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# Exterior Angle of a Triangle
## Exterior Angle of a Triangle
When a side of a triangle is produced, the angle formed outside of the triangle is called an exterior angle of the triangle. In the given figure, ABC is a triangle and ACD is an exterior angle.
The exterior angle of a triangle is always equal to the sum of two opposite interior angles of the triangle. In the above figure, exterior angle ACD is equal to the sum of two opposite interior angles ABC and BAC i.e. ACD = ABC + BAC.
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### Theoretical proof:
Given: ABC is a triangle with exterior angle ACD.
To prove: ABC + BAC = ACD
Proof:
Statements Reasons
1. ABC + BAC + ACB = 180°----------> sum of angles of a Δ.
2. ACB + ACD = 180° ---------> linear pair of angles.
3. ABC + BAC + ACB = ACB + ACD -----> from statements 1 and 2.
4. ABC + BAC = ACD -----> removing ACB from both sides .
Proved.
### Workout Examples
Example 1: Find the values of x from the given figure.
Solution: From the figure,
x + 50° = 110° ----> exterior angle of a Δ is equal to the sum of opposite interior angles.
or, x = 110° – 50°
= 60°
Example 2: Find the values of x and y in the given figure.
Solution: From the figure,
x = 35° + 75° ----> exterior angle of a Δ is equal to the sum of opposite interior angles.
or, x = 110°
x + y = 180° ----------> linear pair of angles.
or, 110° + y = 180°
or, y = 180° – 110°
or, y = 70°
x = 110° and y = 70°
Example 3: Find the values of a and b in the given figure.
Solution: From the figure,
a = 50° ------> base angles of an isosceles Δ.
b = a + 50° ------> exterior angle of a Δ is equal to the sum of opposite interior angles.
= 50° + 50°
= 100°
a = 50° and b = 100°
You can comment your questions or problems regarding the angles of a triangle here. |
## How do you find resistance in parallel and series?
To calculate the total overall resistance of a number of resistors connected in this way you add up the individual resistances. This is done using the following formula: Rtotal = R1 + R2 +R3 and so on.
## What is the resistance of the parallel combination?
This implies that the total resistance in a parallel circuit is equal to the sum of the inverse of each individual resistances. Therefore, for every circuit with n number or resistors connected in parallel, Rn(parallel)=1R1+1R2+1R3…
## How do you find the missing resistance in a parallel circuit?
Resistance for parallel circuits are calculated by the formula: 1/Req= 1/R1 + 1/R2.
## What is the formula of resistance?
R = V ÷ I Question What is the resistance of the lamp? To calculate the resistance of an electrical component, an ammeter is used to measure the current and a voltmeter to measure the potential difference. The resistance can then be calculated using Ohm’s Law.
## What is the formula of resistance in series?
Series circuits The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors: equivalent resistance of resistors in series : R = R1 + R2 + R3 + With a 10 V battery, by V = I R the total current in the circuit is: I = V / R = 10 / 20 = 0.5 A.
## What is the effective resistance?
the resistance to an alternating current, expressed as the ratio of the power dissipated to the square of the effective current.
## Why is resistance different in series and parallel?
In a series circuit, the output current of the first resistor flows into the input of the second resistor; therefore, the current is the same in each resistor. In a parallel circuit, all of the resistor leads on one side of the resistors are connected together and all the leads on the other side are connected together.
## Why do resistors in parallel have less resistance?
Resistors in parallel In a parallel circuit, the net resistance decreases as more components are added, because there are more paths for the current to pass through. The two resistors have the same potential difference across them. The current through them will be different if they have different resistances.
## How do you find the missing resistance?
Calculating Resistance of Unknown resistor, total current and voltage across each resistor- Resistance of R2. Since it is given that total power is 60 watts, and the series circuit has 120v, then we can calculate total resistance to be RT = (120*120)/60 = 240 ohms. – Current in the circuit. – Voltage across each resistor.
You might be interested: Regression equation example
## What happens to current in a parallel circuit?
The current in a parallel circuit splits into different branches then combines again before it goes back into the supply. When the current splits, the current in each branch after the split adds up to the same as the current just before the split.
## How do you calculate unknown resistance?
Each time you can calculate the unknown resistance using the formula:unknown resistance(a)=known resistance(b)*L/M. The value is found to be the same for all values of known resistance.
## What is an example of resistance?
Resistance is defined as a refusal to give in or to something that slows down or prevents something. An example of resistance is a child fighting against her kidnapper. An example of resistance is wind against the wings of a plane.
## What is resistance unit?
Siemens per meterElectrical conductivityOhm meterElectrical resistivity
### Releated
#### Convert polar to rectangular equation
How do you convert from Polar to Rectangular? To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) 😡 = r × cos( θ )y = r × sin( θ ) How do you convert Rectangular to Polar on a calculator? Convert from Rectangular to Polar Coordinates on the TI-83 PlusPress [2nd][APPS] to access the […]
#### Find the real solutions of the equation
What are real solutions of an equation? How To Solve Them? The “solutions” to the Quadratic Equation are where it is equal to zero. There are usually 2 solutions (as shown in this graph). Just plug in the values of a, b and c, and do the calculations. What does it mean to find all […] |
RBSE Maths Class 11 Chapter 6: Important Questions and Solutions
RBSE Solutions for Class 11 Maths Chapter 6 – Linear inequalities are available here. The important questions and solutions of Chapter 6, available at BYJU’S, contain step by step explanations along with the solutions. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 11 solutions.
Chapter 6 of the RBSE Class 11 Maths will help the students to solve problems related to inequalities, algebraic solutions of linear inequalities, graphical solution of linear inequalities, solution of the system of linear inequalities. After referring to these important questions and solutions, you can refer to 2020-2021 solutions for the chapter here.
RBSE Maths Chapter 6: Textbook Important Questions and Solutions
Question 1: Solve the following inequalities.
[i] 4x + 4 < 5x + 7
[ii] 3x – 9 > 5x – 1
Solution:
[i] 4x + 4 < 5x + 7
Subtracting 7 from both the sides, we get
4x + 4 – 7 < 5x + 7 – 7
4x – 3 < 5x
Subtracting 4x from both the sides,
4x – 3 – 4x < 5x – 4x
x > – 3
The solutions of the inequality given are defined by the real numbers which are greater than -3.
The required solution set is (-3, ∞)
[ii] 3x – 9 > 5x – 1
3x – 9 > 5x – 1
Adding 9 to both the sides, we get
3x – 9 + 9 > 5x – 1 + 9
3x > 5x + 8
Subtracting 5x from both the sides,
3x – 5x > 5x + 8 – 5x
-2x > 8
Divide both sides by -2
-2x / -2 < 8 / -2
x < -4
The solutions of the given inequality are defined by all the real numbers which are less than -4.
The required solution set is (-∞, -4).
Question 2: Solve the given inequality for the real value of x: x + x / 2 + x / 4 < 12.
Solution:
x + x / 2 + x / 4 < 12
Taking as common from the terms we get,
x [1 + (1 / 2) + (1 / 4)] < 12
On taking the LCM,
x {[4 + 2 + 1] / 4} < 12
7x / 4 < 12
Dividing by 12 on both sides
[7x / 4 * 12] < [12 / 12] [7x / 48] < 1
7x < 48
Divide by 7 on both sides
[7x / 7] < [48 / 7]
x < [48 / 7]
Question 3: Solve the inequality 3 (x – 2) / 6 ≤ 5 (2 – x) / 4.
Solution:
3 (x – 2) / 6 ≤ 5 (2 – x) / 4
Apply the fraction cross multiplication,
3 (x – 2) * 4 ≤ 6 * 5 (2 – x)
12 (x – 2) ≤ 30 (2 – x)
12x – 24 ≤ 60 – 30x
12x – 24 + 24 ≤ 60 – 30x + 24
12x ≤ 84 – 30x
12x + 30x ≤ 84 – 30x + 30x
42x ≤ 84
Divide both sides by 42
x ≤ 2
Question 4: Graph the following inequalities.
[i] 2x + y < 3
[ii] y < x2
[iii] y < 5x + 1
[iv] 4y + 11 < 9
[v] x < y6
Solution:
[i] 2x + y < 3
x < [-1 / 2] y + [3 / 2]
[ii] y < x2
[iii] y < 5x + 1
[iv] 4y + 11 < 9
(4y + 11) + (-9) < 9 + (-9)
4y + 11 – 9 < 9 – 9
4y + 2 < 0
y < (-1 / 2)
[v] x < y6
Question 5: Solve the following system of inequalities.
[i] 4x + y ≥ 3 , x + 8y ≤ 16
[ii] 3x + y ≥ 6, 9x – y < 9
[iii] 3x – y > 1, x – 4y < – 1
[iv] 4x – y > 1, x – 3y < – 1
Solution:
[i] 4x + y ≥ 3 , x + 8y ≤ 16
4x + y ≥ 3
Subtract 4x on both sides
4x + y – 4x ≥ 3 – 4x
y ≥ 3 – 4x
x + 8y ≤ 16
Subtract x on both sides
x + 8y – x ≤ 16 – x
8y ≤ 16 – x
Divide both sides by 8
[8y / 8] ≤ [16 – x] / [18]
y ≤ [16 – x] / [8]
The solution is unbounded.
[ii] 3x + y ≥ 6, 9x – y < 9
3x + y ≥ 6
Subtract 3x on both sides
3x + y – 3x ≥ 6 – 3x
y ≥ 6 – 3x
9x – y < 9
Subtract 9x on both sides
9x – y – 9x < 9 – 9x
– y < 9 – 9x
Multiply both sides by (-1) to reverse the inequality
y > – 9 + 9x
The solution is unbounded.
[iii] 3x – y > 1, x – 4y < – 1
3x – y > 1
Subtract 3x on both sides
3x – y – 3x > 1 – 3x
– y > 1 – 3x
Multiply both sides by (-1) to reverse the inequality
y < – 1 + 3x
x – 4y < – 1
Subtract x on both sides
x – 4y – x < – 1 – x
– 4y < – 1 – x
Multiply both sides by (-1) to reverse the inequality
4y > (1 + x)
Divide both sides by 4
y > [1 + x] / [4]
The solution is unbounded.
[iv] 4x – y > 1, x – 3y < – 1
4x – y > 1
Subtract 4x on both sides
4x – y – 4x > 1 – 4x
– y > 1 – 4x
Multiply both sides by (-1) to reverse the inequality
y < -1 + 4x
x – 3y < – 1
Subtract x on both sides
x – 3y – x < – 1 – x
– 3y < – 1 – x
Multiply both sides by (-1) to reverse the inequality
3y > 1 + x
The solution is unbounded.
Question 6: Solve the following inequalities using a number line.
[i] 4x + 3 ≤ 23
[ii] 5x + 9 < 8
[iii] 5x – 1 > 26
[iv] 3 (x + 7) > 2 – x
[v] 6 – x > 13 – 2x
Solution:
[i] 4x + 3 ≤ 23
Subtract 3 from both sides.
4x + 3 − 3 ≤ 23 − 3
4x ≤ 20
Divide both sides by
4x / 4 ≤ 20 / 4
x ≤ 5
[ii] 5x + 9 < 8
Subtract 9 from both sides.
5x + 9 − 9 < 8 − 9
5x < −1
Divide both sides by 5.
5x / 5 < −1 / 5
x < −1 / 5
[iii] 5x – 1 > 26
5x − 1 + 1 > 26 + 1
5x > 27
Divide both sides by 5.
5x / 5 > 27 / 5
x > 27 / 5
[iv] 3 (x + 7) > 2 – x
⇒ 3x + 21 > 2 – x
⇒ 3x + x > 2 – 21
⇒ 4x > -19
⇒ x > -19 / 4
[v] 6 – x > 13 – 2x
6 – 13 > – 2x + x
-7 > -x
Multiply both sides by (-1) to reverse the inequality
x > 7 |
# Integers that Represent Different Situations
## Write variable expressions to represent word problems including positives and negatives
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Use Simple Integer Equations to Solve Real-World Problems
### [Figure1] License: CC BY-NC 3.0
A train is hauling 150 freight cars full of grain and corn. Each train car can hold either 90.25 tons of corn or 114 tons of grain. If there are 90 cars full of grain, how much weight is the train hauling?
