
1 \begin{align*}y =\frac {1}{1}=1\end{align*}
\begin{align*}\frac {3}{2}\end{align*} \begin{align*}y=\frac {1}{\frac{3}{2}}=0.67\end{align*}
2 \begin{align*}y =\frac {1}{2}=0.5\end{align*}
3 \begin{align*}y=\frac {1}{3}=0.33\end{align*}
4 \begin{align*}y =\frac {1}{4}=0.25\end{align*}
5 \begin{align*}y =\frac {1}{5}=0.2\end{align*}
10 \begin{align*}y=\frac {1}{10}=0.1\end{align*}
Here is a graph showing these points connected with a smooth curve.
Both the table and the graph demonstrate the relationship between variables in an inverse variation. As one variable increases, the other variable decreases and vice-versa. Notice that when \begin{align*}x= 0\end{align*}, the value of \begin{align*}y\end{align*} is undefined. The graph shows that when the value of \begin{align*}x\end{align*} is very small, the value of \begin{align*}y\end{align*} is very big and it approaches infinity as \begin{align*}x\end{align*} gets closer and closer to zero.
Similarly, as the value of \begin{align*}x\end{align*} gets very large, the value of \begin{align*}y\end{align*} gets smaller and smaller, but never reaches the value of zero. We will investigate this behavior in detail throughout this chapter
## Write Inverse Variation Equations
As we saw an inverse variation fulfills the equation: \begin{align*}y = \left ( \frac{k}{x} \right )\end{align*}. In general, we need to know the value of \begin{align*}y\end{align*} at a particular value of \begin{align*}x\end{align*} in order to find the proportionality constant. After the proportionality constant is known, we can find the value of \begin{align*}y\end{align*} for any given value of \begin{align*}x\end{align*}.
Example 2
If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*} and \begin{align*}y = 10\end{align*} when \begin{align*}x = 5\end{align*}. Find \begin{align*}y\end{align*} when \begin{align*}x = 2\end{align*}.
Solution
\begin{align*}\text{Since} \ y \ \text{is inversely proportional to} \ x, \\ \text{then the general relationship tells us} & & y & =\frac {k}{x}\\ \text{Plug in the values} \ y = 10 \ \text{and} \ x = 5. & & 10 & =\frac {k}{5}\\ \text{Solve for} \ k \ \text{by multiplying both sides of the equation by} \ 5. & & k&=50\\ \text{Now we put} \ k \ \text{back into the general equation.}\\ \text{The inverse relationship is given by} & & y & =\frac {50}{x}\\ \text{When} \ x = 2 & & y& =\frac {50}{2} \ \text{or} \ y=25\end{align*}
Answer \begin{align*}y = 25\end{align*}
Example 3
If \begin{align*}p\end{align*} is inversely proportional to the square of \begin{align*}q\end{align*}, and \begin{align*}p = 64\end{align*} when \begin{align*}q = 3\end{align*}. Find \begin{align*}p\end{align*} when \begin{align*}q = 5\end{align*}.
Solution:
\begin{align*}\text{Since} \ p \ \text{is inversely proportional to} \ q^2, \\ \text{then the general equation is} & & p& =\frac {k}{q^2}\\ \text{Plug in the values} \ p = 64 \ \text{and} \ q = 3. & & 64&=\frac {k}{3^2} \ \text{or} \ 64=\frac {k}{9}\\ \text{Solve for} \ k \ \text{by multiplying both sides of the equation by} \ 9. & & k& =576\\ \text{The inverse relationship is given by} & & p&=\frac {576}{q^2}\\ \text{When} \ q = 5 & & p&=\frac {576}{25} \ \text{or} \ p=23.04\end{align*}
Answer \begin{align*}p=23.04\end{align*}.
## Solve Real-World Problems Using Inverse Variation Equations
Many formulas in physics are described by variations. In this section we will investigate some problems that are described by inverse variations.
Example 4
The frequency, \begin{align*}f\end{align*}, of sound varies inversely with wavelength, \begin{align*}\lambda\end{align*}. A sound signal that has a wavelength of 34 meters has a frequency of 10 hertz. What frequency does a sound signal of 120 meters have?
Solution
\begin{align*}\text{The inverse variation relationship is} & & f&=\frac {k}{\lambda}\\ \text{Plug in the values} \ \lambda = 34 \ \text{and} \ f = 10. & & 10&=\frac {k}{34}\\ \text{Multiply both sides by} \ 34. & & k&=340 \\ \text{Thus, the relationship is given by} & & f&=\frac {340}{\lambda} \\ \text{Plug in} \ \lambda = 120 \ meters. & & f&=\frac {340}{120}\Rightarrow f=2.83\end{align*}
Answer \begin{align*} f=2.83\ Hertz\end{align*}
Example 5
Electrostatic force is the force of attraction or repulsion between two charges. The electrostatic force is given by the formula: \begin{align*}F=\left ( \frac{Kq_1q_2}{d^2}\right )\end{align*} where \begin{align*}q_1\end{align*} and \begin{align*}q_2\end{align*} are the charges of the charged particles, \begin{align*}d’\end{align*} is the distance between the charges and \begin{align*}k\end{align*} is proportionality constant. The charges do not change and are, thus, constants and can then be combined with the other constant \begin{align*}k\end{align*} to form a new constant \begin{align*}K\end{align*}. The equation is rewritten as \begin{align*}F=\left ( \frac{K}{d^2} \right )\end{align*}. If the electrostatic force is \begin{align*}F = 740\end{align*} Newtons when the distance between charges is \begin{align*}5.3 \times 10^{-11} \ meters\end{align*}, what is \begin{align*}F\end{align*} when \begin{align*}d = 2.0 \times 10^{-10} \ meters\end{align*}?
Solution
\begin{align*}\text{The inverse variation relationship is} & & f&=\frac {k}{d^2}\\ \text{Plug in the values} \ F = 740 \ \text{and} \ d = 5.3\times10^{-11}. & & 740&=\frac {k}{\left ( {5.3\times 10^{-11}} \right )^2}\\ \text{Multiply both sides by} \ (5.3\times10^{-11})^2. & & K&=740 \left ( {5.3\times 10^{-11}} \right )^2\\ \text{The electrostatic force is given by} & & F&=\frac {2.08\times 10^{-18}}{d^2}\\ \text{When} \ d = 2.0 \times 10^{-10} & & F&=\frac {2.08\times 10^{-18}}{\left ( 2.0\times 10^{-10} \right )^2}\\ \text{Enter} \ \frac{2.08*10^{(-18)}}{\left (2.0*10^{(-10)} \right )^2} \ \text{into a calculator}. & & F&=52\end{align*}
Answer \begin{align*}F=52 \ Newtons\end{align*}
Note: In the last example, you can also compute \begin{align*}F=\frac {2.08\times 10^{-18}}{\left ( 2.0\times 10^{-10} \right )^2}\end{align*} by hand.
\begin{align*}F & = \frac {2.08 \times 10^{-18}}{\left ( 2.0 \times 10^{-10} \right )^2} \\ & = \frac {2.08 \times 10^{-18}}{4.0 \times 10^{-20}}\\ & = \frac {2.08 \times 10^{20}}{4.0 \times 10^{18}}\\ & = \frac {2.08}{4.0} \left( 10^{2} \right) \\ & = 0.52(100) \\ & =52 \end{align*}
This illustrates the usefulness of scientific notation.
## Review Questions
Graph the following inverse variation relationships.
1. \begin{align*} y=\frac {3}{x}\end{align*}
2. \begin{align*} y=\frac {10}{x}\end{align*}
3. \begin{align*} y=\frac {1}{4x}\end{align*}
4. \begin{align*} y=\frac {5}{6x}\end{align*}
5. If \begin{align*}z\end{align*} is inversely proportional to \begin{align*}w\end{align*} and \begin{align*}z = 81\end{align*} when \begin{align*}w = 9\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}z = 24\end{align*}.
6. If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*} and \begin{align*}y = 2\end{align*} when \begin{align*}x = 8\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x = 12\end{align*}.
7. If \begin{align*}a\end{align*} is inversely proportional to the square root of \begin{align*}b\end{align*}, and \begin{align*}a = 32\end{align*} when \begin{align*}b = 9\end{align*}, find \begin{align*}b\end{align*} when \begin{align*}a = 6\end{align*}.
8. If \begin{align*}w\end{align*} is inversely proportional to the square of \begin{align*}u\end{align*} and \begin{align*}w = 4\end{align*} when \begin{align*}u = 2\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}u = 8\end{align*}.
9. If \begin{align*}x\end{align*} is proportional to \begin{align*}y\end{align*} and inversely proportional to \begin{align*}z\end{align*}, and \begin{align*}x = 2\end{align*}, when \begin{align*}y = 10\end{align*} and \begin{align*}z = 25\end{align*}. Find \begin{align*}x\end{align*} when \begin{align*}y = 8\end{align*} and \begin{align*}z = 35\end{align*}.
10. If \begin{align*}a\end{align*} varies directly with \begin{align*}b\end{align*} and inversely with the square of \begin{align*}c\end{align*} and \begin{align*}a = 10\end{align*} when \begin{align*}b = 5\end{align*} and \begin{align*}c = 2\end{align*}. Find the value of \begin{align*}a\end{align*} when \begin{align*}b = 3\end{align*} and \begin{align*}c = 6\end{align*}.
11. The intensity of light is inversely proportional to the square of the distance between the light source and the object being illuminated. A light meter that is 10 meters from a light source registers 35 lux. What intensity would it register 25 meters from the light source?
12. Ohm’s Law states that current flowing in a wire is inversely proportional to the resistance of the wire. If the current is 2.5 Amperes when the resistance is 20 ohms, find the resistance when the current is 5 Amperes.
13. The volume of a gas varies directly to its temperature and inversely to its pressure. At 273 degrees Kelvin and pressure of 2 atmospheres, the volume of the gas is 24 Liters. Find the volume of the gas when the temperature is 220 kelvin and the pressure is 1.2 atmospheres.
14. The volume of a square pyramid varies jointly as the height and the square of the length of the base. A cone whose height is 4 inches and whose base has a side length of 3 inches has a volume of \begin{align*}12 \ in^3\end{align*}. Find the volume of a square pyramid that has a height of 9 inches and whose base has a side length of 5 inches.
1. \begin{align*} W=\frac {243}{8}\end{align*}
2. \begin{align*} y=\frac {4}{3}\end{align*}
3. \begin{align*}b = 256\end{align*}
4. \begin{align*} w=\frac {1}{4}\end{align*}
5. \begin{align*} x=\frac {8}{7}\end{align*}
6. \begin{align*} a=\frac {2}{3}\end{align*}
7. \begin{align*}I = 5.6 \ lux\end{align*}
8. \begin{align*}R = 10 \ ohms\end{align*}
9. \begin{align*}V = 32.2 \ L\end{align*}
10. \begin{align*}V = 75 \ in^3\end{align*}
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# Determinant of a 3 x 3 Matrix
In matrices, determinants are the special numbers calculated from the square matrix. The symbol used to represent the determinant is represented by vertical lines on either side.
Let A be the matrix, then the determinant of a matrix A is denoted by |A|. To find any matrix such as determinant of 2×2 matrix, determinant of 3×3 matrix, or n x n matrix, the matrix should be a square matrix. It means that the matrix should have an equal number of rows and columns. Find determinants of a matrix are helpful in solving the inverse of a matrix, a system of linear equations, and so on.
## Determinant of a 3 x 3 Matrix Formula
We can find the determinant of a matrix in various ways. If we break the smaller 2 x 2 determinant problems as it is easy to handle, we can find the determinant of a 3 x 3 matrix.
Let’s suppose you are given a square matrix C where
C = $$\begin{bmatrix} a & b &c \\ d& e &f \\ g& h &i \end{bmatrix}$$
Let’s calculate the determinant of matrix C,
Det $$\begin{bmatrix} a & b &c \\ d& e &f \\ g& h &i \end{bmatrix}$$
= a. det $$\begin{bmatrix} e & f\\ h & i \end{bmatrix}$$ – b.det $$\begin{bmatrix} d & f\\ g & i \end{bmatrix}$$ + c . det $$\begin{bmatrix} d & e\\ g & h \end{bmatrix}$$
### Few Important points on 3x 3 Determinant Matrix
• The scalar multipliers to a corresponding 2 x 2 matrix have top row elements a, b and c serving to it.
• The scalar element gets multiplied by 2 x 2 matrix of remaining elements created at the time when vertical and horizontal line segments were drawn through passing through a.
• This is how we construct the 2 by 2 matrices for scalar multipliers b and c.
The determinant of 3 x 3 matrix formula is given by,
$$\begin{bmatrix} a & b &c \\ d& e &f \\ g& h &i \end{bmatrix}$$ = $$\begin{bmatrix} 2 & -3 &9 \\ 2 & 0 & -1\\ 1& 4 & 5 \end{bmatrix}$$
## Determinant of the 3 x 3 Matrix Examples
### Example 1:
Calculate the determinant of the 3 x 3 matrix.
$$\begin{bmatrix} 2 & -3 &9 \\ 2 & 0 & -1\\ 1& 4 & 5 \end{bmatrix}$$
Solution:
Let’s find the correspondence between the generic elements in the formula and elements of real problem.
$$\begin{bmatrix} a & b &c \\ d& e &f \\ g& h &i \end{bmatrix}$$ = $$\begin{bmatrix} 2 & -3 &9 \\ 2 & 0 & -1\\ 1& 4 & 5 \end{bmatrix}$$
Use the 3 x 3 determinant formula:
Applying the formula,
= 2[ 0 – (-4)] + 3 [10 – (-1)] + 3 [10 – (-1)] +1 [8-0]
= 2 (0+4) +3 (10 +1) + 1(8)
= 2(4) +3(11) + 8
= 8+33+8
= 49
Therefore, the determinant of $$\begin{bmatrix} 2 & -3 &9 \\ 2 & 0 & -1\\ 1& 4 & 5 \end{bmatrix}$$ = 49
### Example 2:
Calculate the determinant of the 3 x 3 matrix.
$$\begin{bmatrix} 1 & 3 &2 \\ -3 & -1 & -3\\ 2& 3 & 1 \end{bmatrix}$$
Solution:
Let’s find the correspondence between the generic elements in the formula and elements of real problem.
$$\begin{bmatrix} a & b &c \\ d& e &f \\ g& h &i \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 3 &2 \\ -3 & -1 & -3\\ 2& 3 & 1 \end{bmatrix}$$
Use the 3 x 3 determinant formula:
= 1[ -1 – (-9)] – 3 [-3 – (-6)] + 2 [-9 – (-2)]
= 1 (-1+9) -3 (-3 +6) + 2(-9 + 2)
= 1(8) -3(3) +2(-7)
= 8 -9-14
= -15
Therefore, the determinant of $$\begin{bmatrix} 1 & 3 &2 \\ -3 & -1 & -3\\ 2& 3 & 1 \end{bmatrix}$$ =-15 |
# Natural Numbers Notes - Class 6
## Class 6: Natural Numbers Notes - Class 6
The document Natural Numbers Notes - Class 6 is a part of Class 6 category.
All you need of Class 6 at this link: Class 6
Natural numbers are a part of the number system which includes all the positive integers from 1 till infinity. It should be noted that the natural numbers include only the positive integers i.e. set of all the counting numbers like 1, 2, 3, ………. excluding the fractions, decimals, and negative numbers. In this article, we are going to learn more about natural number with respect to its definition, comparison with whole numbers, representation in the number line, properties, etc.
## Natural Number Definition
As explained in the introduction part, Natural numbers are the numbers which are positive in nature and includes numbers from 1 till infinity(∞). These numbers are countable and are usually used for calculations purpose. The set of natural numbers are represented by the letter “N”.
## Natural Numbers and Whole Numbers
Natural numbers include all the whole numbers excluding the number 0. In other words, all natural numbers are whole numbers, but all whole numbers are not natural numbers. Check out the difference between natural and whole numbers to know more about the differentiating properties of these two sets of numbers.
The above representation of sets shows two regions,
A ∩ B ie. intersection of natural numbers and whole numbers (1, 2, 3, 4, 5, 6, ……..) and the green region showing A-B, i.e. part of the whole number (0).
Thus, a whole number is “a part of Integers consisting of all the Natural number including 0.”
### Is ‘0’ a Natural Number?
The answer to this question is ‘No’. As we know already, natural numbers start with 1 to infinity and are positive in nature. But when we mention 0 with a positive integer such as 10, 20, etc. it becomes a natural number. In fact, 0 is a whole number which has a null value.
### Representing Natural Numbers on a Number Line
Natural numbers representation on a number line is as follows:
The above number line represents natural numbers and the whole numbers on a number line. All the integers on the right-hand side of 0 represent the natural numbers, thus forming an infinite set of numbers. When 0 is included, these numbers become whole numbers which are also an infinite set of numbers.
### Set of Natural Numbers
In set notation, the symbol of natural number is “N” and it is represented as given below.
Statement:
N = Set of all numbers starting from 1.
In Roster Form:
N = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ………………………………}
In Set Builder Form:
N = {x : x is an integer starting from 1}
### Natural Numbers Examples
The natural numbers include the positive integers (also known as non-negative integers) and a few examples include 1, 2, 3, 4, 5, 6, ..∞. In other words, natural numbers are a set of all the whole numbers excluding 0.
### Properties of Natural Numbers
Natural numbers properties are segregated into four main properties which include closure property, commutative property, associative property, and distributive. Each of these properties is explained below in detail.
#### Closure Property
The natural numbers are always closed under addition and multiplication i.e. the addition and multiplication of natural numbers will always yield a natural number. In the case of subtraction and division, natural numbers are not closed which means subtracting or dividing two natural numbers might not give a natural number as a result.
• Addition: 1 + 2 = 3, 3 + 4 = 7, etc. In each of these cases, the resulting number is alwasy a natural number.
• Multiplication: 2 × 3 = 6, 5 × 4 = 20, etc. In this case also, the resultant is always a natural number.
• Subtraction: 9 – 5 = 4, 3 – 5 = -2, etc. In this case, the result may or may not be a natural number.
• Division: 10 ÷ 5 = 2, 10 ÷ 3 = 3.33, etc. In this case also, the resultant number may or may not be a natural number.
#### Associative Property
The associative property holds true in case of addition and multiplication of natural numbers i.e. a + ( b + c ) = ( a + b ) + c and a × ( b × c ) = ( a × b ) × c. On the other hand, for subtraction and division of natural numbers, the associative property does not hold true. An example of this is given below.
• Addition: a + ( b + c ) = ( a + b ) + c => 3 + (15 + 1 ) = 19 and (3 + 15 ) + 1 = 19.
• Multiplication: a × ( b × c ) = ( a × b ) × c => 3 × (15 × 1 ) = 45 and ( 3 × 15 ) × 1 = 45.
• Subtraction: a – ( b – c ) ≠ ( a – b ) – c => 2 – (15 – 1 ) = – 12 and ( 2 – 15 ) – 1 = – 14.
• Disivion: a ÷ ( b ÷ c ) ≠ ( a ÷ b ) ÷ c => 2 ÷( 3 ÷ 6 ) = 4 and ( 2 ÷ 3 ) ÷ 6 = 0.11.
#### Commutative Property
For commutative property,
• Addition and multiplication of natural numbers show the commutative property. For example, x + y = y + x and a × b = b × a.
• Subtraction and division of natural numbers does not show the commutative property. For example, x – y ≠ y – x and x ÷ y ≠ y ÷ x.
#### Distributive Property
• Multiplication of natural numbers is always distributive over addition. For example, a × (b + c) = ab + ac.
• Multiplication of natural numbers is also distributive over subtraction. For example, a × (b – c) = ab – ac.
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# What is the derivative of y=arctan sqrt((1-x)/(1+x))?
Nov 26, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\sqrt{{\left(1 + x\right)}^{5}}}{2 \sqrt{1 - x}}$
#### Explanation:
Let $u = \sqrt{\frac{1 - x}{1 + x}}$. Also observe $\frac{1 - x}{1 + x} = 1 - \frac{2 x}{1 + x}$
then using chain rule $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\frac{1 - x}{1 + x}}} \times \frac{d}{\mathrm{dx}} \left(1 - \frac{2 x}{1 + x}\right)$
= $\frac{\sqrt{1 + x}}{2 \sqrt{1 - x}} \times \frac{d}{\mathrm{dx}} \left(1 - \frac{2 x}{1 + x}\right)$
= $\frac{\sqrt{1 + x}}{2 \sqrt{1 - x}} \times \left(- \frac{2 \left(1 + x\right) - 2 x}{1 + x} ^ 2\right)$
= $\frac{\sqrt{1 + x}}{2 \sqrt{1 - x}} \times \left(- \frac{2}{1 + x} ^ 2\right)$
= $- {\sqrt{1 + x}}^{3} / \sqrt{1 - x}$
Hence $y = \arctan u$ and hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {u}^{2}} \times \frac{\mathrm{du}}{\mathrm{dx}}$
i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + \frac{1 - x}{1 + x}} \times \left(- \frac{\sqrt{{\left(1 + x\right)}^{3}}}{\sqrt{1 - x}}\right)$
= $- \frac{1 + x}{2} \times \frac{\sqrt{{\left(1 + x\right)}^{3}}}{\sqrt{1 - x}}$
= $- \frac{\sqrt{{\left(1 + x\right)}^{5}}}{2 \sqrt{1 - x}}$ |
# Find All Factors of 27 | Prime Factors of 27
Definition of factors of 27: If a number completely divides 27 without leaving a remainder, then that number is called a factor of 27.
By the above definition, we can say that the factors of 27 are the divisors of 27. In this section, we will learn about the factors of 27 and the prime factors of 27.
## Highlights of Factors of 27
• 27=3×3×3 is the prime factorization of 27.
• The factors of 27 are 1, 3, 9 and 27.
• The only prime factor of 27 is 3.
• Negative factors of 27 are –1, -3, -9 and –27.
## What are the factors of 27?
Let us write the number 27 multiplicatively in all possible ways. We definitely have:
27 = 1×27
27 = 3×9
As there are no numbers between 3 and 9 that can divide 27, so we will stop right now. So the factors of 27 in pairs are given as follows:
Thus the pair factors of 27 are (1, 27) and (3, 9). As all the numbers appearing in pair factors are the factors of 27, we conclude that
## Number of factors of 27
From above we have calculated the factors of 27 which are 1, 3, 9, and 27. Thus the total number of factors of 27 is four.
## Prime Factors of 27
Note that the factors of 27 are 1, 3, 9, and 27. Among those factors, we observe that only 3 is a prime number as 3 does not have any proper divisors.
∴ the only prime factor of 27 is 3.
Question: What are the factors of 27?
Video Solution:
## How to find factors of 27?
Now we will determine the factors of 27 by division method. In this method, we will try to find the numbers that can divide 27 with no remainder. See that
27/1=27 and the remainder is 0.
1 and 27 are factors of 27.
27/3=9 and the remainder is 0.
3 and 9 are factors of 27.
Note that no numbers other than the numbers in violet color can divide 27. So the numbers in violet color, that is, 1, 3, 9, and 27 are the complete list of factors of 27.
## FAQs on Factors of 27
Q1: Is 7 a factor of 27?
Ans: As 7 does not divide 27, we say that 7 is not a factor of 27.
Q2: Find all factors of 27.
Answer: All factors of 27 are 1, 3, 9, and 27.
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Equation Solver
Equation Solver is a tool used to solve polynomial equations of any order, such as linear or quadratic equations. It employs methods like adding or subtracting the same variable term and manipulating the equation to find the value(s) of the variable that satisfies the equation.
What is meant by equation and expression?
Equation
It states that the two expressions are equal is known as an equation. Simply say that it is a combination of two expressions that are distributive by the equal sign. Equations are useful for clarifying the connection between various variables and a constant. It plays a crucial role in solving many mathematical expressions and problems.
i.e., 2x + 5 = 10 + x is an equation with the variable “x”.
Expression
It is a mixture of numbers, variables, and mathematical operations (such as addition and subtraction) while not an equal sign. Expressions can be as simple as a single number or variable or more complex, involving multiple terms and operations.
i.e., 3x + 7 is the expression with the variable “x”.
How to solve the equation?
Example 1:
4x2 + 2x2 - 4 = 0 solve the equation.
Solution
Step 1: Collect and add the like terms.
(4x2 + 2x2) - 4 = 0
Step 2: Add the polynomial terms together.
(6x2) - 4 = 0
Step 3: Add the constant term from both sides.
6x2 - 4 + 4 = 0 + 4
Step 4: Simplify the left and right-hand side.
6x2 = 4
Step 5: To find the value of “x” divide by “6” on both sides and take square root.
6x2 /6 = 4/6
x2 = 2/3
x2 = √2/3
x = ∓ 0.8165
Example 2:
4x ^ 2 - 5x - 12 = 0 solve the equation.
Solution
Step 1: Take the given equation and write quadratic formula terms to solve it.
4x ^ 2 - 5x - 12 = 0
a = 4
b = -5
c = -12
Step 2: Now take the quadratic formula.
x = [-b ± √(b2 - 4ac)]/2a
Step 3: Place the value to solve the equation for "x"
x = [-(-5) ± √(-5)2 - 4(4)(-12)]/2(4)
Step 4: Simplify the above equation to find the sum and difference separately.
x = [-(-5) ± √(-5)2 - 4(4)(-12)]/2(4)
x = [5 ± √(25 + 192)]/8
x = [5 ± √217]/8
x = [5 ± 14.73]/8
x = [5 + 14.73]/8
x = 19.73/8
x = 2.4663
For Subtraction
x = [5 - 14.73]/8
x = - 9.73/8
x = -1.2163
How to simplify expression/order equations?
To solve expression/order equations:
• First simplify the terms inside parentheses, brackets, braces, and fractions bars.
• Evaluate all powers.
• Solve all multiplications or divisions from left to right.
• Do all additions and subtractions from left to right.
What is the golden rule for solving equations?
The golden rule of algebra is taking a balanced scale. If any operation you apply on one side of the equal sign then you must do the same thing on the other side. i.e., If you multiply by “-2” on the left side then must multiply by “-2” on the other side.
How to solve the linear equation?
To solve the linear equation, get all variables on one side & constant terms on the other side of the equal sign and simplify the terms using algebraic rules such as addition or subtraction.
How to solve the quadratic equation?
Identify the coefficients “a, b, and c” from the quadratic equation “ax2 + bx + c” then Substitute the values into the quadratic formula and Simplify the operations.
Are linear and quadratic equations same?
No, linear equation has the highest degree is one while the quadratic equation highest degree is two. On the other hand, graphically linear shows a straight line graph, and quadratic produces the parabola graph. |
## Move Two Matches to Get the Largest Number
Puzzle. If you can move exactly two matches, what is the largest possible number you can get?
Many people assume that given that 508 has three digits, the answer also has three digits. They get the number 999 and stop there.
Some students realized they can make an extra 1 out of two matches and gave answers with four digits: 1503, 5051, and 5103.
Another great idea is to take out the two horizontal matches from 0, turning the 0 into 11. For example, one might use the two matches to turn 5 into 8 and get 8118. However, we can combine this idea with the previous one and use the two matches to make a new digit 1 to get 51181.
When I first saw this puzzle several years ago, 51181 was the official answer. But my students went further, much, much further.
Another elegant idea is to rotate the picture and get 81151.
Some students decided that having all the digits be the same height is not very important. One vertical match reads as 1, even if it is half the height. The largest number they got this way was 511811. Combining this idea with flipping the number allows us to get 811511.
However, small-font digits are often used in powers. This train of thought led to a brilliant answer, 511811. This number is way larger than everything we saw before: it has 41 digits. If we combine this idea with rotating the number, we can get 811511, a 43-digit number.
I recently decided to check the Internet again and discovered a chain of videos showing solutions to this puzzle by the same author. He started with a simple solution, 51181, then people left comments in the video with better solutions, so he remade the video. This happened several times. I probably should send him a link to this essay for yet another video.
Here is an interesting solution by one of my students that I haven’t seen anywhere else. The idea is to use scientific notation. The student took two matches from the right side of 0. He used one of them to convert the first digit from 5 to 9 and the second digit from 0 into the letter E. He got 9E8 = 900000000. If we combine this idea with using a small 1, we can get 5E81, an 82-digit number.
Another brilliant idea is using two matches to create a caret symbol representing an exponent. This way, we can get 5^118, which is an 83-digit number. If we combine this idea with the rotation idea, we can get 8^115, a 104-digit number.
If we ignore the relative sizes of the digits, we can have the small digits as the base of the exponent and the larger digits as the power. One of the students suggested 115118, a 5330-digit number. And if we rotate the number, the answer we can get is 118115: the number with 8451 digits. This is a humongous number compared to the mere 3-digit one we started with, to the pathetic 5-digit original solution, and to the pitiful 104-digit previous record. But my students didn’t stop there.
One student suggested snapping two matches and creating a double factorial 505!!, with 575 digits. Combining this with other ideas, we can get 8115!!, a 14102-digit number.
One of my students decided to use two matches from the last digit 8 to create a factorial sign. She glued one of the matches perpendicular to the plane to get 505! in the projection. She even made a picture that you can see below. This number has 1148 digits, and combining this with previous ideas, we can get 5118!, which has 16763 digits. Or even 8115!, which has 28202 digits.
For this puzzle, a factorial works better than a double factorial. Here is my own suggestion: use one of the matches to create a small digit one, and split the other match into a factorial. This way, we can get 81151!, a number with a whooping 363154 digits.
This is an excellent example of a thinking-outside-the-box puzzle. I love this puzzle because it has not one, but many boxes one can think outside off.
Share:
1. #### Ivan:
Let us start from here: “… One student suggested snapping two matches and creating a double factorial …” So we do a similar thing, a left factorial of 815!, i.e. !815!. This is a better solution, but unfortunately its digital length its too hard to find.
2. #### Ivan:
I am sorry, I meant !8115!
3. #### tanyakh:
Cool. I never heard of left factorial before.
4. #### Leo B.:
To allow solutions in the form of expressions, the question must have been worded as “what is the largest possible *value* you can get?”
As posed, the final answer to the question is 811511. All subsequent tricks are not “numbers”, they are arithmetic expressions with a numerical value.
5. #### Calcdave:
I wonder if you could squeeze 2 of the matches into a / symbol to get something like 30/0. Perhaps infinity does not count as a “number” in this case.
6. #### Jamas:
If you are allowed to break matches, I’m thinking up arrows, eg 5↑↑5.. and when combined with other ideas…
7. #### Cristóbal Camarero:
With rotation, scientific notation, a little one, and factorial, we get 1E05!, of 456,573 digits.
8. #### Cristóbal Camarero:
And we can also make an E from 0 to get 5e8!, with 4e9 digits.
9. #### Ivan:
In all of the examples above the numbers are finite. If splitting is allowed, then the infinite splitting of the first match will produce 8115!!!…, which leaves a match to light the cigarette we need after the exhausting thinking.
10. #### tarik:
To allow solutions in the form of expressions, the question must have been worded as “what is the largest possible *value* you can get?”
As posed, the final answer to the question is 811511. All subsequent tricks are not “numbers”, they are arithmetic expressions with a numerical value. |
At times, monomials can have coefficients and/or be raised to a power before you begin the binomial expansion. In this case, you have to raise the entire monomial to the appropriate power in each step. For example, here's how you expand the expression (3x2 – 2y)7:
1. Write out the binomial expansion by using the binomial theorem, substituting in for the variables where necessary.
In case you forgot, here is the binomial theorem:
Replace the letter a in the theorem with the quantity (3x2) and the letter b with (–2y). Don't let those coefficients or exponents scare you — you're still substituting them into the binomial theorem. Replace n with 7. You end up with
2. Find the binomial coefficients.
The formula for binomial expansion is written in the following form:
You may recall the term factorial from your earlier math classes. If not, here is a reminder: n!, which reads as "n factorial," is defined as
Now, back to the problem. Using the combination formula gives you the following:
3. Replace all
4. with the coefficients from Step 2.
