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## Simplifying a sum of consecutive cubes
If you were asked to find the sum of the first four cubes, you would easily be able to do so:
$1^3 + 2^3 + 3^3 + 4^3 = 100$
But here you may notice something interesting:
$1^3 + 2^3 + 3^3 + 4^3 = (1+2+3+4)^2$
It turns out that this holds for the sum of the first $n$ cubes for any $n$ — an identity. This identity can be written also as:
$\displaystyle \sum_{i=1}^n i^3 = \left( \displaystyle \sum_{i=1}^n i \right)^2$
A second way of expressing the same identity simplifies the right side and expresses it as a combination:
$\displaystyle \sum_{i=1}^n i^3 = \binom{n+1}{2}^2$
### A proof by bijection
This identity can be proven combinatorially by noticing that $\binom{n+1}{2}$ is the number of ways to choose 2 elements with repetition from the set {1..n} — the following is a proof by Benjamin and Orrison.
In this proof, we prove the third form of the identity
$\displaystyle \sum_{i=1}^n i^3 = \binom{n+1}{2}^2$
by constructing one set of size $\displaystyle \sum_{i=1}^n i^3$ and constructing another set of size $\binom{n+1}{2}^2$, then constructing a bijection between the two sets.
Let $S$ be the set of ordered 4-pairs $(a,b,c,d)$ such that $1 \leq a,b,c \leq d \leq n$ — that is, the fourth integer of the pair is greater or equal to each of the other three. If we fix $d$, the number of possibilities for the variables $a,b,c$ is $d^3$. As $d$ ranges from 1 to n, the total number of elements in $S$ is equal to $\displaystyle \sum_{i=1}^n i^3$.
Also, since $\binom{n+1}{2}$ is the number of ways to choose 2 elements with repetition from the set {1..n}, there are $\binom{n+1}{2}$ pairs of integers $(h,i)$ satisfying $1 \leq h \leq i \leq n$. So let $T$ be the set of pairs of such pairs $((h,i),(j,k))$ where $1 \leq h \leq i \leq n$ and $1 \leq j \leq k \leq n$. The number of elements in $T$ is then $\binom{n+1}{2}^2$.
We further partition the sets as follows: let $S_1$ be the subset of $S$ where $a \leq b$ and let $S_2$ be the subset of $S$ where $a > b$. Similarly, let $T_1$ be the subset of $T$ where $i \leq k$ and let $T_2$ be the subset of $T$ where $i > k$. We can construct a bijection between $S$ and $T$ by constructing two bijections:
$S_1 \leftrightarrow T_1$
$S_2 \leftrightarrow T_2$
The following is trivially a bijection from $S_1$ to $T_1$ — the case where $a \leq b$:
$(a,b,c,d) \rightarrow ((a,b),(c,d))$
The equivalent bijection from $S_2$ to $T_2$ — the case where $a > b$:
$(a,b,c,d) \rightarrow ((c,d),(b,a-1))$
We can confirm that the two operations we defined are indeed bijections. This proves the identity.
### 3 Responses to Simplifying a sum of consecutive cubes
1. JL says:
Would it not be easier to inductively prove that the sum of the first n integers is n(n+1)/2 and that the sum of the first n cubes is n^2(n+1)^2/4, which equals (n(n+1)/2)^2?
• luckytoilet says:
Shhhh
Besides, any bijective proof beats any boring inductive argument xD
2. Anonymous says:
Bijections are awesome
|
# Radius is half of Diameter
What is a radius of circle?
Radius is the line joining from center to point on circle.
In a particular circle, there can be many radius but they have same length
Given above is the image of a circle with center O
Here OA, OB, OC and OD are the radius since they are straight line joining the center and circle.
Its given that OA = 4 cm
Since, value of all radius are same, we get:
OA = OB = OC = OD = 4 cm
What is diameter of circle?
Diameter is a straight line joining two points on a circle while passing through center O.
In a circle there can be multiple diameters and there length will be same.
There are two important points about diameter
(a) It passes through center
(b) Its end points lie on the circle
Example
Above image is of circle with center O.
Here AB and CD are the diameters of the circle because:
(a) they are straight line
(b) passes through center
(c) end points of the line are on circle’s boundary
Above image is of circle with center O.
Here AB is the radius of the circle.
If you observe carefully you will find that the diameter is made by joining two different radius
Hence,
AB = OA + OB
AB = r + r
AB = 2r
Hence the diameter AB is twice the radius.
Thus;
This is one of the important property of circle.
Remember this formula as it would help us to solve multiple problems in the future
Let us now solve some questions related to the concept
(01) The radius of circle is given 6 cm. Find the length of the diameter.
Solution
We know that:
Diameter = 2 x 6
Diameter = 12 cm
(02) Find the length of line OC
Solution
AB is the diameter of circle with length 8 cm
We know that:
(03) Find the value of radius if diameter is 7 cm
Solution
Diameter = 7 cm
We know that:
(04) Observe the below figure and find diameter of circle
Solution
OA is the radius of circle
OA = 6 cm
We know that
Diameter = 2 x 6
Diameter = 12 cm
(05)Observe the below figure and fill in the blanks
(a) Radius of circle is __________
(b) Diameter of circle is _________
Solution
(a) Radius of circle is 12 cm
(b) Calculating diameter
Diameter = 2 x 12
Diameter = 24 cm
Diameter of circle is 24 cm
(06) Find the length of line CD
Solution
OA is the radius of the circle
OA = 5 cm
CD is the diameter of circle
We know that:
Diameter = 2 x 5
Diameter = 10 cm
Hence length of CD = 10 cm
(07) Find the length of radius of below circle
Here CD and EF are the chords of the circle
AB is the diameter as it passes through center O
We know that:
Hence the radius of circle is 3 cm
(08) Radius of circle is 3.5 cm. Find the diameter
Solution
Diameter = 2 x 3.5
Diameter = 7 cm
Hence diameter is 7 cm
(09) Study the figure below and find the length of segment BC
Solution
The figure is of circle with center O.
Here OA is the radius of the circle with length 3 cm
OA = 3 cm
Segment BC is also the diameter of the circle
We know that:
Diameter = 2 x 3
Diameter = 6 cm
Hence the length of side BC = 6 cm
(10) Find the radius of below circle if the length of following segments are given:
AB = 2.5 cm
EF = 6 cm
CD = 2. cm
Solution
AB & CD are the chords of the circle
EF is the diameter
EF = 6 cm
We know that
|
# Class 10 Maths Books English Medium chapter 1
CBSE Class 10 Maths Books English Medium chapter 1 Ex 1.1 homework guffo.in
Chapter 1: Real Numbers
Example 1 : Use Euclid’s algorithm to find the HCF of 4052 and 12576.
अध्याय 1: रियल नंबर
उदाहरण 1: 4052 और 12576 के एचसीएफ को खोजने के लिए यूक्लिड के एल्गोरिदम का उपयोग करें।
Example 2 : Show that every positive even integer is of the form 2q, and that every
positive odd integer is of the form 2q + 1, where q is some integer.
Example 3 : Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where
q is some integer.
Example – A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to
stack them in such a way that each stack has the same number, and they take up the
least area of the tray. What is the number of that can be placed in each stack for this
purpose?
EXERCISE 1.1
1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is
some integer.
# Class 10 Maths Books English Medium chapter 1
3. An army contingent of 616 members is to march behind an army band of 32 members in
a parade. The two groups are to march in the same number of columns. What is the
maximum number of columns in which they can march?
4. Use Euclid’s division lemma to show that the square of any positive integer is either of
the form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square
each of these and show that they can be rewritten in the form 3m or 3m + 1.]
5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form
9m, 9m + 1 or 9m + 8.
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### Theory:
Consider the quadratic equation $$ax^2 + bx + c = 0$$, where $$a \ne 0$$.
Let us find the roots of this equation by the method of completing the square.
Divide the equation by $$a$$.
${x}^{2}+\frac{b}{a}x+\frac{c}{a}=0$
Move the constant to the right side.
${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
Add the square of one half of coefficient of $$x$$ on both sides.
${x}^{2}+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$
${\left(x+\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+\frac{{b}^{2}}{4{a}^{2}}$
${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4\mathit{ac}}{4{a}^{2}}$
Taking square root on both sides.
$x+\frac{b}{2a}=±\sqrt{\frac{{b}^{2}-4\mathit{ac}}{4{a}^{2}}}$
$x+\frac{b}{2a}=±\frac{\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$
$x=-\frac{b}{2a}±\frac{\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$
$x=\frac{-b±\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$
Therefore, the roots of $$ax^2 + bx + c = 0$$ are $x=\frac{-b+\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$ and $x=\frac{-b-\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$.
The formula for finding the roots of the quadratic equation $$ax^2 + bx + c = 0$$ is:
$x=\frac{-b±\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$
This formula is known as the quadratic formula.
Example:
1. Find the roots of $$2x^2 + 3x - 77 = 0$$ by using quadratic formula.
Solution:
The given equation is $$2x^2 + 3x - 77 = 0$$.
Here, $$a = 2$$, $$b = 3$$ and $$c = -77$$.
$x=\frac{-b±\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$
Substitute the given values in the formula.
$x=\frac{-3±\sqrt{{3}^{2}-4×2×\left(-77\right)}}{2×2}$
$x=\frac{-3±\sqrt{9+616}}{4}$
$x=\frac{-3±\sqrt{625}}{4}$
$x=\frac{-3±25}{4}$
$x=\frac{-3+25}{4}$ or $x=\frac{-3-25}{4}$
$x=\frac{22}{4}$ or $x=\frac{-28}{4}$
$$x =$$ $\frac{11}{2}$ or $$x = -7$$
Therefore, the roots of the given equation are $$-7$$ and $\frac{11}{2}$.
|
LetsPlayMaths.Com
WELCOME TO THE WORLD OF MATHEMATICS
Class 3 Length
Length
Conversion of Meters into Centimeters
Conversion of Centimeters into Meters
Conversion of Kilometers into Meters
Conversion of Meters into Kilometers
Addition of Meters and Centimeters Without Carrying
Addition of Meters and Centimeters With Carrying
Subtraction of Meters and Centimeters Without Borrowing
Subtraction of Meters and Centimeters With Borrowing
Multiplication of Meters and Centimeters
Multiplication of Kilometers and Meters
Division of Meters and Centimeters
Division of Kilometers and Meters
Length Test
Length Worksheet
Length
It is a form of measurement, where we can tell the length of various things like length of a pencil, length of a rope, length of a ship and distance between two cities. In today’s world Centimeters, Meters and Kilometers are used as standard units for measurement. To measure smaller lengths like length of a pen, we use centimeters (cm). For measuring length of garden or swimming pool, we use meter (m). For measuring greater lengths as distance between two cities, we use kilometers.
Conversion of Meters into Centimeters
1 Meter = 100 Centimeters
a) To convert meters into centimeters, multiply the number by 100.
b) To convert meters and centimeters into centimeters, multiply the number of meters by 100 and add
the number of centimeters to it.
Example 1. Convert 65 meters into centimeters
Solution. 65 X 100 = 6500 cm
So, 65 meters is equal to 6500 centimeters
Example 2. Convert 17 m 45 cm into centimeters.
Solution. First, we need to convert 17 meters into centimeters
17 meters = 17 X 100 = 1700 centimeters
Now, add 1700 centimeters with 45 centimeters 1700 cm + 45 cm = 1745 cm
So, 17 m 45 cm is equal to 1745 centimeters
Conversion of Centimeters into Meters
100 cm = 1 m
Example 1. Convert 475 centimeters into meters
Solution. 475 cm = 400 cm + 75 cm = 4 m 75 cm
So, 475 centimeters is equal to 4 m 75 cm
Example 2. Convert 4365 centimeters into meters
Solution. 4365 cm = 4300 cm + 65 cm = 43 m 65 cm
So, 4365 centimeters is equal to 43 m 65 cm
Conversion of Kilometers into Meters
1 km = 1000 meters
a) To convert kilometers into meters, multiply the number of kilometers by 1000.
b) To convert the kilometers and meters into meters, multiply the number of kilometers by 1000 and
then add the number of meters.
Example 1. Convert 11 kilometer into meters
Solution. 11 km = 11 X 1000 = 11000 m
So, 11 km is equal to 11000 meters
Example 2. Convert 24 km 745 m into meters.
Solution. 24 km 745 meters = 24 X 1000 m +745 m = 24000 m + 745m = 24745 m
So, 24 km 745 meters is equal to 24745 meters
Conversion of Meters into Kilometers
1000 meters = 1 kilometers
Example 1. Convert 6000 meters into kilometers
Solution. 6000 m = 6 X 1000 m = 6 X 1 km = 6 km
So, 6000 meters is equal to 6 km
Example 2. Convert 12545 meters into kilometers and meters.
Solution. 12545 m = 12000 m + 545 m = 12 X 1000 m + 545 m = 12 X 1 km + 545 m
= 12 km 545 m
Addition of Meters and Centimeters Without Carrying
Example 1. Add 45 m 25 cm and 14 m 23 cm
Solution. Arrange the numbers in tabular format as shown below.
Step 1. Add centimeters to centimeters first
25 cm + 23 cm = 48 cm
Write 48 cm under centimeter column
Step 2. Add the meters to meters
45 m + 14 m = 59 m
Write 59 m under meter column
So, the result is 59 m 48
Addition of Meters and Centimeters With Carrying
Example 1. Add 25 m 75 cm and 15 m 55 cm
Solution. Arrange the numbers in tabular format as shown below.
Step 1. First add centimeters to centimeters
75 cm + 55 cm = 130 cm = 100 cm + 30 cm = 1 m + 30 cm
Write 30 cm under centimeter column. Carry 1 m to meter column.
Step 2. Add the meters now. 25 m + 15 m + 1 m (Carry over from centimeter addition) = 41 m.
Write 41 m below meter column.
So, the answer is 41 m 30 cm.
Subtraction of Meters and Centimeters Without Borrowing
Example 1. Subtract 15 m 23 cm from 48 m 55 cm
Solution. Arrange the numbers in tabular format as shown below.
Step 1. First subtract centimeters from centimeters.
55 cm – 23 cm = 32 cm. Write 32 cm under centimeter column.
Step 2. Subtract meter from meter
48 m – 15 m = 33 m. Write 33 m under meter column.
So, the answer is 33 m 32 cm.
Subtraction of Meters and Centimeters With Borrowing
Example 1. Subtract 15 m 25 cm from 27 m 7 cm.
Solution. Arrange the numbers in tabular format and mention the centimeters in two digits. That is 7 cm
should be written as 07 cm. Here we are following the normal subtraction.
So, the answer is 11 m 82 cm.
Multiplication of Meters and Centimeters
We can multiply meters and centimeters like normal numbers. We must consider two digits centimeters while multiplication. For example, 5m 7cm should be considered as 5m 07 cm. First two digits on the right of the results gives the number of centimeters, the number formed by the remaining digits gives the meters.
Example 1. Multiply 17 m 25 cm by 5.
Solution. Write the numbers in column format as shown below.
So, the answer is 86 m 25 cm.
Multiplication of Kilometers and Meters
We can multiply kilometers and meters like normal numbers. Meters should always be considered as 3 digits number. For example, 12 km 5 m should be written 12 km 005 m while multiplication.
Example 1. Multiply 24 km 75 m by 8.
Solution.
So, the answer is 192 m 600 cm.
Division of Meters and Centimeters
Division of meters and centimeters can be done by two methods. Both the methods are given below.
Method 1.
Step 1. Convert meters and centimeters into centimeters only.
Step 2. Divide the total number of centimeters by divisor.
Step 3. Convert the quotient (in cm) into meters and centimeters.
Example 1. Divide 9 m 75 cm by 5.
Solution.
Step 1. Convert 9 m 75 cm into centimeters. 9 m + 75 cm = 900 cm + 75 cm = 975 cm
Step 2. Divide 975 cm by 5. 975 ÷ 5 = 195 cm
Step 3. Convert 195 cm into meters and centimeters.
195 cm = 100 cm + 95 cm = 1 m 95 cm
So, the answer is 1 m 95 cm
Method 2.
We can divide meters and centimeters like normal division. In this method, centimeters should be of two digits. For example, if the dividend is 24 m 8 cm then it should be written as 24 m 08 cm. when we will bring the first digit of the centimeters down we must write the remaining quotient numbers in the centimeters column.
Example 1. Divide 9 m 5 cm by 5.
Solution. Write the dividend and divisor in the normal division format as shown below.
Division of Kilometers and Meters
Division of kilometers and meters can be done by two methods. Both the methods are given below.
Method 1.
Step 1. Convert kilometers and meters into meters only.
Step 3. Divide the total number of meters by divisor.
Step 2. Convert the quotient (in m) into kilometers and meters.
Example 1. Divide 6 km 65 m by 5.
Solution.
Step 1. Convert 6 km 65 m into meters. 6 km + 65 m = 6000 m + 65 m = 6065 m
Step 2. Divide 6065 m by 5. 6065 ÷ 5 = 1213 m
Step 3. Convert 1213 m into kilometers and meters.
1213 m = 1000 m + 213 m = 1 km 213 m
So, the answer is 1 km 213 m
Method 2.
We can divide kilometers and meters like normal division. In this method, meters should be of three digits. For example, if the dividend is 24 km 8 m then it should be written as 24 km 008 m. when we will bring the first digit of the meters down we must write the remaining quotient numbers in the meters column.
Example 1. Divide 24 km 5 m by 5.
Solution. Write the dividend and divisor in the normal division format as shown below.
So, the answer is 4 km 801 m.
Length Test - 1
Length Test - 2
Length Worksheet
Length Worksheet - 1
|
# 180 Days of Math for Fourth Grade Day 119 Answers Key
By accessing our 180 Days of Math for Fourth Grade Answers Key Day 119 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fourth Grade Answers Key Day 119
Directions: Solve each problem.
Question 1.
24 + 29 = __________
24
29
53
The sum of 24 and 29 is 53.
Question 2.
Write $$\frac{50}{100}$$ as a percentage.
$$\frac{50}{100}$$ as a percentage is 50%
Question 3.
The quotient is 7 and the remainder is 0.
Question 4.
24 ÷ 6 = ___________
The quotient is 4 and the remainder is 0.
Question 5.
Write 56 in words.
Question 6.
9 + 9 + 9 = × 9
We can see three 9’s from the above
So, it can form three groups of nine.
9 + 9 + 9 = 3 × 9
Question 7.
Which month follows:
March? __________
May? ___________
April comes after March. It comes before May.
Question 8.
Each cube has 1-cm sides. What is the volume of the model?
s = 1 cm
Volume = 1 × 1 × 1
V = 1 cubic cm
There are 12 cubes in the model
Volume of the model = 1 × 12 = 12 cubic centimeter
Question 9.
Fill in the blank with rotate, reflect, or translate.
Patrick has swimming lessons 3 times a week for $$\frac{1}{2}$$ hour. How much time does he spend in 4 weeks at swimming lessons?
Patrick has swimming lessons 3 times a week for $$\frac{1}{2}$$ hour.
4 × $$\frac{1}{2}$$ = 2
|
# AP Statistics Curriculum 2007 Hypothesis Var
(Difference between revisions)
Revision as of 23:12, 6 February 2008 (view source)IvoDinov (Talk | contribs)← Older edit Current revision as of 16:32, 1 May 2008 (view source) (→ General Advance-Placement (AP) Statistics Curriculum - Testing a Claim about a Standard Deviation or Variance) (4 intermediate revisions not shown) Line 1: Line 1: ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Testing a Claim about a Standard Deviation or Variance== ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Testing a Claim about a Standard Deviation or Variance== - Assessing the amount of variation in a process, natural phenomenon or an experiment is of paramount importance in many fields. For instance, a computer manufacturer may dismiss a batch of computer chips if they vary more than certain tollerance levels in their clock-speed, heat emmissions or energy consumptions. + Assessing the amount of variation in a process, natural phenomenon, or an experiment is of paramount importance in many fields. For instance, a computer manufacturer may dismiss a batch of computer chips if they vary by more than certain tolerance levels in their clock-speeds, heat emissions, or energy consumptions. === [[AP_Statistics_Curriculum_2007_Estim_Var | Background]]=== === [[AP_Statistics_Curriculum_2007_Estim_Var | Background]]=== Line 12: Line 12: For Normally distributed random variables, given $H_o: \sigma^2 = \sigma_o^2$ vs. $H_1: \sigma^2 \not= \sigma_o^2$ , then ${(n-1) s^2 \over \sigma_o^2}$ has a $\Chi^2_{(df=n - 1)}$ distribution, where [[AP_Statistics_Curriculum_2007_EDA_Var |$s^2$ is the sample variance]]. For Normally distributed random variables, given $H_o: \sigma^2 = \sigma_o^2$ vs. $H_1: \sigma^2 \not= \sigma_o^2$ , then ${(n-1) s^2 \over \sigma_o^2}$ has a $\Chi^2_{(df=n - 1)}$ distribution, where [[AP_Statistics_Curriculum_2007_EDA_Var |$s^2$ is the sample variance]]. - Notice that the Chi-square distribution is not symmetric (it is positively skewed) and therefore, there are two critical values for each level of confidence $(1-\alpha)$. The value $\chi_L^2$ represents the left-tail critical value and $\chi_R^2$ represents the right-tail critical value. For various degrees of freedom and areas, you can compute all critical values either using the [http://socr.ucla.edu/htmls/SOCR_Distributions.html SOCR Chi-Square Distribution] or using the [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm SOCR Chi-square distribution calculator]. + Notice that the Chi-square distribution is not symmetric (it is positively skewed) and therefore, there are two critical values for each level of confidence $(1-\alpha)$. The value $\chi_L^2$ represents the left-tail critical value and $\chi_R^2$ represents the right-tail critical value. For various degrees of freedom and areas, you can compute all critical values either using the [http://socr.ucla.edu/htmls/SOCR_Distributions.html SOCR Chi-Square Distribution] or using the [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm SOCR Chi-Square Distribution Calculator]. * Example: A random sample of size 30 drawn from a Normal distribution has sample-variance $s^2 = 5$. Test at the $\alpha=0.05$ level of significance if this is consistent with $H_o: \sigma^2 = 2$. * Example: A random sample of size 30 drawn from a Normal distribution has sample-variance $s^2 = 5$. Test at the $\alpha=0.05$ level of significance if this is consistent with $H_o: \sigma^2 = 2$. Line 21: Line 21: === Testing a Claim about the Standard Deviation ($\sigma$)=== === Testing a Claim about the Standard Deviation ($\sigma$)=== - As the standard deviation is just th square root of the variance ($\sigma = |\sqrt{\sigma^2}|$), we do significance testing for the standard deviation anologously. + The standard deviation is just the square root of the variance ($\sigma = |\sqrt{\sigma^2}|$). We do significance testing for the standard deviation analogously. For Normally distributed random variables, given $H_o: \sigma = \sigma_o$ vs. $H_1: \sigma \not= \sigma_o$ , then ${(n-1) s^2 \over \sigma_o^2}$ has a $\Chi^2_{(df=n - 1)}$ distribution, where [[AP_Statistics_Curriculum_2007_EDA_Var |$s^2$ is the square of the sample standard deviation]]. For Normally distributed random variables, given $H_o: \sigma = \sigma_o$ vs. $H_1: \sigma \not= \sigma_o$ , then ${(n-1) s^2 \over \sigma_o^2}$ has a $\Chi^2_{(df=n - 1)}$ distribution, where [[AP_Statistics_Curriculum_2007_EDA_Var |$s^2$ is the square of the sample standard deviation]]. - ===Hands-on activities=== + ===Hands-on Activities=== - * Formulate appropriate hypothesis and assess the significance of the evidence to reject the null hypothesis for the population standard deviation ($\sigma$) assuming the observations below represent a random sample from the liquid content (in fluid ounces) of 16 beverage cans and can be considered as Normally distributed. Use a 90% level of confidence ($\alpha=0.1$). + * Formulate appropriate hypotheses and assess the significance of the evidence to reject the null hypothesis for the population standard deviation ($\sigma$) assuming the observations below represent a random sample from the liquid content (in fluid ounces) of 16 beverage cans and can be considered as Normally distributed. Use a 90% level of confidence ($\alpha=0.1$).
{| class="wikitable" style="text-align:center; width:75%" border="1" {| class="wikitable" style="text-align:center; width:75%" border="1" Line 34: Line 34:
- * Hypotheses: $H_o: \sigma = 0.04$ vs. $H_1: \sigma \not= 0.04$ . + * Hypotheses: $H_o: \sigma = 0.06 (\sigma_o)$ vs. $H_1: \sigma \not= 0.06$ . * Get the sample statistics from [http://socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts] (e.g., Index Plot); Sample-Mean=14.8875; Sample-SD=0.072700298, Sample-Var=0.005285333. * Get the sample statistics from [http://socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts] (e.g., Index Plot); Sample-Mean=14.8875; Sample-SD=0.072700298, Sample-Var=0.005285333.
[[Image:SOCR_EBook_Dinov_Estim_Var_020408_Fig3.jpg|500px]]
[[Image:SOCR_EBook_Dinov_Estim_Var_020408_Fig3.jpg|500px]]
- * Identify the degrees of freedom ($df=n-1=15$) and the level of confidence (${\alpha \over 2}=0.05$, as we are looking for a $(1-\alpha)100% CI(\sigma)$). + * Identify the degrees of freedom ($df=n-1=15$). - * Find the left and right critical values, $\chi_L^2=7.261$ and $\chi_R^2=24.9958, as in the image below. + * Test Statistics: [itex]\Chi_o^2 = {(n-1)s^2 \over \sigma_o^2} \sim \Chi_{(df=n-1)}^2.$ -
[[Image:SOCR_EBook_Dinov_Estim_Var_020408_Fig4.jpg|500px]]
+ - * CI($\sigma^2$) + * Significance Inference: $\chi_o^2 = {15\times 0.005285333 \over 0.06^2}=22.022221$ - : ${15\times 0.0053 \over 24.9958} \leq \sigma^2 \leq {15\times 0.0052 \over 7.261}$ + : P-value=$P(\Chi_{(df=n-1)}^2 > \chi_o^2) = 0.107223$. This p-value does not indicate strong evidence in the data to reject a possible population standard deviation of 0.06. - * CI(\sigma[/itex]) +
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig8.jpg|500px]]
- : $\sqrt{15\times 0.0053 \over 24.9958} \leq \sigma \leq \sqrt{15\times 0.0052 \over 7.261} + - ===More examples=== + ===More Examples=== - * You randomly select and measure the contents of 15 bottles of cough syrup. The results (in fluid ounces) are shown. Use a 95% level of confidence to construct a confidence interval for the standard deviation ([itex]\sigma$) assuming the contents of these cough syrup bottles is Normally distributed. Does this CI($\sigma$) suggest that the variation in the bottles is at an acceptable level if the '''population standard deviation''' of the bottle’s contents should be less than 0.025 fluid ounce? + * You randomly select and measure the contents of 15 bottles of cough syrup. The results (in fluid ounces) are shown. Formulate and test hypotheses about the standard deviation ($\sigma_o=0.025$) assuming the contents of these cough syrup bottles are Normally distributed. Is there data-driven evidence suggesting that the standard-deviation of the fluids in the bottles is not at an acceptable level?
{| class="wikitable" style="text-align:center; width:75%" border="1" {| class="wikitable" style="text-align:center; width:75%" border="1" Line 60: Line 58:
- * The gray whale has the longest annual migration distance of any mammal. Gray whales leave Baja, California, and western Mexico in the spring, migrating to the Bering and Chukchi seas for the summer months. Tracking a sample of 50 whales for a year provided a mean migration distance of 11,064 miles with a standard deviation of 860 miles. Construct a 90% confidence interval for the variance for the migrating whales. Assume that the population of migration distances is Normally distributed. + * The gray whale has the longest annual migration distance of any mammal. Gray whales leave Baja, California, and western Mexico in the spring, migrating to the Bering and Chukchi seas for the summer months. Tracking a sample of 50 whales for a year provided a sample mean migration distance of 11,064 miles with a standard deviation of 860 miles. Assume that the population of migration distances is Normally distributed to formulate and test hypotheses for the population standard deviation ($\sigma_o=500$). - * For the [[SOCR_012708_ID_Data_HotDogs | hot-dogs dataset]] construct 97% CI for the population standard deviation of the calorie and sodium contents, separately. + * Use the [[SOCR_012708_ID_Data_HotDogs | hot-dogs dataset]] to formulate and test hypotheses about the population standard deviation of the sodium content in the poultry hot-dogs ($\sigma_o=70$).
+ ===References=== ===References===
## General Advance-Placement (AP) Statistics Curriculum - Testing a Claim about a Standard Deviation or Variance
Assessing the amount of variation in a process, natural phenomenon, or an experiment is of paramount importance in many fields. For instance, a computer manufacturer may dismiss a batch of computer chips if they vary by more than certain tolerance levels in their clock-speeds, heat emissions, or energy consumptions.
### Background
Recall that the sample-variance (s2) is an unbiased point estimate for the population variance σ2, and similarly, the sample-standard-deviation (s) is a good point estimate for the population-standard-deviation σ.
The sample-variance is roughly Chi-square distributed:
$\chi_o^2 = {(n-1)s^2 \over \sigma^2} \sim \Chi_{(df=n-1)}^2$
### Testing a Claim about the Variance (σ2)
For Normally distributed random variables, given $H_o: \sigma^2 = \sigma_o^2$ vs. $H_1: \sigma^2 \not= \sigma_o^2$ , then ${(n-1) s^2 \over \sigma_o^2}$ has a $\Chi^2_{(df=n - 1)}$ distribution, where s2 is the sample variance.
Notice that the Chi-square distribution is not symmetric (it is positively skewed) and therefore, there are two critical values for each level of confidence (1 − α). The value $\chi_L^2$ represents the left-tail critical value and $\chi_R^2$ represents the right-tail critical value. For various degrees of freedom and areas, you can compute all critical values either using the SOCR Chi-Square Distribution or using the SOCR Chi-Square Distribution Calculator.
• Example: A random sample of size 30 drawn from a Normal distribution has sample-variance s2 = 5. Test at the α = 0.05 level of significance if this is consistent with Ho2 = 2.
Test statistic: $\chi_o^2 = {29\times 5 \over 2} = 72.5$.
Left and Right Chi-Square critical values (for α = 0.05) are $\chi_L^2=16.047$ and $\chi_R^2=45.722$. Since, $\chi_o^2 = 72.5 > \chi_R^2=45.722$, Ho is rejected at the α = 0.05 level of significance.
### Testing a Claim about the Standard Deviation (σ)
The standard deviation is just the square root of the variance ($\sigma = |\sqrt{\sigma^2}|$). We do significance testing for the standard deviation analogously.
For Normally distributed random variables, given Ho:σ = σo vs. $H_1: \sigma \not= \sigma_o$ , then ${(n-1) s^2 \over \sigma_o^2}$ has a $\Chi^2_{(df=n - 1)}$ distribution, where s2 is the square of the sample standard deviation.
### Hands-on Activities
• Formulate appropriate hypotheses and assess the significance of the evidence to reject the null hypothesis for the population standard deviation (σ) assuming the observations below represent a random sample from the liquid content (in fluid ounces) of 16 beverage cans and can be considered as Normally distributed. Use a 90% level of confidence (α = 0.1).
14.816 14.863 14.814 14.998 14.965 14.824 14.884 14.838 14.916 15.021 14.874 14.856 14.86 14.772 14.98 14.919
• Hypotheses: Ho:σ = 0.06(σo) vs. $H_1: \sigma \not= 0.06$ .
• Get the sample statistics from SOCR Charts (e.g., Index Plot); Sample-Mean=14.8875; Sample-SD=0.072700298, Sample-Var=0.005285333.
• Identify the degrees of freedom (df = n − 1 = 15).
• Test Statistics: $\Chi_o^2 = {(n-1)s^2 \over \sigma_o^2} \sim \Chi_{(df=n-1)}^2.$
• Significance Inference: $\chi_o^2 = {15\times 0.005285333 \over 0.06^2}=22.022221$
P-value=$P(\Chi_{(df=n-1)}^2 > \chi_o^2) = 0.107223$. This p-value does not indicate strong evidence in the data to reject a possible population standard deviation of 0.06.
### More Examples
• You randomly select and measure the contents of 15 bottles of cough syrup. The results (in fluid ounces) are shown. Formulate and test hypotheses about the standard deviation (σo = 0.025) assuming the contents of these cough syrup bottles are Normally distributed. Is there data-driven evidence suggesting that the standard-deviation of the fluids in the bottles is not at an acceptable level?
4.211 4.246 4.269 4.241 4.26 4.293 4.189 4.248 4.22 4.239 4.253 4.209 4.3 4.256 4.29
• The gray whale has the longest annual migration distance of any mammal. Gray whales leave Baja, California, and western Mexico in the spring, migrating to the Bering and Chukchi seas for the summer months. Tracking a sample of 50 whales for a year provided a sample mean migration distance of 11,064 miles with a standard deviation of 860 miles. Assume that the population of migration distances is Normally distributed to formulate and test hypotheses for the population standard deviation (σo = 500).
• Use the hot-dogs dataset to formulate and test hypotheses about the population standard deviation of the sodium content in the poultry hot-dogs (σo = 70).
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# 9.1 Ratios and Proportions. You will learn to use ratios and proportions to solve problems. 1) ratio 2) proportion 3) cross products 4) extremes 5) means.
## Presentation on theme: "9.1 Ratios and Proportions. You will learn to use ratios and proportions to solve problems. 1) ratio 2) proportion 3) cross products 4) extremes 5) means."— Presentation transcript:
9.1 Ratios and Proportions
You will learn to use ratios and proportions to solve problems. 1) ratio 2) proportion 3) cross products 4) extremes 5) means
In 2000, about 180 million tons of solid waste was created in the United States. The paper made up about 72 million tons of this waste. The ratio of paper waste to total waste is 72 to 180. This ratio can be written in the following ways. 72 to 18072:18072 ÷ 180 Definition of Ratio A ratio is a comparison of two numbers by division. a to ba:ba ÷ b where b 0
A __________ is an equation that shows two equivalent ratios. proportion Every proportion has two cross products. In the proportion to the right, the terms 20 and 3 are called the extremes, and the terms 30 and 2 are called the means. The cross products are 20(3) and 30(2). The cross products are always _____ in a proportion. equal 30(2) =20(3) 60 = 60
Theorem 9-1 Property of Proportions For any numbers a and c and any nonzero numbers b and d, Likewise,
Solve each proportion: 15(2x) =30(6) 30x = 180 x = 6 3(x) =(30 – x)2 3x = 60 – 2x 5x = 60 x = 12
The gear ratio is the number of teeth on the driving gear to the number of teeth on the driven gear. Driving gear Driven gear If the gear ratio is 5:2 and the driving gear has 35 teeth, how many teeth does the driven gear have? given ratio equivalent ratio = 5 2 35 x = driving gear driven gear driving gear driven gear 5 x = 70 x = 14 The driven gear has 14 teeth.
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# Question Video: Finding the Solution Set of Root Equations Involving Absolute Value Mathematics • 9th Grade
Find the solution set of the equation √(4𝑥² − 28𝑥 + 49) = |𝑥 + 4|.
03:52
### Video Transcript
Find the solution set of the equation the square root of four 𝑥 squared minus 28𝑥 plus 49 is equal to the absolute value of 𝑥 plus four.
In this question, we’re asked to find the solution set of a given equation involving radicals and the absolute value. To do this, we first recall the solution set is the set of all solutions to this equation. So we need to start by solving this equation. That’s finding all the values of 𝑥 such that the left-hand side is equal to the right-hand side of the equation. So to solve this equation, let’s start by taking a look at the equation.
On the left-hand side, we have the square root of four 𝑥 squared minus 28𝑥 plus 49. And usually, the easiest way to solve an equation involving a square root is to square both sides of the equation. However, if we do this, we can introduce extra solutions. So we do need to be careful. This is particularly important because we’re taking the square root of a number. For example, if we take the square root of a negative number, we’ll end up with a complex number.
However, there is one interesting thing worth noting about this example. On the left-hand side of this equation, we’re taking the square root, which means we take the positive value. And on the right-hand side, we’re taking the absolute value. So this is also the positive value. So both sides of the equation are already positive. This can help justify taking the squares of both sides of the equation. This gives us that four 𝑥 squared minus 28𝑥 plus 49 is equal to the absolute value of 𝑥 plus four squared.
And we can simplify the right-hand side of this equation by remembering the absolute value of a number is its magnitude. It doesn’t matter about the sign. But if we’re squaring this value, it doesn’t matter if we take the positive or negative value. In other words, for any real number 𝑎, the magnitude of 𝑎 squared is just equal to 𝑎 squared. Therefore, we can use this to simplify the right-hand side of our equation to be 𝑥 plus four all squared.
And now, we can simplify the right-hand side of this equation further by distributing the exponent over the parentheses. We can do this by using the FOIL method or binomial expansion. In either case, we get 𝑥 squared plus eight 𝑥 plus 16. And remember, this needs to be equal to four 𝑥 squared minus 28𝑥 plus 49. And now, we’re just solving a quadratic equation. We can do this by collecting like terms. So we subtract 𝑥 squared from both sides of the equation, eight 𝑥 from both sides of the equation, and 16 from both sides of the equation and simplify. We get three 𝑥 squared minus 36𝑥 plus 33 is equal to zero.
And now, there’s many different ways of solving a quadratic equation. We’re going to start by noticing all three terms share a factor of three. So we can just divide through by three. This gives us 𝑥 squared minus 12𝑥 plus 11 is equal to zero. And we can solve this by factoring. We need two numbers which multiply to give 11 and add to give negative 12. And of course, negative one times negative 11 is equal to 11, and negative one plus negative 11 is equal to negative 12.
Therefore, we can factor this quadratic to get 𝑥 minus one multiplied by 𝑥 minus 11 is equal to zero. And finally, for the product of two numbers to be equal to zero, one of the two factors must be equal to zero. In other words, either 𝑥 is equal to one or 𝑥 is equal to 11. And in this case, it’s not actually necessary to check whether these two solutions are valid or not because, as we already said, both sides of the equation were already positive. And if we substituted these values into the left-hand side of that equation and got a negative value, there wouldn’t be a solution because the right-hand side of the equation is already positive.
So by being solutions to the equation, we guarantee that the radical is defined at these values of 𝑥. However, this won’t always be the case for all equations of this form. So it can be a good idea to check anyway. We would just substitute 𝑥 is equal to one and 𝑥 is equal to 11 into our original equation to check that they are indeed solutions. And in both cases, we would determine that they are indeed solutions.
Therefore, we were able to determine the solution set of the equation the square root of four 𝑥 squared minus 28𝑥 plus 49 is equal to the absolute value of 𝑥 plus four is the set containing one and 11.
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# How do you add 2 different variables?
## How do you add 2 different variables?
How to Add and Subtract Variables
1. Addition. In this equation, you can add all of the coefficients (2, 5, and 4) because the variables are the same (a). 2a + 5a + 4a = 11a.
2. Subtraction. In this equation, you can subtract all of the coefficients (11, 5, and 4) because the variables are the same (a). 11a – 5a – 4a = 2a.
What is the rule for adding polynomials?
You add polynomials when there are plus signs. You subtract them when there is a minus sign. Remember to only add/subtract like terms within the polynomials.
### Can 2x 1 be a polynomial?
HERE THE EXPONENT OF 2 IS 1/2,WHICH IS NOT A WHOLE NUMBER.SO,IT IS NOT A POLYNOMIAL.
How do you add two terms?
To add two or more like terms, we add the numerical coefficients of the given terms and form another like term with the sum obtained as the numerical coefficient of the resulting term.
#### Can polynomials have two variables?
Polynomials can contain more than one variable and can be evaluated in the same way as polynomials with one variable. To evaluate any polynomial, you substitute the given values for the variable and perform the computation to simplify the polynomial to a numerical value.
What is a polynomial with two terms called?
A binomial is a polynomial with 2 terms.
## How do you classify polynomials?
Polynomials can be classified by the degree of the polynomial. The degree of a polynomial is the degree of its highest degree term. So the degree of 2×3+3×2+8x+5 2 x 3 + 3 x 2 + 8 x + 5 is 3. A polynomial is said to be written in standard form when the terms are arranged from the highest degree to the lowest degree.
The rule here is that only like terms can be added together. Like terms are those terms which contain the same powers of same variables. They can have different coefficients, but that is the only difference.
### When do I add polynomials to a variable?
Any time I have a variable without a coefficient, there is an “understood” 1 as the coefficient. If it’s helpful to me to write that 1 in, then I’ll do so. Now, I’ll do the adding vertically: Either way, I get the same answer. For my hand-in answer, I won’t include the ” +0x ” term.
How to add and subtract polynomials in math?
Adding Polynomials. 1 Place like terms together. 2 Add the like terms Example: Add 2×2 + 6x + 5 and 3×2 – 2x – 1 Start with: 2×2 + 6x + 5 + 3×2 − 2x − 1 Place like terms together: 2×2 +3×2 + 6x−2x +
#### Which is easier to add two polynomials or one polynomial?
We strongly recommend to minimize your browser and try this yourself first. Addition is simpler than multiplication of polynomials. We initialize result as one of the two polynomials, then we traverse the other polynomial and add all terms to the result.
How to add and subtract polynomials in TSI?
Add the coefficients for each set of like terms. The variables carry along: The simplified form of 7×5 + 2×2 − 3×2 + 3 − x5 is 6×5 − x2 + 3. Be sure to change the subtraction to addition before combining like terms.
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# Manipulating Surds
## Manipulating Surds
• Understand the definition: Surds are square roots which cannot be simplified to a rational number. They are ‘irrational’ numbers and are often left in surd form to maintain precision.
• Know how to simplify surds: Simplifying a surd involves removing any-square rootable factors. For instance, √18 can be simplified to 3√2. This is achieved by identifying the highest square number that can divide 18 (which is 9), taking the square root of that number (which is 3), and multiplying it with the square root of the remainder (which is √2).
• Remember surd rules: √a x √b equals √(a x b) (multiplication rule), and √(a/b) equals √a / √b (division rule).
• Be aware of surd addition and subtraction: You can only add or subtract surds if they are ‘like’ or ‘similar’ i.e., if they have the same number under the square root sign (radicand). For instance, 4√3 + 2√3 = 6√3.
• Recognize standard surd forms: The standard form of a surd is to have no square roots in the denominator of fractions. To achieve this, you multiply the numerator and the denominator by the conjugate of the denominator. This is known as ‘rationalizing the denominator’.
• Practice expanding brackets with surds: Use the FOIL method (First, Outside, Inside, Last) to expand brackets with surds. Be cautious of signs, and remember that the product of two identical surds is a rational number (as √a x √a equals a).
• Brush up on solving equations involving surds: To solve such equations, you often need to square both sides at some point. Be careful with negatives - if (√a = -b) is squared, the result is (a = b^2), not (- a = b^2).
• Tackle more complex surds problems: Once confident with basic manipulation and simplifying surds, try tackling problems involving surd expressions in both the numerator and denominator, or equations where the surd quantity is not isolated.
• Practice makes perfect: There are a variety of online platforms and textbooks available offering an array of surd problems that will build familiarity and fluency. It’s important to check your answers and understand any mistakes you may have made along the way.
Remember, although surds can seem complex initially, they follow clear rules and with regular practice, you’ll soon find them straightforward to manipulate.
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# Chapter 8: Quadratic Equations Exercise – 8.10
### Question: 1
The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.
### Solution:
Let the length of one side of the right triangle be x cm then,
the other side be = (x + 5) cm
and given that hypotenuse = 25 cm
By using Pythagoras Theorem,
x2 + (x + 5)2 = 252
x2 + x+ 10x + 25 = 625
2x+ 10x + 25 - 625 = 0
2x2 + 10x - 600 = 0
x2 + 5x - 300 = 0
x2 - 15x + 20x - 300 = 0
x(x - 15) + 20(x -15) = 0
(x - 15)(x + 20) = 0
x = 15 or x = - 20
Since, the side of triangle can never be negative
Therefore, when, x = 15
And, x + 5 = 15 + 5 = 20
Therefore, length of side of right triangle is = 15 cm and other side is = 20 cm
### Question: 2
The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
### Solution:
Let the length of smaller side of rectangle be x metres then, the larger side be (x + 30) metres and
diagonal be = (x + 60) metres
By using Pythagoras theorem,
x2 + (x + 30)2 = (x + 60)2
x2 + x2 + 60x + 900 = x2 + 120x + 3600
2x2 + 60x + 900 - x2 - 120x - 3600 = 0
x2 - 60x - 2700 = 0
x2 - 90x + 30x - 2700 = 0
x(x - 90) + 30(x - 90) = 0
(x - 90)(x + 30) = 0
x = 90 or x = -30
Since, the side of rectangle can never be negative
Therefore, x = 90
x + 30 = 90 + 30 = 120
Therefore, the length of smaller side of rectangle is = 90 metres and larger side is = 120 metres.
### Question: 3
The hypotenuse of a right triangle is 3√10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 9√5 cm. How long are the legs of the triangle?
### Solution:
Let the length of smaller side of right triangle be = x cm then large side be = y cm
By using Pythagoras theorem,
x+ y2 = 90 .... eqn. (1)
If the smaller side is triple and the larger side is doubled, the new hypotenuse is 9√5 cm
Therefore,
9x2 + 4y2 = 405 .... eqn. (2)
From equation (1) we get,
y2 = 90 - x2
Now putting the value of y2 in eqn. (2)
9x2 + 4(90 - x2) = 405
9x2 + 360 - 4X2 - 405 = 0
5x2 - 45 = 0
5(x2 - 9) = 0
x2 - 9 = 0
x2 = 9
x = √9
x = ±3
Since, the side of triangle can never be negative
Therefore, when x = 3
Then, y2 = 90, x= 90 , (3)= 90, 9 = 81
y = √81
y = ±9
Hence, the length of smaller side of right triangle is = 3 cm and larger side is = 9 cm
### Question: 4
A pole has to be erected at a point on the boundary of a circular park of diameter meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
### Solution:
Let P be the required location on the boundary of circular park such that its distance from the gate B is x metres that is BP = x metres
Then, AP = x + 7
In right triangle ABP, by using Pythagoras theorem,
AP2 + BP2 = AB2
(x + 7)2 + x= 132
x+ 14x + 49 + x2 = 169
2x+14x + 49 - 169 = 0
2x2 + 14x - 120 = 0
2(x+ 7x - 60) = 0
x2 + 12x - 5x - 60 = 0
x(x + 12) - 5(x + 12) = 0
(x + 12)(x - 5) = 0
x = - 12 or x = 5
Since, the side of triangle can never be negative
Therefore, P is at a distance of 5 metres from the gate B
### Course Features
• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
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# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-7-ex-7-2/
Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter Chapter 7 Chapter Name Coordinate Geometry Exercise Ex 7.2 Number of Questions Solved 10 Category NCERT Solutions
## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2
Question 1.
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Solution:
Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Solution:
Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in given
figure below. Niharika runs $$\frac { 1 }{ 4 }$$ th the distance AD on the 2nd line and posts a green flag. Preet runs $$\frac { 1 }{ 5 }$$ th distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Solution:
y-coordinate of green flag = $$\frac { 1 }{ 4 }$$ x 100 m = 25 m
Coordinates of green flag are P (2, 25)
y-coordinate of red flag = $$\frac { 1 }{ 5 }$$ x 100 = 20
Coordinates of red flag are Q (8, 20)
The distance between two points is
The blue flag is in the 5th line, at a distance of 22.5 m.
Question 4.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Let the required ratio be k : 1
Question 5.
Find the ratio in which line segment joining A (1, -5) and B (-4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Solution:
Question 6.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Solution:
Question 8.
If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = $$\frac { 3 }{ 7 }$$ AB and P lies on the line segment AB.
Solution:
Question 9.
Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts.
Solution:
Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
[Hint: Area of a rhombus = $$\frac { 1 }{ 2 }$$ (product of its diagonals)]
Solution:
We hope the NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2, drop a comment below and we will get back to you at the earliest.
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# Operations With Real Numbers SOL 7
## Presentation on theme: "Operations With Real Numbers SOL 7"— Presentation transcript:
Operations With Real Numbers SOL 7
Operations With Real Numbers SOL 7.3 by Lisa Beebe Chincoteague Combined School Commutative Property Associative Property Distributive Property Identity Properties Inverse Properties Multiplicative Property of Zero
Commutative Property (ORDER)
The order in which numbers are added or multiplied does not change the answer. Examples: 3 + 5 = 5 + 3 Both sides equal 8. 2 · 7 = 7 · 2 Both sides equal 14.
Commutative Property
Associative Property (GROUPING)
The way in which three numbers are grouped with parentheses in addition or multiplication does not change the answer Examples: (1 + 2) + 3 = 1 + (2 + 3) Both sides equal 6. (2 · 3) · 5 = 2 · (3 · 5) Both sides equal 30.