In this concept, you will write equations and solve real-world problems using simple integer equation
### Applying Integer Equations to Solve Real World Problems
Integers are used in everyday life when you solve real-world problems. Let’s look at a real-life problem using integers.
Kaitlyn borrowed $950 to buy a computer. So far, she has paid back$175. How much does Caitlin still owe?
First, write a simple equation to represent the problem. Let \begin{align*}x\end{align*} be the amount that Kaitlyn still owes.
\begin{align*}950=175+x\end{align*}
Next, rearrange the equation to isolate \begin{align*}x\end{align*}.
\begin{align*}x=950-175\end{align*}
Then, solve for \begin{align*}x\end{align*}.
\begin{align*}x=775\end{align*}
Kaitlyn still owes 775. Here is another real-world example. The population of a certain town in 2002 was 312,980. In 2006, the population increased to 391,740. To the nearest thousand, what was the population increase from 2002 to 2006? First, since you are estimating, round the populations in 2002 and 2006 to the nearest thousand. Round the first number. Since there is a 9 in the hundreds place, 312,980 rounds up to 313,000. Round the second number. Since there is a 7 in the hundreds place, 391,740 rounds up to 392,000. Next, find the difference between the two populations. \begin{align*}392000-313000=79000\end{align*} The answer is 79000. The population increased by 79,000 people. ### Examples #### Example 1 Remember the freight train hauling the grain and corn? There are 150 cars in total, 90 carrying grain and 60 carrying corn. First, write an expression for the weight of each type of load for all of the freight cars. Let \begin{align*}x\end{align*} equal the total weight being hauled by the train. \begin{align*}x = 90.25 \frac{\ \text{tons}}{\ \text{car}} \times 60 \ \text{cars} + 114 \ \frac{\text{tons}}{\ \text{car}} \times 90 \ \text{cars}\end{align*} Next, simplify the equation. \begin{align*}x=5415 \ \text{tons} + 10260 \ \text{tons}\end{align*} Then, solve for \begin{align*}x\end{align*}. \begin{align*}x=15675\end{align*}The answer is 15675. The amount of grain and corn is 15675 tons. #### Example 2 Yuri saved215 each month for six months. About how much has Yuri saved? What is the exact amount that Yuri saved?
First, you need to find an estimate the amount Yuri saved.
Round the first number to a number that is easy to multiply. The number 215 rounds down to 200.
Next, multiply 200 by the 6 months.
\begin{align*}200 \times 6 = 1200\end{align*}
Yuri saved about 1,200. Then, you need to find the exact answer. You need to write a simple equation to represent the problem. Let \begin{align*}x\end{align*} represent the total amount Yuri saved. \begin{align*}\begin{array}{rcl} x &=& 215 \times 6 \\ x &=& 1290 \end{array}\end{align*} The answer is 1290. Yuri saved exactly1,290.
Here are a few for you to try.
#### Example 3
John earned three bonus points on his test. If he started with a 78, what was his final score?
First, write an equation to represent this problem.
\begin{align*}x= 78+3\end{align*}
Next, solve for \begin{align*}x\end{align*}.
\begin{align*}x=81\end{align*}The answer is 81.
John’s final score on his test was 81.
#### Example 4
A football team lost fifteen yards at the 20 yard line. Since football moves backwards, what yard line did the team start the next play on?
First, write an equation to represent this problem.
\begin{align*}x =20 - (-15)\end{align*}
Next, solve for \begin{align*}x\end{align*}.
\begin{align*}\begin{array}{rcl} x &=& 20 + 15 \\ x &=& 35 \end{array}\end{align*}
The next play is from the 35 yard line.
#### Example 5
You are helping to make potato salad for a family picnic. You can peel 2 potatoes every minute. How long will it take you to peel the 50 potatoes you need?
First, write an equation to represent this problem.
\begin{align*}x = \frac{50 \ \text{potatoes}}{2 \ \text{potatoes/min}}\end{align*}
Next, solve for \begin{align*}x\end{align*}.
\begin{align*}x=25\end{align*}
It will take you 25 minutes.
### Review
Use simple integer equations to solve each real-world problem.
1. Karen saved fifteen dollars a week for eight weeks. How much money did she have at the end of this time?
2. Jocelyn spent as much as Karen had saved. Write an integer to show the amount that Jocelyn spent.
3. If a car backs up fifteen feet and then goes forward forty feet. How many feet did the car advance?
4. Tasha owes her brother fifty dollars. She paid five dollars towards the debt. How much does she still owe her brother?
5. The temperature on Monday began at 5 degrees, then went up to 20 degrees and then decreased to 7 degrees. Show the temperature change in an equation.
6. Represent this change in temperature by writing an integer.
7. Joshua spent fifteen dollars, then he spent five more dollars and then he spent three dollars and fifty cents. Write an equation to show his spending.
8. How much money did Joshua spend in all?
9. If Joshua had started shopping with \$30.00, would he have gotten any change?
10. How much change would he have gotten?
11. Jessica is shortening her dress length three inches. Write an integer to show this change.
12. If the length of the dress is 40 inches, how long will the dress be after the alteration?
13. If Jessica is five feet tall, how far will the hem be from the floor?
14. Carly is scuba diving. She descends to fifteen feet and then proceeds to descend another 35 feet.
1. Show this change using an equation.
2. What is her final depth?
To see the Review answers, open this PDF file and look for section 2.14.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Difference The result of a subtraction operation is called a difference.
Integer The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3... |
# Question: How To Solve Square Root Easily?
## How do you calculate square root?
Example: Calculate the square root of 10 ( ) to 2 decimal places.
1. Find the two perfect square numbers it lies between. Solution: 32 = 9 and 42 = 16, so lies between 3 and 4.
2. Divide 10 by 3. 10/3 = 3.33 (you can round off your answer)
3. Average 3.33 and 3. ( 3.33 + 3)/2 = 3.1667.
## How do you solve root equations?
1. Isolate the radical on one side of the equation.
2. Square both sides of the equation.
3. Solve the new equation.
4. Check the answer. Some solutions obtained may not work in the original equation.
## What is the shortcut key for square root?
The Alt code for the symbol for the Square root is Alt +251 or 221A, then Alt+X. Follow these three simple steps to attach the symbol using the Alt code: – Position the pointer in the place where you want the square root symbol inserted. – Press and hold down the Alt key and type 251 from the numeric keypad.
## What is the square of 1 to 30?
Square, Cube, Square Root and Cubic Root for Numbers Ranging 0 – 100
Number x Square x2 Square Root x1/2
28 784 5.292
29 841 5.385
30 900 5.477
31 961 5.568
You might be interested: Often asked: What Is Square Root Of Zero?
## How do you solve a square?
Steps
1. Step 1 Divide all terms by a (the coefficient of x2).
2. Step 2 Move the number term (c/a) to the right side of the equation.
3. Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.
## What is 9 The square root of?
The square root of a number is a number that, when multiplied by itself, equals the desired value. So, for example, the square root of 49 is 7 (7×7=49). List of Perfect Squares.
NUMBER SQUARE SQUARE ROOT
6 36 2.449
7 49 2.646
8 64 2.828
9 81 3.000
## What is a square root of 144?
The value of the square root of 144 is equal to 12. In radical form, it is denoted as √ 144 = 12.
## IS 400 a perfect square?
400 is a perfect square. Because 20 * 20 = 400.
## Why is 9 The square root of 81?
Explanation: 81 = 9 ⋅ 9 then the square root of √ 81 = 9. Because the double multiplication for the same sign is always positive, the square root is also valid with the other sign 81 =(− 9 )⋅(− 9 ) then √ 81 =− 9 and we can say that √ 81 =± 9.
## How can I get root 3 value?
It is not a natural number but a fraction. The square root of 3 is denoted by √ 3. The square root basically, gives a value which, when multiplied by itself gives the original number. Hence, it is the root of the original number. Table of Square Root.
Number Square Root (√)
2 1.414
3 1.732
4 2.000
5 2.236 |
# Fraction Calculator
A simple, but powerful, calculator to solve multiple types of fraction math problems.
+ - × ÷ of ^ + - × ÷
This Fraction calculator simplify fractions of addition, subtraction, multiplication, and division. Perform simplifying fractions effortlessly by entering the numerator and denominator.
Whether you have fractions or mixed numbers, The fraction calculator will reduce them to the most simplified form within seconds.
## Formula Used By Fraction Calculator:
The mixed number calculator displays a fraction addition, subtraction, multiplication, and division. This utilizes the below formulas by taking the hassle out of fractions and gives you accurate results.
• Identify the common denominators
• Find the least common multiple (LCM) or least common denominator (LCD)
• If the denominators are not the same, multiply each fraction by a factor that makes its common denominator equal to the LCM
• Add the numerators and convert the given improper fractions to the mixed.
$$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}$$
Example:
How to add fractions 4/3 & 2/3?
Solution:
4/3 + 2/3
= (4+2)/3 (LCD is 3)
= 6/3
= 2
## How to Subtract Fractions?
• Identify the common denominators
• Find the least common multiple or the least common denominator
• If the denominators are not the same, multiply each fraction by a factor that makes its denominator equal to the LCM
• Subtract the numerators and simplify fractions to the lowest terms
### Formula for Fractions Subtraction:
$$\dfrac{a}{b} – \dfrac{c}{d} = \dfrac{ad – bc}{bd}$$
Example:
How to subtract fractions 2/7 & 8/3?
Solution:
2/7 - 8/3
= 2(3) - 7(8) / 21 (LCD is 21)
= (6-56) / 21
= -50 / 21
## How to Multiply Fractions?
• Multiply numerators of all fractions
• Likewise, go by multiplying denominators as well
• Write the results as a single fraction
• Simplify to reduce the fraction to lowest terms
### Formula for Fractions Multiplication:
$$\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}$$
Example:
How to multiply the numbers 1/7 & 8/9?
Solution:
1/7 & 8/9
= (1*8)/(7*9)
= 8/63
## How to Divide Fractions?
• Find the reciprocal and rewrite the Division as Multiplication
• Flip the second fractions by switching the top and bottom numbers
• Multiply Numerators and Denominators
• Simplify the results to the lowest form
### Formula for Fractions Division:
$$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{ad}{bc}$$
Example:
How to divide fractions 7/3 ÷ 7/2?
Solution:
Here we have:
7/3 ÷ 7/2
The reciprocal of the divisor (7/2) is 2/7. So we have;
7/3 x 2/7
= (7*2)/(3*7)
= 14/21
= 2/3
## Common Fractions to Decimals:
Our fraction calculator helps you to simplify and reduce fractions to the lowest form. It performs fraction operations, and expressions with integers, decimals, and mixed numbers.