1(3x2)7(–2y)0 + 7(3x2)6(–2y)1 + 21(3x2)5(–2y)2 + 35(3x2)4(–2y)3 + 35(3x2)3(–2y)4 + 21(3x2)2(–2y)5 + 7(3x2)1(–2y)6 + 1(3x2)0(–2y)7
5. Raise the monomials to the powers specified for each term.
1(2,187x14)(1) + 7(729x12)(–2y) + 21(243x10)(4y2) + 35(81x8)(–8y3) + 35(27x6)(16y4) + 21(9x4)(–32y5) + 7(3x2)(64y6) + 1(1)(–128y7)
6. Simplify.
2,187x14 – 10,206x12y + 20,412x10y2 – 22,680x8y3 + 15,120x6y4 – 6,048x4y5 + 1,344x2y6 – 128y7 |
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# Квартиль хоорондын зай(IQR)
## Video transcript
- [Instructor] Let's get some practice calculating interquartile ranges and I've taken some exercises from the Khan Academy exercises here. I'm just gonna solve it on my scratch pad. The following data points represent the number of animal crackers in each kid's lunch box. Sort the data from least to greatest and then find the interquartile range of the data set and I encourage you to do this before I take a shot at it. Alright, so let's first sort it and if we were actually doing this on the Khan Academy exercise, you could just drag these, you could just click and drag these numbers around to sort 'em but I'll just do it by hand. So let's see, the lowest number here looks like it's a four. So I have that four then I have another four and then I have another four and let's see, are there any fives? No fives but there is a six. So then there is a six and then there's a seven. There doesn't seem to be an eight or a nine but then we get to a 10 and then we get to 11, 12. No 13 but then we got 14 and then finally we have a 15. So the first thing we wanna do is figure out the median here. So the median's the middle number. I have one, two, three, four, five, six, seven, eight, nine numbers so there's going to be just one middle number. I have an odd number of numbers here and it's going to be the number that has four to the left and four to the right and that middle number, the median is going to be 10. Notice I have four to the left and four to the right and the interquartile range is all about figuring out the difference between the middle of the first half and the middle of the second half. It's a measure of spread, how far apart all of these data points are and so let's figure out the middle of the first half. So we're gonna ignore the median here and just look at these first four numbers and so out of these first four numbers, since I have an even number of numbers, I'm gonna calculate the median using the middle two numbers so I'm gonna look at the middle two numbers here and I'm gonna take their average. So the average of four and six, halfway between four and six is five or you could say four plus six is, four plus six is equal to 10 but then I wanna divide that by two so this is going to be equal to five. So the middle of the first half is five. You can imagine it right over there and then the middle of the second half I'm gonna have to do the same thing. I have four numbers. I'm gonna look at the middle two numbers. The middle two numbers are 12 and 14. The average of 12 and 14 is going to be 13, is going to be 13. If you took 12 plus 14 over two, that's going to be 26 over two which is equal to 13 but an easier way for numbers like this, you say hey, 13 is right exactly halfway between 12 and 14. So there you have it. I have the middle of the first half is five. I have the middle of the second half, 13. To calculate the interquartile range, I just have to find the difference between these two things. So the interquartile range for this first example is going to be 13 minus five. The middle of the second half minus the middle of the first half which is going to be equal to eight. Let's do some more of these. This is strangely fun. Find the interquartile range of the data in the dot plot below. Songs on each album in Shane's collection and so let's see what's going on here and like always, I encourage you to take a shot at it. So this is just representing the data in a different way but we could write this again as an ordered list so let's do that. We have one song or we have one album with seven songs I guess you could say. So we have a seven. We have two albums with nine songs so we have two nines. Let me write those, we have two nines then we have three 10s. Cross those out. So 10, 10, 10 then we have an 11. We have an 11. We have two 12s, two 12s and then finally, we have, I used those already and then we have an album with 14 songs. 14. So all I did here is I wrote this data like this so we could see, okay, this album has seven songs, this album has nine, this album has nine and the way I wrote it, it's already in order so I could immediately get, I can immediately start calculating the median. Let's see, I have one, two, three, four, five, six, seven, eight, nine, 10 numbers. I have an even number of numbers so to calculate the median, I'm gonna have to look at the middle two numbers. So the middle two numbers look like it's these two 10s here because I have four to the left of them and then four to the right of them and so since I'm calculating the median using two numbers, it's going to be halfway between them. It's going to be the average of these two numbers. Well, the average of 10 and 10 is just going to be 10. So the median is going to be 10. Median is going to be 10 and in a case like this where I calculated the median using the middle two numbers, I can now include this left 10 in the first half and I can include this right 10 in the second half. So let's do that. So the first half is going to be those five numbers and then the second half is going to be these five numbers and it makes sense 'cause I'm literally just looking at first half it's gonna be five numbers, second half is gonna be five numbers. If I had a true middle number like the previous example then we ignore that when we look at the first and second half or at least that's the way that we're doing it in these examples but what's the median of this first half if we look at these five numbers? Well, if you have five numbers, if you have an odd number of numbers, you're gonna have one middle number and it's going to be the one that has two on either sides. This has two to the left and it has two to the right. So the median of the first half, the middle of the first half is nine right over here and the middle of the second half, I have one, two, three, four, five numbers and this 12 is right in the middle. You have two to the left and two to the right so the median of the second half is 12. Interquartile range is just going to be the median of the second half, 12 minus the median of the first half, nine which is going to be equal to three. So if I was doing this on the actual exercise, I would fill out a three right over there. |
# Divisibility rules for numbers 1–20
The rules given below transform a given number into a generally smaller number, while preserving divisibility by the divisor of interest. Therefore, unless otherwise noted, the resulting number should be evaluated for divisibility by the same divisor. In some cases the process can be iterated until the divisibility is obvious; for others (such as examining the last n digits) the result must be examined by other means.
For divisors with multiple rules, the rules are generally ordered first for those appropriate for numbers with many digits, then those useful for numbers with fewer digits.
Note: To test divisibility by any number that can be expressed as 2n or 5n, in which n is a positive integer, just examine the last n digits.
Note: To test divisibility by any number that can be expressed as the product of prime factors p 1 n p 2 m p 3 q {\displaystyle p{1}^{n}p{2}^{m}p{3}^{q}} p{1}^{n}p{2}^{m}p{3}^{q}, we can separately test for divisibility by each prime to its appropriate power. For example, testing divisibility by 24 (24 = 83 = 233) is equivalent to testing divisibility by 8 (23) and 3 simultaneously, thus we need only show divisibility by 8 and by 3 to prove divisibility by 24. Divisor Divisibility condition Examples
1 No special condition. Any integer is divisible by 1. 2 is divisible by 1.
2 The last digit is even (0, 2, 4, 6, or 8).[1][2] 1294: 4 is even.
3 Sum the digits. If the result is divisible by 3, then the original number is divisible by 3.[1][3][4] 405 → 4 + 0 + 5 = 9 and 636 → 6 + 3 + 6 = 15 which both are clearly divisible by 3. 16,499,205,854,376 → 1+6+4+9+9+2+0+5+8+5+4+3+7+6 sums to 69 → 6 + 9 = 15 → 1 + 5 = 6, which is clearly divisible by 3. Subtract the quantity of the digits 2, 5, and 8 in the number from the quantity of the digits 1, 4, and 7 in the number. Using the example above: 16,499,205,854,376 has four of the digits 1, 4 and 7 and four of the digits 2, 5 and 8; ∴ Since 4 − 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3.
4 Examine the last two digits.[1][2] 40,832: 32 is divisible by 4. If the tens digit is even, the ones digit must be 0, 4, or 8. If the tens digit is odd, the ones digit must be 2 or 6. 40,832: 3 is odd, and the last digit is 2. Twice the tens digit, plus the ones digit. 40832: 2 × 3 + 2 = 8, which is divisible by 4.
5 The last digit is 0 or 5.[1][2] 495: the last digit is 5.
6 It is divisible by 2 and by 3.[5] 1458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6.
7 Form the alternating sum of blocks of three from right to left.[4][6] 1,369,851: 851 − 369 + 1 = 483 = 7 × 69 Subtract 2 times the last digit from the rest. (Works because 21 is divisible by 7.) 483: 48 − (3 × 2) = 42 = 7 × 6. Or, add 5 times the last digit to the rest. (Works because 49 is divisible by 7.) 483: 48 + (3 × 5) = 63 = 7 × 9. Or, add 3 times the first digit to the next. (This works because 10a + b − 7a = 3a + b − last number has the same remainder) 483: 4×3 + 8 = 20 remainder 6, 6×3 + 3 = 21. Or, add the last two digits to twice the rest. (Works because 98 is divisible by 7.) 483,595: 95 + (2 × 4835) = 9765: 65 + (2 × 97) = 259: 59 + (2 × 2) = 63. Multiply each digit (from right to left) by the digit in the corresponding position in this pattern (from left to right): 1, 3, 2, -1, -3, -2 (repeating for digits beyond the hundred-thousands place). Then sum the results. 483,595: (4 × (-2)) + (8 × (-3)) + (3 × (-1)) + (5 × 2) + (9 × 3) + (5 × 1) = 7.
8 If the hundreds digit is even, examine the number formed by the last two digits. 624: 24. If the hundreds digit is odd, examine the number obtained by the last two digits plus 4. 352: 52 + 4 = 56. Add the last digit to twice the rest. 56: (5 × 2) + 6 = 16. Examine the last three digits.[1][2] 34,152: Examine divisibility of just 152: 19 × 8 Add four times the hundreds digit to twice the tens digit to the ones digit. 34,152: 4 × 1 + 5 × 2 + 2 = 16
9 Sum the digits. If the result is divisible by 9, then the original number is divisible by 9.[1][3][4] 2880: 2 + 8 + 8 + 0 = 18: 1 + 8 = 9.
10 The last digit is 0.[2] 130: the last digit is 0.
11 Form the alternating sum of the digits.[1][4] 918,082: 9 − 1 + 8 − 0 + 8 − 2 = 22. Add the digits in blocks of two from right to left.[1] 627: 6 + 27 = 33. Subtract the last digit from the rest. 627: 62 − 7 = 55. Add the last digit to the hundredth place (add 10 times the last digit to the rest). 627: 62 + 70 = 132. If the number of digits is even, add the first and subtract the last digit from the rest. 918,082: the number of digits is even (6) → 1808 + 9 − 2 = 1815: 81 + 1 − 5 = 77 = 7 × 11 If the number of digits is odd, subtract the first and last digit from the rest. 14,179: the number of digits is odd (5) → 417 − 1 − 9 = 407 = 37 × 11
12 It is divisible by 3 and by 4.[5] 324: it is divisible by 3 and by 4. Subtract the last digit from twice the rest. 324: 32 × 2 − 4 = 60.
13 Form the alternating sum of blocks of three from right to left.[6] 2,911,272: 2 - 911 + 272 = -637 Add 4 times the last digit to the rest. 637: 63 + 7 × 4 = 91, 9 + 1 × 4 = 13. Subtract the last two digits from four times the rest. 923: 9 × 4 - 23 = 13. Subtract 9 times the last digit from the rest. 637: 63 - 7 × 9 = 0.
14 It is divisible by 2 and by 7.[5] 224: it is divisible by 2 and by 7. Add the last two digits to twice the rest. 364: 3 × 2 + 64 = 70. 1764: 17 × 2 + 64 = 98.
15 It is divisible by 3 and by 5.[5] 390: it is divisible by 3 and by 5.
16 If the thousands digit is even, examine the number formed by the last three digits. 254,176: 176. If the thousands digit is odd, examine the number formed by the last three digits plus 8. 3408: 408 + 8 = 416. Add the last two digits to four times the rest. 176: 1 × 4 + 76 = 80.
1168: 11 × 4 + 68 = 112. Examine the last four digits.[1][2] 157,648: 7,648 = 478 × 16.
17 Subtract 5 times the last digit from the rest. 221: 22 − 1 × 5 = 17. Subtract the last two digits from two times the rest. 4,675: 46 × 2 - 75 = 17.
18 It is divisible by 2 and by 9.[5] 342: it is divisible by 2 and by 9.
19 Add twice the last digit to the rest. 437: 43 + 7 × 2 = 57. Add 4 times the last two digits to the rest. 6935: 69 + 35 × 4 = 209.
20 It is divisible by 10, and the tens digit is even. 360: is divisible by 10, and 6 is even. The number formed by the last two digits is divisible by 20.[2] 480: 80 is divisible by 20.
Note by Faizan Khan
5 years, 1 month ago
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## Comments
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Yup, this is just Divisiblity Rules.
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Nice app thank you
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Are your fingers all right , how did you typed this all.. But this all are nice method..
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# 4.17: Solve Problems Involving Rates and Unit Analysis
Difficulty Level: At Grade Created by: CK-12
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“I would LOVE to climb Mount Everest!” Josh exclaimed at breakfast one morning.
“Really?” his Dad said smiling. “Well son, you had better start saving now.”
Josh looked up from his oatmeal with a puzzled look on his face.
“What makes you say that?” Josh asked.
“What makes me say that is that the going rate for one climb on Everest is about $60,000. That’s what makes me say that,” his Dad explained taking a sip of his coffee. “Really? Wow! I had no idea,” Josh said. “Well, I guess I’ll just have to make a lot of money!” Josh said leaving the table. He kept thinking about what his Dad had said all the way to school. Sixty thousand dollars was a lot of money to climb a mountain, but what really amazed Josh was thinking about the numbers of people who had climbed the mountain more than once. When he got to class, he looked up in his book that Apa Sherpa a man from Nepal had successfully climbed Everest 19 times. Now he was often a guide who was paid, but still, Josh couldn’t help thinking about how much money Apa Sherpa would have spent if he had paid to climb Everest 19 times at the rate his father spoke about. How much would it have cost? We can use units, ratios and proportions to solve this problem. By thinking of one trip as a unit, we can look at the proportion and solve for the correct amount of money. ### Guidance A rate refers to speed or a rate can refer to the amount of money someone makes per hour. When we talk about a unit rate, we look at comparing a rate to 1, or how much it would take for 1 of something. It could be one apple, one mile, one gallon. We are comparing a quantity to one. A key word when working with unit rate is the word “per”. We can use ratios and proportions to solve problems involving rates and unit rates. Jeff makes$150.00 an hour as a consultant. What is his rate per minute?
To figure this out, we have to think about the unit rate that Jeff is paid as a consultant. You will see that we have “an hour” written into the problem. This is the unit rate. Also notice that the would "per" is used in the problem.
Let’s write the unit rate as a ratio compared to 1.
150.001\begin{align*}\frac{\ 150.00}{1}\end{align*} Next, we need to think about what the problem is asking for. It is asking for his rate per minute. The given information is in hours, so we need to write a ratio that compares hours to minutes. 1 hour60 minutes\begin{align*}\frac{1 \ hour}{60 \ minutes}\end{align*} Now we can write an expression combining the two ratios.1501 hour1 hour60 minutes\begin{align*}\frac{\ 150}{1 \ hour} \cdot \frac{1 \ hour}{60 \ minutes}\end{align*}
That’s a great question. We don’t compare hours to hours because we aren’t comparing hours. We are comparing money to hours and we need to figure out the rate of money per minute. You always have to think about what is being compared when working with proportions.
Next, we can solve. Notice that because 1 hour is diagonal from 1 hour, we can cross cancel the hours. That leaves us with a ratio that compares money to minutes.
15060 minutes\begin{align*}\frac{\ 150}{60 \ minutes}\end{align*} This also helped us to convert hours to minutes making it easier to figure out the answer to the problem. Now we can divide to figure out our answer. Jeff makes2.50 per minute.
What is unit analysis?
Unit analysis is when we look at how to measure individual units in different measurement amounts and it is used to convert units of measurement.
When we use unit analysis, we convert different measurement units by comparing the units using ratios and proportions. Unit analysis is very helpful when checking results.
Take a look at this dilemma.
Juanita worked for 18 hours. She made $116.00 at the end of her shift. Juanita was sure that her manager had made a mistake and that she should have made more money. Juanita makes$9.00 per hour. Did Juanita make the correct amount of money or was there a mistake?
To work on this problem, we can use unit analysis. Let’s start by writing a ratio to compare how much Juanita made for the hours worked.
18 hours116.00\begin{align*}\frac{18 \ hours}{\ 116.00}\end{align*} Next, we can use her hourly rate to work with. She makes9.00 per hour.
91 hour\begin{align*}\frac{\ 9}{1 \ hour}\end{align*} If Juanita makes9.00 per hour, we can multiply 18×9\begin{align*}18 \times 9\end{align*}. We have an answer of $162.00. Juanita was only paid$116.00, so there definitely was an error in her payment.
Solution: $3.00 #### Example C Two tickets to a ballgame costs$111.50. What is the cost for one ticket?
Solution: $55.75 Now let's go back to the dilemma from the beginning of the Concept. Now let’s look at solving this problem. We know that it costs$60,000 for 1 trip up Mount Everest.
$60,0001x=x19=$1,140,000\begin{align*}\frac{\ 60,000}{1} &= \frac{x}{19}\\ x &= \ 1,140,000\end{align*}
We can also use unit analysis to solve this problem.
60,000 dollars (19x dollars)\begin{align*}\left( \frac{19}{x \ dollars}\right)\end{align*} 60,000×19=1,140,000\begin{align*}60,000 \times 19 = \ 1,140,000\end{align*} is the cost of the nineteen trips.
This is our solution.
### Vocabulary
Rate
a unit that is in relationship with another unit. It could be a pay rate or a rate of speed. It is a unit that is measured.
Unit Rate
a rate compared to 1. A pay rate would be an amount of money per hour. A gasoline unit rate would be the amount of money for one gallon of gasoline.
Unit Analysis
is a method of converting different units of measurement by using ratios and proportions to compare and convert the units.
### Guided Practice
Here is one for you to try on your own.
Solve and then check using unit analysis.
Jesse has a car that holds 14 gallons of gasoline. During the first week of the month, gasoline cost $2.75 per gallon. During the second week of the month, gasoline cost$2.50 per gallon. How much was the total cost for the 28 gallons of gasoline?
Solution
Let’s start by writing a variable expression to work on this problem. We know that the number of gallons of gasoline does not change. That can be our variable.
x=\begin{align*}x = \end{align*} number of gallons of gasoline
The other parts of the expression include the different prices for the gasoline.
2.75x+2.50x\begin{align*}2.75x+2.50x\end{align*}
This expression will help us to determine how much money Jesse spent on 28 gallons of gasoline. Each full tank is 14 gallons. We can substitute 14 for our variable x\begin{align*}x\end{align*}.
2.75(14)$38.50+2.50(14)+$35.00\begin{align*}2.75(14)&+ 2.50(14)\\ \ 38.50 &+ \ 35.00\end{align*}
The total amount of money spent was $73.50. We can check our work by using unit analysis. 2.751 gallon2.501 gallon=x14 gallons=$38.50=x14 gallons=35.00\begin{align*}\frac{2.75}{1 \ gallon} &= \frac{x}{14 \ gallons} = \ 38.50\\ \frac{2.50}{1 \ gallon} & = \frac{x}{14 \ gallons} = \ 35.00\end{align*} The sum of the money spent was73.50.
### Practice
Directions: Use what you have learned to solve each problem.
1. Peter runs at a rate of 10 kilometers per hour. How many kilometers will he cover in 8 hours?
2. A cheetah can run at a speed of 60 miles per hour. What is his distance after 6 hours?
3. What is the distance formula?
4. If a car travels at a rate of 65 miles per hour for 30 minutes, how far will it travel?
5. A train travels at a rate of 50 miles per hour. If it needs to travel 320 miles, how many minutes will it take?
6. A car travels 65 mph for 12 hours. How many miles will it travel?
7. A bus traveled 300 miles at an average speed of 50 miles per hour. How long did this trip take the bus?
8. A car traveled at an average speed of 40 miles per hour through a construction zone. If the car traveled 20 miles at this rate, how many hours did it take to travel the 20 miles?
9. What is velocity?
10. What is the formula for velocity?
11. What is the velocity of an object that travels 500 miles in 2.5 hours?
12. If an object has a velocity of 125 miles per hour, how long will it take to travel 4,375 miles?
13. If an object has a velocity of 7 kilometers per minute, how far will it travel in 2 hours?
14. If an object has a velocity of 4 meters per second, how many kilometers will it travel in 2 days?
15. The formula for density is D=mv\begin{align*}D = \frac{m}{v}\end{align*} where D\begin{align*}D\end{align*} represents the density of an object, m\begin{align*}m\end{align*} represents the mass of the object, and v\begin{align*}v\end{align*} represents the volume of the object. What is the density of a brick that weighs 9 pounds and has a volume of 36 cu. in.?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Rate A rate is a special kind of ratio that compares two quantities.
Unit Analysis Unit analysis is a method of converting units of measurement by using ratios and proportions.
Unit Rate A unit rate is a ratio that compares a quantity to one. The word “per” is a key word with unit rates.
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## BEAR BUILDER [DAY 2]
### CALCULATING PERCENT
Create and draw bears that vary in size
## Intentionality & Unit Overview
### Access each lesson from this unit using the navigation links below
Students will continue exploring the relationship between the area of rectangles and the array model. The array replaces the need to count individual units and makes it possible to calculate an area.
## Intentionality…
The purpose of the Day 2 activities is to reinforce key concepts from Day 1. Students will engage in a string of related problems through a math talk and will have an opportunity to complete independent purposeful practice. The math talk and purposeful practice serve to develop a deeper understanding of the following big ideas:
• Percent is a comparison measured out of 100;
• To solve percent related problems we can use the following strategies:Â
• Analyze Between and Within Ratios (Scaling Up or Down);
• Find a Unit Rate
• Use y = kx
## Math Talk
Present the following percent problems one at a time (use the visual prompts). Encourage students to describe where the values should be placed along a number line.
This first set of problems was selected to emerge benchmark comparisons and scaling in tandem strategies. Consider modelling student thinking using a double number line.
The second set of problems also promotes the use of a benchmark strategy, however the last problem encourages the use of a unit rate strategy.
10% of 30
20% of 30
5% of 30
15% of 30
Â
10% of 75
15% of 75
65% of 75
33% of 75
Consider watching this silent solution animation to help you prepare to facilitate this math talk utilizing a double number line.
## Visual Math Talk Prompt #1
Show students the following visual math talk prompt and be prepared to pause the video where indicated:
How much will the percent-machine output?
Where along the number line should the output be placed?
Where should each input be placed on the numberline?
Similar to Day 1 of the Bear Builder problem based unit, students will use the percentage of 10% to determine 10% of 30.Â
Since we are aiming to build fluency and flexibility working with percent by using multiplication and division we are going to ensure students are not leveraging the calculator as a tool.
As the facilitator, we are going to do our best to work from a double number line to highlight the strategies students use to determine the percent value.
Some students will choose to place benchmarks along their number line like 50% by using a halving strategy. They may then break the bottom half of the number line up into five 10% sections revealing that each 10% is 3 units and each 5% is 1.5 units.Â
Of course, these visuals can support your facilitation of this visual number talk, but it would be most effective if the facilitator draws the models both as arrays or area models and represents symbolically as the thinking emerges.
## Visual Math Talk Prompt #2
Show students the following visual math talk prompt and be prepared to pause the video where indicated:
How much will the percent-machine output?
Where along the number line should the output be placed?
Where will each input be placed on the numberline?
Again, facilitating this based on student thinking and modelling on the board for all to see is helpful here rather than simply playing the silent solution animation.
While we will model based on whatever approach students would like to use to solve this problem, we do want to ensure that we highlight the idea of using benchmark percentages here as a strategy that students may be able to add to their repertoire of strategies for the future.
Similarly to Visual Math Talk Prompt #1 students may use a halving strategy to show 50% and 25% on the numberline and then switch to a division strategy to break 25% into 5 groups of 5% each.Â
When students approach 33% of 75 they may break their 5% group up into five 1% each groups (unit rate) and then scale the 1% group to 33%.Â
Again, we do not want to limit students from leveraging other strategies, however we do want them to recognize that certain strategies may be more helpful than others in particular situations
## While Students Are Practicing…
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## Purposeful Practice
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We suggest collecting this reflection as an additional opportunity to engage in the formative assessment process to inform next steps for individual students as well as how the whole class will proceed.
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## Educator Discussion Area
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## Explore Our 60+ Problem Based Units
This Make Math Moments Lesson was designed to spark curiosity for a multi-day unit of study with built in purposeful practice, number talks and extensions to elicit and emerge strategies and mathematical models.
Dig into our other units of study and view by concept continuum, grade or topic! |
# Number Line Prime Class 7 Solutions Chapter 16
## Number Line Prime Class 7 Solutions Chapter 16 Properties of Triangles
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 16, Properties of Triangles. Here students can easily find step by step solutions of all the problems for Properties of Triangles, Exercise 16A, 16B, 16C, 16D and 16E Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 16 solutions. Here in this post all the solutions are based on latest Syllabus.
Properties of Triangles Exercise 16A Solution :
Question no – (1)
Solution :
(a) ST
(b) ∠R
(C) S
Question no – (2)
Solution :
(a) m, n, o points
(b) S, T Points
(c) u, v, w points.
Question no – (3)
Solution :
(a) All lengths are non equal
∴ It is a scalene triangle.
(b) All lengths are equal.
∴ It is a equilateral triangle.
Question no – (4)
Solution :
(a) All angle is 90°
∴ It is a right triangle.
(b)Here are angle is greater than 90°
∴ It is a obtuse – Angled triangle.
(c) All angle are less than 90°
∴ It is a acute – angled triangle
Question no – (5)
Solution :
∴ In △ABC, AB = BC, BL = LC, AL ⊥ BC
∴ It is a altitude = median = AL
Properties of Triangles Exercise 16C Solution :
Question no – (1)
Solution :
(a)
∴ <ACB = 180° – 110°
∴ <ACB = 70°
∴ <ABC + <BAC + <ACB = 180°
∴ x + 60° + 70° = 180°
∴ x = 180° + 150°
∴ x = 30°
(b
∴ y = 180° – 45°
∴ y = 135°
Question no – (2)
Solution :
∴ <R = 180° – (2P + LQ)
∴ <R = 180° (80° + 65°)
∴ <R = 180° – 145°
∴ <R = 35°
∴ Vertex R° = 180° – 35°
∴ R° = 145°
Question no – (3)
Solution :
∴ LC = 180° – <BCD = 180° – 3x
∴ x + 80° + 180° – 3x = 180°
∴ – 2x = – 80°
∴ x = 40°
Question no – (4)
Solution :
∴ Let, interior angles – x°
∴ 2x° = 120°
∴ x = 60°
∴ Angles of the triangle – 60°, 60°, 60°
Question no – (5)
Solution :
∴ Let the interior opposite angles – x°
∴ Are angle should be,
= 180° – 100°
= 80°
∴ 80° + x + x = 180°
∴ 2x = 100
∴ x = 50°
∴ Opposite angle = 50°, 50°
Properties of Triangles Exercise 16D Solution :
Question no – (1)
Solution :
(a) Given, AB = 6cm, BC = 6 cm, AC = 6 cm
∴ In △ABC,
∴ AB – BC < CA
= 6 – 6 < 6
= 0 < 6,
∴ BC – CA < AB
= 6 – 6 < 6
= 0 < 6,
∴ CA – AB < BC
= 6 – 6 < 6
= 0 < 6
∴ Lengths are sides of triangle.
(b) Given, AB = 4 cm, BC = 7 cm and CA = 13 cm
∴ 4 – 7 < 13
= -3 < 13,
∴ 7 – 13 < 4
= – 6 < 4,
∴ 13 – 4 < 7
= 9 < 7
∴ Lengths are not side of a triangles.
(c) Given, AB = 9 cm, BC = 10 cm and CA = 20 cm
∴ 9 – 10 < 20
= – 1 < 20,
∴ 10 – 20 < 9
= – 10 < 9,
∴ 20 – 9 < 10
= 11 < 10
∴ Lengths are not side of a triangle.
Question no – (2)
Solution :
NO, Sum of any two lengths is always greeter than third angle.
Question no – (3)
Solution :
No, because they will from a straight line.
Properties of Triangles Exercise 16E Solution :
Question no – (1)
Solution :
(a) 9= 81
= 62 + 82
= 36 + 64
∴ 100 ≠ 81
∴ It is not a right angled triangle.
(b) ∴ (13)2
= 169,
∴ 122 + 52
= 144 + 25
= 169
∴ It is a right angled triangle.
Question no – (2)
Solution :
(a) x2 = 102 + 242
∴ x2 = 100 + 576
∴ x2 = 676
∴ x = √676
∴ x = 26
(b) (25)2 = x+ 72
∴ x2 = 252 – 72
∴ x2 = 625 – 49
∴ x = √676
∴ x = 24
Question no – (3)
Solution :
(a) ∴ 152
= 225
∴ 122 + 132
= 144 + 169
= 313
∴ Not satisfied.
(b) ∴ 52
= 25
∴ 32 + 42
= 9 + 16
= 25
∴ Pythagorean satisfied
Question no – (4)
Solution :
Let, the perpendicular – x
∴ (15)2 = (12)2 + x2
∴ x2 = 225 – 144
∴ x= 81
∴ x = √81
∴ x = 9
∴ Perpendicular – 9cm
Question no – (5)
Solution :
∴ PR = √(PQ)2 + (QR)2
= √64 + 225
= √289
= 17 cm
Therefore, PR will be 17 cm.
Question no – (6)
Solution :
∴ (26)2
= 676
∴ (10)2 + (24)2
= 100 + 576
= 676
∴ It is a right- angles triangle.
Question no – (7)
Solution :
∴ Let the distance x
∴ x2 = √144 + 25
∴ x2 = √169
∴ x2 = 13 cm
Therefore, the distance between initial and final position will be 13 cm.
Question no – (8)
Solution :
∴ BD = √(13)2 – (12)2
= √169 – 144
= √25
= 5
∴ DC = √(15)2 – (12)2
= √225 – 144
DC = √81
= 9
∴ BC = BD + DC
= 5 +9
= 14 cm
Therefore, the length of BC will be 14 cm.
Question no – (9)
Solution :
∴ Let the distance – x
∴ x = √ (10)2 + (24)2
∴ x = √100 + 576
∴ x = √676
∴ x = 26 m
Therefore, the distance between starting point and end point will be 26 m.
Next Chapter Solution :
Updated: July 25, 2023 — 6:11 am |
# Elementary Row Operations For Matrices
Elementary Row Operations for Matrices
A. Introduction
A matrix is a rectangular array of numbers - in other words, numbers grouped into rows and
columns. We use matrices to represent and solve systems of linear equations. For example, the
system of equations
8y + 16z = 0
*Make sure to line up all variables and
x
- 3z = 1
leave space if one is missing.
-4x + 14y + 2z = 6
can be represented by what is called an augmented matrix as seen below:
) →
Row 1 (R
0
8
16
0
1
* Place a 0 in the matrix if
) →
the coefficient of a
Row 2 (R
1
0
-3
1
2
variable is 0.
) →
Row 3 (R
-4
14
2
6
3
x
y
z
constant
Coefficients of the three unknown variables ( x, y, and z ) and the constant terms are placed in
their respective places in the matrix.
Solving a system of equations using a matrix means using row operations to get the matrix into the
form called reduced row echelon form like the example below:
1
0
0
3
* Make sure only ones are on the diagonal with
0's every other position except for the last
0
1
0
6
column.
0
0
1
2
This column can have any numbers.
B. Row Operations
We can perform elementary row operations on a matrix to solve the system of linear equations it
represents. There are three types of row operations.
1) Interchanging two rows
R
ows can be moved around by switching any two. In this case, R
and R
have been
1
2
switched.
0
8
16
0
1
0
-3
1
↔ R
1
0
-3
1
R
0
8
16
0
1
2
-4
14
2
6
-4
14
2
6
2) Multiplying a row by a nonzero constant
We can multiply any row by any number except 0. When a row is multiplied by a number,
every element in that row must be multiplied by the same number. Below, R
is multiplied by 2.
2
1
0
-3
1
1
0
-3
1
R
0
8
16
0
0
16
32
0
2 R
2
2
-4
14
2
6
-4
14
2
6 |
# How Do You Add A Percentage To A Percent?
Can I add two percentages together?
Percents can be added directly together if they are taken from the same whole, which means they have the same base amount. You would add the two percentages to find the total amount.
How do I calculate a percentage increase between two percentages?
• First: work out the difference (increase) between the two numbers you are comparing.
• Increase = New Number - Original Number.
• Then: divide the increase by the original number and multiply the answer by 100.
• % increase = Increase ÷ Original Number × 100.
The common error of adding two percentage changes at face value. After the first percentage change, the base changes, and the second percentage does not have the same base. Two percentages that have different base values cannot be directly combined by addition!
## Related Question How do you add a percentage to a percent?
### How do you multiply percentages together?
• Convert. Take the math problem and convert the percentage into a decimal.
• Substitute. Now that you've converted the percentage into a decimal, substitute the new value into the problem.