Associative Property
Distributive Property (Distribute)
States that the product of a number and the sum (or differences) of two other numbers equals the sum (or difference) of the products of the number and each other number For any three numbers a, b, and c – it is true that a(b + c) = ab + ac
Distributive Property
A number plus zero is always itself. Example: = 11
Identity Property of Multiplication
A number multiplied by 1 is always itself Example: 17 · 1 = 17
States that the sum of a number and its additive inverse always equals zero. What you need to add to a number to get the identity Example: = 0
Multiplicative Inverse Property
States that the product of a number and its multiplicative inverse (or reciprocal) always equals one What you need to multiply to a number to get the identity Example: 4 × ¼ = 1
Multiplicative Property of Zero
States that the product of any real number and zero is always zero. Example: 5 × 0 = 0
Identify the following property: a + b = b + a
Commutative Property of Multiplicaton Associative Property Commutative Property of addition Additive Identity Property 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Which property is shown in the equation 26 + 0 = 26?
Inverse property of addition Inverse property of multiplication Identity property of addition Multiplicative property of zero 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Identity the following property: 5 · (x · y) = (5 · x) · y
Distributive Property Associative property of multiplication Associative Property of addition Commutative property of multiplication 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Identify the following property: 5 · (x · y) = 5 · (y · x)
Associative Property of Multiplication Associative Property of Addition Distributive Property Commutative Property of multiplication 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Which equation uses the distributive property?
-3(4 + 7) = -3(4) + -3(7) -3(4 + 7) = -3(11) -3(4 + 7) = -3(4) + 7 -3(4 + 7) = 4(-3) + 4(7) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Which equation shows the multiplicative property of zero?
-35.9(1) = -35.9 -35.9(0) = 0 -35.9(359/10) = 1 -35.9(35.9) = 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Which property is shown in the equation 7 × 0 = 0?
Additive identity property Multiplicative Identity property Multiplicative inverse property Multiplicative property of zero 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
What is the identity for addition?
-1 1 1/0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
What is the identity for multiplication?
-1 1 1/0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Which of the following is the additive inverse of 4?
-4 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Identify the property used. (1 + 3) × (4 + 1) = (1 + 3) × (1 + 4)
Associative property of multiplication Associative property of addition Commutative property of multiplication Commutative property of addition 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Jessica has found that ¼ × 4/1 = 1
Jessica has found that ¼ × 4/1 = 1. She has discovered an example of _____________. the multiplicative identity the additive identity a multiplicative inverse an additive inverse 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
What is the reciprocal, or multiplicative inverse , of -3?
1/3 -1/3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
The expression x + (-x) = 0 demonstrates which property?
Multiplicative inverse Distributive property Additive inverse Multiplicative property of zero 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
If the distributive property is applied to 5(7 + 3), which is the result?
5(7)(3) 5(7) + 3 5(7) + 5(3) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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## Stability in Numbers
• Lesson
3-5
1
In this lesson, students apply their knowledge of addition equations to create number sentences using an electronic balance tool.
To begin the lesson, ask students what an addition sentence or addition equation looks like. They might respond with, 9 + 7 = 16. Students should notice there are two addends, the addition symbol, the equal symbol, and a sum. Students may share other examples, as necessary.
Introduce the Pan Balance - Numbers Tool to the students.
Pan Balance - Numbers
Put 4 onto the left pan by typing in the red space above the pan. Ask students, "What happens? Why?"
Now put 9 onto the blue right pan by typing in the space above the pan. Ask students, "What happens? Why?"
Note: These two steps help introduce students to the mechanics of the electronic tool.
Next, pose the following scenario to students:
What number sentences can you create using the number 12?
First reset the balance. Put 12 into the right pan. Now put a sum (two numbers added together) onto the left pan that will balance with 12. Ask students why the pan balanced.
Next, show another sum that balances with 12. Ask students, "How many sums can you find?" Students may indicate the following sums:
0 + 12
1 + 11
2 + 10
3 + 9
4 + 8
5 + 7
6 + 6
This would also be an appropriate time to review the commutative property of addition, namely 3 + 9 has the same sum as 9 + 3.
Next put 3 + 9 on the left pan. Put something onto the right pan that will make it balance. Then find something else that will balance with 3 + 9. Ask students, "Why does this work? How many answers can you find?" Students may indicate the following sums:
7 + 5
6 + 6
and so on...
As necessary, share more examples with the class. When you have finished the whole-class demonstration and exploration, students should work in pairs to use the the Pan Balance - Numbers Tool.
Pan Balance - Numbers
Distribute the Addition Equations activity sheet to each pair of students.
Students should work together to complete the five activities on the activity sheet.
Assessments
Collect the the Addition Equations activity sheets from each pair of students. Use the responses on the activity sheet to assess student understanding.
Extensions
1. The operations of subtraction and division can be used in the expression balance as well. The shape balance equations can be translated directly into number sentences which emphasize the relationship role of the equals sign.
2. The expression balance can also be used to verify number sentences such as:
3 × (5 - 2) = 15 - 6.
Use * for multiplication, / for division, and - for subtraction on your keypad.
3. Ask students, "What is the largest number of numbers that you can use to balance with 12? What is the smallest number of numbers?"
Questions for Students
Can you add up three numbers to get a sum of 12? Four numbers?
### Learning Objectives
Students will:
• Investigate the equivalence of two numeric expressions
• Create and solve addition equations
### Common Core State Standards – Practice
• CCSS.Math.Practice.MP1
Make sense of problems and persevere in solving them.
• CCSS.Math.Practice.MP4
Model with mathematics.
• CCSS.Math.Practice.MP5
Use appropriate tools strategically.
• CCSS.Math.Practice.MP7
Look for and make use of structure.
|
# Factors of 6
Definition of factors of 6: If a number completely divides 6 without a remainder, then that number is called a factor of 6.
By the above definition, we can say that the factors of 6 are the divisors of 6. In this section, we will learn about the factors of 6 and the prime factors of 6.
## Highlights of Factors of 6
• 6=2×3 is the prime factorization of 6.
• The factors of 6 are 1, 2, 3, and 6.
• Prime factors of 6 are 2 and 3.
• Negative factors of 6 are –1, -2, -3, and –6.
## What are the Factors of 6?
Let us write the number 6 multiplicatively in all possible ways. We have:
6 = 1×6
6 = 2×3
As we can express 6 multiplicatively in no other ways, so we will stop right now. So the factors of 6 in pairs are given as follows:
Thus the pair factors of 6 are (1, 6) and (2, 3). As all the numbers appearing in pair factors are the factors of 6, we conclude that
## Number of factors of 6
From above we have calculated the factors of which are 1, 2, 3, and 6. Thus the total number of factors of 6 is four.
## Prime Factors of 6
Note that the factors of 6 are 1, 2, 3, and 6. Among those factors, we observe that only 2 and 3 are prime numbers as they do not have any proper divisors.
∴ the prime factors of 6 are 2 and 3.
Question: What are the factors of 6?
Video Solution:
## How to find factors of 6?
Now we will determine the factors of 6 by division method. In this method, we will find the numbers that can divide 6 with no remainder. See that
6/1=6 and the remainder is 0.
1 and 6 are factors of 6.
6/2=3 and the remainder is 0.
2 and 3 are factors of 6.
Note that no numbers other than the numbers in violet color can divide 6. So the numbers in violet color, that is, 1, 2, 3, and 6 are the complete list of factors of 6.
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# Solve the following quadratic equation by factorization: $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$
Given:
Given quadratic equation is $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$.
To do:
We have to solve the given quadratic equation.
Solution:
$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$
Multiplying both sides by $(x-1)(x-2)(x-3)(x-4)$, we get,
$\begin{array}{l} ( x-1)( x-2)( x-3)( x-4)\left[\frac{1}{( x-1)( x-2)} +\frac{1}{( x-2)( x-3)} +\frac{1}{( x-3)( x-4)}\right] =( x-1)( x-2)( x-3)( x-4)\left[\frac{1}{6}\right]\\ \\ \frac{( x-1)( x-2)( x-3)( x-4)}{( x-1)( x-2)} +\frac{( x-1)( x-2)( x-3)( x-4)}{( x-2)( x-3)} +\frac{( x-1)( x-2)( x-3)( x-4)}{( x-3)( x-4)} =\frac{( x-1)( x-2)( x-3)( x-4)}{6}\\ \\ ( x-3)( x-4) +( x-1)( x-4) +( x-1)( x-2) =\frac{( x-1)( x-2)( x-3)( x-4)}{6}\\ \\ 6[( x-3)( x-4) +( x-1)( x-4) +( x-1)( x-2)] =( x-1)( x-2)( x-3)( x-4)\\ \\ 6\left[ x^{2} -3x-4x+12+x^{2} -x-4x+4+x^{2} -x-2x+2\right] =( x-1)( x-2)( x-3)( x-4)\\ \\ 6\left[ 3x^{2} -15x+18\right] =( x-1)( x-2)( x-3)( x-4)\\ \\ 6\times 3\left( x^{2} -5x+6\right) =( x-1)( x-2)( x-3)( x-4)\\ \\ 18\left( x^{2} -3x-2x+6\right) =( x-1)( x-2)( x-3)( x-4)\\ \\ 18[ x( x-3) -2( x-3)] =( x-1)( x-2)( x-3)( x-4)\\ \\ 18[( x-3)( x-2)] =( x-1)( x-2)( x-3)( x-4)\\ \\ 18=( x-1)( x-4)\\ \\ x^{2} -x-4x+4=18\\ \\ x^{2} -5x+4-18=0\\ \\ x^{2} -5x-14=0\\ \\ x^{2} -7x+2x-14=0\\ \\ x( x-7) +2( x-7) =0\\ \\ ( x+2)( x-7) =0\\ \\ x+2=0\ or\ x-7=0\\ \\ x=-2\ or\ x=7 \end{array}$
The values of $x$ are $-2$ and $7$.
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Simply Easy Learning
Updated on: 10-Oct-2022
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# How can we prove that a negative number multiplied/divided/subtacted by a negative gives a positive result?
## I've always wondered if we could prove this...
Apr 15, 2018
It's not the case that a negative minus a negative is always positive. For instance, $- 3 - \left(- 2\right) = - 3 + 2 = - 1$, which is still negative.
We can, however, rigorously prove that the product of two negative numbers is always positive. From that conclusion we can additionally prove that the same holds for a negative number divided by a negative number. The proof is not out of reach for an algebra student, either, as long as you are willing to accept that the product of two positive numbers is positive.
Consider that any negative number can be written in the form $- a$, where $a$ is a positive number.
Suppose we have two negative numbers, $x$ and $y$. Because any negative number can be written $- a$, where $a > 0$, $x = - a$ and $y = - b$, where $a , b$ are positive numbers. The product $x y$ is then equal to $\left(- a\right) \left(- b\right) = \left(- 1\right) \left(- 1\right) \left(a b\right) = a b$. The product is equivalent to the product of two positive numbers, which must also be positive. $\square$
We have proved that if $x , y$ are two negative numbers, then $x y$ is positive. This can be extended to division as follows.
Suppose we have negative numbers $x , y$. The quotient $\frac{x}{y}$ is equivalent to $x \left(\frac{1}{y}\right) = x z$, where $z = \frac{1}{y}$. Because $y$ is negative, $y = - a$, where a is some positive number. Then $z = \frac{1}{y} = \frac{1}{- a} = - \frac{1}{a}$. So $z$ is negative. Thus, we again have the product of two negative numbers, which (by above) must be positive.
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Whether mixed numbers have the same denominators or different denominators, adding them is a lot like adding whole numbers: you stack them up one on top of the other, draw a line, and add. For this reason, some students feel more comfortable adding mixed numbers than adding fractions.
Here’s how to add two mixed numbers:
1. Add the fractional parts and if necessary, change this sum to a mixed number and reduce it.
2. If the answer you found in Step 1 is an improper fraction, change it to a mixed number, write down the fractional part, and carry the whole number part to the whole number column.
3. Add the whole number parts (including any number carried).
Your answer may also need to be reduced to lowest terms. The examples that follow show you everything you need to know.
## Add mixed numbers with the same denominators
As with any problem involving fractions, adding is always easier when the denominators are the same. For example, suppose you want to add 3 1/3 + 5 1/3. Doing mixed number problems is often easier if you place one number above the other:
As you can see, this arrangement is similar to how you add whole numbers, but it includes an extra column for fractions. Here’s how you add these two mixed numbers step by step:
1. Add the fractions.
2. Switch improper fractions to mixed numbers; write down your answer.
Because 2/3 is a proper fraction, you don’t have to change it.
3. Add the whole number parts.
3 + 5 = 8
Here’s how your problem looks in column form:
This problem is about as simple as they get. In this case, all three steps are pretty easy. But sometimes, Step 2 requires more attention. For example, suppose you want to add 8 3/5 + 6 4/5. Here’s how you do it:
1. Add the fractions.
2. Switch improper fractions to mixed numbers, write down the fractional part, and carry over the whole number.
Because the sum is an improper fraction, convert it to the mixed number 1 2/5. Write down 2/5 and carry the 1 over to the whole-number column.
3. Add the whole number parts, including any whole numbers you carried over when you switched to a mixed number.
1 + 8 + 6 = 15
Here’s how the solved problem looks in column form. (Be sure to line up the whole numbers in one column and the fractions in another.)
As with any other problems involving fractions, sometimes you need to reduce at the end of Step 1.
## Add mixed numbers with different denominators
The most difficult type of mixed number addition is when the denominators of the fractions are different. This difference doesn’t change Steps 2 or 3, but it does make Step 1 tougher.
For example, suppose you want to add 16 3/5 and 7 7/9.
1. Add the fractions.
Add 3/5 and 7/9:
2. Switch improper fractions to mixed numbers, write down the fractional part, and carry over the whole number.
This fraction is improper, so change it to the mixed number 1 17/45. Fortunately, the fractional part of this mixed number isn’t reducible.
Write down the 17/45 and carry over the 1 to the whole-number column.
3. Add the whole numbers.
1 + 16 + 7 = 24
Here’s how the completed problem looks:
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Online Tution » Important Question » Mean Formula and Meaning in Maths
# Mean Formula and Meaning in Maths
## What is the Definition of Mean?
Mean is the average of all the set of numbers which is used to get the result between the number of numbers. It is calculated by adding together all the numbers in a set and then dividing by the number of items in that set.
## Mean Meaning
Mean is also known as arithmetic mean or mathematical average. Calculating the average from a set of numbers requires at least 2 numbers. This form of average is widely used in both scientific calculation and statistical notation.
The mean is a useful tool for understanding and analyzing data sets. It is often used in statistical analysis and in mathematical modeling. The mean can help you to identify trends and patterns in data sets.
## Mean Formula with Example for Class 10
Now we know that to get the result from a data set, we should use the following formula. In which first all the numbers given in the data set are summed and it is divided by the number of items given in the data set. So now let’s understand it through an example –
if you wanted to find out the mean for 4, 5, 6, 3, and 7 first add up all values (4 + 5 + 6 + 3 + 7 = 25) and then divide that sum by the number of items (25/5 = 5). The answer would be an average or mean value; in this case it would be 5.
## Mean Symbol
The symbol for the mean is usually given as x-bar, or x̄.
Examples of its usage in everyday life include a cricketer’s average number of runs scored in test matches, as well as the mean price of houses in certain geographic areas as estimated by real estate agents.
## Mean Formula
The mean is the average of a set of numbers, and is often used as a measure of central tendency. The formula for mean is very simple: add up all the numbers in a set, and then divide by the total number of values in the set.
Mean = Sum of the number given in Data/ Total number given in Data
Example Question – Find the mean of first 5 prime Number.
Solution – First 5 prime number is – 1,2,3,4,5
Using formula – (1+2+3+4+5) / 5 =
15/5 = 3 Ans.
## How to Find Mean?
This page focuses on two types of mean formula in statistics: those employed when data is ungrouped and those used when data is grouped. As a refresher, the mean formula for ungrouped data is the sum of all observations divided by the number of observations; however, the formula changes when the data is categorized. Let’s dive right into our exploration!
## Mean Formula for Ungrouped Data
Given n observed values x1, x2, x3…xn, the mean of these values can be calculated as follows:
The formula for the ungrouped data mean is the same as the formula for the ordinary mean. But if we want to write it with mean symbol, then we will write this formula like this –
x̄=∑ x/n
Example Question – Find the mean of the following data: 10, 15, 12, 16, 15, 10, 14, 15, 12, 10.
Solution:
Using Formula = (10+15+12+16+15+10+14+15+12+10) / 10
= 129/10
= 12.9
## Mean Formula for Grouped Data
For large data sets, a frequency distribution table is a useful way to organize the values. This type of table provides information about the value and its frequency – that is, how many times each value appears in the dataset. For discrete frequency distributions, there are several methods for computing the arithmetic mean.
Method Formula
#### Direct method
Direct method of mean formula x̄ = Σfixi/Σfi
#### Shortcut method
Mean x̄ = A + ∑fidi/∑fi
#### Step-Deviation method
Mean x̄ = A + h (∑fiui / ∑fi)
## Types of Mean in Mathematics
There are 3 types of Mean which is important in Mathematics –
1. Arithmetic Mean
2. Geometric Mean
3. Harmonic Mean
### 1. Arithmetic Mean
Arithmetic Mean is an important method for calculating the central tendency of a given dataset. It takes into account all of the data points in the set and can have any sign – positive, negative, or zero.
There are two types of Arithmetic Mean: the sample mean and the population mean. The sample mean divides the sum of a set of observations by their count and indicates the centrality of that particular data set;
whereas, the population mean divides total observations by its size in order to ascertain a group’s overall characterizing middle value.
### 2. Geometric Mean
The geometric mean is used to find the nth root of the product of n numbers from the given data. You can find the nth root of a given number by multiplying it through the geometric mean. For example, if you have three numbers X, Y, Z, then their geometric mean will be 3 XYZ.
### Formula of Geometric Mean is – x̄ = n√(x1 · x2 · x3 · … · xn)
Example – Geometric mean of 4 and 2 = √4 x 2 = 2√2
### 3. Harmonic Mean
Harmonics Mean is a powerful way to analyze data. It is used to find the average ratio. This makes it especially useful in time and average analysis, as well as in any situation where large or small numbers would impact the analysis results.
Let there be two numbers X and Y, for which the harmonic mean will be 2xy(x+y) and for the same three numbers X, Y and Z the harmonic mean will be 3xyz(xy+xz+yz).
## What are the Benefits of Learning Mean in Statistics?
We often see that whenever we add any number in mathematics and divide it by the total number of data, then we call it average. But in statistics, the method of calculating this average is called mean. Because in mathematics there is only one formula to find the average, while in statistics there are many formulas and also in many different ways this average can be found.
So, there are many advantages to learning the Mean in statistics.
• The Mean is a very important concept in statistics, and it is used in a wide variety of applications.
• The Mean is also a very useful tool for analyzing data.
• In addition, the Mean is a very important concept in statistics, and it is used in a wide variety of applications.
• By calculating the Mean of each data set, you can easily compare their central tendencies and determine if there are any significant differences between them.
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## FAQs
### How to calculate the mean?
In this article, along with the formula of mean, we have also shown you an example question by solving it, through which you will be able to understand the mean easily and will calculate by yourself.
### What are the three formulas of mean?
The three major formulas of mean are arithmetic mean, Geometric mean, Harmonic mean. In this article we discussed all the three formulas in brief.
### What does Class 10 mean Formula?
The class 10 mean formula is very simple: Add the given number and dividing them the total number is given.
### What is the mean Formula symbol?
The mean formula symbol is x̄ in X-bar.
### TOPICS:
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Common Core Mathematics Standards
### 3.MD.3
Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step “how many more
## View Standards Related to3.MD.3
### 3.OA.1
Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.
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### 3.OA.2
Interpret whole-number quotients of whole numbers, e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 ÷ 8.
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### 3.OA.3
Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.1
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### 3.OA.4
Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 × ? = 48, 5 = ? ÷ 3, 6 × 6 = ?.
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### 3.OA.5
Apply properties of operations as strategies to multiply and divide.2 Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)
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# Multiplication Formula
In this page multiplication formula we can find basic formulas in addition, subtraction, multiplication and division using fraction. By using this method we can easily add, subtract, multiply and divide any two fractions.
Difference:
If we want to multiply two fractions we have to multiply the numerators with numerators and denominator with denominator. If it is possible to simplify we have to do that and show the answer in the simplified form.
We have also given worksheets for each category of topics. By practicing this worksheets you can get more knowledge. These are the most basic topics in math. Students must practice this kind of worksheets. The students who are in the grade of elementary must practice these kinds of worksheets to get more knowledge in math. In this page you can get a clear idea in this topic and you have worksheets which consist of five question with clear steps has given.
Multiplication formula
ab x cd = (a x c) (b x d)
To understand this topic much better we have given some examples using formula.
Example 1:
Multiply (5/6) x (3/8)
Solution:
Now let us compare the given two fractions that is 5/6 with a/b and 3/8 with c/d. The formula to multiply two fractions is ab/cd.
= (5/6) x (3/8)
= (5x3)/(6x8)
= 15/48
We can simplify this using 3 times table. So we will get 5/16.
Example 2:
Multiply (1/17) x (5/8)
Solution:
Now let us compare the given two fractions that is 5/6 with a/b and 3/8 with c/d. The formula to multiply two fractions is ab/cd.
= (1/17) x (5/8)
= (1x5)/(17x8)
= 5/136
We can simplify this. So the final answer is 5/136
Example 3:
Multiply (8/7) x (15/18)
Solution:
Now let us compare the given two fractions that is 5/6 with a/b and 3/8 with c/d. The formula to multiply two fractions is ab/cd.
= (8/7) x (15/18)
= (8x15)/(7x18)
= 120/126
We can simplify this using 2 times table. So the final answer is 60/63
multiplication formulas to math formulas
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## Question: How To Name A Line In Geometry?
When we draw lines in geometry, we use an arrow at each end to show that it extends infinitely.
1. A line can be named either using two points on the line (for example, ↔AB ) or simply by a letter, usually lowercase (for example, line m ).
2. A segment is named by its two endpoints, for example, ¯AB.
## How do you name a straight line?
There are three different types of straight lines: (i) Horizontal lines: The lines drawn horizontally are called horizontal lines. (ii) Vertical lines: The lines drawn vertically are called vertical lines. (iii) Oblique or slanting lines: The lines drawn in a slanting position are called oblique or slanting lines.
## Does AB end in B?
In the figure above, a line segment AB has two end points A and B. It starts from point A and ends at point B. One and only one line-segment can be between two given points A and B. Line segment can also be a part of a line as in the figure below.
## How do you name a line with 3 points?
These three points all lie on the same line. This line could be called ‘Line AB’, ‘Line BA’, ‘Line AC’, ‘Line CA’, ‘Line BC’, or ‘LineCB’.
## What is label line?
Labeled line principle is a hypothesis to explain how different nerves, all of which use the same physiological principles in transmitting impulses along their axons, can generate different sensations. One nerve can have multiple receptors attached to it.
You might be interested: How Was Geometry Created?
## What is a straight line in geometry?
A straight line is an endless one-dimensional figure that has no width. It is a combination of endless points joined on both sides of a point. A straight line does not have any curve in it. It can be horizontal, vertical, or slanted.
## How many ways can a line be named?
line, and line. Therefore, in 12 ways the line l can be named using only these points A, B, C, and D, if different order implies different name. A plane can be named using any three non collinear points on the same plane.
## What is a straight line called in geometry?
A line is a straight one-dimensional figure having no thickness and extending infinitely in both directions. A line is sometimes called a straight line or, more archaically, a right line (Casey 1893), to emphasize that it has no “wiggles” anywhere along its length. Harary (1994) called an edge of a graph a “line.”
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## College Algebra (11th Edition)
$x=\left\{ -\dfrac{14}{3},8 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log_4[(3x+8)(x-6)]=3 ,$ change to exponential form. Then express the resulting equation in the form $ax^2+bx+c=0.$ Use the Quadratic Formula to solve for the values of the variable. Then do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Since $y=b^x$ is equivalent to $\log_b y=x,$ the exponential form of the equation above is \begin{array}{l}\require{cancel} (3x+8)(x-6)=4^3 \\\\ (3x+8)(x-6)=64 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} 3x(x)+3x(-6)+8(x)+8(-6)=64 \\\\ 3x^2-18x+8x-48=64 \\\\ 3x^2+(-18x+8x)+(-48-64)=0 \\\\ 3x^2-10x-112=0 .\end{array} In the equation above, $a= 3 ,$ $b= -10 ,$ and $c= -112 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4(3)(-112)}}{2(3)} \\\\ x=\dfrac{10\pm\sqrt{100+1344}}{6} \\\\ x=\dfrac{10\pm\sqrt{1444}}{6} \\\\ x=\dfrac{10\pm\sqrt{(38)^2}}{6} \\\\ x=\dfrac{10\pm38}{6} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{10-38}{6} \\\\ x=\dfrac{-28}{6} \\\\ x=\dfrac{-14}{3} \\\\ x=-\dfrac{14}{3} \\\\\text{OR}\\\\ x=\dfrac{10+38}{6} \\\\ x=\dfrac{48}{6} \\\\ x=8 .\end{array} Upon checking, $x=\left\{ -\dfrac{14}{3},8 \right\} ,$ satisfy the original equation.
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# Introduction to Proofs : Functions, Relations, & Cardinality
## Example Questions
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### Example Question #1 : Intro To Proofs
If
Find the cardinality of
Explanation:
The question is asking us to find the number of elements in the intersection of B and B .
We simply find the common elements of the two sets, and count the number of elements.
Now we simply count the number of elements in this set.
### Example Question #1 : Intro To Proofs
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the intersection of A and B .
We simply find the common elements of the two sets, and count the number of elements.
Now we simply count the number of elements in this set.
### Example Question #1 : Intro To Proofs
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the union of A and A .
We simply combine the two sets together, and count the number of elements. If there are any elements that are the same, we keep one of them, and not both.
Now we simply count the number of elements in this set.
### Example Question #1 : Functions, Relations, & Cardinality
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the intersection of B and B .
We simply find the common elements of the two sets, and count the number of elements.
Now we simply count the number of elements in this set.
### Example Question #1 : Functions, Relations, & Cardinality
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the intersection of A and B .
We simply find the common elements of the two sets, and count the number of elements.
Now we simply count the number of elements in this set.
### Example Question #3 : Intro To Proofs
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the union of B and A .
We simply combine the two sets together, and count the number of elements. If there are any elements that are the same, we keep one of them, and not both.
Now we simply count the number of elements in this set.
### Example Question #7 : Intro To Proofs
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the union of A and A .
We simply combine the two sets together, and count the number of elements. If there are any elements that are the same, we keep one of them, and not both.
Now we simply count the number of elements in this set.
### Example Question #8 : Intro To Proofs
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the intersection of A and B .
We simply find the common elements of the two sets, and count the number of elements.
Now we simply count the number of elements in this set.
### Example Question #1 : Functions, Relations, & Cardinality
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the intersection of A and A .
We simply find the common elements of the two sets, and count the number of elements.
Now we simply count the number of elements in this set.
### Example Question #10 : Intro To Proofs
If
Find the cardinality of .
Explanation:
The question is asking us to find the number of elements in the union of A and A .
We simply combine the two sets together, and count the number of elements. If there are any elements that are the same, we keep one of them, and not both.
Now we simply count the number of elements in this set.
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### Standard Deviation
Consider the following data about the heights of plants in Jonathan's garden:
3cm, 4cm, 5cm, 7cm, 11cm
Now, let's calculate the mean - μ - of these values.
μ = (3 + 4 + 5 + 7 + 11)/5 = 6cm
If we use this value to describe the mean height of plants, we immediately run into difficulties; because, it does not represent the true nature of heights of these plants - some are as short as 3 cm and some are as tall as 11 cm.
Therefore, the mean in this case, to say the least, is a bit misleading. This leads to a need of another value that helps us to understand the distribution of data in a given situation.
Now let's see how much each value of data has deviated ( going away ) from the mean:
x μ (x - μ) 3 4 5 7 11 6 6 6 6 6 -3 -2 -1 1 5
Let's find the average of these deviations from the mean value:
Σ(x - μ) / 5 = (-3 + -2 + -1 + 1 + 5 )/5 = 0
The deviations turned out to be zero, not because of lack of deviations; it is because, the deviations turned out to be negative and positive which in the end led to be cancelled out.
Now, in order to deal with issue, let's square the deviations to remove the negative signs, which is as follows:
x μ (x - μ) (x - μ)2 3 4 5 7 11 6 6 6 6 6 -3 -2 -1 1 5 9 4 1 1 25
Since we squared the deviations, just to deal with negative values, it's time we reversed the process: let's find the square root of the following result:
√(Σ(x - μ)2)/5 = √(40/5) = 2.8
This is called the standard deviation of the above set of data representing the heights of plants in Jonathan's garden. It gives us a clearer picture of data distribution along with the mean. With the value of the standard deviation, the data can be described in the following way:
The mean height of the plants in Jonathan's garden is 6cm and the standard deviation is 2. 8. That means the heights of most plants falls into the range from (6-2.8) = 3.2cm to (6+2.8)=8.8cm.
The example shows how important the Standard deviation is to get a clear picture about a set of data. Without it, talking about data is like, recalling the fate of Titanic without the iceberg!!
So, the formula for standard deviation is as follows:
σ = √Σ(x - μ)2/n
where n is the total frequency.
Calculator-friendly formula for Standard Deviation
σ = √(Σ(x - μ)2)/n
σ = √(Σ(x2 - 2xμ + μ2)/n
σ = √(Σ(x2 - Σ2xμ + Σμ2)/n
σ = √(Σ(x2 - 2μΣx + Σμ2)/n
σ = √(Σ(x2 - 2μnμ + nμ2)/n
σ = √(Σ(x2 - 2nμ2 + nμ2)/n
σ = √(Σ(x2 - nμ2)/n
σ = √(Σx2/n) - μ2
σ = √(Σx2/n) - μ2
To find the standard deviation in grouped data, we change the method
slightly - σ = √(Σf(x - μ)2)/n, where f is the frequency of each class and n is the total frequency.
E.g.
The frequency of shoe sizes of students in a certain class is as follows:
shoe-size(x)frequency(f)
33
45
510
68
74
μ = Σfx/n = 5.2
σ = √(Σf(x - μ)2)/n = √(Σf(x - 5.2;)2)/30 = 2.3
E.g.
The marks obtained by a group of students for maths are as follows:
Marks(x)frequency(f)
0 - 203
21 - 406
41 - 609
61 - 808
81 - 1004
μ = Σfx/n = 52.7
σ = √(Σf(x - μ)2)/n = √(Σf(x - 52.7)2)/30 = 2.55
I am sure, you have got a good understanding of the concept of standard deviation by now.
Now, in order to complement what you have just learnt, work out the following questions:
1. The time taken by 10 engineers to install a satellite dish, in minutes, is as follows:
51, 49, 56, 60, 52, 58, 49, 56, 52, 57
Find the mean and the standard deviation.
2. When a die is thrown, the numbers turn out as follows:
Numberfrequency (f)
13
27
310
414
58
62
Find the mean and the standard deviation.
3. The weights of some chicks obtained by a farmer are as follows:
weight (x)frequency (f)
0 - 207
21 - 4011
41 - 603
61 - 807
81 - 1002
Find the mean and the standard deviation.
4. The standard deviation of a certain set of data is 4.2. What would be the next standard deviation, if each data was increased by 5 ?
5. The standard deviation of a certain set of data is 4.2. What would be the next standard deviation, if each data was multiplied by 5 ?
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# Problem of the Week Problem B and Solution 'Temp'ting Crickets
## Problem
Crickets can help determine the temperature, in degrees Celsius. One possible way to make this calculation is to follow the steps below.
• Step 1: Count the number of chirps in $$25$$ seconds.
• Step 2: Divide the number from Step 1 by $$3$$.
• Step 3: Add $$4$$ to the number from Step 2.
1. By filling in each $$\underline{\ \ \ \ \ }$$ in the following equation with either a variable or a number, write an equation to show how to get the temperature, $$t$$, based on a certain number of chirps, $$c$$, in $$25$$ seconds. $t = \underline{\ \ \ \ \ }\div \underline{\ \ \ \ \ }+\underline{\ \ \ \ \ }$
2. Fill in the second column of the following table.
Chirps ($$c$$) in $$25$$ seconds Temperature ($$t$$) in degrees Celsius
$$60$$
$$54$$
$$66$$
3. Fill in the first column of the following table.
Chirps ($$c$$) in $$25$$ seconds Temperature ($$t$$) in degrees Celsius
$$18$$
$$20$$
$$16$$
## Solution
1. To determine the temperature, $$t$$, we take the number of chirps in $$25$$ seconds, $$c$$, divide by $$3$$, then add $$4$$. That is, $$t = \underline{\,c\,}\div\underline{\,3\,}+\underline{\,4\,}$$.
2. You may use the given steps or the equation from part (a) to fill in the table.
For example when there are $$60$$ chirps, we divide by $$3$$ to get $$20$$, and then add $$4$$ to get $$24$$ degrees Celsius.
Or we may use the equation $$t = 60 \div 3 + 4 = 20 + 4 = 24$$.
Chirps ($$c$$) in $$25$$ seconds Temperature ($$t$$) in degrees Celsius
$$60$$ $$24$$
$$54$$ $$22$$
$$66$$ $$26$$
3. To find the number of chirps for a given temperature, we work backwards, reversing the steps as we go. That is, we subtract $$4$$ from the given temperature, and then multiply by $$3$$.
For example when the temperature is $$18$$ degrees Celsius, we subtract $$4$$ to get $$14$$, and then multiply $$14$$ by $$3$$ to get $$42$$ chirps.
The equation to calculate chirps, $$c$$, given temperature, $$t$$, is $$c=(t-4)\times 3$$.
Chirps ($$c$$) in $$25$$ seconds Temperature ($$t$$) in degrees Celsius
$$42$$ $$18$$
$$48$$ $$20$$
$$36$$ $$16$$
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Question 1: Find the modulus and argument of the following complex numbers and hence express each one of them in polar form:
$\displaystyle \text{i) } 1+i \hspace{1.0cm} \text{ii) } \sqrt{3}+i \hspace{1.0cm} \text{iii) } 1-i \hspace{1.0cm} \text{iv) } \frac{1-i}{1+i} \hspace{1.0cm} \text{v) } \frac{1}{1+i}$
$\displaystyle \text{vi) } \frac{1+2i}{1-3i} \hspace{1.0cm} \text{vii) } \sin 120^{\circ} - i \cos 120^{\circ} \hspace{1.0cm} \text{viii) } \frac{-16}{1+i \sqrt{3}}$
$\displaystyle \text{i) } z = 1+i$
$\displaystyle r = |z| = \sqrt{(1)^2+(1)^2} = \sqrt{2}$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{1}{1} \Big| \Rightarrow \alpha = \frac{\pi}{4}$
Since the point $\displaystyle (1, 1)$ lies in the first quadrant, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = \alpha = \frac{\pi}{4}$
$\displaystyle \text{Polar form of } 1 + i \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = \sqrt{2} \big( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \big)$
$\displaystyle \text{ii) } z = \sqrt{3}+i$
$\displaystyle r = |z| = \sqrt{(\sqrt{3})^2+(1)^2} = 2$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{1}{\sqrt{3}} \Big| \Rightarrow \alpha = \frac{\pi}{6}$
Since the point $\displaystyle (\sqrt{3}, 1)$ lies in the first quadrant, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = \alpha = \frac{\pi}{6}$
$\displaystyle \text{Polar form of } \sqrt{3}+i \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = 2 \big( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \big)$
$\displaystyle \text{iii) } z = 1-i$
$\displaystyle r = |z| = \sqrt{(1)^2+(-1)^2} = \sqrt{2}$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{-1}{1} \Big| \Rightarrow \alpha = \frac{\pi}{4}$
Since the point $\displaystyle (1, -1)$ lies in the fourth quadrant, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = -\alpha = - \frac{\pi}{4}$
$\displaystyle \text{Polar form of } 1 - i \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = \sqrt{2} \big( \cos \frac{-\pi}{4} + i \sin \frac{-\pi}{4} \big)$
$\displaystyle = \sqrt{2} \big( \cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \big)$
$\displaystyle \text{iv) } z = \frac{1-i}{1+i} = \frac{1-i}{1+i} \times \frac{1-i}{1-i} = \frac{-2i}{2} =-i$
$\displaystyle z = -i$
$\displaystyle r = |z| = \sqrt{(0)^2+(1)^2} = 1$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{-1}{0} \Big| \Rightarrow \alpha = \frac{-\pi}{2}$
Since the point $\displaystyle (0, -1)$ lies on the negative direction of the imaginary axis, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = \alpha = \frac{3\pi}{2}$
$\displaystyle \text{Polar form of } \frac{1-i}{1+i} \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = 1 \big( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \big)$
$\displaystyle = \big( \cos \frac{\pi}{2} - i \sin \frac{\pi}{2} \big)$
$\displaystyle \text{v) } z = \frac{1}{1+i} = \frac{1}{1+i} \times \frac{1-i}{1-i} = \frac{1-i}{2} = \frac{1}{2} - i \frac{1}{2}$
$\displaystyle z = \frac{1}{2} - i \frac{1}{2}$
$\displaystyle r = |z| = \sqrt{(\frac{1}{2})^2+(\frac{-1}{2})^2} = \sqrt{\frac{1}{2}}$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{\frac{1}{2}}{-\frac{1}{2}} \Big| = 1 \Rightarrow \alpha = \frac{\pi}{4}$
Since the point $\displaystyle (\frac{1}{2}, -\frac{1}{2})$ lies in the fourth quadrant, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = -\alpha = - \frac{\pi}{2}$
$\displaystyle \text{Polar form of } \frac{1}{1+i} \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = \frac{1}{\sqrt{2}} \big( \cos \frac{-\pi}{4} + i \sin \frac{-\pi}{4} \big)$
$\displaystyle = \big( \cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \big)$
$\displaystyle \text{vi) } z = \frac{1+2i}{1-3i} = \frac{1+2i}{1-3i} \times \frac{1+3i}{1+3i} = \frac{-5+5i}{10} = - \frac{1}{2} + i \frac{1}{2}$
$\displaystyle z = - \frac{1}{2} + i \frac{1}{2}$
$\displaystyle r = |z| = \sqrt{(\frac{-1}{2})^2+(\frac{1}{2})^2} = \sqrt{\frac{1}{2}}$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{\frac{1}{2}}{-\frac{1}{2}} \Big| = 1 \Rightarrow \alpha = \frac{\pi}{4}$
Since the point $\displaystyle \big(- \frac{1}{2} , \frac{1}{2} \big)$ lies in the second quadrant, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = \pi -\alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
$\displaystyle \text{Polar form of } \frac{1+2i}{1-3i} \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = \frac{1}{\sqrt{2}} \big( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \big)$
$\displaystyle \text{vii) } \sin 120^{\circ} - i \cos 120^{\circ} = \frac{\sqrt{3}}{2} + i \frac{1}{2}$
$\displaystyle r = |z| = \sqrt{(\frac{\sqrt{3}}{2})^2+(\frac{1}{2})^2} = \sqrt{1} = 1$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \Big| = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \frac{\pi}{6}$
Since the point $\displaystyle \big( \frac{\sqrt{3}}{2} , \frac{1}{2} \big)$ lies in the first quadrant, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = \alpha = \frac{\pi}{6}$
$\displaystyle \text{Polar form of } \sin 120^{\circ} - i \cos 120^{\circ} \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = \big( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \big)$
$\displaystyle \text{viii) } z = \frac{-16}{1+i \sqrt{3} } = \frac{-16}{1+i \sqrt{3} } \times \frac{1-i\sqrt{3}}{1-i \sqrt{3} } = \frac{-16+16\sqrt{3} i}{4} = -4 + 4 \sqrt{3} i$
$\displaystyle z = -4 + 4 \sqrt{3} i$
$\displaystyle r = |z| = \sqrt{(-4)^2+(4 \sqrt{3})^2} = \sqrt{64} = 8$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{4\sqrt{3}}{-4} \Big| = \sqrt{3} \Rightarrow \alpha = \frac{\pi}{3}$
Since the point $\displaystyle (-4 , 4 \sqrt{3})$ lies in the third quadrant, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = \pi - \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\displaystyle \text{Polar form of } \frac{-16}{1+i \sqrt{3} } \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = 8 \big( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \big)$
$\displaystyle \\$
Question 2: Write $\displaystyle (i^{25})^3$ in polar form
$\displaystyle z = (i^{25})^3 = i^{75} = i^{4 \times 18 + 3} = i^3 = -i$
$\displaystyle z = - i$
$\displaystyle r = |z| = \sqrt{(0)^2+(-1)^2} = 1$
$\displaystyle \text{Let } \tan \alpha = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \Rightarrow \tan \alpha = \Big| \frac{1}{0} \Big| \Rightarrow \alpha = \frac{\pi}{2}$
Since the point $\displaystyle (0, -1)$ lies in the fourth quadrant, the $\displaystyle \text{arg}(z)$ is given by
$\displaystyle \theta = - \alpha = - \frac{\pi}{2}$
$\displaystyle \text{Polar form of } (i^{25})^3 \text{ is given by } = r ( \cos \theta + i \sin \theta)$
$\displaystyle = \big( \cos \frac{-\pi}{2} + i \sin \frac{-\pi}{2} \big)$
$\displaystyle = \big( \cos \frac{\pi}{2} - i \sin \frac{\pi}{2} \big)$
$\displaystyle \\$
Question 3: Express the following complex numbers in the form $\displaystyle r ( \cos \theta + i \sin \theta)$:
$\displaystyle \text{i) } 1+ i \tan \alpha \hspace{1.0cm} \text{ii) } \tan \alpha - i \hspace{1.0cm} \text{iii) } 1 - \sin \alpha + i \cos \alpha \hspace{1.0cm} \text{iv) } \frac{1-i}{\cos \frac{\pi}{3}+ i \sin \frac{\pi}{3} }$
$\displaystyle \text{i) } \text{Let } z = 1+ i \tan \alpha$
$\displaystyle \tan \alpha$ is a periodic function with period $\displaystyle \pi$.
Hence we take $\displaystyle \alpha \in \Big[ 0, \frac{\pi}{2} \Big) \cup \Big( \frac{\pi}{2} , \pi \Big]$
$\displaystyle \underline{ \text{Case I:}} \alpha \in \Big[ 0, \frac{\pi}{2} \Big)$
$\displaystyle z = 1 + i \tan \alpha$
$\displaystyle \Rightarrow |z| = \sqrt{1 + \tan^2 \alpha} = \sqrt{\sec^2 \alpha} = | \sec \alpha | = \sec \alpha$
$\displaystyle \text{Let } \beta \text{ be an acute angle given by } \tan \beta = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \tan \beta = | \tan \alpha | = \tan \alpha \Rightarrow \beta = \alpha$
We can see that $\displaystyle Re(z) > 0 \text{ and } Im(z) > 0$, $\displaystyle \text{Hence } z$ lies in first quadrant, $\displaystyle arg(z) = \beta = \alpha$.
Thus $\displaystyle z$ in the polar form is given by $\displaystyle z = \sec \alpha ( \cos \alpha + i \sin \alpha)$
$\displaystyle \underline{ \text{Case II:}} \alpha \in \Big( \frac{\pi}{2}$ , $\displaystyle \pi \Big]$
$\displaystyle z = 1 + i \tan \alpha$
$\displaystyle \Rightarrow |z| = \sqrt{1 + \tan^2 \alpha} = \sqrt{\sec^2 \alpha} = | \sec \alpha | = -\sec \alpha$
$\displaystyle \text{Let } \beta \text{ be an acute angle given by } \tan \beta = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \tan \beta = | \tan \alpha | = - \tan \alpha \Rightarrow \tan \beta = \tan ( \pi - \alpha) \Rightarrow \beta = \pi - \alpha$
We can see that $\displaystyle Re(z) > 0 \text{ and } Im(z) < 0$, $\displaystyle \text{Hence } z$ lies in fourth quadrant, $\displaystyle arg(z) = -\beta = \alpha - \pi$.
Thus $\displaystyle z$ in the polar form is given by $\displaystyle z = -\sec \alpha \big( \cos (\alpha - \pi) + i \sin (\alpha - \pi) \big)$
$\displaystyle \text{ii) } \text{Let } z = \tan \alpha - i$
$\displaystyle \tan \alpha$ is a periodic function with period $\displaystyle \pi$.
Hence we take $\displaystyle \alpha \in \Big[ 0, \frac{\pi}{2} \Big) \cup \Big( \frac{\pi}{2} , \pi \Big]$
$\displaystyle \underline{ \text{Case I:}} \alpha \in \Big[ 0, \frac{\pi}{2} \Big)$
$\displaystyle z = \tan \alpha - i$
$\displaystyle \Rightarrow |z| = \sqrt{\tan^2 \alpha + 1} = \sqrt{\sec^2 \alpha} = | \sec \alpha | = \sec \alpha$
$\displaystyle \text{Let } \beta \text{ be an acute angle given by } \tan \beta = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \tan \beta = \Big| \frac{1}{\tan \alpha} \Big| = | \cot \alpha | = \cot \alpha$
$\displaystyle \Rightarrow \tan \beta = \tan \big( \frac{\pi}{2} - \alpha \big) \Rightarrow \beta = \frac{\pi}{2} - \alpha$
We can see that $\displaystyle Re(z) > 0 \text{ and } Im(z) < 0$, $\displaystyle \text{Hence } z$ lies in fourth quadrant, therefore $\displaystyle arg(z) = -\beta = \alpha - \frac{\pi}{2}$ .
Thus $\displaystyle z$ in the polar form is given by
$\displaystyle z = \sec \alpha \Big\{ \cos (\alpha - \frac{\pi}{2} \big) + i \sin \big( \alpha - \frac{\pi}{2} ) \Big\}$
$\displaystyle \underline{ \text{Case II:}} \alpha \in \Big( \frac{\pi}{2}$ , $\displaystyle \pi \Big]$
$\displaystyle z = \tan \alpha - i$
$\displaystyle \Rightarrow |z| = \sqrt{\tan^2 \alpha+1 } = \sqrt{\sec^2 \alpha} = | \sec \alpha | = -\sec \alpha$
$\displaystyle \text{Let } \beta \text{ be an acute angle given by } \tan \beta = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \tan \beta = \Big| \frac{1}{\tan \alpha} \Big| = | \cot \alpha | = - \cot \alpha$
$\displaystyle \Rightarrow \tan \beta = \tan \big( \alpha - \frac{\pi}{2} \big) \Rightarrow \beta = \alpha - \frac{\pi}{2}$
We can see that $\displaystyle Re(z) < 0 \text{ and } Im(z) < 0$, $\displaystyle \text{Hence } z$ lies in third quadrant, therefore $\displaystyle arg(z) = \pi + \beta = \frac{\pi}{2} + \alpha$.