The following table is packed with fractions values that you may encounter on a daily basis and may help you simplify fractions value in seconds:
64th 32nd 16th 8th 4th 2nd Decimal
1/64 0.015625
2/64 1/32 0.03125
3/64 0.046875
4/64 2/32 1/16 0.0625
5/64 0.078125
6/64 3/32 0.09375
7/64 0.109375
8/64 4/32 2/16 1/8 0.125
9/64 0.140625
10/64 5/32 0.15625
11/64 0.171875
12/64 6/32 3/16 0.1875
13/64 0.203125
14/64 7/32 0.21875
15/64 0.234375
16/64 8/32 4/16 2/8 1/4 0.25
17/64 0.265625
18/64 9/32 0.28125
19/64 0.296875
20/64 10/32 5/16 0.3125
21/64 0.328125
22/64 11/32 0.34375
23/64 0.359375
24/64 12/32 6/16 3/8 0.375
25/64 0.390625
26/64 13/32 0.40625
27/64 0.421875
28/64 14/32 7/16 0.4375
29/64 0.453125
30/64 15/32 0.46875
31/64 0.484375
32/64 16/32 8/16 4/8 2/4 1/2 0.5
33/64 0.515625
34/64 17/32 0.53125
35/64 0.546875
36/64 18/32 9/16 0.5625
37/64 0.578125
38/64 19/32 0.59375
39/64 0.609375
40/64 20/32 10/16 5/8 0.625
41/64 0.640625
42/64 21/32 0.65625
43/64 0.671875
44/64 22/32 11/16 0.6875
45/64 0.703125
46/64 23/32 0.71875
47/64 0.734375
48/64 24/32 12/16 6/8 3/4 0.75
49/64 0.765625
50/64 25/32 0.78125
51/64 0.796875
52/64 26/32 13/16 0.8125
53/64 0.828125
54/64 27/32 0.84375
55/64 0.859375
56/64 28/32 14/16 7/8 0.875
57/64 0.890625
58/64 29/32 0.90625
59/64 0.921875
60/64 30/32 15/16 0.9375
61/64 0.953125
62/64 31/32 0.96875
63/64 0.984375
64/64 32/32 16/16 8/8 4/4 2/2 1
## References:
Wikipedia: Fractions, Reciprocals and the "invisible denominator", Ratios, Decimal fractions and percentages, Historical notions, Arithmetic with fractions. |
Introduction to Puncture Points (Holes)
# Introduction to Puncture Points (Holes)
Rational functions can exhibit puncture points (also called ‘holes’).
As you'll learn in Calculus, puncture points are an example of a ‘removable discontinuity’.
Here's the idea.
The function $\,\displaystyle R(x) := \frac{x^3-2x^2}{x-2}\,$ certainly looks like a typical rational function.
Upon closer inspection, though, we see that there's an extra factor of $\,1\,$ in the formula:
\begin{align} \cssId{s6}{\frac{x^3 - 2x^2}{x - 2}}\ &\cssId{s7}{=\ \frac{x^2(x-2)}{x-2}}\cr\cr &\cssId{s8}{=\ x^2\cdot\frac{x-2}{x-2}} \end{align}
Therefore, $\,\displaystyle R(x) = \frac{x^3-2x^2}{x-2}\,$ has exactly the same outputs as the much simpler function, $\,P(x) := x^2\,,$ except that the function $\,R\,$ isn't defined when $\,x = 2\,.$
The graphs of both $\,P\,$ and $\,R\,$ are shown below—the puncture point (hole) in $\,R\,$ is caused by that extra factor of $\,1\,.$
$P(x) = x^2$
$$\cssId{s13}{R(x)} \cssId{s14}{= \frac{x^3 - 2x^2}{x-2}} \cssId{s15}{= x^2\cdot\frac{x-2}{x-2}}$$
Puncture points are studied in more detail in a future section. |
## Tuesday, June 30, 2015
### Links to triangle classification with angles
Other problems are included, but you can find problems like the homework due on 1 July 2015 on these pages.
6x + 5y = 18
2x - y = -4
3x + 7y = 12
2x + 5y = -18
### 100 coin (or ticket) problems
a) 100 coins, all quarters and pennies, total = \$13.72
b) 100 tickets, all children (\$6) and adult (\$12), total =\$912
c) 100 coins, all quarters and nickels, total = \$8.20
d) 100 coins, all dimes and pennies, total = \$3.79
e) 100 coins, all dimes and nickels, total = \$7.85
f) 100 coins, all quarters and dimes, total = \$14.65
## Sunday, June 28, 2015
### The Richter scale: From two readings, the relative strength and vice versa
In class last week, we learned how to find out how much stronger one quake is compared to another given the two Richter scale readings. For example, on June 28, the strongest quake in the U.S. was a 3.4 in Oklahoma, while the strongest in North America was a 5.4 in Niltepec, Mexico. What is the difference in levels of energy? Here are the steps.
Step 1: Subtract little from big. In our case, 5.4 - 3.4 = 2.0.
Step 2: multiply difference by 1.5. 2.0 * 1.5 = 3.0.
Step 3: Raise 10 to the power of the answer from Step 2: 10^3.0 = 1,000. The Mexican quake was 1,000 times stronger than the Oklahoma quake.
Let's ask the question in the opposite direction. Let's say we have a reading for a quake and we know another quake was x times stronger. Again, it will be a three step process, but now we will take the inverse of our three steps above in reverse order. Let's say we have a quake 350 times stronger than the one in Niltepec. Here are our steps.
Step 1: Take the log of the strength multiplier. Log is the inverse of raising 10 to a power, just like addition is the inverse of subtraction and division is the inverse of multiplication. log(350) = 2.544...,
Step 2: divide the answer from Step 1 by 1.5 and round this answer to the nearest tenth. 2.544/1.5 = 1.696..., which rounds to 1.7. We round to the nearest tenth because the Richter scale rounds to the nearest tenth.
Step 3: Add the answer from Step 2 to the Richter reading we know. In this case, it would be 5.4+1.7 = 7.1, the reading of the stronger quake. If instead we were told a quake was 350 times weaker than Niltepec, it would be 5.4-1.7 = 3.7
1. 16 times stronger than a 6.1
2. 250 times stronger than a 6.7
3. 8 times weaker than a 5.8
## Wednesday, June 24, 2015
### Link to the computer science biographies
Click on this link for the biographies of the five computer scientists. There will be questions on them on the midterm tomorrow.
## Tuesday, June 16, 2015
### Links to homework 1, due Wednesday 17 June 2015
Link for prime factorization and all factors, with practice examples.
Roman and Hindu-Arabic numerals, fractions and decimals.
Practice for Roman to Hindu-Arabic and vice versa, fractions and decimals.
Practice for time conversion.
The following problems are in the form minutes:seconds. If the answer gives more than 60 minutes, write the answer as hours:minutes:seconds. Answers to the question below are in the comments.
18:34
- 6:58
42:23
41:07
43:01
42:18
+42:17
## Friday, June 12, 2015
### Syllabus for Summer 2015
Math 15: Math for Liberal Arts Summer 2015 (L1 30451)
Instructor: Matthew Hubbard
Email: mhubbard@peralta.edu
Text: no required text. If you want a text, personal recommendations can be made
Class website: http://mathlibarts.blogspot.com/
Class hours MTWTh: 10:00 am - 12:05 pm, G-211
Office hours: Math lab G-201
TTh 9:25-9:55 am 3:05-3:35 pm (also available by appointment)
Scientific calculator required (TI-30IIXs, TI-83 or TI-84 recommended)
Last date to add, if class is not full: Sat., June 21
Last date to drop class without a "W": Sat. June 21
Last date to withdraw from class: Tues., July 23
Holidays: No holidays this session
Midterm and Finals schedule:
Midterm 1__________Thursday, June 25
Midterm 2__________Thursday, July 9
Comprehensive Final ___Thursday, July 23
Quiz schedule (most Tuesdays and Thursdays) no make-up quizzes given
First week: 6/16 and 6/18
Second week: 6/23
Third week: 6/30 and 7/2
Fourth week: 7/7
Fifth week: 7/14 and 7/16
Sixth week: 7/21
Homework to be turned in: Assigned every Tuesday and Thursday, due the next class
(late homework accepted at the beginning of next class period, 2 points off grade)
If arranged at least a week in advance, make-up midterm can be given.
The lowest two scores from homework and the lowest two scores from quizzes will be removed from consideration before grading.
Quizzes 25%* best 2 out of three of these grades
Midterm 1 25%* best 2 out of three of these grades
Midterm 2 25%* best 2 out of three of these grades
Homework 20%
Lab 5%
Final 25%
Anyone who misses less than two homework assignments and gets a higher percentage score on the final than the weighted average of all grades combined will get the final percentage instead deciding the final grade.
Anyone with a class grade of 97% out of all work before the final (this grade will be given on the next to last day) does not have to take the final. That grade is worth an A.
Academic honesty: Your homework, exams and quizzes must be your own work. Anyone caught cheating on these assignments will be punished, where the punishment can be as severe as failing the class or being put on college wide academic probation. Working together on homework assignments is allowed, but the work you turn in must be your own, and you are responsible for checking its accuracy. If I see multiple homework assignments turned in with the exact same wrong answers, I will give a warning. If it happens a second time, the student will get a 0 on the assignment and it will be counted towards the grade.
Class rules: Cell phones and beepers turned off, no headphones or text messaging during class
You will need your own calculator and handout sheets for tests and quizzes. Do not expect to be able to borrow these from someone else.
Student Learning Outcomes
• Analyze an argument for validity using simple rules of logic, and if invalid identify the type of mistake made.
• Compute, with sophisticated formulas, such quantities as interest payments for amortized loans.
• Interpret patterns and draw inferences from them.
Students with disabilities
The Disabled Students Program Services (DSPS) should have your academic accommodation with the instructor. After the first day, I will accept these accommodations electronically or by hard copy on paper. If you need academic accommodation and have not yet applied, please call 510-464-3428 for an appointment.
Exam policies
Quizzes will be closed book and closed notes. Some information you will be expected to remember, other formulas and information will be provided. No sharing of calculators is allowed. You are responsible for knowing how to use your calculator to find answers.
The reciprocal relationship
The teacher will be on time and prepared to teach the class.
The students will be on time and prepared to learn.
The teacher will present the material to the best of his ability.
The students will absorb the material to the best of their ability. They will ask questions when topics are not clear.
The teacher will do his best to answer the questions the students ask about the material, either by repeating an answer with more details included or by taking a different approach to the material that might be clearer to some students.
The students will understand if the teacher feels a topic has been covered enough for the majority of the class and will accept questions being answered outside the class, either in extra time or through written communication.
The teacher will do his best to keep the class about the material. Personal details and distractions that are not germane to the class should not be part of the class.
The students will do their best to keep the class about the material. Questions that are not about the topic should be avoided. Distractions like cell phones and texting are not welcome when the class is in session.
The teacher will give assignments that will help the students master the skills required to pass the course.
The students will put in their best efforts to complete the assignments.
When the assignments are completed, the teacher will make every effort to get the assignments graded and back to the students in a timely manner, by the next class session whenever possible.
The teacher will present real life situations where the skills being learned will be used when they exist. In math, sometimes a particular skill is needed in general to solve later problems that will have real life applications. Other skills have the application of “learning how to learn”, of committing an idea to memory so that committing other ideas to memory becomes easier in the long run.
The student has the right to ask “When will I use this?” when dealing with mathematical topics. Sometimes, the answer is “We need this skill for the next skill we will learn.” Other times, the answer is “We are learning how to learn.” Both of these answers are as valid in their way as “We will need this to understand perspective” or “We use this to balance our checkbooks” or “Ratios can be used to figure out costs” or other real life applications. |
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# Newton's Second Law
## F = ma ; Force = mass * acceleration
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Practice Newton's Second Law
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Newton's Second Law
The acceleration experienced by an object will be proportional to the applied force and inversely proportional to its mass. If there are multiple forces, they can be added as vectors and it is the net force that matters.
### Key Equations
Newton's Second Law describes his famous equation for the motion of an object
The change of motion is proportional to the motive force impressed; and is made in the direction of the right (straight) line in which that force is impressed.
The "motion" Newton mentions in the Second Law is, in his language, the product of the mass and velocity of an object --- we call this quantity momentum --- so the Second Law is actually the famous equation : $\vec{F} = \frac{\Delta(m\vec{v})}{\Delta t} = \frac{m\Delta \vec{v}}{\Delta t} =m\vec{a} && \text{[1]}$
$\text{Force Sums} \begin{cases}\vec{F_{\text{net}}} = \sum_{i} F_i = m \vec{a} & \text{Net force is the vector sum of all the forces}\\F_{\text{net,}x} = \sum_{i} {F_{ix}} = m{a}_{x} & \text{Horizontal components add also}\\F_{\text{net,}y} = \sum_{i} {F_{iy}} = m{a}_{y} & \text{As do vertical ones}\end{cases}$
Guidance
To calculate the net force on an object, you need to calculate all the individual forces acting on the object and then add them as vectors. This requires some mathematical skill.