• Solve. The last step is to do the multiplication operation and multiply the two numbers together for a final answer.
• ### Can you calculate percent change of percentages?
First Step: find the difference between two percentages, in this case, it's 15% - 5% = 10%. Second: Take 10 percent, and divide by 2nd percentage: 10/5 = 2. Now multiply this number by 100: 2*100 = 200%. You're done!
### Can you multiply a percentage by a percentage?
Percentage of percentage allows you to multiply % by % and get the result. The percentage of a percentage calculator will multiply one % by a second % to get the cumulative value.
### How do I get a percentage from two numbers?
Answer: To find the percentage of a number between two numbers, divide one number with the other and then multiply the result by 100.
### What is the easiest way to calculate percentage?
Generally, the way to figure out any percentage is to multiply the number of items in question, or X, by the decimal form of the percent. To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1.
### What is percentage and its formula?
Percentage Formula
To determine the percentage, we have to divide the value by the total value and then multiply the resultant to 100. Percentage formula = (Value/Total value)×100. Example: 2/5 × 100 = 0.4 × 100 = 40 per cent.
### How do I calculate 20% of a total?
Multiply the original price by 0.2 to find the amount of a 20 percent markup, or multiply it by 1.2 to find the total price (including markup). If you have the final price (including markup) and want to know what the original price was, divide by 1.2.
### What is the percentage of 200 is 50?
50 is 25% of 200.
Posted in FAQ |
# Matlab Matrix Operations
If you ever tried to work with huge matrices, you will know how unpleasant and tedious. Here is where Matlab come to play, it makes working with Matrices easier.
With Matlab, one of the major problem for beginners is to understand how the software works and what the software need in order to help them accomplish their goal using it.
In this read, we will hand over to you some basic Matlab Matrix operation and how to use them to get what you want.
Content:
## Matlab Matrix Operations
### Write a Matrix in Matlab
we will write
`A=[1 1 -2;2 2 1;2 1 1]`
after pressing ENTER, here is how it will look in Matlab window
### Find the size of a Matrix
The size of a Matrix is its number of rows and columns. To find the size of a Matrix, use the following code
`size(A)`
Note A here is the matrix we created in the previous step.
Here is its size
Meaning, A has 3 rows and 3 columns.
Let’s try a second example.
If we type
`size(B)`
We will see the following
To add two matrices A and B, we need size(A) to be identical to size(B)
So, let’s create a new Matrix C with the same size as A
Now we can add A and C using the following code
`A+C`
### Divide Matrices element by element
To divide two Matrices element by element use the following
`A./C`
Remember both matrices need to have the same size.
### Find the inverse of a Matrix
To find the inverse of a Matrix, use the code:
`inv(A)`
### Find the determinant of a Matrix
To find the determinant of a Matrix in Matlab, use the following code
`det(A)`
### Define a Matrix with Random elements
To create a Matrix with Random element in Matlab, use
`rand(3,2)`
Where (3,2) is the size of the Matrix
### Find the diagonal of a Matrix
DIAG help access diagonals of Matrices in Matlab.
To find the main diagonal of A, we will use
`diag(A)`
to find the first upper diagonal use
`diag(A,1)`
to find the first lower diagonal use
`diag(A,-1)`
Here is how Matlab read matrix diagonals
### Compute the Transpose of a Matrix
To find the transpose of a Matrix, use the following
`A'`
or
`transpose(A)`
Here is the transpose of A
### Extract an element in a Matrix
You can individually access element of a Matrix or a whole vector.
Let’s consider the following Matrix
If I need to access the first row of the Matrix, I will use the following code
`C(1,:)`
I will use the following to access the element on the first rows-second column.
`C(1,2)`
The following will help access element of the third column
`C(:,3)`
### Multiply Matrices
To multiply A X B, A and B being two distinct matrices, A and B has to obey these conditions.
To multiply A by B in Matlab, use the code
`A*B`
### Multiply 2 Matrices element by element
To multiply a Matrices element by element, remember the size of the two matrices has to be the same.
Use the following line
`A.*B`
### Create a Matrix with all elements equal to zero
To create a Matrix with all elements equal to zero, use the following code
`G=zeros(3,4)`
where (3,4) is the size of the Matrix
### Create a Matrix with All elements equal to one
To create a Matrix with all elements equal to one, use the following code
`O=ones(4,5)`
where (4,5) is the size of the Matrix
## Summary of Matrix functions
size: Size of a matrix
det: Determinant of a square matrix
inv: Inverse of a matrix
rank: Rank of a matrix
rref: Reduced row echelon form
eig: Eigenvalues
poly: Characteristic polynomial
norm: Norm of matrix
lu: LU factorization
svd: Singular value decomposition
eye: Identity matrix
zeros: Matrix of zeros
ones: Matrix of ones
diag: extract/create diagonal of a matrix
rand: randomly generated matrix |
# Sampling Distributions
## Presentation on theme: "Sampling Distributions"— Presentation transcript:
Sampling Distributions
How likely are the possible values of a statistic?
Part 1: the Sampling Distribution of the Sample Mean
Briefly: What have we covered? We use statistical analysis to make inferences about a population. 2. Sample statistics can be used to make such inferences. 3. We also learned that probability distributions can be used to construct models of a population
Question Who recalls what a sample statistic is?
In practice, sample statistics are numerical summaries of sample data such as mean, variance, standard deviation, and binomial proportion which are used to estimate population parameters. What was the definition of a population parameter? It is a numerical summary of a population which is almost always unknown.
1. We want to develop the notion that a sample statistic is a random variable with a probability distribution. 2. Define a sampling distribution for a sample statistic. Link the sampling distribution of the sample statistic to the normal probability distribution. I remember that:
Question Before we proceed,
does anyone know what a sampling distribution is or the definition? The concept of a sampling distribution is a little difficult for some students to understand. Basically, we have a population in which we could draw many different samples from the population. Sample 1 Sample 2 population Sample 3 Sample 4
Conjecture What is the result of being able to choose different samples in which to get a sample statistic? The sample statistic itself is a random variable. Thus, the sampling distribution of a sample statistic calculated from a sample of n measurements is the probability distribution of the statistic, that is, it is the probability distribution that specifies probabilities for the possible values the statistic can take. Moreover, sampling distributions describe the variability that occurs from study to study using statistics to estimate population parameters.
Sampling Distribution of the Sample Mean,
is the probability distribution of all possible values of the random variable computed from a sample of size n from a population with mean and standard deviation . IMPORTANT: Even though we depend on sampling distribution models, we never actually get to see them. We never actually take repeated samples from the same population and make a histogram. We only imagine or simulate them.
Can we look at a simulation Wilber?
Are you confused YET? Can we look at a simulation Wilber? You will find screen shots for the simulation on the next few slides.
Screen shots 1 Sample of size 5 drawn from population.
Here are the five drawn from the population Here is their mean. Here are the means of the ten trials done. One trial of drawing a sample of size 5. Ten trials of drawing a sample of size 5.
Screen shots 2 Sample of size 30 drawn from population.
One trial of drawing a sample of size 30. Notice how the means are more clustered for the trials that contained 30 subjects in each trial verses the ten trials in which the sample size was 5. Ten trials of drawing a sample of size 30.
Screen shots 3 Sample of size 30 drawn from population. Sample of size 5 drawn from population. 10000 trials of drawing a sample of size 5. 10000 trials of drawing a sample of size 30. Notice that the sampling distribution is more squashed in for the sample sizes of 30 verses 5
Thoughts What can you conclude when we take larger sample sizes?
As we take larger sample sizes, the larger values are offset by smaller values giving us less spread in the sample means. In fact, the larger the sample size n, the more approximately normal the shape of the sample mean becomes. Why is it important for us to have a normal distribution? To be able to use previous results we have studied such as z-scores and the standard normal distribution.
Deviation in the Sampling Distribution
Does anyone know what the standard deviation is called for a sampling distribution? The sampling distribution of has a standard deviation called the standard error in this case, the standard error of the sample mean, which gives us a mechanism to understand how much variability to expect in sample statistics that occur by chance. The standard error of the sample mean is given by: Where is the population standard deviation and n is the sample size. This holds for any size sample. Now do you understand why the size of n matters? As the size of n increases, so does the denominator which makes the standard error decrease! Moreover, the sample mean is more likely to fall closer to the population mean with a larger n.
Mean and Shape of the Sampling Distribution
What about the sampling distribution mean? The sampling distribution of the sample mean will have mean: = µ Where µ is the population mean What about a population that is not normally distributed, how will that affect the sampling distribution of the mean ? This is when the Central Limit Theorem comes in.
Central Limit Theorem The Central Limit Theorem says that for a random sampling with a large size n, the sampling distribution of the sample mean is approximately normal. This result holds no matter what the shape of the distribution the samples were taken from. HOWEVER: The sampling distribution of the sample mean becomes more bell-shaped as the random sample size n increases. [Recall the example from earlier when n was 5 then 30.] The more skewed the population distribution, the larger the n must be for the shape of the sampling distribution is close to normal. Usually, the shape of the sampling distribution is usually close to normal when the sample size is at least 30.
Pause and Think Why is it important for us to be able to have a normal distribution for the sampling distribution when the population is not normally distributed? This enables us to make inferences about population means regardless of the shape of the population distribution. Let’s revisit the applet: Does the distribution to the right match any from the previous table?
Example 1 Suppose existing houses for sale average 2200 square feet in size, with a standard deviation of 250 square feet. What is the probability that a randomly selected house will have at least 2300 square feet? Strategy: Connect: Do you recall anything we have done that can help you set up this problem? We have used the standard normal distribution to find the probability that a given value is a specific amount. So we must standardize the value of 2300 square feet. Example 1
Calculate Which here is:
Here we have the value of x being greater than or equal to 2300 square feet. So we need to standardize this in order to use the standard normal distribution. We know the population mean and standard deviation, so we can find the z-value for x = 2300 square feet as we have done previously: Which here is: HOWEVER, this question is asking us to find the probability that x ≥ or: Example 1
Question What is the relevance of finding the z-value for the given value of ? Recall that the z-values tell us how many standard deviations away a value is from the mean. Here we are questioning the probability that is the lower bound for the size of a house randomly selected from a population whose mean size is with a standard deviation of Thus we need to find:
Visualize What are we trying to calculate?
Recall that this area is .5. So to find the area you desire we must subtract the area for the z-value from table 4 from .5. What are we trying to calculate? This is the area I want to find. This is the probability that z ≥. 40 By table 4 this area for z = .40 is .1554
Calculate and Summarize
Thus, by using table 4, we have that the Summarize: If a house is chosen at random from a group in which the average square footage is 2200 square feet with a standard deviation of 250 square feet, the probability that the house is greater than 2300 square feet is or 34.46%.
Key Think of the as an x value like we have dealt with previously. Then, as the sample size increases, by the Central Limit Theorem the sampling distribution that comes from becomes approximately normal. Thus we can use the z-value and normal distribution values (table 4) to find the probability that does….
Example 2 What is the probability that a randomly selected sample of 16 houses will average at least 2300 square feet? Strategy: Connect: How do we connect this problem to the previous problem? This is a similar problem BUT in this case we are asking the probability for a randomly selected sample of houses not just one house. Question: What do we know that can help to solve this problem? We know how to find the z-value of a given x, but here we are asked about the mean of one randomly selected sample of 16 houses that were chosen from the population of houses. Thus is a value that will fall within the sampling distribution of the sample mean. Thus by the Central Limit Theorem, I can find the z-value for .
Caution What is the one difference between calculating the z-value for and the z-value for a specific x? The difference is that instead of dividing by the population standard deviation, you have to divide by the standard error of which is the standard deviation divided by the square root of n. That is: = Think: What are we trying to find?
Calculate and Summarize
: Thus for = 2300 with µ = 2200 and = = : Again by table 4 we must subtract the area of associated with the z-value of 1.60, to get: Summarize: The probability that a randomly selected sample of 16 houses will average a size greater than 2300 square feet given that the population of houses average 2200 square feet with a standard deviation of 250 square feet is or 5.48%.
You Try Water taxies have a safe capacity of 3500 lbs. Given that the population of men has normally distributed weights with a mean of 172 lb and a standard deviation of 29 lb, a) If one man is randomly selected, find the probability that his weight is greater than175 lb. Solution: Connect: How do the previous examples connect to this example? Question: What do I know that will help me? Think: Visualize what area or probability I am trying to find.
Calculate and Summarize
The appropriate z-value, divide by the right quantity, i.e. Find the area for this z from table 4 which is Summarize: For a man chosen at random from the population of men with mean weight 172 lbs. and a standard deviation of 29lbs, the probability that the randomly chosen man weighs more than 172 lbs. is or 46.02%.
Second Part b) If 20 different men are randomly selected, find the probability that their mean weight is greater than 175 lb (so that their total weight exceeds the safe capacity for the water taxi of 3500 pounds). Strategy: Connect: How do the previous examples connect to this example? Question: What do I know that will help me? Think: Visualize what area or probability I am trying to find.
Calculate and Summarize
The appropriate z-value, make sure you divide by the right quantity, i.e. Area for z = .46 from table 4 Summarize: Given that the safe capacity of the water taxi is 3500 pounds, there is a fairly good chance (with probability ) that it will be overloaded with 20 randomly selected men. Also notice that it is much easier for an individual to deviate from the mean than it is for a group of 20 to deviate from the mean.
You Try IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability a random sample of 20 people have a mean IQ score greater than 110? Mozart and Einstein were hypothesized to have IQs of about 4 standard deviations above the mean of 100.
Strategy Connect: How do the previous examples connect to this example? Question: What do I know that will help me? Think: Visualize what area or probability I am trying to find. Calculate: The appropriate z-value, divide by the right quantity, i.e Summarize: Answer:
Part 2: Sampling Distribution for the Sample Proportion
. Part 2: Sampling Distribution for the Sample Proportion What is the sampling distribution for the sample proportion? Like previously for the sampling distribution for the sample mean, it is a probability distribution of the sample proportion. The sample proportion is found by measuring if an individual either has or does not have a specific characteristic, this is a binomial variable. How is the sample proportion found? We find a variable “p-hat” which is the proportion of the individuals in the sample with a specific characteristic we are interested in, x, divided by the number of individuals in the sample, n, The sample proportion estimates the population proportion p.
Simulation AGAIN: This statistic will vary depending upon the sample taken from the population. Thus, this statistic is a binomial random variable as well. Each sample will vary with the number of individuals having the characteristic. Using the simulation we would have: This proportion is set so that 50% of the population has the interested characteristic so 50% does not. We selected 5 individuals randomly at a time In this random sample 2 individuals have the characteristic. I ran 1 trial. I ran another trial of 5 randomly selected individuals and only 1 had the characteristic. You can see the sampling distribution on the bottom now has 2 entries.
Screen shots 2 Notice what is happening as we take larger sample sizes and more trials.
Screen Shots 3 I set the probability of the population to 70 % has the characteristic in this case. IN ALL CASES no matter what the is as long as the sample size is large and enough trials are done, the sampling distribution of the sample proportion becomes approximately normal!!!
Summary: As the size of the sample, n, increases, the shape of the sampling distribution of the sample proportion becomes approximately normal. The mean of the sampling distribution of the sample proportion equals the population proportion, p. That is, the mean of the sample proportions is the population proportion. The expected value of the sample proportion is equal to the population proportion. The standard deviation (standard error) of the sampling distribution of the sample proportion decreases as the sample size, n, increases. Why is it important to be normal?! So we can use the z-values and normal distribution values (table 4).
Standard Error and Mean
For the mean For the standard error Sampling distribution of the sample proportion will be approximately normal if np(1 - p) ≥ 10.
Example In a 2008 study : 85% of college students with cell phones use text messaging. 1136 college students surveyed; 84% reported that they text on their cell phone. Assume the value 0.85 given in the study is the proportion p of college students that text message; that is 0.85 is the population proportion p Compute the probability that in a sample of 1136 students, 84% or less, use text messaging.
Solution By table 4, z = -.94 has area .3264. Thus .5 - .3264 = .1736
Hence there is a 17.36% probability that 84% or less of college students use text message.
Summary of Sampling Distributions
This is the probability distribution of a sample statistic. With random sampling, the sampling distribution provides probabilities for all the possible values of the statistic. The sampling distribution provides the key for telling us how close a sample statistic falls to the corresponding unknown parameter. Its standard deviation is called the standard error. For random sampling with a large sample size n, by The Central Limit Theorem the sampling distribution of the sample mean is approximately a normal distribution. This result applies no matter what the shape of the probability distribution from which the samples are taken. In practice, the sampling distribution is usually close to normal when the sample size n is at least about 30, and for sample proportions np(1-p)≥10. If the population distribution is approximately normal, then the sampling distribution is approximately normal for all sample sizes. |
# How to Simplify a Square Root
Co-authored by wikiHow Staff
Updated: January 22, 2020
Simplifying a square root isn't as hard as it looks. To simplify a square root, you just have to factor the number and pull the roots of any perfect squares you find out of the radical sign. Once you've memorized a few common perfect squares and know how to factor a number, you'll be well on your way to simplifying the square root.
### Method 1 of 3: Simplifying a Square Root by Factoring
1. 1
Understand factoring. The goal of simplifying a square root is to rewrite it in a form that is easy to understand and to use in math problems. Factoring breaks down a large number into two or more smaller factors, for instance turning 9 into 3 x 3. Once we find these factors, we can rewrite the square root in simpler form, sometimes even turning it into a normal integer. For example, √9 = √(3x3) = 3. Follow the steps below to learn this process for more complicated square roots.[1]
2. 2
Divide by the smallest prime number possible. If the number under the square root is even, divide it by 2. If your number is odd, try dividing it by 3 instead. If neither of these gives you a whole number, move down this list, testing the other primes until you get a whole number result. You only need to test the prime numbers, since all other numbers have prime numbers as their factors. For example, you don't need to test 4, because any number divisible by 4 is also divisible by 2, which you already tried.[2]
• 2
• 3
• 5
• 7
• 11
• 13
• 17
3. 3
Rewrite the square root as a multiplication problem. Keep everything underneath the square root sign, and don't forget to include both factors. For example, if you're trying to simplify √98, follow the step above to discover that 98 ÷ 2 = 49, so 98 = 2 x 49. Rewrite the "98" in the original square root using this information: √98 = √(2 x 49).[3]
4. 4
Repeat with one of the remaining numbers. Before we can simplify the square root, we keep factoring it until we've broken it down into two identical parts. This makes sense if you think about what a square root means: the term √(2 x 2) means "the number you can multiply with itself to equal 2 x 2." Obviously, this number is 2! With this goal in mind, let's repeat the steps above for our example problem, √(2 x 49):
• 2 is already factored as low as it will go. (In other words, it's one of those prime numbers on the list above.) We'll ignore this for now and try to divide 49 instead.
• 49 can't be evenly divided by 2, or by 3, or by 5. You can test this yourself using a calculator or long division. Because these don't give us nice, whole number results, we'll ignore them and keep trying.
• 49 can be evenly divided by seven. 49 ÷ 7 = 7, so 49 = 7 x 7.
• Rewrite the problem: √(2 x 49) = √(2 x 7 x 7).
5. 5
Finish simplifying by "pulling out" an integer. Once you've broken the problem down into two identical factors, you can turn that into a regular integer outside the square root. Leave all other factors inside the square root. For example, √(2 x 7 x 7) = √(2)√(7 x 7) = √(2) x 7 = 7√(2).[4]
• Even if it's possible to keep factoring, you don't need to once you've found two identical factors. For example, √(16) = √(4 x 4) = 4. If we kept on factoring, we'd end up with the same answer but have to do more work: √(16) = √(4 x 4) = √(2 x 2 x 2 x 2) = √(2 x 2)√(2 x 2) = 2 x 2 = 4.
6. 6
Multiply integers together if there are more than one. With some large square roots, you can simplify more than once. If this happens, multiply the integers together to get your final problem. Here's an example:
• √180 = √(2 x 90)
• √180 = √(2 x 2 x 45)
• √180 = 2√45, but this can still be simplified further.
• √180 = 2√(3 x 15)
• √180 = 2√(3 x 3 x 5)
• √180 = (2)(3√5)
• √180 = 6√5
7. 7
Write "cannot be simplified" if there are no two identical factors. Some square roots are already in simplest form. If you keep factoring until every term under the square root is a prime number (listed in one of the steps above), and no two are the same, then there's nothing you can do. You might have been given a trick question! For example, let's try to simplify √70:[5]
• 70 = 35 x 2, so √70 = √(35 x 2)
• 35 = 7 x 5, so √(35 x 2) = √(7 x 5 x 2)
• All three of these numbers are prime, so they cannot be factored further. They're all different, so there's no way to "pull out" an integer. √70 cannot be simplified.
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### Method 2 of 3: Knowing the Perfect Squares
1. 1
Memorize a few perfect squares. Squaring a number, or multiplying it by itself, creates a perfect square. For example, 25 is a perfect square because 5 x 5, or 52, equals 25. Memorizing at least the first ten perfect squares can help you recognize and quickly simplify perfect square roots. Here are the first ten perfect squares:
• 12 = 1
• 22 = 4
• 32 = 9
• 42 = 16
• 52 = 25
• 62 = 36
• 72 = 49
• 82 = 64
• 92 = 81
• 102 = 100
2. 2
Find the square root of a perfect square. If you recognize a perfect square under a square root symbol, you can immediately turn it into its square root and get rid of the radical sign (√). For example, if you see the number 25 under the square root sign, you know that the answer is 5 because 25 is a perfect square. Here's the same list as above, going from the square root to the answer:
• √1 = 1
• √4 = 2
• √9 = 3
• √16 = 4
• √25 = 5
• √36 = 6
• √49 = 7
• √64 = 8
• √81 = 9
• √100 = 10
3. 3
Factor numbers into perfect squares. Use the perfect squares to your advantage when following the factor method of simplifying square roots. If you notice a way to factor out a perfect square, it can save you time and effort. Here are some tips:[6]
• √50 = √(25 x 2) = 5√2. If the last two digits of a number end in 25, 50, or 75, you can always factor out 25.
• √1700 = √(100 x 17) = 10√17. If the last two digits end in 00, you can always factor out 100.
• √72 = √(9 x 8) = 3√8. Recognizing multiples of nine is often helpful. There's a trick to it: if all digits in a number add up to nine, then nine is always a factor.
• √12 = √(4 x 3) = 2√3. There's no special trick here, but it's usually easy to check whether a small number is divisible by 4. Keep this in mind when looking for factors.
4. 4
Factor a number with more than one perfect square. If the number's factors contain more than one perfect square, move them all outside the radical symbol. If you found multiple perfect squares during your simplification process, move all of their square roots to the outside of the √ symbol and multiply them together. For example, let's simplify √72:
• √72 = √(9 x 8)
• √72 = √(9 x 4 x 2)
• √72 = √(9) x √(4) x √(2)
• √72 = 3 x 2 x √2
• √72 = 6√2
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### Method 3 of 3: Knowing the Terminology
1. 1
Know that the radical symbol (√) is the square root symbol. For example, in the problem, √25, "√" is the radical symbol.[7]
2. 2
Know that the radicand is the number inside the radical symbol. You will need to find the square root of this number. For example, in the problem √25, "25" is the radicand.[8]
3. 3
Know that the coefficient is the number outside the radical symbol. This is the number that the square root is being multiplied by; this sits to the left of the √ symbol. For example, in the problem, 7√2, "7" is the coefficient.
4. 4
Know that a factor is a number that can be evenly divided out of another number. For example, 2 is a factor of 8 because 8 ÷ 4 = 2, but 3 is not a factor of 8 because 8÷3 doesn’t result in a whole number. As another example, 5 is a factor of 25 because 5 x 5 = 25.
5. 5
Understand the meaning of simplifying a square root. Simplifying a square root just means factoring out any perfect squares from the radicand, moving them to the left of the radical symbol, and leaving the other factor inside the radical symbol. If the number is a perfect square, then the radical sign will disappear once you write down its root. For example, √98 can be simplified to 7√2.
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## Community Q&A
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Add New Question
• Question
How do you simplify square roots with variables?
wikiHow Staff Editor
Staff Answer
Use the same procedure you would for simplifying square roots with numerical radicands. Bring any identical pairs of factors out of the radical to become the coefficient. For example, if you need to simplify √25a^3, change it to √5×5×a×a×a. Factor out 5 and a^2 to get 5a^2√a.
• Question
What is the simplest radical form?
wikiHow Staff Editor
Staff Answer
If something is written in its simplest radical form, that means that you have already found all possible roots and eliminated any radicals from the denominator of a fraction. A square root can’t be simplified any further if there are no 2 identical factors remaining and every term under the radical symbol is a prime number.
• Question
How do you simplify radical fractions?
wikiHow Staff Editor
Staff Answer
Start by multiplying the numerator and denominator by a radical that will eliminate the radical in the denominator. For example, if your fraction is 2/√7, multiply by √7/√7 to get 2√7/√49. This will simplify to 2√7/7. If possible, simplify the fraction by dividing out any common factors in the numerator and denominator.
• Question
What is the square root of 18 minus the square root of 8?
Donagan
Top Answerer
√18 = √(2x3x3) = 3√2. √8 = √(2x2x2) = 2√2. Then (3√2) - (2√2) = 1√2 = √2.
• Question
What is the square root of 841?
Community Answer
The square root of 841 is 29.
• Question
How do you simplify 4sqrt2 - 2 sqrt 8 + 3 sqrt 2?
Community Answer
You have to factor 2sqrt8 further. To simplify the numbers you listed, the numbers in the root have to be equivalent so you can perform addition/subtraction. Try this: sqrt8 is 2 root 2. 2*2=4 plus the root. Your equation now is 4sqrt2- 4sqrt 2 + 3sqrt 2. 4sqrt2-4sqrt2=0. Your answer is 3sqrt 2.
• Question
How do you simplify the square root of 25?
Community Answer
25 is a whole number square root. X*X=25 What is X? 5*5=25. That means the square root of 25 is 5.
• Question
Why is 48 not a perfect square?
Donagan
Top Answerer
A "perfect" square is the square of a whole number, but 48 is the square of 6.928.
• Question
How can I write a simplest form of root 20?
Community Answer
You have to make a factor tree. Move from top to bottom like this: 20 2-10 *-5-2 * means no further splits. what number(s) are repeated? 2 is repeated twice, 5 is only written once. Since 2 is repeated twice, the answer is 2sqrt5. However, if you have an odd number of repeats of one number, go up the tree once. EX: 40 2-20 *-2-10 *-*-2-5 2 repeated 3 times, 5 repeated once. That means your answer is 2sqrt10
• Question
How do I solve √16n^2?
Donagan
Top Answerer
√(16n²) = √[(4²)(n²)] = (4)(n) = 4n.
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## Tips
• One way to find perfect squares that factor into a number is to look through the list of perfect squares, beginning with the one that is the next smallest compared to your radicand, or the number under the square root sign. For example, when looking for the perfect square that goes into 27, you might start at 25 and go down the list to 16 and stop at 9, when you found the one that divides into 27.
Thanks!
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## Warnings
• Calculators can be useful for large numbers, but the more you practice working this out on your own, the easier this will get.
Thanks!
• Simplifying is not the same as evaluating. At no point in this process should you get a number with a decimal point in it!
Thanks!
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## About This Article
Co-Authored By:
wikiHow Staff Editor
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Together, they cited information from 8 references. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article meets our high standards.
292 votes - 63%
Co-authors: 76
Updated: January 22, 2020
Views: 1,118,470
Article SummaryX
To simplify a square root, start by dividing the square root by the smallest prime number possible. For example, if you're trying to find the square root of 98, the smallest prime number possible is 2. If you divide 98 by 2, you get 49. Then, rewrite the square root as a multiplication problem under the square root sign. In this case, you'd rewrite the square root as 2 × 49 under the square root sign. From there, keep factoring the numbers until you have 2 identical factors. In this example, 49 divided by 7 is 7. Rewrite the square root as 2 × 7 × 7. Finally, once you have two identical numbers, move them outside of the square root to make them a regular integer. So the simplified square root of 98 is 7 × the square root of 2. However, if you factor the numbers inside the square root as much as you can without getting two identical numbers, then your square root can't be simplified! To learn other ways you can simplify a square root, keep reading!
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# Mathematical Functions: Do You Know What They Are?
Mathematical functions are like magical mathematical machines that help us to connect numbers in a special way. Let’s think of the functions as secret recipes that take a number and then add an interesting touch to it. Within a mathematical function there are combinations of numbers, variables and mathematical operations, such as addition, subtraction, multiplication and division, i.e. algebraic expressions.
Index
## What Is a Function?
A function is like a rule that tells us what to do with numbers. In mathematics, we write a function as f(x) or y = f(x) to show that it is related to a number that we call x. We can use another letter to represent this number. If we use the letter m the function would be written as f(m) or if we use the letter p, the function would be written as f(p).
When we put a number in the function, we get another number as a result. Let’s take a look at an example!
In this Smartick exercise, we have a machine that doubles any number that goes into the machine because it multiplies it by 2. For example, if we put the number 3 into the machine, we get 6! So, “double(3)” equals 6. This is the double function.
Can you guess what the “half” function does? The “half” function is a mathematical function that performs a very simple operation: it takes a number and divides it by 2. In other words, if you put a number x into the “half” function, you will get half of that number as a result. For example, if you put the number 8 into the “half” function, you get 4.
Now, in more formal mathematics, a function is defined as a relation between two variable magnitudes. This means that for each value of the first magnitude, which in our case we call x, there is a corresponding single value of the second magnitude, which is its image.
The variable that we consider first is the independent variable. The other variable is called the dependent variable because its value depends on the value of the first variable.
A function f between the variables x and y is expressed as follows:
y = f(x)
f(x) is the image of x, i.e., the value that corresponds to x by the function f.
## Linear Mathematical Functions
These mathematical functions have the form f(x) = mx + n where:
• m is the slope of the graph of the function. The greater the m the steeper the straight line of the graph.
• n is the origin of the ordinate, which is the value of y when x is equal to 0.
They are also called direct proportionality functions because they express a relationship in which the variables change in the same proportion, which is a key characteristic of these types of mathematical functions.
Example:
y = 2x + 3
This is a linear function where the slope is 2 and the ordinate is 3. The graph of this function is a straight line that passes through the point (0, 3) and has a slope of 2.
There are other types of mathematical functions like quadratic functions that have the form f(x) = ax²+ bx + c or constant functions that have the form f(x) = c, where c is a constant. I invite you to uncover more types such as absolute value, exponential, logarithmic and trigonometric functions among others.
### How Are Mathematical Functions Represented?
The graph of a mathematical function is the representation on the coordinate axes of all the pairs (x,y), where x is a value of the independent variable and y is the corresponding value of the dependent variable.
To represent a function, we have to follow several steps. For example, we are going to represent the “double” function which is a linear mathematical function with the slope being 2:
y = 2x
1. Draw a coordinate axis: Start by drawing a pair of perpendicular axes on a piece of paper. One axis will be the horizontal axis (called the x-axis) and the other will be the vertical axis (called the y-axis).
2. Mark the points on the axes: On the horizontal axis (x-axis), label the values you want to represent. We can mark the following: -4, -3, -2, -1, 0, 1, 2, 3 and the 4. On the vertical axis (y-axis), do the same. Remember that there must be the same distance between the points.
3. Make a table of function values: The equation y = 2x tells us how to relate the x-value to the y-value. To find points on the graph, choose several values of x and calculate the corresponding values of y:
• If x=-2 then y=2(-2)=-4. We obtain the point (-2, -4).
• If x=-1 then y=2(-1)=-2. We obtain the point (-1, -2).
• If x=0 then y=2(0)=0. We obtain the point (0, 0).
• If x=1 then y=2(1)=2. We obtain the point (1, 2).
• If x=2 then y=2(2)=4. We obtain the point (2, 4).
4. Draw the points: On the graph, place a point at each of the coordinates you found in the previous step.
5. Draw the line through the points: Now, connect the points you have drawn with a straight line. The line will pass through all the points you calculated and represent the function y = 2x.
6. Label the graph: Add a title to your y = 2x graph.
To represent mathematical functions, we can use the GeoGebra or Desmos graphing calculator. You can choose the one you like the most.