Thus $\displaystyle z$ in the polar form is given by
$\displaystyle z = -\sec \alpha \Big\{ \cos ( \frac{\pi}{2} + \alpha \big) + i \sin ( \frac{\pi}{2} + \alpha \big) \Big\}$
$\displaystyle \text{iii) } \text{Let } z = (1 - \sin \alpha) + i \cos \alpha$
Both Sine and Cosine functions are periodic function with period $\displaystyle 2\pi$
Hence let us take $\displaystyle \alpha \in [ 0, 2\pi ]$
$\displaystyle \Rightarrow |z| = \sqrt{ (1- \sin \alpha)^2 + \cos^2 \alpha } = \sqrt{2- 2 \sin \alpha} = \sqrt{2} \sqrt{1- \sin \alpha}$
$\displaystyle \Rightarrow |z| = \sqrt{2} \sqrt{ (\cos \frac{\alpha}{2} -\sin \frac{\alpha}{2})^2 } = \sqrt{2} \Big| \cos \frac{\alpha}{2} -\sin \frac{\alpha}{2} \Big|$
$\displaystyle \text{Let } \beta \text{ be an acute angle given by } \tan \beta = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \tan \beta = \frac{| \cos \alpha |}{|1- \sin \alpha |} = \Big| \frac{\cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2}}{( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2})^2} \Big| = \Big| \frac{\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2}}{ \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}} \Big| = \Big| \frac{1 + \tan \frac{\alpha}{2}}{ 1 - \tan \frac{\alpha}{2}} \Big|$
$\displaystyle \Rightarrow \tan \beta = \Big| \tan ( \frac{\pi}{4} + \frac{\alpha}{2} ) \Big|$
$\displaystyle \underline{ \text{Case I:}} 0 \leq \alpha < \frac{\pi}{2}$
In this case $\displaystyle \cos \frac{\alpha }{2} > \sin \frac{\alpha }{2} \text{ and } \frac{\pi}{4} + \frac{\alpha}{2} \in \Big[ \frac{\pi }{4} , \frac{\pi}{2} \Big)$
$\displaystyle \Rightarrow |z| = \sqrt{2} \Big( \cos \frac{\alpha }{2} - \sin \frac{\alpha }{2} \Big)$
$\displaystyle \text{and } \tan \beta = \Big| \tan ( \frac{\pi}{4} + \frac{\alpha}{2} ) \Big| = \tan ( \frac{\pi}{4} + \frac{\alpha}{2} )$
$\displaystyle \Rightarrow \beta = \frac{\pi}{4} + \frac{\alpha}{2}$
$\displaystyle \text{Clearly, } z$ lies in the first quadrant, therefore $\displaystyle arg( z) = \frac{\pi}{4} + \frac{\alpha}{2}$
Hence the $\displaystyle \text{Polar form of } z$ is
$\displaystyle \sqrt{2} \Big( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} \Big) \Big\{ \cos ( \frac{\pi}{4} + \frac{\alpha}{2} ) + i \sin ( \frac{\pi}{4} + \frac{\alpha}{2} ) \Big\}$
$\displaystyle \underline{ \text{Case II:}} \frac{\pi}{2} < \alpha < \frac{3\pi}{2}$
In this case $\displaystyle \cos \frac{\alpha }{2} < \sin \frac{\alpha }{2} \text{ and } \frac{\pi}{4} + \frac{\alpha}{2} \in \Big( \frac{\pi }{2} , \pi \Big)$
$\displaystyle |z| = \Big|\sqrt{2} \Big( \cos \frac{\alpha }{2} - \sin \frac{\alpha }{2} \Big) \Big| = - \sqrt{2} \Big( \cos \frac{\alpha }{2} - \sin \frac{\alpha }{2} \Big)$
$\displaystyle \text{and } \tan \beta = \Big| \tan ( \frac{\pi}{4} + \frac{\alpha }{2} ) \Big| = - \tan ( \frac{\pi}{4} + \frac{\alpha }{2} ) = \tan \{ \pi - ( \frac{\pi}{4} + \frac{\alpha }{2} ) \}$
$\displaystyle \tan \beta = \tan ( \frac{3\pi}{4} - \frac{\alpha}{2} )$
$\displaystyle \Rightarrow \beta = \Big( \frac{3\pi}{4} - \frac{\alpha}{2} \Big)$
$\displaystyle \text{Clearly, } z$ lies in the fourth quadrant, therefore
$\displaystyle arg( z) = - \beta = - \Big( \frac{3\pi}{4} - \frac{\alpha}{2} \Big) = \frac{\alpha}{2} - \frac{3\pi}{4}$
Hence the $\displaystyle \text{Polar form of } z$ is
$\displaystyle -\sqrt{2} \Big( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} \Big) \Big\{ \cos ( \frac{\alpha}{2} - \frac{3\pi}{2} ) + i \sin ( \frac{\alpha}{2} - \frac{3\pi}{4} ) \Big\}$
$\displaystyle \underline{ \text{Case III:}} \frac{3\pi}{2} < \alpha < 2\pi$
In this case $\displaystyle \cos \frac{\alpha }{2} < \sin \frac{\alpha }{2} \text{ and } \frac{\pi}{4} + \frac{\alpha}{2} \in \Big( \pi , \frac{5\pi}{4} \Big)$
$\displaystyle |z| = \Big|\sqrt{2} \Big( \cos \frac{\alpha }{2} - \sin \frac{\alpha }{2} \Big) \Big| = - \sqrt{2} \Big( \cos \frac{\alpha }{2} - \sin \frac{\alpha }{2} \Big)$
$\displaystyle \text{and } \tan \beta = \Big| \tan ( \frac{\pi}{4} + \frac{\alpha }{2} ) \Big| = \tan ( \frac{\pi}{4} + \frac{\alpha }{2} ) = - \tan \{ \pi - ( \frac{\pi}{4} + \frac{\alpha }{2} ) \}$
$\displaystyle \tan \beta = \tan ( \frac{\alpha}{2} - \frac{3\pi}{4} )$
$\displaystyle \Rightarrow \beta = \Big( \frac{\alpha}{2} - \frac{3 \pi}{4} \Big)$
$\displaystyle \text{Clearly, } z$ lies in the first quadrant, therefore
$\displaystyle arg( z) = \beta = \Big( \frac{\alpha}{2} - \frac{3\pi}{4} \Big)$
Hence the $\displaystyle \text{Polar form of } z$ is
$\displaystyle -\sqrt{2} \Big( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} \Big) \Big\{ \cos ( \frac{\alpha}{2} - \frac{3\pi}{4} ) + i \sin ( \frac{\alpha}{2} - \frac{3\pi}{4} ) \Big\}$
$\displaystyle \text{iv) } \text{Let } z = \frac{1-i}{\cos \frac{\pi}{3}+ i \sin \frac{\pi}{3} }$
$\displaystyle = \frac{1 - i }{\frac{1}{2} + i \frac{\sqrt{3}}{2} } = \frac{2-2i}{1 + i \sqrt{3}} \times \frac{1 - i \sqrt{3}}{1 - i \sqrt{3}} = \frac{2 - 2i - i 2 \sqrt{3} + 2 \sqrt{3} i^2}{1+3}$
$\displaystyle = \frac{2 -2\sqrt{3}-2i ( 1 + \sqrt{3})}{4} = \frac{(1-\sqrt{3}) - i ( 1 + \sqrt{3})}{2} = \frac{(1-\sqrt{3})}{2} + i \frac{( -1 - \sqrt{3})}{2}$
$\displaystyle \therefore z = \frac{(1-\sqrt{3})}{2} + i \frac{( -1 - \sqrt{3})}{2}$
$\displaystyle |z| = \sqrt{ \Big( \frac{1-\sqrt{3}}{2} \Big)^2 + \Big( \frac{-1-\sqrt{3}}{2} \Big)^2 } = \sqrt{\frac{1+3-2\sqrt{3}}{4}+ \frac{1+3 + 2\sqrt{3}}{4}} = \sqrt{\frac{8}{4}} = \sqrt{2}$
$\displaystyle \text{Let } \beta \text{ be an acute angle given by } \tan \beta = \Big| \frac{Im(z)}{Re(z)} \Big|$
$\displaystyle \tan \beta = \Big| \frac{\frac{1+\sqrt{3}}{2}}{\frac{1-\sqrt{3}}{2}} \Big| = \Big| \frac{1+\sqrt{3}}{1-\sqrt{3}} \Big| = \Big| \frac{\tan \frac{\pi}{4} + \tan \frac{\pi}{3} }{1 - \tan \frac{\pi}{4} \tan \frac{\pi}{3}} \Big| = \tan ( \frac{\pi}{4} + \frac{\pi}{3} ) = \tan \frac{7\pi}{12}$
$\displaystyle \Rightarrow \beta = \frac{7\pi}{12}$
$\displaystyle \text{Clearly, } z$ lies in the fourth quadrant, therefore $\displaystyle arg( z) = \beta = - \frac{7\pi}{12}$
Hence the $\displaystyle \text{Polar form of } z$ is $\displaystyle \sqrt{2} \Big( \cos \frac{7\pi}{12} - \sin \frac{7\pi}{12} \Big)$
$\displaystyle \\$
Question 4: If $\displaystyle z_1 \text{ and } z_2$ are two complex numbers such that $\displaystyle |z_1|= |z_2| \text{ and } arg(z_1)+arg(z_2)= \pi$ , then show that $\displaystyle z_1 = - \overline{z_2}$
$\displaystyle \text{Let } \theta_1$ by $\displaystyle arg(z_1) \text{ and } \theta_2$ by $\displaystyle arg(z_2)$
$\displaystyle \text{Given } |z_1| = |z_2|$; $\displaystyle \text{and } arg(z_1)+arg(z_2)= \pi$
$\displaystyle \text{Since } z_1$ is a complex number
$\displaystyle z_1 = |z_1| ( \cos \theta_1 + i \sin \theta_1)$
$\displaystyle = |z_2| [ \cos (\pi - \theta_2) + i \sin (\pi - \theta_2) ]$
$\displaystyle = |z_2| [ - \cos \theta_2 + i \sin \theta_2 ]$
$\displaystyle = - |z_2|[\cos \theta_2 - i \sin \theta_2 ]$
$\displaystyle = - \overline{z_2}$
$\displaystyle \text{Hence } z_1 = - \overline{z_2}$
$\displaystyle \\$
Question 5: If $\displaystyle z_1, z_2 \text{ and } z_3, z_4$ are two pairs of conjugate complex numbers, prove that $\displaystyle arg \Big( \frac{z_1}{z_4} \Big) + arg \Big( \frac{z_2}{z_3} \Big) = 0$
$\displaystyle z_1, z_2 \text{ and } z_3, z_4$ are two pairs of conjugate complex numbers.
$\displaystyle \therefore z_1 = r_1 e^{i \theta_1}, z_2 = r_1 e^{-i \theta_1}, z_3 = r_2 e^{i \theta_2} , z_4 = r_2 e^{-i \theta_2}$
$\displaystyle \therefore \frac{z_1}{z_4} = \frac{r_1 e^{i \theta_1}}{r_2 e^{-i \theta_2}} = \Big( \frac{r_1}{r_2} \Big) e^{i( \theta_1 - \theta_2)}$
$\displaystyle \Rightarrow arg \Big( \frac{z_1}{z_4} \Big) = \theta_1 - \theta_2$ … … … … … i)
$\displaystyle \therefore \frac{z_2}{z_3} = \frac{r_1 e^{-i \theta_1}}{r_2 e^{i \theta_2}} = \Big( \frac{r_1}{r_2} \Big) e^{i( -\theta_1 + \theta_2)}$
$\displaystyle \Rightarrow arg \Big( \frac{z_2}{z_3} \Big) = \theta_2 - \theta_1$ … … … … … ii)
$\displaystyle arg \Big( \frac{z_1}{z_4} \Big) + arg \Big( \frac{z_2}{z_3} \Big) = \theta_1 - \theta_2 + \theta_2 - \theta_1$
$\displaystyle \Rightarrow arg \Big( \frac{z_1}{z_4} \Big) + arg \Big( \frac{z_2}{z_3} \Big) = 0$. Hence proved.
$\displaystyle \\$
Question 6: Express $\displaystyle \sin \frac{\pi}{5} + i \Big( 1 - \cos \frac{\pi}{5} \Big)$ in polar form.
$\displaystyle \text{Let } z = \sin \frac{\pi}{5} + i \Big( 1 - \cos \frac{\pi}{5} \Big)$
$\displaystyle |z| = \sqrt{ \big( \sin \frac{\pi}{5} \big)^2 + \big(1 - \cos \frac{\pi}{5} \big)^2 }$
$\displaystyle = \sqrt{ \sin^2 \frac{\pi}{5} + 1 +\cos^2 \frac{\pi}{5} - 2 \cos \frac{\pi}{5} }$
$\displaystyle = \sqrt{2} \Big( \sqrt{1 -\cos \frac{\pi}{5} } \Big)$
$\displaystyle = \sqrt{2} \sqrt{ 2 \sin^2 \frac{\pi}{10}}$
$\displaystyle = 2 \sin \frac{\pi}{10}$
$\displaystyle \tan \beta = \frac{|1 - \cos \frac{\pi}{5} |}{|\sin \frac{\pi}{5} |} = \Big| \frac{2 \sin^2 \frac{\pi}{10}}{ 1 \sin \frac{\pi}{10} \cos \frac{\pi}{10}} \Big| = | \tan \frac{\pi}{10} |$
$\displaystyle \Rightarrow \beta = \frac{\pi}{10}$
$\displaystyle \text{Clearly, } z \text{ lies in the first quadrant. Therefore } arg(z) = \frac{\pi}{10} . \\ \\ \text{ Hence the } \text{Polar form of } z \text{ is } 2 \sin \frac{\pi}{10} \Big( \cos \frac{\pi}{10} + i \sin \frac{\pi}{10} \Big)$
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1. ## Dice session :-)
10 dice is rolled, what is the probability of:
a) getting one & two once and all other numbers exactly twice
b) getting one & two exactly 3 times
c) getting four even and six odd numbers
d) getting one & three two times, if we know that two appeared once and five 3 times
2. The correct answer to part a) is $\frac{10!}{(2^4)(6^{10})}$.
That is the number of ways to rearrange the string $1233445566$ times the probability of any string.
3. I'm not sure I got it, which distribution is used in this solution?
4. Originally Posted by losm1
I'm not sure I got it, which distribution is used in this solution?
See this Wikipedia article on multinomial coefficient.
Let E = the event that we get one 1, one 2, and two of every other face value.
Let M = {1, 2, 3, 3, 4, 4, 5, 5, 6, 6}
Say the dice are labeled from 1 to 10, and that after rolling them you put them in ascending order from left to right. This is the same as rolling a single die 10 times in a row and writing down the result from left to right.
There are 6^10 possible combinations, as Plato says, strings.
The number of ways to get E is the number of permutations of M, which is
$\frac{10!}{1!1!2!2!2!2!} = \frac{10!}{2^4}$
Then we simply write
$P(E)=\frac{\text{number of ways to get E}}{\text{number of ways to roll dice}}=\frac{10!}{2^4\cdot6^{10}}$
5. What about b)? I'm not sure how to apply multinomial distribution formula on this one?
Formula is $\frac{n!}{k_1! k_1! \cdots k_n!}\times p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}$
$\frac{10!}{3!3!\cdot 6^{10}}$ doesn't feel right.
6. Originally Posted by losm1
What about b)? I'm not sure how to apply multinomial distribution formula on this one?
Formula is $\frac{n!}{k_1! k_1! \cdots k_n!}\times p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}$
$\frac{10!}{3!3!\cdot 6^{10}}$ doesn't feel right.
You really need to learn to model the outcome space.
For part (b) the desired outcomes are rearrangements of the string $111222XXXX$.
Where each $X$ can be $3,4,5,\text{ or }6$.
The string can be arranged in $\frac{10!}{(3!)^2(4!)}$ ways.
But each $X$ can be replaced in four ways that gives $\frac{(10!)(4^4)}{(3!)^2(4!)}$.
Now each of those strings has probability $\frac{1}{6^{10}}$.
7. Thanks Plato!
8. Originally Posted by Plato
You really need to learn to model the outcome space.
For part (b) the desired outcomes are rearrangements of the string $111222XXXX$.
Where each $X$ can be $3,4,5,\text{ or }6$.
The string can be arranged in $\frac{10!}{(3!)^2(4!)}$ ways.
But each $X$ can be replaced in four ways that gives $\frac{(10!)(4^4)}{(3!)^2(4!)}$.
Now each of those strings has probability $\frac{1}{6^{10}}$.
I don't know if this post will be of value to anyone, but here is another method which is much more complicated than Plato's and gives the same result. I outlined it originally because I was having trouble following Plato's reasoning, even though it seems really easy to follow now.
Consider each possible type of string, by which I mean:
111222XXXX {4}
111222XXXY {1,3}
111222XXYY {2,2}
111222XYZZ {1,1,2}
111222XYZW {1,1,1,1}
where X, Y, Z, and W can be 3, 4, 5, or 6.
This corresponds with the ways 4 can be partitioned as a sum of positive integers. (For partition function P, P(4) = 5.)
Then you count the ways to get each one and multiply by the corresponding ways to arrange the string.
4 ways to get {4} resulting in $\frac{(4)(10!)}{(3!)^2(4!)^1}$.
$\binom{4}{2}\cdot2$ ways to get {1,3} resulting in $\frac{(12)(10!)}{(3!)^3(1!)^1}$.
$\binom{4}{2}$ ways to get {2,2} resulting in $\frac{(6)(10!)}{(3!)^2(2!)^2}$.
$\binom{4}{3}\cdot3$ ways to get {1,1,2} resulting in $\frac{(12)(10!)}{(3!)^2(1!)^2(2!)^1}$.
1 way to get {1,1,1,1} resulting in $\frac{(1)(10!)}{(3!)^2(1!)^4}$.
Which can be summarized as
$P=\left(\frac{10!}{(3!)^2 6^{10}}\right)\left(\frac{4}{(4!)^1}+\frac{12}{(1! )^1(3!)^1} + \frac{6}{(2!)^2} + \frac{12}{(1!)^2(2!)^1} + \frac{1}{(1!)^4}\right)$
We get $P=\frac{350}{19683} \approx 0.01778$.
By the way, I also programmed a Monte Carlo to verify the result. (Because initially there was a discrepancy between the two answers, due to my writing a 12 instead of a 6 in one spot.)
Code:
import java.util.Random;
public class diceRollsProblem() {
Random g = new Random();
public void monteCarlo() {
int successes = 0;
for (int i = 0; i < Integer.MAX_VALUE; i++) {
if (i % 500 == 0)
System.out.println(i + " " + successes/(double)i);
int[] rolls = new int[10];
for (int j = 0; j < 10; j++)
rolls[j] = g.nextInt(6)+1;
int num1s = 0, num2s = 0;
for (int j = 0; j < 10; j++) {
if (rolls[j] == 1) num1s++;
if (rolls[j] == 2) num2s++;
}
if (num1s == 3 && num2s == 3) successes++;
}
}
}
Edit: This program's a bit better.
Code:
import java.util.Random;
public class diceRollsProblem() {
Random g = new Random();
public void monteCarlo() {
int successes = 0;
for (int i = 0; i < Integer.MAX_VALUE; i++) {
if (i % 100000 == 0)
System.out.println(i + " " + successes/(double)i);
int[] distr = new int[6];
for (int j = 0; j < 10; j++)
distr[g.nextInt(6)]++;
if (distr[0] == 3 && distr[1] == 3) successes++;
}
}
}
9. Great explanation, undefined! You're right, when you put it this way it's much easier to comprehend.
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### AVERAGE
1. Explanation :
There are five prime numbers between 30 and 50. They are 31,37,41,43 and 47.Therefore the required average=$$\frac{(31+37+41+43+47)}{5}$$=$$\frac{199}{5}$$ =39.8.
1. Explanation :
sum of first n natural numbers=n(n+1)/2; So,sum of 40 natural numbers=$$\frac{(40*41)}{2}$$=820.
Therefore the required average=$$\frac{820}{40}$$=20.5.
1. Explanation :
Required average =7*$$\frac{(1+2+3+…….+20)}{20}$$=$$\frac{(7*20*21)}{(20*2)}$$=$$\frac{147}{2}$$=73.5.
1. Explanation :
let the numbers be x,x+2,x+4 and x+6. Then,
$$\frac{(x+(x+2)+(x+4)+(x+6))}{4}$$ = 27
$$\frac{(4x+12)}{4}$$= 27
x+3=27
x=24. Therefore the largest number=(x+6)=24+6=30.
1. Explanation :
Total weight of(36+44) students=(36*40+44*35)kg =2980kg. Therefore weight of the total class=(2980/80)kg =37.25kg.
1. Explanation :
Let the average expenditure of all nine be Rs.x Then 12*8+(x+8)=9x or 8x=104 or x=13. Total money spent = 9x = Rs.(9*13)= Rs.117.
1. Explanation :
Let the third number be x.
Then second number = 3x.
First number=3x/2.
Therefore x+3x+(3x/2)=(44*3) or x=24 So largest number= 2nd number=3x=72.
1. Explanation :
Clearly 13th result=(sum of 25 results)-(sum of 24 results)
=(18*25)-(14*12)+(17*12)
=450-(168+204)
=450-372
=78
1. Explanation :
6th result = (58*6+63*6-60*11)=66
1. Explanation :
Let A,B,c represent their individual wgts.
Then, A+B+C=(45*3)Kg=135Kg
A+B=(40*2)Kg=80Kg & B+C=(43*2)Kg=86Kg
B=(A+B)+(B+C)-(A+B+C) =(80+86-135)Kg =31Kg.
1. Explanation :
Total age of 39 persons = (39 * 15) years = 585 years.
Average age of 40 persons= 15 yrs 3 months = 61/4 years.
Total age of 40 persons = ((61/4 )*40) years= 610 years.
Age of the teacher = (610 - 585) years=25 years.
1. Explanation :
Total weight increased =(1.8 x 10) kg =18 kg.
Weight of the new man =(53 + 18) kg =71 kg.
1. Explanation :
Let the original average expenditure be Rs. x. Then, 42 (x - 1) - 35x = 42 --> 7x = 84 --> x = 12 --> Original expenditure = Rs. (35 x 12) = Rs. 420.
1. Explanation :
Let the average after 17th inning = x. Then, average after 16th inning = (x - 3) 16 (x - 3) + 87 = 17x or x = (87 - 48) = 39
1. Explanation :
Required average speed = ((2xy)/(x+y)) km / hr
=(2 x 84 x 56)/(84+56)km/hr
= (2*84*56)/140 km/hr
=67.2 km/hr.
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The additive inverse, or opposite, of a number n is the number which, when added to n, yields zero. The additive inverse of n is denoted −n.
For example:
• The additive inverse of 7 is −7, because 7 + (−7) = 0;
• The additive inverse of −0.3 is 0.3, because −0.3 + 0.3 = 0.
Thus by the last example, −(−0.3) = 0.3.
The additive inverse of a number is its inverse element under the binary operation of addition. It can be calculated using multiplication by −1; that is, −n = −1 × n.
Types of numbers with additive inverses include:
Types of numbers without additive inverses (of the same type) include:
But note that we can construct the integers out of the natural numbers by formally including additive inverses. Thus we can say that natural numbers do have additive inverses, but because these additive inverses are not themselves natural numbers, the set of natural numbers is not closed under taking additive inverses.
Contents
## General definition
The notation '+' is reserved for commutative binary operations, i.e. such that x + y = y + x, for all x,y. If such an operation admits a neutral element o (such that x + o (= o + x) = x for all x), then this element is unique (o' = o' + o = o). If then, for a given x, there exists x' such that x + x' (= x' + x) = o, then x' is called an additive inverse of x.
If '+' is associative ((x+y)+z = x+(y+z) for all x,y,z), then an additive inverse is unique
( x" = x" + o = x" + (x + x') = (x" + x) + x' = o + x' = x' )
and denoted by (– x), and one can write x – y instead of x + (– y).
## Other examples
All the following examples are in fact abelian groups:
• addition of real valued functions: here, the additive inverse of a function f is the function –f defined by (– f)(x) = – f(x), for all x, such that f + (–f) = o, the null function (constantly equal to zero, for all arguments).
• more generally, what precedes applies to all functions with values in an abelian group ('zero' meaning then the neutral element of this group):
• complex valued functions,
• vector space valued functions (not necessarily linear),
## Universal construction
to do: symmetrization of an abelian semigroup
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy
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# 7.07 Dot plots
Lesson
The dot plot is a useful way to express discrete data in a visually simple manner. The main advantages of the dot plot are that we can find the mode and range very easily, as well as quickly see how the data is distributed. The main disadvantages are that we need to count each dot when finding the median and it is often easier to convert to a table to find the mean.
The dot plot is particularly suited to discrete data where the frequency of results are often greater than one.
Dot plot
In a dot plot, each dot represents one data point belonging to the result that it is placed above.
The mode(s) of a dot plot will be the result(s) with the most dots. Since a dot plot stacks vertically, the greatest column(s) will belong to the mode(s).
Let's have a look at an example of a dot plot.
#### Worked Example
Consider the dot plot below.
What is the mode, range, median and mean of the data set represented by this dot plot?
Think: We have just learned that the mode will be the result with the greatest column. To find the range, median, and mean, we'll need to find the numerical values of each piece of data displayed in the dot plot.
Do: Since the stack of dots above "$1$1" is the greatest, we know that the mode is "$1$1".
The range will be the difference between the greatest score, $6$6, and the least score, $0$0. Since $6-0=6$60=6, the range is "$6$6".
By counting the total number of dots in the dot plot, we find that there are $19$19 dots. Since there is an odd number of scores, the median will be the middle score, in this case the tenth score. By counting the dots moving top to bottom and left to right, we find that the tenth dot is in the "$2$2" column. So the median is "$2$2".
To find the mean, we want to add all the scores together and then divide that sum by $19$19 (the number of scores). We recall that we can add our scores together more easily by adding the product of each result and its frequency. For this, we want to convert our dot plot into a frequency table by counting how many dots in each column.
No. children Frequency $fx$fx
$0$0 $3$3 $0$0
$1$1 $6$6 $6$6
$2$2 $4$4 $8$8
$3$3 $2$2 $6$6
$4$4 $3$3 $12$12
$5$5 $0$0 $0$0
$6$6 $1$1 $6$6
We obtained the values in the third column by multiplying each result by its frequency. Adding up all the numbers in the third column is equivalent to finding the sum of all the scores. So we can find the mean by summing the numbers in the third column and then dividing that sum by $19$19. This gives us:
Mean = $\frac{0+6+8+6+12+0+6}{19}$0+6+8+6+12+0+619 = $\frac{38}{19}$3819 = $2$2
So our summary of the data in the dot plot would be:
• Mode = $1$1
• Range =$6$6
• Median = $2$2
• Mean = $2$2
Reflect: As we saw, the mode and range are easy to identify on a dot plot, while the median takes a bit more work and the mean takes a lot more work.
The other feature of the data that we can see clearly with the dot plot is that the data is concentrated around "$1$1" and "$2$2", so it makes sense that our mean, median and mode would be close to these results.
#### Practice questions
##### Question 1
The dot plot shows the temperature ($^\circ C$°C) in a town over a several week period. Identify the temperature that is an outlier.
##### Question 2
The responses were:
$22,17,17,17,19,21,17,22,21,18,18,17,18,22,18$22,17,17,17,19,21,17,22,21,18,18,17,18,22,18
Represent the responses with a dot plot and answer the questions below.
1. What is the range of this data set?
2. What is the mode of this data set?
3. What is the median of this data set?
4. How many people passed their driving test on or after their $19$19th birthday?
##### Question 3
A supermarket manager takes a note every time an employee is late to work, and how late they were (rounded to the nearest half hour). The dot plot below shows their results for the last month:
1. What was the median amount of time that employees were late?
2. What fraction of late employees were later than the median amount?
3. If an employee is more than $1$1 hour late, the manager fines them $\$1010. How much money did the manager collect in fines over the last month?
### Outcomes
#### 6.SP.4
Display numerical data in plots on a number line, including dot plots (line plots), histograms, and box plots. (GAISE Model, step 3)
#### 6.SP.5
Summarize numerical data sets in relation to their context.
#### 6.SP.5a
Report the number of observations.
#### 6.SP.5b
Describe the nature of the attribute under investigation, including how it was measured and its units of measurement.
#### 6.SP.5c
Find the quantitative measures of center (median and/or mean) for a numerical data set and recognize that this value summarizes the data set with a single number. Interpret mean as an equal or fair share. Find measures of variability (range and interquartile range) as well as informally describe the shape and the presence of clusters, gaps, peaks, and outliers in a distribution.
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# Cofunction identity of sin function
## Formula
Sexagesimal System
$\sin{(90^\circ-\theta)} \,=\, \cos{\theta}$
$\sin{\Big(\dfrac{\pi}{2}-x\Big)} \,=\, \cos{x}$
In this formula, the angle in degrees is denoted by Theta ($\theta$) and the angle in radians is denoted by $x$.
The angle of sin function is an allied angle in first quadrant. So, it is called as first quadrant’s allied angle identity of sin function. The angle of sin function is complement of the angle of the cos function. Therefore, it is also called as cofunction identity of sin function.
### Proof
Consider first quadrant. A line segment is rotated to some angle and it is named as $\overline{PQ}$.
Draw perpendicular lines to both horizontal and vertical lines from $Q$ and they interest them at points $S$ and $R$ respectively. Thus, it forms two right angled triangles $\Delta QPS$ and $\Delta QPR$ in first quadrant.
Take $\angle QPR = \theta$, then $\angle QPS = 90^\circ-\theta$.
Take $PS = g$, then $RQ = g$. Similarly, take $RP = h$, then $QS = h$.
#### Express sin of allied angle
The angle of $\Delta QPS$ is an allied angle and it is $90^\circ-\theta$. Now, find the value of sine of first quadrant’s allied angle by expressing it in terms of ratio of the sides of the triangle.
$\sin{(90^\circ-\theta)} = \dfrac{SQ}{PQ}$
The length of the opposite side ($\overline{QS}$) is $h$ but the length of the hypotenuse is unknown. However, the length of the adjacent side is known. Therefore, the length of the hypotenuse can be calculated by Pythagorean theorem.
$\implies \sin{(90^\circ-\theta)} = \dfrac{h}{\sqrt{g^2+h^2}}$
#### Find equivalent value of the fraction
In previous step, sin of allied angle of first quadrant is expressed as a fraction. Now, calculate the value of the fraction in the form a trigonometric function.
It can be done by considering $\Delta QPR$.
$\dfrac{h}{\sqrt{g^2+h^2}} = \dfrac{RP}{PQ}$
As per $\Delta QPR$, the angle of the triangle is theta. The ratio of lengths of $\overline{RP}$ to $\overline{PQ}$ is cos of angle theta.
$\implies \dfrac{h}{\sqrt{g^2+h^2}} = \cos{\theta}$
#### Cofunction identity of sin function
Geometrically, the two steps have proved that
$\sin{(90^\circ-\theta)}$ $=$ $\dfrac{h}{\sqrt{g^2+h^2}}$ $=$ $\cos{\theta}$
$\therefore \,\,\,\,\,\, \sin{(90^\circ-\theta)} \,=\, \cos{\theta}$
Sin of allied angle of first quadrant is equal to the cos of angle. So, it is known first quadrant’s allied angle identity of sin function. The two angles of both trigonometric functions are complementary angles. So, the trigonometric identity is called as cofunction identity of sin function.
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Part 3: Multiplication
# 3.4: Word problems for multiplying fractions
Notes
In the following, “A/B” represents the fraction “A over B.” For example, “2/3” refers to the fraction “2 over 3” or “two-thirds.”
Word problems for the multiplication problem, A/B x C/D, have the form “A/B of C/D of a unit is how much of a unit?” For example,
• Arithmetic problem: Solve 3/5 x 1/3 (three-fifths times one-third).
• Word problem: If 1/3 of the students in a class have a pet and 3/5 of the students with a pet have a dog, then what fraction of the students in the class have a dog as a pet?
• 3/5 x 1/3 = 1/5, so 1/5 of the students in the class have a dog as a pet.
Here’s a couple more examples:
• Arithmetic problem: Solve 1/4 x 2/3 (one-quarter times two-thirds).
• Word problem: If a recipe calls for 2/3 of a cup of soy sauce, then how much soy sauce is needed for 1/4 of the recipe?
• 1/4 x 2/3 = 1/6, so 1/6 of a cup of soy sauce is needed for 1/4 of the recipe.
• Arithmetic problem: Solve 1/6 x 3/4 (one-sixth times three-quarters).
• Word problem: If Alastair used 3/4 of a cup of butter to make a cake and he ate 1/6 of the cake, then how much butter did he eat?
• 1/6 x 3/4 = 1/8, so Alastair ate 1/8 of a cup of butter.
Pay close attention to the wording for this type of problem:
• Word problem: If 1/5 of the students in a class have a dog and 1/6 of the students in the class have a cat, then what fraction of the students in the class have a cat and a dog?
• Can we solve this by calculating 1/6 x 1/5?
• No, because this assumes that 1/6 of the students in the class that have a dog also have a cat (and we don’t know that). It could be that no students in the class have a cat and a dog!
• Word problem: If Carina used 6/7 of a bag of chocolate chips to make a batch of brownies and then she ate 2/3 of the batch of brownies, then how many chocolate chips did she eat?
• Can we solve this by calculating 2/3 x 6/7?
• No, because this calculates the fraction of a bag of chocolate chips that Carina ate, not the number of chocolate chips she ate.
• Word problem: If Sebastián ate 1/3 of a whole pizza from 3/4 of a pizza left over in the fridge, then how much of a whole pizza did he eat?
• Can we solve this by calculating 1/3 x 3/4?
• No, because this would calculate the fraction of a whole pizza that Sebastián ate if he ate 1/3 of the leftover pizza. But he ate 1/3 of a whole pizza, which gives us the answer to the question directly with no calculation needed.
The video below works through some examples of word problems for multiplying fractions.
Practice Exercises
Do the following exercises to practice matching fraction multiplication problems and word problems.
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# rhombus perimeter formula
° rad. The formula for rhombus perimeter is as given perimeter = 4a, where a is the side length as shown in Figure 1. Since all four sides of a rhombus are equal, much like a square, the formula for the perimeter is the product of the length of one side with 4 $$P = 4 \times \text{side}$$ Angles of a Rhombus. The diagonals are not the same length, but they are perpendicular to one another and halve one another. In a rhombus, each diagonal of a rhombus divides it into two congruent triangles. P=2sqrt(2*2*17*17) P=2sqrt(34*34) P=2xx34 P=68 9. Area of a Rhombus Formula. Find the perimeter of a rhombus whose sides measure 9 cm each. There is one using the altitude and side, another using the side and angle, and one for the diagonals. Area and Perimeter of Rhombus: Area = 1/2 (d1 + d2) Perimeter = 4a; Where d1 and d2 are the diagonals of the rhombus a is the side of the rhombus. In a rhombus, if one angle is right, then all angles are right. Perimeter eines Rhombus, Perimeter Formel Rechner Finden einen Umfang eines Trapezes, das durch die Formel, alle Trapezseiten unter Verwendung der Länge. The perimeter of rhombus formula – Rhombus has: All sides of equal length. Both squares and rhombuses have perpendicular diagonal bisectors that split each diagonal into 2 equal pieces, and also splitting the quadrilateral into 4 … In the case of a rhombus, all four sides are the same length by definition, so the perimeter is four times the length of a side. Another way to calculate the area of the rhombus be divide the product of the two diagonals. The total distance covered around the edge of the rhombus. From the properties we know that diagonals of rhombus bisect each other into equal parts. Example 1: On the figure below the diagonals of the rhombus are $\displaystyle AO=2x+1$ and $\displaystyle OC=\frac{1}{2}x+10$. Area = ah. . Download the set (5 Worksheets) How to find the perimeter of a rhombus. $$4 = "number of sides in a rhombus"$$ The grey space is the area of the rhombus in the diagram below. Perimeter of Rhombus = 4(a) where, a = side, b = base, h = height, d1, d2 are diagonals Rhombus Definition: A rhombus is a four-sided polygon in which every side has the same length. So if one of the angles is 60°, then the adjacent angle will be 120°. Find x. Using trigonometry $\displaystyle P=a+a+a+a=4a$ Where $\displaystyle a$ is one side of the rhombus. The altitude is the distance at right angles between two parallel sides. Solution: All the sides of a rhombus are congruent, so HO = (x + 2).And because the diagonals of a rhombus are perpendicular, triangle HBO is a right triangle.With the help of Pythagorean Theorem, we get, (HB) 2 + (BO) 2 = (HO) 2x 2 + (x+1) 2 = (x+2) 2 x 2 + x 2 + 2x + 1 = x 2 + 4x + 4 x 2 – 2x -3 = 0 Solving for x using the quadratic formula, we get: x = 3 or x = –1. => AO=OC=1/2AC, and BO=OD=1/2BD Now back to our question : Given that the two diagonals are 30, and 16, => AO=30/2=15, BO=16/2=8, angleAOB=90^@ From Pythagorean … Please update your bookmarks accordingly. Rhombus consists of all sides equal and its opposite angles are equal in measure. In rhombus all sides are equal. Perimeter of a Rhombus | Decimals. Problem 1: Find the perimeter of a rhombus with a side length of 10. It is a geometric figure with 4 sides, all the sides have equal length. Geometry Formula. © 2017 WikiFormulas.com Trapezoid Worksheets. Perimeter of a rhombus; Perimeter of a trapezoid; Circumference of a circle; Length of an arc; Length of an arc, the Huygens formula ; All formulas for perimeter of geometric figures; Volume of geometric shapes. XY = XY = First, input 2 values between edges(a, b), diagonal (d1, d2), or with alpha / beta angle in the rhombus 2. They are both quadrilaterals but a rhombus doesn’t have right angles at the corners but a rhombus does have four corners that are the same so the perimeter formula is the same as a … Example 1: On the figure below the diagonals of the rhombus are $\displaystyle AO=2x+1$ and $\displaystyle OC=\frac{1}{2}x+10$. Answer: The perimeter of rhombus WXYZ is . Post Views: 36. 10. The perimeter of a circle, often called the circumference, is proportional to its diameter and its radius.That is to say, there exists a constant number pi, π (the Greek p for perimeter), such that if P is the circle's perimeter and D its diameter then, = ⋅. Al copiarlas a tu sitio acuerdas dar attribución a "© WikiFormulas.com". Worksheets. Circle. Base Times Height Method : Area of Rhombus = b * h. Diagonal Method : Area of Rhombus = ½ * d1 * d2 . All rhombus can be parallelograms. By Devendra Vishwakarma Math Formulas formula, of, perimeter, RHOMBUS 0 Comments. PERIMETER OF RHOMBUS FORMULA. Calculating the Rhombus Area and perimeter pg Area Sides Perimeter 68 24 5 20 1 3 1.5 1.58114 … Area $$= P \frac{Q}{2}$$ Perimeter of the rhombus $$= 4 \times$$ side of the rhombus. Example 7 : In the rhombus ABCD shown below, if the lengths of the diagonals AC and BD are 10 units and 8 units respectively, find its perimeter. The perimeter of a rhombus = 4 × a The perimeter of a rhombus = 4 × 5 The perimeter of a rhombus = 20 cm α2 =. Length of the sides are equal of a rhombus. The diagonals of the rhombus bisect each other at 90 degrees. The perimeter of a Rhombus is the total length of all sides. In the case of a circle, the perimeter is called a circumference. (Jump to Area of a Rhombus or Perimeter of a Rhombus) A Rhombus is a flat shape with 4 equal straight sides. When a Rhombus is concerned, it is a flat shape with four same sides and four angles that are NOT necessarily 90 degrees. Formulas. There is one using the altitude and side, another using the side and angle, and one for the diagonals. Solved Example. The formula to calculate the perimeter of a rhombus is: P = 2 x (b + h) where... P = perimeter of rhombus. (the same length). ✉ wikiformulas@gmail.com Study of mathematics online. Perimeter of a Rhombus Formulas & Calculator. Math Open Reference. In the case of a rhombus, all four sides are the same length by definition, so the perimeter is four times the length of a side. Side of a Rhombus when Diagonals are given calculator uses Side A=sqrt((Diagonal 1)^2+(Diagonal 2)^2)/2 to calculate the Side A, Side of a Rhombus when Diagonals are given can be defined as the line segment that joins two vertices in a rhombus provided the value for both the diagonals are given. What is the length of one of the table’s sides? The formula of the perimeter of a rhombus is given as p = 4 × a, where ‘a’ is the length of the side of the rhombus. Practice. A parallelogram whose diagonals intersect at right angles is a rhombus, A parallelogram whose diagonals are the bisectors of its angles is a rhombus. Trigonometry Method : Area of Rhombus = a² * SinA. Triangle. A rhombus is a quadrilateral with 4 equal sides, a pair of opposing equal acute angles, and a pair of opposing equal obtuse angles.The formula for perimeter of a rhombus is given as:P = 4sWhere P is the perimeter and s is the side length. This website uses cookies to ensure you get the best experience. Problem 1:Find the perimeter of a rhombus with a side length of 10. The formula for perimeter of a rhombus is given as: P = 4s Where P is the perimeter and s is the side length. Next, you can choose what decimal places you need. 68 Some of the rhombus' properties: a) The sides of a rhombus are all congruent. A perimeter is defined by the outer path of a shape. The perimeter is the sum of the length of all the 4 sides. Formula for perimeter of a rhombus : = 4s Substitute 2√5 for s. = 4(2√5) = 8 √5. where. Using diagonals. Everything about Rhombus. Given the sample output in Figure 3, your program should be able to produce the exact copy of the output. Diagonals of a rhombus bisect each other and also perpendicular to each other. Area formula. A rhombus … As mentioned in the properties of a rhombus, the sum of two adjacent angles is 180°. $\displaystyle P=a+a+a+a=4a$ Where $\displaystyle a$ is one side of the rhombus. These angles need necessarily be of 90 degrees. The total distance traveled along the border of a rhombus is the perimeter of a rhombus … Volume of a cube; Volume of a rectangular prism; Volume of a sphere; Volume of a spherical segment; Volume of a spherical zone; Volume of a spherical … Perimeter of a rhombus; Perimeter of a trapezoid; Circumference of a circle; Length of an arc; Length of an arc, the Huygens formula; All formulas for perimeter of geometric figures ; Volume of geometric shapes. Properties of a rhombus. The diagonals of a rhombus bisect each other as it is a parallelogram, but they are also perpendicular to each other. We therefore have four congruent right triangles. Comments are closed. The area is half the product of the diagonals. Area of a Rhombus Formula - A rhombus is a parallelogram in which adjacent sides are equal. Calculate area & perimeter calculator - calculate area & perimeter of 192 cm given data: P=2sqrt 256+900. D1 * d2 a 2-dimensional shape that has four equal sides of quadrilaterals. 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(Analysis by Benjamin Qi)
Subtask 1: $P=1$
Let's start by computing the following quantity $Q$: the minimum number of moves to move all of liquid $1$ to tube $1$ and all of liquid $2$ to tube $2$. Once we know how to compute this quantity, we can compute $Q'$, the minimum number of moves to move all of liquid $1$ to tube $2$ and all of liquid $2$ to tube $1$, in a similar way. The answer for this subtask will then be $\min(Q,Q')$.
The current state of liquid in the tubes and beakers can be represented by three binary strings $f$, $s$, and $b$, with $f$ and $s$ defined the same way as the problem statement, and $b$ denoting the state of the beaker from bottom to top.
First, observe that $Q$ does not change if we make the following simplifications:
1. Compress runs of equal characters in all three strings. For example, 1221 becomes 121 and 2211 becomes 21.
2. Remove the first character of $f$ if it is $1$, since this liquid never needs to leave tube $1$
3. Similarly, remove the first character of $s$ if it is $2$.
Next, observe that $Q$ must be at least the number of characters after simplifying, because after performing any move and simplifying again, the number of characters will have stayed the same or gone down by one. The final state corresponds to all of $f$, $s$, and $b$ simplifying to empty strings.
Is $Q$ exactly equal to the number of characters after simplifying? It is when we don't need to utilize the beaker. For example, consider the following input:
f = "111112"
s = "222222"
After simplifying,
f = "2"
s = ""
So the number of characters after simplifying is one, and $Q=1$ because we only need one pour to separate the liquid (tube 1 into tube 2). However, when the number of characters after simplifying is greater than one, we need to utilize the beaker, and when we first pour liquid into the beaker the number of characters after simplifying does not change. In such cases, $Q$ is at least the number of characters after simplifying plus one. For example, the first test case of the sample input after simplifying becomes:
f = "21"
s = "1"
There are three characters after simplifying, and $Q=4$.
Is it true that $Q$ is always at most the number of characters after simplifying plus one? Assuming this allows you to solve the first subtask. At this point, it's not obvious why this assumption is correct, but we will prove that it works later by providing a construction using exactly this number of moves.
Implementation (which computes $\min(Q,Q')$ directly rather than $Q$ and $Q'$ separately):
def to_reduced_list(s):
"""Compress runs of equal chars in a string s
>>> to_reduced_list('2211')
['2', '1']
"""
l = []
for c in s:
if len(l) > 0 and l[-1] == c:
continue
l.append(c)
return l
def solve():
N, P = map(int, input().split())
assert P == 1
tubes = [to_reduced_list(input()) for _ in range(2)]
tubes.append([])
ans = len(tubes[0]) + len(tubes[1]) - 2
# ^ num chars after simplifying, if tubes[0] and tubes[1] start with different chars and we choose the better of Q and Q'
if tubes[0][0] == tubes[1][0]:
ans += 1
if ans > 1: # require extra pour into beaker
ans += 1
print(ans)
T = int(input())
for _ in range(T):
solve()
Subtask 2: $P=2$
Here we provide a construction that moves all of liquid $i$ into tube $i$ and achieves a number of moves at most the number of characters after compressing runs plus two. From the previous subtask, we know that the answer must be at least the number of characters after compressing runs minus two. Therefore, our solution is at most four away from optimal.
1. While tube $2$ is nonempty, pour it into tube $1$ if tube $2$ has liquid $2$ on top, or to the beaker otherwise.
2. While tube $1$ is nonempty, pour it into tube $2$ if tube $1$ has liquid $2$ on top, or to the beaker otherwise.
3. Now, tube $2$ contains all of liquid $2$ and the beaker contains all of liquid $1$. The final step is to pour from the beaker into tube $1$.
Implementation of the strategy described above (using 0- instead of 1-indexing):
def to_reduced_list(s):
"""Compress runs of equal chars in a string s
>>> to_reduced_list('2211')
['2', '1']
"""
l = []
for c in s:
if len(l) > 0 and l[-1] == c:
continue
l.append(c)
return l
def solve():
N, P = map(int, input().split())
assert P == 2
tubes = [to_reduced_list(input()) for _ in range(2)]
tubes.append([])
moves = []
def move(src, dst):
moves.append((src, dst))
if len(tubes[dst]) == 0 or tubes[dst][-1] != tubes[src][-1]:
tubes[dst].append(tubes[src][-1])
tubes[src].pop()
while len(tubes[1]) > 0:
if tubes[1][-1] == "2":
move(1, 0)
else:
move(1, 2)
while len(tubes[0]) > 0:
if tubes[0][-1] == "2":
move(0, 1)
else:
move(0, 2)
move(2, 0)
print(len(moves))
for a, b in moves:
print(1 + a, 1 + b)
T = int(input())
for _ in range(T):
solve()
Subtask 3: $P=3$
Our construction achieving the optimal number of moves is similar to the one for the previous subtask in the sense that we first empty out most of one tube, then most of the other, and finally the beaker.
Before describing the strategy we use the following trick to ensure that we never need to deal with a tube being empty, which makes the implementation simpler. Suppose we choose to move all of liquid $1$ to tube $1$ and liquid $2$ to tube $2$; then let's insert an extra $1$ in front of $f$ and an extra $2$ in front of $s$ before compressing runs (or the other way around if we choose to move liquid $1$ to tube $2$ and liquid $2$ to tube $1$).
Our strategy is as follows. We assume that before every move we perform step $1$ only of the simplification process described in subtask 1 (compressing runs of equal characters).
1. If the last characters of $f$ and $s$ are equal, choose any tube with a string of length greater than one and pour it into the other tube.
2. If both $f$ and $s$ now have length one, we're done. Otherwise, we're in a case where the beaker must be used. Choose any tube with a string of length greater than one and pour it into the beaker. The last characters of both tube strings are now equal.
3. Suppose the beaker now contains liquid $i$. Choose the tube whose string has first character not equal to $i$, and repeatedly pour from it until its string has length equal to one. We should pour it into either the beaker or the other tube depending on which pour reduces the number of characters after compressing by one.
4. Next, pour from the other tube until its string has length one.
5. Finally, pour from the beaker into one of the tubes.
Example:
Initial:
f = 12121
s = 2121
After step 1:
f = 1212
s = 2121
After step 2:
f = 1212
s = 212
b = 1
After step 3:
f = 1212
s = 2
b = 1
After step 4:
f = 1
s = 2
b = 1
After step 5:
f = 1
s = 2
Every move besides the first move into the beaker reduces the number of characters after compressing by one, so this construction matches the lower bound we described in subtask 1.
Implementation:
def to_reduced_list(s):
"""Compress runs of equal chars in a string s, and converts to int
>>> to_reduced_list('2211')
[2, 1]
"""
l = []
for c in s:
c = int(c)
if len(l) > 0 and l[-1] == c:
continue
l.append(c)
return l
def solve():
N, P = map(int, input().split())
tubes = [to_reduced_list(input()) for _ in range(2)]
tubes.append([])
if tubes[0][0] == tubes[1][0]: # ensure f and s start with different chars
tubes[0].insert(0, tubes[0][0] ^ 3)
ans = len(tubes[0]) + len(tubes[1]) - 2
if ans > 1:
ans += 1
print(ans)
if P == 1:
return
moves = []
def move(src, dst):
moves.append((src, dst))
if len(tubes[dst]) == 0 or tubes[dst][-1] != tubes[src][-1]:
tubes[dst].append(tubes[src][-1])
tubes[src].pop()
if tubes[0][-1] == tubes[1][-1]: # step 1: if equal last chars
if len(tubes[0]) > len(tubes[1]):
move(0, 1)
else:
move(1, 0)
for i in range(2):
if len(tubes[i]) > 1:
move(i, 2) # step 2: move from any tube with string length > 1 to beaker
idx_to_empty = 0 # step 3: choose a tube to (almost) empty first
if tubes[idx_to_empty][0] == tubes[2][0]:
idx_to_empty ^= 1
while len(tubes[idx_to_empty]) > 1:
if tubes[idx_to_empty][-1] == tubes[2][0]:
move(idx_to_empty, 2)
else:
move(idx_to_empty, idx_to_empty ^ 1)
idx_to_empty ^= 1 # step 4: next, (almost) empty the other tube
while len(tubes[idx_to_empty]) > 1:
if tubes[idx_to_empty][-1] == tubes[2][0]:
move(idx_to_empty, 2)
else:
move(idx_to_empty, idx_to_empty ^ 1)
move(2, idx_to_empty) # step 5: finish
break
assert len(moves) == ans
for a, b in moves:
print(1 + a, 1 + b)
T = int(input())
for _ in range(T):
solve()
|
Mister Exam
# Graphing y = y=(0.2x^3-0.3x^2+0.1)(-0.1x^5+0.4x^2+4)+0.3
v
from to
does show?