#### Example 1
A 175-g bluebird slams into a window with a force of 19.0 N. What is the bird’s acceleration?
Question: $a = ? [m/s^2]$
Given: $m = 175\ grams = 0.175\ kg$
${\;}\qquad \quad F = 19.0 \ N$
Equation: $a = \frac{F_{net}}{m}$
Plug n’ Chug: $a = \frac{F_{net}}{m} = \frac{19.0 \ N}{0.175 \ kg} = \frac{19.0 \frac{kg \cdot m}{s^2}}{0.175\ kg} = 109 \frac{m}{s^2}$
Answer: $\boxed{\mathbf{109 \ m/s^2}}$
#### Example 2
Calculate the acceleration of a rocket that has 500N of thrust force and a mass of 10kg.
Question: $a = ? [m/s^2]$
Given: $m = 10\ kg$
${\;} \qquad \quad F_{\text{thrust}} = 500\ N$
${\;} \qquad \quad g = 10.0\ m/s^2$
Equations: $\sum F_{\text{individual forces}} = ma$
or, in this case, $\sum F_{y-\text{direction forces}} = ma_y$
Plug n Chug: Use FBD to “fill in” Newton’s second law equation:
$\sum F_{y-\text{direction forces}} &= ma_y \\F - Mg & = Ma \\500N - 10\ kg(10\ m/s^2) & = 10kg (a) \\a & = 40\ m/s^2$
### Time for Practice
1. During a rocket launch, the rocket’s acceleration increases greatly over time. Explain, using Newton’s Second Law. (Hint: most of the mass of a rocket on the launch pad is fuel).
2. When pulling a paper towel from a paper towel roll, why is a quick jerk more effective than a slow pull?
3. You pull a wagon with a force of 20 N. The wagon has a mas of 10 kg. What is the wagon's acceleration?
4. The man is hanging from a rope wrapped around a pulley and attached to both of his shoulders. The pulley is fixed to the wall. The rope is designed to hold 500 N of weight; at higher tension, it will break. Let’s say he has a mass of 80 kg. Draw a free body diagram and explain (using Newton’s Laws) whether or not the rope will break.
5. Now the man ties one end of the rope to the ground and is held up by the other. Does the rope break in this situation? What precisely is the difference between this problem and the one before?
6. A crane is lowering a box of mass 50 kg with an acceleration of $2.0 \ m/s^2$ .
1. Find the tension $F_T$ in the cable.
2. If the crane lowers the box at a constant speed, what is the tension $F_T$ in the cable?
7. A physics student weighing 500 N stands on a scale in an elevator and records the scale reading over time. The data are shown in the graph below. At time $t = 0$ , the elevator is at rest on the ground floor.
1. Draw a FBD for the person, labeling all forces.
2. What does the scale read when the elevator is at rest?
3. Calculate the acceleration of the person from 5-10 sec.
4. Calculate the acceleration of the person from 10-15 sec. Is the passenger at rest?
5. Calculate the acceleration of the person from 15-20 sec. In what direction is the passenger moving?
6. Is the elevator at rest at $t=25 \ s$ . Justify your answer.
Answers (using $g = 10 \ m/s^2$ ):
1. According to Newton's second law: the acceleration of an object is inversely proportional to it's mass, so if you decrease it's mass while keeping the net force the same, the acceleration will increase.
2. When you jerk the paper towel, the paper towel that you are holding onto will accelerate much more quickly than the entire roll causing it to rip. Again, acceleration is inversely proportional to the mass of the object.
3. $2 m/s^2$
4. The rope will not break because his weight of 800 N is distributed between the two ropes.
5. Yes, because his weight of 800 N is greater than what the rope can hold.
6. a. 400 N b. 500 N
7. b. 500 N c. $6 \ m/s^2$ d. 0 e. $-4 \ m/s^2$ |
Cereal Sorting
In this lesson, children will sort loop cereal such as Fruit Cheerios or Fruit Loops by color and then make patterned bracelets.
Math Lesson for:
Toddlers/Preschoolers
(See Step 5: Adapt lesson for toddlers or preschoolers.)
Content Area:
Algebra
Numbers and Operations
Learning Goals:
This lesson will help toddlers and preschoolers meet the following educational standards:
• Understand numbers, ways of representing numbers and relationships among numbers and number systems
• Understand patterns, relations and functions
Learning Targets:
After this lesson, toddlers and preschoolers should be more proficient at:
• Counting with understanding and recognizing “how many” in sets of objects
• Sorting, classifying and ordering objects by size, number and other properties
• Recognizing, describing and extending patterns such as sequences of sounds and shapes or simple numeric patterns and translating from one representation to another
• Analyzing how both repeating and growing patterns are generated
Cereal Sorting
Lesson plan for toddlers/preschoolers
Step 1: Gather materials.
• Chenille stems
• Loop cereal such as Fruit Cheerios or Fruit Loops
• Egg carton or muffin tin
Note: Small parts pose a choking hazard and are not appropriate for children age five or under. Be sure to choose lesson materials that meet safety requirements.
Step 2: Introduce activity.
Many educators wear “themed” jewelry on special occasions, such as spider earrings and a bat necklace for Halloween or snowflake earrings for the first snowfall. For this particular lesson, you can wear a beaded necklace that has a repeating pattern of colors and shapes.
2. Tell the children that they will be making their own patterned bracelets out of cereal. Let the children know that, while the cereal is yummy to eat, first they will be sorting the cereal by color and then using the cereal to create beautiful bracelets. After they are done with the activity, the children may eat their bracelets, if you deem this appropriate.
Step 3: Engage children in lesson activities.
1. Break the children into groups of three or four. Give them an egg carton or a muffin tin and ask them to sort the cereal into the separate containers. This is also a great activity for developing fine motor skills. While the children are sorting the cereal into their containers, fold the tips of the chenille stems inward to make sure the wire does not poke the children.
2. After the children are done sorting, give each child a chenille straw. Ask the children to thread the cereal onto the chenille stems in a pattern of green, green, orange, orange. Depending on the children, you can do patterns of A-B-A-B, A-A-B-B (like the given example), A-B-C-A-B-C or any other pattern, depending on how many colors are available in the cereal.
4. To make it a neat bracelet, simply wrap the chenille stem around the child’s wrist and twist the ends together. Now the children can finish up their fun learning activity with a great snack. The best thing about this snack is that it is portable and can be taken along for the fun.
• Tell the children to create their own patterns for the bracelets. It is usually best for the children to create their patterns on the table before stringing it onto their chenille straws. That way, if they make an error or change their minds about the patterns, they can easily make adjustments without having to unstring and restring the loop cereal.
• Create cards with patterns. For example, a card might have a purple-green-yellow-blue-green color pattern. Using the card as a guide, the children can create bracelets that replicate the color pattern.
Step 4: Math vocabulary.
• Pattern: A repeated design or recurring sequence (e.g.,”The pattern of my necklace is a yellow round bead, a blue square bead, orange oval beads and again, a yellow round bead.”)
• Sort: To separate and put (people or things) in a particular order (e.g., “Sort the cereal by color into the individual compartments.”)
Step 5: Adapt lesson for toddlers or preschoolers.
Toddlers may:
• Not be ready for patterns
Home child care providers may:
• Allow the children to simply thread the cereal onto the stem. This activity alone provides fine motor skill and hand-eye coordination practice.
Home child care providers may:
• Have the children create their own patterns for the bracelets. It is usually best for the children to create their patterns on the table before stringing them onto the chenille straw. That way, if they make an error or change their mind about the pattern, they can easily make adjustments without having to unstring and restring the loop cereal.
• Create cards with patterns. For example, a card might have a purple-green-yellow-blue-green color pattern. Using the card as a guide, the children can create bracelets that replicate the color pattern.
Suggested Books
• Pattern by Henry Arthur Pluckrose (New York: Scholastic Library Publishing, 1995)
• Pattern Fish by Trudy Harris (Minneapolis, MN: Lerner Pub Group, 2007)
Outdoor Connections
The children can collect different types of leaves and sort, categorize and create patterns using the leaves that they have collected. Glue the leaves onto a piece of construction paper in the pattern that the children have created. They can create patterns by leaf color, type (maple, oak) or size. |
Adding Real Numbers Worksheet | Problems & Solutions
• Page 1
1.
Find the sum.
(- $\frac{5}{9}$) + $\frac{2}{9}$ + (- $\frac{7}{9}$)
a. - $\frac{10}{9}$ b. - $\frac{1}{9}$ c. - $\frac{14}{9}$
Solution:
(- 5 / 9) + 2 / 9+ (- 7 / 9)
[Original Expression.]
= (- 5 / 9) + (- 7 / 9) + 2 / 9
[Use commutative property.]
= ((- 5 / 9 ) + (- 7 / 9)) + (2 / 9)
[Use associative property.]
= (- 12 / 9) + (2 / 9)
= - 10 / 9
2.
Find the sum.
$2\frac{1}{5}$ + $1\frac{2}{5}$ + $4\frac{1}{5}$
a. $7\frac{1}{5}$ b. $7\frac{4}{5}$ c. $7\frac{5}{5}$ d. $1\frac{4}{5}$
Solution:
= 115 + 75 + 215
21 / 5 + 12 / 5 + 41 / 5
[Write mixed fraction as improper fraction.]
= (115 + 75 ) + (215)
[Use associative property.]
= 185 + 215
= 395
[Write improper fraction as mixed fraction.]
The sum is 74 / 5.
3.
Find the sum.
(- $\frac{5}{7}$) + $\frac{2}{7}$ + $\frac{9}{7}$
a. $\frac{6}{7}$ b. - $\frac{7}{6}$ c. - $\frac{6}{7}$ d. $\frac{7}{6}$
Solution:
(- 5 / 7) + 2 / 7 + 9 / 7
[Original Expression.]
= (- 5 / 7) + 9 / 7 + 2 / 7
[Use commutative property.]
= (- 5 / 7 + 9 / 7) + (2 / 7)
[Use associative property.]
= (4 / 7) + (2 / 7)
= 6 / 7
4.
Find the sum.
15 + 39
a. 44 b. 74 c. 54 d. 64
Solution:
15 + 39
[To add two numbers with same sign, first add their absolute values.]
= 54
[Attach the common sign.]
5.
Find the sum.
16 + (- 64)
a. 48 b. - 48 c. 58 d. 63
Solution:
To add two numbers with opposite sign, first subtract the smaller absolute value from the larger one and attach the sign of the larger absolute value.
16 + (- 64)
= - ( | - 64 | - | 16 | )
= - (64 - 16)
[Find the absolute value of the numbers.]
= - 48
6.
Find the sum.
- 52 + (- 19)
a. - 71 b. - 76 c. 76 d. 71
Solution:
To add two numbers with same sign, first add their absolute values and attach the common sign.
- 52 + (- 19)
= - (| - 52 | + | - 19 |)
= - (52 + 19)
[Find absolute value of the numbers.]
= - 71
7.
(- 8 + 6) + 13
a. 14 b. 16 c. 8 d. 11
Solution:
(- 8 + 6) + 13 = - 8 + (6 + 13)
[Use associative property.]
= - 8 + 19
= 11
[Subtract their absolute values and attach the sign of the number with larger absolute value.]
8.
- 0.9 + 8 + 0.9
a. 7.1 b. 8 c. 9.8 d. 8.9
Solution:
- 0.9 + 8 + 0.9 = - 0.9 + 0.9 + 8
[Use commutative property.]
= (- 0.9 + 0.9) + 8
[Use associative property.]
= 0 + 8
[Use inverse property.]
= 8
[Use identity property.]
9.
8 + (- $\frac{5}{6}$) + (- $\frac{1}{6}$)
a. 7 b. 8 c. 10 d. 4
Solution:
8 + (- 56) + (- 16) = 8 + [(- 56) + (- 16)]
[Use associative property.]
= 8 + (- 1)
[Add - 5 / 6 and - 1 / 6.]
= 7
10.
Find the sum.