## Exercise on Linear Mathematical Functions Applied to Real Life
At Leo’s fruit stand, a kilogram of oranges costs $1.85. Answer the following questions: • Draw up a table of values relating the quantity of oranges to their price. • Are the kilograms of oranges bought and the price proportional? • Make the graph of the function obtained. ### Solution to the Proposed Exercise • Draw up the table of values that relates the quantity of oranges to their price. x y 1 1,85 2 3,7 3 5,55 4 7,4 • Are the kilograms of oranges purchased and the price proportional? • Yes, the kilograms of oranges purchased and the price are proportional in this case. The reason is that the price per kilogram of oranges is constant ($1.85 per kilogram). This means that if you double the quantity of oranges (for example, from 2 to 4 kilograms), the price will also double. This constant ratio indicates direct proportionality.
• Make the graph of the function obtained.
I hope you found the post interesting, feel free to share it or leave comments below with your questions or the topics you would like to learn more about. To find out more, sign up for Smartick, the online method for learning math for children ages 4 to 14. |
# Solving Simple Equations: Linear Algebra Introduction
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## 12 Questions
### What is the first step in solving a linear equation?
Write the equation
### After moving constants and coefficients, what should be done next when solving a linear equation?
Isolate the variable
### What should you do if the solution of a linear equation does not satisfy the original equation?
Try another method or value
### When solving multiple linear equations with two variables, what method can be used to find the values of the variables?
Gaussian elimination
### In linear equations with more than two variables, which method involves transforming the augmented matrix into row-echelon form?
Gaussian elimination method
### What fundamental skill in mathematics is applicable in various areas of life, according to the text?
Solving simple linear equations
### What is the defining characteristic of a linear equation?
Consists of constants and variables raised to powers at most one
### Which of the following is NOT an example of a linear equation?
$y = 3x^2 + 2$
### What should be done to nonlinear terms in a linear equation before solving it?
Convert them to linear terms by removing their exponents or fractions
### Which step is typically the first step in solving a linear equation?
Set the equation equal to zero
### If a linear equation has multiple nonlinear terms, what should be done before solving the equation?
Simplify or remove the nonlinear terms
### In linear equations, what must each term have to maintain linearity?
Same degree or power of the variable
## Solving Simple Equations - An Introduction
Solving simple equations is one of the fundamental concepts in mathematics. These equations represent various real-life situations, such as business transactions, physical processes, scientific experiments, and more. In this article, we will focus on solving simple linear equations, which form a significant part of algebraic equations. We will explore the steps involved in solving these equations and learn how to apply these methods to solve other types of linear equations.
## Understanding Linear Equations
Linear equations are mathematical expressions consisting of constants and variables raised to powers at most one. They can also involve addition and multiplication operations between the constants and variables, including division by a constant. Examples of linear equations include:
• y = 3x + 2
• x^2 + 7x + 6 = 0
• 2y - 3 = 5
• 4x + 2 = 9
In a linear equation, every term must have the same degree or power of the variable. This means that any nonlinear terms, such as squares or fractions, need to be removed before trying to find a solution. For instance, the equation y = 3x^2 + 2 would become y = 3x and the equation x^2 + 7x + 6 = 0 would become 2x + 3 = 0 upon simplifying the square and removing the constant term.
## Steps to Solve Linear Equations
To solve a linear equation, you typically follow these four main steps:
1. Write the equation: Start with the given linear equation and set it equal to zero.
2. Move the constants and coefficients to one side of the equals sign: Combine like terms, if necessary, to move all the constant and variable terms to one side of the equation. For example, x - 5 = 0 becomes x = 5.
3. Isolate the variable: If there is only one variable, divide both sides of the equation by its coefficient. Otherwise, solve for one variable using cross-multiplication or elimination.
4. Check your solution: Substitute the value you found in the original equation and see if it is true. If not, try another method or value.
Let's consider the equation 2x + 6 = 0 as an example.
1. Write the equation: 2x + 6 = 0.
2. Move the constants and coefficients to one side of the equals sign: 2x = -6.
3. Isolate the variable: Divide both sides of the equation by 2: x = -3.
4. Check your solution: Substitute x = -3 in the original equation: 2(-3) + 6 = 0. This holds true, which confirms that our solution is correct.
## Variation and Combination of Linear Equations
When solving multiple linear equations, you can use the methods mentioned above to find the values of the variables. If you have two linear equations with two variables, you can solve them simultaneously by using the elimination method, where you add or subtract the equations to eliminate one variable. If you have more than two equations, you can use the Gaussian elimination method, which involves transforming the augmented matrix into row-echelon form and then back-substituting to find the values of the variables.
## Conclusion
Solving simple linear equations is a fundamental skill in mathematics that is applicable in various areas of life, from business to science. By understanding the concept and following the steps outlined above, you can confidently tackle different types of linear equations and find their solutions.
Learn the fundamental concepts of solving simple linear equations, a crucial aspect of algebra. Understand how to write, manipulate, and solve linear equations step by step. Explore methods for handling single and multiple linear equations with examples.
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# Introduction to Sequences and Series
Posted by on October 16, 2018
## Arithmetic Sequences
An arithmetic sequence is a sequence in which the difference between each consecutive term is constant. An arithmetic sequence can be defined by an explicit formula in which an = d (n - 1) + c, where d is the common difference between consecutive terms, and c = a1. An arithmetic sequence can also be defined recursively by the formulas a1 = can+1 = an + d, in which d is again the common difference between consecutive terms, and c is a constant.
The sum of an infinite arithmetic sequence is either ∞, if d > 0, or - ∞, if d < 0.
There are two ways to find the sum of a finite arithmetic sequence. To use the first method, you must know the value of the first term a1 and the value of the last term an. Then, the sum of the first n terms of the arithmetic sequence is Sn = n(). To use the second method, you must know the value of the first term a1 and the common difference d. Then, the sum of the first n terms of an arithmetic sequence is Sn = na1 + (dn - d ).
2 + 5 + 8 + 11 + 14 = 40 14 + 11 + 8 + 5 + 2 = 40 16 + 16 + 16 + 16 + 16 = 80
Computation of the sum 2 + 5 + 8 + 11 + 14. When the sequence is reversed and added to itself term by term, the resulting sequence has a single repeated value in it, equal to the sum of the first and last numbers (2 + 14 = 16). Thus 16 × 5 = 80 is twice the sum.
The sum of the members of a finite arithmetic progression is called an arithmetic series. For example, consider the sum:
$\left\{\displaystyle 2+5+8+11+14\right\}$
This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2:
$\left\{\displaystyle \left\{\frac \left\{n\left(a_\left\{1\right\}+a_\left\{n\right\}\right)\right\}\left\{2\right\}\right\}\right\}$
In the case above, this gives the equation:
$\left\{\displaystyle 2+5+8+11+14=\left\{\frac \left\{5\left(2+14\right)\right\}\left\{2\right\}\right\}=\left\{\frac \left\{5\times 16\right\}\left\{2\right\}\right\}=40.\right\}$
This formula works for any real numbers
$\left\{\displaystyle a_\left\{1\right\}\right\}$
and
$\left\{\displaystyle a_\left\{n\right\}\right\}$
. For example:
$\left\{\displaystyle \left\left(-\left\{\frac \left\{3\right\}\left\{2\right\}\right\}\right\right)+\left\left(-\left\{\frac \left\{1\right\}\left\{2\right\}\right\}\right\right)+\left\{\frac \left\{1\right\}\left\{2\right\}\right\}=\left\{\frac \left\{3\left\left(-\left\{\frac \left\{3\right\}\left\{2\right\}\right\}+\left\{\frac \left\{1\right\}\left\{2\right\}\right\}\right\right)\right\}\left\{2\right\}\right\}=-\left\{\frac \left\{3\right\}\left\{2\right\}\right\}.\right\}$
Sequence – a function whose domain is a set of consecutive integers. If a domain is not specified, it is understood that the domain starts at one
finite sequence – has a last term
infinite sequence – continues without stopping
(346) |
## Precalculus (6th Edition) Blitzer
Let us consider the given inequalities \begin{align} & 3x+y\le 9 \\ & 2x+3y\ge 6 \\ & x\ge 0 \\ & y\ge 0 \end{align} Put the equals symbol in place of the inequality and rewrite the equation as given below: By finding any two solutions of the linear equation, plot the graph of the linear equation $3x+y=9$: To find the value of the x-intercept, put the value of y = 0 as given below: \begin{align} & 3x+0=9 \\ & x=\frac{9}{3} \\ & x=3 \end{align} To find the value of the y-intercept, put the value of x = 0 as given below: \begin{align} & 3\left( 0 \right)+y=9 \\ & y=9 \end{align} Plot the intercepts $\left( 3,0 \right)\text{ and }\left( 0,9 \right)$ and draw a solid line passing through these points. In the inequality there is the $\le$ symbol in which the equality is included. Now, this solid line divides the plane in three regions -- the line itself and two half-planes. Now, take the origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade: Check if the test point satisfies the inequality. \begin{align} 3x+y\le 9 & \\ 3\left( 0 \right)+0\overset{?}{\mathop{\le }}\,9 & \\ 0\le 9 & \\ \end{align} Since the test point satisfies the inequality, shade the half-plane containing that test point towards the origin. Plot the graph using the intercepts as given below: By finding any two solutions of the linear equation, plot the graph of the linear equation $2x+3y\ge 6$: To find the value of the x-intercept, put the value of y = 0 as given below: \begin{align} & 2x+3\left( 0 \right)=6 \\ & 2x=6 \\ & x=\frac{6}{2} \\ & x=3 \end{align} To find the value of the y-intercept, put the value of x = 0 as given below: \begin{align} & 2\left( 0 \right)+3y=6 \\ & 3y=6 \\ & y=\frac{6}{3} \\ & y=2 \end{align} Plot the intercepts $\left( 3,0 \right)\text{ and }\left( 0,2 \right)$ and draw a solid line passing through these points; since the inequality contains the $\le$ symbol, the equality is included. Now, this dashed line divides the plane in 3 region: the line itself, and the two half planes. Then, take the origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade: Check if the test point satisfies the inequality. \begin{align} & 2x+3y\ge 6 \\ & 0+0\overset{?}{\mathop{>}}\,9 \\ & 0<9 \\ \end{align} Since, the test point does not satisfy the inequality, shade the half-plane not containing that test point that is away from the origin. And to plot the equation $x\ge 0$, graph the line $x=0$ and shade the right part of the line; that is, shade the region for which x values will be positive, as the inequality contains the $\ge$ symbol. Similarly, to plot the inequality $y\ge 0$, graph the line $y=0$ and shade the right part above the line; that is, shade the region for which y values will be positive, as the inequality contains the $\ge$ symbol. See the final graph below. |
Learning Statistics
Holy Angel University
## Grouped Frequency Distribution of Interval Data
To organize the interval- level scores into a grouped frequency distributions, we condense the separate scores into a fewer number of categories or groups. Each group contains more than one score value, called the class interval. The class interval contains the number of score values.
Examples: Given the following scores is a psychology test, make a frequency table. Raw Scores
Step 1. Compute the range. In the given data, the highest score is 82 and the lowest score is 28. The range (H – L) is 82 – 28 = 54 Step 2. Divide the range by 10 to 15 to determine the acceptable size of the interval. Let the lowest interval begin with a number which is a multiple of the interval size. In applying this step, we should decide how large each of the interval is going to be. Generally, the most acceptable number of intervals is as few as 5 to as many as 20. In estimating the number of intervals, a trial and error fashion is suggested. If we try 2 as the size of our class interval, we reject this because it results in too many categories of intervals.
If we try 12 as the size of our class interval, we also reject this because there are only 4 categories of intervals. The number 4 is not within the acceptable range of the number of class intervals or categories of class intervals. Since we need between 5 and 20 categories of intervals in our frequency distribution, we can take 10 to 15 and divide this into the range. In this case we may divide 54 by 10,11,12,13,14, and 15. If we select 10, it results in a value of 5.4 or 5. This value could be used as our interval size.
As we may have observed, most frequency distributions have odd numbers as the size of the interval. The advantage of using old numbers is that, the midpoints of the intervals will be whole numbers.
Step 3. Organize the class interval. See to it that the lowest interval begins with a number that is a multiple of the interval size. Since the lowest score is 28 and the size of your interval is 5, the lowest interval would begin with 25 and end at 29. These are the interval limits. You take note that the upper and lower limits (the exact or real limits)
Step 4. Tally each score to the category of class interval it belongs to
Step 5. Count the tally column and summarize it under column f. Then add your frequency which is the total number of cases (N)
Step 6. Compute the midpoint (M) for each class interval and put it under column M. You can obtain the midpoint by the formula below.Step 6. Compute the midpoint (M) for each class interval and put it under column M. You can obtain the midpoint by the formula below.
Step 6. Compute the midpoint (M) for each class interval and put it under column M. Where: M = the midpoint LS = the lowest score in the class interval HS = the highest score in the class interval
Step 7. Compute cumulative distributions for “less than” and “greater than”. Then put them under column “less than” cumulative and “more than” cumulative distribution. Cumulative frequencies can be obtained by adding the frequency for any class interval or category to the total frequency for all categories above and below it.
From the given less than (CF<) and greater than (CF>) cumulative frequency distribution, we can interpret that there is only 1 student whose score is less than 30, just as there are 4 students who scored less than 40. Those who scored less than a given score is evident in the CF less than column. Also, you can interpret this as that only 1 student has a score which is greater than 79, 2 students obtained a score that is greater than 74, and so forth.
Step 8. Compute the relative frequency distribution. Where: RF = the relative frequency CF = the class frequency TF = the total frequency
(Back to the Top)
Exercise:
Below are scores in Statistics examination. Convert the following distribution of scores into a grouped frequency table and (a) determine the size of class intervals, (b) indicate the upper and lower limits of each class interval, (c) identify the midpoints of each class interval, (d) find the cumulative frequency “less than” and “greater than” for each class interval and (e) find the relative frequency for each class interval
(Back to the Top)
Presentation and Analysis
The table reveals the respondents’ length of service. From among the school heads, 5 or 20.83% had served the system from 31-40 years; likewise, 11 or 45.83 % served for 21-30 years, and another 5 0r 20.83% served from 11-20 years while 3 or 12.5% served from 1-10 years.
From the teacher-participants, Nine (9) or 13.24 % had served for 21-30 years while twenty-seven (27) or 39.71% had served from 11-20 years and the remaining thirty-two (32) or 47.05 had served from 1-10 years.
In general, school heads had a mean of 23 years of service while teachers had a mean of 12 years.
Interpretation
It could be gleaned that if compared with the length of service of the school head-participants, others could have made it in the school heads’ positions. But it should be understood that length of service or seniority does not necessarily mean an automatic qualification for headship position.
It could be said that the trend is to place qualified teachers as school heads regardless of the length of service. It could be said also that other factors other than length of service or eligibility or educational attainment are necessary to weigh the credentials of those who would like to assume higher positions than a classroom teacher.
(Back to the Top) |
### Calculations with Significant Figures
```Sig Figs meet Rounding!
In Science we take measurements, but those
measurements are sometimes needed to find the
values that we really want.
For example:
Volume (Length x Width x Height)
Volume by Difference using a graduated cylinder
(Height of Water with object – Height of water initially)
Density (Mass ÷ Volume)
Calculations cannot be more precise than the
measurements.
The least precise measurement dictates the precision
The least precise will be the measurement with the
LEAST decimal places.
When adding or subtracting – pretend that you are
back in elementary school and line those numbers up!
Example: Two different pieces of metal are massed
separately using different balances. The first piece has
a mass of 19.473 g. The second piece has a mass of 3.82
g. What is the total mass? Don’t forget to use
significant figures.
Step 1 – Line them up and do the math.
19.473
+ 3.82
23.293
Step 2 – Find the LAST significant figure. Start from
the right. Look for the first column that has a
significant figure from both measurements. Underline
the number in the answer that is in that column.
19.473
+ 3.82
23.293
Step 3 – Round to that place – do not forget to LOOK
right!
19.473
+ 3.82
23.293
Answer : 23.29 g – Don’t forget the Unit!!!
Subtraction is the SAME!!!
Example – The mass of a paper cup is 1.284 g. The
mass of the paper cup plus a sample of Manmium is
38.2 g. What is the mass of the Manmium sample?
Step 1 – What is it?
Yep – Line them up!
38.2
- 1.284
36.916
Step 2 – What is it?
Yep – Underline the last significant figure!
38.2
- 1.284
36.916
Step 3 – What is it?
Yep – Look right and ROUND!
38.2
- 1.284
36.916
489.2 g + 8.03 g = ?
0.2800 L + 4.8 L =?
285.0 kg – 3.82 kg = ?
0.01963 g + 0.290 g =?
926.028 kg – 30.02 kg = ?
3902 L + 284.5 L = ?
489.2 g + 8.03 g = 497.2 g
0.2800 L + 4.8 L = 5.1 L
285.0 kg – 3.82 kg = 281.2 kg
0.01963 g + 0.290 g = 0.310 g
926.028 kg – 30.02 kg = 896.01 kg
3,902 L + 284.5 L = 4,186 L
The least precise measurement dictates the precision
The least precise will be the measurement with the
LEAST number of significant figures.
The answer should be rounded to the least number of
significant figures
Example: Find the area of a rectangle with a length of
24.35 m and a width of 4.09 m.
Step 1 – Do the math!
Area = Length x width
24.35 m x 4.09 m = 99.5915 m2
Step 2 – Determine the number of significant figures
in the original measurements
24.35 m (4 sig figs)
4.09 m (3 sig figs)
Step 3 – Determine the least number of sig figs
3 sig figs is less than 4 sig figs
Step 4 – round the answer to the least number of sig
figs that you determined in Step 3
99.5915 m2
Final Answer – 99.6 m2
Division is the same!
Example – An object has a mass of 48.309 g and a
volume of 9.28 mL. What is the density of this object?
What is Step 1?
Yep – do the math!
Density = mass/volume
Density = 48.309 ÷ 9.28 = 5.20571121 g/mL
What is Step 2?
Yep – determine the number of significant figures in
each of the original measurements
48.309 g (5 sig figs)
9.28 mL ( 3 sig figs)
What is Step 3?
Yep – determine the least number of sig figs in the
original measurements
3 sig figs
What is Step 4?
Yep – Round to the least number of sig figs found in
Step 3
5.20571121 g/mL round to three sig figs
Final Answer = 5.21 g/mL
40.38 cm x 3.2903 cm = ?
0.00382 m x 0.08291 m = ?
293.0 g ÷ 3.0023 cm3 = ?
3.0193 x 104 kg ÷ 4.93 L = ?
293.0 cm x 3.28 cm = ?
Challenge: 3.927 cm x 12.736 cm x 4.000 cm = ?
40.38 cm x 3.2903 cm = 132.9 cm2
0.00382 m x 0.08291 m = 0.000317 m2 or 3.17 x 10-4 m2
293.0 g ÷ 3.0023 cm3 = 97.59 g/cm3
3.0193 x 104 kg ÷ 4.93 L = 6,120 kg/L or 6.12 x 103 kg/L
293.0 cm x 3.28 cm = 89.3 cm2
Challenge: 3.927 cm x 12.736 cm x 4.000 cm = 200.1 cm3
``` |
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How many 5 digit numbers are there that are the same when the order of their digits is reversed (example: 14341)?
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Hint: By using the example given in the question, think about the number in general sense. Think about how the digits should be placed so that the number does not change after reversing its order. Then think about the possibilities of choosing any number in place of each digit.
We have to find the number of 5 digit numbers that are the same when the order of their digits is reversed.
Let us say that the five-digit number will look like the following.
$abcba$
This number must look like above because the first digit must be the same as that of the unit place digit. The second digit must be the same as that of the ten’s place digit. Then only, they would have the same position after reversing their order.
Since, the third digit, $c$ , is in the middle. Its position will not change by changing the order. Therefore, we do not need to consider any possibilities for it.
Now, let us consider the possibilities for each digit from the left hand side.
Digit one, i.e. $a$ cannot be zero. Therefore, there are nine possibilities to allocate a digit to $a$ .
Since, other digits can be zero, there are 10 possibilities to allocate a digit to both $b$ and $c$ , respectively.
Fourth and fifth digit, i.e. $b$ and $a$ will be the same as that of the first and the second digit. Therefore, there is only one possibility of allocating any digits to them.
Hence, the total possibilities of choosing a five-digit number such that it will not change after reversing its order will be the product of all the possibilities of selecting all the digits. i.e.
$\Rightarrow 9 \times 10 \times 10 \times 1 \times 1 = 900$
Therefore, number of possibilities of selecting a five digit number such that the number will not change after reversing its order is $900$
So, the correct answer is “ $900$ ”.
Note: Key point in this question is to understand that $a$ cannot be zero and hence it would have only nine possibilities. Because if $a = 0$ , then the given number will be a four digit number. Also, you need to understand that there is only one possibility to choose the units place and ten’s place digit as they must be the same as that of the first and the second number from the left. |
Working rule to Divide a Polynomial by Another Polynomial: Step 1: First arrange the term of dividend and the divisor in the decreasing order of their degrees. Step 3: To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor. Step 4: Continue this process till the degree of remainder is less than the degree of divisor. Hence, all its zeroes are $$\sqrt{\frac{5}{3}}$$, $$-\sqrt{\frac{5}{3}}$$, –1, –1. Since its proof is very similar to the corresponding proof for integers, it is worthwhile to review Theorem 2.9 at this point. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). x − 1. (For some of the following, it is su cient to choose a ring of constants; but in order for the Division Algorithm for Polynomials … Example 5: Obtain all the zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are $$\sqrt{\frac{5}{3}}$$ and $$-\sqrt{\frac{5}{3}}$$. Start New Online test. Polynomials are represented as hash-maps of monomials with tuples of exponents as keys and their corresponding coefficients as values: e.g. • Solved Examples based on Division Algorithm for Polynomials So, 3x4 + 6x3 – 2x2 – 10x – 5 = (3x2 – 5) (x2 + 2x + 1) + 0 Quotient = x2 + 2x + 1 = (x + 1)2 Zeroes of (x + 1)2 are –1, –1. If and are polynomials in, with 1, there exist unique polynomials … Dividend = Divisor × Quotient + Remainder . Example 6: On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Now, we apply the division algorithm to the given polynomial and 3x2 – 5. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. The division algorithm for polynomials has several important consequences. It can be done easily by hand, because it separates an otherwise complex division problem into smaller ones. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. Performance & security by Cloudflare, Please complete the security check to access. The Euclidean algorithm can be proven to work in vast generality. Please enable Cookies and reload the page. For example, if we were to divide $2{x}^{3}-3{x}^{2}+4x+5$ by $x+2$ using the long division algorithm, it would look like this: We have found What are Addition and Multiplication Theorems on Probability? Zeros of a Quadratic Polynomial. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. In the following, we have broken down the division process into a number of steps: Step-1 Cloudflare Ray ID: 60064a20a968d433 We rst prove the existence of the polynomials q and r. The Division Algorithm in F[x] Let F be a eld and f;g 2F[x] with g 6= 0F. This will allow us to divide by any nonzero scalar. Since two zeroes are $$\sqrt{\frac{5}{3}}$$ and $$-\sqrt{\frac{5}{3}}$$ x = $$\sqrt{\frac{5}{3}}$$, x = $$-\sqrt{\frac{5}{3}}$$ $$\Rightarrow \left( \text{x}-\sqrt{\frac{5}{3}} \right)\left( \text{x +}\sqrt{\frac{5}{3}} \right)={{\text{x}}^{2}}-\frac{5}{3}$$ Or 3x2 – 5 is a factor of the given polynomial. You may need to download version 2.0 now from the Chrome Web Store. The Division Algorithm states that, given a polynomial dividend $$f(x)$$ and a non-zero polynomial divisor $$d(x)$$ where the degree of $$d(x)$$ is less than or equal to the degree of $$f(x)$$, there exist unique polynomials $$q(x)$$ and $$r(x)$$ such that (For some of the following, it is sufficient to choose a ring of constants; but in order for the Division Algorithm for Polynomials … Consider dividing x 2 + 2 x + 6 x^2+2x+6 x 2 + 2 x + 6 by x − 1. x-1. The Division Algorithm for Polynomials over a Field. Sol. The Extended Euclidean Algorithm for Polynomials The Polynomial Euclidean Algorithm computes the greatest common divisor of two polynomials by performing repeated divisions with remainder. Step 2: To obtain the first term of quotient divide the highest degree term of the dividend by the highest degree term of the divisor. If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x). 2xy + 3x + 5y + 7 is represented as {[1 1] 2, [1 0] 3, [0 1] 5, [0 0] 7}. Division Algorithm for Polynomials. Theorem 17.6. i.e When a polynomial divided by another polynomial Dividend = Divisor x Quotient + Remainder, when remainder is zero or polynomial of degree less than that of divisor Table of Contents. Division of polynomials Just like we can divide integers to get a quotient and remainder, we can also divide polynomials over a field. Dividing two numbersQuotient Divisor Dividend Remainder Which can be rewritten as a sum like this: Division Algorithm is Dividend = Divisor × Quotient + Remainder Quotient Divisor Dividend Remainder Dividing two Polynomials Let’s divide 3x2 + x − 1 by 1 + x We can write Dividend = Divisor × Quotient + Remainder 3x2 + x – 1 = (x + 1) (3x – 2) + 1 What if…We don’t divide? 1. In algebra, an algorithm for dividing a polynomial by another polynomial of the same or lower degree is called polynomial long division. With steps shown coefficient then compare the next least degree ’ s coefficient and proceed with division. * x gives you temporary access to the digits ( and place )! In general, you can skip the multiplication sign, so 5x is equivalent . Trapezoidal rule and Simpson ’ s rule in Numerical Integration exponents as keys and their corresponding coefficients values! Introduce division algorithms for multi- the Euclidean algorithm for polynomials: Let be a field getting page! Do you find the Minimum and Maximum values of a Function security Check to access is... Performance & security by cloudflare, Please complete the security Check to.. Complete the security Check to access gives you temporary access to the corresponding proof for integers it! Part here is that you can use the fact that naturals are well ordered by looking at degree... 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Addition of Real Numbers
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# Addition of Real Numbers - PowerPoint PPT Presentation
Addition of Real Numbers. Section 1.6 (42). Objectives (41). Add real numbers using the number line Add fractions Identify opposites (additive inverses) Add using absolute values Note: We don’t cover Using Calculators. 1.6.1 Add Real Numbers Using a Number Line (41).
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### Addition of Real Numbers
Section 1.6
(42)
Objectives(41)
Add real numbers using the number line
Identify opposites (additive inverses)
Add using absolute values
Note: We don’t cover Using Calculators
1.6.1 Add Real Numbers Using a Number Line(41)
We first locate the first point.
If we add a positive number, we move over that number of units to the right.
If we add a negative number, we move over that number of units to the left.
Example: -2 + ( +3 )
-4 -3 -2 -1 0 1 2 3 4
o
- - - 3 - -x
1
Examples
+3 + (-5)
-4 -3 -2 -1 0 1 2 3 4
o
x - - - - 5 - - - - -
-2
-2 + (+2)
-4 -3 -2 -1 0 1 2 3 4
o
- - - - x
0
Example
A miner descents 120 feet down a mine shaft. Later, he descends another 145 feet. Find the depth of the descent.
Given: Initial 120 feet down (-120)
Later 145 feet further down (-145)
Find: Total distance down
How: Add initial and distance down
Solve: -120 + ( -145 )
If we add two negative numbers, we add the absolute values and put a minus in front.
-120
+ -145
-265
The miner would have descended 265 feet (-265)
When you add fractions, you MUST have the same denominators.
You have to find the LCD and create equivalent fractions with that denominator.
Then just add the numerators.
Example:
1/2 + 2/3 LCD = 6
1/2 (3/3) + 2/3 (2/2)
3/6 + 4/6 = (3 + 4)/6 = 7/6 = 1 1/6
Examples
5/8 + (-6/7) LCD 56
5/8 (7/7) + (-6/7) (8/8)
35/56 + (-48/56)
[ +35 + (-48) ] / 56
-13/56
1/8 + ( -7/12 ) LCD 24
1/8 (3/3) + (-7/12) (2/2)
3/24 + (-14/24)
[ +3 + (-14)] / 24
-11/24
1.6.3 Identify Opposites(44)
The purpose for opposites is to determine which value to add to make the sum equal 0.
They are better known as additive inverses.
The additive inverse is the same value with the opposite sign.
Examples:
the opposite of 3.5 is -3.5
the opposite of -3/4 is +3/4
the opposite of 0 is 0
1.6.4 Add Using Absolute Values(44)
Same sign:
To add two numbers with the same sign (either both positive or both negative), add their absolute values and use the same sign.
Examples:
( +10 ) + ( +5 ) = +15
( -9 ) + ( +5 ) = -4
Using Absolute Values (cont)(44)
Different signs.
To add two numbers with different signs (one positive, the other negative), subtract the absolute value of the smaller from the absolute value of the larger and use the sign of the larger.
Examples:
( -9 ) + ( +4 ) = -5
( -2 ) + ( +8 ) = +6
Examples
-1/4 + ( +3/4)
-2/4 = -1/2
5/6 + ( -1/9 ) LCD 18
5/6 ( 3/3) + ( -1/9 ) ( 2/2)
15/18 + ( -2/18 ) = +13/18
-2/5 + ( +1/4 ) LCD 20
-2/5 ( 4/4) + ( +1/4 ) ( 5/5 )
-8/20 + ( + 5/20 ) = -3/20
Example
A bakery had a profit of \$450,567 for the first five months of the year, and a loss of \$52,987 for the remainder of the year. Find the net-profit or loss for the year.
Given: \$450,567 profit Jan – May
\$52,987 loss Jun – Dec
Find: Net-profit or loss for the year
How: Add the values from each set
Solve: 450,567 + ( -52,987 )
450,567 Since they are different signs,
-52,987 subtract absolute values and
397,580 use the sign of the larger
Solution: There was a net-profit of \$397,580 for the year.
Objectives(41)
Add real numbers using the number line |
Games
Problems
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9th grader Akhil from Kerala asks the following: "The sides of a quadrilateral are extended to form exterior angles. Find the sum of these exterior angles."
Hi Akhil, instead of just answering the question you asked, I'd like to explore the question in a little more depth, and explain why the answer is what it is. After all, if I just told you the answer, that would make for a pretty boring blog post!
So let's start off with a quadrilateral. I'm going to show you a picture of one below, and ask a question about it. How many degrees are there in the measures of all the interior angles of the quadrilateral?
Now, you might already know the answer to that question; the sum of all the interior angles is 360 degrees. But can you prove that to be true? It's actually not hard to understand why that quadrilateral (and all quadrilaterals, for that matter) have an interior angle sum of 360 degrees. To see why, just look at this picture, in which I've divided the quadrilateral into triangles:
How many triangles did it take to divide up that quadrilateral? It took two, right? And how many degrees does each of those triangles have? 180 degrees. And so altogether, how many degrees do they have in total? 180 x 2 = 360 degrees.
Understanding that the total of the interior angles is 360 degrees is important to finding the number of degrees in the exterior angles. Here's a picture that shows what you described - the sides being extended to form exterior angles:
By extending these sides, we've created linear pairs - pairs of angles that form a straight line. Each linear pair includes an interior angle and an exterior angle. Each linear pair adds to 180 degrees (a straight line), and there are four linear pairs (because there are four sides).
So if there are 4 linear pairs, and each linear pair is 180 degrees, that's a total of 720 degrees. But 360 degrees of that is interior angles, so the exterior angles have to add up to 720 - 360 = 360 degrees! And that's your answer.
Now, it might be tempting to think that the interior angles of a polygon and the exterior angles of a polygon will always match, since they do in the case of a quadrilateral, but that is not the case. Let's take a look at a pentagon to see why. Here I'll give you a pentagon divided into triangles, and with sides extended:
In the case of a pentagon, we can divide the polygon into 3 triangles, which means the interior angles add to 3 x 180 = 540 degrees.
We've created 5 linear pairs, which total 5 x 180 = 900 degrees.
Since there are 900 degrees total, and the interior angles add to 540 degrees, the sum of the exterior angles is 900 - 540 = 360 degrees!
It turns out that the sum of the exterior angles is 360 degrees regardless of whether it's a quadrilateral or a pentagon.
What do you think...can you generalize it even further? How many degrees are there in the interior of an n-gon? And how many degrees in the exterior angles of an n-gon? I'll leave it to you to figure that out. Thanks for asking, Akhil!
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1. ## HELP with INEQUALITIES
I'm having issues figuring out an equation to figure this out...any help would be greatly appreciated!
Applications The perimeter of a rectangle is to be between 170 and 230 inches. Find the range of values for its length when its width is 45 inches.