### The solution
You have entered [src]
/ 3 2 \ / 5 2 \
|x 3*x 1 | | x 2*x | 3
f(x) = |-- - ---- + --|*|- -- + ---- + 4| + --
\5 10 10/ \ 10 5 / 10
$$f{\left(x \right)} = \left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10}$$
f = (x^3/5 - 3*x^2/10 + 1/10)*(-x^5/10 + 2*x^2/5 + 4) + 3/10
The graph of the function
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
$$\left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10} = 0$$
Solve this equation
The points of intersection with the axis X:
Analytical solution
$$x_{1} = \operatorname{CRootOf} {\left(2 x^{8} - 3 x^{7} - 7 x^{5} + 12 x^{4} - 80 x^{3} + 116 x^{2} - 70, 0\right)}$$
$$x_{2} = \operatorname{CRootOf} {\left(2 x^{8} - 3 x^{7} - 7 x^{5} + 12 x^{4} - 80 x^{3} + 116 x^{2} - 70, 1\right)}$$
Numerical solution
$$x_{1} = 2.2997370756655$$
$$x_{2} = -0.634585226081237$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x^3/5 - 3*x^2/10 + 1/10)*(-x^5/10 + 2*x^2/5 + 4) + 3/10.
$$\frac{3}{10} + \left(\frac{0^{3}}{5} - \frac{3 \cdot 0^{2}}{10} + \frac{1}{10}\right) \left(- \frac{0^{5}}{10} + \frac{2 \cdot 0^{2}}{5} + 4\right)$$
The result:
$$f{\left(0 \right)} = \frac{7}{10}$$
The point:
(0, 7/10)
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} =$$
the first derivative
$$\left(\frac{3 x^{2}}{5} - \frac{3 x}{5}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \left(- \frac{x^{4}}{2} + \frac{4 x}{5}\right) \left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
$$x_{2} = 1$$
$$x_{3} = 1.98199343607655$$
The values of the extrema at the points:
(0, 7/10)
(1, 3/10)
(1.98199343607655, 1.5028345683614)
Intervals of increase and decrease of the function:
Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:
Minima of the function at points:
$$x_{1} = 1$$
Maxima of the function at points:
$$x_{1} = 0$$
$$x_{1} = 1.98199343607655$$
Decreasing at intervals
$$\left(-\infty, 0\right] \cup \left[1, \infty\right)$$
Increasing at intervals
$$\left(-\infty, 1\right] \cup \left[1.98199343607655, \infty\right)$$
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} =$$
the second derivative
$$\frac{- 3 x^{2} \left(x - 1\right) \left(5 x^{3} - 8\right) + \frac{3 \cdot \left(2 x - 1\right) \left(- x^{5} + 4 x^{2} + 40\right)}{2} - \left(5 x^{3} - 2\right) \left(2 x^{3} - 3 x^{2} + 1\right)}{25} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0.518009371996229$$
$$x_{2} = 1.67997462933324$$
Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
$$\left[0.518009371996229, 1.67997462933324\right]$$
Convex at the intervals
$$\left(-\infty, 0.518009371996229\right] \cup \left[1.67997462933324, \infty\right)$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (x^3/5 - 3*x^2/10 + 1/10)*(-x^5/10 + 2*x^2/5 + 4) + 3/10, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10}}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10}}{x}\right) = -\infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
$$\left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10} = \left(- \frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(\frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10}$$
- No
$$\left(\frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(- \frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) + \frac{3}{10} = - \left(- \frac{x^{3}}{5} - \frac{3 x^{2}}{10} + \frac{1}{10}\right) \left(\frac{x^{5}}{10} + \frac{2 x^{2}}{5} + 4\right) - \frac{3}{10}$$
- No
so, the function
not is
neither even, nor odd
The graph
|
## Friday, November 1, 2013
### Counting I : Ways to Avoid Over Counting or Under Counting
Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.
Question #1: How many ways to count a triple of natural numbers whose sum adds up to 12
if A. order doesn't matter? B. if order matters?
Solution I:
A. Find a systematic way to solve this type of problem in an organized manner.
We can start with the smallest natural number "1".
(1, 1, 10), (1, 2, 9), (1, 3, 8), (1, 4, 7), (1, 5, 6) Since (1, 5, 6) is the same as (1, 6, 5) so stop.
(2, 2, 8), (2, 3, 7), (2, 4, 6), (2, 5, 5)
(3, 3, 6), (3, 4, 5)
(4, 4, 4) so total there are 12 ways.
B. Since in this case order matters so if you add up all the possible ways, for example, there are "3" ways to arrange (1, 1, 10) -- 3!/2! = 3
This is similar to "how many ways to arrange the letters 'odd'.
If all letters/numbers are different, you have 3! ways to arrange them; however, in this case, the two numbers 1, 1 are indistinguishable and there are 2! ways to arrange them, thus 3!/2! = 3
There are 3! or 6 ways to arrange triples such as (1, 2, 9).
Add all the possible ways and there are total 55 ways.
Solution II:
B.There is a much easier way to tackle this question, using bars and stars or sticks and stones method.
There are 11 spaces between 12 stones. @__@__@__@__@__@__@__@__@__@__@__@
If you places the two sticks on two of the spaces, you'll split the stones into three groups.
For example, if you have @ | @@@@@ | @@@@@@
The triples are 1, 5, and 6.
If it's @@@@@@@@ | @@ | @@ , the triples are 8, 2, and 2.
So 11C2 = 55 ways
Let me know if you are still confused. Have fun problem solving !!
|
## Precalculus (6th Edition) Blitzer
$(1,-1,1)$
Step 1. Writing a matrix based on the given system of equations, we have $\begin{bmatrix} 1 & -2 & -1 & | & 2 \\ 2 & -1 & 1 & | & 4 \\ -1 & 1 & -2 & | & -4 \end{bmatrix} \begin{array} ..\\-2R1+R2\to R2\\ R1+R3\to R3 \end{array}$ Step 2. Perform row operations given to the right of the matrix: $\begin{bmatrix} 1 & -2 & -1 & | & 2 \\ 0 & 3 & 3 & | & 0 \\ 0 & -1 & -3 & | & -2 \end{bmatrix} \begin{array} ..\\..\\ R2+3R3\to R3 \end{array}$ Step 3. Perform row operations given to the right of the matrix: $\begin{bmatrix} 1 & -2 & -1 & | & 2 \\ 0 & 3 & 2 & | & 0 \\ 0 & 0 & -6 & | & -6 \end{bmatrix} \begin{array} ..\\..\\ .. \end{array}$ Step 4. The last equation gives $-6z=-6$ or $z=1$ Step 5. Use back-substitution to get $3y+3z=0$ and $y=-1$, $x-2y-z=2$ and $x=2+2(-1)+1=1$ Step 6. The solution set for the system is $(1,-1,1)$
|
Tamil Nadu Board of Secondary EducationHSC Commerce Class 12th
Determine an initial basic feasible solution to the following transportation problem by using least cost method Destination Supply D1 D2 D3 S1 9 8 5 25 Source S2 6 8 4 35 S3 7 6 9 40 Requirement - Business Mathematics and Statistics
Chart
Sum
Determine an initial basic feasible solution to the following transportation problem by using least cost method
Destination Supply D1 D2 D3 S1 9 8 5 25 Source S2 6 8 4 35 S3 7 6 9 40 Requirement 30 25 45
Solution
Total supply = 25 + 35 + 40 = 100 = sum"a"_"i"
Total requirement = 30 + 25 + 45 = 100 = sum"b"_"j"
Since sum"a"_"i" = sum"b"_"j"
The given transportation problem is balanced and we can find an initial basic feasible solution.
Least cost method (LCM)
First allocation:
D1 D2 D3 (ai) S1 9 8 5 25 S2 6 8 (35)4 35/0 S3 7 6 9 40 (bj) 30 25 45/10
Second allocation:
D1 D2 D3 (ai) S1 9 8 (10)5 25/15 S3 7 6 9 40 (bj) 30 25 10/0
Third allocation:
D1 D2 (ai) S1 9 8 15 S3 7 (25)6 40/15 (bj) 30 25/0
Fourth allocation:
D1 (ai) S1 (15)9 15/0 S3 (15)7 15/0 (bj) 30/15/0
We first allow 15 units to cell (S3, D1) since it has the least cost.
Then we allow the balance 15 units to cell (S1, D1).
The final allotment is given as follows.
D1 D2 D3 Supply S1 (15)9 8 (10)5 25 S2 6 8 (35)4 35 S3 (15)7 (25)6 9 40 Requirement 30 25 45
Transportation schedule:
S1 → D1
S1 → D3
S2 → D3
S3 → D1
S3 → D2
i.e x11 = 15
x13 = 10
x23 = 35
x31 = 15
x32 = 25
Total cost is = (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)
= 136 + 50 + 140 + 105 + 150
= 580
The optimal cost by LCM is ₹ 580.
Concept: Transportation Problem
Is there an error in this question or solution?
|
Utah Math 2 - 2020 Edition
7.05 Proving triangles congruent
Lesson
Now that we know the criteria for identifying congruent triangles, the next step is to find a way to communicate that knowledge in the form of a proof. Recall two commonly used proof forms, the two-column proof and the paragraph proof.
### Two-column proofs
A two-column proof might be considered the most structured form, where every line of work is placed on a row with an accompanying reason. A generic shell for the proof looks like this:
To prove: $\left[\text{Statement to be proved}\right]$[Statement to be proved] Statement Reason Statement 1 Reason 1 Statement 2 Reason 2 Statement 3 Reason 3 $\ldots$… $\ldots$… $\left[\text{Statement to be proved}\right]$[Statement to be proved] Reason $n$n
We use as many lines of reasoning as we need, and make sure the last line always has the statement we want to prove at the end. Any information provided to us should have the reason Given written next to it, to indicate that no reasoning actually took place to arrive at the statement. Let's try a few examples where we prove triangles congruent using a two-column proof.
#### Worked examples
##### Question 1
Show the following two triangles are congruent using a two-column proof:
Think: Before writing anything down, it's a good idea for us to think for a minute about which congruence test we will use. Since none of the angles have been marked, we will have to rely on the sides. The next step (even if you haven't decided on the test yet!) will be to translate the information provided to us into the start of a two-column proof.
Do: We write down what we want to prove, followed by the information obtained from the markings on the diagram:
To prove: $\Delta ADC\cong\Delta CBA$ΔADC≅ΔCBA Statement Reason $\overline{AD}\cong\overline{CB}$AD≅CB Given $\overline{DC}\cong\overline{BA}$DC≅BA Given
We have two sides, and by the SSS rule, we only need to show one more pair of sides (one from each triangle) are congruent. In this case the two triangles share a side in common, so we use the reflexive property of congruence to note this:
To prove: $\Delta ADC\cong\Delta CBA$ΔADC≅ΔCBA Statement Reason $\overline{AD}\cong\overline{CB}$AD≅CB Given $\overline{DC}\cong\overline{BA}$DC≅BA Given $\overline{AC}\cong\overline{AC}$AC≅AC Reflexive property of congruence
Now we have demonstrated that three sides are the same, we can write our conclusion at the end:
To prove: $\Delta ADC\cong\Delta CBA$ΔADC≅ΔCBA Statement Reason $\overline{AD}\cong\overline{CB}$AD≅CB Given $\overline{DC}\cong\overline{BA}$DC≅BA Given $\overline{AC}\cong\overline{AC}$AC≅AC Reflexive property of congruence $\Delta ADC\cong\Delta CBA$ΔADC≅ΔCBA Side-side-side congruency (SSS)
##### Question 2
Show the following two triangles are congruent using a two-column proof:
Think: Make sure to take note of every marking on the diagram, and think about what rule to use.
Do: Once again we translate the information provided to us, both what we want to show and what the markings tell us:
To prove: $\Delta AMB\cong\Delta CMD$ΔAMB≅ΔCMD Statement Reason $\overline{AM}\cong\overline{CM}$AM≅CM Given $\overline{BM}\cong\overline{DM}$BM≅DM Given
We don't know anything about the third pair of sides, so we will have to use an angle instead. To use the SAS rule, we need to show that the angles lying between the two pairs of congruent sides are also congruent. Here we can use the fact that the angles are vertical, and reach our conclusion:
To prove: $\Delta AMB\cong\Delta CMD$ΔAMB≅ΔCMD Statement Reason $\overline{AM}\cong\overline{CM}$AM≅CM Given $\overline{BM}\cong\overline{DM}$BM≅DM Given $\angle AMB\cong\angle CMD$∠AMB≅∠CMD Vertical angles $\Delta AMB\cong\Delta CMD$ΔAMB≅ΔCMD Side-angle-side congruency (SAS)
##### Question 3
Show the following two triangles are congruent using a two-column proof:
Think: As usual we make sure to take note of every marking on the diagram, and think about what test would be best to use. Since we are only shown information about a single pair of sides, we need to concentrate on finding two pairs of congruent angles.
Do: Translating what we want to prove and each of the markings in the diagram into the two-column format gives us the following:
To prove: $\Delta PXQ\cong\Delta RXS$ΔPXQ≅ΔRXS Statement Reason $\overline{PQ}\cong\overline{RS}$PQ≅RS Given $\overleftrightarrow{PQ}\parallel\overleftrightarrow{SR}$›‹PQ∥›‹SR Given
Having established that $\overleftrightarrow{PQ}$PQ and $\overleftrightarrow{SR}$SR are parallel, we can use the alternate interior angles theorem to show that the angles lying on either end of $\overline{PQ}$PQ and $\overline{RS}$RS are congruent, and reach our conclusion:
To prove: $\Delta PXQ\cong\Delta RXS$ΔPXQ≅ΔRXS Statement Reason $\overline{PQ}\cong\overline{RS}$PQ≅RS Given $\overleftrightarrow{PQ}\parallel\overleftrightarrow{SR}$›‹PQ∥›‹SR Given $\angle XPQ\cong\angle XRS$∠XPQ≅∠XRS Alternate interior angles theorem $\angle XQP\cong\angle XSR$∠XQP≅∠XSR Alternate interior angles theorem $\Delta PXQ\cong\Delta RXS$ΔPXQ≅ΔRXS Angle-side-angle congruency (ASA)
### Paragraph proofs
The paragraph proof describes the same information, and goes through the same process, as a two-column proof, but we write it all out in more natural language and style. Here are some examples.
#### Worked examples
##### question 4
Show the following two triangles are congruent using a paragraph proof:
Think: Just like in a two-column proof we want to make sure we know what we want to prove, know what each marking means, and have an idea for a congruence test we want to use in the proof.
Do: We want to prove that $\Delta XAY\cong\Delta XBY$ΔXAYΔXBY. First we can see that $\angle XAY$XAY and $\angle XBY$XBY are both given as right-angles, so both $\Delta XAY$ΔXAY and $\Delta XBY$ΔXBY are right triangles. We can also see that $\overline{XY}$XY is the hypotenuse of both triangles, so by the reflexive property of congruence, the hypotenuses of both triangles are congruent. Finally, the segments $\overline{XA}$XA and $\overline{XB}$XB are congruent legs, as they are both radii of the circle centered at $X$X. We have therefore shown by Hypotenuse-leg congruency (HL) that $\Delta XAY\cong\Delta XBY$ΔXAYΔXBY, QED.
Did you know?
You may have seen proofs conclude with the letters QED, an abbreviation for the Latin phrase "quod erat demonstrandum". This roughly translates to "I have proved what I set out to prove", but it really means "I delivered on my promise and no-one can argue with me".
It is commonly mispronounced as "quod erat demonstraTum", and modern mathematicians usually use the symbol in its place. Still, it is a powerful phrase and powerful idea, which is why it shows up outside of mathematics - in other academic learning areas, as well as books, films, and other popular media.
##### Question 5
Show the following two triangles are congruent using a paragraph proof:
Think: If you're more comfortable with two-column proofs, you might want to write out a list of statements first in that style before writing it in full sentences.
Do: We wish to demonstrate that $\Delta PAC\cong\Delta QBD$ΔPACΔQBD. We are given that $\overline{PA}$PA and $\overline{QB}$QB are parallel, that $\angle APC$APC and $\angle BQD$BQD are congruent, and that the segments $\overline{AB}$AB$\overline{BC}$BC, and $\overline{CD}$CD are all congruent. Since congruent segments have equal measure, we know that $AB=CD$AB=CD, and by the addition property of equality we can conclude that $AB+BC=BC+CD$AB+BC=BC+CD. This then means that the segments $\overline{AC}$AC and $\overline{BD}$BD have equal measure, so they must be congruent. The angles $\angle PAC$PAC and $\angle QBD$QBD are congruent since they are corresponding angles in the parallel lines $\overleftrightarrow{PA}$PA and $\overleftrightarrow{QB}$QB. We have therefore shown using angle-angle-side congruency (AAS) that $\Delta PAC$ΔPAC and $\Delta QBD$ΔQBD are congruent, QED.
No matter which method of proof you use, showing that two triangles are congruent involves preparing carefully justified statements that line up with one of the tests for congruency. Make sure you know them all and are ready to use them before trying to learn these two new proof methods.
#### Practice questions
##### Question 6
This two-column proof shows that $\Delta ABC\cong\Delta XYZ$ΔABCΔXYZ in the attached diagram, but it is incomplete.
Statements Reasons
$\overline{AC}\cong\overline{XZ}$ACXZ Given
$\angle CBA\cong\angle ZYX$CBAZYX
Given
$\angle CAB\cong\angle ZXY$CABZXY
Given
$\Delta ABC\cong\Delta XYZ$ΔABCΔXYZ
$\left[\text{____}\right]$[____]
1. Select the correct reason to complete the proof.
Angle-angle-side congruence (AAS)
A
Side-side-side congruence (SSS)
B
Side-angle-side congruence (SAS)
C
Side-side-angle congruence (SSA)
D
Angle-side-angle congruence (ASA)
E
Angle-angle-side congruence (AAS)
A
Side-side-side congruence (SSS)
B
Side-angle-side congruence (SAS)
C
Side-side-angle congruence (SSA)
D
Angle-side-angle congruence (ASA)
E
##### Question 7
This two-column proof shows that $\Delta PQR\cong\Delta RSP$ΔPQRΔRSP in the attached diagram, but it is incomplete.
To prove: $\Delta PQR\cong\Delta RSP$ΔPQRΔRSP
Statements Reasons
$\left[\text{__________}\right]$[__________] $\left[\text{__________}\right]$[__________]
$\overline{PR}\cong\overline{RP}$PRRP Reflexive property of congruence
$\angle QRP\cong\angle SPR$QRPSPR
Given
$\Delta PQR\cong\Delta RSP$ΔPQRΔRSP
Side-angle-side congruence (SAS)
1. Select the correct reason to complete the proof.
$\overline{PS}\cong\overline{RQ}$PS≅RQ Transitive property of congruence
A
$\overline{SR}\cong\overline{QP}$SR≅QP Given
B
$\overline{PS}\cong\overline{RQ}$PS≅RQ Given
C
$\overline{SR}\cong\overline{QP}$SR≅QP Transitive property of congruence
D
$\overline{PS}\cong\overline{RQ}$PS≅RQ Transitive property of congruence
A
$\overline{SR}\cong\overline{QP}$SR≅QP Given
B
$\overline{PS}\cong\overline{RQ}$PS≅RQ Given
C
$\overline{SR}\cong\overline{QP}$SR≅QP Transitive property of congruence
D
##### Question 8
This two-column proof shows that $\Delta DEH\cong\Delta FEG$ΔDEHΔFEG in the attached diagram, but it is incomplete.
Statements Reasons
$E$E is the midpoint of $\overline{DF}$DF Given
$\overline{DH}\cong\overline{FG}$DHFG Given
$\overline{EH}\cong\overline{EG}$EHEG
Given
$\left[\text{___}\right]$[___] $\left[\text{___}\right]$[___]
$\Delta DEH\cong\Delta FEG$ΔDEHΔFEG
$\left[\text{___}\right]$[___]
1. Select the correct pair of reasons to complete the proof.
$\angle EDH\cong\angle FEG$∠EDH≅∠FEG Corresponding angles theorem and $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG Side-side-angle congruence (SSA)
A
$\overline{DE}\cong\overline{EF}$DE≅EF Reflexive property of congruence and $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG Side-side-side congruence (SSS)
B
$\angle DHE\cong\angle FGE$∠DHE≅∠FGE Triangle hinge theorem and $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG Side-angle-side congruence (SAS)
C
$\overline{DE}\cong\overline{EF}$DE≅EF Properties of a midpoint and $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG Side-side-side congruence theorem (SSS)
D
$\angle EDH\cong\angle FEG$∠EDH≅∠FEG Corresponding angles theorem and $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG Side-side-angle congruence (SSA)
A
$\overline{DE}\cong\overline{EF}$DE≅EF Reflexive property of congruence and $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG Side-side-side congruence (SSS)
B
$\angle DHE\cong\angle FGE$∠DHE≅∠FGE Triangle hinge theorem and $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG Side-angle-side congruence (SAS)
C
$\overline{DE}\cong\overline{EF}$DE≅EF Properties of a midpoint and $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG Side-side-side congruence theorem (SSS)
D
### Outcomes
#### II.G.SRT.5
Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
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# For a linear function f, f(-1)=3 and f(2)=4 find an equation for fplease answer soon, I need help badly
mcetner | Certified Educator
Another way to answer this question is that you know any linear function can always be written y = mx + b, where m is the slope.
Find m first, (4 - 3)/(2 - -1)=1/3.
Then plug in one point to solve for b, 4 = (1/3)(2) + b, solve for b and b = 10/3.
Replace m and b with their values, and y = (1/3)x + (10/3).
Here is another example where we find the equation of a line from two points:
giorgiana1976 | Student
The general form of a linear function is
f(x)=ax + b
So, f(-1)=3 means that we hhave to substitute, in the general form, the unknown x with the value -1.
-1xa + b = 3
We'll do the same thing with the following f(2)=4
2a +b = 4
Now, in order to find the function f, we have to determine the unknown coefficients a and b.
SO, we'll subtract from the last relation, 2a +b = 4, the anterior one ,-a + b= 3.
2a +b - (-a) +b = 4-3
3a = 1
a = 1/3
With the known value of a, we are going in any of the both anterior relation, substitute a and find out the value of b unknown.
We choose the relation -a + b =3, beccause it's more simple to calculate b, after substituting a value.
-1/3 + b = 3
b = 3 + 1/3
The common denominator is 3, so we have to multiply the first term 3, with the common denominator 3
b = ( 9 +1)/3
b = 10/3
Now we could write down the linear function f.
f(x) = (1/3)x + 10/3
revolution | Student
General equation of linear function f is:
f(x)=mx+c
Since f(-1)=3, so we have to subsitute x=-1
-m+c=3 -eqn 1
Also, f(2)=4, so we have to sub. x=2
2m+c=4 -eqn 2
Subtract eqn 1 from eqn 2
2m-(-m)+c-c=4-3
3m=1
m=1/3
subsitute m=1/3 into eqn 2
2(1/3)+c=4
2/3+c=4
c=4-2/3
=12/3-2/3
= 10/3 or 3 1/3
The equation of f(x): 1/3 x+10/3 (mx+c)
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# Binomial Theorem Aptitude Important Formulas, Definitions, & Examples:
#### Overview:
Topic Included: Formulas, Definitions & Exmaples. Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Permutation and Combination Aptitude Notes & Questions. Questions for practice: 10 Questions & Answers with Solutions.
Binomial Theorem: It is the sum of combinations from zero to $$n$$ different objects.$$nC_0 + nC_1 + nC_2.....nC_n = 2^n$$
Example: If there are three balls and a boy is asked to select balls out of three, then find how many ways the boy can select the balls?
Solution: If no ball is selected by the boy$$= 3C_0$$
If one ball is selected by the boy$$= 3C_1$$
If two balls are selected by the boy$$= 3C_2$$
If three balls are selected by the boy$$= 3C_3$$
According to the binomial theorem.$$nC_0 + nC_1 + nC_2.....nC_n = 2^n$$ $$3C_0 + 3C_1 + 3C_2 + 3C_3 = 2^3$$ $$= 2^3 = 8 \ ways$$
Important Cases:
Case(1): If there are $$n$$ total number of objects and the objects are subdivided into 4 types a, b, c, and d, then the objects can be arranged in $$\frac{n!}{a! \ b! \ c! \ d!}$$ ways.
Example: How many different words can be formed by using letters of SCHOOL?
Solution: In the word SCHOOL, the letter "O" is used 2 times and other letters are used 1 time each, hence $$= \frac{n!}{a! \ b! \ c! \ d!}$$ $$= \frac{6!}{2! \ 1! \ 1! \ 1! \ 1!}$$ $$= \frac{6!}{2!}$$ $$= \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$$ $$= 360 \ words$$
Case(2): If there are n total number of objects and have taken all objects at a time, whereas repetition is not allowed then the objects can be arranged in $$n!$$ ways.
Example: If there are 4 chairs for 4 persons A, B, C, and D, then find how many ways of sitting arrangement are possible?
Solution: People can sit in $$4!$$ ways.$$4! = 4 \times 3 \times 2 \times 1$$ $$= 24 \ ways$$
Case(3): If there are n total number of objects and have taken $$r$$ objects at a time, whereas repetition is allowed then the objects can be arranged in $$n^r$$ ways.
Example: How many three digit numbers can be formed by 1, 2, 3, and 4 if repetition is allowed?
Solution: $$n^r = 4^3 = 64$$
Case(4): Circular Permutation: If there are $$n$$ number of objects and need to arrange in a circular way, then the objects can be arranged in $$(n - 1)!$$ ways.
Note: In the case of flowers in the garland and beads in a neckless, the flowers or beads can be arranged in $$\frac{(n - 1)!}{2}$$ ways.
Example: If there are 5 chairs around a dining table in a circular way for five persons, then how many ways of sitting arrangement are possible?
Solution: For circular arrangement it can be done in $$(n - 1)!$$ ways, $$= (5 - 1)!$$ $$= 4! = 4 \times 3 \times 2 \times 1$$ $$= 24 \ ways$$
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# Suppose ${a_2},{a_3},{a_4},{a_5},{a_6},{a_7}$ are integers such that,$\dfrac{5}{7} = \dfrac{{{a_2}}}{{2!}} + \dfrac{{{a_3}}}{{3!}} + \dfrac{{{a_4}}}{{4!}} + \dfrac{{{a_5}}}{{5!}} + \dfrac{{{a_6}}}{{6!}} + \dfrac{{{a_7}}}{{7!}}$ Where, $0 \leqslant a < j$ for $j = 2,3,4,5,6,7$ . The sum ${a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7}$ is?
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Hint: Here we are given some fractions, which have factorial in its denominator. For the relation we are given above, we have to find the sum of all the unknown variables ${a_2},{a_3},{a_4},{a_5},{a_6},{a_7}$. To do so, we see that the denominator values are the product of the previous term denominator and just the next value of the previous value under factorial of the previous term. We use this pattern to solve this question.
Complete step-by-step solution:
We are given that,
$\dfrac{5}{7} = \dfrac{{{a_2}}}{{2!}} + \dfrac{{{a_3}}}{{3!}} + \dfrac{{{a_4}}}{{4!}} + \dfrac{{{a_5}}}{{5!}} + \dfrac{{{a_6}}}{{6!}} + \dfrac{{{a_7}}}{{7!}}$
Since we know that $a! = a \times (a - 1) \times ... \times 1$, we can write the denominators of RHS as,
$\Rightarrow \dfrac{5}{7} = \dfrac{{{a_2}}}{{2 \times 3}} + \dfrac{{{a_3}}}{{3 \times 2 \times 1}} + \dfrac{{{a_4}}}{{4 \times 3 \times 2 \times 1}} + \dfrac{{{a_5}}}{{5 \times 4 \times 3 \times 2 \times 1}} + \dfrac{{{a_6}}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1}} + \dfrac{{{a_7}}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1!}}$
We now take common $\dfrac{1}{2}$ from the RHS, $\dfrac{1}{3}$from all terms of RHS except first term and so on,
$\Rightarrow \dfrac{5}{7} = \dfrac{1}{2}\left[ {{a_2} + \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right]} \right]$
We now solve this as,
$\Rightarrow \dfrac{{5 \times 2}}{7} = {a_2} + \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right] \\ \Rightarrow \dfrac{{10}}{7} = {a_2} + \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right] \\ \Rightarrow 1 + \dfrac{3}{7} = {a_2} + \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right] \\$
We know that terms after ${a_2}$ have values less than $1$, so we get
${a_2} = 1$
Now on solving further,
$\Rightarrow \dfrac{3}{7} = \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right] \\ \Rightarrow \dfrac{9}{7} = {a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right] \\ \Rightarrow 1 + \dfrac{2}{7} = {a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right] \\$
We can see that terms after ${a_3}$ have value less than $1$, so we get
${a_3} = 1$
On solving further we get the values of other unknown integers as,
${a_4} = 1$
${a_5} = 0$
${a_6} = 4$
${a_7} = 2$
Now we are asked to find the value of ${a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7}$, we find it as,
${a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7} = 1 + 1 + 1 + 0 + 4 + 2 \\ \Rightarrow {a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7} = 9$
Hence we get the answer as $9$.
Note: Factorial of a number is a way to write the multiplication of consecutive terms in a decreasing way of that number. It is highly used in Permutation Combination, Binomial theorem, Probability theory and many more areas. It is important to be comfortable with its properties to be able to study math and statistics, especially the topics mentioned above further.
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# 2017 AMC 8 Problems/Problem 18
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$, $BC=4$, $CD=3$, and $AD=13$. What is the area of quadrilateral $ABCD$?
$[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("B", (0, 0), SW); label("A", (12, 0), ESE); label("C", (2.4, 3.6), SE); label("D", (0, 5), N);[/asy]$
$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$
## Solution 1
We first connect point $B$ with point $D$.
$[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label("B", (0, 0), SW); label("A", (12, 0), ESE); label("C", (2.4, 3.6), SE); label("D", (0, 5), N);[/asy]$
We can see that $\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\triangle BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $\triangle BDA$ is $\frac{5\cdot 12}{2}$, and the area of the smaller 3-4-5 triangle is $\frac{3\cdot 4}{2}$. Thus, the area of quadrialteral $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$
## Solution 2
$\triangle BCD$ is a 3-4-5 right triangle. So the area of $\triangle BCD$ is 6. Then we can use Heron's formula to compute the area of $\triangle ABD$ whose sides have lengths 5,12,and 13. The area of $\triangle ABD$ = $\sqrt{s(s-5)(s-12)(s-13)}$ , where s is the semi-perimeter of the triangle, that is $s=(5+12+13)/2=15.$ Thus, the area of $\triangle ABD$ =30, so the area of $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$ ---LarryFlora
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Evaluate the integral or show that it is divergent?
Jan 4, 2018
The improper integral converges and equals $\frac{32}{3}$.
Explanation:
First note that
$\int \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right) = \int {\left(x + 2\right)}^{- \frac{1}{4}} \mathrm{dx} = \frac{4}{3} {\left(x + 2\right)}^{\frac{3}{4}} + C$.
Next, note that the "impropriety" in the definite integral occurs at $x = - 2$ because the function has a vertical asymptote there. Hence,
${\int}_{- 2}^{14} \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right) = {\lim}_{a \to - 2 +} {\int}_{a}^{14} \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right)$
assuming this limit exists, where the positive sign to the right of $- 2$ in the limit indicates that $a$ is approaching $- 2$ from the right.
${\int}_{a}^{14} \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right) = \frac{4}{3} {\left(x + 2\right)}^{\frac{3}{4}} {|}_{a}^{14}$
$= \frac{4}{3} \cdot {16}^{\frac{3}{4}} - \frac{4}{3} {\left(a + 2\right)}^{\frac{3}{4}} = \frac{32}{3} - \frac{4}{3} {\left(a + 2\right)}^{\frac{3}{4}}$.
Since ${\lim}_{a \to - 2 +} {\left(a + 2\right)}^{\frac{3}{4}} = {0}^{\frac{3}{4}} = 0$, it follows that the improper integral converges to $\frac{32}{3}$.
We therefore write
${\int}_{- 2}^{14} \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right) = \frac{32}{3}$.
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# How do you solve the inequality 16 - 2c < 14?
Dec 30, 2015
Step by step solution of solving the linear inequality is given below. Please go through the explanation.
#### Explanation:
To solve the inequality $16 - 2 c < 14$
Start by isolating the term containing the variable. This part is done as we do for equality.
$16 - 2 c < 14$ we see $16$ is added to $- 2 c$
Using inverse or opposite operation to addition to remove the $16$
$16 - 2 c - 16 < 14 - 16$
$- 2 c < - 2$
Now we have $2 c$ term isolated. We move on to the next step of isolating $c$.
Note of caution : When we multiply or divide both sides of inequality by a negative sign, the inequality flips. The most common error is ' not flipping the inequality ' so watchout.
$\frac{- 2 c}{-} 2 > - \frac{2}{-} 2$ See how the inequality is flipped.
$c > 2$ is the solution.
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Explain the Hamming Codes in Error Correction
The Hamming code can be used to the data units of any length when we consider the two codewords which have the same length. The Hamming distance between two codewords is represented as the number of positions in which their specific item differs.
Example
Hamming Distance is 3.
Hamming code is also known as linear block code. The family of (n, k) hamming codes for m is defined.
• Block length n=2m-1.
• The number of message bits k=2m-m-1.
• Number of Parity Bits = n-k=m.
Where m≥3
• Minimum distancemin=3.
• Code rate = Code efficiency = $\frac{k}{n} = \frac{2^{m}-m-1}{2^{m}-1} = 1 - \frac{m}{2^{m}-1}$ of m>> 1
then, code rate r=1 where k is message bits and n is number of transmitted bits per block.
Structure
The parity (redundant) bits are inserted between the data units or the end of data units.
To evaluate, several redundancy bits are needed to accurately represent the given number of data bits m. We should find relationships between m and r.
If a total transmission is m + r, r must indicate m + r + 1 different states. One state defines no error, and m + r states denote the position of an error in each of m + r positions.
Therefore, m + r + 1 states should be discovered by r bits, and r bits can denote 2r multiple states. So 2r ≥m+r+1.
Let us suppose m = 7, then the minimal r-value that satisfies this equation is 4.
11 10 9 8 7 6 5 4 3 2 1 d D r4 D D D D r3 D r2 r1
In this r bit is put on 2n position. For example, 20 = 1, 21 = 2, 22 = 4 ect positions
In Hamming code, each r bit is VRC (Vertical Redundancy Check).
r1 bit is calculated by using all bit positions whose binary representation contains 1 in the rightmost position, 1, 3, 5, 7, 9, 11.
r2 bit is computed by using all positions with a 1 in 2, 3, 6, 7, 10, 11 positions and so on. r3 bit will take care of bits position at 4, 5, 6. r4 is checked at 8, 9, 10, 11.
Example
1011
11 1001 0111 0101 0011 0001
d D D r4 d d d r3 d r2 r1
r2 will take care of it
1011 1010
11 10 0111 0110 0111 0010
d D D r6 d d d r4 d r2 r1
and so on.
Calculate r values
We locate each bit of original characters in its suitable position in the 11-bit unit. We can calculate even parities for the multiple bit combinations. The parity value for each consolidation is the value of the correlating r bit.
Example
is calculated to support the even parity for merging bits 3, 5, 7, 9, 11, etc. The last 11-bit code is sent through the transmitted line.
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Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
At that time people will see the Son of Man coming in clouds with great power and glory. Mark 13:26 NIV
# 4-10 Applications of Right Triangle Trigonometry
Summary: In this section, you will:
• Solve problems with right triangles and trigonometry
SDA NAD Content Standards (2018): PC.7.3
Right triangles can be used to solve many problems. One example is calculating the angle a camera would have to be set at to capture a model rocket at it's apogee, or highest altitude.
## Solve Problems with Right Triangles
For problems that can be solved with right triangle trigonometry, draw a triangle as described in the problem. Then use trigonometry to solve for the unknown.
#### Example 1: Model Rockets
It is estimated that a certain model rocket will reach an altitude of 200 ft. A photographer is setting up a camera 50 ft away from the launch pad. At what angle should he set the tripod to get a picture at the maximum altitude?
###### Solution
Draw a right triangle to model the problem.
The hypotenuse and the angle with the ground are known. The opposite side is the unknown. Use sine to solve the problem.
$$\tan u = \frac{opp}{adj}$$
$$\tan u = \frac{200 \textrm{ ft}}{50 \textrm{ ft}}$$
$$u = \tan^{−1} \left(\frac{200}{50} \right) \approx 76.0°$$
##### Try It 1
If a boat is tied to a pier 10 ft below the deck of the boat with a rope that is 15 ft long, what angle does it make with the dock?
41.8°
#### Example 2: Find an Acute Angle
In areas that get a lot of snow, roofs must be inclined at a certain angle to meet building code. That way the snow will slide off the roof and not crush the house. In one town the incline must be at least 20° above the horizontal. A builder is making a roof with a rise of 4 feet for every 12 feet of run. Will this roof meet building code?
###### Solution
Draw a right triangle to represent the roof.
The opposite side is 4 ft and the adjacent side is 12 ft. Tangent has this ratio.
$$\tan u = \frac{4 \textrm{ ft}}{12 \textrm{ ft}}$$
$$u = \tan^{–1} \frac{4}{12} \approx 18.4°$$
This is less than the required 20°, so the roof is not steep enough and will not meet building code.
##### Try It 2
A bird is sitting on top of a 10 m high tower that is 15 m away from an avid birder. What angle should the birder use to aim his small telescope to see the bird?
33.7°
## Solve Right Triangles
Solving a triangle means to find the measures of all unknown sides and angles. To do this use the trigonometric functions and inverse functions, the Pythagorean theorem, and the triangle sum theorem. The triangle sum theorem states that the sum of the angles of a triangle equal 180°.
###### Solve a Right Triangle
Find the lengths of all the sides and angles using
• Trigonometric functions
• Inverse trigonometric functions
• Pythagorean theorem
• Triangle sum theorem (sum of angle measures = 180°)
The sides are labeled with a lowercase letter to match the opposite angle.
#### Example 3: Solve a Right Triangle
Solve the triangle.
###### Solution
Since two sides are known, start by finding the third side using the Pythagorean theorem with the hypotenuse, b.
b2 = 82 + 122
b2 = 208
$$b = 4\sqrt{13}$$
Now find the angles. Because the side lengths are known, inverse trigonometric functions should be used.
$$A = \tan^{-1} \frac{8}{12}$$
A ≈ 33.7°
Angle C can be found using either inverse tangent or the triangle sum theorem.
A + B + C = 180°
33.7° + 90° + C = 180°
C ≈ 56.3°
##### Try It 3
Solve the triangle.
a ≈ 13.2, b ≈ 5.6, C = 65°
##### Lesson Summary
###### Solve a Right Triangle
Find the lengths of all the sides and angles using
• Trigonometric functions
• Inverse trigonometric functions
• Pythagorean theorem
• Triangle sum theorem (sum of angle measures = 180°)
The sides are labeled with a lowercase letter to match the opposite angle.
## Practice Exercises
1. A 12-foot ladder is leaning against a rain gutter 10 feet above the ground. What angle, in radians, does the ladder make with the ground?
2. Two people climb 220 feet up the side of a sand dune so that the change in elevation is 120 feet. What is the angle of elevation of the side of the sand dune?
3. The congruent legs of an isosceles triangle are 10 cm, and the base is 6 cm. What is the measure of a base angle of the triangle?
4. Without using a calculator, estimate the value of tan−1(1,000,000). Explain your reasoning.
5. A guy-wire is a cable that attaches to the top of an electrical pole at an angle to hold it upright. It forms a right triangle with the pole and the ground. If the pole is 13 feet tall and the guy-wire attaches to the ground 5 feet from the pole, what angle does the wire make with the pole?
6. What is the angle that the line $$y = \frac{2}{3}x$$ makes with the positive x-axis?
7. What is the angle that the line $$y = \frac{5}{2}x$$ makes with the positive x-axis?
8. The percent grade of a road is the change in height over a 100-foot horizontal distance. What is the percent grade of a road with a 3° angle of elevation?
9. One of the trusses on a railroad bridge is shaped like a right triangle. If the vertical leg is 20 feet and the horizontal leg is 12 feet, what angle does the hypotenuse make with the horizontal leg?
10. Frank is building a chicken coop. The frame for the roof will be an isosceles triangle with a base of 4 feet and a height of 1.5 feet. What angle should he cut the wood at the end of the base to get a tight fit?
11. Solve the Right Triangle
12. Mixed Review
13. (4-09) Evaluate $$\sin\left(\cos^{–1} \frac{4}{5}\right)$$.
14. (4-09) Evaluate $$\tan^{–1}\left(\tan \frac{π}{3}\right)$$.
15. (4-08) Evaluate $$\arcsin \frac{\sqrt{3}}{2}$$.
16. (4-07) Graph $$y = \frac{\cos x}{\sin x}$$ and y = cot x on the same graph. What is the relationship between the two functions?
17. (4-05) If sec θ = −3 and sin θ < 0, find a) tan θ and b) csc θ.
1. 0.985
2. 33.1°
3. 72.5°
4. $$\frac{\pi}{2}$$
5. 21.0°
6. 33.7°
7. 68.2°
8. 5%
9. 59.0°
10. 36.9°
11. $$a = \sqrt{11}$$, $$A \approx 33.6°$$, $$B \approx 56.4°$$
12. $$a = \sqrt{41}$$, $$B \approx 51.3°$$, $$C \approx 38.7°$$
13. $$c = \sqrt{15}$$, $$B \approx 61.0°$$, $$C \approx 29.0°$$
14. $$c = \sqrt{65}$$, $$A \approx 29.7°$$, $$B \approx 60.3°$$
15. $$a = \sqrt{17}$$, $$A \approx 27.3°$$, $$C \approx 62.7°$$
16. $$\frac{3}{5}$$
17. $$\frac{π}{3}$$
18. $$\frac{π}{3}$$
19. They are the same graph.
20. $$2\sqrt{2}$$; $$−\frac{3\sqrt{2}}{4}$$
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# Find the derivative of the function.$y = \sqrt {x + \sqrt {x + \sqrt {x}}}$
## $$\frac{1}{2 \sqrt{x}}+1\\\frac{2 \sqrt{x+\sqrt{x}}+1}{2 \sqrt{x+\sqrt{x+\sqrt{x}}}}$$
Derivatives
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##### Heather Z.
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##### Kristen K.
University of Michigan - Ann Arbor
##### Michael J.
Idaho State University
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### Video Transcript
all right here we have quite the extensive function that we're going to differentiate using the chain rule. And every time I have a square root function, I like to change it to a 1/2 power function before I differentiate. That allows me to use the product or the power rule. Excuse me. So that means I'm going to rewrite this as the quantity X plus the quantity X plus X to the 1/2 to the 1/2 to the 1/2 which really doesn't make anything look better. Um, and it's going to look pretty bad once I take the derivative, but then we'll kind of clean it up after that. So using the chain rule, we have white prime equals. We're going to work on the derivative of the outside function. So the outer 1/2 power function bring down the 1/2 and then take the entire inside and raise that to the negative 1/2. Now we move on to multiply by the derivative of the inside, and this is the insight. So the inside is a some and to find the derivative of a some you find the derivative of each term in the sun. So we start with the derivative of X, and that's one. And now we move on to adding the derivative of the second part of the some, and we're going to need the chain rule for that. So again, a 1/2 power. So bring down the 1/2 and raised the inside to the negative 1/2 and then we need to multiply by its inside derivative. So inside his X plus extra, the 1/2 which is also a some so it's derivative will be the the some of the derivatives. So the derivative of X is one plus the derivative of X to the 1/2 is 1/2 X to the negative 1/2. Now we made it to the very inside of everything. So the next thing we want to do is simplify our answer and change things back into square root signs. So we have this too, which is going to go in the denominator of the entire thing. And because this is to the negative power, it's also going to go in the denominator of the whole thing, and we're going to change it back to a square root sign. So all the 1/2 powers will be changed back to square root signs. So the denominator of the whole big answer is going to have a two in it. And then it's going to have the square root of X plus the quantity X whoops. How about if I go back and change that to a square root as well? I'm gonna make that notation look a little bit better while I'm erasing. Okay, so we're rewriting this thing right here. We're rewriting that. So it's a square root of X plus, the square root of X plus the square root of X. Okay, now we're gonna work on simplifying this thing, and it's our numerator. So it's actually, um Well, we've got the plus one. We can start with that, and then we're gonna work on this part in a very similar way. We just worked on the previous part, so we have the two on the bottom. We have a negative 1/2 power, so that part's gonna go on the bottom as well. So we're going to have the two times the square root of X plus square root X. So that takes care of all of this and then this part will be in the numerator. So in the numerator we have one plus. But now deja vu. This part is going to have a two on the bottom. And because of the negative power, the square roots going to be on the bottom. So one over two square root X So how is that for a tower of incredible numbers and algebra?
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##### Heather Z.
Oregon State University
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University of Michigan - Ann Arbor
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Idaho State University
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# The angle of depression of the top and the bottom of a 7m tall building from the top of tower are ${45^ \circ }$ and ${60^ \circ }$ respectively, find the height of the tower.
Last updated date: 09th Aug 2024
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Hint: By looking at the given figure we can clearly see the right angled triangle AEC and ABD. And we have to find the length of AB, also the length of the CD is given to us.
We will use suitable trigonometric functions for each of the triangles and then solving the equation formed, we will get required length.
Given angle of depression of top of the building from top of tower is ${45^ \circ }$,
And the angle of depression of the bottom of the building from top of tower is ${60^ \circ }$.
Let the height of the tower be h.
Length of the building is $CD = 7m$ -(1)
Now,
$AE = AB - EB$
As, $EB = CD$,
$AE = h - CD$
Using (1),
$AE = h - 7$ -(2)
Now from the above figure we can see a right-angled triangle AEC-
Therefore,
$\tan \theta = \dfrac{{perpendicular}}{{base}} \\ \tan {45^ \circ } = \dfrac{{AE}}{{CE}} \\$
Using $\tan {45^ \circ } = 1$ and (2) equation,
$1 = \dfrac{{h - 7}}{{BD}}$
$BD = h - 7$ -(3)
Now we can also see the right-angled triangle ABD-
So,
$\tan {60^ \circ } = \dfrac{{AB}}{{BD}}$
Using $\tan {60^ \circ } = \sqrt 3$ and (3) equation,
$\sqrt 3 = \dfrac{h}{{h - 7}} \\ \sqrt 3 h - 7\sqrt 3 = h \\ \sqrt 3 h - h = 7\sqrt 3 \\ h\left( {\sqrt 3 - 1} \right) = 7\sqrt 3 \\ h = \dfrac{{7\sqrt 3 }}{{\sqrt 3 - 1}}m \\$
Therefore, the height of the tower is $\dfrac{{7\sqrt 3 }}{{\sqrt 3 - 1}}$m.
Note: In the above question we have used $\tan \theta$ because if we see the diagram there are two right-angled triangles formed with bases of equal length, perpendicular is required and the corresponding angles are given.
Hence, we need a trigonometric function including base and perpendicular which is $\tan \theta$.
Also, we can rationalize the final answer and we will obtain height of tower equal to-
$\dfrac{{7\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{2}$.
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# Step deviation Method for Finding the Mean with Examples
Statistics is a discipline of mathematics that uses quantified models and representations to gather, review, analyze, and draw conclusions from data. The most commonly used statistical measures are mean, median, and mode. Variance and standard deviation are measures of dispersion in statistics and various measures of concentration including quartiles, quintiles, deciles, and percentiles. Statistics is a way more beyond the topics mentioned, but here we stop for the “Mean” by Step Deviation method. In general, there are 3 types of mean:
1. Arithmetic mean
2. Geometric mean
3. Harmonic mean
This article is about the Arithmetic mean by Step Deviation method. The arithmetic mean, also called the average or average value is the quantity obtained by summing two or more numbers or variables and then dividing by the number of numbers or variables. The arithmetic mean is important in statistics. For example, Let’s say there are only two quantities involved, the arithmetic mean is obtained simply by adding the quantities and dividing by 2. Mean or Arithmetic Mean is the average of the numbers i.e. a calculated central value for a set of numbers. General Formulae for Mean is,
Mean = Sum of observation / Number Of Observation
Example:
The marks obtained by 5 students in a class test are 7, 9, 6, 4, 2 out of 10. Find the mean marks for the class?