(- $\frac{2}{5}$) + (- $\frac{3}{5}$) + $\frac{2}{5}$
a. - $\frac{1}{3}$ b. - $\frac{3}{5}$ c. - $\frac{6}{7}$ d. - $\frac{4}{5}$
Solution:
(- 2 / 5) + (- 3 / 5) + 2 / 5
[Original Expression.]
= - 2 / 5 + 2 / 5 - 3 / 5
[Use commutative property.]
= (- 2 / 5 + 2 / 5) + (- 3 / 5)
[Use associative property.]
= 0 + (- 3 / 5)
[Use inverse property.]
= - 3 / 5
[Use identity property.] |
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# Find the Coordinates of Circumcentre and Radius of Circumcircle of ∆Abc If A(7, 1), B(3, 5) and C(2, 0) Are Given. - Geometry
ConceptConcepts of Coordinate Geometry
#### Question
Find the coordinates of circumcentre and radius of circumcircle of ∆ABC if A(7, 1), B(3, 5) and C(2, 0) are given.
#### Solution
Let the circumcentre be $P\left( a, b \right)$.
The given points are A(7, 1), B(3, 5) and C(2, 0).
The circumcircle passes through the points A, B and C and the thus,
PA = PB = PC
$\Rightarrow {PA}^2 = {PB}^2 = {PC}^2$
$P A^2 = P B^2$
$\Rightarrow \left( 3 - a \right)^2 + \left( 5 - b \right)^2 = \left( 7 - a \right)^2 + \left( 1 - b \right)^2$
$\Rightarrow 9 + a^2 - 6a + 25 + b^2 - 10b = 49 + a^2 - 14a + 1 + b^2 - 2b$
$\Rightarrow a - b = 2 . . . . . \left( 1 \right)$
$P A^2 = P C^2$
$\Rightarrow \left( 7 - a \right)^2 + \left( 1 - b \right)^2 = \left( 2 - a \right)^2 + \left( 0 - b \right)^2$
$\Rightarrow 49 + a^2 - 14a + 1 + b^2 - 2b = 4 + a^2 - 4a + b^2$
$\Rightarrow 5a + b = 23 . . . . . \left( 2 \right)$
$\left( 1 \right) + \left( 2 \right)$
$a = \frac{25}{6}, b = \frac{13}{6}$
$= \sqrt{\left( \frac{25}{6} - 2 \right)^2 + \left( \frac{13}{6} - 0 \right)^2}$
$= \sqrt{\left( \frac{13}{6} \right)^2 + \left( \frac{13}{6} \right)^2}$
$= \frac{13}{6}\sqrt{2}$ |
Group Isomorphism: Definition, Properties, Examples
An isomorphism of groups is a special kind of group homomorphisms. It preserves every structure of groups. In this article, we will learn about isomorphism between groups, related theorems, and applications.
Definition of Isomorphism
A map Φ: (G, 0) → (G′, *) between two groups is called an isomorphism if the following conditions are satisfied:
• Φ is a group homomorphism, that is, Φ(ab)=Φ(a)Φ(b) ∀ a, b ∈ G.
• Φ is one-to-one.
• Φ is onto.
A bijective group homomorphism between groups is called an isomorphism.
For example, the identity map i: Z → Z defined by i(n)=n ∀ n ∈ Z is an example of an isomorphism. Below are a few more examples of isomorphism of groups.
The map Φ: (Z5, +) → (Z5, +) defined by Φ($\bar{x}$)=3$\bar{x}$ ∀ $\bar{x}$ ∈ Z5 is an example of group isomorphism.
The map Φ: (Z, +) → (2Z, +) defined by Φ(n)=2n ∀ n ∈ Z is an isomorphism.
Properties of Isomorphism
Property 1: If Φ: (G, 0) → (G′, *) is a group isomorphism, then the kernel of the map is trivial, that is, ker(Φ)={eG}.
Proof: For a proof, visit the page: Injectivity criteria for homomorphism.
Property 2: If Φ: (G, 0) → (G′, *) is a group isomorphism, then we have:
1. order of a = order of Φ(a) ∀ a ∈ G
2. Both G and G′ have the same cardinality.
Property 3: Let Φ: (G, 0) → (G′, *) be a group isomorphism. Then the following are true:
1. G is abelian if and only if G′ is abelian
2. G is cyclic if and only if G′ is cyclic
Remark:
• We see that both abelian and cyclic properties are preserved by a group isomorphism.
• If a is a generator of G, then Φ(a) is a generator of G′.
Property 4: If Φ: (G, 0) → (G′, *) is a group isomorphism, then the inverse map Φ-1: (G′, *) → (G, 0) is also an isomorphism.
Property 5: The composition of two isomorphisms is an isomorphism.
Non Isomorphic Groups
Example 1: The groups (Z, +) and (Q, +) are not isomorphic.
Solution:
We know that (Z, +) is a cyclic group whereas (Q, +) is a non-cyclic group, see the page on cyclic groups. As the cyclic property is preserved by an isomorphism, we conclude that both the additive groups Z and Q are not isomorphic.
Example 2: The groups (Q, +) and (R, +) are not isomorphic.
Solution:
If there is an isomorphism between the additive groups Q and R, then they must have the same cardinality. But one knows that both Q and R have different cardinalities. So (Q, +) and (R, +) cannot be isomorphic.
Example 3: The groups (Q, +) and (Q+, .) are not isomorphic.
Solution:
Example 4: The groups (R×, .) and (R, +) are not isomorphic.
Solution:
Click below to Read Isomorphism Theorems:
First Isomorphism Theorem
Second Isomorphism Theorem
Third Isomorphism Theorem
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Free Algebra Tutorials!
# Multiplying and Dividing Fractions
## Multiplication
Multiply straight across: Multiply the numerators to get the new numerator. Multiply the denominators to get the new denominator.
## Division
The first fraction is said to be “divided by” the second fraction. The fraction you are “dividing by” is called the “divisior.” To divide, “invert” the divisor (turn it upside down), then multiply.
(When a fraction is inverted we call the resulting fraction the “reciprocal” of the original number. Any whole number can be thought of as a fraction with 1 in the denominator. Since , the reciprocal of 2 is .)
## Chain Multiplication and Division
There are times (especially in science classes in high school and beyond) when you need to multiply and divide a whole series of fractions and whole numbers. This is actually easy to do. Think of making a shish kabob: a skewer loaded up with chunks of meat, tomatoes, onions, bell pepper, mushrooms, etc. and cooked on a grill. Think of one long fraction bar as a skewer. To multiply a fraction in the list, simply skewer it on. To divide by a fraction, flip it over and skewer it on. Any whole number can be thought of as a fraction with 1 in the denominator, so can treat it like any other fraction. Multiplying by a whole number puts the whole number into the numerator. Dividing by a whole number puts the whole number into the denominator. (Once everything is in place we can ignore the 1’s.) |
# A can complete 1/2 of the work in 20 days. B can complete 25% of the same work in 15 days. They started working together and B left the work after 7 days. If C joined A 3 days after B left the work and total work is completed after 18 days of start, then in how many days B and C together will complete 3/4 of the work?
1. $$11\frac{{10}}{{17}}$$
2. $$10\frac{{10}}{{17}}$$
3. $$12\frac{{13}}{{17}}$$
4. $$9\frac{{11}}{{17}}$$
Option 2 : $$10\frac{{10}}{{17}}$$
## Detailed Solution
A can complete 1/2 of the work in 20 days
⇒ A can complete the work in 40 days
B can complete 25% of same work in 15 days
⇒ B can complete the work in 60 days
Let total work be (LCM of 40, 60) = 120 units
Efficiency of A = 120/40 = 3 units
Efficiency of A = 120/60 = 2 units
⇒ Work done by A and B together in 7 days = 7 × (3 + 2) = 35 units
⇒ Remaining work = 120 – 35 = 85 units
C joined A 3 days after B left the work
Therefore, A worked alone for 3 days
⇒ Work done by A in 3 days = 3 × 3 = 9 units
⇒ Remaining work = 85 – 9 = 76 units
⇒ Work done by A in next 8 days = 3 × 8 = 24 units
⇒ Remaining work = 76 – 24 = 52 units
⇒ Remaining work is done by C in 8 days
Efficiency of C = 52/8 = 6.5 units
⇒ Work done by B and C together in 1 day = 2 + 6.5 = 8.5 units
∴ B and C together will complete 3/4 of the work in $$= \frac{{\frac{3}{4} \times 120\;}}{{8.5}} = 10\frac{{10}}{{17}}\;days$$ |
# How do you solve abs(3x+2)-5<0?
Aug 1, 2016
$\text{ } - \frac{7}{3} < x < 1$
#### Explanation:
$| 3 x + 2 | < 5$
Note that $| 3 x + 2 |$ is always positive so there is a limit to the negative value of $3 x + 2$
Suppose $3 x + 2 = 5 \implies x = 1$
Then the maximum value attributable to $x$ is such that $x < 1$
Suppose $3 x + 2 = - 5 \implies x = - \frac{7}{3}$
So the minimum value of $x$ is such that $- \frac{7}{3} < x$
So $\text{ } - \frac{7}{3} < x < 1$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{In summery:}}$
solve for $3 x + 2 = | \pm 5 |$ and make the appropriate value of $x$ the $\textcolor{m a \ge n t a}{\underline{\text{not inclusive}}}$ upper and lower bounds. $\text{ } - \frac{7}{3} \textcolor{m a \ge n t a}{<} x \textcolor{m a \ge n t a}{<} 1$
'.............................................................................................
$\textcolor{b r o w n}{\text{Just for information:}}$
If they had been $\textcolor{b l u e}{\underline{\text{inclusive}}}$ then we would have $- \frac{7}{3} \le x \le 1$ |
# 30MSC 2023 #14: Squaring
Squaring any number or polynomial can be done using the traditional method of multiplication. However, in Vedic Maths, there are several techniques, some specific and 1 general, available to easily perform squaring. The general method of squaring is called the duplex method, which is based on the Vertically and Crosswise sutra.
The duplex, denoted as D of numbers are as follows:
1. D(a) = a^2
2. D(ab) = 2 x (a x b)
3. D(abc) = 2 x (a x c) + b^2
4. D(abcd) = 2 x (a x d + b x c)
5. D(abcde) = 2 x (a x e + b x d) + c^2
In general, for numbers with an even number of digits, their duplex is twice the sum of the products of the digits equidistant from the center. For numbers with an odd number of digits, their duplex, is twice the sum of the products of the digits equidistant from the center plus the square of the middle digit.
The square of a number is the sum of its duplexes.
In determining the duplexes of polynomials, we simply replace the digits by the terms of polynomials in the formulae.
In our featured example which was taken in the 1st Math2Shine International Vedic Mathematics Competition(open division) the square of the polynomial (2x^2 – 3x + 4) is computed as:
1. D(2x^2) = (2x^2)^2 = 4x^4
2. D(2x^2 – 3x) = 2 x (2x^2)(– 3x) = – 12 x^3
3. D(2x^2 – 3x + 4) = 2 x (2x^2)(4) + (3x)^2 = 16x^2 + 9x^2 = 25x^2
4. D( – 3x + 4) = 2 x (-3x)(4) – 24x
5. D(4) = (4)^2 = 16
(2x^2 – 3x + 4)^2 = 4x^4 – 12x^3 + 25x^2 – 24x + 16
Other squaring techniques are discussed in Chapter 4 of our forthcoming book 30 Master Strategies in Computing, Vol II.