< length <
Applications The perimeter of a square is to be from 45 meters to 90 meters. Find the range of values for its area.
< area <
2. Hello, epetrik!
The perimeter of a rectangle is to be between 170 and 230 inches.
Find the range of values for its length when its width is 45 inches.
We know that: . $P \:=\:2L + 2W$
Since $W = 45$, we have: . $P \:=\:2L + 90$
The perimeter is between 170 and 230 inches.
. . $\begin{array}{cccccc}
\text{So we have:} & 170 & < & 2L + 90 & <& 230 \\ \\[-3mm]
\text{Subtract 90:} & 80 &<& 2L &<& 140 \\ \\[-3mm]
\text{Divide by 2:} & 40 &<& L &<& 70 \end{array}$
Therefore, the length $L$ must be between 40 and 70 inches.
The perimeter of a square is to be from 45 meters to 90 meters.
Find the range of values for its area.
Let $x$ = side of the square.
The perimeter of a square is: . $4x$
The perimeter is between 45 and 90: . $45 \;<\;4x \;<\;90$
Divide by 4: . $\frac{45}{4} \;<\:x\;<\;\frac{45}{2}$
The length of the side is between 11¼ and 22½ meters.
The area of the square is: . $x^2$
So we have: . $\left(\frac{45}{4}\right)^2 \;<\;x^2 \;<\:\left(\frac{45}{2}\right)^2$
Therefore: . $\frac{2025}{16} \;<\;\text{Area} \;<\;\frac{2025}{4}$
The area is between: . $126\frac{9}{16}\,\text{ and }\,506\frac{1}{4}\text{ square meters.}$ |
5.12 Fraction patterns
Lesson
To be able to identify patterns with fractions, it will help us if we can add and subtract fractions . Let's try this problem to review.
Examples
Example 1
What is \dfrac{1}{12} + \dfrac{1}{12}?
Worked Solution
Create a strategy
Use area models.
Apply the idea
\dfrac{1}{12} means means one twelfth or 1 part out of 12 parts should be shaded in an area model.
Here is the area model of \dfrac{1}{12} + \dfrac{1}{12}.
We can see that adding the two shaded parts means we get 2 shaded parts out of 12 parts.
This can be written as the fraction \dfrac{2}{12}.
\dfrac{1}{12}+\dfrac{1}{12}= \dfrac{2}{12}
Idea summary
We can add or subtract fractions by using area models.
Patterns with tenths and hundredths
Just like we did with whole numbers, we can explore, continue and create patterns with fractions. Let's start by looking at tenths and hundredths.
Examples
Example 2
Create a pattern by adding \dfrac{1}{10} each time.
\dfrac{3}{10}, \,\dfrac{4}{10}, \,⬚, \, ⬚, \,⬚
Worked Solution
Create a strategy
Add \dfrac{1}{10} to the last given value to complete the pattern.
Apply the idea
This picture shows \dfrac{4}{10} blocks shaded blue. If another 1 block were shaded, 5 out of 10 would be shaded in total. So the next number is \dfrac{5}{10}.
\dfrac{4}{10} + \dfrac{1}{10} = \dfrac{5}{10}
We can see that the numerator increased by 1 and the denominator stayed the same. Similarly doing this for the following numbers we can complete the pattern:\dfrac{3}{10}, \,\dfrac{4}{10}, \,\dfrac{5}{10}, \, \dfrac{6}{10}, \,\dfrac{7}{10}
Idea summary
We can create patterns with fractions by adding or subtracting the same fraction each time.
Patterns with area models
We can also use the area models to represent fraction patterns, this video shows us how.
Examples
Example 3
Which fraction comes next in the pattern?
A
B
C
D
Worked Solution
Create a strategy
Notice how the number of shaded parts changes each time.
Apply the idea
In the first shape, 3 out of 3 parts are shaded.
In the second shape, 2 out of 3 parts are shaded.
In the third shape, 1 out of 3 parts are shaded.
The number of shaded parts are decreasing by 1 each time. So the next shape should have \\ 1-1=0 parts shaded.
So option C is the answer.
Idea summary
Area models can be used to represent fractions in a pattern.
Create and continue fraction patterns
This video shows an example of what to look for when trying to determine the pattern in a sequence of fractions.
Examples
Example 4
Complete the pattern:
\dfrac{1}{7},\,\dfrac{2}{7},\,\dfrac{3}{7},\,⬚,\,⬚,\,⬚
Worked Solution
Create a strategy
Find how much the numbers are increasing by each time and add that to the last given value to complete the pattern.
Apply the idea
To work out how much the numbers are increasing by each time, you can count up or find the difference between two fractions.
So we need to add \dfrac{1}{7} to each number to complete the pattern.
The complete pattern is \dfrac{1}{7},\,\dfrac{2}{7},\,\dfrac{3}{7},\,\dfrac{4}{7},\,\dfrac{5}{7},\,\dfrac{6}{7}
Idea summary
So to complete a pattern with fractions, we need to:
• Find the pattern - which will either be given or which will need to be found by looking at the list of numbers.
• Continue the pattern - using the pattern that you found.
Outcomes
MA3-7NA
compares, orders and calculates with fractions, decimals and percentages
MA3-8NA
analyses and creates geometric and number patterns, constructs and completes number sentences, and locates points on the Cartesian plane |
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## Related Articles
• RD Sharma Class 8 Solutions for Maths
# Class 8 RD Sharma Solutions – Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) – Exercise 22.1 | Set 2
### Question 11: Find the cost of plastering the inner surface of a well at Rs. 9.50 m2. If it is 21 m deep and diameter of its top is 6 m.
Solution:
The details given about well are –
Height of well = 21 m
Diameter of well = 6 m
So radius of well = 6/2 = 3 m
Curved surface area of cylinder = 2 * (22 /7) * r * h
= 2 * (22/7) * 3 * 21
= 396 m2
Cost of plastering the inner surface of a well = 396 * 9.50
= Rs 3762
### Question 12: A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of tin – plating it onthe inside at the rate of 50 paise per hundred square centimeters.
Solution:
The details given about cylinder are –
Diameter of cylinder = 20 cm
So, radius = 20/2 = 10 cm
Height of cylinder = 14 cm
Total surface area of cylinder = 2 * (22/7) * r * h + (22/7) * r2
= 2 * (22/7) * 10 * 14 + (22/7) * 14 * 14
= 880 + 2200/7
= ((880 * 7) + 2200)/7
= (6160 + 2200)/7
= 8360/7 cm2
It is given that code per 100 cm2 = 50 paise
So, cost per 1 cm2 = Rs 0.005
Cost of tin-plating the area inside the vessel = (8360/7) * 0.005
= Rs 5.97
### Question 13: The inner diameter of circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curvedsurface at Rs 4 square meter.
Solution:
The details given about well are –
Inner diameter of circular well = 3.5 m
Height of well = 10 m
Curved surface area of well = 2 * (22/7) * r * h
= 2 * (22/7) * (3.5/2) * 10
= 110 m2
Cost of painting 1 m2 = Rs. 4
Cost of painting 110 m2 = 4 * 110
= Rs 440
### Question 14: The diameter of roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions moving once overto level a playground. What is the area of the playground?
Solution:
The details given about roller are –
Diameter of roller = 84 cm
So radius = 84/2 = 42 cm
Length of roller = 120 cm
Curved surface area of roller = 2 * (22/7) * r * h
= 2 * (22/7) * 42 * 120
= 31680 cm2
Area of playground = Number of revolution * Curved surface area of roller
= 500 * 31680
= 15840000 cm2
= 1584 m2 (1 m = 100 cm)
### Question 15: Twenty-one cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50m high and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2. 50 per square meter?
Solution:
The details given about pillar are –
Diameter of pillar = 0.50 m
So radius = 0.50/2 = 0.25 m
Height of roller = 4 m
Curved surface area of pillar = 2 * (22/7) * r * h
= 2 * (22/7) * 0.25 * 4
= 44/7 m2
Curved surface area of pillar = (21 * 44)/7
= 132 m2
Cost of cleaning the pillars = 2.50 * 132
= Rs 330
### Question 16: The total surface area of a hollow cylinder which is open from both sides if 4620 sq cm, area of base ring is 115.5 sq cm and height 7 cm. Find the thickness of the cylinder.
Solution:
The details given about cylinder are –
Total surface area of hollow cylinder = 4620 cm2
Area of base ring = 115.5 cm2
Height of cylinder = 7 cm
Area of hollow cylinder = 2 * (22/7) * (R2 – r2) + 2 * (22/7) * R * h + 2 * (22/7) * r * h
= 2 * (22/7) * (R + r) (R – r) + 2 * (22/7) * r * h (R + r)
= 2 * (22/7) * (R + r) * (h + R – r)
Area of base = (22/7) * R * R – (22/7) * r * r
= (22/7) (R + r) * (R – r)
Total Surface area/Area of base = 4620/115.5
(2 * (22/7) (R + r) * (h + R – r))/((22/7) (R + r) (R – r)) = 4620/115.5
2 * (h + R – r)/(R – r) = 4620/115.5
R – r = thickness of cylinder
Let thickness of cylinder = t
2 * (h + t)/t = 4620/115.5
2 * h + 2 * t = 40 * t
2h = 40t – 2t
2 * 7 = 38t
14/38 = t
t = 7/19 cm
### Question 17: The sum of the radius of the base and height of a solid cylinder is 37 cm. If the total surface area of the solidcylinder is 1628 cm2, find the circumference of its base.
Solution:
The details about the cylinder are –
Sum of radius of base and height of cylinder = 37 cm
Total surface area pf cylinder = 1628 cm2
Total surface area of cylinder = 2 * (22/7) * r * (h + r)
1628 = 2 * (22/7) * r * 37
(1628 * 7)/(2 * 22 * 37) = r
11396/1628 = r
r = 7 m
Circumference of base = 2 * (22/7) * r
= 44 m
### Question 18: Find the ratio between the total surface area of a cylinder to its curved surface area, given that its heightand radius are 7.5 cm and 3.5 cm.
Solution:
The details given about the cylinder are –
Height of cylinder = 7.5 cm
Radius of cylinder = 3.5 cm
Total surface area of cylinder/Curved surface area of cylinder = (2 * (22/7) * r * (h + r))/(2 * (22/7) * r * h)
= (h + r)/h
= (7.5 + 3.5)/7.5
= 11/7.5
= 22/15
### Question 19: A cylindrical vessel, without lid, has to be tin – coated on its both sides. If the radius of the base is 70 cm andits height is 1.4 m, calculate the cost of tin coating at the rate of Rs 3.50 per 1000 cm2.
Solution:
The details given about cylinder are –
Radius of base = 70 cm
Height of cylinder = 1.4 m = 140 cm (1 m = 100 cm)
Total surface area of vessel = Area of outer side of base + Area of inner and outer curved surface
= 2 ((22/7) * r * r + 2 * (22/7) * r * h)
= 2 * (22/7) * r * (r + 2 * h)
= 2 * (22/7) * 70 * (70 + 280)
= 2 * (22/7) * 70 * 350
= 154000 cm2
Cost of tin coating at the rate of Rs 3.50 per 1000 cm2 = (3.50 * 154000)/1000
= Rs 539
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Section 2.3 #35: The values of $\sin(\theta)$ and $\cos(\theta)$ are given. Find the value of the four remaining trigonometric functions: $\sin(\theta)=-\dfrac{3}{5}$ and $\cos(\theta)=\dfrac{4}{5}$.
Solution: Calculate $$\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{-\frac{3}{5}}{\frac{4}{5}}=\left( -\dfrac{3}{5} \right) \left( \dfrac{5}{4} \right) = -\dfrac{3}{4} ,$$ $$\cot(\theta) = \dfrac{1}{\tan(\theta)} = \dfrac{\cos(\theta)}{\sin(\theta)} = \dfrac{\frac{4}{5}}{-\frac{3}{5}} = \left( \dfrac{4}{5} \right) \left( - \dfrac{5}{3} \right) = -\dfrac{4}{3},$$ $$\csc(\theta) = \dfrac{1}{\sin(\theta)} = \dfrac{1}{-\frac{3}{5}} = -\dfrac{5}{3},$$ and $$\sec(\theta) = \dfrac{1}{\cos(\theta)} = \dfrac{1}{\frac{4}{5}} = \dfrac{5}{4}.$$
Section 2.3 #44: Find the value of the remaining five trigonometric functions given that $\cos(\theta) = \dfrac{3}{5}$ and $\theta$ is in quadrant IV.
Solution: We will first find the value of $\sin(\theta)$ using the Pythagorean identity for $\sin$ and $\cos$, $\sin^2(\theta)+\cos^2(\theta)=1$. Plugging into this the known value $\cos(\theta)=\dfrac{3}{5}$ yields the equation $$\sin^2(\theta) + \left(\dfrac{3}{5} \right)^2 = 1.$$ Isolate $\sin^2(\theta)$ to get $$\sin^2(\theta) = 1 - \dfrac{9}{25} = \dfrac{16}{25}.$$ Take the square root of each side of the equation (recall when doing this you must consider both the positive and negative square root): $$\sin(\theta) = \pm \sqrt{\dfrac{16}{25}} = \pm \dfrac{4}{5}.$$ But we are told in the statement of the problem that $\theta$ is in quadrant IV and so $\sin(\theta)$ must be negative. Hence we take the value $$\sin(\theta) = -\dfrac{4}{5}.$$ From this we may compute $$\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)}=\dfrac{-\frac{4}{5}}{\frac{3}{5}}=\left( - \dfrac{4}{5} \right) \left( \dfrac{5}{3} \right) = -\dfrac{4}{3},$$ $$\cot(\theta) = \dfrac{1}{\tan(\theta)} = \dfrac{\cos(\theta)}{\sin(\theta)} =\dfrac{\frac{3}{5}}{-\frac{4}{5}} = \left( \dfrac{3}{5} \right) \left( - \dfrac{5}{4} \right) = -\dfrac{3}{4},$$ $$\csc(\theta) = \dfrac{1}{\sin(\theta)} = \dfrac{1}{-\frac{4}{5}} = -\dfrac{5}{4},$$ and $$\sec(\theta) = \dfrac{1}{\cos(\theta)} = \dfrac{1}{\frac{3}{5}} = \dfrac{5}{3}.$$
Section 2.3 #56: Find the value of the remaining five trigonometric functions given that $\cot(\theta) = \dfrac{4}{3}$ and $\cos(\theta)<0$.
Solution: We use the Pythagorean identity for cotangent and cosecant: $\cot^2(\theta) + 1 = \csc^2(\theta)$. Substitue the value we know for $\cot(\theta)$ to get the equation $$\left( \dfrac{4}{3} \right)^2 + 1 = \csc^2(\theta).$$ Simplify the left-hand side to get $$\dfrac{25}{9} = \csc^2(\theta),$$ so take the square root of both sides (be careful to include the $\pm$!) to get $$\csc(\theta) = \pm \dfrac{5}{3}.$$ How do we determine which to take? We were told that $\cos(\theta)<0$ and so $\theta$ must lie in quadrants II or III. But we also know that $\cot(\theta)=\dfrac{4}{3}$, and so since cotangent is positive only in quadrants I and III, we must deduce that $\theta$ lies in quadrant III. Since sine is negative in quadrant III and $\csc(\theta) = \dfrac{1}{\sin(\theta)}$, we must take the value of $\csc(\theta)$ to be $-\dfrac{5}{3}$. Therefore $$\sin(\theta) = \dfrac{1}{\csc(\theta)} = -\dfrac{3}{5},$$ $$\tan(\theta) = \dfrac{1}{\cot(\theta)} = \dfrac{3}{4}.$$ To find $\cos(\theta)$ and $\sec(\theta)$ note that we know $\sin(\theta)$ and $\cos(\theta)$ and since $\cot(\theta) =\dfrac{\cos(\theta)}{\sin(\theta)}$ we may multiply both sides of that equation by $\sin(\theta)$ to get the formula $\cot(\theta) \sin(\theta) = \cos(\theta)$. Therefore we see $$\cos(\theta) = \cot(\theta) \sin(\theta) = \dfrac{4}{3} \cdot \left( - \dfrac{3}{5} \right) = -\dfrac{12}{15}$$ and $$\sec(\theta) = \dfrac{1}{\cos(\theta)} = -\dfrac{15}{12}.$$
Section 2.3 #72: Use the even-odd properties to find the exact value of $\sin \left( -\dfrac{3\pi}{2} \right)$.
Solution: Recall that sine is an odd function and so $\sin(-x)=-\sin(x)$. We know from the unit circle that $\sin \left( \dfrac{3\pi}{2} \right)=-1$. Therefore we may compute $$\sin \left( - \dfrac{3\pi}{2} \right) = -\sin \left( \dfrac{3\pi}{2} \right) = -(-1) = 1.$$
Section 2.3 #113: If $f(\theta)=\sin(\theta)$ and $f(a)=\dfrac{1}{3}$, find the exact value of
a.) $f(-a)$, and
b.) $f(a)+f(a+2\pi)+f(a+4\pi)$.
Solution: For part a.), since sine is an odd function, we may compute $$f(-a)=-f(a)=-\dfrac{1}{3}.$$ For part b.), since sine is periodic with period $2\pi$, we may compute $$f(a)+f(a+2\pi)+f(a+4\pi) = f(a)+f(a)+f(a+2\pi) = 2f(a)+f(a) = 3f(a) = 1.$$ |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
The simplified form of a complex rational expression $\frac{\frac{a}{a-b}}{\frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}}$ is $\frac{a+b}{a}$.
$\frac{\frac{a}{a-b}}{\frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}}$ The numerator and denominator of the complex rational expression are single. Divide the numerator by the denominator, $\frac{\frac{a}{a-b}}{\frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}}=\frac{a}{a-b}\div \frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}$ \begin{align} & \frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C} \\ & =\frac{AD}{BC} \end{align} where $\frac{A}{B},\left( B\ne 0 \right)$, $\frac{C}{D},\left( D\ne 0 \right)$ are rational expressions with $\frac{C}{D}\ne 0$ The reciprocal of $\frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}$ is $\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}$ So, multiply the reciprocal of the divisor, $\frac{\frac{a}{a-b}}{\frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}}=\frac{a}{a-b}\cdot \frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}$ Simplify the terms, \begin{align} & \frac{\frac{a}{a-b}}{\frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}}=\frac{a\cdot \left( {{a}^{2}}-{{b}^{2}} \right)}{\left( a-b \right)\cdot \left( {{a}^{2}} \right)} \\ & =\frac{a}{{{a}^{2}}}\cdot \frac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( a-b \right)} \end{align} Take the factor for the numerator term, \begin{align} & \frac{\frac{a}{a-b}}{\frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}}=\frac{a}{{{a}^{2}}}\cdot \frac{\left( a+b \right)\left( a-b \right)}{\left( a-b \right)} \\ & =\frac{1}{a}\cdot \left( a+b \right) \\ & =\frac{a+b}{a} \end{align} |
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The polar coordinates are defined using the distance, r, and the angle, θ. On the other hand, rectangular coordinates, also known as Cartesian coordinates, are defined by x and by y. We can find equations that relate these coordinates using a right triangle and the trigonometric functions sine and cosine.
Here, we will learn about the formulas that we can use to transform from polar to rectangular coordinates. Then, we will apply these formulas by solving some practice problems.
##### TRIGONOMETRY
Relevant for
Learning to transform from polar to rectangular coordinates.
See examples
##### TRIGONOMETRY
Relevant for
Learning to transform from polar to rectangular coordinates.
See examples
## How to transform from polar coordinates to rectangular coordinates?
Polar coordinates have the form , where r is the distance of the point from the origin and θ is the angle formed by the line and the x-axis. Rectangular coordinates or Cartesian coordinates have the form .
To transform from polar coordinates to rectangular coordinates, we use trigonometry and relate these two coordinates.
Let’s consider the following diagram:
Clearly, we see that we can find the x-coordinates using the cosine function and we can find the y-coordinates using the sine function. Therefore, we have the formulas:
## Area of an isosceles triangle – Examples with answers
The following examples are solved by applying the transformation formulas from polar coordinates to rectangular coordinates. Try to solve the problems yourself before looking at the answer.
### EXAMPLE 1
If we have a point with polar coordinates , what are its rectangular coordinates?
We can observe the values and . We use the formulas found above to convert to rectangular coordinates. Therefore, the value of x is found using the cosine function:
The value of y is found using the sine function:
Therefore, the rectangular coordinates are (2.5, 4.33).
### EXAMPLE 2
A point has the polar coordinates . What are its rectangular coordinates?
We start by recognizing the values and . We convert these coordinates by applying the formulas seen above. Therefore, we find the value of x using the cosine function:
Now, we find the value of y using the sine function:
Therefore, the rectangular coordinates are (-6, -10.4).
### EXAMPLE 3
What are the rectangular coordinates of the point that is written in polar coordinates?
From the given coordinates, we have the values and . We substitute these values in the transformation formulas to find the rectangular coordinates. Therefore, the value of x is found using the cosine function:
The value of y is found using the sine function:
Therefore, the rectangular coordinates are (-7.78, -7.78).
### EXAMPLE 4
If a point has polar coordinates , what are its rectangular coordinates?
We have the values and . We find the value of x by using the cosine function and substituting these values:
We find the value of y using the sine function:
Therefore, the rectangular coordinates are (16.18, 11.76).
## Polar to rectangular coordinates – Practice problems
Practice what you have learned about transforming polar to rectangular coordinates by solving the following problems. If you need help with this, you can look at the solved examples above. |
# How do you integrate int e^x cos ^2 x dx using integration by parts?
Jun 7, 2016
$\frac{1}{10} {e}^{x} \cos 2 x + \frac{1}{5} {e}^{x} \sin 2 x + \frac{1}{2} {e}^{x} + C$
#### Explanation:
First, consider the identity $\cos 2 x = 2 {\cos}^{2} x - 1$. Use this identity to transform the integral
$\int \left({e}^{x} {\cos}^{2} x\right) \mathrm{dx}$
into the integral
$\int \frac{{e}^{x} \left(\cos 2 x + 1\right)}{2} \mathrm{dx} = \frac{1}{2} \int {e}^{x} \cos 2 x \mathrm{dx} + \frac{1}{2} \int {e}^{x} \mathrm{dx}$
Finding $\int {e}^{x} \mathrm{dx}$ is straightforward i.e. $\int {e}^{x} \mathrm{dx} = {e}^{x} + C$
We use integration by parts to find $\int {e}^{x} \cos 2 x \mathrm{dx}$. By LIATE, we integrate ${e}^{x}$ and differentiate $\cos 2 x$:
int (e^x cos 2x) dx = e^x cos 2x - int e^x (–2 sin 2x) dx = e^x cos 2x + 2 int (e^x sin 2x) dx = e^x cos 2x + 2[(e^x sin 2x)-int e^x (2 cos 2x) dx] = e^x cos 2x + 2e^x sin 2x-4 int (e^x cos 2x) dx
Thus,
$\int \left({e}^{x} \cos 2 x\right) \mathrm{dx} = {e}^{x} \cos 2 x + 2 {e}^{x} \sin 2 x - 4 \int \left({e}^{x} \cos 2 x\right) \mathrm{dx}$
$5 \int \left({e}^{x} \cos 2 x\right) \mathrm{dx} = {e}^{x} \cos 2 x + 2 {e}^{x} \sin 2 x \mathrm{dx}$
$\int \left({e}^{x} \cos 2 x\right) \mathrm{dx} = \frac{1}{5} {e}^{x} \cos 2 x + \frac{2}{5} {e}^{x} \sin 2 x \mathrm{dx}$
Therefore,
$\int \frac{{e}^{x} \left(\cos 2 x + 1\right)}{2} \mathrm{dx} = \frac{1}{2} \left(\frac{1}{5} {e}^{x} \cos 2 x + \frac{2}{5} {e}^{x} \sin 2 x\right) + \frac{1}{2} {e}^{x} + C = \frac{1}{10} {e}^{x} \cos 2 x + \frac{1}{5} {e}^{x} \sin 2 x + \frac{1}{2} {e}^{x} + C$ |
# Joint probability distribution
Joint probability distribution
In the study of probability, given two random variables X and Y that are defined on the same probability space, the joint distribution for X and Y defines the probability of events defined in terms of both X and Y. In the case of only two random variables, this is called a bivariate distribution, but the concept generalizes to any number of random variables, giving a multivariate distribution.
## Example
Consider the roll of a die and let A = 1 if the number is even (i.e. 2,4, or 6) and A = 0 otherwise. Furthermore, let B = 1 if the number is prime (i.e. 2,3, or 5) and B = 0 otherwise. Then, the joint distribution of A and B is
$\mathrm{P}(A=0,B=0)=P\{1\}=\frac{1}{6},\; \mathrm{P}(A=1,B=0)=P\{4,6\}=\frac{2}{6}$
$\mathrm{P}(A=0,B=1)=P\{3,5\}=\frac{2}{6},\; \mathrm{P}(A=1,B=1)=P\{2\}=\frac{1}{6}$
## Cumulative distribution
The cumulative distribution function for a pair of random variables is defined in terms of their joint probability distribution;
$F(x,y)=P(X \le x, Y \le y) .$
## Discrete case
The joint probability mass function of two discrete random variables is equal to
\begin{align} \mathrm{P}(X=x\ \mathrm{and}\ Y=y) & {} = \mathrm{P}(Y=y \mid X=x) \cdot \mathrm{P}(X=x) \\ & {} = \mathrm{P}(X=x \mid Y=y) \cdot \mathrm{P}(Y=y). \end{align}
In general, the joint probability distribution of n discrete random variables X1,...,Xn is equal to
$\mathrm{P}(X_1=x_1,\dots,X_n=x_n)=\mathrm{P}(X_1=x_1)\cdot \mathrm{P}(X_2=x_2|X_1=x_1)\cdot \mathrm{P}(X_3=x_3|X_1=x_1,X_2=x_2) \cdot ... \cdot P(X_n=x_n|X_1=x_1,\dots,X_{n-1}=x_{n-1})$
This identity is known as the chain rule of probability.
Since these are probabilities, we have
$\sum_x \sum_y \mathrm{P}(X=x\ \mathrm{and}\ Y=y) = 1.\;$
## Continuous case
Similarly for continuous random variables, the joint probability density function can be written as fX,Y(xy) and this is
$f_{X,Y}(x,y) = f_{Y|X}(y|x)f_X(x) = f_{X|Y}(x|y)f_Y(y)\;$
where fY|X(y|x) and fX|Y(x|y) give the conditional distributions of Y given X = x and of X given Y = y respectively, and fX(x) and fY(y) give the marginal distributions for X and Y respectively.
Again, since these are probability distributions, one has
$\int_x \int_y f_{X,Y}(x,y) \; dy \; dx= 1.$
## Mixed case
In some situations X is continuous but Y is discrete. For example, in a logistic regression, one may wish to predict the probability of a binary outcome Y conditional on the value of a continuously-distributed X. In this case, (X, Y) has neither a probability density function nor a probability mass function in the sense of the terms given above. On the other hand, a "mixed joint density" can be defined in either of two ways:
\begin{align} f_{X,Y}(x,y) &= f_{X|Y}(x|y)\mathrm{P}(Y=y)\\ &= \mathrm{P}(Y=y \mid X=x) f_X(x) \end{align}
Formally, fX,Y(x, y) is the probability density function of (X, Y) with respect to the product measure on the respective supports of X and Y. Either of these two decompositions can then be used to recover the joint cumulative distribution function:
\begin{align} F_{X,Y}(x,y)&=\sum\limits_{t\le y}\int_{s=-\infty}^x f_{X,Y}(s,t)\;ds \end{align}
The definition generalizes to a mixture of arbitrary numbers of discrete and continuous random variables.
## General multidimensional distributions
The cumulative distribution function for a vector of random variables is defined in terms of their joint probability distribution;
$F(x_1,\dots,x_n)=P(X_1 \le x_1,\dots, X_n \le x_n) .$
The joint distribution for two random variables can be extended to many random variables X1, ... Xn by adding them sequentially with the identity
\begin{align} f_{X_1, \ldots X_n}(x_1, \ldots x_n) =& f_{X_n | X_1, \ldots X_{n-1}}( x_n | x_1, \ldots x_{n-1}) f_{X_1, \ldots X_{n-1}}( x_1, \ldots x_{n-1} )\\ =& f_{X_1} (x_1) \\ & \cdot f_{X_2|X_1} (x_2|x_1)\\ & \cdot \dots \\ & \cdot f_{X_{n-1}| X_1 \ldots X_{n-2}}(x_{n-1}| x_1, \ldots x_{n-2} ) \\ & \cdot f_{X_n | X_1, \ldots X_{n-1}}( x_n | x_1, \ldots x_{n-1}),\end{align}
where
\begin{align} f_{X_i| X_1, \ldots X_{i-1}}(x_i | x_1, \ldots x_{i-1})= &\frac{f_{X_1, \dots X_i}(x_1,\dots x_i)}{\int f_{X_1, \dots X_i}(x_1,\dots x_{i-1},u_i) \mathrm{d} u_i}\\ = &\frac{\int \dots \int f_{X_1, \dots X_n}(x_1,\dots x_i,u_{i+1}, \dots u_n) \mathrm{d} u_{i+1}\dots \mathrm{d}u_n}{\int \dots \int \int f_{X_1, \dots X_n}(x_1,\dots x_{i-1},u_i, \dots u_n) \mathrm{d} u_i \,\mathrm{d} u_{i+1}\dots \mathrm{d}u_n} \end{align}
and
$f_{X_1,\dots X_i}(x_1,\dots x_i) = \int \dots \int f_{X_1,\dots X_n}(x_1,\dots x_i,x_{i+1},\dots x_n) \mathrm{d} x_{i+1} \dots \mathrm{d} x_n$
(notice, that these latter identities can be useful to generate a random variable $(X_1, \dots X_n)$ with given distribution function $f(x_1,\dots x_n)$); the density of the marginal distribution is
$f_{X_i}(x_i) = \int \dots \int \int \dots \int f_{X_1,\dots X_n}(x_1,\dots x_{i-1},x_i,x_{i+1},\dots x_n) \mathrm{d} x_1\dots \mathrm{d}x_{i-1} \, \mathrm{d}x_{i+1} \dots \mathrm{d}x_n.$
The joint cumulative distribution function is
$F_{X_1,\dots X_n}\left( x_1, \dots x_n\right)= \int_{-\infty}^{x_1} \dots \int_{-\infty}^{x_n} f_{X_1,\dots X_n}\left(u_1,\dots u_n\right) \mathrm{d} u_1 \dots \mathrm{d}u_n,$
and the conditional distribution function is accordingly
\begin{align} F_{X_i| X_1, \ldots X_{i-1}}(x_i| x_1, \ldots x_{i-1})= &\frac{\int_{-\infty}^{x_i}f_{X_1, \dots X_i}(x_1,\dots x_{i-1},u_i)\mathrm{d}u_i}{\int_{-\infty}^\infty f_{X_1, \dots X_i}(x_1,\dots x_{i-1},u_i) \mathrm{d} u_i}\\ = &\frac{\int_{-\infty}^\infty \dots \int_{-\infty}^\infty \int_{-\infty}^{x_i} f_{X_1, \dots X_n}(x_1,\dots x_{i-1},u_i, \dots u_n) \mathrm{d} u_i\dots \mathrm{d}u_n}{\int_{-\infty}^\infty \dots \int_{-\infty}^\infty \int_{-\infty}^\infty f_{X_1, \dots X_n}(x_1,\dots x_{i-1},u_i,\dots u_n) \mathrm{d} u_i \dots \mathrm{d} u_n}. \end{align}
$\mathbb{E}\left[h(X_1,\dots X_n) \right]=\int_{-\infty}^\infty \dots \int_{-\infty}^\infty h(x_1,\dots x_n) f_{X_1,\dots X_n}(x_1,\dots x_n) \mathrm{d} x_1 \dots \mathrm{d} x_n;$
suppose that h is smooth enough and $h(u_1,\dots u_n)=h(x_1,\dots x_n)$ for $u_1 \ge x_1, \dots u_n\ge x_n$, then, by iterated integration by parts,
\begin{align}\mathbb{E}\left[h(X_1,\dots X_n) \right]=& h(x_1,\dots x_n)+ \\ & (-1)^n \int_{-\infty}^{x_1} \dots \int_{-\infty}^{x_n} F_{X_1,\dots X_n}(u_1,\dots u_n) \frac{\partial^n}{\partial x_1 \dots \partial x_n} h(u_1,\dots u_n) \mathrm{d} u_1 \dots \mathrm{d} u_n.\end{align}
## Joint distribution for independent variables
If for discrete random variables $\ P(X = x \ \mbox{and} \ Y = y ) = P( X = x) \cdot P( Y = y)$ for all x and y, or for absolutely continuous random variables $\ f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)$ for all x and y, then X and Y are said to be independent.