According to the formula mean marks of the class are:
Average marks = Sum of observation / Number Of Observation
Here average marks = (7 + 9 + 6 + 4 + 2) / 5 = 28 / 5 = 5.6
Hence the mean marks for the class is 5.6
### Derivation of Formula for Mean by Step Deviation Method
The general formula for mean in statistics is:
Mean = Σfixi / Σfi
Where,
Σfixi: the weighted sum of elements and
Σfi: the number of elements
In the case of grouped data, assume that the frequency in each class is centered at its class-mark. If there are n classes and fi denotes the frequency and yi denotes the class-mark of the ith class the mean is given by,
Mean = Σfiyi / Σfi
When the number of classes is large or the value of fi and yi is large, an approximate (assumed) mean is taken near the middle, represented by A and deviation (di) is taken into consideration. Then mean is given by,
Mean = A + Σfidi / Σf
In the problems where the width of all classes is the same, then further simplify the calculations of the mean by computing the coded mean, i.e. the mean of u1, u2, u3, …..un where,
ui = (yi – A) / c
Then the mean is given by the formula,
### Mean = A + c x (Σfiui / Σfi)
This method of finding the mean is called the Step Deviation Method.
### Examples
Question 1: Find the mean for the following frequency distribution?
Solution:
Applying the Standard Deviation Method,
We take the assumed mean to A = 99, and here the width of each class(c) = 6
Mean = A + c x (Σfiui / Σfi)
= 99 + 6 x (-8/50)
= 99 – 0.96
= 98.04
Question 2: Find the mean for the following frequency distribution?
Solution:
Applying the Standard Deviation Method,
Construct the table as under, taking assumed mean A = 45, and width of each class(c) = 10.
Mean = A + c x (Σfiui / Σfi)
= 45 + 10 x (23/50)
= 45 + 4.6
= 49.6
Question 3: The weight of 50 apples was recorded as given below
Calculate the mean weight, to the nearest gram?
Solution:
Construct the table as under, taking assumed mean A = 97.5. Here width of each class(c) = 5
Mean = A + c x (Σfiui / Σfi)
= 97.5 + 5 x (-16/50)
= 97.5 – 1.6
= 95.9
Hence the mean weight to the nearest gram is 96 grams.
Question 4: The following table gives marks scored by students in an examination:
Calculate the mean marks correct to 2 decimal places?
Solution:
Construct the table as under, taking assumed mean A = 17.5. Here width of each class(c) = 5
Mean = A + c x (Σfiui / Σfi)
= 17.5 + 5 x (17/80)
= 17.5 + 1.06
= 18.56
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# How can you simplify this expression?
$$\frac{(n+1)!}{k! \cdot (n-k+1)!}$$
I don't have any clue. Can $(n-k+1)!$ be simplified into $(n+1)!(k)!$
I am not too familiar with this symbol: !.
• Relevant reading: Factorial and Binomial_coefficient – user2468 Nov 1 '12 at 16:33
• The symbol $!$ refers to factorial. – Sasha Nov 1 '12 at 16:34
• i get n!(n+1)/k! (n-k)! (n-k+1) how do I get rid of the k in the denominator? – Martin Desjardins Nov 1 '12 at 16:47
This is the binomial coefficient $\binom{n+1}{k}$.
• The formula for $\binom{n}{k}$ is $\frac{n!}{k!(n-k)!}$. If you substitute $n+1$ for $n$, you get the above. – ngn Nov 1 '12 at 16:53
The symbol $!$ is the factorial, it is defined as $$n! = n\cdot(n-1)\cdot(n-2)\cdots 2\cdot 1 = \prod_{j=1}^n j.$$ The expression $\frac{n!}{k!\cdot(n-k)!}=\binom nk$ is also known as the binomial coefficient. There is a number of interesting properties of this quantity. Most intuitively, it is the number of ways to pick $k$ elements out of a set that contains $n$ elements.
A concrete example might be helpful. Let $n = 4$ and $k = 3$. The expression can be expanded to $$\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }{(3 \cdot 2 \cdot 1)(2 \cdot 1)}.$$
Notice both the expressions in the denominator are shorter factorials, so you can quickly cancel the "tail" of the factorial in the numerator as $$\frac{5 \cdot 4}{2 \cdot 1}$$ or $$\frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1}.$$ (Of course, more cancellation can be done with specific numbers, but I'm trying to mimic the general case where $n$ and $k$ are not specified.)
Back to general $n$ and $k$, we can simplify your original expression as either $$\frac{(n+1) \cdot n \cdots (k+1)}{(n+1-k)!}$$ or $$\frac{(n+1) \cdot n \cdots (n+2-k)}{k!}.$$ In the former, I cancelled the $k!$ "tail". In the latter, I cancelled the $(n+1-k)!$ "tail". Without more information about $n$ and $k$, this is the most cancellation one can hope for.
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# Integers and their Opposites
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Integers and their Opposites
CCSS.MATH.CONTENT.6.NS.C.6.A
## Integers and Their Opposites
Understanding integers and their opposites is like having a superpower in math! Imagine being able to represent any whole number, positive or negative, and knowing its exact opposite. From calculating temperatures to understanding bank account balances, knowing about integers and their opposites helps us make sense of the world around us. Once you understand how integers and their opposites work, you can easily learn subtracting integers and adding integers
Integers are whole numbers that can be positive, negative, or zero. They are used to represent both quantities and positions on the number line.
## Understanding Integers and Their Opposites – Explanation
An integer is a number without any fractional or decimal part. It can be positive, negative, or zero. Integers are often used to represent situations involving counting, direction, temperature, and more.
Is the number $\frac{1}{5}$ an integer?
Is zero an integer?
Is negative twenty an integer?
Is the fraction $\frac{3}{4}$ an integer?
Is six and a half an integer?
Is the number negative two thousand three hundred fifty-six an integer?
## What is the opposite of an integer?
The opposite of an integer is the number that, when added to it, gives zero. For example, the opposite of 5 is -5, and the opposite of -3 is 3.
Integer Opposite
0 0
5 -5
-3 3
10 -10
-7 7
2 -2
## Integers and Their Opposites – Examples
Let’s practice understanding integers and their opposites:
What is the opposite of -18?
• The opposite of -18 is 18.
Identify the opposite of -7.
Find the opposite of 9.
What is the opposite of 0?
What is the opposite of -12?
Determine the opposite of 15.
What is the opposite of -3?
Find the opposite of 20.
Identify the opposite of -100.
## Integers and Their Opposites – Summary
Key Learnings from this Text:
• Integers are whole numbers that include positive, negative, and zero values.
• The opposite of an integer is the number that, when added to it, results in zero.
• Understanding integers and their opposites is essential for various mathematical operations and real-life applications.
## Integers and Their Opposites – Frequently Asked Questions
What are integers, and how do they differ from other types of numbers?
Can you provide real-life examples where integers and their opposites are used?
Why is it important to understand the concept of opposites when dealing with integers?
Are all whole numbers considered integers?
What happens when you add an integer to its opposite?
Do integers have a specific place on the number line?
Can you have an opposite for zero?
What happens when you multiply or divide two integers?
How do integers relate to the concept of direction?
Can you have fractions or decimals as integers?
### TranscriptIntegers and their Opposites
"Hey Kai, what's under that cover?" "That button is very tempting!" "Go on Kai, you should totally push it." "Oh, that goes without saying, June!" Welcome to the Integer Universe! To make it back to your universe, you will need to win the game by using integers and their opposites. Buckle up and pay close attention to the tutorial! An integer is any whole number. The opposite of an integer is a number that is the same distance from a set point on a number line. This is called equidistant. Take a look at this number line. Going left from zero are the negative integers, and going right from zero are the positive integers. Zero is a special case here, since it is neither positive or negative, which makes zero it's own opposite! To find the opposite of an integer, you always move in the opposite direction from zero the same number of values. If we move two places to the right from zero we end up on two. To find the opposite, we must move two places to the left from zero to end up on negative two. In this case, the opposite of two is negative two! Now, you probably noticed we counted the same value from zero both ways, two. This is what is known as the absolute value. The absolute value of two and negative two is two, since we moved the same distance from zero each time. Absolute value is the integer's distance from zero, whether it is a positive integer or negative integer. That concludes the tutorial, get ready to play! Level one, begin! Find the opposite integer of seven! The opposite must be equidistant, or equal distance, from zero, so we count seven values to the left of the zero, which is negative seven. The opposite of seven is negative seven. Now find the opposite integer of negative eleven! The opposite must be equidistant from zero, so we count eleven values to the right of the zero, which is eleven. The opposite of negative eleven is eleven. Level one complete! Level two, begin! Find the opposite of six, and name the absolute value! The opposite of six is negative six. Since we counted six values either direction from zero both times, the absolute value is six. Now find the opposite of negative ten, and name the absolute value! The opposite of negative ten is ten. Since we counted ten values either direction from zero, the absolute value is ten. For a bonus, in this problem, what is the opposite of the opposite? Since the opposite of negative ten is ten, the opposite of the opposite is the original value, negative ten. Phew! You won the game, but before you return, let's summarize. All integers have a positive value and a negative value, except for zero. Zero is neither positive or negative, and is it's own opposite. The opposite of any integer is equidistant from zero on a number line. The distance positive and negative integers are from zero is known as the absolute value. "Woah. I feel very negative about that experience!" "Well I'm absolutely positive that we shouldn't do that again. Let's cover it back up Kai!"
## Integers and their Opposites exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Integers and their Opposites.
• ### Connect the terms and definitions.
Hints
1, -1, 2, -2 and so on are all integers.
1.5, -1.5, 2.75, -2.75 are NOT integers.
What is the difference in these groups of numbers?
5, 6,10, and 20 are all positive integers.
-5, -6, -10, and -20 are all negative integers.
The absolute value of 10 and -10 is 10.
The absolute value of 25 and -25 is 25.
The absolute value of 42 and - 42 is 42.
What is the definition of absolute value?
Solution
• Two or more things that are the same distance from a set point are equidistant.
• Another name for a whole number is integer.
• Whole numbers with a positive value more than zero are positive integers.
• Whole numbers with a negative value less than zero are negative integers.
• The distance from an integer to zero on a number line is the integer's absolute value.
• ### Identify the true sentences.
Hints
The absolute value is the distance from an integer to zero on a number line.
Example: The absolute value of 10 and -10 is 10.
Both 10 and -10 are integers.
10.5 and -12.6 are NOT.
-12 and 12 are opposite integers.
They are both 12 away from zero on a number line.
There are 3 correct choices and 2 false.
Solution
TRUE
• All integers have a positive or negative value except for zero.
• The opposite of any integer is equidistant from zero on a number line.
• Both positive and negative integers have the same absolute value.
FALSE
• All integers have a positive value.
• Only positive integers have an absolute value.
• ### Find opposite integers.
Hints
The opposite of any integer is equidistant from zero on a number line.
Integers are whole numbers.
There are 2 correct choices and 4 false.
Solution
CORRECT
These are opposite integers.
• 10 and -10
• 15 and -15
FALSE
• 5, -10 are not equidistant from zero.
• 20, 20 are the same number and cannot be opposites.
• -17, 7 are not equidistant from zero.
• 2.5, -2.5 are not whole numbers and therefore not integers.
• ### Find the opposite integer and absolute value.
Hints
The opposite of any integer is equidistant from zero on a number line.
The absolute value is the distance from an integer to zero on a number line.
Example: The absolute value of 10 and -10 is 10.
If an integer is positive, its opposite integer must be negative.
If an integer is negative, its opposite integer must be positive.
Solution
5
The opposite integer is -5 because this is equidistant from zero on the number line.
The absolute value is 5 because this is the distance from zero.
17
The opposite integer is -17 because this is equidistant from zero on the number line.
The absolute value is 17 because this is the distance from zero.
-99
The opposite integer is 99 because this is equidistant from zero on the number line.
The absolute value is 99 because this is the distance from zero.
-22
The opposite integer is 22 because this is equidistant from zero on the number line.
The absolute value is 22 because this is the distance from zero.
• ### Find the absolute value.
Hints
Absolute value is a number's distance from zero on a number line.
Both negative and positive numbers will have the same absolute value if they are opposite.
Example: 20, and -20 have the same absolute value.
Solution
The absolute value of 5 and -5 is 5.
This is because they are both 5 away from zero on the number line.
• ### Determine the opposite integer and absolute value.
Hints
If an integer is positive, its opposite integer must be negative.
If an integer is negative, its opposite integer must be positive.
The absolute value is the distance from an integer to zero on a number line.
Example: The absolute value of 10 and -10 is 10
The opposite of any integer is equidistant from zero on a number line.
Solution
43
The opposite integer is - 43 because this is equidistant from zero on the number line.
The absolute value is 43 because this is the distance from zero.
- 5
The opposite integer is 5 because this is equidistant from zero on the number line.
The absolute value is 5 because because this is the distance from zero.
38
The opposite integer is - 38 because this is equidistant from zero on the number line.
The absolute value is 38 because this is the distance from zero.
- 47
The opposite integer is 47 because this is equidistant from zero on the number line.
The absolute value is 47 because this is the distance from zero.
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Browse Questions
# Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 black balls. One ball is transferred from bag I to bag II and then a ball is drawn from bag II at random. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Toolbox:
• According to Bayes Theorem, if $E_1, E_2, E_3.....E_n$ are a set of mutually exclusive and exhaustive events, then $P\left(\large \frac{E_i}{E}\right ) = \Large \frac{P\left(\frac{E}{E_i}\right ). P(E_i)} {\sum_{i=1}^{n} (P\left(\frac{E}{E_i}\right ).P(E_i))}$
Step 1:
Given Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball drawn is red in color.
Let $E_1$ be the event that a red ball is transferred from Bag I to Bag II, and $E_2$ be the event that a black ball is transferred from Bag I to Bag II. Let A be the event that the ball drawn is red.
$P (E_1) = \large\frac{3}{3+4} = \frac{3}{7}$
$P(E_2) = \large\frac{4}{4+3} = \frac{4}{7}$
Step 2:
$P \large(\frac{A}{E_1}) = \frac{5}{10} = \frac{1}{2}$
$P \large(\frac{A}{E_2}) = \frac{4}{10} = \frac{2}{5}$
$P\left(\large \frac{E_2}{E}\right ) = \Large \frac{P\left(\frac{E}{E_2}\right ). P(E_2)} {\sum_{i=1}^{4} (P\left(\frac{E}{E_i}\right ).P(E_i))}$.
$\Rightarrow \large\;\frac{\frac{4}{7}\;\times\;\frac{2}{5}}{\large\frac{3}{7}\;\times\;\frac{1}{2}\;+\;\frac{4}{7}\;\times\;\frac{2}{5}}=\large\;\frac{16}{31}$
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# Lesson Percentage Word Problems (discount)
Algebra -> Algebra -> Percentage-and-ratio-word-problems -> Lesson Percentage Word Problems (discount) Log On
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Word Problems: Problems on percentages, ratios, and fractions Solvers Lessons Answers archive Quiz In Depth
This Lesson (Percentage Word Problems (discount)) was created by by mathick(4) : View Source, Show
A typical word problem involving percentages might look something like this:
Ned got a 12% discount when he bought his new jacket. If the original price, before the discount, was \$50, how much was the discount?
Word problems tend to be even wordier than this one. The solution process involves making the problem simpler and simpler, until it's a math problem with no words.
Step 1. Identify what they're asking for, and call it x.
x = amount of the discount.
Step 2. Use the information given to write an equation that relates the quantities involved.
12% of 50 dollars = the amount of the discount (x).
Step 3. Translate into Math:
(12/100) * 50 = x.
Step 4. Solve for x:
6 = x.
This means that Ned's 12% off amounted to a \$6 discount.
A few examples of some word problems using this method are in the table below. You can try a sample problem if you'd like to practice and see if you're doing it correctly.
Problems: A. The original price was \$160, and Ned got a 20% discount? How much was the discount?
B. The original price was \$90, and Ned got a \$36 discount. How many percent off was the discount?
C. Ned's discount was 20% off the original price, which meant a \$40 discount. What was the original price?
Write an equation: A. What is 20% of 160?
B. 36 is what percent of 90?
C. 40 is 20% of what number?
Translate into math:
is = “=”
of = “*”
What = “x”
5% = “5/100”, etc.
A.
What is 20% of 160? x = (20/100) * 160.
B.
36 is what % of 90? 36 = (x/100) * 90.
C.
40 is 20% of what number? 40 = (20/100) * x.
Simplify & solve for x:
(This may involve solving a linear equation -
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> > > Geometric Mean Formula
# Geometric Mean Formula
We have used the arithmetic mean in many data-related problems. Here we will see another such term frequently used with data analysis. A geometric mean formula is used to calculate the geometric mean of a set of numbers. It is a type of mean that indicates the central tendency of a set of numbers by using the product of their values. It is also defined as the nth root of the product of n numbers. The geometric mean is properly defined only for a positive set of real numbers. In this article, we will discuss the geometric mean formula with examples. Let us begin learning!
## Geometric Mean Formula
### What is the Geometric Mean?
The geometric mean is the mean value of a set of products. Its calculation is commonly used to determine the performance results of an investment or portfolio. It can be stated as “the nth root value of the product of n numbers.”
The geometric mean should be used when working with percentages, which are derived from values. The geometric mean is a very useful tool for calculating portfolio performance. It is because it takes into account the effects of compounding.
### The formula for Geometric Mean
The geometric mean is used as a proportion in geometry and therefore it is sometimes called the “mean proportional”. The mean proportional of two positive numbers a and b, will e the positive number x, so that:
$$\frac{a}{x}=\frac{x}{b}$$
i.e. after doing cross multiplication we get
x=$$\sqrt{a \times b}$$
In general for n multiple numbers as a_1,a_2, a_3,…..,a_n then geometric mean GM will be the nth root of the product of the numbers. In terms of formula it is:
GM = $$\sqrt[n]{a_1 \times a_2 \times a_3 \times …….\times a_n}$$
### Some real-life uses of geometric mean:
1. Aspect Ratios:
The geometric mean has been used in film and video also to find the appropriate aspect ratios i.e. the proportion of the width to the height of a screen or image. It is used to find an appropriate balancing between the two aspect ratios as well as for distorting or cropping both ratios equally.
1. Computer Science:
Computers use mind-boggling amounts of large data which generally requires the summarization for many applications using various statistical measurements.
1. Medicine:
The Geometric Mean has many applications in the medical industry also. It is known as the “gold standard” for some measurements, including for the calculation of gastric emptying time.
1. Proportional Growth:
It is very useful in finding the growth rate. The geometric mean is used for calculating the proportional growth as well as demand growth.
## Solved Example
1: Find the geometric mean of 4 and 3?
Solution: Using the formula for G.M.,
a=4 and b=3
Geometric Mean will be:
x= √(4×3)
= 2√3
So, GM will be 3.46
Example-2: Find the geometric mean of 5 numbers as 4, 8, 3, 9 and 17?
Solution:
Here multiple numbers are taken.
n = 5
Find geometric mean using the formula:
GM = $$\sqrt[n]{a_1 \times a_2 \times a_3 \times …….\times a_n}$$
Putting values of numbers,
We get
GM = $$\sqrt[5]{4 \times 8 \times 3 \times 9 \times 17}$$
i.e. GM = $$\sqrt[5]{4 \times 8 \times 3 \times 9 \times 17}$$
i.e. GM= $$\sqrt[5]{14688}$$
So, geometric mean = 6.81
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# Mastering multiplication. Expert tips for fractions with whole numbers
## Crush Math Class with These Expert Tips on Multiplying Fractions with Whole Numbers
Are you struggling to understand how to multiply fractions with whole numbers in your math class? Don't worry, you're not alone! This topic can be tricky for many students, but with the right tips and tricks, you can crush your math class and become a pro at multiplying fractions with whole numbers. In this blog post, we'll cover some expert tips that will help you master this concept and ace your next math test. So let's dive in and conquer those fractions!
## Understanding the Basics: What are Fractions and Whole Numbers?
Fractions and whole numbers are fundamental concepts in mathematics. A fraction represents a part of a whole, with a numerator (the number on top) representing the number of parts we have and a denominator (the number on the bottom) representing the total number of equal parts in the whole. Whole numbers, on the other hand, are integers without any fractional or decimal parts. They are numbers like 1, 2, 3, and so on. Understanding the difference between fractions and whole numbers is crucial when it comes to multiplying them together, as it forms the foundation for solving more complex mathematical problems. So, let's delve deeper into this topic and uncover the secrets behind multiplying fractions with whole numbers.
## Breaking Down the Process: Step-by-step Guide to Multiply Fractions with Whole Numbers
To successfully multiply fractions with whole numbers, follow these simple steps. First, write the whole number as a fraction by placing it over 1. Next, multiply the numerators (the top numbers) to get the new numerator. Then, multiply the denominators (the bottom numbers) to get the new denominator. Simplify the fraction if possible by dividing the numerator and denominator by their greatest common factor. Finally, if the fraction can be simplified further, do so. Remember, practice makes perfect, so try solving different problems to reinforce your understanding of this process. With these steps, you'll be multiplying fractions with whole numbers like a pro in no time!
## Practical Examples: Solving Real-Life Problems Using Fraction and Whole Number Multiplication
Are you wondering how multiplying fractions with whole numbers applies to real-life situations? Let's explore some practical examples to illustrate its relevance. Imagine you are baking cookies and the recipe calls for 3/4 cup of flour. However, you want to double the recipe. By multiplying the fraction (3/4) by the whole number (2), you find that you need 1 1/2 cups of flour. This shows how multiplying fractions with whole numbers can be used to scale measurements in recipes. Another example is calculating discounts. If an item is on sale for 25% off, multiplying the original price by the fraction (1/4) will give you the discount amount. These real-life scenarios demonstrate how mastering the skill of multiplying fractions with whole numbers can be beneficial in everyday situations.
## Common Mistakes to Avoid When Multiplying Fractions with Whole Numbers
When it comes to multiplying fractions with whole numbers, there are some common mistakes that students often make. One common mistake is forgetting to convert the whole number into a fraction by placing it over 1. This step is crucial because it allows you to properly multiply the numerators and denominators. Another mistake is not simplifying the fraction if possible. Remember to always divide the numerator and denominator by their greatest common factor to simplify the fraction. Additionally, make sure to double-check your calculations and pay attention to signs when dealing with negative numbers. By avoiding these common mistakes, you'll be well on your way to mastering the skill of multiplying fractions with whole numbers.
## Handy Tips and Tricks for Fraction Multiplication Mastery
To become a true master of multiplying fractions with whole numbers, here are some handy tips and tricks to keep in mind. First, always simplify the fraction before multiplying, if possible. This will make the calculations easier and the answer more precise. Another tip is to pay attention to the placement of negative signs. When multiplying a negative fraction with a negative whole number, the result will be positive. Additionally, practice estimating the answer before calculating it to ensure you are on the right track. Lastly, remember to always double-check your work and use scratch paper to avoid errors. By applying these tips and tricks, you'll be multiplying fractions with whole numbers like a pro in no time!
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# Algebra Worksheets
In this page you can get free algebra worksheets for the topic algebra. It will be very useful to students who are in7th,8th and 9th grade students. The students must practice this kind of worksheets to get clear knowledge in algebra. You can find many different questions in these worksheets.
One of the most important topic in algebra is algebraic identities.This formulas will be useful not only in school days in college days too.Students must clear about these kinds of topis.By using these formulas even we can multiply the large numbers also.Let us consider the example. Multiply 256 x 244 many of them will multiply these two number directly otherwise they will use calculators. There is another way to multiply this.
256 x 244 can be written in the form of (250+6) (250-6). This is the first step to be done. Now we have to remember the formula (a+b) (a-b). The formula is a^2-b^2. Instead of "a" we have 250 and instead of "b" we have 6. So now let us write them in the form of (250 )^2 . We can easily find 25 x 25 that is 625 if two zeros will be added at the end of the 625 it will become 62500 . Now we need to find the value of 6^2 that is 36.if we subtract 36 from 62500 it will become 62464.This is the final answer we will get if we multiply 256 x 244.
### Algebra worksheets-Related topics
1. Worksheets for (a + b) 2
2. Worksheets for (a - b) 2
3. Worksheets for (a + b) 3
4. Worksheets for (a - b) 3
5. Worksheets for (a3 + b3)
6. Worksheets for (a3 - b3)
7. Worksheets for (a2 - b2)
8. Worksheets for (a + b + c) 2)
9. Quote on Mathematics
“Mathematics, without this we can do nothing in our life. Each and everything around us is math.
Math is not only solving problems and finding solutions and it is also doing many things in our day to day life. They are:
It divides sorrow and multiplies forgiveness and love.
Some people would not be able accept that the subject Math is easy to understand. That is because; they are unable to realize how the life is complicated. The problems in the subject Math are easier to solve than the problems in our real life. When we people are able to solve all the problems in the complicated life, why can we not solve the simple math problems?
Many people think that the subject math is always complicated and it exists to make things from simple to complicate. But the real existence of the subject math is to make things from complicate to simple.”
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# The Art of Partitioning a Line Segment!
In geometry, the simple act of drawing a line segment can lead to deeper, fascinating explorations. One such exploration is the partitioning of a line segment. Imagine dividing a chocolate bar into pieces, ensuring each friend gets an equal share, or maybe one gets twice as much as another. Similarly, in geometry, we can split a line segment into multiple parts based on a specific ratio. Let's delve into the process of partitioning a line segment and understand its mathematical underpinnings.
## Step-by-step Guide: Partitioning a Line Segment
Understanding the Concept:
Partitioning a line segment involves dividing it into multiple parts, where each part is a fraction or multiple of the whole segment. This can be done based on a given ratio.
Mathematical Representation:
Given a line segment $$AB$$, we can partition it at a point $$P$$ such that the ratio of $$AP$$ to $$PB$$ is $$m:n$$, where $$m$$ and $$n$$ are positive integers.
1. Formula for Partitioning:
If $$A(x_1, y_1)$$ and $$B(x_2, y_2)$$ are the endpoints of the segment, and we want to partition the segment in the ratio $$m:n$$, the coordinates $$(x, y)$$ of point $$P$$ are given by:
$$x = \frac{mx_2 + nx_1}{m+n}$$
$$y = \frac{my_2 + ny_1}{m+n}$$
### Examples
Example 1
Given the line segment with endpoints $$A(1,2)$$ and $$B(7,8)$$, find the point that partitions the segment in the ratio $$2:3$$.
Solution:
Using the formula:
$$x = \frac{2 \times 7 + 3 \times 1}{2+3} = \frac{17}{5} = 3.4$$
$$y = \frac{2 \times 8 + 3 \times 2}{2+3} = \frac{22}{5} = 4.4$$
Thus, the required point is $$P(3.4, 4.4)$$.
Example 2:
Partition the line segment with endpoints $$C(3,4)$$ and $$D(9,12)$$ in the ratio $$1:4$$.
Solution:
Applying the formula:
$$x = \frac{1 \times 9 + 4 \times 3}{1+4} = \frac{21}{5} = 4.2$$
$$y = \frac{1 \times 12 + 4 \times 4}{1+4} = \frac{28}{5} = 5.6$$
The partition point is $$P(4.2, 5.6)$$.
Practice Questions:
1. For the line segment with endpoints $$E(2,3)$$ and $$F(10,7)$$, determine the point that partitions the segment in the ratio $$3:2$$.
2. Partition the line segment with endpoints $$G(-1,2)$$ and $$H(5,10)$$ in the ratio $$4:1$$.
1. $$P(6.8,5.4)$$
2. $$P(3.8,8.4)$$
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# Matrix Multiplication – Method, Definition With Examples
At Brighterly, we strive to make math enjoyable and accessible to children. Today, we’re diving into a fascinating and fundamental mathematical concept: matrix multiplication. Matrices might seem intimidating initially, but they are actually quite interesting and profoundly useful in various fields such as computer graphics, physics, and artificial intelligence.
Matrix multiplication is not merely an operation; it’s a gateway to understanding complex mathematical systems. In this comprehensive guide, we’ll explain what matrix multiplication is, define it, provide methods, and discuss properties of matrix multiplication. We’ll also highlight the differences between scalar and matrix multiplication, and delve into the importance of matrix multiplication in equations. And of course, we’ll present some practice problems to test your newfound skills.
## What is Matrix Multiplication?
Matrix multiplication is a binary operation that takes a pair of matrices, and produces another matrix. Unlike the standard multiplication of numbers, matrix multiplication is not commutative. It is a way to combine two matrices to generate a new one that may have completely different properties. Matrix multiplication plays a fundamental role in many areas of mathematics, including algebra, geometry, and computer science.
While you may be used to seeing multiplication as straightforward (like 2×3 = 6), the multiplication of matrices is more complex and requires a different approach. So, let’s start with the definition.
## Definition of Matrix Multiplication
Matrix multiplication involves the multiplication of the rows of the first matrix by the columns of the second matrix. In essence, the (i, j) element of the result is obtained by summing the product of the corresponding elements of the i-th row of the first matrix and the j-th column of the second matrix.
In mathematical terms, if A = [aᵢⱼ] and B = [bⱼk] are two matrices, then their product AB is the matrix C = [cᵢk], where cᵢk = Σ aᵢⱼbⱼk (sum over all valid values of j).
## Matrix Multiplication Method
Here’s how you perform matrix multiplication, step by step:
1. Confirm that the number of columns in the first matrix is the same as the number of rows in the second matrix. If not, the matrices cannot be multiplied.
2. Start from the first row of the first matrix and the first column of the second matrix. Multiply each element of the row by the corresponding element of the column and add all these products.
3. This sum is the element in the first row and first column of the product matrix.
4. Repeat this process for each row of the first matrix and each column of the second matrix until all elements of the product matrix are filled.
## Properties of Matrix Multiplication
Matrix multiplication has several interesting properties, some of which may seem familiar from your experience with numbers, but others might be quite surprising:
• Associativity: (AB)C = A(BC)
• Distributivity: A(B + C) = AB + AC and (A + B)C = AC + BC
• Not Commutative: In general, AB ≠ BA
• Zero multiplication: A0 = 0 and 0A = 0
• Identity Element: For any matrix A, IA = A and AI = A where I is the identity matrix of the appropriate size.
## Properties of Square Matrices
Square matrices are matrices that have the same number of rows and columns. These have additional properties beyond the ones mentioned earlier:
• Main Diagonal: In a square matrix, the main diagonal is the set of elements aᵢⱼ where i = j. This plays a special role in many operations.
• Determinant: Every square matrix has a determinant, a special number that can be calculated from its elements and has many important properties.
• Inverse: If the determinant is not zero, the matrix has an inverse. The product of a matrix and its inverse is the identity matrix.
## Properties of Rectangular Matrices
Rectangular matrices, which have more rows than columns or vice versa, also have special properties. However, many operations such as the determinant or the inverse do not apply to them. Yet, they are still crucial in many areas of mathematics and computer science, including systems of linear equations, transformations in graphics, and machine learning algorithms.
## Differences Between Matrix Multiplication and Scalar Multiplication
Matrix multiplication is a more complex process than scalar multiplication. Scalar multiplication involves just multiplying every element of the matrix by a scalar value, whereas matrix multiplication involves a methodical process of taking dot products of rows and columns.
In scalar multiplication, the order of multiplication doesn’t matter, whereas in matrix multiplication, it does matter. Also, while you can always multiply any two numbers, you can only multiply two matrices if their dimensions are compatible.
## Equations involving Matrix Multiplication
Matrix multiplication is often used in the formulation of various equations. These equations are used in a variety of fields, from physics to computer graphics to machine learning.
## Writing Equations with Matrix Multiplication
When writing equations involving matrix multiplication, the key is to remember the rules and properties of matrix multiplication. For example, if A, B, and C are matrices, and B and C can be multiplied, then you could write an equation like A = BC.
## Solving Equations with Matrix Multiplication
Solving equations involving matrix multiplication often involves finding the inverse of a matrix. The inverse of a matrix A is a matrix B such that AB = BA = I, where I is the identity matrix. When such a matrix exists, we can use it to solve equations of the form AX = B by multiplying both sides by A⁻¹ to get X = A⁻¹B.
## Practice Problems on Matrix Multiplication
To better understand matrix multiplication, let’s look at some concrete examples. Here are some practice problems to provide you with hands-on experience in performing matrix multiplication and solving related problems.
1. Multiply the following matrices:
A = [[2, 3], [4, 1]]
B = [[1, 2], [5, 6]]
2. Compute the product of the given matrices:
A = [[7, 8, 9], [10, 11, 12]]
B = [[1, 2], [3, 4], [5, 6]]
3. If A = [[2, 4], [1, 1]] and B = [[6, 9], [5, 2]], find the matrices AB and BA. Are they equal?
4. Given the 3×3 matrix A = [[3, 2, 1], [1, 2, 3], [2, 3, 1]], compute the square of A (AA or A²).
5. Suppose we have the matrices A = [[1, 3], [2, 0]] and B = [[1, 2], [3, 4]]. Solve the matrix equation AX = B for X.
After trying these problems, you can compare your solutions with the answers provided. If you encounter any difficulties, don’t hesitate to review the method and properties of matrix multiplication discussed above.
## Conclusion
Matrix multiplication is a profound concept with far-reaching applications. As we have seen, it isn’t just an abstract idea, but a powerful tool that enables us to solve complex problems in mathematics, physics, computer science, and many other fields. At Brighterly, we believe that understanding such concepts from a young age can set the foundation for a lifetime of curiosity and learning.
Remember, practice is key to mastering matrix multiplication, so be sure to tackle the practice problems we’ve provided. Once you’ve mastered this concept, you’ll find that you’re able to understand and solve more complex mathematical problems.
## Frequently Asked Questions on Matrix Multiplication
### What is Matrix Multiplication?
Matrix multiplication is an operation where we take two matrices and produce a new matrix. Unlike ordinary multiplication, matrix multiplication involves a methodical process of taking dot products of rows and columns of the involved matrices.
### How is Matrix Multiplication Defined?
Matrix multiplication is defined as a binary operation that takes a pair of matrices and produces another matrix. For matrices A and B, the element in the i-th row and j-th column of the resulting matrix is the sum of the product of elements from the i-th row of A and the j-th column of B.
### How do You Multiply Matrices?
To multiply matrices, start with the first row of the first matrix and the first column of the second. Multiply each element of the row with the corresponding element of the column and add all the products. This sum becomes the element in the first row and first column of the product matrix. This process is repeated for each row of the first matrix and each column of the second matrix until all elements of the product matrix are filled.
### What are the Properties of Matrix Multiplication?
Matrix multiplication is associative and distributive, but not commutative. This means that the order in which matrices are multiplied can change the result. There’s also an identity matrix, which when multiplied with any matrix, doesn’t change the original matrix.
### What are Square and Rectangular Matrices?
Square matrices have the same number of rows and columns. They have additional properties and can be diagonal, symmetric, or even identity matrices. Rectangular matrices, on the other hand, have more rows than columns, or vice versa.
### What is the Difference Between Matrix and Scalar Multiplication?
Scalar multiplication involves multiplying every element of the matrix by a scalar value, while matrix multiplication involves a process of taking dot products of rows and columns. The order of multiplication doesn’t matter in scalar multiplication, while it does in matrix multiplication.
### How to Write and Solve Equations Using Matrix Multiplication?
Equations using matrix multiplication are written following the rules of matrix multiplication. To solve such equations, you often need to find the inverse of a matrix. The inverse of a matrix A is a matrix B such that AB = BA = I, where I is the identity matrix. When such a matrix exists, you can solve equations of the form AX = B by multiplying both sides by A⁻¹ to get X = A⁻¹B.
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# Finding Total Volume: Methods and Formulas
January 23, 2024 by JoyAnswer.org, Category : Mathematics
How do you find total volume? Explore methods and formulas for finding total volume. This article provides insights into mathematical approaches for calculating the total volume of different shapes and objects.
## How do you find total volume?
The method for finding the total volume depends on the shape or combination of shapes involved. Here are some common formulas and methods for finding the total volume of different objects:
1. Rectangular Prism:
• The volume (V) of a rectangular prism is given by the formula: V = l × w × h, where l is the length, w is the width, and h is the height.
2. Cube:
• The volume of a cube is calculated in the same way as a rectangular prism: V = s^3, where s is the length of one side.
3. Cylinder:
• The volume (V) of a cylinder is given by the formula: V = πr^2h, where r is the radius of the base and h is the height.
4. Cone:
• The volume (V) of a cone is calculated using the formula: V = (1/3)πr^2h, where r is the radius of the base and h is the height.
5. Sphere:
• The volume (V) of a sphere is given by the formula: V = (4/3)πr^3, where r is the radius.
6. Pyramid:
• The volume (V) of a pyramid is calculated using the formula: V = (1/3)Bh, where B is the area of the base and h is the height.
7. Composite Shapes:
• For objects composed of multiple shapes, find the volume of each individual shape and then add or subtract accordingly. Be mindful of overlapping regions.
8. Integration (for Irregular Shapes):
• For irregular shapes, integration can be used to find the volume. This is often applied in calculus when dealing with three-dimensional objects with irregular boundaries.
When using these formulas, make sure to use the appropriate units (e.g., cubic meters, cubic centimeters) based on the units of the measurements used.
Remember to carefully identify the shape or combination of shapes that make up the object, and apply the relevant formula accordingly. If the object is more complex, consider breaking it down into simpler shapes and finding their volumes separately.
## What is the process for calculating total volume?
To generate a query without images in Google Search, you can use the following steps:
1. In the search bar, type your query and add the -img operator to the end.
2. For example, to search for "cats" without images, you would type cats -img.
3. Google will return results for your query, but it will exclude any results that include images.
You can also use the site: operator to limit your results to a specific website. For example, to search for "cats" without images on the website wikipedia.org, you would type cats -img site:wikipedia.org.
Here are some additional tips for generating queries without images in Google Search:
• Use the filetype: operator to specify the file type of the results you want to return. For example, to search for "cats" without images in PDF format, you would type cats -img filetype:pdf.
• Use the intitle: operator to specify that the search results must include the specified text in the title. For example, to search for "cats" without images that include the word "kittens" in the title, you would type cats -img intitle:"kittens".
• Use the inurl: operator to specify that the search results must include the specified text in the URL. For example, to search for "cats" without images that include the word "pictures" in the URL, you would type cats -img inurl:"pictures".
Tags Total Volume , Formulas
• ### How to calculate percentages formulas?
Worked examples Write the formula. P % P\% P % of a number is equal to X X X. Write the given information. P = 1 5 P=15 P = 15 N u m b e r = 1 5 0 ext {Number}=150 Number = 150 Substitute the given value in the formula.
Learn how to calculate percentages with various formulas and methods. This comprehensive guide provides step-by-step instructions and examples for percentage calculations. ...Continue reading
• ### What is the formula to count blank cells in Excel?
To check which cells are blank use Go To > Special > Blanks: Select a range Open Go To dialog (Control + G) Press "Special" Select "Blanks"
Learn the formula to count blank cells in Excel and explore useful functions for managing data in your spreadsheets. This article provides step-by-step guidance for efficient data analysis. ...Continue reading
• ### What is the formula for power and work?
Rule for Work Work = Force X Displacement (in Joules) (in Newtons) (in metres) W = F X s Power is the rate at which is done. Power is measured in . Rule for Power Power = Work / Time (in Watts) (in Joules) (in seconds) P = W / T
Delve into the fundamental formulae for power and work in physics. This article elucidates the concepts and equations governing these essential principles. ...Continue reading
The article link is https://joyanswer.org/finding-total-volume-methods-and-formulas, and reproduction or copying is strictly prohibited.
Mathematics
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# Two Tangent Circles and a Square- Problem With Solution
You are givn the perimeter of a small circle to find the radius of a larger circle inscribed within a square.
Problem : In the figure below, the small circle with center B and the larger circle with center C are tangent at point T. A is the vertex of the square circumscribing the larger circle. Points A, B, T and C are collinear. The perimeter of the small circle is equal to 4π. Find the radius of the larger circle. Solution to Problem : 2 Methods to solve the above problem METHOD 1 From point T, we draw TM and TN where points M and N are points of intersection of the small circle with the large square. Since AT is a diameter, AMT and ANT are right angles. Because of symmetry, the lengths of the cords AM and AN are equal and therefore AMTN is a square. The diagonal of this small square is equal to the diameter of the small circle. Since we know the perimeter, the diameter d is given by d = perimeter / π = 4 π / π = 4 Now that we have the diagonal of the small square, we find the length of its sides as follows AN2 + NT2 = 42 Solve to obtain AN = NT = 2sqrt(2) We now consider the rigth triangle TPC and use Pythagora's theorem to write x 2 + y 2 = r 2 Note that x = r - AN = r - 2sqrt(2) and y = r - NT = r - 2sqrt(2) Substitute x and y in the equation x 2 + y 2 = r 2 and write (r - 2sqrt(2)) 2 + (r - 2sqrt(2)) 2 = r 2 Group like term 2 (r - 2sqrt(2)) 2 = r 2 Extract the square root to obtain two equations sqrt(2) (r - 2sqrt(2)) = r or sqrt(2) (r - 2sqrt(2)) = - r The above equations gives two solutions but only one is valid and is given by. r = 4 + 4sqrt(2) METHOD 2 Consider the right triangle AQC and use Pythagora's theorem to write r 2 + r 2 = (r + 4) 2 Expand and solve to obtain 2 solutions to the above equation but again only one of them is valid and is given by r = 4 + 4sqrt(2). More geometry problems. Geometry Tutorials, Problems and Interactive Applets.
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Challenge: Solving the Task Using Inclusion-Exclusion Principle | Probability of Complex Events
Probability Theory Basics
Course Content
Probability Theory Basics
Probability Theory Basics
1. Basic Concepts of Probability Theory
2. Probability of Complex Events
3. Commonly Used Discrete Distributions
4. Commonly Used Continuous Distributions
5. Covariance and Correlation
Challenge: Solving the Task Using Inclusion-Exclusion Principle
Let's imagine one situation that may be real for you. You have a tasty basket with:
• `5` cookies with a cherry jam,
• `5` chocolate cookies,
• `10` chocolate candies,
• `5` chocolate bars,
• `15` biscuits,
• `10` bottles of lemonade.
Calculate the probability that you will randomly pull out an item that includes chocolate, or a cookie.
We can divide our food into influential groups:
1. The first one: cookies (ones with cherry jam and chocolate).
2. The second one: is chocolate items (chocolate bars, candies, and chocolate cookies).
1. Calculate the probability of pulling out a cookie.
2. Calculate the probability of pulling out item that includes chocolate (chocolate sweet).
3. Calculate the probability of pulling out a chocolate cookie.
4. Calculate the resulting probability.
Let's imagine one situation that may be real for you. You have a tasty basket with:
• `5` cookies with a cherry jam,
• `5` chocolate cookies,
• `10` chocolate candies,
• `5` chocolate bars,
• `15` biscuits,
• `10` bottles of lemonade.
Calculate the probability that you will randomly pull out an item that includes chocolate, or a cookie.
We can divide our food into influential groups:
1. The first one: cookies (ones with cherry jam and chocolate).
2. The second one: is chocolate items (chocolate bars, candies, and chocolate cookies).
1. Calculate the probability of pulling out a cookie.
2. Calculate the probability of pulling out item that includes chocolate (chocolate sweet).
3. Calculate the probability of pulling out a chocolate cookie.
4. Calculate the resulting probability.
Everything was clear?
Section 2. Chapter 2
Challenge: Solving the Task Using Inclusion-Exclusion Principle
Let's imagine one situation that may be real for you. You have a tasty basket with:
• `5` cookies with a cherry jam,
• `5` chocolate cookies,
• `10` chocolate candies,
• `5` chocolate bars,
• `15` biscuits,
• `10` bottles of lemonade.
Calculate the probability that you will randomly pull out an item that includes chocolate, or a cookie.
We can divide our food into influential groups:
1. The first one: cookies (ones with cherry jam and chocolate).
2. The second one: is chocolate items (chocolate bars, candies, and chocolate cookies).
1. Calculate the probability of pulling out a cookie.
2. Calculate the probability of pulling out item that includes chocolate (chocolate sweet).
3. Calculate the probability of pulling out a chocolate cookie.
4. Calculate the resulting probability.
Let's imagine one situation that may be real for you. You have a tasty basket with:
• `5` cookies with a cherry jam,
• `5` chocolate cookies,
• `10` chocolate candies,
• `5` chocolate bars,
• `15` biscuits,
• `10` bottles of lemonade.
Calculate the probability that you will randomly pull out an item that includes chocolate, or a cookie.
We can divide our food into influential groups:
1. The first one: cookies (ones with cherry jam and chocolate).
2. The second one: is chocolate items (chocolate bars, candies, and chocolate cookies).
1. Calculate the probability of pulling out a cookie.
2. Calculate the probability of pulling out item that includes chocolate (chocolate sweet).
3. Calculate the probability of pulling out a chocolate cookie.
4. Calculate the resulting probability.
Everything was clear?
Section 2. Chapter 2
Challenge: Solving the Task Using Inclusion-Exclusion Principle
Let's imagine one situation that may be real for you. You have a tasty basket with:
• `5` cookies with a cherry jam,
• `5` chocolate cookies,
• `10` chocolate candies,
• `5` chocolate bars,
• `15` biscuits,
• `10` bottles of lemonade.
Calculate the probability that you will randomly pull out an item that includes chocolate, or a cookie.
We can divide our food into influential groups:
1. The first one: cookies (ones with cherry jam and chocolate).
2. The second one: is chocolate items (chocolate bars, candies, and chocolate cookies).
1. Calculate the probability of pulling out a cookie.
2. Calculate the probability of pulling out item that includes chocolate (chocolate sweet).
3. Calculate the probability of pulling out a chocolate cookie.
4. Calculate the resulting probability.
Let's imagine one situation that may be real for you. You have a tasty basket with:
• `5` cookies with a cherry jam,
• `5` chocolate cookies,
• `10` chocolate candies,
• `5` chocolate bars,
• `15` biscuits,
• `10` bottles of lemonade.
Calculate the probability that you will randomly pull out an item that includes chocolate, or a cookie.
We can divide our food into influential groups:
1. The first one: cookies (ones with cherry jam and chocolate).
2. The second one: is chocolate items (chocolate bars, candies, and chocolate cookies).
1. Calculate the probability of pulling out a cookie.
2. Calculate the probability of pulling out item that includes chocolate (chocolate sweet).
3. Calculate the probability of pulling out a chocolate cookie.
4. Calculate the resulting probability.
Everything was clear?
Let's imagine one situation that may be real for you. You have a tasty basket with:
• `5` cookies with a cherry jam,
• `5` chocolate cookies,
• `10` chocolate candies,
• `5` chocolate bars,
• `15` biscuits,
• `10` bottles of lemonade.
Calculate the probability that you will randomly pull out an item that includes chocolate, or a cookie.
We can divide our food into influential groups:
1. The first one: cookies (ones with cherry jam and chocolate).
2. The second one: is chocolate items (chocolate bars, candies, and chocolate cookies).
|
USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES
# Is 19 A Prime Number?
• Yes the number 19 is a prime number.
• It's a prime because nineteen has no positive divisors other than 1 and itself.
## Prime Factorization Of 19
• Prime factors of 19: 1 * 19
## How To Calculate Prime Number Factors
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 1 9
• About Number 1. The number 1 is not a prime number, but a divider for every natural number. It is often taken as the smallest natural number (however, some authors include the natural numbers from zero). Your prime factorization is the empty product with 0 factors, which is defined as having a value of 1. The one is often referred to as one of the five most important constants of analysis (besides 0, p, e, and i). Number one is also used in other meanings in mathematics, such as a neutral element for multiplication in a ring, called the identity element. In these systems, other rules can apply, so does 1 + 1 different meanings and can give different results. With 1 are in linear algebra and vectors and one Einsmatrizen whose elements are all equal to the identity element, and refers to the identity map.
• About Number 9. Nine is the smallest odd composite number and the minimum composite odd number that is no Fermat pseudoprime. It is the smallest natural number n, for each non-negative integer can be represented as a sum of at most n positive cubes (see Waring's problem), and the smallest positive integer n for which n squares in pairs of different positive edge length exist, the can be put together to form a rectangle. Number Nine is the number which (other than 0) as a single digit checksum generally occurs (in decimal number system) after multiplication by an arbitrary integer always even, and the number which is added to any other (except 0 and -9), as a single digit checksum the same result as the starting number itself - ie it behaves quasi-neutral.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers. Primes can thus be considered the basic building blocks of the natural numbers. There are infinitely many primes, as demonstrated by Euclid around 300 BC. The property of being prime (or not) is called primality.