Follow us on our Facebook page, MATH-Inic Philippines at https://www.facebook.com/MATHInicPhils to see tomorrow’s 30MSC 2023 #15: Base Division |
Calculus Derivative Rules for Powers & +/-Constants Math Help Fun Game Tips: - A derivative in Calculus can be described as an instantaneous rate of change. The slope of a curve at a point P is an example of a derivative. The speed of a moving particle at a time T is an example of a derivative. - A sample curve or motion can be represented by a relation such as y=x2 or the equivalent f(x)=x2. The derivative corresponding to y=x2 can be represented as dy/dx = 2x or y' = 2x or f'(x) = 2x. The three derivative symbols dy/dx, y' and f'(x) are in common use by various authors. - Similarly, the relation represented as s=t3 or s(t)=t3 has it's corresponding derivative represented as ds/dt = 3t2 or s'(t) = 3t2. - A derivative power rule is used to get dy/dx = 2x from the given relation y=x2. The same derivative power rule is used to get ds/dt = 3t2 from the given relation s(t)=t3. - The following 3 derivative rules are used in this game, where c and n can be positive or negative. 1) In general, the derivative power rule for y=xn yields the derivative dy/dx = nxn-1. For example, the derivatives of y=x4 and s(t)=t-6 are respectively dy/dx=4x3 and s'(t)=-6t-7. 2) In general, the derivative constant rule for y=c yields the derivative dy/dx = 0. For example, the derivatives of y=-3 and s(t)=8 are respectively dy/dx=0 and s'(t)=0. 3) The derivative rule for a constant times a power such as y=cxn yields the derivative dy/dx = cnxn-1. For example, the derivatives of y=-8x-4 and g(t)=-3t6 are respectively y'=32x-5 and g'(t)=-18t5. - The score report automatically appears after you have made 15 choices. - The Math score is based on your choices only and does not count the fish hits. - The Game score is reduced by the number of fish hits. - The game can be played using the mouse by itself or using the keyboard by itself. - If the game doesn't respond to keyboard input, click inside the game area to reset the game's focus. - Adjust the game's speed by pressing the + or - key repeatedly. - Refresh/Reload the webpage to start over again. More SliderMath Copyright © All Rights Reserved.
E0 |
If you desire to discover more, climate please keep reading, and also you won"t be disappointed.
You are watching: 120 is what percent of 90?
## Step through step method for calculating what percent of 90 is 120
We currently have our first value 90 and also the second value 120. Let"s assume the unknown worth is Y i m sorry answer us will discover out.
As we have all the forced values we need, now we deserve to put them in a an easy mathematical formula together below:
STEP 1Y = 120/90
By multiply both numerator and denominator through 100 we will get:
STEP 2Y = 120/90 × 100/100 = 133.333/100
STEP 3Y = 133.333
Finally, we have found the worth of Y which is 133.333 and that is our answer.
You can use a calculator to uncover what percent that 90 is 120, just get in 120 ÷ 90 × 100 and you will get your answer i m sorry is 133.333
Here is a calculator to solve percent calculations such together what percent that 90 is 120. You have the right to solve this type of calculation with your values by entering them into the calculator"s fields, and also click "Calculate" to gain the an outcome and explanation.
What percent of
is
Calculate
## Sample questions, answers, and how to
Question: her uncle had 90 share of his own agency a few years earlier, and also now he has 120 of them. What percent of the shares of his agency he has now?
Answer: He has 133.333 percent of shares of his firm now.
How To: The an essential words in this problem are "What Percent" because they allow us recognize that it"s the Percent that is missing. So the 2 numbers that it gives us must be the "Total" and the "Part" we have.
Part/Total = Percent
In this case, it"s the total that our uncle owned. For this reason we placed 90 ~ above the bottom of the portion and 120 top top top. Currently we"re prepared to figure out the component we don"t know; the Percent.
120/90 = Percent
To uncover the percent, all we must do is transform the portion into the percent form by multiplying both top and also bottom part by 100 and also here is the method to figure out what the Percent is:
120/90 × 100/100 = 133.333/100
133.333 = Percent
And that means he has actually 133.333 percent of the share of his company now.
## Another step by action method
Step 1: Let"s fix the equation for Y by an initial rewriting it as: 100% / 90 = Y% / 120
Step 2: fall the percent marks to leveling your calculations: 100 / 90 = Y / 120
Step 3: main point both political parties by 120 to isolate Y on the best side the the equation: 120 ( 100 / 90 ) = Y |
Practice with the help of Spectrum Math Grade 5 Answer Key Chapter 5 Pretest regularly and improve your accuracy in solving questions.
Check What You Know
Write each fraction in simplest form.
Question 1.
a.
Answer: $$\frac{7}{8}$$
To add fractions with like denominators, add the numerators and use the common denominator.
$$\frac{1}{8}$$ + $$\frac{6}{8}$$ = $$\frac{1+6}{8}$$ = $$\frac{7}{8}$$
b.
Answer: $$\frac{6}{7}$$
To add fractions with like denominators, add the numerators and use the common denominator.
$$\frac{3}{7}$$ + $$\frac{3}{7}$$ = $$\frac{3+3}{7}$$ = $$\frac{6}{7}$$
c.
Answer: $$\frac{3}{6}$$
To add fractions with like denominators, add the numerators and use the common denominator.
$$\frac{2}{6}$$ + $$\frac{1}{6}$$ = $$\frac{2+1}{6}$$ = $$\frac{3}{6}$$
d.
Answer: $$\frac{7}{9}$$
To add fractions with like denominators, add the numerators and use the common denominator.
$$\frac{4}{9}$$ + $$\frac{3}{9}$$ = $$\frac{4+3}{9}$$ = $$\frac{7}{9}$$
Question 2.
a.
Answer: 1 $$\frac{1}{20}$$
To add fractions with like denominators, add the numerators and use the common denominator.
$$\frac{4}{5}$$ + $$\frac{2}{8}$$
LCD is 40.
$$\frac{32}{40}$$ + $$\frac{10}{40}$$
= $$\frac{32+10}{40}$$ = $$\frac{42}{10}$$ = 1 $$\frac{1}{20}$$
b.
$$\frac{3}{6}$$ + $$\frac{2}{4}$$
LCD is 12.
$$\frac{6}{12}$$ + $$\frac{6}{12}$$
= $$\frac{6+6}{12}$$ = $$\frac{12}{12}$$ = 1
c.
Answer: 11 $$\frac{8}{9}$$
4$$\frac{2}{3}$$ + 7$$\frac{2}{9}$$
4 + $$\frac{2}{3}$$ + 7 + $$\frac{2}{9}$$
4 + 7 = 11
$$\frac{2}{3}$$ + $$\frac{2}{9}$$
LCD is 9.
$$\frac{6}{9}$$ + $$\frac{2}{9}$$ = $$\frac{8}{9}$$
11 + $$\frac{8}{9}$$ = 11$$\frac{8}{9}$$
d.
Answer: 7 $$\frac{37}{40}$$
1$$\frac{4}{5}$$ + 6$$\frac{1}{8}$$
1 + $$\frac{4}{5}$$ + 6 + $$\frac{1}{8}$$
1 + 6 = 7
$$\frac{4}{5}$$ + $$\frac{1}{8}$$
LCD is 40.
$$\frac{32}{40}$$ + $$\frac{5}{40}$$ = $$\frac{37}{40}$$
7 + $$\frac{37}{40}$$ = 7$$\frac{37}{40}$$
Question 3.
a.
To subtract fractions with like denominators, subtract the numerators and use the common denominator.
$$\frac{7}{8}$$ – $$\frac{1}{8}$$ = $$\frac{7-1}{8}$$ = $$\frac{6}{8}$$ = $$\frac{3}{4}$$
b.
To subtract fractions with like denominators, subtract the numerators and use the common denominator.
$$\frac{5}{9}$$ – $$\frac{4}{9}$$ = $$\frac{5-4}{9}$$ = $$\frac{1}{9}$$
c.
To subtract fractions with like denominators, subtract the numerators and use the common denominator.
$$\frac{8}{10}$$ – $$\frac{3}{10}$$ = $$\frac{8-3}{10}$$ = $$\frac{5}{10}$$
d.
To subtract fractions with like denominators, subtract the numerators and use the common denominator.
$$\frac{7}{12}$$ – $$\frac{1}{12}$$ = $$\frac{7-1}{12}$$ = $$\frac{6}{12}$$
Question 4.
a.
Answer: $$\frac{1}{8}$$
$$\frac{7}{8}$$ – $$\frac{3}{4}$$
LCD is 8.
$$\frac{7}{8}$$ – $$\frac{6}{8}$$
$$\frac{7-6}{8}$$ = $$\frac{1}{8}$$
b.
Answer: $$\frac{2}{35}$$
$$\frac{6}{7}$$ – $$\frac{4}{5}$$
LCD is 35.
$$\frac{30}{35}$$ – $$\frac{28}{35}$$
$$\frac{30-28}{8}$$ = $$\frac{2}{35}$$
c.
Answer: $$\frac{22}{63}$$
$$\frac{4}{7}$$ – $$\frac{2}{9}$$
LCD is 63.
$$\frac{36}{63}$$ – $$\frac{14}{63}$$
$$\frac{36-14}{63}$$ = $$\frac{22}{63}$$
d.
Answer: 4 $$\frac{1}{12}$$
6 $$\frac{1}{4}$$ – 2$$\frac{1}{6}$$
6 + $$\frac{1}{4}$$ – 2 – $$\frac{1}{6}$$
6 – 2 = 4
$$\frac{1}{4}$$ – $$\frac{1}{6}$$
LCD is 12.
$$\frac{3}{12}$$ – $$\frac{2}{12}$$ = $$\frac{1}{12}$$
4 + $$\frac{1}{12}$$ = 4$$\frac{1}{12}$$
Solve each problem. Show your work.
Question 5.
Julianne needs 7 yards of string for her kite. She has $$\frac{5}{8}$$ yards. How many more yards does Julianne need for her kite?
Julianne needs _____________ more yards of string.
Given,
Julianne needs 7 yards of string for her kite.
She has $$\frac{5}{8}$$ yards.
7 – $$\frac{5}{8}$$ = 6$$\frac{3}{8}$$ yards
Julianne needs 6$$\frac{3}{8}$$ more yards of string.
Question 6.
Mrs. Thompson’s cookie recipe includes $$\frac{1}{3}$$ cup sugar and 4 cups flour. How many cups of sugar and flour does Mrs. Thompson need for her cookies?
Mrs. Thompson needs ______________ cups of ingredients.
Given,
Mrs. Thompson’s cookie recipe includes $$\frac{1}{3}$$ cup sugar and 4 cups flour.
$$\frac{1}{3}$$ + 4 = 4$$\frac{1}{3}$$
Mrs. Thompson needs 4$$\frac{1}{3}$$ cups of ingredients.
Question 7.
Marlon watched a movie 1$$\frac{8}{9}$$ hours long. Jessie watched a movie 2$$\frac{2}{7}$$ hours long. How much longer was Jessie’s movie than Marlon’s?
Jessie’s movie was _______________ hours longer.
Given,
Marlon watched a movie 1$$\frac{8}{9}$$ hours long. Jessie watched a movie 2$$\frac{2}{7}$$ hours long.
2$$\frac{2}{7}$$ – 1$$\frac{8}{9}$$ = $$\frac{25}{63}$$
Jessie’s movie was $$\frac{25}{63}$$ hours longer.
Question 8.
Carrie is running in a track meet. In one race she must run $$\frac{1}{4}$$ mile, and in a second race she must run 1$$\frac{2}{5}$$ miles. How many miles must Carrie run in all?
Carrie must run _____________ miles.
Given,
Carrie is running in a track meet. In one race she must run $$\frac{1}{4}$$ mile, and in a second race she must run 1$$\frac{2}{5}$$ miles.
$$\frac{1}{4}$$ + 1$$\frac{2}{5}$$ = 1$$\frac{13}{20}$$
Carrie must run 1$$\frac{13}{20}$$ miles.
Question 9.
David practiced soccer twice last week, On Monday, he practiced 2$$\frac{1}{3}$$ hours. On Wednesday, he practiced 1$$\frac{7}{9}$$ hours. How much longer did David practice on Monday?
David practiced _____________ hours longer on Monday.
David practiced soccer twice last week, On Monday, he practiced 2$$\frac{1}{3}$$ hours. On Wednesday, he practiced 1$$\frac{7}{9}$$ hours.