## Joint Distribution for conditionally independent variables
If a subset A of the variables $X_1,\cdots,X_n$ is conditionally independent given another subset B of these variables, then the joint distribution P(X1,...,Xn) is equal to $P(B)\cdot P(A|B)$. Therefore, it can be efficiently represented by the lower-dimensional probability distributions P(B) and P(A | B). Such conditional independence relations can be represented with a Bayesian network.
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# COMPARING AND ORDERING WHOLE NUMBERS
## About "Comparing and ordering whole numbers"
Comparing and ordering whole numbers :
The set of all non negative integers is whole numbers.
That is,
W = { 0, 1, 2, 3, 4, .........................}
We use number line to compare and order whole numbers.
That is, if we want to compare two whole numbers, first we have to locate the numbers on the number line and mark them.
The number which comes to the right of the other number is greater. The number which comes to the left of the other number is smaller.
To order the whole numbers, we have to locate the given numbers on the number line and mark them. Then, write the numbers from left to right to list them in order from least to greatest.
## Comparing whole numbers - Examples
Example 1 :
Compare the whole numbers 3 and 6
Solution :
Let us locate the two whole numbers 3 and 6 on a number line and mark them.
Here, 6 comes to the right of 3.
Therefore "6" is greater than "3"
And 3 comes to the left of 6.
Therefore "3" is smaller than "6"
Example 2 :
Compare the whole numbers 6 and 9
Solution :
Let us locate the two whole numbers 6 and 9 on a number line and mark them.
Here, 9 comes to the right of 6.
Therefore "9" is greater than "6"
And 6 comes to the left of 9.
Therefore "6" is smaller than "9"
## Ordering whole numbers - Example
Question :
Order the whole numbers 12, 5, 9, 6, 1, 3 from least to greatest.
Solution :
Let us locate the whole numbers 12, 5, 9, 6, 1, 3 on a number line and mark them.
From the above number line, write the whole numbers from left to right to list them in order from least to greatest.
Thus, we get
1, 3, 5, 6, 9, 12
## Comparing and ordering whole numbers - Real world problems
Problem 1 :
In 2010, Sacramento, California, received 23 inches in annual precipitation. In 2011, the city received 17 inches in annual precipitation. In which year was there more precipitation ?
Solution :
Locate the two whole numbers 23 and 17 on a number line and mark them.
23 is to the right of 17 on the number line.
This means that 23 is greater than 17.
We can write the above situation in terms of inequality as 23 > 17.
17 is to the left of 23 on the number line.
This means that 17 is less than 23.
We can write the above situation in terms of inequality as 17 < 23.
There was more precipitation in 2010.
Problem 2 :
John recorded the following golf scores during his first week at a golf academy. In golf, a lower score beats a higher score.
Graph John’s scores on the number line, and then list the numbers in order from least to greatest.
Solution :
Step 1 :
Graph the scores on the number line.
Step 2 :
Read from left to right to list the scores in order from least to greatest.
The scores listed from least to greatest are
0, 1, 3, 5, 6, 9, 12
After having gone through the stuff given above, we hope that the students would have understood "Comparing and ordering whole numbers".
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Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
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# Intersection
We can say that a Set is the collection of different objects grouped together. Any set is the most basic element used in mathematics. In everyday life we observe that Sets are used to make groups of same type of elements. When we study sets theory, we need to re- organize the group of the sets which always helps us to get the solutions. One of the main applications of the sets theory is to make the Relations between two sets and then observe the elements of the new set. For making these relations, certain operations can be performed on the Numbers. We will discuss these operations one by one.
Intersection of two sets: We can also form a new set by determining the common elements of the given two sets. This is called intersection of the two sets. If A∩ B = ∅, then ‘A’ and ‘B’ are said as disjoint setsIntersection of a set is a basic operation performed in the sets Algebra. Here we must first know that as Union represents “ OR ” operation, similarly Intersection represents “AND” operation, which means the element exist in both the sets, will belong to the intersection sets. It means that in order to form the new set with the intersection sets ( A and B ), the new set will have to coincide in both or more sets.
Here are some of the examples to show the intersection of sets. Let A= 1, 2 and B= red, black then ‘A’ intersection B = φ, Null set. We observe that there is no element common in both the sets, so the result is the null set (a set with no element).
Similarly if A= 1, 2, 3, 4 and B = 4, 5, 6, 7.
Here both sets ‘A’ and ‘B’ have 4 as common, so ‘A’ intersection ‘B’ will be = 4
Again we observe that if set ‘A’ and set ‘B’ have all elements same, then ‘A’ intersection B = A.
Note:
Intersection of two sets is the common term in the given set, if we have two set ‘A’ and ‘B’ and ‘A’ has elements as 1,2,3,4 and set ‘B’ has 3,4,5,6 so intersection will be 3,4, intersection is denoted by the symbol ‘∪’, and if two sets have nothing in common than the set is called as Nullary intersection.
Note:
Arbitrary Intersections of two sets ‘X’ and ‘Y’ refers to the set that contains all the elements of ‘X’ and also belong to ‘Y’ but not the other elements. It is abbreviated as X ∩ Y. If X ∩ Y belongs to ‘R’ then it means that ‘X’ belongs to ‘R’ and ‘Y’ belongs to ‘R’.
## What do you mean by Intersection of the two sets? Explain with an example.
Intersection of Sets, represented by A ∩ B. Word Intersection means the elements of the two Sets which are common. So if we write Let A = 1, 2, 3, 4 and B = 4, 5, 6 then we can say that
A ∩ B = 4 which includes all the common elements of Set A and set B. Here we observe that only element 4 is common in both the sets A and B.
## Is A ∩ φ = A True or False? Give reason in support of your answer.
A ∩ φ = A is not a True statement. As when we write A ∩ φ, it means the intersecting elements of Set ‘A’ and a null set φ. Let if A = 1, 3, 5 and we know that null set has no element in it. Then the Intersection of two such Sets is always a null set, because intersection means the common elements of the two Sets. So we come to a conclusion that A ∩ φ = φ.
## Further Read
FAQ of Intersection
Math Topics
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# What is the common ratio in the geometric sequence 6 18 54?
Definition of Geometric Sequences For example, the sequence 2,6,18,54,⋯ 2 , 6 , 18 , 54 , ⋯ is a geometric progression with common ratio 3 . Similarly 10,5,2.5,1.25,⋯ 10 , 5 , 2.5 , 1.25 , ⋯ is a geometric sequence with common ratio 12 . for every integer n≥1. n ≥ 1.
## What is the common ratio of the sequence 6 Negative 18 54?
For example, the sequence 2, 6, 18, 54, is a geometric progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, is a geometric sequence with common ratio 1/2.
## What is the common ratio?
Definition of common ratio : the ratio of each term of a geometric progression to the term preceding it.
## How do you find the common ratio of a geometric sequence?
To calculate the common ratio in a geometric sequence, divide the n^th term by the (n – 1)^th term. Start with the last term and divide by the preceding term. Continue to divide several times to be sure there is a common ratio.
## What is the sum of the first 6 terms?
The sum of the first 6 terms of a geometric sequence is 9 times the sum of its first 3 terms. Find the common ratio. Sum of n terms of a geometric series is given by a ( r^n – 1) / r – 1 where a, r and n are the first term, ratio and number of terms of the series respectively.
## What is the common difference in the arithmetic sequence 10 8 6 4?
The common difference of the arithmetic sequence 10, 8, 6, 4, 2,… is -2. Note: By considering the formula of arithmetic sequence we verify the common difference which we obtained. We have to check the common difference for all the terms.
## What is the value of in the following number sequence 6/18 54 N 486?
6,18,54,___,486,1458. So the missing number is 162. Solution:- in this pattern, we are multiplying the numbers by 3. You see, if we multiply 6 by 3( 6×3 =18).
## What are the next two terms of the sequence 6 18 54?
2, 6, 18, 54, 162, 486, 1458, 4374. we multiply 3 to each no for next no. Therefore the 8th term in the sequence will be 2*3^(8–1) = 2*3^7 = 2*2187 = 4374.
## What is the common ratio formula?
From the formula for the sum for n terms of a geometric progression, Sn = a(rn − 1) / (r − 1) where a is the first term, r is the common ratio and n is the number of terms. Therefore, for the n th term of the above sequence, we get: 4 n + 1 − 1 4 − 1 = 4 n + 1 − 1 3 .
## What is the common ratio of the sequence 16 8 4 2?
This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by 12 gives the next term. In other words, an=a1⋅rn−1 a n = a 1 ⋅ r n – 1 . This is the form of a geometric sequence.
## How do you find the sum of a sequence?
To find the sum of an arithmetic sequence, start by identifying the first and last number in the sequence. Then, add those numbers together and divide the sum by 2. Finally, multiply that number by the total number of terms in the sequence to find the sum.
## How do you find the sum of the first 6 terms of a geometric sequence?
Use the formula for the sum of a geometric series: S(n)=[a(1-r^n)]/(1-r). Here we have by observation; a=2, n=6, r=3. Plugging in we find; S(6)=[2(1–3^6)]/(1–3). S(6)=728.
## What is the common difference?
Definition of common difference : the difference between two consecutive terms of an arithmetic progression.
## What is the common difference of the arithmetic sequence?
The common difference is the value between each successive number in an arithmetic sequence. Therefore, the formula to find the common difference of an arithmetic sequence is: d = a(n) – a(n – 1), where a(n) is the last term in the sequence, and a(n – 1) is the previous term in the sequence.
## How do you find the common difference in harmonic progression?
The common difference can be found by subtracting any two adjacent terms. Each term after the first can be found by adding recursively the common difference d to the preceding term. The sum of the first n terms of arithmetic progression is n times the average of the first term and the last term.
## Is 13 a Fibonacci number?
Fibonacci Numbers (Sequence): 1,1,2,3,5,8,13,21,34,55,89,144,233,377,This sequence of numbers was first created by Leonardo Fibonacci in 1202 .
## What kind of sequence is 6 18 54?
A geometric sequence (also known as a geometric progression) is a sequence of numbers in which the ratio of consecutive terms is always the same. For example, in the geometric sequence 2, 6, 18, 54, 162, …, the ratio is always 3. This is called the common ratio. |
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# A square garden is surrounded by a 2 feet wide walkway on
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A square garden is surrounded by a 2 feet wide walkway on [#permalink]
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26 Jul 2012, 05:34
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A square garden is surrounded by a 2 feet wide walkway on each side.The area of the garden and the walkway is 225 square feet.If the garden is broadened by increasing side of the garden by 5 units then what will be the area of the garden?
A. 121
B. 169
C. 225
D. 256
E. 324
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Re: A square garden is surrounded by a 2 feet wide walkway [#permalink]
### Show Tags
26 Jul 2012, 07:50
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Hi,
Area of garden & path = 225
Side = $$\sqrt{225}$$ = 15 ft
As per the below diagram,
Attachment:
square.jpg [ 11.57 KiB | Viewed 7918 times ]
Side of garden = 11 ft
When sides of garden are increased by 5 ft,
new dimension = 11+5 = 16 ft
Area = $$16^2$$ = 256
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Re: A square garden is surrounded by a 2 feet wide walkway [#permalink]
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26 Jul 2012, 09:49
Thanks.
I missed to take 2 feet on both sides. so took the side of the garden as 15-2 =13.
so 13 +5 after increasing and the area of garden as 18^2 = 324.
Have to be careful . Thanks for the solution.
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Re: A square garden is surrounded by a 2 feet wide walkway on [#permalink]
### Show Tags
27 May 2017, 04:17
1
Let 'x' be the side of garden excluding the walkway.
Thus side of garden including the walkway = (x+4) (since its 2 feet on both sides)
Area = (x+4)^2 = 225 Or (x+4) = 15 Or x = 11
Right now side of square garden is 11 feet. New side = 11+5 = 16 feet
New Area = 16*16 = 256 sq feet
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A square garden is surrounded by a 2 feet wide walkway on [#permalink]
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28 Oct 2018, 15:09
Solution
Given:
• A square garden is surrounded by a 2 feet wide walkway on each side.
• The area of the garden and the walkway is 225 square feet.
• Garden is broadened by increasing side of the garden by 5 units.
To find:
• Area of the garden if the garden is broadened by increasing side of the garden by 5 units.
Approach and Working
• Let the side of the garden is x feet.
o Hence, the side of garden +walkway= l+4 feet
• $$(x+4)^2$$= 225
o x= 11
• x+5= 16
• $$(x+5)^2$$= 256 feet.
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A square garden is surrounded by a 2 feet wide walkway on [#permalink] 28 Oct 2018, 15:09
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# Congruent triangles|Class 9 Maths notes
## What is Congruence
• In geometry, two figures or objects are congruent if they have the same shape and size, or if one has the same shape and size as the mirror image of the other.
• A combination of rigid motions, namely a translation, a rotation, and a reflection is also permitted in Congruence. Translation means sliding,Rotation means turning and Reflection means Flipping. If two figures can be overlapped using translation,rotation or reflection then they are congurence
• It has come from Latin word congruere "agree, correspond with"
## What is the difference between Congurence and Similarity?
Two Figures are similar if the objects differ in size but not in shape while in congurence size and shape remains same
## Why do we need to Learn it?
• We need to learn congurence in order to identify and describe many congurent object in real life. It may be some sculpture, some old monuments,
• In Pyramid, all the three triangular surface are congurent.AROUND 2600 B.C., construction of the Great Pyramid of Khufu began. It took the ancient Egyptians about 30 years to transform 6.5 million tons of stone into a pyramid with a square base and four congruent triangular faces
• We need several identical shirts button ,various factory part and many numerous objects. All these need the concept of Congurence
## Congruence
Two Geometric figure are said to be congruence if they are exactly same size and shape
• Symbol used is ≅
• Two angles are congruent if they are equal
• Two circle are congruent if they have equal radii
• Two squares are congruent if the sides are equal
• Two line segments are congruent if they have equal length
## Triangle Congruence
• Two triangles are congruent if three sides and three angles of one triangle is congruent to the corresponding sides and angles of the other
• Corresponding sides are equal AB=DE , BC=EF ,AC=DF
• Corresponding angles are equal
• We write this as
ABC ≅ DEF
• The above six equalities are between the corresponding parts of the two congruent triangles. In short form this is called C.P.C.T
• We should keep the letters in correct order on both sides
## Properties of Congurent triangles
1) Reflexive Property of Congurent Triangle: Every Triangle is congurent to itself
ABC ≅ ABC
2) Commutative or symmetric property of Congurent Triangle:
if ABC ≅ DEF then DEF ≅ ABC
3) Transitive Property Of congurent Triangle
if ABC ≅ DEF and DEF ≅ LKM then ABC ≅ LKM
Since Congurence relation satisfy Reflexive ,symmetric and Transitive ,it is an equivalence relation
## Different Criterion for Congruence of the triangles
### RHS Congurence (Theorem)
Caveat
AAA( Angle Angle Angle) is not the right condition to Prove congurence.
## How to Prove the congurence of Two Triangle
1) We have already studied that two triangles are congurent when all the sides and all the angles are equal. But we dont need prove all these while solving the Problem
2) We just need to prove the congurence using the different criterio like SSS,ASA,SAS,RHS,AAS
3) Dont use AAA
4) Use the theorem learn in previous Geometry chapter like vertically opposite angles,alternate interior angles,corresonding angles
5)Write down the corresponding angles and corresponding sides carefully
6) We need to be careful with the labelling when our Triangles are in different positions
## Inequality of Triangle
1) In a triangle angle opposite to longer side is larger
2) In a triangle side opposite to larger angle is larger
3) The sum of any two sides of the triangle is greater than the third side
In triangle ABC
AB +BC > AC
4) Difference of any two sides of triangle is less than the third side
AC-AB < BC where AC > AB
## Important defination for Triangles
### Median
A line Segment joining the corner of the triangle to the mid point of the opposite side of the triangle. All the median of the triangle passes through the same point
### Centroid
Point of intersection of the three median of the triangle is called the centroid of the triangle. The centroid divides each of the median in the ration of 2:1
### Altitude
A line Segment from the corner of the triangle and perpendicular to the opposite side of the triangle. All the altitude of the triangle passes through the same point
### Ortho center
Point of intersection of the three altitude of the triangle is called the orthocenter of the triangle.
### Incenter
All the angle bisector of the triangle passes through same point .It is called incenter.
### Circumcenter
The perpendicular bisector of the sides of the triangles passes through same point. It is called circumcenter
## Solved Example
What is the measure of the angles P,M,R,N?
Write down the corresponding congurent sides?
Solution:
By Congurence, All the corresponding angles and corresponding sides are equal
LMN ≅ PQR
Steps to find the All the corresponding angles and corresponding sides
1) For angles, The First alphabet on right hand side corresponds to First alphabet on Left hand side,Similary second and third alphabet
So ∠ L=∠ P
∠M=∠ Q
∠ N=∠ R
Now ∠ L=1050
∠ Q=45, So ∠ M=45 0
∠P=1050
Now ∠ R can be found from the theorem
"Sum of the all the angles in a traingle is 1800"
So ∠ R=180-105-45=300
∠ N=30 0
2) For sides,The first two alphabet on right hand side corresponds to First two alphabet on Left hand side,Similary 2,3 and 3,1
LM=PQ
MN=QR
LN=PR |
Paul's Online Notes
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### Section 3.3 : Differentiation Formulas
8. Find the derivative of $$\displaystyle R\left( z \right) = \frac{6}{{\sqrt {{z^3}} }} + \frac{1}{{8{z^4}}} - \frac{1}{{3{z^{10}}}}$$ .
Show Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this section.
Remember that you’ll need to rewrite the terms so that each of the $$z$$’s are in the numerator with negative exponents and rewrite the root as a fractional exponent before taking the derivative. Here is the rewritten function.
$R\left( z \right) = 6{z^{ - \,\,\frac{3}{2}}} + \frac{1}{8}{z^{ - 4}} - \frac{1}{3}{z^{ - 10}}$
The derivative is,
$R'\left( z \right) = 6\left( { - \frac{3}{2}} \right){z^{ - \,\,\frac{5}{2}}} + \frac{1}{8}\left( { - 4} \right){z^{ - 5}} - \frac{1}{3}\left( { - 10} \right){z^{ - 11}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 9{z^{ - \,\,\frac{5}{2}}} - \frac{1}{2}{z^{ - 5}} + \frac{{10}}{3}{z^{ - 11}}}}$ |
# 8.1 Solve equations using the subtraction and addition properties (Page 5/6)
Page 5 / 6
## Practice makes perfect
Solve Equations Using the Subtraction and Addition Properties of Equality
In the following exercises, determine whether the given value is a solution to the equation.
Is $y=\frac{1}{3}$ a solution of $4y+2=10y?$
yes
Is $x=\frac{3}{4}$ a solution of $5x+3=9x?$
Is $u=-\frac{1}{2}$ a solution of $8u-1=6u?$
no
Is $v=-\frac{1}{3}$ a solution of $9v-2=3v?$
In the following exercises, solve each equation.
$x+7=12$
x = 5
$y+5=-6$
$b+\frac{1}{4}=\frac{3}{4}$
$b=\frac{1}{2}$
$a+\frac{2}{5}=\frac{4}{5}$
$p+2.4=-9.3$
p = −11.7
$m+7.9=11.6$
$a-3=7$
a = 10
$m-8=-20$
$x-\frac{1}{3}=2$
$x=\frac{7}{3}$
$x-\frac{1}{5}=4$
$y-3.8=10$
y = 13.8
$y-7.2=5$
$x-15=-42$
x = −27
$z+5.2=-8.5$
$q+\frac{3}{4}=\frac{1}{2}$
$q=-\frac{1}{4}$
$p-\frac{2}{5}=\frac{2}{3}$
$y-\frac{3}{4}=\frac{3}{5}$
$y=\frac{27}{20}$
Solve Equations that Need to be Simplified
In the following exercises, solve each equation.
$c+3-10=18$
$m+6-8=15$
17
$9x+5-8x+14=20$
$6x+8-5x+16=32$
8
$-6x-11+7x-5=-16$
$-8n-17+9n-4=-41$
−20
$3\left(y-5\right)-2y=-7$
$4\left(y-2\right)-3y=-6$
2
$8\left(u+1.5\right)-7u=4.9$
$5\left(w+2.2\right)-4w=9.3$
1.7
$-5\left(y-2\right)+6y=-7+4$
$-8\left(x-1\right)+9x=-3+9$
−2
$3\left(5n-1\right)-14n+9=1-2$
$2\left(8m+3\right)-15m-4=3-5$
−4
$-\left(j+2\right)+2j-1=5$
$-\left(k+7\right)+2k+8=7$
6
$6a-5\left(a-2\right)+9=-11$
$8c-7\left(c-3\right)+4=-16$
−41
$8\left(4x+5\right)-5\left(6x\right)-x=53$
$6\left(9y-1\right)-10\left(5y\right)-3y=22$
28
Translate to an Equation and Solve
In the following exercises, translate to an equation and then solve.
Five more than $x$ is equal to $21.$
The sum of $x$ and $-5$ is $33.$
x + (−5) = 33; x = 38
Ten less than $m$ is $-14.$
Three less than $y$ is $-19.$
y − 3 = −19; y = −16
The sum of $y$ and $-3$ is $40.$
Eight more than $p$ is equal to $52.$
p + 8 = 52; p = 44
The difference of $9x$ and $8x$ is $17.$
The difference of $5c$ and $4c$ is $60.$
5 c − 4 c = 60; 60
The difference of $n$ and $\frac{1}{6}$ is $\frac{1}{2}.$
The difference of $f$ and $\frac{1}{3}$ is $\frac{1}{12}.$
$f-\frac{1}{3}=\frac{1}{12};\phantom{\rule{0.2em}{0ex}}\frac{5}{12}$
The sum of $-4n$ and $5n$ is $-32.$
The sum of $-9m$ and $10m$ is $-25.$
−9 m + 10 m = −25; m = −25
Translate and Solve Applications
In the following exercises, translate into an equation and solve.
Pilar drove from home to school and then to her aunt’s house, a total of $18$ miles. The distance from Pilar’s house to school is $7$ miles. What is the distance from school to her aunt’s house?
Jeff read a total of $54$ pages in his English and Psychology textbooks. He read $41$ pages in his English textbook. How many pages did he read in his Psychology textbook?
Let p equal the number of pages read in the Psychology book 41 + p = 54. Jeff read pages in his Psychology book.
Pablo’s father is $3$ years older than his mother. Pablo’s mother is $42$ years old. How old is his father?
Eva’s daughter is $5$ years younger than her son. Eva’s son is $12$ years old. How old is her daughter?
Let d equal the daughter’s age. d = 12 − 5. Eva’s daughter’s age is 7 years old.
Allie weighs $8$ pounds less than her twin sister Lorrie. Allie weighs $124$ pounds. How much does Lorrie weigh?
For a family birthday dinner, Celeste bought a turkey that weighed $5$ pounds less than the one she bought for Thanksgiving. The birthday dinner turkey weighed $16$ pounds. How much did the Thanksgiving turkey weigh?
21 pounds
The nurse reported that Tricia’s daughter had gained $4.2$ pounds since her last checkup and now weighs $31.6$ pounds. How much did Tricia’s daughter weigh at her last checkup?
Connor’s temperature was $0.7$ degrees higher this morning than it had been last night. His temperature this morning was $101.2$ degrees. What was his temperature last night?
100.5 degrees
Melissa’s math book cost $\text{22.85}$ less than her art book cost. Her math book cost $\text{93.75}.$ How much did her art book cost?
Ron’s paycheck this week was $\text{17.43}$ less than his paycheck last week. His paycheck this week was $\text{103.76}.$ How much was Ron’s paycheck last week?
$121.19 ## Everyday math Baking Kelsey needs $\frac{2}{3}$ cup of sugar for the cookie recipe she wants to make. She only has $\frac{1}{4}$ cup of sugar and will borrow the rest from her neighbor. Let $s$ equal the amount of sugar she will borrow. Solve the equation $\frac{1}{4}+s=\frac{2}{3}$ to find the amount of sugar she should ask to borrow. Construction Miguel wants to drill a hole for a $\frac{5}{\text{8}}\phantom{\rule{0.1em}{0ex}}\text{-inch}$ screw. The screw should be $\frac{1}{12}$ inch larger than the hole. Let $d$ equal the size of the hole he should drill. Solve the equation $d+\frac{1}{12}=\frac{5}{8}$ to see what size the hole should be. $d=\frac{13}{24}$ ## Writing exercises Is $-18$ a solution to the equation $3x=16-5x?$ How do you know? Write a word sentence that translates the equation $y-18=41$ and then make up an application that uses this equation in its solution. Answers will vary. ## Self check After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. If most of your checks were: …confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific. …with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved? …no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need. #### Questions & Answers where we get a research paper on Nano chemistry....? Maira Reply nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? Maira Reply There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn Google da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery Hafiz Reply revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials Jyoti Reply ya I also want to know the raman spectra Bhagvanji I only see partial conversation and what's the question here! Crow Reply what about nanotechnology for water purification RAW Reply please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm Brian Reply is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? LITNING Reply what is a peer LITNING Reply What is meant by 'nano scale'? LITNING Reply What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? Bob Reply write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? Damian Reply what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? Stoney Reply why we need to study biomolecules, molecular biology in nanotechnology? Adin Reply ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. Adin why? Adin what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe how did you get the value of 2000N.What calculations are needed to arrive at it Smarajit Reply Privacy Information Security Software Version 1.1a Good A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
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I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? |
# Geometry: Theorems
## Contents
#### Triangle Inequalities
Triangles are governed by two important inequalities. The first is often referred to as the triangle inequality. It states that the length of a side of a triangle is always less than the sum of the lengths of the other two sides. Can you see why this must be true? Were one side of a triangle longer than the sum of the lengths of the other two, the triangle could not exist. As one side grows, the other two collapse toward that side until the altitude from the vertex opposite the growing side eventually becomes zero. This (an altitude of zero) would happen if the length of the one side was equal to the sum of the lengths of the other two. For this reason, the length of any side must be less than the sum of the lengths of the other sides.
The second inequality involving triangles has to do with opposite angles and sides. It states that when a pair of angles are unequal, the sides opposite them are also unequal. The converse is true also: when a pair of sides are unequal, so are their opposite angles. In essence, this theorem complements the theorem involving isosceles triangles, which stated that when sides or angles were equal, so were the sides or angles opposite them. The theorem about unequal pairs, though, goes a little farther. Given unequal angles, the theorem holds that the longer side of the triangle will stand opposite the larger angle, and that the larger angle will stand opposite the longer side. This inequality is helpful to prove triangles aren't congruent.
Figure %: The larger of two unequal angles is opposite the longer of two unequal sides, and vice versa.
Notice the symbols in the figure above. When angles or sides are equal, the same number of tick marks, or small dashes, can be drawn on them. In a case where sides or angles are unequal, this can be symbolized by different numbers of tick marks on the angles or sides. More tick marks signifies a greater measure.
#### Exterior Angles of a Triangle
A triangle's exterior angle is just like that of any polygon; it is the angle created when one side of the triangle is extended past a vertex. The exterior angle has two interesting properties that follow from one another. 1) The exterior angle at a given vertex is equal in measure to the sum of the two remote interior angles. These remote interior angles are those at the other two vertices of the triangle. 2) Knowing this, it follows that the measure of any exterior angle is always greater than the measure of either remote interior angle. The first fact (1), the equality, is useful for proving congruence; the second fact (2), the inequality, is useful for disproving congruence.
Figure %: Angle 4 is greater than angle 2 and angle 3; angle 4 = angle 2 + angle 3 |
# Integration of Tan Square x
Created by: Team Maths - Examples.com, Last Updated: May 15, 2024
## Integration of Tan Square x
Integration of tan²(x) involves techniques that encompass a broad array of mathematical concepts. From exploring rational and irrational numbers to delving into the realms of algebra and statistics, the process links various foundational elements like integers and square and square roots. Employing methods such as the least squares method in data fitting, this topic extends beyond simple calculus, bridging the gap between numerical theory and real-world applications. Such integrations often challenge our understanding of numbers and their properties, underscoring the interconnected nature of mathematical disciplines.
## What is Integration of Tan Square x?
The integration of tan² can be computed by using a trigonometric identity and substitution. First, recall the identity tan²(𝑥) = sec²(𝑥)−1. Thus, the integral of tan² becomes ∫tan²(𝑥) 𝑑𝑥 = ∫(sec²(𝑥)−1) 𝑑𝑥, which simplifies to tan(𝑥)−𝑥+𝐶, where 𝐶 is the constant of integration.
## Integration of Tan Square x Formula
∫ tan²x dx = tan x – x + C
The integral ∫tan²(𝑥) 𝑑𝑥 equals tan(𝑥)−𝑥+𝐶 by utilizing the trigonometric identity tan²(𝑥) = sec²(𝑥)−1. This transformation allows the integration of sec²(x) and −1 separately, simplifying to tan(𝑥)−𝑥, with 𝐶 as the integration constant representing any constant value added to the indefinite integral.
## Integration of Tan Square x Proof
To provide a proof for the integral ∫tan²(𝑥) 𝑑𝑥, we start by utilizing a known trigonometric identity and proceed with basic integration techniques. Here’s a detailed step-by-step explanation:
### Step 1: Use the Trigonometric Identity
First, we employ the identity for tan²(x):
tan²(𝑥) = sec²(𝑥)−1
### Step 2: Express the Integral Using the Identity
Using the identity, we rewrite the integral:
∫tan²(𝑥) 𝑑𝑥 = ∫(sec²(𝑥)−1) 𝑑𝑥
### Step 3: Separate the Integral
The integral can now be separated into two simpler integrals:
∫tan²(𝑥) 𝑑𝑥 = ∫sec²(𝑥) 𝑑𝑥−∫1 𝑑𝑥
### Step 4: Integrate Each Term
The integral of sec²(x) is straightforward:
∫sec²(𝑥) 𝑑𝑥 = tan(𝑥)
The integral of 1 with respect to 𝑥 is simply 𝑥:
∫1 𝑑𝑥 = 𝑥
### Step 5: Combine the Results
Subtracting the integral of 1 from the integral of sec²(x) gives:
∫tan²(𝑥) 𝑑𝑥 = tan(𝑥)−𝑥+𝐶
Where 𝐶 is the constant of integration.
## Integration of Tan Square x From 0 to Pi by 4
To find the definite integral of tan²(x) from 0 to 𝜋/4, we can use the previously established integral formula for tan²(x), which is ∫tan²(𝑥) 𝑑𝑥 = tan(𝑥)−𝑥. Applying the limits, we evaluate:
### Integral Setup
∫tan²(𝑥) 𝑑𝑥𝜋/⁴₀ = [tan(𝑥)−𝑥]𝜋/⁴₀
### Evaluate at the Upper Limit 𝜋/4
At 𝑥 = 𝜋/4, tan(𝜋/4)=1, so:
tan(𝜋/4)−𝜋/4 = 1−𝜋/4
### Evaluate at the Lower Limit 0
At 𝑥 = 0, tan(0) = 0, so:
tan(0)−0 = 0
### Compute the Integral
Subtract the value at the lower limit from the value at the upper limit:
[1−𝜋/4]−[0] = 1−𝜋/4
## Integration of Tan Square x Examples
### Example 1: Integration from 𝜋/6 to 𝜋/3
Problem: Compute the integral of tan²(x) from 𝜋/6 to 𝜋/3.
Solution: Using the identity tan²(𝑥) = sec²(𝑥)−1 and the integral formula:
∫tan²(𝑥) 𝑑𝑥 = tan(𝑥)−𝑥
Evaluating from 𝜋/6 to 𝜋/3:
[tan(𝑥)−𝑥]𝜋/³𝜋/₆ = (tan(𝜋/3)−𝜋/3)−(tan(𝜋/6)−𝜋/6)
Given tan(𝜋/3) = 3 and tan(𝜋/6) = 1/3, the calculation becomes:
(3−𝜋/3)−(1/3−𝜋/6)
Result: The integral evaluates to a specific numerical value, simplifying the expression by combining terms.
### Example 2: Integration over a Full Period of tan²(x)
Problem: Compute the integral of tan²(x) over one full period, from −𝜋/4 to 3𝜋/4.