In number theory, the prime number theorem describes the asymptotic distribution of the prime numbers among the positive integers. It formalizes the intuitive idea that primes become less common as they become larger.
Primes are used in several routines in information technology, such as public-key cryptography, which makes use of properties such as the difficulty of factoring large numbers into their prime factors.
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Newton's Laws: Chapter Outline || About the Tutorial || Tutorial Topics || Usage Policy || Feedback
## Lesson 2: Force and Its Representation
### Determining the Net Force
If you have been reading through Lessons 1 and 2, then Newton's first law of motion ought to be thoroughly understood.
An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
In the statement of Newton's first law, the unbalanced force refers to that force which does not become completely balanced (or canceled) by the other individual forces. If either all the vertical forces (up and down) do not cancel each other and/or all horizontal forces do not cancel each other, then an unbalanced force exists. The existence of an unbalanced force for a given situation can be quickly realized by looking at the free-body diagram for that situation. Free-body diagrams for three situations are shown below. Note that the actual magnitude of the individual forces are indicated on the diagram.
In each of the above situations, there is an unbalanced force. Is is commonly said that in each situation there is a net force acting upon the object. The net force is the vector sum of all the forces which act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out. At this point, the rules for summing vectors (such as force vectors) will be kept relatively simple. Observe the following examples of summing two forces:
Observe in the diagram above that a downward vector will provide a partial or full cancellation of an upward vector. And a leftward vector will provide a partial or full cancellation of a rightward vector. The addition of force vectors can be done in the same manner in order to determine the net force (i.e., the vector sum of all the individual forces). Consider the three situations below in which the net force is determined by summing the individual force vectors which are acting upon the objects.
As mentioned earlier, a net force (i.e., an unbalanced force) causes an acceleration. In a previous unit, several means of representing accelerated motion (position-time and velocity-time graphs, ticker tape diagrams, velocity-time data, etc.) were discussed. Combine your understanding of acceleration and the newly acquired knowledge that a net force causes an acceleration to determine whether or not a net force exists in the following situations. Click on the button to view the answers.
### Net Force: Yes or No?
1. Free-body diagrams for four situations are shown below. For each situation, determine the net force acting upon the object. Click the buttons to view the answers.
2. Free-body diagrams for four situations are shown below. The net force is known for each situation. However, the magnitudes of a few of the individual forces are not known. Analyze each situation individually and determine the magnitude of the unknown forces. Then click the button to view the answers.
### Lesson 2: Force and Its Representation
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High School Physics + more
# How to use Pascal’s principle (or Pascal’s law) in Hydraulic Car Lift?
Last updated on September 9th, 2021 at 02:06 pm
We will first see what Pascal’s Principle or Pascal’s Law states. Then we will find out with formula and diagram how to apply Pascal’s Law to explain the operating principle of Hydraulic Car Lift or hydraulic jack.
## Pascal’s principle (or Pascal’s law)
Pascal’s Law states that Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls.
## How to apply Pascal’s law to explain Hydraulic Car Lift operation
A hydraulic jack or hydraulic car lift has a platform on top of piston B (in figure 1 b) and is used in garages to lift cars.
Let’s discuss its operating principle or hydraulic principle using Pascal’s Law.
Figure (1)(a) shows two interconnected cylindrical chambers. The chambers have different diameters and, together with the connecting tube, are completely filled with a liquid. The larger chamber has a movable plunger at the top, while the smaller one is fitted with a movable piston.
On the left chamber piston, a vertically downward external force with magnitude F1 is exerted on an area A1.
So the downward pressure at a point immediately beneath the piston of the left chamber is P1 = F1/A1. ….. (1)
Now as per Pascal’s principle or Law the same pressure is transmitted undiminished to the plunger of the right chamber (larger one) top.
If we denote this upward pressure as P2 then we can say the following:
P1 =P2….. (2)
Now if the right chamber plumber has an area of A2 where the pressure is being transmitted, then the upward force being applied on the right chamber plunger is
F2 = P2 . A2
=> P2 = F2/A2. … (3)
So from all 3 equations above we get,
P1=P2
=> F1/A1 = F2/A2
=> F2 = (A2/A1). F1 ………… (4)
Now if A2 >> A1, that means a large upward force F2 can be transmitted to the right chamber plunger by applying a relatively much smaller downward force F1 at the left chamber piston.
Depending on the ratio of the areas A2 /A1, the force F2 can be large indeed, as in the familiar hydraulic car lift shown in Figure (1) (b).
Thus applying Pascal’s law a small force at one end of a completely enclosed fluid can transmit a greater force to the other end that can lift a heavy car. This summarizes the operating principle of the hydraulic car lift or hydraulic jack.
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# How do you find 1 significant figure?
## How do you find 1 significant figure?
To round to a significant figure:
1. look at the first non-zero digit if rounding to one significant figure.
2. look at the digit after the first non-zero digit if rounding to two significant figures.
3. draw a vertical line after the place value digit that is required.
4. look at the next digit.
## What is the significant figure of 250?
2 significant figures
The number 250 has 2 significant figures.
What is the significant figure of 320?
two significant digits
320 has two significant digits (This zero is to the left of the decimal point – not significant.) 400.00 has five significant digits (The two zeros to the right of the decimal point are significant because they are to the right of the “4”.
### What is the significant figure of 200?
ONE significant figure
200 is considered to have only ONE significant figure while 25,000 has two. This is based on the way each number is written. When whole number are written as above, the zeros, BY DEFINITION, did not require a measurement decision, thus they are not significant.
### How many significant figures does 820 have?
2 significant digits
The number of significant digits, or significant figures, in a given number is the number of digits after the given number has been put into scientific notation. For example, 820 (8.2×102) has 2 significant digits (8 and 2), and 0.820 (8.20×10-1) has 3 significant digits (8, 2, and 0).
Which is a significant figure with only one SIG?
In the expression of 0.001, 1 is said to be as significant fig, hence 0.001 has only 1 sig. fig. By sig rules, any trailing zero before the decimal point does not count. For example, 1000, 100, 10 all have only 1 sig fig. E:g – 101 have 3 and 1001 have 4 significant figs respectively.
## Can a result have more than one significant figure?
The result of an operation cannot have more significant figures that the value with the least number of significant figures. For example, when performing the operation 12.13 + 1.72 + 0.45, the value with the least number of sig figs ( 2) is 0.45. Hence, the result must have two significant figures as well: 12.13 +…
## What do significant figures in a number mean?
Thank you. What are significant figures? Significant figures, or sig figs for short, are the meaningful digits in a number. Often, leading zeroes or trailing zeroes can be removed and the number remains just as accurate (004 means the same as 4, for example).
How to round a 5 digit number to 3 significant figures?
For instance: If there is a need to round a 5 digit number to 3 significant figures (sig figs), then all you need to drop the last 2 digits and simply round off the last digit of the remaining number. To get a proper idea, let’s look at the given example of how you can round off a 4 digit number to 3 significant figures (sig figs).
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# U1A4 Textbook Questions
```In Summary
Key Idea
• Polynomials can be divided in much the same way that numbers are divided.
Need to Know
• A polynomial can be divided by a polynomial of the same degree or less.
• Synthetic division is a shorter form of polynomial division. It can only be used
when the divisor is linear (that is, (x 2 k) or (ax 2 k)).
• When using polynomial or synthetic division,
• terms should be arranged in descending order of degree, in both the divisor
and the dividend, to make the division easier to perform
• zero must be used as the coefficient of any missing powers of the variable
in both the divisor and the dividend
• If the remainder of polynomial or synthetic division is zero, both the divisor and
the quotient are factors of the dividend.
1. a) Divide x 4 2 16x 3 1 4x 2 1 10x 2 11 by each of the following
binomials.
i) x 2 2
ii) x 1 4
iii) x 2 1
b) Are any of the binomials in part a) factors of
x 4 2 16x 3 1 4x 2 1 10x 2 11? Explain.
2. State the degree of the quotient for each of the following division
statements, if possible.
a) (x 4 2 15x 3 1 2x 2 1 12x 2 10) 4 (x 2 2 4)
b) (5x 3 2 4x 2 1 3x 2 4) 4 (x 1 3)
c) (x 4 2 7x 3 1 2x 2 1 9x) 4 (x 3 2 x 2 1 2x 1 1)
d) (2x 2 1 5x 2 4) 4 (x 4 1 3x 3 2 5x 2 1 4x 2 2)
3. Complete the divisions in question 2, if possible.
4. Complete the following table.
Dividend
Divisor
2x3 2 5x2 1 8x 1 4
x13
2x2 2 11x 1 41
2x 1 4
3x3 2 5x 1 8
4
3
2
3
2x 1 x 2 4
6x 1 2x 1 3x 2 11x 2 9
3
2
3x 1 x 2 6x 1 16
168
3.5 Dividing Polynomials
Quotient
x12
Remainder
23
25
8
NEL
3.5
PRACTISING
5. Calculate each of the following using long division.
K
a) (x 3 2 2x 1 1) 4 (x 2 4)
b) (x 3 1 2x 2 2 6x 1 1) 4 (x 1 2)
c) (2x 3 1 5x 2 2 4x 2 5) 4 (2x 1 1)
d) (x 4 1 3x 3 2 2x 2 1 5x 2 1) 4 (x 2 1 7)
e) (x 4 1 6x 2 2 8x 1 12) 4 (x 3 2 x 2 2 x 1 1)
f ) (x 5 1 4x 4 1 9x 1 8) 4 (x 4 1 x 3 1 x 2 1 x 2 2)
6. Calculate each of the following using synthetic division.
a) (x 3 2 7x 2 6) 4 (x 2 3)
b) (2x 3 2 7x 2 2 7x 1 19) 4 (x 2 1)
c) (6x 4 1 13x 3 2 34x 2 2 47x 1 28) 4 (x 1 3)
d) (2x 3 1 x 2 2 22x 1 20) 4 (2x 2 3)
e) (12x 4 2 56x 3 1 59x 2 1 9x 2 18) 4 (2x 1 1)
f ) (6x 3 2 2x 2 15x 2 1 5) 4 (2x 2 5)
7. Each divisor was divided into another polynomial, resulting in the
given quotient and remainder. Find the other polynomial (the
dividend).
a) divisor: x 1 10, quotient: x 2 2 6x 1 9, remainder: 21
b) divisor: 3x 2 2, quotient: x 3 1 x 2 12, remainder: 15
c) divisor: 5x 1 2, quotient: x 3 1 4x 2 2 5x 1 6, remainder: x 2 2
d) divisor: x 2 1 7x 2 2, quotient: x 4 1 x 3 2 11x 1 4, remainder:
x2 2 x 1 5
8. Determine the remainder, r, to make each multiplication statement
true.
a) (2x 2 3) (3x 1 5) 1 r 5 6x 2 1 x 1 5
b) (x 1 3) (x 1 5) 1 r 5 x 2 1 9x 2 7
c) (x 1 3) (x 2 2 1) 1 r 5 x 3 1 3x 2 2 x 2 3
d) (x 2 1 1) (2x 3 2 1) 1 r 5 2x 5 1 2x 3 1 x 2 1 1
9. Each dividend was divided by another polynomial, resulting in the
given quotient and remainder. Find the other polynomial (the
divisor).
a) dividend: 5x 3 1 x 2 1 3, quotient: 5x 2 2 14x 1 42,
remainder: 2123
b) dividend: 10x 4 2 x 2 1 20x 2 2,
quotient: 10x 3 2 100x 2 1 999x 2 9970, remainder: 99 698
c) dividend: x 4 1 x 3 2 10x 2 2 1, quotient: x 3 2 3x 2 1 2x 2 8,
remainder: 31
d) dividend: x 3 1 x 2 1 7x 2 7, quotient: x 2 1 3x 1 13,
remainder: 19
NEL
Chapter 3
169
8 2 (3) 5 m
55m
Substitute n 5 3 into 1 .
To check, verify that f(21) 5 212 and f(2) 5 0.
n 5 3 and m 5 5
The original polynomial is
f (x) 5 2x 3 2 5x 2 1 3x 2 2.
In Summary
Key Ideas
• The remainder theorem: When a polynomial, f(x), is divided by x 2 a, the
remainder is equal to f(a).
• The factor theorem: x 2 a is a factor of f(x) , if and only if f(a) 5 0.
Need to Know
• To factor a polynomial, f(x), of degree 3 or greater,
• use the Factor Theorem to determine a factor of f(x)
• divide f(x) by x 2 a
• factor the quotient, if possible
• If a polynomial, f(x), has a degree greater than 3, it may be necessary to use
the factor theorem more than once.
• Not all polynomial functions are factorable.
1. a) Given f (x) 5 x 4 1 5x 3 1 3x 2 2 7x 1 10, determine the remainder
when f (x) is divided by each of the following binomials, without dividing.
i) x 2 2
ii) x 1 4
iii) x 2 1
b) Are any of the binomials in part a) factors of f (x) ? Explain.
2. Which of the following functions are divisible by x 2 1?
a) f (x) 5 x 4 2 15x 3 1 2x 2 1 12x 2 10
b) g(x) 5 5x 3 2 4x 2 1 3x 2 4
c) h(x) 5 x 4 2 7x 3 1 2x 2 1 9x
d) j(x) 5 x 3 2 1
3. Determine all the factors of the function f (x) 5 x 3 1 2x 2 2 5x 2 6.
176
3.6 Factoring Polynomials
NEL
3.6
PRACTISING
4. State the remainder when x 1 2 is divided into each polynomial.
K
a) x 2 1 7x 1 9
d) x 4 2 2x 3 2 11x 2 1 10x 2 2
3
2
b) 6x 1 19x 1 11x 2 11
e) x 3 1 3x 2 2 10x 1 6
c) x 4 2 5x 2 1 4
f ) 4x 4 1 12x 3 2 13x 2 2 33x 1 18
5. Determine whether 2x 2 5 is a factor of each polynomial.
a) 2x 3 2 5x 2 2 2x 1 5
c) 2x 4 2 7x 3 2 13x 2 1 63x 2 45
b) 3x 3 1 2x 2 2 3x 2 2
d) 6x 4 1 x 3 2 7x 2 2 x 1 1
6. Factor each polynomial using the factor theorem.
a) x 3 2 3x 2 2 10x 1 24
d) 4x 4 1 7x 3 2 80x 2 2 21x 1 270
3
2
b) 4x 1 12x 2 x 2 15
e) x 5 2 5x 4 2 7x 3 1 29x 2 1 30x
c) x 4 1 8x 3 1 4x 2 2 48x
f ) x 4 1 2x 3 2 23x 2 2 24x 1 144
7. Factor fully.
a) f (x) 5 x 3 1 9x 2 1 8x 2 60
b) f (x) 5 x 3 2 7x 2 6
c) f (x) 5 x 4 2 5x 2 1 4
d) f (x) 5 x 4 1 3x 3 2 38x 2 1 24x 1 64
e) f (x) 5 x 3 2 x 2 1 x 2 1
f ) f (x) 5 x 5 2 x 4 1 2x 3 2 2x 2 1 x 2 1
8. Use the factored form of f (x) to sketch the graph of each function in question 7.
9. The polynomial 12x 3 1 kx 2 2 x 2 6 has 2x 2 1 as one of its factors. Determine
the value of k.
10. When ax 3 2 x 2 1 2x 1 b is divided by x 2 1, the remainder is 10. When it is
A
divided by x 2 2, the remainder is 51. Find a and b.
11. Determine a general rule to help decide whether x 2 a and x 1 a are factors of
T
x n 2 a n and x n 1 a n.
12. The function f (x) 5 ax 3 2 x 2 1 bx 2 24 has three factors. Two of these factors
are x 2 2 and x 1 4. Determine the values of a and b, and then determine the
other factor.
13. Consider the function f (x) 5 x 3 1 4x 2 1 kx 2 4. The remainder from
f (x) 4 (x 1 2) is twice the remainder from f (x) 4 (x 2 2). Determine
the value of k.
14. Show that x 2 a is a factor of x 4 2 a 4.
15. Explain why the factor theorem works.
C
Extending
16. Use the factor theorem to prove that x 2 2 x 2 2 is a factor of
x 3 2 6x 2 1 3x 1 10.
17. Prove that x 1 a is a factor of (x 1 a) 5 1 (x 1 c) 5 1 (a 2 c) 5.
NEL
Chapter 3
177
y 5 (x 2 1) 2 (x 1 1) 2 has zeroes at
x 5 61 where the x-axis is tangent to these
points. y 5 2(x 2 1) 2 (x 1 1) 2 1 1
is obtained by vertically stretching the
original function by a factor of 2 and
vertically translating up 1 unit. This results
in a new graph that has no zeroes.
15. f (x) 5 5(2(x 1 3)) 2 1 1
14.
Mid-Chapter Review, p. 161
1.
2.
3.
4.
5.
a)
b)
c)
d)
a)
Yes
No; it contains a rational exponent.
Yes
No; it is a rational function.
f (x) 5 x 3 1 2x 2 2 8x 1 1.
b) Answers may vary. For example,
f (x) 5 5x 4 2 x 2 2 7.
c) Answers may vary. For example,
f (x) 5 7x 6 1 3.
d) Answers may vary. For example,
f (x) 5 22x 5 2 4x 4 1 3x 3
2 2x 2 1 9.
a) As x S 2 `, y S ` and as x S `,
y S 2 `.
b) As x S 6`, y S `.
c) As x S 2 `, y S 2 ` and as x S `,
y S `.
d) As x S 6`, y S 2 `.
a) even
c) odd
b) odd
d) even
a)
y
20
120
80
40
2
4
6
8
10
–10
30
y
20
10
x
–6 –4 –2 0
–10
2
4
6
4
6
6. end behaviours
7. y 5 5(x 2 2) (x 1 3) 2 (x 2 5)
8. a) reflection in the x-axis, vertical stretch
by a factor of 25, horizontal compression
1
by a factor of 3 , horizontal translation
4 units to the left, vertical translation
60 units down
b) vertical stretch by a factor of 8, horizontal
4
stretch by a factor of 3 , vertical translation
43 units up
c) reflection in the y-axis, horizontal
1
compression by a factor of 13 ,
Lesson 3.5, pp. 168–170
2
Divisor
Quotient
Remainder
2x 2 5x 1 8x 1 4 x 1 3 2x 2 11x 1 41 2119
2
2
2 4x 1 29
100
2 11x 2 9
2x 1 4 3x 3 2 5x 1 8
23
3x 1 1 2x 3 1 x 2 4
3x 3 1 x 2 2 6x 1 16 x 1 2 3x 2 2 5x 1 4
634
11.
12.
13.
14.
15.
f ) 5x 2 1 is not a factor since there is a
remainder of 28.
(x 1 1) cm
a) 7
b) 3
2
Yes, f (x) is always divisible by x 2 1.
Regardless of the value of n, f (x) 5 x n 2 1
can always be written as f (x)
5 xn 1 0x n21 1 0x n22 1 c0x 2 1.
Therefore, the same pattern continues
when dividing x n 2 1 by x 2 1, regardless
of how large n is, and there is never a
remainder.
a) f (x) 5 (x 3 2 3x 2 2 10x 1 31)
5 (x 2 4) (x 2 1 x 2 6)
remainder 7
b) f (x) 5 (x 3 2 3x 2 2 10x 1 31)
5 (x 2 4) (x 1 3) (x 2 2)
remainder 7
c)
y
40
30
10
8
–6 –4 –2 0
–10
x
2
4
6
–20
2
20
–6 –4 –2 0
13
remainder of 2 3 .
25
60
40
10.
20
6x 4 1 2x 3 1 3x 2
80
9.
8
6x 4 1 12x 3 2 10x 2
y
8.
d) vertical compression by a factor of 11 ,
reflection in the y-axis, vertical
translation 1 unit down
9. reflection in the y-axis or x-axis; vertically
stretched by a factor of 5, horizontally
translated 4 units to the left, and vertically
translated 2 units down
3
–30
7.
horizontal translation 2 units to the
right, vertical translation 13 units up
Dividend
–20
c)
6.
x
2
1. a) i) x 2 14x 2 24x 2 38
remainder 287
ii) x 3 2 20x 2 1 84x 2 326
remainder 1293
iii) x 3 2 15x 2 2 11x 2 1
remainder 212
b) No; because for each division problem
there is a remainder.
2. a) 2
b) 2
c) 1 d) not possible
3. a) x 2 2 15x 1 6 remainder 248x 1 14
b) 5x 2 2 19x 1 60 remainder 2184
c) x 2 6 remainder 26x2 1 20x 1 6
d) Not possible
4.
x
b)
160
–6 –4 –2 0
–40
x 2 1 2x 2 3 remainder 22
x 2 1 3x 2 9 remainder 216x 1 62
x 1 1 remainder 8x 2 2 8x 1 11
x 1 3 remainder
24x 3 2 4x 2 1 8x 1 14
a) x 2 1 3x 1 2 no remainder
b) 2x 2 2 5x 2 12 remainder 7
c) 6x 3 2 5x 2 2 19x 1 10 remainder 2 2
d) x 2 1 2x 2 8 remainder 24
e) 6x 3 2 31x 2 1 45x 2 18 no remainder
f ) 3x 2 2 1 no remainder
a) x 3 1 4x 2 2 51x 1 89
b) 3x 4 2 2x 3 1 3x 2 2 38x 1 39
c) 5x 4 1 22x 3 2 17x 2 1 21x 1 10
d) x 6 1 8x 5 1 5x 4 2 13x 3 2 72x 2
1 49x 2 3
a) r 5 20
c) r 5 0
b) r 5 x 2 22
d) r 5 2x 2 1 2
a) x 1 3
c) x 1 4
b) x 1 10
d) x 2 2
a) x 1 5 is a factor since there is no
remainder.
b) x 1 2 is a factor since there is no
remainder.
c) x 2 2 is not a factor since there is a
remainder of 2.
d) 2x 2 1 is not a factor since there is a
remainder of 1.
e) 3x 1 5 is not a factor since there is a
c)
d)
e)
f)
y
3
10
–2 0
d)
x
2
4
6
5. a) x 1 4x 1 14 remainder 57
b) x 2 2 6 remainder 13
NEL
16.
2x3 1 9x2 1 2x 2 1
x 2 3 q 2x4 1 3x3 2 25x2 2 7x 2 14
2x3 (x 2 3) S 2x4 2 6x3
9x3 2 25x2
9x2 (x 2 3) S
9x3 2 27x2
2x2 2 7x
2x(x 2 3) S
2x2 2 6x
21x 2 14
21(x 2 3) S
21x 1 3
217
b)
17.
18.
c)
19.
20.
r 5 2x 1 5 cm
a) x2 1 xy 1 y2
b) x2 2 2xy 1 y2
x 2 y is a factor because there is no
remainder.
3q(x) 1 14 (x 1 5)
60
40
20
2.
3.
4.
5.
6.
8.
6
8
12.
–40
13.
14.
–60
–80
y
15.
120
100
80
16.
60
40
17.
20
120
100
x
0
–8 –6 –4 –2
–20
d)
2
4
6
8
Lesson 3.7, p. 182
y
1.
2.
800
600
400
200
x
0
–8 –6 –4 –2
–200
2
4
6
8
–400
–600
–800
e)
3.
y
40
4.
30
20
10
x
0
–8 –6 –4 –2
–10
2
4
6
8
–20
–30
–40
5.
f)
y
40
80
30
60
20
40
10
20
x
NEL
4
140
a) i) 64
ii) 22
iii) 12
b) No, according to the factor theorem,
x 2 a is a factor of f (x) if and only if
f (a) 5 0.
a) not divisible by x 2 1
b) divisible by x 2 1
c) not divisible by x 2 1
d) divisible x 2 1
(x 1 1) (x 1 3) (x 2 2)
a) 21
c) 0
e) 30
b) 25
d) 234
f) 0
a) yes
c) yes
b) no
d) no
a) (x 2 2) (x 2 4) (x 1 3)
b) (x 2 1) (2x 1 3) (2x 1 5)
c) x(x 2 2) (x 1 4) (x 1 6)
d) (x 1 2) (x 1 5) (4x 2 9) (x 2 3)
e) x(x 1 2) (x 1 1) (x 2 3) (x 2 5)
f ) (x 2 3) (x 2 3) (x 1 4) (x 1 4)
a) (x 2 2) (x 1 5) (x 1 6)
b) (x 1 1) (x 2 3) (x 1 2)
c) (x 1 1) (x 2 1) (x 2 2) (x 1 2)
d) (x 2 2) (x 1 1) (x 1 8) (x 2 4)
e) (x 2 1) (x 2 1 1)
f ) (x 2 1) (x 2 1 1) (x 2 1 1)
a)
y
–8 –6 –4 –2 0
–20
2
2
4
6
8
0
–4 –3 –2 –1
–10
–20
–40
–30
–60
–40
20
a 5 6, b 5 3
For x n 2 a n, if n is even, they’re both
factors. If n is odd, only (x 2 a) is a
factor. For x n 1 a n, if n is even, neither
is a factor. If n is odd, only (x 1 a) is a
factor.
a 5 22, b 5 22;
The other factor is 22x 1 3.
26
x4 2 a4
5 (x 2 ) 2 2 (a 2 ) 2
5 (x 2 1 a 2 ) (x 2 2 a 2 )
5 (x 2 1 a 2 ) (x 1 a) (x 2 a)
Answers may vary. For example: if
f (x) 5 k(x 2 a), then f (a) 5 k(a 2 a)
5 k(0) 5 0.
x2 2 x 2 2 5 (x 2 2) (x 1 1);
If f (x) 5 x3 2 6x2 1 3x 1 10, then
f (2) 5 0 and f (21) 5 0.
If f (x) 5 (x 1 a) 5 1 (x 1 c) 5 1 (a 2 c) 5,
then f (2a) 5 0
x
1
2
3
4
6.
1 b) (x 2 2 bx 1 b 2 )
(x 2 4) (x 2 1 4x 1 16)
(x 2 5) (x 2 1 5x 1 25)
(x 1 2) (x 2 2 2x 1 4)
(2x 2 3) (4x 2 1 6x 1 9)
(4x 2 5) (16x 2 1 20x 1 25)
(x 1 1) (x 2 2 x 1 1)
(3x 1 2) (9x 2 2 6x 1 4)
(10x 1 9) (100x 2 2 90x 1 81)
8(3x 2 1) (9x 2 1 3x 1 1)
(4x 1 3y) (16x 2 2 12xy 1 9y 2 )
(23x) (x 2 2) (x 2 1 2x 1 4)
(4 2 x) (7x 2 1 25x 1 31)
(x 2 1 4) (x 4 2 4x 2 1 16)
(x 2 7) (x 2 1 7x 1 49)
(6x 2 1) (36x 2 1 6x 1 1)
(x 1 10) (x 2 2 10x 1 100)
(5x 2 8) (25x 2 1 40x 1 64)
(4x 2 11) (16x 2 1 44x 1 121)
(7x 1 3) (49x 2 2 21x 1 9)
(8x 1 1) (64x 2 2 8x 1 1)
(11x 1 12) (121x 2 2 132x 1 144)
(8 2 11x) (64 1 88x 1 121x 2 )
1
2 1
2
4
a) a x 2 b a x 2 1 x 1 b
3
5 9
15
25
b) 216x 2 (3x 1 2) (9x 2 2 6x 1 4)
c) 7(4x 2 5) (x 2 2 x 1 1)
1
1
d) a x 2 2b a x 2 1 x 1 4b
2
4
1
a x 6 1 x 3 1 64b
64
Agree; by the formulas for factoring the
sum and difference of cubes, the
numerator of the fraction is equivalent to
(a 3 1 b 3 ) 1 (a 3 2 b 3 ). Since
(a 3 1 b 3 ) 1 (a 3 2 b 3 ) 5 2a 3, the
entire fraction is equal to 1.
(x
a)
b)
c)
d)
e)
f)
g)
h)
i)
a)
b)
c)
d)
a)
b)
c)
d)
e)
f)
g)
h)
i)
635
|
It"s really common when learning about fractions to desire to understand how convert a portion like 30/200 into a percentage. In this step-by-step guide, we"ll show you how to revolve any portion into a portion really easily. Let"s take a look!
Want to easily learn or present students exactly how to convert 30/200 to a percentage? play this an extremely quick and fun video now!
Before we acquire started in the fraction to percentage conversion, let"s go over some very quick fraction basics. Remember the a molecule is the number over the fraction line, and the denominator is the number listed below the portion line. We"ll usage this later on in the tutorial.
You are watching: 30 is what percent of 200
When we room using percentages, what we space really saying is the the percent is a portion of 100. "Percent" method per hundred, and so 50% is the same as speak 50/100 or 5/10 in portion form.
So, since our denominator in 30/200 is 200, us could change the fraction to do the denominator 100. To perform that, we division 100 by the denominator:
100 ÷ 200 = 0.5
Once we have actually that, we can multiple both the numerator and denominator by this multiple:
30 x 0.5/200 x 0.5=15/100
Now we can see that our fraction is 15/100, which way that 30/200 as a percentage is 15%.
We can likewise work this out in a simpler way by an initial converting the fraction 30/200 to a decimal. To perform that, we simply divide the numerator by the denominator:
30/200 = 0.15
Once we have actually the prize to the division, we deserve to multiply the answer by 100 to do it a percentage:
0.15 x 100 = 15%
And there you have it! Two different ways to convert 30/200 to a percentage. Both room pretty straightforward and easy to do, however I personally like the convert to decimal an approach as it takes less steps.
I"ve seen a many students get confused whenever a question comes up around converting a fraction to a percentage, however if you follow the procedures laid out below it must be simple. The said, you may still need a calculator because that more facility fractions (and girlfriend can constantly use our calculator in the kind below).
If you desire to practice, grab you yourself a pen, a pad, and also a calculator and shot to transform a couple of fractions come a portion yourself.
Hopefully this tutorial has actually helped you come understand how to convert a fraction to a percentage. You can now go forth and convert fountain to percentages as lot as your little heart desires!
If you discovered this content helpful in her research, please perform us a great favor and use the tool listed below to make sure you properly reference united state wherever you usage it. We really appreciate your support!
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See more: Always Be Faster Than The Boys, : Mikaela Shiffrin Vies For Gold
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# How to calculate a ladder?
Blondel's rule for determining the rise and tread of a staircase The best known ratio is that: 2 a + p = 62/65 cm. This ratio is also known as Blondel's rule and is the one that is commonly used the most.
## How do you calculate the length of a stair?
This will be the number of stair treads. Now calculate the total length of the stair by multiplying this number by the depth of a single tread. So, if you are using 25 '' treads and need 20 risers, you will multiply 25cm by 19, giving you a total length of 475cm (4,75m).
## How is an internal staircase made?
To build an internal staircase, a regular design of the step is necessary, consequently it is advisable to respect the ratio between rise (a) and tread (p): 2a + p = 63 cm. From this formula it can be deduced that the ideal rise is less than 19 cm, while the ideal tread is 30 cm.
## How to calculate an 8 step ladder?
Blondel's law establishes the following equation as the correct formula for those who use the ladder to perform the act of climbing the ladder: 2 rises + 1 tread = 64 cm. The ideal relationship is rise = 18cm and tread = 28cm.
## How to calculate a steel staircase?
Example of calculation of a 2,60 meter high staircase made according to the Bondel formula
1. 260/15 = 17,33 cm high for each upright.
2. Calculate the width of the step.
3. Apply Blondel's formula:
4. (2 x 17,33 cm) + (1 x tread) = 64.
5. Each step will measure 29,34cm.
## Find 18 related questions
### How do you calculate the slope of a staircase?
Note both values, simply divide the riser by the tread and multiply by one hundred. The value obtained is equivalent to the percentage of inclination of the scale.
### How to heal an internal staircase?
If, on the other hand, the changes are more substantial and concern, for example, structural elements or changes in the building elevations (e.g. new internal staircase, moving windows, balconies, etc.), a SCIA (Certified Notification of Commencement of Activities) must be submitted in amnesty.
### What slope must an internal staircase have?
The flight of internal stairs must be at least 80 cm wide, while in the case of a spiral staircase the diameter must be at least 110 cm. With respect to the floor, the inclination of the stair ramp must be between 30 and 60 degrees.
### How to divide a masonry staircase?
HOW TO DRAW STAIRS AND INSTALL THE STEPS CORRECTLY
1. Find the base point. ...
2. Find the finished point on the top floor. ...
3. Divide the maximum height of the staircase by the number of steps. ...
4. Check that the risers are the right size by subtracting the total height of the step from the thickness of the tread.
### How can I cover an internal staircase?
Marble. Among the most common materials for covering internal stairs, marble is an example of a classic and at the same time modern style. The main features that make this material one of the most used are undoubtedly its resistance to scratches and impacts and its durability over time.
### How do you calculate the scaling?
At a scale of 1: 200, the value that you report on the paper will be the real value of the length divided by 200, at a scale of 1:50 it will be divided by 50, at a scale of 1: 1000 it will be divided by 1000. We have obtained that a length of 8,40, 4,2 m, corresponds to a 1 cm segment on the paper, in a 200: XNUMX scale drawing.
### How many steps to make a floor?
Sixty steps, to be precise.
### What does scale 1 to 10000 mean?
The way to do it is simple: just separate the last two digits of the scale with a comma: So in scale we will know that: 1: 10000 a graphic centimeter is equivalent to 100 meters on the ground; ... 1: 100000 a graph centimeter equals 1000 meters on the ground.
### How much space does it take for a staircase?
For a staircase with side-by-side ramps, there will be a minimum overall width of 170 cm, considering the central 10 cm between the two ramps. Finally, in the case of a spiral staircase, the diameter must be at least 110 cm.
### Where to place the internal staircase?
Internal stairs, if not between two walls, must necessarily have a railing and a handrail. The horizontal or vertical elements that make up the railing must be at a maximum distance of 10 cm from each other, while the handrail must be placed 90 cm from the steps.
### How much does it cost to repair a staircase?
Its service consists in evaluating the building intervention, ascertaining compliance with the town planning and building regulations in force and drawing up a report to be attached to the request for amnesty. Usually, the cost is between 500 euros and 1.500 euros.
### How to heal window opening?
Having said that, the opening of an opening on the prospectus (and therefore modification of the same) is a building abuse of "MEDIUM" severity which in some cases can be remedied, precisely, with the presentation of a SCIA in amnesty (fine: variable from a minimum of 1.000 euros if the property does not fall within the fabrics of the historic city, ...
### How to heal pre-1967 building abuse?
It will be enough to present the Cila in which it is certified that the work was carried out before 1967, even in an inhabited center, and to register the building under this title. From that moment the building has been sanitized and marketable.
### How is the slope of an inclined plane calculated?
An inclined plane (for example an uphill or downhill road) can be characterized by its percentage slope, defined as the ratio between the two legs of the triangle ABC, multiplied by 100: percentage slope = AC / BC Β· 100.
### How to calculate scale 1 to 50?
For the scale of 1:50 we have: 10 meters (measured in reality), divided by the number 50, correspond on the sheet to 0,2 meters (ie 20 centimeters). For the scale of 1:20 we have: 10 meters (measured in reality), divided by the number 20, correspond on the sheet to 0,5 meters (i.e. 50 centimeters).
### How high must a step be?
The legislation states that the tread can vary between 23 and 30 cm, while the rise ranges from 16 to 20 cm. The first step of the ramp, called an invitation, can be slightly higher than the others, while the last step is called the "landing step". Normally each flight should have about 10-12 consecutive steps.
### How much does a steel ladder weigh?
Open staircases are those with the most variable weight, this depends on the type of structure, development and of course on the materials, in a completely indicative way, consider that a staircase with 15 metal and wooden steps has a weight ranging from 350 kg to 400 kg. kg, a steel and glass ladder from 500 kg to 600 kg.
|
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# ZingPath: Representing Linear Inequalities and Two Quantities
Searching for
## Representing Linear Inequalities and Two Quantities
Learn in a way your textbook can't show you.
Explore the full path to learning Representing Linear Inequalities and Two Quantities
### Lesson Focus
#### Graphing Linear Inequalities in One Variable
Math Foundations
Graphing linear inequalities in one variable on the number line is explained.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• Translate verbal sentences into linear inequalities in one variable
• Graph linear inequalities in one variable on the number line.
• Write linear inequalities in one variable represented by a graph.
### Everything You'll Have Covered
In this Activity Object, learners will explore single or compound inequalities. Note that double inequalities are also called compound inequalities.
Single Inequalities
An inequality is a mathematical sentence that uses symbols such as <, ?, >, or ? to compare two quantities. The inequality sign expresses that the quantity on the left hand side is the less than, greater than, less than or equal to, or greater than or equal to the quantity on the right hand side.
x < 3: x is less than 3.
x ? 3: x is less than or equal to 3.
x > 3: x is greater than 3.
x ? 3: x is greater than or equal to 3.
Compound inequalities
Compound inequalities are two separate inequalities joined by "and" or "or."
?3 < x < 10: All numbers between -3 and 10 OR all numbers greater than ?3 and less than 10. (In the Activity Object, these inequalities are called double inequalities.)
x > 0 and x ? 2: All numbers that are greater 0 and greater than or equal to2.
x > 0 or x ? ?2: All numbers that are greater 0 or less than or equal to ?2.
Open circle vs. closed circle
When we graph inequalities on a number line, circles are used to show if a number is included or not. An open circle shows that the number is not included, while a closed circle includes the number.
For example,
Interval notation
When we write inequalities with interval notation, parenthesis and square brackets are used.
For example,
Consider the inequality x > 5: Using interval notation, the group of numbers would be written as .
The parenthesis on the left means the set of numbers starts at the real number which is immediately to the right of 5 on the number line; that is, 5 is excluded from the group. Because this group of numbers continues all the way to positive infinity, the positive infinity symbol is used, which is always accompanied by a parenthesis.
Now consider the inequality x ? 5: Using interval notation, the group of numbers would be written as .
The square bracket on the left means that the set of numbers starts on the number line with 5 and that 5 is included in the solution set. Again, the infinity symbol is always accompanied by a parenthesis.
Next we can examine the inequality 3 ? x < 6.
Using interval notation, the group of numbers would be written as .
The square bracket on the left means that the set of numbers starts on the number line with 3 and that 3 is included in the solution set. The parenthesis on the right means that the set of numbers finishes close to 6, without including 6.
### Tutorial Details
Approximate Time 30 Minutes Pre-requisite Concepts Students should know how to solve linear equations. Course Math Foundations Type of Tutorial Skills Application Key Vocabulary compound inequality, inequality, linear inequality
|
# 23 Times Table
In 23 times table we will learn how to read and write multiplication table of 23.
We read twenty three times table as:
One time twenty three is 23
Two times twenty three are 46
Three times twenty three are 69
Four times twenty three are 92
Five times twenty three are 115
Six times twenty three are 138
Seven times twenty three are 161
Eight times twenty three are 184
Nine times twenty three are 207
Ten times twenty three are 230
Eleven times twenty three are 253
Twelve times twenty three are 276
We write 23 times table as:
1 × 23 = 23
2 × 23 = 46
3 × 23 = 69
4 × 23 = 92
5 × 23 = 115
6 × 23 = 138
7 × 23 = 161
8 × 23 = 184
9 × 23 = 207
10 × 23 = 230
11 × 23 = 253
12 × 23 = 276
Let us follow the how to read and write multiplication of 23 times table chart:
23 Times Table
This is the easiest way to follow 23 times table in the chart.
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Jul 22, 24 03:27 PM
Decimal numbers can be expressed in expanded form using the place-value chart. In expanded form of decimal fractions we will learn how to read and write the decimal numbers. Note: When a decimal is mi…
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Mathematics » Introducing Integers » Solve Equations Using Integers, The Division Property of Equality
# Modeling the Division Property of Equality
## Modeling the Division Property of Equality
All of the equations we have solved so far have been of the form $$x+a=b$$ or $$x-a=b.$$ We were able to isolate the variable by adding or subtracting the constant term. Now we’ll see how to solve equations that involve division.
We will model an equation with envelopes and counters in the figure below.
Here, there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope?
To determine the number, separate the counters on the right side into $$2$$ groups of the same size. So $$6$$ counters divided into $$2$$ groups means there must be $$3$$ counters in each group (since $$6÷2=3\right).$$
What equation models the situation shown in the figure below? There are two envelopes, and each contains $$x$$ counters. Together, the two envelopes must contain a total of $$6$$ counters. So the equation that models the situation is $$2x=6.$$
We can divide both sides of the equation by $$2$$ as we did with the envelopes and counters.
We found that each envelope contains $$\text{3 counters.}$$ Does this check? We know $$2·3=6,$$ so it works. Three counters in each of two envelopes does equal six.
The figure below shows another example.
Now we have $$3$$ identical envelopes and $$\text{12 counters.}$$ How many counters are in each envelope? We have to separate the $$\text{12 counters}$$ into $$\text{3 groups.}$$ Since $$12÷3=4,$$ there must be $$\text{4 counters}$$ in each envelope. See the figure below.
The equation that models the situation is $$3x=12.$$ We can divide both sides of the equation by $$3.$$
Does this check? It does because $$3·4=12.$$
## Example
Write an equation modeled by the envelopes and counters, and then solve it.
### Solution
There are $$\text{4 envelopes,}$$ or $$4$$ unknown values, on the left that match the $$\text{8 counters}$$ on the right. Let’s call the unknown quantity in the envelopes $$x.$$
Write the equation. Divide both sides by 4. Simplify.
There are $$\text{2 counters}$$ in each envelope.
Want to suggest a correction or an addition? Leave Feedback
|
## Engage NY Eureka Math 4th Grade Module 4 Lesson 9 Answer Key
### Eureka Math Grade 4 Module 4 Lesson 9 Problem Set Answer Key
Question 1.
Complete the table.
Refer to the below table.
Explanation:
In pattern block A, the total number that fit around one vertex is 4 and the interior angle measurement is 360° ÷ 4 which is 90° and the sum of the angles around the vertex is 90°+90°+90°+90°= 360°.
In pattern block B, the total number that fit around one vertex is 6 and the interior angle measurement is 360° ÷ 6 which is 60° and the sum of the angles around the vertex is 60°+60°+60°+60°+60°+60°= 360°.
In pattern block C, the total number that fit around one vertex is 3 and the interior angle measurement is 360° ÷ 3 which is 120° and the sum of the angles around the vertex is 120°+120°+120°= 360°.
In pattern block D, the total number that fit around one vertex is 6 and the interior angle measurement is 360° ÷ 6 which is 60° and the sum of the angles around the vertex is 60°+60°+60°+60°+60°+60°= 360°.
In pattern block E, the total number that fit around one vertex is 3 and the interior angle measurement is 360° ÷ 3 which is 120° and the sum of the angles around the vertex is 120°+120°+120°= 360°.
In pattern block F, the total number that fit around one vertex is 12 and the interior angle measurement is 360° ÷ 12 which is 30° and the sum of the angles around the vertex is 30°+30°+30°+30°+30°+30°+30°+30°+30°+30°+30°+30°= 360°.
Question 2.
Find the measurements of the angles indicated by the arcs.
Refer to the below table.
Explanation:
In pattern block A, the total number angle measurement which 150° and the sum of the angles around the vertex is 60°+90° which is 150°.
In pattern block B, the total number angle measurement which 180° and the sum of the angles around the vertex is 60°+120° which is 180°.
In pattern block C, the total number angle measurement which 210°, and the sum of the angles around the vertex is 120°+90° which is 210°.
Question 3.
Use two or more pattern blocks to figure out the measurements of the angles indicated by the arcs.
Refer to the below table.
Explanation:
In pattern block A, the total number angle measurement which 60°, and the sum of the angles around the vertex is 30°+30° which is 60°.
In pattern block B, the total number angle measurement which 210°, and the sum of the angles around the vertex is 120°+90° which is 210°.
In pattern block C, the total number angle measurement which 120°, and the sum of the angles around the vertex is 90°+30° which is 120°.
### Eureka Math Grade 4 Module 4 Lesson 9 Exit Ticket Answer Key
Question 1.
Describe and sketch two combinations of the blue rhombus pattern block that create a straight angle.
Refer to the below image.
Explanation:
Here, we have constructed two combinations of the blue rhombus pattern block that create a straight angle.
Question 2.
Describe and sketch two combinations of the green triangle and yellow hexagon pattern block that create a straight angle.
Refer to the below image.
Explanation:
Here, we have constructed two combinations of the green triangle and yellow hexagon pattern block that create a straight angle.
### Eureka Math Grade 4 Module 4 Lesson 9 Homework Answer Key
Sketch two different ways to compose the given angles using two or more pattern blocks. Write an addition sentence to show how you composed the given angle.
Question 1.
point A, B, and C form a straight line.
Refer to the below image.
Explanation:
Here, we have constructed three points A, B, C, and formed a straight line, and formed two different ways to compose the given angles using two or more pattern blocks. In the first image, we have divided 180° angle as 90°+30°+60°, so the addition sentence is 90°+30°+60°= 180°. And in the second image, we have constructed three points A, B, C, and formed a straight line we have divided 180° angle as 120°+60°, so the addition sentence is 120°+60°= 180°.
Question 2.
∠DEF = 90°
The addition sentence is 30°+60°= 90°.
Explanation:
Here, we have constructed <DEF a and formed two different ways to compose the given angles using two or more pattern blocks. In the first image, we have divided 90° angle as 30°+60°, so the addition sentence is 30°+60°= 90°. And in the second image, we have constructed <DEF and we have divided 90° angle as 30°+60°, so the addition sentence is 30°+60°= 90°.
Question 3.
∠GHI = 120°
The addition sentence for the <GHI is 30°+90°= 120°.
The addition sentence for the <GHI is 30°+90°= 120°
Explanation:
Here, we have constructed <GHI a and formed two different ways to compose the given angles using two or more pattern blocks. In the first image, we have divided 120° angle as 30°+90°, so the addition sentence is 30°+90°= 120°. And in the second image, we have constructed <GHI and we have divided 90° angle as 60°+60°, so the addition sentence is 60°+60°= 120°.
Question 4.
x° = 270°
The addition sentence for the first image is 60°+90°+120°= 270°.
The addition sentence for the second image is 60°+60°+30°+60°+60°= 270°.
Explanation:
Here, we have constructed <LKJ a and formed two different ways to compose the given angles using two or more pattern blocks. In the first image, we have divided 270° angle as 60°+90°+120°, so the addition sentence is 60°+90°+120°= 270°. And in the second image, we have constructed <LKJ and we have divided 270° angle as 60°+60°+30°+60°+60°, so the addition sentence is 60°+60°+30°+60°+60°= 270°.
Question 5.
Micah built the following shape with his pattern blocks. Write an addition sentence for each angle indicated by an arc and solve. The first one is done for you.
a. y° = 120° + 90°
y° = 210°
b. z° = _____________
z° = _____________
z° = 60°+30°
z° = 90°
Explanation:
Here, the value of z is 60°+30° which is 90°
c. x° = _____________
x° = _____________
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# Question
Abi spent of her money on 2 identical bookmarks and some identical pens. The cost of each bookmark to the cost of each pen was 2:3. She bought some more bookmarks with 2/5 of her remaining money and had 24 bookmarks in the end How many pens did she buy at first?
Source: Ang Mo Kio Primary
Here is how I would do it with block diagrams.
Since the price ratio is 2:3, let’s assume a bookmark cost \$2 and a pen cost \$3.
• The remaining money is bottom 5 blocks.
• 2/5 of that is two green blocks.
• Two green blocks are used to pay for 24 – 2 = 22 bookmarks.
• Each green block is the cost of 11 bookmarks = 2 x 11 = \$22.
• Orange block is cost of 2 bookmarks = \$4.
• Blue block is 22 – 4 = \$18.
• So number of pens bought = \$18/\$3 = 6.
To be more general, we can say a bookmark costs 2u and a pen cost 3u. But really, a \$ is just another unit. So we can use \$ itself as the unit. Makes it easier to think about. You can replace the \$ above with “u” if you like. 🙂
0 Replies 1 Like
Book mark Pen
Cost of each 2u 3u Total spent = 1/6 of money
2/5 of remaining money(1-1/6=5/6) = 2/5 x 5/6 = 1/3
1/3 of money buys (24-2)= 22 bookmarks which is 22 x 2u = \$44u
1/6 of money = \$44u / 2 = \$22u
Cost of 2 bookmarks = 2 x 2u = 4u
So cost of pens = 22u-4u = 18u
Hence number of pens bought = 18u / 3u =6
0 Replies 1 Like
Assuming Abi had 6U of money at first.