2$$\frac{1}{3}$$ – 1$$\frac{7}{9}$$ = $$\frac{5}{9}$$
David practiced $$\frac{5}{9}$$ hours longer on Monday. |
Tuesday, June 30, 2009
Vedic Mathematics Lesson 2: 10's Complements
You can read about my interest in Vedic Mathematics in this earlier post and find a spectacular application of mental computation using Vedic Mathematics principles in Lesson 1 of this series of posts.
Today's lesson is going to be quite short. I am going to explain how to calculate the 10's complement of any number. Essentially, the 10's complement of a number tells you how far the number is below the next higher power of 10. For instance, 89 is 11 below 100. I call 11 the 10's complement of 89. Some texts refer to 11 as the "deficit" of 89 also.
The 10's complement of a number is very useful in a variety of Vedic Mathematical computations. Thus, learning to work out the 10's complement quickly and (preferably) mentally will come in handy when we proceed to later lessons in Vedic Mathematics.
The sutra that tells us how to compute the 10's complement of a number reads Nikhilam Navatascaramam Dasataha. Literally translated, it means All From 9 And The Last From 10.
The practical application of it is actually quite easy to follow directly from the translation of the sutra. Simply put, take your number and subtract each number from 9 as you go from left to right (All From 9). Put down the answers you get as the digits of the 10's complement going from left to right. When you get to the last digit (right-most digit) of the number for which you are finding the complement, subtract it from 10 (The Last From 10) and write this answer down as the last digit (right-most) of the answer.
Note that since all the subtractions are of single digits from 9 and the value being subtracted from is 9 (which is the highest single-digit number in the decimal system), there arises no question of borrowing digits or doing other mental gymnastics to get the individual digits of the answer. Hopefully, subtracting a single-digit number from 10 should not involve any extraordinary mental gymnastics either. However, that last subtraction can lead to a minor problem we deal with later in this lesson.
Let us apply this lesson to a simple example. Let us take 389,384,753 as the number for which we need the 10's complement. Note that the number has 9 digits, so we are looking for the difference between 1,000,000,000 (1 followed by 9 zeroes, making it a 10-digit number) and the given number (10 raised to the power of 9 is the next higher power of 10 for the given number).
Taking the first digit of the given number from left, we get 3. Subtract it from 9 to get 6. 6 is the first digit of the answer. The next digit of the answer is 9-8 = 1. The third digit is 9-9 = 0, and so on. When we get to the right-most digit of the given number, we find that the right-most digit of the answer has to be 10 - 3 = 7. Remember to subtract the last number from 10 rather than 9 to complete the answer. The answer in this case turns out to be 610,615,247. You can verify the answer in any calculator that can handle 10 digits or more. But the method will work for numbers with any number of digits, even numbers that can not be handled by any calculator because they have too many digits.
Note that we defined the 10's complement as the deficit from the next higher power of 10. This is not necessary for the method to work. This method can be put to work to find the difference between the given number and any power of 10 that has more digits than the given number. Let me illustrate with another example.
Suppose we need to find the difference between 1,000,000 and 98,567. Note that next power of 10 that is larger than the given number is 100,000. Thus, to simply find its 10's complement, we would apply the formula illustrated above and find the answer to be 1,433 (the left-most digit computes to a zero, and has therefore been dropped).
To find the deficit from a higher power of 10, first find the 10's complement. This time do not drop zeroes from the left of the answer. We find the 10's complement of the given number to be 01,433. Now pad the 10's complement to the left with 9's until it has one less digit than the power of 10 from which we are trying to find the deficit. Another way to express this is as follows: pad the 10's complement to the left with 9's until it has the same number of digits as the power of 10 has zeroes. In our case, the power of 10 we are finding the deficit from has six zeroes. So, padding out the 10's complement with 9's to the left so that the answer is 6 digits long gives us 901,433.
Yet another way to express this that may be more intuitive is: pad the given number with zeroes to the left so that it has as many digits as the power of 10 has zeroes. Then find the 10's complement of the padded number using the same rule as before. By this method, first we get our padded number as 098,567. Finding the 10's complement of this number using the rule we explained in the beginning, we get 901,433.
The only trouble you might encounter in the application of this method is if the last digit of the number is a zero. In that case, subtracting it from 10 gives you a 2-digit answer (10) rather than a single-digit answer. The way to deal with this is to then put down zero as the last digit of the answer and carry over the 1 to the left hand side (add it to the number you found earlier for that digit). If that carryover leads to the second digit becoming 10, repeat the procedure, carrying over extra digits to the left as long as is necessary.
That may sound confusing, so the easier way to deal with this is as follows: if the number consists of n zeroes at the end, leave them off initially. Find the 10's complement of the remaining number with respect to the power of 10 just above the left-over number. Then add n zeroes back to the right of the answer you get.
Let me illustrate by finding the 10's complement of 89,000. n, in this case, is 3. By dropping the 3 zeroes from the end of 89,000, we get 89. The 10's complement of 89 with respect to the next higher power of 10 (100) is 11. Therefore, the 10's complement of 89,000 is 11,000 (which is 11 with 3 zeroes added back to its right).
That is all there is to it! It should be easy to reel the 10's complements of any number off in seconds using the mental trick illustrated here. Remember to practice! Happy computing and good luck!!
Note that there are applications out there that define the 10's complement as one more than the 10's complement we have computed in this lesson. That kind of 10's complement is useful in some computations involving subtraction and addition. Remember to not get confused by this distinction between the different definitions of 10's complements.
------------------------------------------
I taught my kids the trick to finding the reciprocal of any 2-digit number ending in 9. I thought it would be confusing, given their ages, but they picked it up right away and had no trouble applying either method 1 or method 2 to any given problem. They then went and demonstrated their new-found talent to their mother, who was very proud of them and very happy with me! Now, I have to test them in a few days and make sure they remember what I taught them. That will also give me a chance to make sure I remember what I taught them!
In other news, it looks like I will be taking two weeks off to visit and take care of my parents after my father's surgery starting sometime next week. I have convinced my wife that nothing will go wrong when I am not around. That does not mean that she will be entirely at peace, but for now, she seems confident about handling things during my absence.
Our plan to visit my brother-in-law to help with his mother's surgery and its aftermath turned into a bit of a disastrous fiasco because of various problems. We planned to go out there on Friday. But the aircraft on the flight we were supposed to take was replaced with one that had 70 fewer seats. So, we changed our plans and were accommodated on a flight that would take us where we wanted to go with a connection. But at the last minute, this aircraft developed a mechanical problem that delayed the flight so the connection would not work. We had to give up that day.
To make a long story short, we ended up trying repeatedly to get out during the weekend, and then my wife and kids tried on Monday, and today also. Ultimately, the airline managed to accommodate them today evening on a flight and they are off. Since she is going to her brother's later than originally planned, we are still not sure when she is going to come back. We had originally planned on her being back before I left to visit my parents. But now we might have to reevaluate that option.
The weed-killer application using the hose-end sprayer actually worked. The weeds shrivelled up and died, and the lawn looks much better now. So, I might be switching to the hose-end sprayer permanently and ditching the pressure sprayer altogether.
This last weekend saw me doing more yardwork as I had to dig a 1 foot x 1 foot x 1 foot hole in the ground to plant a rosebush. The rosebush unfortunately sat in a friend's car in the hot sun for a while after it was bought, and came to us with shriveled leaves. We watered it (in the pot) for a few days, but it still did not show much signs of life. I tried arguing that it was a waste of effort digging such a large hole for a plant that may already be dead. But my wife convinced me that all the plant needed was abundant sunshine and natural soil to thrive. So, the plant is in the ground, under more than abundant sunshine, and has been watered for a couple of days now. Still no signs of life, leave alone thriving. Oh well, at least I got some exercise...
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# Find a fraction which becomes ( 1/2 ) when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes ( 1/3 ) when 7 is subtracted from the numerator and 2 is subtracted from the denominator.
• 0
An important and exam oriented question from linear equations in two variables as it was already asked in various examinations in which we have given that when 1 is subtracted from the numerator, the fraction becomes 1/2 and when 2 is added to the denominator, and the fraction becomes ( 1/3 ) when 7 is subtracted from the numerator and 2 is subtracted from the denominator. And we asked to find the fraction
Kindly solve the above problem
RS Aggarwal, Class 10, chapter 3E, question no 22
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1. Let the required fraction be x/y . Then, we have:
x−1/ y+2 = 1/2
⇒ 2(x – 1) = 1(y + 2)
⇒2x – 2 = y + 2
⇒2x – y = 4 ……(i)
Again, x−7/ y−2 = 1/3
⇒3(x – 7) = 1(y – 2)
⇒3x – 21 = y – 2
⇒ 3x –y = 19 ……(ii)
On subtracting (i) from (ii), we get: x = (19 – 4) = 15
On substituting x = 15 in (i), we get: 2 × 15 – y = 4
⇒ 30 – y = 4
⇒y = 26
∴ x = 15 and y = 26
Hence, the required fraction is 15/26
• 1 |
How do you simplify (2times10^-6)(6times10^7)?
Feb 13, 2017
See the entire simplification process below:
Explanation:
First, we can rewrite this expression as:
$\left(2 \times 6\right) \left({10}^{-} 6 \times {10}^{7}\right) = 12 \left({10}^{-} 6 \times {10}^{7}\right)$
We can now use this rule for exponents to combine the 10s terms:
${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$
$12 \left({10}^{\textcolor{red}{- 6}} \times {10}^{\textcolor{b l u e}{7}}\right) = 12 \times {10}^{\textcolor{red}{- 6} + \textcolor{b l u e}{7}} = 12 \times {10}^{1} = 12 \times 10 = 120$ |
# Ratio and Proportion Practice Set
1. A bag contains 50 p, 25 p and 10 p coins in the ratio 5 : 9 : 4, amounting to Rs.206. Find the number of coins of each type respectively. — एक बैग में अनुपात 5:9:4 में क्रमशः 50P, 25P और 10P के सिक्के हैं, जो 206 रुपये है। प्रत्येक प्रकार के सिक्के की संख्या बताऐं। A. 200, 305, 106 B. 210, 350, 148 C. 200, 360, 160 D. 200, 160, 180 Answer: Option C Explanation: Let the number of 50 p, 25 p and 10 p coins be 5?, 9? and 4? respectively. Then, 5?⁄2 + 9?⁄4 + 4?⁄10 = 206 50? + 45? + 8? = 4120 103? = 4120 ? = 103⁄4120 ? = 40 Number of 50 p coins = (5 × 40) = 200; number of 25 p coins = (9 × 40) = 360; number of 10 p coins = (4 × 40) = 160 View / Hide Answer
2. The ratio of length to width of a rectangular sheet of paper is 5 : 3. If the width of the sheet is 18 cm, find its length? — कागज की आयताकार शीट की लंबाई से चौड़ाई का अनुपात 5: 3. है यदि शीट की चौड़ाई 18 सेमी है, तो इसकी लंबाई बताऐं? A. 10 cm B. 20 cm C. 25 cm D. 30 cm Answer: Option D Explanation:Let the length of sheet of paper be x cm. Then the ratio of length to width = ? : 18 Thus, ? : 18 = 5 : 3 ? × 3 = 18 × 5 ? = 30 Hence the length of paper = 30 cm View / Hide Answer
3. The ratio between the number of men and women in an office is 5 : 7. If the number of women working in the office is 56. Find the number of men working in the office? — कार्यालय में पुरुषों और महिलाओं की संख्या के बीच अनुपात 5: 7 है। यदि कार्यालय में काम करने वाली महिलाओं की संख्या 56 है। कार्यालय में काम करने वाले पुरुषों की संख्या बताऐं? A. 40 B. 50 C. 45 D. 60 Answer: Option A Explanation:5 : 7 = ? : 56 (suppose number of men = ?) ? = 40 (by the first property) Therefore, number of men in the office = 40 View / Hide Answer
4. If (a + b) : (a – b) = 15 : 1, then the value of ?2 – ?2 is : A. 56 B. 15 C. 112 D. 8 Answer: Option B Explanation:(a + b)⁄(a − b) = 15⁄1 ?⁄? = 8⁄7 (by componendo and dividend) Therefore a2 – b2 = 64 – 49 = 15 View / Hide Answer
5. The students in three classes are in the ratio of 2 : 3 : 4. If 40 students are added in each class, the ratio becomes 4 : 5 : 6. Find the total number of students in all the three classes is : — तीन वर्गों में छात्र 2 : 3 : 4 के अनुपात में हैं। यदि प्रत्येक वर्ग में 40 छात्र जोड़े जाते हैं, तो अनुपात 4 : 5 : 6 हो जाता है। सभी तीन वर्गों में छात्रों की कुल संख्या प्राप्त करें: A. 270 B. 180 C. 126 D. 135 Answer: Option B Explanation:2x + 40 = 4y … (i) 3x + 40 = 5y …. (ii) 4x + 40 = 6y ….. (iii) Therefore 2x = 40 x = 20 Hence, total number of students = 2x + 3x + 4x = 9x = 9 × 20 = 180 View / Hide Answer
6. Two equal containers are filled with the mixture of milk and water. The concentration of milk in each of the container is 20% and 25% respectively. What is the ratio of water in both the containers respectively? — दो बराबर कंटेनर दूध और पानी के मिश्रण से भरे हुए हैं। प्रत्येक कंटेनर में दूध की एकाग्रता क्रमशः 20% और 25% है। क्रमशः दोनों कंटेनरों में पानी का अनुपात क्या है? A. 15:16 B. 16:15 C. 4:5 D. 5:4 Answer: Option B Explanation:Milk 20% 25% Water 80% 75% Therefore, required ratio = 80⁄75 = 16⁄15 or 16 : 15 View / Hide Answer |
# Matrices 3D
### Definición:
La multiplicación se define como:
$$(AB)C$$
Ejemplo 1:Multiplicar:
$$\left [ \begin{matrix} 1 \\\ 2 \\\ 3 \end{matrix} \right ] \left [ \begin{matrix} 4 & 5 & 6 \end{matrix} \right ] \left [ \begin{matrix} 7 & 8 & 9 \end{matrix} \right ]$$
Primero se multiplican $A$ y $B$:
$$\left [ \begin{matrix} 1 \\\ 2 \\\ 3 \end{matrix} \right ] \left [ \begin{matrix} 4 & 5 & 6 \end{matrix} \right ] = \left [ \begin{matrix} 4 & 5 & 6 \\\ 8 & 10 & 12 \\\ 12 & 15 & 18 \end{matrix} \right ]$$
Luego se multiplican todas las columnas de $AB$ con $C$.