Solution: Again using the integral formula:
∫tan²(𝑥) 𝑑𝑥 = tan(𝑥)−𝑥
Evaluating from −𝜋/4 = to 3𝜋/4:
[tan(𝑥)−𝑥]³𝜋/⁴−𝜋/₄ = (tan(3𝜋/4)−3𝜋/4)−(tan(−𝜋/4)−(−𝜋/4))
Since tan(3𝜋/4) = −1 and tan(−𝜋/4)= −1, the calculation becomes:
(−1−3𝜋/4)−(−1+𝜋/4)=−𝜋
Result: The result demonstrates the behavior of the tan²(x) function over its periodic interval.
### Example 3: Evaluating Zero to 𝜋
Problem: Evaluate ∫𝜋₀tan²(𝑥) 𝑑𝑥.
Solution: Direct computation is challenging due to the undefined nature of tan(𝑥) at 𝜋/2. This integral technically diverges as tan²(x) approaches infinity near 𝑥 = 𝜋/2.
Result: The integral does not converge due to the asymptotic behavior at 𝜋/2.
## What is the significance of integrating tan²(x) in practical applications?
Integrating tan²(x) is often encountered in physics and engineering, particularly in problems involving angular motion and oscillations where tangent functions model behaviors varying over time.
## How does the periodicity of tan(𝑥) affect its square’s integration?
The periodic nature of tan(𝑥) implies that integrals of tan²(x) over its full periods or multiples thereof simplify the evaluation, but care must be taken to avoid points of discontinuity where the tangent function is undefined.
## How does the trigonometric identity help in integrating tan²(x)?
By substituting tan²(x) with sec²(𝑥)−1, the integral breaks into simpler terms, sec²(x) and 1, which are straightforward to integrate individually.
Text prompt |
Section 04 Part 09 – The DIVU & DIVS Instructions “When there is no enemy within, the enemies outside cannot hurt you.” ~Winston Churchill
Introduction
These instructions are for division, and there are two of them. Just like MULU and MULS, one is for unsigned and the other is for signed.
Before we go any further, you must understand that there are multiple ways of answering a divide equation. For example:
10 ÷ 3
When dividing 10 into 3 parts, you could say:
1. The answer is 3.3 recurring (3.3333333... etc).
2. The answer is 3 r1 (3 remainder 1).
While the first answer is the most accurate, the second answer is valid as well. It is of course, the second answer that DIVU and DIVS will use. As far as the 68k is concerned:
10 ÷ 3 = 3 r1.
With 3 r1 being the answer, 3 is what we call the “quotient”, and 1 is what we call the “remainder” (hence the letter “r”). So now, when I say the words “quotient” or “remainder”, you’ll know what I’m talking about.
The DIVU Instruction
DIVU – Unsigned DIVide
This instruction will divide the long-word of the destination operand by the word of the source operand. The result is split into the quotient, and the remainder. The remainder is saved in the upper word of the destination operand, while the quotient is saved in the lower word of the destination operand.
If the quotient is larger than a word, the destination operand, remains unchanged.
Examples
divu.w #\$0002,d0
We’ll pretend that d0 contains 00000803 for this example. So we have the long-word of d0, divided by the word 0002.
00000803 ÷ 0002 = 0401 r0001
The answer is 401 r1 (quotient = 0401, remainder = 0001), and so d0 will now contain 00010401.
The destination operand must be a data register; the source operand however, can be an immediate number, a data register, a memory address, or even a memory address using an address register. A few examples:
divu.w d1,d0 divu.w \$00000010,d0 divu.w (a0),d0
The only thing you cannot use for the source operand is an address register directly:
divu.w a0,d0
Finally, only word can be used for the size.
The DIVS Instruction
DIVS – Signed DIVide
This is exactly the same as the DIVU instruction. Except the source and destination are both treated as “signed” instead of “unsigned”. So this instruction can divide negative numbers as well as positive.
Examples
We’ll pretend that d0 contains FFFFFFF8 in this example.
divs.w #\$0002,d0
So, the long-word FFFFFFF8 from d0 is divided by the word 0002.
FFFFFFF8 ÷ 0002 = FFFC r0000
FFFFFFF8 is treated as negative, and the quotient result FFFC is also negative. So what you really have is:
-00000008 ÷ +0002 = -00000004
And so, d0 now contains 0000FFFC (-4 r0).
Please note, if the destination operand is positive to begin with, then the remainder is also positive. If the destination operand is negative, then the remainder is also negative. Unless of course, the remainder is 0000, then it stays as... Well 0000.
Apart from the signed/unsigned differences, DIVS is the same as DIVU.
Invalid situations
The division process is relatively simple; however, there are a few situations where divide will simply not work. For example, if we pretend that d0 contains 00040000, and we perform this instruction:
divu.w #\$0004,d0
We then have:
00040000 ÷ 0004 = 10000 r0000
Notice how the quotient is 10000. The maximum a quotient result can be for unsigned is FFFF (a word size). If you get a quotient result that is higher than FFFF, then the divide instruction will not put the answer inside d0. Instead, d0 will stay as 00040000, and the 68k will continue as if nothing happened.
The same applies for signed numbers that result in; a number higher than 0000FFFF for positive, or lower than FFFF0000 for negative.
Here is another interesting situation where the divide instruction cannot divide:
divu.w #\$0000,d0
In the rules of mathematics, you cannot divide a number by 0; it is defined as impossible to calculate an answer. If you attempt to divide by 0, the 68k will halt, and run a “divide by zero exception routine” (we’ll be getting to exception routines later on in time, so don’t worry about them for now).
It would seem logical to me if they simply made it so that; if a divide by 0 occurs, make the result automatically 0. Despite it going against the rules of mathematics, from a simple programming perspective, it may very well have proved to be useful. Well, that’s my rant over. But you get the idea.
Homework
By now, you should know about:
• Signed and unsigned numbers
• Bit shifting
• Bit rotation
• Multiplication & division instructions
So, here’s a list of instructions:
move.w #\$0010,d0 mulu.w #\$0003,d0 neg.w d0 ext.l d0 asr.l #\$01,d0 ror.w #\$01,d0 ext.l d0 divu.w #\$0002,d0
d0 will start with 00000000. See if you can work out what d0 will contain at the end of this. Once again, you are free to use a hex calculator at any time. These are to ensure that you understand how the instructions work, they’re not here to test your mathematical skills. |
# CBSE Class 10 Mathematics Exam 2019: Important 4 Marks Questions with Solutions
This article brings the CBSE Class 10 Mathematics: Important 4 Marks Questions to prepare for board exams 2019. All the questions are provided with appropriate solutions to give you an idea to present answers in the best way in exam.
Jan 11, 2019 13:35 IST
CBSE Class 10 Mathematics Important 4 Marks Questions
In this article we are providing the CBSE Class 10 Mathematics: Important 4 Marks Questions to prepare for the board exam 2019. All these questions are completely solved.
In CBSE Class 10 Mathematics Board Exam 2019, Section - D will comprise 8 questions of 4 marks each. Thus almost 40 percent marks in class 10 Maths paper can be secured by attempting the the 4 marks questions correctly. So, if one wishes to excel in the examination, a good practice of 3 marks question is essential.
CBSE Class 10 Mathematics Board Exam 2018: Latest Exam Pattern
Given below are some sample questions for CBSE Class 10 Mathematics: Important 4 Marks Questions:
Q. For any positive integer n, prove that n3 – n is divisible by 6.
Solution.
n3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = product of three consecutive positive integers.
Now, we have to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer n is of the form 3q, 3q + 1 or 3q + 2 for some positive integer q.
Now three consecutive positive integers are n, n + 1, n + 2.
Case I. If n = 3q.
n(n + 1) (n + 2) = 3q(3q + 1) (3q + 2)
But we know that the product of two consecutive integers is an even integer.
∴ (3q + 1) (3q + 2) is an even integer, say 2r.
n(n + 1) (n + 2) = 3q × 2r = 6qr, which is divisible by 6.
Case II. If n = 3n + 1.
∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3)
= (even number say 2r) (3) (q + 1)
= 6r (q + 1),
which is divisible by 6.
Case III. If n = 3q + 2.
∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4)
= multiple of 6 for every q
= 6r (say),
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.
CBSE Class 10 Mathematics Exam 2018: Important 3 Marks Questions
Q. The first and the last terms of an AP are 10 and 361 respectively. If its common difference is 9 then find the number of terms and their total sum?
Sol.
Given, first term, a = 10
Last term, al = 361
And, common difference, d = 9
al = a + (n −1)d
361 = 10 + (n − 1)9
361 = 10 + 9n − 9
361 = 9n + 1
9n = 360
n = 40
Therefore, total number of terms in AP = 40
Now, sum of total number of terms of an AP is given as:
Sn = n/2 [2a + (n − 1)d]
S40 = 40/2 [2 x 10 + (40 − 1)9]
= 20[20 + 39 x 9]
=20[20 + 351]
=20 x 371 = 7420
Thus, sum of all 40 terms of AP = 7420
To get the complete set of questions, click on the following link:
All these Class 10 Mathematics: Important 4 Marks Questions have been prepared after carrying the thorough analysis of the trend followed in previous years’ Class 10 Maths CBSE question papers and the latest syllabus. Questions have been picked from the most important topics of class 10 Maths. Thus practicing these questions will help to cover the significant part of the curriculum thus making an effective preparation for the CBSE Class 10 Board Exam 2019. Moreover, the solutions provided here will give an idea to present your answers in the best way to grab maximum scores.
To get more of such useful articles for CBSE Class 10 Board Exam preparations, click on the following links:
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Question
# Two conducting spheres of radii R1 and R2 are kept widely separated from each other. What are their individual capacitances? If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series−parallel connections.
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Solution
## We need to calculate the capacitance of an isolated charged sphere. Let us assume that the charge on the sphere is Q and its radius is R. Capacitance of the charged sphere can be found by imagining a concentric sphere of infinite radius consisting of −Q charge. Potential difference between the spheres = $\frac{1}{4\pi {\in }_{0}}\frac{Q}{R}$ − 0 = $\frac{1}{4\pi {\in }_{0}}\frac{Q}{R}$ Capacitance is the ratio of the magnitude of the charge on each conductor to the potential difference between them. $C=\frac{Q}{\frac{1}{4\pi {\in }_{0}}\frac{Q}{R}}=4\pi {\in }_{0}R$ Therefore, the capacitances of spheres of radii R1 and R2 are C1 and C2, respectively. They are given by ${C}_{1}=4\pi {\in }_{0}{R}_{1}\phantom{\rule{0ex}{0ex}}{C}_{2}=4\mathrm{\pi }{\in }_{0}{R}_{2}$ If the spheres are connected by a metal wire, the charge will flow from one sphere to another till their potentials become the same. As there potentials become the same, the potential difference between the conductors for both the capacitors also becomes the same. Thus, it can be concluded that the capacitors are connected in parallel. Thus, the capacitance of the combination is given by ${C}_{\mathrm{eq}}$ = C1 + C2 $=4\mathrm{\pi }{\in }_{0}\left({R}_{1}+{R}_{2}\right)$
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Algebra Worksheets - Page 4 | Problems & Solutions
# Algebra Worksheets - Page 4
Algebra Worksheets
• Page 4
31.
The formula for the volume of a rectangular pyramid is $V$ = ($\frac{1}{3}$)$l$$w$$h$. Select a formula that would help you to find the height ($h$) of the pyramid.
a. $\frac{3V}{l}$ units b. $\frac{V}{lw}$ units c. $\frac{3V}{w}$ units d. $\frac{3V}{lw}$ units
#### Solution:
V = (13) × lwh
[Formula.]
3V = (13) × lwh × (3)
[Multiply by 3 on each side.]
3V = lwh
[Simplify.]
3Vlw = lwhlw
[Divide each side by lw.]
3Vlw = h
[Simplify.]
So, the height of the rectangular pyramid is 3Vlw units.
32.
The formula F = $\frac{9}{5}$C + 32 is used to convert temperature from Celsius (C) scale to Fahrenheit (F) scale. Which of the following choices best suits the conversion of temperature from Fahrenheit scale to Celsius scale?
a. C = $\frac{5}{9}$(F - 12) b. C = $\frac{5}{9}$(F + 32) c. C = $\frac{5}{9}$(F - 22) d. C = $\frac{5}{9}$(F - 32)
#### Solution:
F = 95C + 32
[Formula.]
F - 32 = (95) × C + 32 - 32
[Subtract 32 from each side.]
F - 32 = (95) × C
[Simplify.]
5(F - 32) = 5 × (95) × C
[Multiply by 5 on each side.]
5(F - 32) = 9C
[Simplify.]
59(F - 32) = 9C9
[Divide by 9 on each side.]
59(F - 32) = C
[Simplify.]
The formula C = 5 / 9(F - 32) is used to convert temperature from Fahrenheit scale to Celsius scale.
33.
The formula S = $\frac{n\left(n+1\right)}{2}$ is used to find the sum of the first $n$ natural numbers. Find the sum of the first 14 natural numbers.
a. 105 b. 112 c. 98 d. 119
#### Solution:
S = n(n+1)2
[Formula.]
S = 14(14+1)2
[Replace n with 14.]
S = 14(15)2
S = 7(15)
[Divide.]
S = 105
[Multiply.]
The sum of the first 14 natural numbers is 105.
34.
Total surface area of a cube is given by the formula, L = 6$s$2, where $s$ is the side of the cube. Find the side of the cube, if the total surface area of the cube is known.
a. $\sqrt{6L}$ b. $\sqrt{\frac{L}{6}}$ c. $\sqrt{\frac{L}{2}}$ d. 2$\sqrt{L}$
#### Solution:
L = 6s2
L / 6 = s2
[Divide both sides by 6.]
L6 = s2
[Take square root on both sides.]
L6 = s
So, the side of the cube = L6
35.
Find the base of a triangle, if area of the triangle A and height $h$ are known.
a. $\frac{h}{2A}$ b. $\frac{2h}{A}$ c. $\frac{2A}{h}$ d. 2A$h$
#### Solution:
Area of the triangle with base b and height h is A = 1 / 2 × bh.
2A = bh
[Multiply by 2 on both sides.]
2Ah = b
[Divide both sides by h.]
So, the base of the triangle is 2Ah. |
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Maria and Nate graphed lines on a coordinate plane. Maria's line is represented by the equation y = -5/6x +8. Nate's line is perpendicular to Maria's line. Which of the following could be an equation for Nate's line?
a 5x-6y=15
b 5x+6y=15
c 6x-5y=15
d 6x+5y=15
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c) 6x - 5y = 15
Step-by-step explanation:
Slope-intercept form of a linear equation: $$y=mx+b$$
(where m is the slope and b is the y-intercept)
Maria's line: $$y=-\dfrac{5}{6}x+8$$
Therefore, the slope of Maria's line is $$-\frac{5}{6}$$
If two lines are perpendicular to each other, the product of their slopes will be -1.
Therefore, the slope of Nate's line (m) is:
\begin{aligned}\implies m \times -\dfrac{5}{6} &=-1\\m & =\dfrac{6}{5}\end{aligned}
Therefore, the linear equation of Nate's line is:
$$y=\dfrac{6}{5}x+b\quad\textsf{(where b is some constant)}$$
Rearranging this to standard form:
$$\implies y=\dfrac{6}{5}x+b$$
$$\implies 5y=6x+5b$$
$$\implies 6x-5y=-5b$$
Therefore, option c could be an equation for Nate's line.
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# Liverpoololympia.com
Just clear tips for every day
# What are the basic formulas for geometry?
## What are the basic formulas for geometry?
Basic geometry formulas where the mathematical constant π is used are,
• Area of a Circle = A = π×r.
• Circumference of a Circle = 2πr.
• The curved surface area of a Cylinder = 2πrh.
• Total surface area of a Cylinder = 2πr(r + h)
• Volume of a Cylinder = V = πr2h.
• The curved surface area of a cone = πrl.
What are the important formulas?
Algebra
• Completing the square: x2+bx+c=(x+b2)2−b24+c.
• Quadratic formula: the roots of ax2+bx+c are −b±√b2−4ac2a.
• Circle: circumference=2πr, area=πr2.
• Sphere: vol=4πr3/3, surface area=4πr2.
• Cylinder: vol=πr2h, lateral area=2πrh, total surface area=2πrh+2πr2.
### Does geometry have a formula?
Geometry Formulas are crucial elements to calculate the perimeter, length, area, and volume of geometric shapes and figures. There are loads of geometric formulas that are concerned with height, length, width, radius, perimeter, area, volume or surface area.
What is a geometry calculation?
Geometry calculations often involve determining the perimeter and area of polygons and the volume of solid figures. Perimeter measures the length around a flat shape while area measures the surface of the shape. Volume measures the capacity of a solid figure.
## What is famous math formula?
Whether one is versed in mathematics or physics, or knows nothing of the vocabulary of maths, everyone knows Albert Einstein’s famous formula: E = mc².
What is K in geometry?
Since k is constant (the same for every point), we can find k when given any point by dividing the y-coordinate by the x-coordinate. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = = 3. Thus, the equation describing this direct variation is y = 3x.
### What is the best app for geometry?
Best Geometry Apps
• Shapes Match Game.
• Dragon Shapes: Geometry Challenge.
• CyberChase Shape Quest.
• DragonBox Elements.
• Math 8: Talk math with Leon!
• SAT Math : Geometry and Measurement Lite.
• GeoBoard Game.
• King of Maths.
How many math formulas are there?
Class 6 to 12 Formulas….List of Maths Formulas (for All Concepts)
2cosacosb Formula 30-60-90 Formulas
Inverse Matrix Formula Inverse Tangent Formula
Inverse Trigonometric Formulas Inverse Variation Formula
Isosceles Trapezoid Formula Isosceles Triangle Perimeter Formula
Lagrange Interpolation Formula Lateral Area Formula
## What is J in math?
J letter is used to represent the imaginary number. The imaginary number, root of -1 is represented by ‘i’ in mathematics where in electrical engineering ‘j’ is used for the same. Imaginary numbers are the complex numbers when squared gives a negative result value.
What is H in geometry?
h: h is an abbreviation for height. Height: The distance from bottom to top. Heptagon: A polygon that has seven sides.
### Is there a geometry calculator?
Calculators covering formulas for standard 2D plane and 3D solid geometric shapes and trigonometric functions. |
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# Check if the value of $\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \ldots \ldots + \infty } \right)$ is $\dfrac{1}{2}$
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Hint: Try to take out common terms and make the series simpler to solve.
Given series: $\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \ldots \ldots + \infty } \right)$
$\Rightarrow \dfrac{1}{4}\left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right]{\text{ }} \ldots \left( 1 \right)$
Here, we assume the above equation as the sum of terms in a Geometric Progression with $a = 1,{\text{ }}r = \dfrac{{\dfrac{1}{2}}}{1} = 1$ .
Now, applying the sum of a Geometric Progression:
$\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{a}{{1 - r}} \\ \Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{1}{{\left( {1 - \dfrac{1}{2}} \right)}} \\ \Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{1}{{\left( {\dfrac{1}{2}} \right)}} \\ \Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = 2 \\$
Putting this value in equation (1), we get
$\Rightarrow \dfrac{1}{4}\left( 2 \right) \\ \Rightarrow \dfrac{1}{2} \\$
Note: Whenever we are supposed to find the sum of a series, always try to make the series in the form of Arithmetic Progression or Geometric Progression or Harmonic Progression and then simply use the sum of series formula which is already defined for these progressions. |
Working with Linear Functions
# Working with Linear Functions: Finding a New Point, Given a Point and a Slope
Suggested review of lines from the Algebra I curriculum:
Recall:
\begin{align} &\cssId{s7}{\text{slope of line through points } (x_1,y_1) \text{ and } (x_2,y_2)}\cr\cr &\qquad \cssId{s8}{=\ \ m}\cr\cr &\qquad \cssId{s9}{=\ \ \frac{\text{rise}}{\text{run}}}\cr\cr &\qquad\cssId{s10}{=\ \ \frac{\text{change in } y}{\text{change in } x}}\cr\cr &\qquad\cssId{s11}{=\ \ \frac{\Delta y}{\Delta x}}\cr\cr &\qquad\cssId{s12}{=\ \ \frac{y_2 - y_1}{x_2 - x_1}} \end{align}
A common scenario is (see diagram below):
• You have a line with known slope $\,m\,.$
• You know the coordinates of one point on the line. Call this known point $(x_{\text{old}},y_{\text{old}})\,.$
• There is another point on the line whose coordinates are needed. Call this desired point $(x_{\text{new}},y_{\text{new}})\,.$
• You know the change in $\,x\,$ between the old and new point: $$\cssId{s20}{\Delta x = x_{\text{new}} - x_{\text{old}}}$$
$$\begin{gather} \cssId{s21}{\Delta x \gt 0}\cr \cssId{s22}{\text{is equivalent to}}\cr \cssId{s23}{\text{the new point lies to the right of the old point}}\cr\cr\cr \cssId{s24}{\Delta x \lt 0}\cr \cssId{s25}{\text{is equivalent to}}\cr \cssId{s26}{\text{the new point lies to the left of the old point}} \end{gather}$$
• You want the $y$-coordinate, $\,y_{\text{new}}\,,$ of the new point.
• Solving for $\,y_{\text{new}}\,$ in terms of the known quantities: \begin{align} &\cssId{s29}{m = \frac{y_{\text{new}} - y_{\text{old}}}{x_{\text{new}} - x_{\text{old}}}}\cr &\qquad \cssId{s30}{\Rightarrow}\ \ \cssId{s31}{y_{\text{new}} - y_{\text{old}} = m \overbrace{(x_{\text{new}} - x_{\text{old}})}^{\Delta x}}\cr\cr &\qquad \cssId{s32}{\Rightarrow}\ \ \cssId{s33}{y_{\text{new}} = y_{\text{old}} + \overbrace{m\Delta x}^{\Delta y}} \end{align}
## Example
You have a known point $\,(1,-5)\,$ on a line with slope $\,7.5\,.$ When $\,x = 1.4\,,$ what is the $y$-value of the point on the line?
Solution: The change in $\,x\,$ in going from the known point ($\,x = 1\,$) to the new point ($\,x = 1.4\,$) is:
$$\cssId{s39}{\Delta x = 1.4 - 1 = 0.4}$$ \begin{align} \cssId{s40}{y_{\text{new}}}\ &\cssId{s41}{= y_{\text{old}} + m\Delta x}\cr &\cssId{s42}{= -5 + (7.5)(0.4)}\cr &\cssId{s43}{= -2} \end{align} |
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# Class 12 Maths NCERT Solutions of exercise 7.3 of the chapter 7-Integrals
Class 12 Maths NCERT solutions of exercise 7.3 of the chapter 7-Integrals are the solutions of exercise 7.3 of the chapter 7-Integrals of class 12 NCERT maths textbook. Class 12 Maths NCERT Solutions of exercise 7.3 of the chapter 7-Integrals are compulsory to be studied for every maths student of class 12 for clearing the concept of the methods used for solving the questions based on the integration of the functions. In this exercise 7.3, you will study the integration of different types of complex trigonometric functions. The solutions of exercise 7.3 are required the inputs of class 11 chapter 3- trigonometric functions because for simplification of complex trigonometric functions the trigonometric identities are required to use for the integration of the functions.
As an example, we can not integrate the function directly Cos²2x, so using the trigometric identity cosA = 2cos²A – 1⇒ cos²A = (1+cosA)/2, so cos²2x = (1+cos4x)/2
Now, integrating the simplified function ((1+cos4x)/2), we have
1/2∫(1+cos4x)dx = 1/2[∫1dx +∫cos4xdx] = x/2 +(1/2)(sin4x)/4 +C= x/4 + (1/8)sin4x +C
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## Class 12 Maths NCERT solutions of Chapter 7 Integrals
Exercise 7.1- Integrals
Exercise 7.2-Integral
Exercise 7.3-Integrals
Exercise 7.4 -Integral
## Class 12 Maths NCERT Solutions of exercise 7.3
Q1.Integrate sin²(2x +5)
Ans.We are given the function
sin²(2x +5)
Applying the trigonometric identity
cos2A=1 – 2sin²A
Now, integrating it
Integrating each terms individually
Q2.Integrate sin3x cos4x.
Using the trigonometric identity
2sinAcosB = sin(A+B)+ sin (A-B)
Now, integrating it
Integrating each term individually
Q3.Integrate cos2x cos4x cos6x
Ans. We are given the function
cos2x cos4x cos6x
Using the trigonometric identity
Now, integrating the given function
Again,applying the identities cos²2x= (1 + cos4x)/2 and
Q4.Integrate sin³(2x +1)
Ans. We are given the function
sin³(2x +1)
Integrating the function
Rewriting the given function
Applying the trigonometric identity
sin²(2x+1) = 1 – cos²(2x+1)
Using the substitution method
Let t = cos(2x+1)
Substituting cos(2x+1) =t and sin(2x+1)dx=-dt/2
Now,substituting back the value of t= cos(2x+1)
Q5.cos³xsin³x
Ans. We are given
cos³xsin³x
Integrating it
Using the identy sin²x = 1-cos²x
Let t= cosx
-dt = sinx dx
Substituting cosx and sinxdx by t and -dt respectively
Substituting back the value of t= cosx
Q6.sinx sin 2x sin3x
Ans.We are given the function
sinx sin 2x sin3x
Integrating it
Using the trigonometric identity
Using cos(-x) = cosx
Substituting the value of sin2x sin3x
Using the identity 2sinx cosx = sin2x, and sinx cos5x = 1/2[sin(x+5x) +sin(x-5x)], we get
Using sin(x-5x) = sin(-4x) = -sin4x
### Class 12 Maths NCERT Solutions of exercise 7.3 of the chapter 7-Integrals
Q7.Integrate sin4x sin8x
Ans. We are given
sin4x sin8x
Integrating it
Using the trigonometric identity
sinA.sinB = 1/2(cos(A-B) -cos(A+B)
Q8. Integrate
Ans. We are given
Applying the identity 1-cosx = 2sin²x/2 and 1+cox = 2cos²x/2
Also applying tan²x/2 =sec²x/2 – 1
Now, integrating it
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The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.
## NCERT Solutions of Science and Maths for Class 9,10,11 and 12
### NCERT Solutions for class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
### NCERT Solutions for class 11 maths
Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
### NCERT Solutions for Class 11 Physics
Chapter 1- Physical World
chapter 3-Motion in a Straight Line
### NCERT Solutions for Class 11 Chemistry
Chapter 1-Some basic concepts of chemistry
Chapter 2- Structure of Atom
### NCERT Solutions for Class 11 Biology
Chapter 1 -Living World
### NCERT solutions for class 12 maths
Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution
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Today, let’s dig into sequences on the GMAT. Let’s first understand what a sequence is (from Wikipedia):
A sequence is an ordered list of objects. The number of terms it contains (possibly infinite) is called the length of the sequence. Unlike a set, order matters in a sequence, and exactly the same elements can appear multiple times at different positions in the sequence. Since order matters, (A, B, C) and (B, C, A) are two different sequences. (A series is the sum of the terms of a sequence but we will not deal with series today.)
There are some special sequences e.g. arithmetic progressions and geometric progressions. We will deal with these in subsequent weeks. Today we will look at some generic sequence questions and will learn how to approach them. I will start with a very basic question. Mind you, most sequence questions will be higher level questions since sequence questions look complicated (even though they are very straight forward, believe me!). Let me show you using some questions from external sources:
A note on notation: The first term of a sequence will be denoted by x(1), second term by x(2) and nth term by x(n). (If I want to show multiplication e.g. multiply x by 2, I will show it by writing x*2)
Question 1: In a certain sequence, the term x(n) is given by the formula x(n) = 2*x(n-1) – (1/2)*x(n-2) for all n>= 2. If x(0) = 3 and x(1) = 2, what is the value of x(3)?
(A) 2.5
(B) 3.125
(C) 4
(D) 5
(E) 6.75
Solution: This is a straight forward question event though the formula given is discomforting. Whenever you have a generic formula for the nth term of a sequence, plug in some numbers to see what pattern you get.
x(0) = 3 (given)
x(1) = 2 (given)
If n = 2, x(2) = 2*x(1) – (1/2)*x(0) = 2*2 – (1/2)*3 = 5/2
If n = 3, x(3) = 2*x(2) – (1/2)*x(1) = 2*(5/2) – (1/2)*2 = 4
I hope you agree that this was very simple. For x(3), you needed x(2). For x(2), you needed x(1) and x(0), both of which you had! So it was a simple matter of quick substitution. Now let’s look a teeny bit complicated question
Question 2: The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2) = -2, a(3) = 6, a(4) = -1, and a(n) = a(n-4) for n > 4. If T = a(10) + a(11) + a(12) + … a(84) + a(85), what is the value of T?
(A) 119
(B) 120
(C) 121
(D) 126
(E) 133
Solution: We know the first four terms: a(1) = 4, a(2) = -2, a(3) = 6, a(4) = -1
Also it is given that a(n) = a(n-4) i.e. the nth term is equal to the (n-4)th term e.g. 5th term is equal to the 1st term. 6th term is equal to the 2nd term. 7th term is equal to the 3rd term etc.
Hence, the sequence becomes: 4, -2, 6, -1, 4, -2, 6, -1, 4, -2, 6, -1 … (It is always helpful to write down the first few terms of the sequence. It helps you see the pattern.)
The sequence has a cyclicity of 4 i.e. the terms repeat after every 4 terms (go back to the definition of sequence above – it says that the same element can appear multiple times at different positions). Therefore, first to fourth terms will form the first cycle, fifth to eighth terms will form the second cycle, ninth to twelfth terms will form the third cycle and so on…
The sum of each group of 4 terms = 4 – 2 + 6 – 1 = 7
What will be the tenth term, a(10)? A new cycle starts from a(9) so a(9) = 4. Then, a(10) must be -2.
a(10) + a(11) + a(12) is the sum of last three terms of a cycle so this sum must be – 2 + 6 – 1 = 3
a(13) to a(16) is a complete cycle, a(17) to a(20) is another complete cycle and so on… The sum of each of the complete cycles is 7.
How many such complete cycles will there be? The first complete cycle will end at a(16), the second one at a(20), the third one at a(24) etc (i.e. at multiples of 4). The last complete cycle will end at a(84). How many complete cycles do we have here then?
16 = 4*4 and 84 = 4*21 so you start from the fourth multiple to the 21st multiple i.e. you have (21 – 4 + 1) = 18 total cycles. If you are confused about the ‘+1’ here, hang on – I will take it up at the end of this post.
The sum of these 18 cycles will be 7*18 = 126 (I know the multiplication table of 18 as should you!)
We still haven’t accounted for a(85), which will be the first term of the next cycle. The first term is 4.
a(10) + a(11) + a(12) + … a(84) + a(85) = 3 + 126 + 4 = 133 = T
or you could just consider this:
You have 18 complete cycles except for the first 3 terms and the last term of the sequence. The last term of the sequence is the first term of a cycle and the first three terms of the sequence are the last three terms of the cycle. So these four terms make one complete cycle. Therefore, instead of 18, you have 19 complete cycles.
T = 7 * 19 = 133
These were some basic sequences questions. I want to leave you with a sequence question from GMAT prep now. Try and work it out. We will look at its solution next week.
Question 3: For every integer m from 1 to 10 inclusive, the mth term of a certain sequence is given by [(-1)^(m+1)]*[(1/2)^m]. If T is the sum of the first 10 terms in the sequence, then T is:
(A) greater than 2
(B)between 1 and 2
(C) between 0.5 and 1
(D) between 0.25 and 0.5
(E) less than 0.25
Note on the ‘+1’ above: How many numbers are there from 11 to 25, both inclusive? If your answer is 25 – 11 = 14, then you are wrong. When we say 25 – 11, we are saying that we have 25 numbers and we are throwing away 11 of them. But we want to keep the 11 (since we have both inclusive); we want all the numbers starting from 11 and ending at 25. We need to add 1 to the result to ensure that the 11 that we threw out, is retained. Consequently, the number of numbers from 11 to 25, both inclusive is 14 + 1 = 15.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
### 2 Responses
1. Huyen Le says:
Hi Karishma,
I have been following you and thank you so much for many approaches and methods I learned from you so far.
I have a question. Regarding question 2, after calculating the sum of a group of 4 ( 1 cycle – 7), I calculated the no of cycles ( 85-10/1+1)/4 = 19. => The sum = 19*7
Is this approach correct?
• Karishma says:
Yes, absolutely!