After spending 1/6 of money or 1U, she was left with 5U
Then she used 2U for 24-2=22 bookmarks
So 1U can buy 11 bookmarks
Cost of a pen was 3/2 times of bookmarks or 1.5 bookmarks
Initially she spent 1U of money and can buy 11 bookmarks
11-2=9
9/1.5=6
So she bought 6 pens initially.
0 Replies 1 Like
Edited and reposted. Thank you
0 Replies 0 Likes
The question seems incomplete. Missing the fraction of money spent, etc. May need to repost.
0 Replies 1 Like
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# How do you solve 7( x - 1) + 5= - 2( 3- 4x ) + 5?
Aug 1, 2017
$x = - 1$
Refer to the explanation for the process.
#### Explanation:
Solve:
$7 \left(x - 1\right) + 5 = - 2 \left(3 - 4 x\right) + 5$
Expand.
$7 x - 7 + 5 = - 6 + 8 x + 5$
Simplify.
$7 x - 2 = - 1 + 8 x$
Subtract $7 x$ from both sides.
$- 2 = - 1 + 8 x - 7 x$
Add $1$ to both sides.
$- 2 + 1 = 8 x - 7 x$
Simplify.
$- 1 = x$
Switch sides.
$x = - 1$
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# Quant Data Sufficiency for IBPS RRB Officer and Assistant Mains Exam
The Quantitative Aptitude section for IBPS RRB Mains examination consists of 40 questions which constitutes twenty-five percentage of the total weightage in the exam. Data sufficiency is an important topic of this section which tests a candidate’s knowledge regarding mathematical concepts. Quant Data Sufficiency for IBPS RRB Mains is discussed below.
## Data Sufficiency for IBPS RRB Mains Exam
In data sufficiency questions, the candidate will be provided with a question followed by a few statements and he/she will have to decide which among the given statements is necessary to answer the question. The uniqueness of these data sufficiency questions is that the candidate is not required to solve for the final answer, they just have to recognize the conditions necessary to solve the question. A few practice questions for IBPS RRB Mains Exam are given below.
## Practice questions of Data Sufficiency for IBPS RRB Mains Exam
1. The question below consists of three statements numbered I, II and III given below it. You have to read all the statements and decide which all statement/s is/are sufficient to answer the question.
What is the speed of the train?
Statement I: The train crosses a signal pole in 18 seconds
Statement II: The train crosses a platform of equal length in 36 seconds
Statement III: Length of the train is 300 metres
(a) Statements I and III together is sufficient to answer the question
(b) Statements II and III together is sufficient to answer the question
(c) Statements I and II together is sufficient to answer the question
(d) Any two of the three statements is sufficient to answer the question
(e) Statement III and either statement I or statement II together is sufficient to answer the question
Solution:
From I and III
Length of train = 300 m
Speed of train = 300/18 = 16.67 m/s
From II and III
Length of train = Length of platform = 300 m
Speed of train = 600/36 = 16.67 m/s
From I and II:
Since the length of the train is unknown. We cannot determine the speed of the train
Hence, statement III and either statement I or statement II is sufficient
Ace the Numerical Ability section with the help of Oliveboard.
1. The question below consists of two statements numbered I and II given below it. You have to read all the statements and decide which all statement/s is/are sufficient to answer the question.
Find the profit percentage earned by Sonam on selling the article
Statement I: Sonam sold the article for Rs.150. Had she sold it for Rs.6 more, she would have earned a profit of 30%.
Statement II: She marked the price of the article 40% above the cost price and sold it for Rs.150 at a discount of Rs.18 on the marked price.
(a) If statement I alone is sufficient to answer the question
(b) If statement II alone is sufficient to answer the question
(c) If both statements I and II together are sufficient to answer the question
(d) If either statement I alone or statement II alone is sufficient to answer the question
(e) If both statements I and II together are not sufficient to answer the question
Solution:
Let the cost price of the article be Rs. P
From I:
P x 130/100 = 150 + 6
=> P = 156 x 100/130
=> P = Rs.120
profit percentage = (150 – 120)/120 x 100 = 25%
From II:
P x 140/100 – 18 = 150
=> P x 140/100 = 168
=> P = Rs.120
profit percentage = (150 – 120)/120 x 100 = 25%
Hence, either statement I alone or statement II alone is sufficient
Ace your preparation with free mock tests and topic tests.
1. The question below consists of three statements numbered I, II and III given below it. You have to read all the statements and decide which all statement/s is/are sufficient to answer the question.
What is the rate of interest?
Statement I: Compound interest on a sum at the end of three years becomes ₹1356.48 more than simple interest.
Statement II: Sum amounts to ₹37632 in two years and ₹42147.84 at the end of three years when interest is compounded annually.
Statement III: Sum amounts to 44987.53 Rs at the end of 4th year.
(a) If statement II alone sufficient to answer the question
(b) If all the three statements together are sufficient to answer the question
(c) If either statement II alone or statement III alone is sufficient to answer the question
(d) If statements II and III together are sufficient to answer the question
(e) If all the three statement together are not sufficient to answer the question.
Solution:
Let the principal amount be Rs. P
From I:
CI – SI = 1356.48
CI – SI = P x (R/100)2 x (300 + R)/100
Since, principal value is not known we cannot calculate interest rate.
Hence, statement I is not sufficient.
From II:
Amount at the end of two years = P x (1 + R/100)2 = 37632 …….. (i)
Amount at the end of three years = P x (1 + R/100)3 = 42147.84 …….. (ii)
Dividing (ii) by (i) we get:
(1 + R/100) = 1.12
Rate = 12%
Hence, statement II alone sufficient.
From III:
Whether it is simple or compound interest was not mentioned. Hence, we cannot find interest rate.
From I and III together:
I and III together are not sufficient to find interest rate, as we don’t have principle amount.
Attempt a FREE mock tests of IBPS RRB Mains
1. The question below consists of two statements numbered I and II given below it. You have to read all the statements and decide which all statement/s is/are sufficient to answer the question.
Nikita, Kailash and Manish entered into a partnership with investment in the ratio 7:4:5 respectively. After one year, Kailash doubled his investment and Manish invested Rs.25000 more. At the end of three years, they earned a profit of Rs.234000. Find Nikita’s share in the profit.
Statement I: Profit ratio of Kailash and Manish is 8:7 respectively.
Statement II: Manish’s share in profit is Rs.70000
(a) If statement I alone is sufficient to answer the question
(b) If statement II alone is sufficient to answer the question
(c) If both statements I and II together are sufficient to answer the question
(d) If either statement I alone or statement II alone is sufficient to answer the question
(e) If both statements I and II together are not sufficient to answer the question
Solution:
Let the initial amounts invested by Nikita, Kailash and Manish be Rs. 7k, Rs. 4k and Rs. 5k respectively.
Profit Ratio
Nikita : Kailash : Manish = (7k * 3) : (4k + 8k * 2) : 5k + (5k + 25000) * 2
= 21k : 20k: (15k + 50000) ————————————-(i)
From I:
20k/(15k + 50000) = 8/7
=> 140k = 120k + 400000
=> k = 20000
From (i)
Nikita : Kailash : Manish = (21*20000) : (20*20000) : (15*20000 + 50000)
= 42 : 40 : 35
Share of Nikita in the profit = 42/(42 + 40 + 35)* 234000
= Rs. 84000
From II:
(15k + 50000)/[21k + 20k + (15k + 50000)] * 234000 = 70000
=> (15k + 50000)/ (56k + 50000) *117 = 35
=> 1755k + 5850000 = 1960k + 1750000
=> k = 20000
From (i)
Nikita : Kailash : Manish = (21*20000) : (20*20000) : (15*20000 + 50000)
= 42 : 40 : 35
Share of Nikita in the profit = 42/(42 + 40 + 35)* 234000
= Rs. 84000
Hence either statement I alone or statement II alone is sufficient to answer the question.
All the best for the examination! For more such practice questions of Quant Data Sufficiency for IBPS RRB Mains, click here.
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How do you find the equation of the XY plane?
XY plane is perpendicular to z-axis and passes through origin, So its equation is , (r −0 ).( z )=0.YZ plane is perpendicular to x-axis and passes through origin, So its equation is , (r −0 ).( x )=0.ZX plane is perpendicular to y-axis and passes through origin, So its equation is ,
What is the equation of the XZ plane?
Similarly, the y-z-plane has standard equation x = 0 and the x-z-plane has standard equation y = 0. A plane parallel to the x-y-plane must have a standard equation z = d for some d, since it has normal vector k. A plane parallel to the y-z-plane has equation x = d, and one parallel to the x-z-plane has equation y = d.
How do you find the parametric equation of a plane?
To find a parametrization, we need to find two vectors parallel to the plane and a point on the plane. Finding a point on the plane is easy. We can choose any value for x and y and calculate z from the equation for the plane. Let x=0 and y=0, then equation (1) means that z=18−x+2y3=18−0+2(0)3=6.
What is a vector equation of a plane?
From the video, the equation of a plane given the normal vector n = [A,B,C] and a point p1 is n . p = n . p1, where p is the position vector [x,y,z].
How do you find the normal of a plane?
Thus for a plane (or a line), a normal vector can be divided by its length to get a unit normal vector. Example: For the equation, x + 2y + 2z = 9, the vector A = (1, 2, 2) is a normal vector. |A| = square root of (1+4+4) = 3.
What is the equation of Z axis?
Solution : The equation of Z – axis, x=0 and y=0.
What does YZ plane mean?
The xy-plane is the plane that contains the x- and y-axes; the yz-plane contains the y- and z-axes; the xz-plane contains the x- and z-axes. These three coordinate planes divide space into eight parts, called octants.
Is the Z axis vertical?
The coordinate surfaces of the Cartesian coordinates (x, y, z). The z-axis is vertical and the x-axis is highlighted in green.
What is the distance between two planes?
Definition. The distance between two planes is equal to length of the perpendicular lowered from a point on a plane.
How do you find the distance from a point to a plane?
Therefore, the distance from point P to the plane is along a line parallel to the normal vector, which is shown as a gray line segment. If we denote by R the point where the gray line segment touches the plane, then R is the point on the plane closest to P. The distance from P to the plane is the distance from P to R.
How do you find the angle between two planes?
The angle between planes is equal to a angle between their normal vectors. Definition. The angle between planes is equal to a angle between lines l1 and l2, which lie on planes and which is perpendicular to lines of planes crossing.
What is a symmetric equation?
The symmetric form of the equation of a line is an equation that presents the two variables x and y in relationship to the x-intercept a and the y-intercept b of this line represented in a Cartesian plane. The symmetric form is presented like this: xa+yb=1, where a and b are non-zero.
Releated
Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […]
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## Inequalities
7.1 Inequalities
Definition The relationship between two quantities that do not have the same value.
Symbol Meaning $$\gt$$ Greater than $$\lt$$ Less than
Example From the number line above, $$–2$$ lies to the left of $$3$$. So, $$–2$$ is less than $$3$$. Thus, the inequality is $$–2 < 3$$.
Example From the number line above, $$–2$$ lies to the right of $$-7$$. So, $$–2$$ is greater than $$-7$$. Thus, the inequality is $$–2 > -7$$.
Describe inequality and form algebraic inequality:
Example From the number line above, $$x$$ is less than $$8$$. Thus, $$x\lt 8$$.
Identify relationship:
Symbol Meaning $$\geq$$ Greater than or equal to $$\leq$$ Less than or equal to
Properties of inequalities:
Converse property of inequality
• If $$a \lt b$$, then $$b \gt a$$.
Example State the converse property of inequality of $$-23\gt-32$$. Answer: $$-32\lt-23$$
Transitive property of inequality
• If $$a \lt b \lt c$$, then $$a \lt c$$.
Example State the transitive property of inequality of $$-15\lt-8\lt0$$. Answer: $$-15\lt0$$
• The inequality symbol remains unchanged when adding or subtracting a positive or negative number to or from both sides of the inequality.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a+c&\lt b+c.\\\\ \end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a-c&\lt b-c. \\\\\end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a+(-c)&\lt b+(-c).\\\\ \end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a-(-c)&\lt b-(-c).\end{aligned}
• The inequality symbol remains unchanged when multiplying or dividing both sides of the inequality by a positive number.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a\times c&\lt b\times c.\\\\ \end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,\dfrac{a}{c}&\lt \dfrac{b}{c}.\end{aligned}
• The direction of the inequality symbol is reversed when multiplying or dividing both sides of the inequality by a negative number.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a\times (-c)&\gt b\times (-c).\\\\ \end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,\dfrac{a}{-c}&\gt \dfrac{b}{-c}.\end{aligned}
• When both sides of the inequality are multiplied by $$-1$$, the direction of the inequality symbol is reversed.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,-a&\gt -b.\end{aligned}
Multiplicative inverse
• When performing reciprocal of both numbers on both sides of the inequality, the direction of the inequality symbol is reversed.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,\dfrac{1}{a}&\gt \dfrac{1}{b}.\end{aligned}
## Inequalities
7.1 Inequalities
Definition The relationship between two quantities that do not have the same value.
Symbol Meaning $$\gt$$ Greater than $$\lt$$ Less than
Example From the number line above, $$–2$$ lies to the left of $$3$$. So, $$–2$$ is less than $$3$$. Thus, the inequality is $$–2 < 3$$.
Example From the number line above, $$–2$$ lies to the right of $$-7$$. So, $$–2$$ is greater than $$-7$$. Thus, the inequality is $$–2 > -7$$.
Describe inequality and form algebraic inequality:
Example From the number line above, $$x$$ is less than $$8$$. Thus, $$x\lt 8$$.
Identify relationship:
Symbol Meaning $$\geq$$ Greater than or equal to $$\leq$$ Less than or equal to
Properties of inequalities:
Converse property of inequality
• If $$a \lt b$$, then $$b \gt a$$.
Example State the converse property of inequality of $$-23\gt-32$$. Answer: $$-32\lt-23$$
Transitive property of inequality
• If $$a \lt b \lt c$$, then $$a \lt c$$.
Example State the transitive property of inequality of $$-15\lt-8\lt0$$. Answer: $$-15\lt0$$
• The inequality symbol remains unchanged when adding or subtracting a positive or negative number to or from both sides of the inequality.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a+c&\lt b+c.\\\\ \end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a-c&\lt b-c. \\\\\end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a+(-c)&\lt b+(-c).\\\\ \end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a-(-c)&\lt b-(-c).\end{aligned}
• The inequality symbol remains unchanged when multiplying or dividing both sides of the inequality by a positive number.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a\times c&\lt b\times c.\\\\ \end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,\dfrac{a}{c}&\lt \dfrac{b}{c}.\end{aligned}
• The direction of the inequality symbol is reversed when multiplying or dividing both sides of the inequality by a negative number.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,a\times (-c)&\gt b\times (-c).\\\\ \end{aligned} \begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,\dfrac{a}{-c}&\gt \dfrac{b}{-c}.\end{aligned}
• When both sides of the inequality are multiplied by $$-1$$, the direction of the inequality symbol is reversed.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,-a&\gt -b.\end{aligned}
Multiplicative inverse
• When performing reciprocal of both numbers on both sides of the inequality, the direction of the inequality symbol is reversed.
\begin{aligned} \text{If }\,a&\lt b, \\\\\text{then }\,\dfrac{1}{a}&\gt \dfrac{1}{b}.\end{aligned}
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# Lesson 14
Surface Area of Right Prisms
### Lesson Narrative
In grade 6, students used nets made up of rectangles and triangles to find the surface area of three-dimensional figures. In this lesson they find surface areas of prisms, and see that structure of a prism allows for shortcuts in adding up the areas of the faces. They see that if the prism is sitting on its base, then the vertical sides can be unfolded into a single rectangle whose height is the height of the prism and whose length is the perimeter of the base. The purpose of the lesson is not to come up with a formula for the surface area of a prism, but to help students see and make use of the structure of the prism to find surface area efficiently (MP7).
### Learning Goals
Teacher Facing
• Calculate the surface area of a prism, and explain (in writing) the solution method.
• Comprehend that surface area and volume are two different attributes of three-dimensional objects and are measured in different units.
• Interpret different methods for calculating the surface area of a prism, and evaluate (orally and in writing) their usefulness.
### Student Facing
Let’s look at the surface area of prisms.
### Required Preparation
Assemble the net from the blackline master to make a prism with a base in the shape of a plus sign. Make sure to print the blackline master at 100% scale so the dimensions are accurate. This prism will be used for both the warm-up and the following activity.
### Student Facing
• I can find and use shortcuts when calculating the surface area of a prism.
• I can picture the net of a prism to help me calculate its surface area.
Building On
Building Towards
### Glossary Entries
• surface area
The surface area of a polyhedron is the number of square units that covers all the faces of the polyhedron, without any gaps or overlaps.
For example, if the faces of a cube each have an area of 9 cm2, then the surface area of the cube is $$6 \boldcdot 9$$, or 54 cm2.
### Print Formatted Materials
For access, consult one of our IM Certified Partners.
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Breaking News
# What's Half Of 27?
## Introduction
When it comes to math, we often encounter problems that require us to divide numbers. One of the most common questions that people ask is “what’s half of 27?” While this may seem like a simple question, it can sometimes be confusing, especially for those who are not familiar with basic math concepts. In this article, we will explore the answer to this question and provide you with tips on how to solve similar problems in the future.
## What is Half?
Before we dive into the answer to the question “what’s half of 27?”, let’s first define what “half” means. Half is a fraction that represents one of two equal parts of a whole. In other words, if we divide a whole into two equal parts, each part will be half of the whole. The symbol for half is ½.
## How to Find Half of 27?
To find half of 27, we need to divide 27 by 2. This can be done using long division or a calculator. When we divide 27 by 2, we get the answer 13.5. Therefore, half of 27 is 13.5.
## Why is the Answer 13.5?
The answer to the question “what’s half of 27?” is 13.5 because 27 divided by 2 equals 13.5. When we divide 27 by 2, we are essentially splitting 27 into two equal parts. Each part is 13.5, which means that half of 27 is 13.5.
## Other Examples
Now that we know how to find half of 27, let’s look at some other examples. Suppose we want to find half of 10. To do this, we simply divide 10 by 2. The answer is 5. Therefore, half of 10 is 5.
### Tip:
When you’re dividing a number by 2 to find half of it, you can simply divide the number by 2 or multiply it by 0.5. Both methods will give you the same answer.
## Conclusion
In conclusion, half is a fraction that represents one of two equal parts of a whole. To find half of 27, we need to divide 27 by 2, which gives us the answer 13.5. We can use the same method to find half of any other number. Hopefully, this article has helped you understand the concept of finding half and how to solve similar problems in the future.
## What Is 50 Of 55?
Introduction When it comes to solving mathematical problems, we all have faced some difficulties in …
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UPSC > Introduction, Basic Concepts & Formulae: Mensuration- 2
# Introduction, Basic Concepts & Formulae: Mensuration- 2 Video Lecture - CSAT Preparation - UPSC
## CSAT Preparation
197 videos|151 docs|200 tests
## FAQs on Introduction, Basic Concepts & Formulae: Mensuration- 2 Video Lecture - CSAT Preparation - UPSC
1. What are the basic concepts of mensuration?
Ans. The basic concepts of mensuration include measuring the length, area, and volume of various geometric shapes. It involves understanding concepts such as perimeter, area, surface area, and volume. The formulas for these calculations depend on the shape being measured.
2. How do you calculate the perimeter of a rectangle?
Ans. To calculate the perimeter of a rectangle, you add the lengths of all the sides. The formula is: Perimeter = 2 * (Length + Width), where Length is the length of the rectangle and Width is the width of the rectangle.
3. What is the formula to find the area of a triangle?
Ans. The formula to find the area of a triangle is: Area = (Base * Height) / 2, where Base is the length of the base of the triangle and Height is the perpendicular distance from the base to the opposite vertex.
4. How do you calculate the surface area of a cube?
Ans. To calculate the surface area of a cube, you find the sum of the areas of all its faces. Since all the faces of a cube are squares, the formula is: Surface Area = 6 * (Side Length)^2, where Side Length is the length of one side of the cube.
5. What is the formula to calculate the volume of a cylinder?
Ans. The formula to calculate the volume of a cylinder is: Volume = π * (Radius)^2 * Height, where π is a constant approximately equal to 3.14159, Radius is the radius of the circular base of the cylinder, and Height is the height of the cylinder.
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Double 47 is 94.
This Math quiz is called 'Addition and Subtraction 2' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is an enjoyable way to learn if you are in the 3rd, 4th or 5th grade - aged 8 to 11.
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Addition and subtraction is adding numbers together or taking them away from each other. Knowing how to work out addition and subtraction problems comes in very handy when dealing with money in everyday life. Imagine that you have \$10 and wish to buy two items that cost \$3.47 and \$6.91... do you have enough money? Well, \$3.47 + \$6.91 = \$10.38 so you'd be 38c short! See how important it is to be able to add and subtract?
1.
Which is the best strategy to calculate 38 + 37?
Count on 37 from 38
Double 38 and subtract 1
Add 50 to 38 then subtract 13
Multiply 38 by 3 subtract 39
2 x 8 = 16 and 2 x 30 = 60
60 + 16 = 76
76 - 1 = 75
2.
Which is a good strategy for adding 19?
Add 20 and take away 11
Add 20 and take away 1
e.g. 46 + 19 so:
46 + 20 = 66
then 66 - 1 = 65
3.
What is the sum of 37 + 28?
9
63
65
75
Add the 10s first and then the units
30 + 20 = 50
then 7 + 8 = 15
then 50 + 15
4.
If we subtract a large number from a small number the total will be?
A fraction
A decimal
A multiple
A negative number
e.g. 8 - 10 = -2
5.
What is double 47?
84
87
92
94
40 x 2 = 80 and 7 x 2 = 14
80 + 14 = 94
6.
What is the best method of subtracting 21?
Subtract 20 subtract 1
e.g. 56 - 21 so:
56 - 20 = 36
then 36 - 1 = 35
7.
What would we add to 56 to total 100?
44
45
54
64
100 - 56 = 44
8.
Which strategy would be best to calculate 827 + 6?
Add 20 to 827 then subtract 14
Count 827 on from 6
Subtract 10 from 827 then add 4
Most math problems can be made simpler
9.
What is the missing number in 350 + ___ = 1,000?
450
550
650
750
100 - 35 = 65
so 1,000 - 350 = 650
10.
Which number sentence is incorrect?
28 + 35 = 63
63 - 35 = 28
35 + 28 = 63
28 - 35 = 63
The other correct number sentence would be 63 - 28 = 35
Author: Amanda Swift
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# Exponential Growth and Decay
### Exponential growth can be amazing!
Let us say we have this special tree.
It grows exponentially , following this formula (e is Euler's number):
Height (in mm) = ex
• At 1 year old it is: e1 = 2.7 mm high ... really tiny!
• At 5 years it is: e5 = 148 mm high ... as high as a cup
• At 10 years: e10 = 22 m high ... as tall as a building
• At 15 years: e15 = 3.3 km high ... 10 times the height of the Eiffel Tower
• At 20 years: e20 = 485 km high ... up into space!
No tree could ever grow that tall.
So when people say "it grows exponentially" ... just think what that means.
## Growth and Decay
But sometimes things can grow (or the opposite: decay) exponentially, at least for a while.
So we have a generally useful formula:
y(t) = a × ekt
Where y(t) = value at time "t"
a = value at the start
k = rate of growth (when >0) or decay (when <0)
t = time
### Example: 2 months ago you had 3 mice, you now have 18.
Assuming the growth continues like that What is the "k" value? How many mice 2 Months from now? How many mice 1 Year from now?
y(t) = a × ekt
We know a=3 mice, t=2 months, and right now y(2)=18 mice:
18 = 3 × e2k
Now some algebra to solve for k:
Divide both sides by 3:6 = e2k
Take the natural logarithm of both sides:ln(6) = ln(e2k)
ln(ex)=x, so:ln(6) = 2k
Swap sides:2k = ln(6)
Divide by 2:k = ln(6)/2
The step where we used ln(ex)=x is explained more at Exponents and Logarithms.
Note: we can calculate k ≈ 0.896, but it is best to keep it as k = ln(6)/2 until we do our final calculations.
We can now put k = ln(6)/2 into our formula (also a=3):
y(t) = 3 e(ln(6)/2)t
Now let's calclulate the population in 2 more months (at t=4 months):
y(4) = 3 e(ln(6)/2)×4 = 108
And in 1 year from now (t=14 months):
y(14) = 3 e(ln(6)/2)×14 = 839,808
That's a lot of mice! I hope you will be feeding them properly.
## Exponential Decay
Some things "decay" (get smaller) exponentially.
### Example: Atmospheric pressure (the pressure of air around you) decreases as you go higher.
It decreases about 12% for every 1000 m: an exponential decay.
The pressure at sea level is about 1013 hPa (depending on weather).
• Write the formula (with its "k" value),
• Find the pressure on the roof of the Empire State Building (381 m),
• and at the top of Mount Everest (8848 m)
y(t) = a × ekt
We know
• a (the pressure at sea level) = 1013 hPa
• t is in meters (distance, not time, but the formula still works)
• y(1000) is a 12% reduction on 1013 hPa = 891.44 hPa
So:
891.44 = 1013 ek×1000
Now some algebra to solve for k:
Divide both sides by 1013:0.88 = e1000k
Take the natural logarithm of both sides:ln(0.88) = ln(e1000k)
ln(ex)=x, so:ln(0.88) = 1000k
Swap sides:1000k = ln(0.88)
Divide by 1000:k = ln(0.88)/1000
Now we know "k" we can write:
y(t) = 1013 e(ln(0.88)/1000)×t
And finally we can calculate the pressure at 381 m, and at 8848 m:
y(381) = 1013 e(ln(0.88)/1000)×381 = 965 hPa
y(8848) = 1013 e(ln(0.88)/1000)×8848 = 327 hPa
(In fact pressures at Mount Everest are around 337 hPa ... good calculations!)
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## Introduction
### Introduction
There are three main characteristics of a geometric experiment.
1. Repeating independent Bernoulli trials until a success is obtained. Recall that a Bernoulli trial is a binomial experiment with number of trials n = 1. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bull's-eye until you hit the bull's-eye. The first time you hit the bull's-eye is a success, so you stop throwing the dart. It might take six tries until you hit the bull's-eye. You can think of the trials as failure, failure, failure, failure, failure, success, stop.
2. In theory, the number of trials could go on forever. There must be at least one trial.
3. The probability, p, of a success and the probability, q, of a failure do not change from trial to trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is $1616$. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = $5656$, the probability of a failure. The probability of getting a three on the fifth roll is $(56)(56)(56)(56)(16)(56)(56)(56)(56)(16)$ = .0804.
X = the number of independent trials until the first success.
p = the probability of a success, q = 1 – p = the probability of a failure.
There are shortcut formulas for calculating mean μ, variance σ2, and standard deviation σ of a geometric probability distribution. The formulas are given as below. The deriving of these formulas will not be discussed in this book.
$μ=1p,σ2=(1p)(1p−1),σ=(1p)(1p−1)μ=1p,σ2=(1p)(1p−1),σ=(1p)(1p−1)$
### Example 4.16
Suppose a game has two outcomes, win or lose. You repeatedly play that game until you lose. The probability of losing is p = 0.57.
If we let X = the number of games you play until you lose (includes the losing game), then X is a geometric random variable. All three characteristics are met. Each game you play is a Bernoulli trial, either win or lose. You would need to play at least one game before you stop. X takes on the values 1, 2, 3, . . . (could go on indefinitely). Since we are measuring the number of games you play until you lose, we define a success as losing a game and a failure as winning a game. The probability of a success $p=.57p=.57$ and the probability of a failure q= 1-p = 1–0.57 = 0.43. Both p and q remain the same from game to game.
If we want to find the probability that it takes five games until you lose, then the probability could be written as P(x = 5). We will explain how to find a geometric probability later in this section.
Try It 4.16
You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on?
### Example 4.17
A safety engineer feels that 35 percent of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions.
If we let X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions, then X is a geometric random variable. All three characteristics are met. Each accident report she reads is a Bernoulli trial: the accident was either caused by failure of employees to follow instructions or not. She would need to read at least one accident report before she stops. X takes on the values 1, 2, 3, . . . (could go on indefinitely). Since we are measuring the number of reports she needs to read until one that shows an accident caused by failure of employees to follow instructions, we define a success as an accident caused by failure of employees to follow instructions. If an accident was caused by another reason, the report is defined as a failure. The probability of a success p = .35 and the probability of a failure $q=1−p=1−.35=.65q=1−p=1−.35=.65$. Both p and q remain the same from report to report.
If we want to find the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions, then the probability could be written as $p=.35p=.35$. If we want to find how many reports, on average, the safety engineer would expect to look at until she finds a report showing an accident caused by employee failure to follow instructions, we need to find the expected value E(x). We will explain how to solve these questions later in this section.
Try It 4.17
An instructor feels that 15 percent of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least 10 exams until she finds one with a grade below a C. What is the probability question stated mathematically?
### Example 4.18
Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55 percent of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people?
This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student).
a. Let X = the number of ________ you must ask ________ one says yes.
Solution 4.18
a. Let X = the number of students you must ask until one says yes.
b. What values does X take on?
Solution 4.18
b. 1, 2, 3, . . ., (total number of students)
c. What are p and q?
Solution 4.18
c. p = .5, q = .45
d. The probability question is P(_______).
Solution 4.18
d. P(x = 4)
Try It 4.18
You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10 percent of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q?
## Notation for the Geometric: G = Geometric Probability Distribution Function
### Notation for the Geometric: G = Geometric Probability Distribution Function
X ~ G(p)
Read this as X is a random variable with a geometric distribution. The parameter is p; p = the probability of a success for each trial.
### Example 4.19
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?
Let X = the number of computer components tested until the first defect is found.
X takes on the values 1, 2, 3, . . . where p = .02. X ~ G(.02)
Find P(x = 7). There is a formula to define the probability of a geometric distribution $P(x)P(x)$. We can use the formula to find $P(x=7)P(x=7)$. But since the calculation is tedious and time consuming, people usually use a graphing calculator or software to get the answer. Using a graphing calculator, you can get $P(x=7)=.0177P(x=7)=.0177$. The instruction of TI83, 83+, 84, 84+ is given below.
### Using the TI-83, 83+, 84, 84+ Calculator
Go into 2nd DISTR. The syntax for the instructions are as follows:
To calculate the probability of a value P(x = value): use geometpdf(p, number). Here geometpdf represents geometric probability density function. It is used to find the probability that a geometric random variable is equal to an exact value. p is the probability of a success and number is the value.
To calculate the cumulative probability P(x ≤ value): use geometcdf(p, number). Here geometcdf represents geometric cumulative distribution function. It is used to determine the probability of “at most” type of problem, the probability that a geometric random variable is less than or equal to a value. p is the probability of a success and number is the value.
To find $P(x=7)P(x=7)$, enter 2nd DISTR, arrow down to geometpdf(. Press ENTER. Enter .02,7). The result is $P(x=7)=.0177P(x=7)=.0177$.
If we need to find $P(x≤7)P(x≤7)$ enter 2nd DISTR, arrow down to geometcdf(. Press ENTER. Enter .02,7). The result is $(x≤=7)=.1319(x≤=7)=.1319$.
The graph of X ~ G(.02) is
Figure 4.2
The previous probability distribution histogram gives all the probabilities of X. The x-axis of each bar is the value of X = the number of computer components tested until the first defect is found, and the height of that bar is the probability of that value occurring. For example, the x value of the first bar is 1 and the height of the first bar is 0.02. That means the probability that the first computer components tested is defective is .02.
The expected value or mean of X is $E(X)=μ=1p=1.02=50E(X)=μ=1p=1.02=50$.
The variance of X is $σ2=(1p)(1p−1)=(1.02)(1.02−1)=(50)(49)=2,450σ2=(1p)(1p−1)=(1.02)(1.02−1)=(50)(49)=2,450$.
The standard deviation of X is $σ=σ2=2,450=49.5σ=σ2=2,450=49.5$.
Here is how we interpret the mean and standard deviation. The number of components that you would expect to test until you find the first defective one is 50 (which is the mean). And you expect that to vary by about 50 computer components (which is the standard deviation) on average.
Try It 4.19
The probability of a defective steel rod is .01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer.
### Example 4.20
The lifetime risk of developing pancreatic cancer is about one in 78 (1.28 percent). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G$(178)(178)$ or X ~ G(.0128).
1. What is the probability that you ask 10 people before one says he or she has pancreatic cancer?
2. What is the probability that you must ask 20 people?
3. Find the (i) mean and (ii) standard deviation of X.
Solution 4.20
1. P(x = 10) = geometpdf(.0128, 10) = .0114
2. P(x = 20) = geometpdf(.0128, 20) = .01
1. Mean = μ = $1p1p$ = $1.01281.0128$ = 78
2. $σ=σ2=(1p)(1p−1)=(1.0128)(1.0128−1)=(78)(78−1)=6,006=77.4984≈77σ=σ2=(1p)(1p−1)=(1.0128)(1.0128−1)=(78)(78−1)=6,006=77.4984≈77$
The number of people whom you would expect to ask until one says he or she has pancreatic cancer is 78. And you expect that to vary by about 77 people on average.
Try It 4.20
The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12 percent. Let X = the number of Afghani women you ask until one says that she is literate.
1. What is the probability distribution of X?
2. What is the probability that you ask five women before one says she is literate?
3. What is the probability that you must ask 10 women?
4. Find the (i) mean and (ii) standard deviation of X.
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# Work
## There are three key ingredients to work - force, displacement, and cause.
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Calculating Work
Credit: CK-12 Foundation
In order for the roller coaster to run down the incline by gravitational attraction, it first must have work done on it towing it up to the top of the hill.
### Work
The word work has both an everyday meaning and a specific scientific meaning. In the everyday use of the word, work would refer to anything, which required a person to make an effort. In physics, however, work is defined as the force exerted on an object multiplied by the distance the object moves due to that force.
W=Fd\begin{align*}W = Fd\end{align*}
In the scientific definition of the word, if you push against an automobile with a force of 200 N for 3 minutes but the automobile does not move, then you have done NO work. Multiplying 200 N times 0 meters yields zero work. If you are holding an object in your arms, the upward force you are exerting is equal to the object’s weight. If you hold the object until your arms become very tired, you have still done no work. When you lift an object, you exert a force equal to the object’s weight and the object moves due to that lifting force. If an object weighs 200. N and you lift it 1.50 meters, then your work is, W=Fd=(200. N)(1.50 m)=300. N m\begin{align*}W = Fd = (200. \ \text{N})(1.50 \ \text{m}) = 300. \ \text{N m}\end{align*}.
Example Problem: A boy lifts a box of apples that weighs 185 N. The box is lifted a height of 0.800 m. How much work did the boy do?
Solution: W=Fd=(185 N)(0.800 m)=148 N m=148 Joules\begin{align*}W = Fd = (185 \ \text{N})(0.800 \ \text{m}) = 148 \ \text{N m} = 148 \ \text{Joules}\end{align*}
Work is done only if a force is exerted in the direction of motion. If the force and motion are perpendicular to each other, no work is done because there is no motion in the direction of the force. If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done.
Example Problem: Suppose a 125 N force is applied to a lawnmower handle at an angle of 25° with the ground and the lawnmower moves along the surface of the ground. If the lawnmower moves 56 m, how much work was done?
Solution: The solution requires that we determine the component of the force that was in the direction of the motion of the lawnmower. The component of the force that was pushing down on the ground does not contribute to the work done.
Fparallel=(Force)(cos25)=(125 N)(0.906)=113 N\begin{align*}F_{\text{parallel}} = (\text{Force})(\cos 25^\circ) = (125 \ \text{N})(0.906) = 113 \ \text{N}\end{align*}
W=Fparalleld=(113 N)(56 m)=630 J\begin{align*}W = F_{\text{parallel}} d = (113 \ \text{N})(56 \ \text{m}) = 630 \ \text{J}\end{align*}
#### Summary
• In physics, work is defined as the force exerted on an object multiplied by the distance the object moves due to that force.
• The unit for work is called the joule.
• If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done.
#### Practice
The following website contains practice questions with answers on the topic of work.
#### Review
1. How much work is done by the force of gravity when a 45 N object falls to the ground from a height of 4.6 m?
2. A workman carries some lumber up a staircase. The workman moves 9.6 m vertically and 22 m horizontally. If the lumber weighs 45 N, how much work was done by the workman?
3. A barge is pulled down a canal by a horse walking beside the canal. If the angle of the rope is 60.0°, the force exerted is 400. N, and the barge is pulled 100. M, how much work did the horse do?
### Notes/Highlights Having trouble? Report an issue.
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Welcome back! We’re glad you made it to Part 2 of this series. Let’s dive right in.
One of the most important statistical concepts is the distribution curve. A distribution curve is a graph that shows the different frequencies of a certain variable. For example, if I were to make a distribution curve of scores received on a test, the graph would show how many people received each score. The area under the distribution curve is the probability that you see a particular instance occur. Going back to our previous example, if I wanted to find the probability of a student receiving a score between 80 and 90, I would find the area under the curve in-between those two test scores. The total area under the curve is equal to 1. These concepts will be important later when discussing more about p-value in hypothesis testing.
In later posts we will frequently be referring to a specific type of distribution called a normal distribution, often referred to as a bell curve. It is referred to as a “bell curve” because it simply looks like a bell! A normal distribution curve is special because it is perfectly symmetrical.
Sample size actually plays a large role in distribution, especially normal distribution. First, let’s give sample size a brief definition: the number of data points or subjects in your sample, typically denoted as “n.” Another way to think about it is simply how much data you have. For example, if you conducted a survey of 50 students in a high school, your sample size would be 50. The larger the sample size, the more normal your distribution becomes as posited by the Central Limit Theorem. Since many variables are naturally distributed normally, the more normal your sample distribution becomes, the better it is at approximating population parameters. This is easier to conceptualize if you can actually see if for yourself. Below we have put together a demonstration that shows how increasing the sample size causes the distribution curve to be more normal.
For this demonstration, let’s use our example data set and look at the living room size. Let’s assume that the true population size is 21,613. We began by randomly drawing a sample of x number of living room sizes from the population and then graphed the sampled data in the form of a histogram to show the distribution. Note that the x axis represents the living room size in square feet, and the y axis represents the frequency in which it occurs.
SAMPLE SIZE DISTRIBUTION CURVE With 100 randomly selected sample data, our distribution curve has a wide spread and does not resemble a normal distribution curve at all. This indicates that a sample size of 100 is inadequate. With 500 randomly selected sample data, the distribution curve has decreased spread and begins to look more normal than that sample of 100. This is a better candidate than the sample of 100, but we can still get a better model by using more data points. With 1,000 randomly selected sample data, the data decreases its spread once again, as well as becomes approximately more normal. This is a better candidate than a sample of 500, but can become more normal with more data points. With 10,000 randomly selected sample data, this is where we see a normal distribution curve begin to truly emerge. Since the spread is decreasing, we know that the number of outliers is also decreasing. With 20,000 randomly selected data points, the curve looks about the same as the sample of 10,000, except it looks slightly more normal and has slightly less spread. This makes it a better candidate than 10,000 points, but likely only marginally. Here is the distribution curve of the “population” of 21613 data points. Obviously, this has the most minimal spread and the most normal distribution. The sample of 20,000 points most closely resembles this curve.
As we can see, as our sample size increased, the distribution curve resembled our population curve more accurately. Once we gathered a sample size of 10,000, an approximately normal distribution curve emerges and improved as we increased the sample size to 20,000. This shows why it is so important to have a large enough sample size if we want to draw valid conclusions about the population from our sample. Something to note, however, is that the sample of 10,000 data points produces a curve that looks very similar to the 20,000 and 21,613 curves; although the 20,000 technically best represents the population, we can say that we will have very similar, if not identical, results when cutting that sample size in half to 10,000 since their respective curves are mostly the same. This helps to save both time and resources since you will have to collect significantly less data.
More broadly, it also shows us why we do not necessarily need to use the full population if we want to accurately answer our questions. Why would I spend all of the time and money collecting 21,613 data points, when I could find the same conclusions with only, say, 10,000 data points? This is exactly why we use sample sizes and not populations for statistical exploration. As long as is is large enough, we can use a sample size that will accurately represent our target population.
But how do we know exactly what our sample size should be, and what is deemed large enough? Obviously we want as large of a sample size as possible, but due to finite time and resources, it may be necessary to simply find the minimum number of subjects required for valid conclusions about the population. Finding this value depends heavily on how accurate you want your results to reflect the population, which will largely be dictated by your margin of error. The smaller your margin of error, the larger your sample size will have to be. After setting a margin of error, you can use the following formulas to calculate the necessary sample size based on the type of method in use:
• If you are looking at the percentage / proportion of a population that contains a certain characteristic:
• p = the estimated percent of the population that contains the attribute in question
• q = p-1 = the estimated percent of the population that does not contain that attribute
• z* = z multiplier
• ex: 95% confidence level → = Z.INV(0.975) = 1.96 (don’t worry, we go through an example of how to do this step later in the section)
• If you are looking at a regression model, this one is a little bit more flexible and subjective
• Green (1991) → n >= 104 + k
• k = number of predictors
• Rule of thumb → 10 subjects for every one predictor
Note that if the calculated sample size has decimals, it is standard to round up to the nearest whole number, regardless of the value of the decimal. For example, a calculated sample size of 145.129 would be rounded up to 146, even though traditional rounding rules would indicate not to. Also note that margin of error must be expressed in decimal form when used in the formula. (ex: 5% → 0.05)
These formulas might seem like a foreign language if you’re non-technical, but don’t despair! They’re actually much more simple than they look. You have all of the information, you just need to plug it in and voila! All of the variables that you are plugging in are either numbers you already have or numbers you are determining yourself. If you are still a little bit confused, don’t worry! Before going on to the next section, we will run through an example for you. Let’s set the scene: you are trying to find out what percentage of houses in King County, WA have been renovated, but you are confused about how many houses are necessary to collect data on for accurate results. We will begin by finding each value we need, and then putting it together by plugging the numbers into the formula and finding our sample size. Let’s begin!
• p → we can estimate that approximately 50% (0.5) of the houses have been renovated; this is just an estimate, and does not need to be accurate
• q → 1-p → 1-0.5 = 0.5; we can estimate that approximately 50% (0.5) of the houses have not been renovated; once again, just an estimate, so don’t get too caught up in figuring out this value
• z* → let’s say we decided we want 95% confidence in our results, we would find our z* multiplier by calculating =Z.INV(0.975) in Excel, which gives us 1.96; 95% is a pretty standard confidence level to use
• Refer the Excel Guide for an explanation of why we input 0.975
• Margin of error → let’s say we decide we want our results to be within 3 percentage points of the true population parameters, then our margin of error would be 3%, but would be put in as 0.03 in the formula.
• NOTE: When we set our confidence level and margin of error, we are saying: “We want to have x% confidence that our results are within y percentage points of the true population parameter where x represents the confidence level and y represents the margin of error.”
• n = 1067.11 → round up → 1068 housing points
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NCERT Solutions for Class 11 Chapter 7- Permutations and Combinations
*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 6.
NCERT Solutions Class 11 Maths Chapter 7 Permutations and Combinations prepared in accordance with the latest update on CBSE Syllabus for 2023-24 have been provided here. A Permutation is nothing but each of several possible ways in which a set or number of things can be ordered or arranged. On the other hand, Combinations can be defined as a selection of all or part of a set of objects, irrespective of the order in which objects are selected. The NCERT Solutions of this chapter, designed by the subject matter experts at BYJU’S, cover all the exercise questions included in the book.
In this chapter, students learn about the topics mentioned earlier, Permutation and Combination, in detail. Students can easily score full marks on the questions from this chapter by practising the NCERT Class 11 Maths solutions for all the problems present in the NCERT textbook.
Access Answers of Maths NCERT Class 11 Chapter 7 – Permutations and Combinations
Access the NCERT Solutions for Class 11 Maths Chapter 7
Exercise 7.1 Page No: 138
1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5, assuming that
(i) Repetition of the digits is allowed?
(ii) Repetition of the digits is not allowed?
Solution:
(i)Â Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.
Now, when repetition is allowed,
The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.
Hence, the total number possible 3-digit numbers =5 × 5 × 5 =125
(ii)Â Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.
Now, when repetition is not allowed,
The number of digits possible at C is 5. Suppose one of 5 digits occupies place C; now, as the repletion is not allowed, the possible digits for place B are 4, and similarly, there are only 3 possible digits for place A.
Therefore, the total number of possible 3-digit numbers=5 × 4 × 3=60
2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, and 6 if the digits can be repeated?
Solution:
Let the 3-digit number be ABC, where C is at the unit’s place, B at the tens place and A at the hundreds place.
As the number has to be even, the digits possible at C are 2 or 4 or 6. That is, the number of possible digits at C is 3.
Now, as repetition is allowed, the digits possible at B is 6. Similarly, at A, also, the number of digits possible is 6.
Therefore, The total number of possible 3-digit numbers = 6 × 6 × 3 = 108
3. How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Solution:
Let the 4-letter code be 1234.
In the first place, the number of letters possible is 10.
Suppose any 1 of the ten occupies place 1.
Now, as repetition is not allowed, the number of letters possible at place 2 is 9. Now, at 1 and 2, any 2 of the 10 alphabets have been taken. The number of alphabets left for place 3 is 8, and similarly, the number of alphabets possible at 4 is 7.
Therefore, the total number of 4-letter codes=10 × 9 × 8 × 7=5040
4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Solution:
Let the five-digit number be ABCDE. Given that the first 2 digits of each number are 67. Therefore, the number is 67CDE.
As repetition is not allowed and 6 and 7 are already taken, the digits available for place C are 0,1,2,3,4,5,8,9. The number of possible digits at place C is 8. Suppose one of them is taken at C; now the digits possible at place D is 7. And similarly, at E, the possible digits are 6.
∴ The total five-digit numbers with given conditions = 8 × 7 × 6 = 336
5. A coin is tossed 3 times, and the outcomes are recorded. How many possible outcomes are there?
Solution:
Given A coin is tossed 3 times, and the outcomes are recorded.
The possible outcomes after a coin toss are head and tail.
The number of possible outcomes at each coin toss is 2.
∴ The total number of possible outcomes after 3 times = 2 × 2 × 2 = 8
6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Solution:
Given 5 flags of different colours.
We know the signal requires 2 flags.
The number of flags possible for the upper flag is 5.
Now, as one of the flags is taken, the number of flags remaining for the lower flag in the signal is 4.
The number of ways in which signal can be given = 5 × 4 = 20
Exercise 7.2 Page No: 140
1. Evaluate
(i) 8!
(ii) 4! – 3!
Solution:
(i)Â Consider 8!
We know that 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320
(ii)Â Consider 4!-3!
4!-3! = (4 × 3!) – 3!
The above equation can be written as
= 3! (4-1)
= 3 × 2 × 1 × 3
= 18
2. Is 3! + 4! = 7!?
Solution:
Consider LHS 3! + 4!
Computing the left-hand side, we get
3! + 4! = (3 × 2 × 1) + (4 × 3 × 2 × 1)
= 6 + 24
= 30
Again, considering RHS and computing, we get
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
Therefore, LHS ≠RHS
Therefore, 3! + 4! ≠7!
3. ComputeÂ
Solution:
4. If find x.
Solution:
5. Evaluate
 ,
When
(i) n = 6, r = 2
(ii) n = 9, r = 5
Solution:
Exercise 7.3 Page No: 148
1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Solution:
2. How many 4-digit numbers are there with no digit repeated?
Solution:
To find the four-digit number (digits do not repeat),
We will have 4 places where 4-digits are to be put.
So, at the thousand’s place = There are 9 ways as 0 cannot be at the thousand’s place = 9 ways
At the hundredth’s place = There are 9 digits to be filled as 1 digit is already taken = 9 ways
At the ten’s place = There are now 8 digits to be filled as 2 digits are already taken = 8 ways
At unit’s place = There are 7 digits that can be filled = 7 ways
The total number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways
So, a total of 4536 four-digit numbers can be there with no digits repeated.
3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution:
An even number means that the last digit should be even.
The number of possible digits at one’s place = 3 (2, 4 and 6)
⇒ Number of permutations=
One of the digits is taken at one’s place; the number of possible digits available = 5
⇒ Number of permutations=
Therefore, the total number of permutations =3 × 20=60
4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, if no digit is repeated. How many of these will be even?
Solution:
Total number of digits possible for choosing = 5
Number of places for which a digit has to be taken = 4
As there is no repetition allowed,
⇒ Number of permutations =
The number will be even when 2 and 4 are in one’s place.
The possibility of (2, 4) at one’s place = 2/5 = 0.4
The total number of even numbers = 120 × 0.4 = 48
5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman, assuming one person cannot hold more than one position?
Solution:
Total number of people in committee = 8
Number of positions to be filled = 2
⇒ Number of permutations =
6. Find n if n-1P3: nP3 = 1: 9.
Solution:
7. Find r if
(i)5Pr = 26Pr-1Â
(ii) 5Pr = 6Pr-1
Solution:
8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Solution:
Total number of different letters in EQUATION = 8
Number of letters to be used to form a word = 8
⇒ Number of permutations =
9. How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ii) All letters are used at a time,
(iii) All letters are used, but the first letter is a vowel.
Solution:
(i) Number of letters to be used =4
⇒ Number of permutations =
(ii) Number of letters to be used = 6
⇒ Number of permutations =
(iii) Number of vowels in MONDAY = 2 (O and A)
⇒ Number of permutations in vowel =
Now, the remaining places = 5
Remaining letters to be used =5
⇒ Number of permutations =
Therefore, the total number of permutations = 2 × 120 =240
10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Solution:
Total number of letters in MISSISSIPPI =11
Letter Number of occurrence
M 1 I 4 S 4 P 2
⇒ Number of permutations =
We take that 4 I’s come together, and they are treated as 1 letter,
∴ Total number of letters=11 – 4 + 1 = 8
⇒ Number of permutations =
Therefore, total number of permutations where four I’s don’t come together = 34650-840=33810
11. In how many ways can the letters of the word PERMUTATIONS be arranged if the
(ii) Vowels are all together,
(iii) There are always 4 letters between P and S?
Solution:
(i) Total number of letters in PERMUTATIONS =12
The only repeated letter is T; 2times
The first and last letters of the word are fixed as P and S, respectively.
Number of letters remaining =12 – 2 = 10
⇒ Number of permutations =
(ii) Number of vowels in PERMUTATIONS = 5 (E, U, A, I, O)
Now, we consider all the vowels together as one.
Number of permutations of vowels = 120
Now, the total number of letters = 12 – 5 + 1= 8
⇒ Number of permutations =
Therefore, the total number of permutations = 120 × 20160 = 2419200
(iii) The number of places is as 1 2 3 4 5 6 7 8 9 10 11 12
There should always be 4 letters between P and S.
Possible places of P and S are 1 and 6, 2 and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12
Possible ways =7,
Also, P and S can be interchanged,
No. of permutations =2 × 7 =14
The remaining 10 places can be filled with 10 remaining letters,
∴ No. of permutations =
Therefore, the total number of permutations = 14 × 1814400 =25401600
Exercise 7.4 Page No: 153
1. If nC8 = nC2, find nC2.
Solution:
2. Determine n if
(i) 2nC3: nC3 = 12: 1
(ii) 2nC3: nC3 = 11: 1
Solution:
Simplifying and computing
⇒ 4 × (2n – 1) = 12 × (n – 2)
⇒ 8n – 4 = 12n – 24
⇒ 12n – 8n = 24 – 4
⇒ 4n = 20
∴ n = 5
⇒ 11n – 8n = 22 – 4
⇒ 3n = 18
∴ n = 6
3. How many chords can be drawn through 21 points on a circle?
Solution:
Given 21 points on a circle.
We know that we require two points on the circle to draw a chord.
∴ The number of chords is are
⇒ 21C2=
∴ The total number of chords that can be drawn is 210
4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Solution:
Given 5 boys and 4 girls in total.
We can select 3 boys from 5 boys in 5C3 ways.
Similarly, we can select 3 boys from 54 girls in 4C3 ways.
∴ The number of ways a team of 3 boys and 3 girls can be selected is 5C3 × 4C3
⇒ 5C3 × 4C3 =
⇒ 5C3 × 4C3 = 10 × 4 = 40
∴ The number of ways a team of 3 boys and 3 girls can be selected is 5C3 × 4C3 = 40 ways
5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution:
Given 6 red balls, 5 white balls and 5 blue balls.
We can select 3 red balls from 6 red balls in 6C3 ways.
Similarly, we can select 3 white balls from 5 white balls in 5C3 ways.
Similarly, we can select 3 blue balls from 5 blue balls in 5C3 ways.
∴ The number of ways of selecting 9 balls is 6C3 ×5C3 × 5C3
∴ The number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour is 6C3 ×5C3 × 5C3 = 2000
6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Solution:
Given a deck of 52 cards.
There are 4 Ace cards in a deck of 52 cards.
According to the question, we need to select 1 Ace card out of the 4 Ace cards.
∴ The number of ways to select 1 Ace from 4 Ace cards is 4C1
⇒ More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards)
∴ The number of ways to select 4 cards from 48 cards is 48C4
∴ The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination is 778320.
7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Solution:
Given 17 players, in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers.
There are 5 players that can bowl, and we can require 4 bowlers in a team of 11.
∴ The number of ways in which bowlers can be selected is: 5C4
Now, other players left are = 17 – 5(bowlers) = 12
Since we need 11 players in a team and already 4 bowlers have been selected, we need to select 7 more players from 12.
∴ The number of ways we can select these players is: 12C7
∴ The total number of combinations possible is: 5C4 × 12C7
∴ The number of ways we can select a team of 11 players where 4 players are bowlers from 17 players is 3960.
8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Solution:
Given a bag contains 5 black and 6 red balls
The number of ways we can select 2 black balls from 5 black balls is 5C2
The number of ways we can select 3 red balls from 6 red balls is 6C3
The number of ways 2 black and 3 red balls can be selected is 5C2× 6C3
∴ The number of ways in which 2 black and 3 red balls can be selected from 5 black and 6 red balls is 200.
9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Solution:
Given 9 courses are available and 2 specific courses are compulsory for every student.
Here, 2 courses are compulsory out of 9 courses, so a student needs to select 5 – 2 = 3 courses
∴ The number of ways in which 3 ways can be selected from 9 – 2(compulsory courses) = 7 are 7C3
∴ The number of ways a student selects 5 courses from 9 courses where 2 specific courses are compulsory is 35.
Miscellaneous Exercise Page No: 156
1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Solution:
The word DAUGHTER has 3 vowels A, E, and U and 5 consonants D, G, H, T and R.
The three vowels can be chosen in 3C2 as only two vowels are to be chosen.
Similarly, the five consonants can be chosen in 5C3Â ways.
∴ The number of choosing 2 vowels and 5 consonants would be 3C2 ×5C3
= 30
∴ The total number of ways of is 30.
Each of these 5 letters can be arranged in 5 ways to form different words =Â 5P5
Total number of words formed would be = 30 × 120 = 3600
2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Solution:
In the word EQUATION, there are 5 vowels (A, E, I, O, U) and 3 consonants (Q, T, N).
The numbers of ways in which 5 vowels can be arranged are 5C5
…………… (i)
Similarly, the numbers of ways in which 3 consonants can be arranged are 3P3
…………….. (ii)
There are two ways in which vowels and consonants can appear together.
(AEIOU) (QTN) or (QTN) (AEIOU)
∴ The total number of ways in which vowel and consonant can appear together are 2 × 5C5 × 3C3
∴ 2 × 120 × 6 = 1440
3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) Exactly 3 girls?
(ii) At least 3 girls?
(iii) At most 3 girls?
Solution:
(i) Given exactly 3 girls.
The total numbers of girls are 4.
Out of which, 3 are to be chosen.
∴ The number of ways in which choice would be made = 4C3
Numbers of boys are 9 out of which 4 are to be chosen which is given by 9C4
Total ways of forming the committee with exactly three girls.
= 4C3 × 9C4
=Â
(ii) Given at least 3 girls.
There are two possibilities for making a committee choosing at least 3 girls.
There are 3 girls and 4 boys, or there are 4 girls and 3 boys.
Choosing three girls we have done in (i)
Choosing four girls and 3 boys would be done in 4C4 ways.
And choosing 3 boys would be done in 9C3
Total ways = 4C4 ×9C3
The total number of ways of making the committee are
504 + 84 = 588
(iii) Given at most 3 girls
In this case, the numbers of possibilities are
0 girl and 7 boys
1 girl and 6 boys
2 girls and 5 boys
3 girls and 4 boys
Number of ways to choose 0 girl and 7 boys = 4C0 × 9C7
The number of choosing 3 girls and 4 boys has been done in (1)
= 504
The total number of ways in which a committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632
4. If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starts with E?
Solution:
In a dictionary, words are listed alphabetically, so to find the words
Listed before E should start with the letter either A, B, C or D.
But the word EXAMINATION doesn’t have B, C or D.
The remaining 10 places are to be filled in by the remaining letters of the word EXAMINATION which are E, X, A, M, 2N, T, 2I, 0
Since the letters are repeating, the formula used would be
Where n is the remaining number of letters, p1Â and p2 are the number of times the repeated terms occurs.
The number of words in the list before the word starting with E
= words starting with letter A = 907200
5. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9, which are divisible by 10 and no digit is repeated?
Solution:
The number is divisible by 10 if the unit place has 0 in it.
The 6-digit number is to be formed out of which unit place is fixed as 0.
The remaining 5 places can be filled by 1, 3, 5, 7 and 9.
Here, n = 5
And the numbers of choice available are 5.
So, the total ways in which the rest of the places can be filled are 5P5
6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Solution:
We know that there are 5 vowels and 21 consonants in the English alphabet.
Choosing two vowels out of 5 would be done in 5C2 ways.
Choosing 2 consonants out of 21 can be done in 21C2 ways.
The total number of ways to select 2 vowels and 2 consonants
= 5C2 × 21C2
Each of these four letters can be arranged in four ways 4P4
Total numbers of words that can be formed are
24 × 2100 = 50400
7. In an examination, a question paper consists of 12 questions divided into two parts, i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Solution:
The student can choose 3 questions from part I and 5 from part II
Or
4 questions from part I and 4 from part II
5 questions from part 1 and 3 from part II
8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Solution:
We have a deck of cards that has 4 kings.
The numbers of remaining cards are 52.
Ways of selecting a king from the deck =Â 4C1
Ways of selecting the remaining 4 cards from 48 cards=Â 48C4
The total number of selecting the 5 cards having one king always
= 4C1 × 48C4
9. It is required to seat 5 men and 4 women in a row so that the women occupy even places. How many such arrangements are possible?
Solution:
Given there is a total of 9 people.
Women occupy even places, which means they will be sitting in 2nd, 4th, 6th and 8th place where as men will be sitting in 1st, 3rd, 5th,7th and 9th place.
4 women can sit in four places and ways they can be seated=Â 4P4
5 men can occupy 5 seats in 5 ways.
The number of ways in which these can be seated = 5P5
The total numbers of sitting arrangements possible are
24 × 120 = 2880
10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Solution:
In this question, we get 2 options, which are
(i) Either all 3 will go
Then, the remaining students in the class are: 25 – 3 = 22
The number of students remained to be chosen for party = 7
Number of ways to choose the remaining 22 students = 22C7
=
(ii) None of them will go
The students going will be 10.
Remaining students eligible for going = 22
The number of ways in which these 10 students can be selected are 22C10
The total number of ways in which students can be chosen is
= 170544 + 646646 = 817190
11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Solution:
In the given word ASSASSINATION, there are 4 ‘S’. Since all the 4 ‘S’ have to be arranged together, let us take them as one unit.
The remaining letters are= 3 ‘A’, 2 ‘I’, 2 ‘N’, T
The number of letters to be arranged is 9 (including 4 ‘S’).
Using the formula
where n is the number of terms and p1, p2Â p3Â are the number of times the repeating letters repeat themselves.
Here, p1= 3, p2= 2, p3Â = 2
Putting the values in formula we get
NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations
The major concepts of Maths covered in Chapter 7 – Permutations and Combinations of NCERT Solutions for Class 11 include:
7.1 Introduction
7.2 Fundamental Principle of Counting
7.3 Permutations
7.3.1 Permutations when all the objects are distinct
7.3.2 Factorial notation
7.3.3 Derivation of the formula for nPr
7.3.4 Permutations when all the objects are not distinct objects
7.4 Combinations
Exercise 7.1 Solutions 6 Questions
Exercise 7.2 Solutions 5 Questions
Exercise 7.3 Solutions 11 Questions
Exercise 7.4 Solutions 9 Questions
Miscellaneous Exercise on Chapter 7 Solutions 11 Questions
NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations
The chapter Permutations and Combinations belongs to the unit Algebra, which adds up to 30 marks of the total 80 marks. There are 4 exercises along with a miscellaneous exercise in this chapter to help students understand the concepts related to Permutations and Combinations clearly. Some of the topics discussed in Chapter 7 of NCERT Solutions for Class 11 Maths are as follows:
1. The fundamental principle of counting is if an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrences of the events in the given order is m × n.
2. The number of permutations of n different things taken r at a time, where repetition is not allowed, is denoted by nPr
3. n! = 1 × 2 × 3 × …×n
4. n! = n × (n – 1) !
5. The number of permutations of n different things, taken r at a time, where repetition is allowed, is nr
Hence, we can conclude that studying the Permutations and Combinations of Class 11 enables the students to understand the Fundamental principle of counting, Factorial (n!), Permutations and combinations, derivation of Formulae for nPr and nCr and their connections and the simple applications of permutations and combinations.
Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 7
Q1
What are the crucial topics covered in the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations?
The crucial topics covered in the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations are
1. Introduction
2. Fundamental Principle of Counting
3. Permutations
4. Combinations
Q2
Discuss the fundamental concepts of Permutations and Combinations in the NCERT Solutions for Class 11 Maths Chapter 7.
Permutation is the several ways in which a number or a set of things can be arranged, whereas combination is a specific arrangement of different elements. Based on some of the primary concepts, the fundamental theorem of counting is the first one. For example, if an event is occurring in ‘n’ different ways and another event is happening in ‘m’ different ways, the net occurrences can be written in the order of n x m.
The second one explains that the number of permutations ‘n’ separate things at ‘r’ time, where the repetition does not take place, can be written as nPr and when repetition takes place is denoted as nr.
Q3
Can I get the NCERT Solutions for Class 11 Maths Chapter 7 in PDF format?
Yes, students can download the NCERT Solutions for Class 11 Maths Chapter 7 in PDF format from BYJU’S website. The solutions for the exercise-wise problems are designed by the highly experienced faculty based on the latest CBSE Syllabus. Students can also cross-check their answers with the solutions PDF in order to get a clear idea about the other methods of solving complex problems effortlessly.
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# Exponents and Roots Practice Problems: GED Math (page 2)
By
Updated on Mar 23, 2011
1. e. 72 = 7 × 7, and 7 × 7 = 49.
2. d. When there is a negative exponent, take the reciprocal of the base and raise it to the positive power. 2–3, = , and .
3. b. is the cube root of 27; what number multiplied by itself twice equals 27? Because 3 × 3 × 3 = 27, the answer is 3.
4. e. An exponent of means the second root or square root of 25. Because 5 × 5 = 25, the answer is 5.
5. a. When you multiply two numbers with the same base, you keep the base and add the exponents. 22 × 23 = 22 + 3 = 25.
6. c. Ten with a negative exponent of 4 dictates that you move the decimal point four places to the left.
7. b. Change the large number to be a decimal number between 1 and 10, followed by multiplication by a power of 10. By doing this, you have moved the decimal point eight places to the left.
8. c. Divide: 5.4 ÷ 9 = 0.6. Then, use the law of exponents: 1016 ÷ 1014 = 1016 – 14 = 102. So this is 0.6 × 102 = 60 by moving the decimal point two places to the right.
9. d. Evaluate the exponent first and then evaluate the negative sign: 52 = 5 × 5, which is 25, so the answer is –25.
10. c. For order of operations, parentheses would have been evaluated first, but there are no parentheses. Evaluate exponents next to get 700 + 25 – 25 × 2. Multiplication is performed next: 700 + 25 – 50. Now, addition and subtraction are done left to right for a result of 725 – 50 = 675.
1. b. For this problem, the exponent is handled first. Eleven squared is 11 times 11, which is 121, and then the negative sign is evaluated to get the answer of –121.
2. b. The exponent of means the fourth root of 81; what factor multiplied by itself three times will yield 81? Trial and error will show that 3 × 3 × 3 × 3 = 81. The root is 3.
3. a. The problem is asking for the cube root of 64; what number multiplied by itself twice will give the product of 64? Because 4 × 4 × 4 = 64, the cube root is 4.
4. a. The problem is asking for the square root of 169. Because 13 times 13 equals 169, the root is 13.
5. c. Combine these radicals by multiplying the radicands together: . The square root of 144 is 12, because 12 times 12 equals 144.
6. e. Break this fractional radicand up into two separate radicals and then simplify what can be simplified. . The is simplified, and = 3. So, the answer is .
7. b. The exponent of 7 on the power of 10 dictates that you move the decimal point in 2.701 seven places to the right. Three of the places will be taken up by the digits 7, 0, and 1, and then four more zeros will follow to result in 27,010,000.
8. c. The negative exponent, –5, on the power of 10 means that you must move the decimal point five places to the left. The number 4.09 has only one digit to the left of the decimal point. Four leading zeros must be added as placeholders: 0.0000409.
9. c. For this problem, you use the commutative and associative properties of multiplication and change the order of the factors to get (2.5 × 3.0) × (10–4 × 108). 2.5 times 3 equals 7.5. For the powers of 10, when you multiply two powers with the same base, you keep the base and add the exponents, which results in 10–4 × 108 = 10–4 + 8 = 104. Now, 7.5 × 104 = 75,000 because you move the decimal point four places to the right, one place being the 5 (in the tenths place) followed by three trailing zeros for placeholders.
10. c. First, perform the addition enclosed in the parentheses: 5 – (–10)2 × 3. Now, take negative 10 and square it, which is –10 × –10 = 100, so the problem becomes 5 – 100 × 3. Now, multiply: 5 – 300. Finally, subtract to get –295.
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## Trigonometry (11th Edition) Clone
The solution set is $$\{\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\}$$
$$\sin x\cos x=\frac{1}{4}$$ over interval $[0,2\pi)$ This equation features both sine and cosine of the same functions of $x$, and they are in the position of a multiplication. In fact, it recalls the identity: $$2\sin x\cos x=\sin 2x$$ $$\sin x\cos x=\frac{\sin 2x}{2}$$ Therefore, the given equation would now become $$\frac{\sin2x}{2}=\frac{1}{4}$$ $$\sin2x=\frac{1}{2}$$ Using the identity $2\sin x\cos x=\sin2x$ helps solve this exercise, since if we keep both sine and cosine, there is no known way to solve it (except probably graphing). 1) Find corresponding interval for $2x$: The interval for $x$ is $[0,2\pi)$. In other words, $$0\le x\lt2\pi$$ Thus, with $2x$, the inequality would be $$0\le2x\lt4\pi$$ So the interval for $2x$ is $[0,4\pi)$ 2) Now consider back the equation $$\sin2x=\frac{1}{2}$$ Over interval $[0,4\pi)$, there are 4 values whose $\sin$ equals $\frac{1}{2}$, which are $\{\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\}$ Therefore, $$2x\in\{\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\}$$ That means $$x\in\{\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\}$$
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# What is Factorial of Hundred
Factorial of hundred, denoted as 100!, is the product of all positive integers from 1 to 100. This number is quite large, with 158 digits in total. Factorial of hundred is often used in mathematics and computer science, and it has various applications in statistics, probability, and combinatorics.
In this article, we will explore the concept of factorial of hundred in detail, and provide you with the necessary information to understand its properties, uses, and calculation methods. We will also answer some frequently asked questions about factorial of hundred, and give you some tips on how to optimize your website’s content for search engines.
## What is Factorial of Hundred
### Properties of Factorial of Hundred
Factorial of hundred is an example of a factorial, which is a function that assigns to each positive integer n the product of all positive integers up to and including n. The factorial function is denoted by the symbol !, and it has the following properties:
1. Factorial of one is equal to one, i.e., 1! = 1.
2. Factorial of zero is equal to one, i.e., 0! = 1.
3. Factorial of a positive integer n is equal to n times factorial of (n-1), i.e., n! = n * (n-1)!.
Using these properties, we can calculate the factorial of hundred as follows:
100! = 100 * 99 * 98 * … * 3 * 2 * 1
This calculation is not practical to perform by hand, due to the large number of digits involved. However, there are various algorithms and software programs that can calculate the factorial of hundred and other large numbers with high accuracy.
### Applications of Factorial of Hundred
Factorial of hundred has various applications in mathematics, computer science, and other fields. Some of its most common uses include:
• Combinatorics: Factorial of hundred represents the number of ways in which 100 distinct objects can be arranged in a line, or the number of permutations of 100 objects.
• Probability: Factorial of hundred is used in the calculation of binomial coefficients, which represent the number of ways in which k objects can be chosen from a set of n objects.
• Number theory: Factorial of hundred has various properties related to prime numbers, divisors, and modular arithmetic.
• Algorithms: Factorial of hundred is used in the analysis of time and space complexity of algorithms that involve permutations or combinations of objects.
### Calculating Factorial of Hundred
As mentioned earlier, calculating the factorial of hundred by hand is not practical due to the large number of digits involved. However, there are various methods and tools that can be used to calculate the factorial of hundred and other large numbers.
One common method is to use the Stirling’s approximation, which is an asymptotic formula that approximates the factorial of a large number n as follows:
n! ≈ sqrt(2πn) * (n/e)^n
Using this formula, we can approximate the factorial of hundred as follows:
100! ≈ sqrt(2π*100) * (100/e)^100
100! ≈ 9.324847 * 10^157
This approximation is quite accurate, with an error of less than 1% compared to the actual value of 100! Therefore, it can be used in various applications that require an estimation of the factorial of hundred.
## FAQs about Factorial of Hundred
What is the largest factorial that can be calculated?
Answer: The largest factorial that can be calculated depends on the computing power and memory of the system used for the calculation. However, with modern computers and software, factorials of several hundred or thousand can be calculated efficiently.
What is the significance of factorial of hundred in statistics and probability?
Answer: Factorial of hundred is used in the calculation of binomial coefficients, which represent the number of ways in which k objects can be chosen from a set of n objects. This is a fundamental concept in statistics and probability theory, and it has various applications in data analysis, hypothesis testing, and decision making.
How is the factorial function related to permutations and combinations?
Answer: The factorial function represents the number of ways in which n distinct objects can be arranged in a line, or the number of permutations of n objects. The number of combinations of k objects chosen from a set of n objects can be calculated as n! / (k! * (n-k)!), which is a binomial coefficient.
Are there any real-world applications of factorial of hundred?
Answer: Factorial of hundred has various applications in mathematics, computer science, and other fields, as mentioned earlier. However, it is not a number that commonly arises in everyday life or practical applications, due to its extremely large size.
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On a particular day, the rainfall recorded in the terrace is 6m long and 5m board is $15cm$. The quantity of water collected in then terrace isA) $300{\text{ }}litres$B) $450{\text{ }}litres$C) $3000{\text{ }}litres$D) $4500{\text{ }}litres$s
Last updated date: 20th Sep 2024
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Hint: In the given question, we are asked that on a particular day, the rainfall recorded in a terrace having some specific dimensions. Therefore we are to find out the quantity of water recorded in the terrace. Firstly we will check the dimensions and shape of the terrace and then use the formula to find out the quantity of water collected in the terrace.
In the given question, we are given that terrace is having dimensions than on a particular day, rainfall is recorded or collected on the terrace which is found to be$\;6m$long, $5m$ broad and high which means the length of rainfall recorded in the terrace is $\;6m$ long, breadth of rainfall recorded is $5m$ broad and height of rainfall recorded is $15cm$ high.
Now, we have to find out the quantity of water collected on the terrace. Since length, breadth and height are given. Therefore we have to find out the quantity or volume of the rainfall recorded.
Given $length{\text{ }} = {\text{ }}6m,{\text{ }}breadth{\text{ }} = {\text{ }}5m{\text{ }}and{\text{ }}height{\text{ }} = {\text{ }}5cm$. Firstly we have to convert height from $cm$ to $m$
Also $1m{\text{ }} = {\text{ }}100cm$
Therefore $1cm{\text{ }} = {\text{ }}\dfrac{1}{{100}}m$
Now $15cm$ in height
Therefore $15cm = \dfrac{{15}}{{100}} = 0.15m$
Now, Volume = $length \times breadth \times height$
$\Rightarrow \left( {6m \times 5m \times 0.15m} \right)$
$\Rightarrow 4.5{m^3}$
Since quantity to find out. Therefore ${m^3}$ should be converted to litre. Also we know the relationship between ${m^3}$ and litre.
So, $1{m^3} = {\text{ }}1000{\text{ }}litre$
$\Rightarrow Volume{\text{ }} = 4.5 \times 1000{\text{ }}litre$
$\Rightarrow 4500{\text{ }}litres$
Therefore, a quantity of water collected in the terrace is 4500 litre
So option (D) is correct.
Note: The rainfall water is collected in the terrace since the terrace is in the form of a cuboid. So rainwater collected on the terrace also took the shape of a cuboid. This can be explained as water collected in the cylinder took the shape of a cylinder. Also, the same water collected in the spherical bowl took the shape of a sphere.
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# Find the area of the quadrant of a circle whose circumference is 44 cm.
Last updated date: 23rd May 2024
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Hint: In circles, if we have a circle of radius r then the circumference of this circle is given by the formula 2$\pi$r and the area of this circle is given by the formula $\pi {{r}^{2}}$.Here,quadrant means $\dfrac{1}{4}$ of circle.Using this, we can solve this question.
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
Let us consider a circle of radius r. The circumference of this circle is given by the formula,
$C=2\pi r$ . . . . . . . . . . . . (1)
The area of the circle is given by the formula,
$A=\pi {{r}^{2}}$ . . . . . . . . . . . . . . . (2)
In this question, we are given a circle having its circumference = 44 cm. Let us assume r be the radius of this circle. Substituting C = 44 in formula (1), we get,
$44=2\pi r$
$\Rightarrow r=\dfrac{22}{\pi }$
Substituting this value of r in the formula (2), we get the area of this circle equal to,
\begin{align} & A=\pi {{\left( \dfrac{22}{\pi } \right)}^{2}} \\ & \Rightarrow A=\dfrac{{{22}^{2}}}{\pi } \\ \end{align}
Substituting $\pi =\dfrac{22}{7}$, we get,
\begin{align} & A=\dfrac{{{22}^{2}}}{\dfrac{22}{7}} \\ & \Rightarrow A=22\times 7 \\ & \Rightarrow A=154c{{m}^{2}} \\ \end{align}
Since we are required to find the area of the quadrant of this circle, we have to divide the above area by 4. So, the area of the quadrant of this circle is equal to $\dfrac{1}{4}\left( 154 \right)=38.5c{{m}^{2}}$.
Hence, the answer is $38.5c{{m}^{2}}$.
Note: There is a possibility that one may write the area of the whole circle instead of the area of the quadrant of this circle as the answer. So, in order to avoid such types of mistakes, one must read the question carefully.
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# How do you find the derivative of f(x)= x/(x-1)?
Apr 26, 2016
$f ' \left(x\right) = - \frac{1}{x - 1} ^ 2$
#### Explanation:
We can use the Quotient Rule to find the derivative, which states:
$y ' = \frac{g \left(x\right) h ' \left(x\right) - h \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$
Where in our case,
$h \left(x\right) = x$ and $g \left(x\right) = x - 1$
We then need to find the derivative of each of these:
$h ' \left(x\right) = 1$ and $g ' \left(x\right) = 1$
Reassembling this we find our derivative:
$f ' \left(x\right) = \frac{\left(x - 1\right) - x}{x - 1} ^ 2 = - \frac{1}{x - 1} ^ 2$
Apr 26, 2016
$f ' \left(x\right) = - \frac{1}{x - 1} ^ 2$
#### Explanation:
In a previous answer, we used the quotient rule which is a simplification of the product rule. Some people, including me, don't like remembering more than one rule, so let's do that again using the product rule , which states:
$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$
in our case
$g \left(x\right) = x$ and $h \left(x\right) = {\left(x - 1\right)}^{- 1}$
Now we need to find the derivatives of these functions:
$g ' \left(x\right) = 1$ and $h ' \left(x\right) = - {\left(x - 1\right)}^{- 2}$
So reassembling we find our solution:
$f ' \left(x\right) = \frac{1}{x - 1} - \frac{x}{x - 1} ^ 2 = - \frac{1}{x - 1} ^ 2$
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# Revision Notes on Areas Related to Circles
## Perimeter and Area of a Circle
The perimeter of all the plain figures is the outer boundary of the figure. Likewise, the outer boundary of the circle is the perimeter of the circle. The perimeter of circle is also called the Circumference of the Circle.
Circumference = 2πr = πd (π = 22/7)
r = Radius and d = 2r
The area is the region enclosed by the circumference.
Area of the circle = πr2
### Example
If a pizza is cut in such a way that it divides into 8 equal parts as shown in the figure, then what is the area of each piece of the pizza? The radius of the circle shaped pizza is 7 cm.
### Solution
The pizza is divided into 8 equal parts, so the area of each piece is equal.
Area of 1 piece = 1/8 of area of circle
## Perimeter and Area of the Semi-circle
The perimeter of the semi-circle is half of the circumference of the given circle plus the length of diameter as the perimeter is the outer boundary of the figure.
Area of the semi-circle is just half of the area of the circle.
## Area of a Ring
Area of the ring i.e. the coloured part in the above figure is calculated by subtracting the area of the inner circle from the area of the bigger circle.
Where, R = radius of outer circle
r = radius of inner circle
## Areas of Sectors of a Circle
The area formed by an arc and the two radii joining the endpoints of the arc is called Sector.
### Minor Sector
The area including ∠AOB with point C is called Minor Sector. So OACB is the minor sector. ∠AOB is the angle of the minor sector.
### Major Sector
The area including ∠AOB with point D is called the Major Sector. So OADB is the major sector. The angle of the major sector is 360° – ∠AOB.
Area of Major Sector = πr2 - Area of the Minor Sector
Remark: Area of Minor Sector + Area of Major Sector = Area of the Circle
## Length of an Arc of a Sector of Angle θ
An arc is the piece of the circumference of the circle so an arc can be calculated as the θ part of the circumference.
### Areas of Segments of the Circle
The area made by an arc and a chord is called the Segment of the Circle.
### Minor Segment
The area made by chord AB and arc X is the minor segment. The area of the minor segment can be calculated by
Area of Minor Segment = Area of Minor Sector – Area of ∆ABO
### Major Segment
The other part of the circle except for the area of the minor segment is called a Major Segment.
Area of Major Segment = πr2 - Area of Minor Segment
Remark: Area of major segment + Area of minor segment = Area of circle
## Areas of Combinations of Plane Figures
As we know how to calculate the area of different shapes, so we can find the area of the figures which are made with the combination of different figures.
### Example
Find the area of the colored part if the given triangle is equilateral and its area is 17320.5 cm2. Three circles are made by taking the vertex of the triangles as the centre of the circle and the radius of the circle is the half of the length of the side of the triangle. (π = 3.14 and √3 = 1.73205)
Solution
Given
ABC is an equilateral triangle, so ∠A, ∠B, ∠C = 60°
Hence the three sectors are equal, of angle 60°.
Required
To find the area of the shaded region.
Area of shaded region =Area of ∆ABC – Area of 3 sectors
Area of ∆ABC = 17320.5 cm2
Side = 200 cm
As the radius of the circle is half of the length of the triangle, so
Area of 3 Sectors = 3 × 15700/3 cm2 cm2
Area of shaded region = Area of ∆ABC – Area of 3 sectors
= 17320.5 - 15700 cm2
= 1620.5 cm2
### Example
Find the area of the shaded part, if the side of the square is 8 cm and the 44 cm.
### Solution
Required region = Area of circle – Area of square
= πr2 – (side)2
Circumference of circle = 2πr = 44
Radius of the circle = 7 cm
Area of circle = πr2
Area of square = (side) 2 = (8)2 = 64 cm2
Area of shaded region = Area of circle – Area of square
= 154 cm2 - 64 cm2
= 90 cm2
### Course Features
• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
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# HiSET Math Practice Test Questions
Taking the HiSET Math test? Try the following FREE HiSET Math sample questions. Reviewing practice questions is the best way to brush up on your Math skills and prepare for the HiSET Math test. Here, we walk you through solving 10 common HiSET Math practice problems covering the most important math topics on the HiSET test.
These HiSET Math practice questions are designed to be similar to those found on the real HiSET Math test. They will assess your level of preparation and will give you a better idea of what to study on your exam.
## 10 Sample HiSET Math Practice Questions
1- Which of the following could be the product of two consecutive prime numbers?
A. 2
B. 10
C. 14
D. 15
E. 20
2- Sophia purchased a sofa for $530.40. The sofa is regularly priced at$624. What was the percent discount Sophia received on the sofa?
A. $$12\%$$
B. $$15\%$$
C. $$20\%$$
D. $$25\%$$
E. $$40\%$$
3- The score of Emma was half as that of Ava and the score of Mia was twice that of Ava. If the score of Mia was 60, what is the score, Emma?
A. 12
B. 15
C. 20
D. 30
E. 40
4- A bag contains 18 balls: two green, five black, light blue, a brown, a red, and one white. If 17 balls are removed from the bag at random, what is the probability that a brown ball has been removed?
A. $$\frac{1}{9}$$
B. $$\frac{1}{6}$$
C. $$\frac{16}{17}$$
D. $$\frac{17}{18}$$
E. $$\frac{1}{2}$$
5- The average of five consecutive numbers is 38. What is the smallest number?
A. 38
B. 36
C. 34
D. 12
E. 8
6- How many tiles of 8 cm$$^2$$ is needed to cover a floor of dimension 6 cm by 24 cm?
☐A. 6
☐B. 12
☐C. 18
☐D. 24
☐E. 36
7- A rope weighs 600 grams per meter in length. What is the weight in kilograms of 12.2 meters of this rope? (1 kilograms $$=$$ 1000 grams)
A. 0.0732
B. 0.732
C. 7.32
D. 7320
E. 73200
8- A chemical solution contains $$4\%$$ alcohol. If there are 24 ml of alcohol, what is the volume of the solution?
A. 240 ml
B. 480 ml
C. 600 ml
D. 1200 ml
E.2400 ml
9-The average weight of 18 girls in a class is 60 kg and the average weight of 32 boys in the same class is 62 kg. What is the average weight of all the 50 students in that class?
A. 60
B. 61.28
C. 61.68
D. 61.90
E. 62.20
10- The price of a laptop is decreased by $$10\%$$ to $360. What is its original price? A.$320
B. $380 C.$400
D. $450 E.$500
## Best HiSET Math Prep Resource for 2021
1- D
Some of prime numbers are: $$2, 3, 5, 7, 11, 13$$
Find the product of two consecutive prime numbers:
$$2 × 3 = 6$$ (not in the options)
$$3 × 5 = 15$$ (bingo!)
$$5 × 7 = 35$$ (not in the options)
2- B
The question is this: 530.40 is what percent of 624?
Use percent formula:
$$part =\frac{percent}{100}× whole$$
$$530.40 =\frac{percent}{100}× 624$$
$$530.40 =\frac{percent× 624}{100}$$
$$⇒53040 = percent ×624$$
$$percent = \frac{53040}{624}= 85$$
530.40 is $$85 \%$$ of 624. Therefore, the discount is: $$100\% – 85\% = 15\%$$
3- B
If the score of Mia was 60, therefore the score of Ava is 30. Since, the score of Emma was half as that of Ava, therefore, the score of Emma is 15.
4- D
If 17 balls are removed from the bag at random, there will be one ball in the bag.
The probability of choosing a brown ball is 1 out of 18. Therefore, the probability of not choosing a brown ball is 17 out of 18 and the probability of having not a brown ball after removing 17 balls is the same.
5- B
Let $$x$$ be the smallest number. Then, these are the numbers:
$$x, x+1, x+2, x+3, x+4$$
$$average = \frac{sum \space of \space terms}{number \space of \space terms}$$
$$38 = \frac{x+(x+1)+(x+2)+(x+3)+(x+4)}{5}$$
$$38=\frac{5x+10}{5}$$
$$⇒ 190 = 5x+10 ⇒ 180 = 5x ⇒ x=36$$
6- C
The area of the floor is: $$6$$cm $$× 24$$ cm $$= 144$$ cm
The number is tiles needed $$= 144 ÷ 8 = 18$$
7- C
The weight of 12.2 meters of this rope is: $$12.2 × 600 g = 7320 g$$
$$1 kg = 1000 g$$, therefore, $$7320 g ÷ 1000 = 7.32 kg$$
8- C
$$4\%$$ of the volume of the solution is alcohol. Let $$x$$ be the volume of the solution.
Then: $$4\%$$ of $$x = 24$$ ml $$⇒ 0.04 x = 24 ⇒ x = 24 ÷ 0.04 = 600$$
9- B
$$average =\frac{sum \space of \space terms}{number \space of \space terms}$$
The sum of the weight of all girls is: $$18 × 60 = 1080$$ kg
The sum of the weight of all boys is: $$32 × 62 = 1984$$ kg
The sum of the weight of all students is: $$1080 + 1984 = 3064$$ kg
$$average =\frac{3064}{50}= 61.28$$
10- C
Let $$x$$ be the original price.
If the price of a laptop is decreased by $$10\%$$ to \$360, then: $$90 \%$$ of $$x=360⇒ 0.90x=360 ⇒ x=360÷0.90=400$$
## High School Equivalency Tests
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What is the solution of the differential equation dy/dx = 2y(5-3y) with y(0)=2?
Dec 21, 2017
$y = \frac{10 {e}^{10 x}}{6 {e}^{10 x} - 1}$
Explanation:
We have:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \left(5 - 3 y\right)$
This is a First Order Separable ODE, so we can "separate the variables" to get
$\int \setminus \frac{1}{y \left(5 - 3 y\right)} \setminus \mathrm{dy} = \int \setminus 2 \setminus \mathrm{dx}$
The RHS is trivial, and the LHS can be integrated by decomposing the integrand into partial fractions:
$\frac{1}{y \left(5 - 3 y\right)} \equiv \frac{A}{y} + \frac{B}{5 - 3 y}$
$\text{ } = \frac{A \left(5 - 3 y\right) + B y}{y \left(5 - 3 y\right)}$
$1 \equiv A \left(5 - 3 y\right) + B y$
Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:
Put $y = 0 \implies 1 = 5 A \implies A = \frac{1}{5}$
Put $x = \frac{5}{3} \implies 1 = \frac{5 B}{3} \implies B = \frac{3}{5}$
So we can now write:
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \int \setminus \frac{\frac{1}{5}}{y} + \frac{\frac{3}{5}}{5 - 3 y} \setminus \mathrm{dy} = \int \setminus 2 \setminus \mathrm{dx}$
$\therefore \frac{1}{5} \int \setminus \frac{1}{y} - \frac{3}{3 y - 5} \setminus \mathrm{dy} = \int \setminus 2 \setminus \mathrm{dx}$
And integrating we get:
$\frac{1}{5} \left\{\ln | y | - \ln | 3 y - 5 |\right\} = 2 x + C$
Using the initial condition $y \left(0\right) = 2$ then:
$\frac{1}{5} \left\{\ln 2 - \ln 1\right\} = C \implies C = \frac{1}{5} \ln 2$
So we have:
$\frac{1}{5} \left\{\ln | y | - \ln | 3 y - 5 |\right\} = 2 x + \frac{1}{5} \ln 2$
$\therefore \ln | y | - \ln | 3 y - 5 | = 10 x + \ln 2$
$\therefore \ln | \frac{y}{3 y - 5} | = 10 x + \ln 2$
$\therefore \frac{y}{3 y - 5} = {e}^{10 x + \ln 2}$
$\therefore y = \left(3 y - 5\right) 2 {e}^{10 x}$
$\therefore y = 6 y {e}^{10 x} - 10 {e}^{10 x}$
$\therefore 6 y {e}^{10 x} - y = 10 {e}^{10 x}$
$\therefore \left(6 {e}^{10 x} - 1\right) y = 10 {e}^{10 x}$
$\therefore y = \frac{10 {e}^{10 x}}{6 {e}^{10 x} - 1}$
Dec 21, 2017
$y \left(x\right) = \frac{10}{6 - {e}^{- 10 x}}$
Explanation:
This is a separable differential equation:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \left(5 - 3 y\right)$
$\frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \mathrm{dx}$
$\left(1\right) \text{ } \int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \int \mathrm{dx}$
Solve the integral in $y$ using partial fractions:
$\frac{1}{2 y \left(5 - 3 y\right)} = \frac{A}{2 y} + \frac{B}{5 - 3 y}$
$\frac{1}{2 y \left(5 - 3 y\right)} = \frac{A \left(5 - 3 y\right) + 2 B y}{2 y \left(5 - 3 y\right)}$
$y \left(2 B - 3 A\right) + 5 A = 1$
$\left\{\begin{matrix}2 B - 3 A = 0 \\ 5 A = 1\end{matrix}\right.$
$\left\{\begin{matrix}A = \frac{1}{5} \\ B = \frac{3}{10}\end{matrix}\right.$
$\int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \frac{1}{10} \int \frac{\mathrm{dy}}{y} + \frac{3}{10} \int \frac{\mathrm{dy}}{5 - 3 y}$
$\int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \frac{1}{10} \left(\int \frac{\mathrm{dy}}{y} - \int \frac{d \left(5 - 3 y\right)}{5 - 3 y}\right)$
$\int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \frac{1}{10} \left(\ln \left\mid y \right\mid - \ln \left\mid 5 - 3 y \right\mid\right) + C$
$\int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = - \frac{1}{10} \ln \left\mid \frac{5}{y} - 3 \right\mid + C$
Substituting in $\left(1\right)$:
$x = - \frac{1}{10} \ln \left\mid \frac{5}{y} - 3 \right\mid + C$
$- 10 x + C = \ln \left\mid \frac{5}{y} - 3 \right\mid$
$c {e}^{- 10 x} = \frac{5}{y} - 3$
$\frac{5}{y} = 3 + c {e}^{- 10 x}$
$y = \frac{5}{3 + c {e}^{- 10 x}}$
For $x = 0$ the initial condition is $y \left(0\right) = 2$, so:
$2 = \frac{5}{3 + c}$
$c = - \frac{1}{2}$
The required solution is then:
$y \left(x\right) = \frac{10}{6 - {e}^{- 10 x}}$
|
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To better understand certain problems involving rockets and propulsion it is necessary to use some mathematical ideas from trigonometry, the study of triangles. On another page we have introduced the trigonometric functions sine, cosine, and tangent of the angles of a right triangle and described how these functions relate the magnitude of the sides of a triangle. On this page we will further explore the relations between the sides of a right triangle and demonstrate that the trigonometric functions depend only the angles of the triangle and not on the size of the triangle. On this page we have constructed three right triangles of different size, but with the same acute angle c at the lower left. Using the terminology from the sine, cosine, and tangent page, we have made the side opposite the angle c equal to 1.0 for the red triangle. And we have made the side adjacent to angle c equal to 2.0. The ratio of the opposite to the adjacent for any right triangle is defined to be the tangent (tan) of the angle. For the red triangle the value of the tangent is: tan(c) = 1 / 2 = .5 For the blue triangle, we keep the angle c the same, but we have doubled the size of the opposite side and the adjacent side. Mathematicians would say that the blue and the red triangles are similar triangles; the angles are equal, but the size is different. What is the ratio of the opposite to the adjacent for the blue triangle? tan(c) = 2 / 4 = .5 The same value as for the red triangle. For the yellow triangle, we have tripled the size of the red triangle. The opposite side is now 3 units long and the adjacent side is now 6 units long. The yellow triangle is mathematically similar to the red and the blue triangle. What is the ratio of the opposite to the adjacent for the yellow triangle? tan(c) = 3 / 6 = .5 The same value as the red and the blue triangle. We could repeat this process for any multiple (m) of the sides of the red triangle and the result would be the same: tan(c) = (1 * m) / (2 * m) = .5 The value of the tangent (and the sine and cosine) depends on the size of the angle, not on the size of the triangle. The initial geometry of the red triangle was arbitrary. If we had picked other numbers for the ratio of the sides of the red triangle, the value of angle c would be different and the value of the tangent would be different. But the same value of the tangent would occur for the blue and yellow triangles as long as the angle c is kept the same. Guided Tours Trigonometry: Maximum Altitude: Activities: Related Sites: Rocket Index Rocket Home Beginner's Guide Home
+ Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility Certification Editor: Tom Benson NASA Official: Tom Benson Last Updated: May 13 2021 + Contact Glenn
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# 12 HOUR CLOCK ARITHMETIC
In modular arithmetic, if the divisor is 12, it is called as 12 hour clock arithmetic.
Because the divisor is 12, the remainders must be less than or equal to 11.
So, the remainder are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
"0" corresponds to the hour "12"
"1" corresponds to the hour "1"
"2" corresponds to the hour "2"
"3" corresponds to the hour "3"
"4" corresponds to the hour "4"
"5" corresponds to the hour "5"
"6" corresponds to the hour "6"
"7" corresponds to the hour "7"
"8" corresponds to the hour "8"
"9" corresponds to the hour "9"
"10" corresponds to the hour "10"
"11" corresponds to the hour "11"
## How to solve problems on 12 hour clock arithmetic?
Case 1 :
Let us say the time now is 3 hours in 12 hours clock. Find the time after 15 hours from now.
If you want to find the time after 15 hours from now, you have to add 15 to 3 and divide the result by 12.
3 + 15 = 18
When 18 is divided by 12, the remainder is 6.
So, time after 15 hours from now (3 hours) is 6 hours.
Case 2 :
Let us say the time now is 2 hours in 12 hours clock. Find the time after 8 hours from now.
If you want to find the time after 8 hours from now, you have to add 8 to 2 and divide the result by 12.
2 + 8 = 10
Here 10 is less than 12. So, we can not divide 10 by 12.
So, time after 8 hours from now (2 hours) is 10 hours.
Case 3 :
Let us say the time now is 5 hours in 12 hours clock. Find the time 32 hours back.
Subtract 32 from 5.
5 - 32 = -27
Here, we get a negative value.
In the case of getting a negative value, we have to find the next integer which is exactly divisible by 12.
That is, we have to get the next integer after 27 which is exactly divisible by 12.
It is 36.
Now, we have to write -27 in terms of -36.
So, -27 can be written as
-27 = -36 + 9
Therefore, the time 32 hours back was 9 hours.
## Solved Problems
Problem 1 :
It is 9 hours now in a 12 hour clock. What will be the time after 18 hours.
Solution :
Now, the time is 9 hours. We want to know the time after 18 hours.
To get answer for our question, we have to do the following steps.
Step 1 :
9 + 18 = 27
Step 2 :
Divide 27 by 12.
27 / 12
Step 3 :
Take the remainder, when 27 is divided by 12.
The remainder is 3.
So, the time after 18 hours will be 3 hours.
Problem 2 :
If the time now in a 12 hour clock is is 9 hours, what was the time 71 hours back?
Solution :
Now, the time is 9 hours. We want to know the time 71 hours back.
To get answer for our question, we have to do the following steps.
Step 1 :
Subtract 71 from 9.
That is,
9 - 71 = -62
Step 2 :
We get negative value in step 1 and also 62 is not divisible by 12.
So, find the next integer after 62 which is exactly divisible by 12.
That is 72.
Step 3 :
Write -62 in terms of -72.
So, -62 can be written as
-62 = -72 + 10
Therefore, the time 71 hours back was 10 hours.
Problem 3 :
Lily was assigned a work when the time was 7 hours in a 12 hours clock. She completed the work after 72 hours. At what time did she complete the work?
Solution :
From the question, we want to know the time 72 hours after 7 hours.
To get answer for our question, we have to do the following steps.
Step 1 :
72 + 7 = 79
Step 2 :
Divide 79 by 12.
79 / 12
Step 3 :
Take the remainder, when 79 is divided by 12.
The remainder is 7.
So, the time after 72 hours will be 7 hours.
Problem 4 :
Mr. Johnson started a work at 4 hours in a 12 hours clock. He was given 38 hours time to complete the work. But he took 3 hours more to complete the work. At what time did he complete the work.
Solution :
From the question, we want to know the time 41 hours (38 + 3) after 4 hours.
To get answer for our question, we have to do the following steps.
Step 1 :
41 + 4 = 45
Step 2 :
Divide 45 by 12.
Step 3 :
Take the remainder, when 45 is divided by 12.
The remainder is 9.
So, he completed the work at 9 hours.
Problem 5 :
Now the time is 6 hours and Mr. Lenin is at home. Lenin was in Washington 81 hours before. At what time was he in Washington ?
Solution :
Now, the time is 6 hours. We want to know the time 81 hours before.
To get answer for our question, we have to do the following steps.
Step 1 :
Subtract 81 from 6.
That is,
6 - 81 = -75
Step 2 :
We get negative value in step 1 and also 75 is not divisible by 12.
So, find the next integer after 75 which is exactly divisible by 12.
That is 84.
Step 3 :
Write -75 in terms of -84.
So, -75 can be written as
-75 = -84 + 9
Therefore, Mr. Lenin was in Washington at 9 hours.
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