Look at the note above:
“or you could just consider this:
You have 18 complete cycles except for the first 3 terms and the last term of the sequence. The last term of the sequence is the first term of a cycle and the first three terms of the sequence are the last three terms of the cycle. So these four terms make one complete cycle. Therefore, instead of 18, you have 19 complete cycles.”
What you have done is just the mathematical interpretation of this tip. |
FAQ: How To Say 3/4?
What is 3/4th called?
3/4 or ¾ may refer to: The fraction (mathematics) three quarters (3⁄4) equal to 0.75.
How do you pronounce the fraction of a number?
Basically, there are three ways to pronounce a fraction. Let’s take 2/3 as an example. We say it as: two-thirds, two over three, and two divided by three. “Over” is used in a more casual manner or for larger and complex numbers.
What does it mean when you say 3 4?
Whole fractions For example, 3/4 means you have three parts out of four parts total. If the top number and the bottom number of a fraction are the same, then the fraction is equal to 1. That’s because you have every part of the fraction, or one whole thing. This is sometimes known as a whole fraction.
How do you say 3/4 in English?
You call 3/4 ” three fourths” or “three quarters”, and 3/5 “three fifths”. Just to point out – “Three fourths” is quite an American way of saying it.
How do you write 1 3rd?
Most people will write it down as 0.33,0.333,0.3333, etc. In practice use 13 as 0.333 or 0.33, depending on the level of accuracy required. 13 is exact and therefore accurate.
You might be interested: Question: How To Say Names?
What is 1/4th called?
When a whole is divided into 4 equal parts, and each part is called one-quarter. One-quarter is one of four equal parts. It is written as 14. It is read as one-quarter or one-fourth.
Can 3 be divided by 4?
We can write 3 divided by 4 as 3/4. Since 3 is a prime number and 4 is an even number. Therefore, the GCF or the greatest common factor of 3 and 4 is 1. So, to simplify the fraction and reduce it to its simplest form we will divide both numerator and denominator by 1.
What is 3/4 as a percentage?
Answer: 3/4 is expressed as 75% in terms of percentage.
How do you say 3/8 in English?
New Member. Actually, it’s ” three-eighths of an inch.”
How do you express fractions?
To express the fraction in words, write the numerator, add a hyphen and then spell out the denominator. In word form, the fraction 3/10 would be spelled out as three-tenths.
Is three fourths the same as three quarters?
Three-fourths is the same as three-quarters. |
# A deck of cards is well shuffled. The cards are dealt one by one, until the first time an ace appears
A deck of cards is well shuffled. The cards are dealt one by one, until the first time an ace appears.
1. Find the probability that no kings, queens or jacks appear before the first ace.
2. Find the probability that exactly one king, exactly one queen and exactly one jack appear (in any order) before the ace first.
## One: Find the probability that no kings, queens or jacks appear before the first ace.
Notation:
• Let b cards precede the first ace
• Let A be the first Ace
• Let a cards follow the first ace
Conditions:
• b will range from 0 to 36 ($52 - (12 + 4)$)
• A will be one of four Ace cards
• a will range from 51 to 15 ($12 + 3$)
In order to maintain equal likeliness of all configurations (an assumption), we will find the number of valid permutations and divide by the total number of permutations which will result in the probability of a valid configuration, denoted P(A).
$$P(A) = \frac{\textrm{number of valid permutations}}{\textrm{number of all permutations}}$$
$$P(A) = \frac{1}{52!}\sum_{i=0}^{36}\left(\frac{36!}{(36-i)!}\frac{4!}{(4-1)!}(51-i)!\right)$$ where within the summation, the first expression selects and permutes the number of cards preceding the first ace, the second expression selects the first ace, the third expression selects and permutes the cards which follow the first ace.
The expression reduces to: $$P(A)=\frac{4\times36!}{52!}\sum_{i=0}^{36} \frac{(51-i)!}{(36-i)!}$$ $$P(A) = 0.25$$
## Two: Find the probability that exactly one king, exactly one queen and exactly one jack appear (in any order) before the ace first.
There are $\frac{4!}{(4-1)!}$ ways to select one of four elements, be it an ace, jack, queen or king.
Proceed in a similar fashion to part 1. by finding the quotient of the total number of permissible permutations and the total number of permutations, labelling it $P(B)$.
$$P(B) = \frac{1}{52!} \sum_{i=0}^{36}\left(\frac{4!}{(4-1)!}\frac{4!}{(4-1)!}\frac{4!}{(4-1)!}\frac{1}{3!}\times\frac{36!}{(36-i)!}\frac{4!}{(4-1)!}(48-i)! \right)$$ Where the first three expressions in the summation are the selection of a king, queen and jack, the fourth expression allows these three cards to be chosen in any order. The expressions following the multiplication symbol are:
• selection of a number of cards ( two, three, ... , ten) to follow the face cards and precede the first ace.
• selection of the first ace
• selection of the remaining cards
The expression collapses to: $$P(B) = \frac{4^4 \times 36!}{52!\times 3!} \sum_{i=0}^{36}\frac{(48-i)!}{(36-i)!}$$ $$P(B) = 2.4752\times{10}^{-5}$$
• This seems all well and good, but perhaps you might find it easier to work with a smaller deck. The only cards that matter at all are the aces and face cards, so how about throwing out all other cards from the deck. After all, their location in the deck doesn't affect the events we are interested in. This leaves a deck of just $16$ cards with four aces, four jacks, four queens, and four kings. What is the probability the first card is an ace? Do you see why this probability should be the same as the probability you tried to calculate in the first problem above? – JMoravitz Feb 22 '17 at 17:41
• Your first answer is overkill, however, because you can rephrase the question as "What is the probability that the first honor card (ace,king,queen, or jack) is an ace." Then it is clearly 1/4 - each is equally likely – Thomas Andrews Feb 22 '17 at 17:52
• Others have already said most of what I was going to say, but on your final calculation, first you should be multiplying by $3!$ not dividing, and you also need to take into account that the previous face cards, and the other cards before the ace, can be chosen in any order. This adds a binomial coefficient which complicates things further, @JMoravitz recommendation is one that I would second. – Pepe Silvia Feb 22 '17 at 18:01
• Thank you all. Your comments have helped tremendously. – DWD Feb 22 '17 at 20:14
• @DipeshGupta I think I did it computationally in Python3 or WXMaxima – DWD Sep 20 '17 at 19:31
You can rephrase the question as "What is the probability that the first honor card (ace,king,queen, or jack) is an ace." Then it is clearly 1/4 - each is equally likely.
Note that all the probabilities are the same if you remove the cards not in question - so that you can think of it as a deck of 16 cards. There are $16!$ orders for the deck. There are $3!4^4$ different ways to have the first four honors be all different and the fourth be an ace.
So the probability for the second case is:
$$\frac{3!4^412!}{16!}\approx 0.035$$
Which is significantly higher than your probability.
Did you assume that there were no cards other than the king, queen, and jack before the ace? Then you need all 52 cards, and you'd get:
$$\frac{3!4^448!}{52!}=\frac{3!4^4}{52\cdot 51\cdot 50\cdot 49}\approx 0.000236$$
Which is closer to your result, but still different. |
### Two Cubes
Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to prove you have found all possible solutions.]
### Common Divisor
Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.
### Novemberish
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
##### Stage: 4 Challenge Level:
We received a number of good solutions to this problem. Most of you noticed that the key to making rectangles that work for all bases is factorising quadratic equations. Luke from London Oratory School gave a good explanation of this:
The green section is a square, so its area is equal to $x^2$.
The red section consists of two rectangles with dimensions $a, x$ and $b, x$. Therefore the
red area is equal to $ax+bx$, which equals $x(a+b)$.
We can see that the area of the blue section will always have dimensions $a$ and $b$, so its area is equal to $ab$, if it ʻfillsʼ the gap created by the red area.
The total area is equal to the sum of these component areas.
Thus you can make a rectangle for all bases for equations of the form $x^2 + x(a+b) + ab$ where $a$ and $b$ are positive integers. The rectangle has dimensions $(x+a)$ by $(x+b)$.
So for the rectangle $x^2 + 7x + 12$, we can see that 1) $7=a+b$ and 2) $12=ab$.
$a=3 , b=4$ satisfies these equations.
So for any base we can make the rectangle with dimensions $(x+3)$ by $(x+4)$.
Vanessa and Annie sent us this solution:
Using 1 square and 12 sticks: You can make 7 different rectangles which work in all bases. These are:
$x(x+12)$, $(x+1)(x+11)$, $(x+2)(x+10)$, $(x+3)(x+9)$, $(x+4)(x+8)$, $(x+5)(x+7)$, $(x+6)(x+6)$.
You can make rectangles for all bases with dimensions $(x+A)(x+B)$ where $A$ and $B$ are positive integers and $A+B=12$.
Using the same logic, if you have one square and 100 sticks, you can make 51 rectangles that work for all bases:
$x(x+100)$, $(x+1)(x+99)$, $(x+2)(x+98)$, ..., $(x+50)(x+50)$.
Using 1 square and 12 units: You can make 3 different rectangles which work in all bases:
$(x+1)(x+12)$, $(x+2)(x+6)$, $(x+3)(x+4)$.
You can make rectangles for all bases with dimensions $(x+A)(x+B)$ where $A$ and $B$ are positive integers and $AB=12$.
Using the same logic, if you have one square and 100 sticks, you can make 5 rectangles which work in all bases: $(x+1)(x+100)$, $(x+2)(x+50)$, $(x+4)(x+25)$, $(x+5)(x+20)$, $(x+10)(x+10)$.
Using one square, $p$ sticks, $q$ units, you can only make a rectangle which works for all bases if $p=A+B$ and $q=AB$ for $A$ and $B$ positive integers.
Extension: Using $n$ squares, we can make rectangles which work in all bases of dimensions $(ax+b)(cx+d)$ when $a$,$b$,$c$,$d$ are positive integers and $ac=n$. Then we will use $ad+bc$ sticks and $bd$ units. This will give rectangles which have the squares arranged in a rectangle. But we could arrange the squares in different ways, like L-shapes.
Well done! |
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# Number Systems (Representation)
## Introduction and Inter-conversion of
– Binary
“There are 10 types of
– Octal
people, those who
– Decimal understand binary and
– BCD
those who don’t”
## Binary Addition, Subtraction (Complement's method)
Common Number systems
## Decimal: -> base10
-> 10 symbols (0,1,2,3,4,5,6,7,8,9)
## Binary: -> base2
-> 2 symbols (0,1)
-> 16 symbols
(0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F)
Decimal Numbers
When we write decimal (base 10) numbers, we use a positional notation system. Each
digit is multiplied by an appropriate power of 10 depending on its position in the number:
For example:
843 = 8 x 102 + 4 x 101 + 3 x 100
= 8 x 100 + 4 x 10 + 3 x 1
= 800 + 40 + 3
For whole numbers, the rightmost digit position is the one’s position (100 = 1). The
numeral in that position indicates how many ones are present in the number.
The next position to the left is ten’s, then hundred’s, thousand’s, and so on. Each digit
position has a weight that is ten times the weight of the position to its right.
In the decimal number system, ten possible values can appear in each digit position,
The decimal numerals are the familiar zero through nine (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).
In a positional notation system, the number base is called the radix. Thus, the base ten
system that we normally use has a radix of 10. Thus, in a case where the radix isn’t
understood, decimal numbers would be written like this:
## 12710 1110 567310
BINARY NUMBER SYSTEM
The binary number system is also a positional notation numbering system, but in this
case, the base is not ten(10), but is instead two(2).
In the binary number system, there are only two possible values (i.e. 0 and 1) that can
appear in each digit position rather than the ten that can appear in a decimal number.
Each digit position in a binary number represents a power of two. So, each binary digit
is multiplied by an appropriate power of 2 based on the position in the number:
For example:
101101 = 1 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20
= 1 x 32 + 0 x 16 + 1 x 8 + 1 x 4 + 0 x 2 + 1 x 1
= 32 + 8 + 4 + 1
= 45
The term ‘bit’ is a contraction of the words ‘binary’ and ‘digit’, and the number in the
above example is a 6 bit number.
## 1011012 112 101102
Conversion between Binary to Decimal
Converting a number from binary to decimal is quite easy. All that is required is to find
the decimal value of each binary digit position containing a 1 and add them up.
## example: to convert 101102 to decimal.
Decimal Hex Binary
10110 0 0 0000
\ \ \___________1 x 21 =2 1 1 0001
\ \_____________1 x 22 =4 2 2 0010
\_______________1 x 24 = 16 3 3 0011
--------- 4 4 0100
22
5 5 0101
## example: to convert 110112 to decimal 6 6 0110
11011 7 7 0111
\ \ \ \___________1 x 20 =1 8 8 1000
\ \ \____________1 x 21 =2 9 9 1001
\ \______________1 x 23 = 8 10 A 1010
\________________1 x 24 = 16
11 B 1011
------------------
27 12 C 1100
13 D 1101
14 E 1110
15 F 1111
Conversion between Decimal and Binary
The method for converting a decimal number to binary involves using successive
division by the radix until the dividend reaches 0.
At each division, the remainder provides a digit of the converted number, starting with
the least significant digit.
## 37 / 2=18 remainder 1 (least significant digit)
18 / 2 = 9 remainder 0
9 / 2=4 remainder 1
4 / 2=2 remainder 0
2 / 2=1 remainder 0
## The resulting binary number is: 1001012
Conversion between Decimal and Binary
## 93 / 2 = 46 remainder 1 (least significant digit)
46 / 2 = 23 remainder 0
23 / 2 = 11 remainder 1
11 / 2 = 5 remainder 1
5/2=2 remainder 1
2/2=1 remainder 0
## The resulting binary number is: 1011101
Another number system used in digital systems with base 16 is called hexadecimal, and
each digit position represents a power of 16.
For any number base greater than ten, a problem occurs because there are more than
ten symbols needed to represent the numerals for that number base.
It is customary in these cases to use the ten decimal numerals followed by the letters of
the alphabet beginning with A to provide the needed numerals.
Since the hexadecimal system is base 16, there are sixteen numerals required. The
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
## The following are some examples of hexadecimal numbers:
1016 4716 3FA16 A03F16
The reason for the common use of hexadecimal numbers is the relationship between
the numbers 2 and 16. Sixteen is a power of 2 (16 = 24).
Because of this relationship, four digits in a binary number can be represented with a
single hexadecimal digit. Hence, hexadecimal can be used to write large binary
numbers with much fewer digits.
By using hexadecimal, the numbers can be written with fewer digits and much less likelihood of
error.
To convert a binary number to hexadecimal, divide it into groups of four digits starting with the
rightmost digit.
If the number of digits isn’t a multiple of 4, prefix the number with 0’s so that each group contains
4 digits.
For each four digit group, convert the 4 bit binary number into an equivalent hexadecimal digit.
(See the Binary, BCD, and Hexadecimal Number Tables at the end of this Slide)
## Divide into groups for 4 digits 1011 0101
Convert each group to hex digit i.e. B 5 = B516
## Another example: Convert the binary number 0110101110001100 to hexadecimal
Divide into groups of 4 digits 0110 1011 1000 1100 Convert each
group to hex digit 6 B 8 C = 6B8C16
To convert a hexa-decimal number to a binary number, convert each hexadecimal digit into a
group of 4 binary digits.
Example: Convert the hex number 374F16 into binary
3 7 4 F
Convert the hex digits to binary 0011 0111 0100 1111
00110111010011112
TRY THIS. Convert the following hexadecimal numbers to their decimal equivalents:
(a) 2 4 3 F16
(b) B E E F16
TRY THIS: Convert the following decimal numbers to their hexadecimal equivalents:
(a) 6949810
(b) 11426710
Decimal to Hex
## Convert Decimal to hex number: E.g. 250010 = ?????16
Work right to left and keep dividing by 16 and writing down the remainder.
## Step1: 2500/16 = 156 remainder 4 = (416)
Step2: 156/16 = 9 remainder 12 = (C16)
Step3: 9/16 =0 remainder 9 = (916 )
## Converting Hex to Decimal E.g. BAD16 = ?????10
keep multiplying by descending power of 16
## B A D = B x 162 + A x 161 + D x 160 [160 = 1]
= 11 x 256 + 10 x 16 + 13 x 1
= 298910
Number Systems: EXTRA RESOURCES
To use this table, simply follow the directions used in this example − Add 4A616 and
1B316. Locate 4, A & 6 in the X column then correspondingly locate 1, B & 3 in the Y
column. The point in 'sum' area where these two columns intersect is the sum of two
numbers.
1) 4A616 + 1B316 = ?????16
1
4 A 616 = 1 1 9 010
+ 1 B 316 = + 4 3 510
------------ ----------------
6 5 916 = 1 6 2 510
## 2) (B A 3)16 + (5 D E)16 = ????16
1 1 3 + E = 11
BA3 A + D = 17
+5 D E 17 + 1 (carry) = 18
---------------- B + 5 = 10
1181 10 + 1 (carry) = 11
## Hence the required sum is 1181 in hexadecimal.
Number Systems: EXTRA RESOURCES
1. Add one column at a time.
2. Convert to decimal and add the numbers.
3 (a). If the result of step two is 16 or larger subtract the result from 16 and carry 1 to
the next column.
3 (b). If the result of step two is less than 16, convert the number to hexadecimal.
1 1 1
A C 5 A 9 (9 + 4 = 13 < than 16 hence use ‘D’)
+ B D 6 9 4 (A + 9 = 19 > than OR =16 then 19 – 16 = 3 carry a 1)
--------------------- (1 + 5 + 6 = 12 < than 16 hence use ‘C’)
1 6 9 C 3 D (C + D = 25 > than OR =16 then 25 – 16 = 9 carry a 1)
(1 + A + B = 22 > than OR =16 then 22 – 16 = 6 carry a 1)
(1 + 0 + 0 = 1 < than 16 hence use ‘1’)
TRY:
-------------------------------------------
The Octal Number System
The same principles of positional number systems we applied to the decimal and binary
number systems can be applied to the octal number system.
However, the base of the octal number system is eight, so each position of the octal
number represents a successive power of eight.
From right to left, the successive positions of the octal number are weighted 1, 8, 64,
512, etc.
## A list of the first several powers of 8 follows:
80 =1 81 =8 82 =64 83 =512 84 =4096 85 =32768
## Converting an Octal Number to a Decimal Number
To determine the value of an octal number (3678, for example), we can expand the
number using the positional weights as follows:
3 6 78
\ \ \_________________7 * 80 = 7 * 1 =7
\ \__________________6 * 81 = 6 * 8 = 48
\___________________6 * 82 = 6 * 64 = 192
-----------------
24710
The Octal Number System
Here's another example to determine the value of the octal number 16018:
1 6 0 18
\ \ \_______________ 1 * 80 =1*1 =1
\ \__________________6 * 82 = 6 * 64 = 384
\___________________1 * 83 = 1 * 512 = 512
-------------------
89710
## Converting a Decimal Number to an Octal Number
To convert a decimal number to its octal equivalent, the remainder method (the same method
used in converting a decimal number to its binary equivalent) can be used. To review, the
remainder method involves the following four steps:
(1) Divide the decimal number by the base (in the case of octal, divide by 8).
## (2) Indicate the remainder to the right.
(3) Continue dividing into each quotient (and indicating the remainder) until the divide
operation produces a zero quotient.
(4) The base 8 number is the numeric remainder reading from the last division to the first
(if you start at the bottom, the answer will read from top to bottom).
The Octal Number System
Example 1: Convert the decimal number 46510 to its octal equivalent:
465 / 8 = 58 remainder 1
58 / 8 = 7 remainder 2
7/8=0 remainder 7
## The resulting decimal number is: 721. So, 46510 = 7218
Divide 8 into 465. The quotient is 58 with a remainder of 1; indicate the 1 on the right.
Divide 8 into 58 (the quotient from the previous division). The quotient is 7 with a remainder of 2,
indicated on the right.
Divide 8 into 7. The quotient is 0 with a remainder of 7, as indicated. Since the quotient is 0, stop
here.
The answer, reading the remainders from top to bottom, is 721, so 46510 = 7218
Binary to Octal / Octal to Binary
Binary to Octal
Step 1 - Divide the binary digits into groups of three (starting from the right).
Step 2 - Convert each group of three binary digits to one octal digit.
## Example: Binary Number : 101012 = ?????8
Step 1: 101012 = 010 101
Step 2: 101012 = 28 58
Step 3: 101012 = 258
## Binary Number : 101012 = Octal Number : 258
Octal to Binary
Step 1 - Convert each octal digit to a 3 digit binary number (the octal digits may be
treated as decimal for this conversion).
Step 2 - Combine all the resulting binary groups into a single binary number.
## Example: Octal Number : 258 = ?????2
Step 1: 258 = 210 510
Step 2: 258 = 0102 1012
Step 3: 258 = 0101012
## Octal Number : 258 = Binary Number : 101012
Decimal to Octal
Example 2: Convert the decimal number 254810 to its octal equivalent:
## 2458 / 8 = 318 remainder 4
318 / 8 = remainder 6
?/8=? remainder ?
?/8=? remainder ?
?/8=? remainder ?
## The resulting decimal number is: 721. So, 254810 = ?8
TRY THIS: Convert the following decimal to octal and equivalent octal to binary:
(a) 300210
(b) 651210
It is a key for binary subtraction, multiplication, division. There four rules of the binary
In fourth case, a binary addition is creating a sum of (1+1=10) i.e. 0 is write in the given
column and a carry of 1 over to the next column.
## Example – Addition 0011010 + 001100 = 00100110
11 CARRY
0011010 = 2610
+ 0001100 = 1210
--------------------------------------------
0100110 = 3810
Binary Subtraction
Subtraction and Borrow, these two words will be used very frequently for the binary
subtraction. There are four rules of the binary subtration
## Example – Subtraction 0011010 - 001100 = 00001110
01
01 BORROW
0011010 = 2610
- 0001100 = 1210
--------------------------------------------
0001110 = 1410
Binary Subtraction
Subtract 19 from 6.
0000 0110 6
−0001 0011 19
============== ====
11111 0011 −12 — An end-around borrow is produced, and the sign bit of
the intermediate result is 1.
−0000 0001 1 — Subtract the end-around borrow from the result.
============== ====
1111 0010 −13 — The correct result (6 − 19 = -13)
## Subtract −3 from 19.
00010011 19
−11111100 −3
============== ====
10001 0111 23 — An end-around borrow is produced.
−0000 0001 1 — Subtract the end-around borrow from the result.
============== ====
0001 0110 22 — The correct result (19 − (−3) = 22).
COMPLIMENTS
Binary system complements
As the binary system has base radix, r = 2. So the two types of complements for the
binary system are 2's complement and 1's complement.
1's complement
The 1's complement of a number is found by changing all 1's to 0's and all 0's to 1's.
This is called as taking complement or 1's complement. Example of 1's Complement is
as follows.
2’s Compliment
The 2's complement of binary number is obtained by adding 1 to the Least Significant
Bit (LSB) of 1's complement of the number.
## Example of 2's Complement is as follows.
2’s Compliment
For example: Find the 2’s complement of the 8 bit number '0 0 1 0 1 0 0 1'
## 11010110 First, invert the bits
--------------------------------
= 11010111
## TRY THIS: Find the 2’s complement of the following numbers
(a) 10110101
(b) 1010101010111111
Addition and Subtraction using 1's and 2's complement
Let's consider problem of subtracting 110 = 00012 from 710 = 01112 using 1's
complement.
## First, we need to convert 00012 to its negative equivalent in 1's complement.
0111 (7)
- 0 0 0 1 - (1)
To do this we change all the 1's to 0's and 0's to 1's. Notice that the most-significant digit
is now 1 since the number is negative. 0001 → 1110
## Next, we add the negative value we computed to 01112. This gives us
0111 (7)
+1110 + (-1)
-------------------------
10101 (6)
Notice that our addition caused an overflow bit. Whenever we have an overflow bit in 1's
complement, we add this bit to our sum to get the correct answer. If there is no overflow
bit, then we leave the sum as it is.
0101
+ 1
-----------
0 1 1 0 (6) This gives us a final answer of 01102 (or 610).
Addition and Subtraction using 1's and 2's complement
Now let's look at problem that does not generate an overflow bit. We will subtract 710
from 110 using 1's complement. First, we state our problem in binary.
0001 (1)
-0111 - (7)
Next, we convert 01112 to its negative equivalent and add this to 00012.
0001 (1)
+1000 +(-7)
--------------------------
1001 (-6)
This time our results does not cause an overflow, so we do not need to adjust the sum.
Notice that our final answer is a negative number since it begins with a 1. Remember
that our answer is in 1's complement notation so the correct decimal value for our
answer is -610 and not 910.
## -0 = 1111 -1 = 1110 -2 = 1101 -3 = 1100 -4 = 1011 -5 = 1010 -6 = 1001
Addition and Subtraction using 1's and 2's complement
## For example, 5 + (-3) = 2 0101 0000 0101 = +5
+ 1111 1101 = -3
3 = 0011 ------------------------------------------------
1's complement -3 = 1100 1 0 0 0 0 0 0 1 0 = +2
2's complement = 1100 + 0001 = 1101
Then discard the carry (indicates overflow), and you have your result: 0000 0010 which
equals to 2 as expected.
## 2's Complement Subtraction
adding a negative number is the same as subtracting a positive one.
## For example, 7 - 12 = (-5) 0 0 0 0 0 1 1 1 = +7
+ 1 1 1 1 0 1 0 0 = -12
-----------------------------------------------
1 1 1 1 1 0 1 1 = -5
12 = 0000 1100
-12 = 1111 0011 + 0000 0001 = 1111 0100
- 5 = 0000 0101 = 1111 1010 + 0000 0001 = 1111 1011
UNIT 1.5 : Number Systems
TRY THIS: Substraction using 1's and 2's complement
(a) 10 -8
(b) 11- 17
(c) 23 – 89
(d) 22 - 22
UNIT 1.5 : Number Systems
Powers of 2:
20 ......................1
21 ......................2
22 ......................4
23 ......................8
24 ....................16
25 ....................32
26 ....................64
27 ..................128
28 ..................256
29 ..................512
210 ................ 1024
211 ................ 2048
212 ................ 4096
213 ................ 8192
214 .............. 16384
215 .............. 32768
216 .............. 65536
UNIT 1.5 : Number Systems
0 ...................... 0 0000
1 ...................... 1 0001
2 ...................... 2 0010
3 ...................... 3 0011
4 ...................... 4 0100
5 ...................... 5 0101
6 ...................... 6 0110
7 ...................... 7 0111
8 ...................... 8 1000
9 ...................... 9 1001
10 ..................... A 1010
11 ..................... B 1011
12 ..................... C 1100
13 ..................... D 1101
14 ..................... E 1110
15 ......................F 1111
Equivalent Numbers in Decimal, Binary and Hexadecimal Notation:
UNIT 1.5 : Number Systems
TRY THIS: Convert the following numbers to their decimal equivalents:
(a) 1 1 0 0 1 1 02
(b) 1 1 1 1 1 0 0 12
(c) 4910
(d) 2110
(e) 5 68
(f) 11638
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Rounding (nearest 10 and 100) Math Lesson Plan & PDF Worksheet | Grades 3-5
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WHAT IS ROUNDING TO THE NEAREST 10 AND 100?
Rounding means choosing a simpler number close to the original number. These simpler numbers may be easier to solve problems with.
To better understand Rounding (NEAREST 10 AND 100)…
WHAT IS ROUNDING TO THE NEAREST 10 AND 100?. Rounding means choosing a simpler number close to the original number. These simpler numbers may be easier to solve problems with. To better understand Rounding (NEAREST 10 AND 100)…
LET’S BREAK IT DOWN!
Toy Cars
Let’s say you have 57 toy cars and you want to round the number of cars you have to the nearest ten. You can draw a number line from 50 to 60 to help you round. After you draw the number line, mark the halfway point at 55. Then you can locate 57 on the number line. 57 is greater than 55 so you know that the point for 57 is to the right of 55. Now you can see that 57 is closer to 60 than to 50, so you round up to 60. You have about 60 toy cars. Try this one yourself. You bought 42 flowers for a party. Round the number of flowers you bought to the nearest ten.
Toy Cars Let’s say you have 57 toy cars and you want to round the number of cars you have to the nearest ten. You can draw a number line from 50 to 60 to help you round. After you draw the number line, mark the halfway point at 55. Then you can locate 57 on the number line. 57 is greater than 55 so you know that the point for 57 is to the right of 55. Now you can see that 57 is closer to 60 than to 50, so you round up to 60. You have about 60 toy cars. Try this one yourself. You bought 42 flowers for a party. Round the number of flowers you bought to the nearest ten.
Bouncy Balls
Let’s say you have 85 bouncy balls and you want to round the number of bouncy balls you have to the nearest ten. 85 bouncy balls is between 80 and 90. You make a number line from 80 to 90 and mark the halfway point at 85. Since the number of bouncy balls you have is halfway between 80 and 90, you follow the rounding rule and round up to 90. You have about 90 bouncy balls. There is a pattern. You don’t have to always draw a number line. If you want to round to the nearest ten, you can look at the digit in the ones place. If it’s 4 or less, you round down. If it’s 5 or greater, you round up. Since there is a 5 in the ones place, you round up to 90 and get the same answer: about 90 bouncy balls. Try this one yourself. You have 64 bouncy balls. Round the number of bouncy balls you have to the nearest ten.
Bouncy Balls Let’s say you have 85 bouncy balls and you want to round the number of bouncy balls you have to the nearest ten. 85 bouncy balls is between 80 and 90. You make a number line from 80 to 90 and mark the halfway point at 85. Since the number of bouncy balls you have is halfway between 80 and 90, you follow the rounding rule and round up to 90. You have about 90 bouncy balls. There is a pattern. You don’t have to always draw a number line. If you want to round to the nearest ten, you can look at the digit in the ones place. If it’s 4 or less, you round down. If it’s 5 or greater, you round up. Since there is a 5 in the ones place, you round up to 90 and get the same answer: about 90 bouncy balls. Try this one yourself. You have 64 bouncy balls. Round the number of bouncy balls you have to the nearest ten.
Cows
Let’s say you’re a farmer and you have 142 cows. You want to round the number of cows you have to the nearest ten. To round to the nearest ten, look at the digit to the right of the tens place. If it’s 5 or more, you round up. If it’s 4 or less, you round down. 142 has a 2 in the ones place, so you round down to 140. You have about 140 cows. Try this one yourself. A farmer has 178 pigs. Round the number of pigs they have to the nearest ten.
Cows Let’s say you’re a farmer and you have 142 cows. You want to round the number of cows you have to the nearest ten. To round to the nearest ten, look at the digit to the right of the tens place. If it’s 5 or more, you round up. If it’s 4 or less, you round down. 142 has a 2 in the ones place, so you round down to 140. You have about 140 cows. Try this one yourself. A farmer has 178 pigs. Round the number of pigs they have to the nearest ten.
Jelly Beans
Let’s say you have 329 jelly beans and you want to round the number of jelly beans you have to the nearest hundred. To round to the nearest hundred, look at the digit to the right of the hundreds place. If it’s 5 or more, you round up. If it’s 4 or less, you round down. 329 has a 2 in the tens place, so you round down to 300. You have about 300 jelly beans. Try this one yourself. You have 378 jelly beans. Round the number of jelly beans you have to the nearest hundred.
Jelly Beans Let’s say you have 329 jelly beans and you want to round the number of jelly beans you have to the nearest hundred. To round to the nearest hundred, look at the digit to the right of the hundreds place. If it’s 5 or more, you round up. If it’s 4 or less, you round down. 329 has a 2 in the tens place, so you round down to 300. You have about 300 jelly beans. Try this one yourself. You have 378 jelly beans. Round the number of jelly beans you have to the nearest hundred.
ROUNDING (NEAREST 10 AND 100) VOCABULARY
Rounding
Makes numbers less exact but more convenient to work with.
Nearest ten
The multiple of ten that is closest to a given number.
Nearest hundred
The multiple of one hundred that is closest to a given number.
Number line
A straight line with numbers placed at equal intervals along its length.
Halfway point
A point that is halfway between two numbers on a number line.
Ones place
The place in a number that tells how many ones are in the number. In 247, there are 7 ones in the ones place.
The place in a number that tells how many tens are in the number. In 247, there are 4 tens in the tens place.
The place in a number that tells how many hundreds are in the number. In 247, there are 2 hundreds in the hundreds place.
Round to a number that is greater than the original number.
Round to a number that is less than the original number. |
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