$$\left [ \begin{matrix} 4 \\\ 8 \\\ 12 \end{matrix} \right ] \left [ \begin{matrix} 7 & 8 & 9 \end{matrix} \right ] = \left [ \begin{matrix} 28 & 32 & 36 \\\ 56 & 64 & 72 \\\ 84 & 96 & 108 \end{matrix} \right ]$$$$\left [ \begin{matrix} 5 \\\ 10 \\\ 15 \end{matrix} \right ] \left [ \begin{matrix} 7 & 8 & 9 \end{matrix} \right ] = \left [ \begin{matrix} 35 & 40 & 45 \\\ 70 & 80 & 90 \\\ 105 & 120 & 135 \end{matrix} \right ]$$$$\left [ \begin{matrix} 6 \\\ 12 \\\ 18 \end{matrix} \right ] \left [ \begin{matrix} 7 & 8 & 9 \end{matrix} \right ] = \left [ \begin{matrix} 42 & 48 & 54 \\\ 84 & 96 & 108 \\\ 126 & 144 & 162 \end{matrix} \right ]$$
$$\left [ \begin{matrix} 1 \\\ 2 \\\ 3\ \end{matrix} \right ] \left [ \begin{matrix} 4 & 5 & 6 \end{matrix} \right ] \left [ \begin{matrix} 7 & 8 & 9 \end{matrix} \right ] = \left [ \left [ \begin{matrix} 28 & 32 & 36 \\\ 56 & 64 & 72 \\\ 84 & 96 & 108 \end{matrix} \right ] \left [ \begin{matrix} 35 & 40 & 45 \\\ 70 & 80 & 90 \\\ 105 & 120 & 135 \end{matrix} \right ] \left [ \begin{matrix} 42 & 48 & 54 \\\ 84 & 96 & 108 \\\ 126 & 144 & 162 \end{matrix} \right ] \right ]$$
### Ejemplo gráfico:
In [1]:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.text(1, 0, 0, "1", color='red')
ax.text(2, 0, 0, "2", color='red')
ax.text(3, 0, 0, "3", color='red')
ax.text(0, 1, 0, "4", color='blue')
ax.text(0, 2, 0, "5", color='blue')
ax.text(0, 3, 0, "6", color='blue')
ax.text(0, 0, 1, "7", color='green')
ax.text(0, 0, 2, "8", color='green')
ax.text(0, 0, 3, "9", color='green')
ax.set_xlim3d(0, 5)
ax.set_ylim3d(0, 5)
ax.set_zlim3d(0, 5)
ax.set_xlabel('Eje x')
ax.set_ylabel('Eje y')
ax.set_zlabel('Eje z')
ax.view_init(17, 24)
plt.show()
In [2]:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.text(1, 0, 0, "1", color='red')
ax.text(2, 0, 0, "2", color='red')
ax.text(3, 0, 0, "3", color='red')
ax.text(0, 1, 0, "4", color='blue')
ax.text(0, 2, 0, "5", color='blue')
ax.text(0, 3, 0, "6", color='blue')
ax.text(0, 0, 1, "7", color='green')
ax.text(0, 0, 2, "8", color='green')
ax.text(0, 0, 3, "9", color='green')
## 4 5 6
## 8 10 12
##12 15 18
ax.text(1, 1, 0, "4", color='magenta')
ax.text(2, 1, 0, "8", color='magenta')
ax.text(3, 1, 0, "12", color='magenta')
ax.text(1, 2, 0, "5", color='orange')
ax.text(2, 2, 0, "10", color='orange')
ax.text(3, 2, 0, "15", color='orange')
ax.text(1, 3, 0, "6", color='purple')
ax.text(2, 3, 0, "12", color='purple')
ax.text(3, 3, 0, "18", color='purple')
ax.set_xlim3d(0, 5)
ax.set_ylim3d(0, 5)
ax.set_zlim3d(0, 5)
ax.set_xlabel('Eje x')
ax.set_ylabel('Eje y')
ax.set_zlabel('Eje z')
ax.view_init(17, 24)
plt.show()
In [3]:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.text(1, 0, 0, "1", color='red')
ax.text(2, 0, 0, "2", color='red')
ax.text(3, 0, 0, "3", color='red')
ax.text(0, 1, 0, "4", color='blue')
ax.text(0, 2, 0, "5", color='blue')
ax.text(0, 3, 0, "6", color='blue')
ax.text(0, 0, 1, "7", color='green')
ax.text(0, 0, 2, "8", color='green')
ax.text(0, 0, 3, "9", color='green')
##28 32 36
##56 64 72
##84 96 108
##
## 35 40 45
## 70 80 90
##105 120 135
##
## 42 48 54
## 84 96 108
##126 144 162
ax.text(1, 1, 0, "28", color='magenta')
ax.text(2, 1, 0, "56", color='magenta')
ax.text(3, 1, 0, "84", color='magenta')
ax.text(1, 1, 1, "32", color='magenta')
ax.text(2, 1, 1, "64", color='magenta')
ax.text(3, 1, 1, "96", color='magenta')
ax.text(1, 1, 2, "36", color='magenta')
ax.text(2, 1, 2, "72", color='magenta')
ax.text(3, 1, 2, "108", color='magenta')
ax.text(1, 2, 0, "35", color='orange')
ax.text(2, 2, 0, "70", color='orange')
ax.text(3, 2, 0, "105", color='orange')
ax.text(1, 2, 1, "40", color='orange')
ax.text(2, 2, 1, "80", color='orange')
ax.text(3, 2, 1, "120", color='orange')
ax.text(1, 2, 2, "45", color='orange')
ax.text(2, 2, 2, "90", color='orange')
ax.text(3, 2, 2, "135", color='orange')
ax.text(1, 3, 0, "42", color='purple')
ax.text(2, 3, 0, "84", color='purple')
ax.text(3, 3, 0, "126", color='purple')
ax.text(1, 3, 1, "48", color='purple')
ax.text(2, 3, 1, "96", color='purple')
ax.text(3, 3, 1, "144", color='purple')
ax.text(1, 3, 2, "54", color='purple')
ax.text(2, 3, 2, "108", color='purple')
ax.text(3, 3, 2, "162", color='purple')
ax.set_xlim3d(0, 5)
ax.set_ylim3d(0, 5)
ax.set_zlim3d(0, 5)
ax.set_xlabel('Eje x')
ax.set_ylabel('Eje y')
ax.set_zlabel('Eje z')
ax.view_init(17, 24)
plt.show() |
# § 2.4 The Slope of a Line.
## Presentation on theme: "§ 2.4 The Slope of a Line."— Presentation transcript:
§ 2.4 The Slope of a Line
Slope Slope of a Line
Slope Example: Find the slope of the line through (4, -3) and (2, 2)
If we let (x1, y1) be (4, -3) and (x2, y2) be (2, 2), then Note: If we let (x1, y1) be (2, 2) and (x2, y2) be (4, -3), then we get the same result.
Slope-Intercept Form Slope-Intercept Form of a line
y = mx + b has a slope of m and has a y-intercept of (0, b). This form is useful for graphing, since you have a point and the slope readily visible.
Slope-Intercept Form Example:
Find the slope and y-intercept of the line –3x + y = -5. First, we need to solve the linear equation for y. By adding 3x to both sides, y = 3x – 5. Once we have the equation in the form of y = mx + b, we can read the slope and y-intercept. slope is 3 y-intercept is (0, – 5)
Slope-Intercept Form Example:
Find the slope and y-intercept of the line 2x – 6y = 12. First, we need to solve the linear equation for y. – 6y = – 2x Subtract 2x from both sides. y = x – Divide both sides by – 6. Since the equation is now in the form of y = mx + b, slope is y-intercept is (0, –2)
Slope of a Horizontal Line
For any 2 points, the y values will be equal to the same real number. The numerator in the slope formula = 0 (the difference of the y-coordinates), but the denominator 0 (two different points would have two different x-coordinates). So the slope = 0.
Slope of a Vertical Line
For any 2 points, the x values will be equal to the same real number. The denominator in the slope formula = 0 (the difference of the x-coordinates), but the numerator 0 (two different points would have two different y-coordinates), So the slope is undefined (since you can’t divide by 0).
Summary of Slope of Lines
If a line moves up as it moves from left to right, the slope is positive. If a line moves down as it moves from left to right, the slope is negative. Horizontal lines have a slope of 0. Vertical lines have undefined slope (or no slope).
Parallel Lines Two lines that never intersect are called parallel lines. Parallel lines have the same slope unless they are vertical lines, which have no slope. Vertical lines are also parallel.
Parallel Lines Example:
Find the slope of a line parallel to the line passing through (0,3) and (6,0) So the slope of any parallel line is also –½
Perpendicular Lines Two lines that intersect at right angles are called perpendicular lines Two nonvertical perpendicular lines have slopes that are negative reciprocals of each other The product of their slopes will be –1 Horizontal and vertical lines are perpendicular to each other
Perpendicular Lines Example:
Find the slope of a line perpendicular to the line passing through (-1,3) and (2,-8) So the slope of any perpendicular line is
Parallel and Perpendicular Lines
Example: Determine whether the following lines are parallel, perpendicular, or neither. –5x + y = –6 and x + 5y = 5 First, we need to solve both equations for y. In the first equation, y = 5x – Add 5x to both sides. In the second equation, 5y = –x Subtract x from both sides. y = x Divide both sides by 5. The first equation has a slope of 5 and the second equation has a slope of , so the lines are perpendicular.
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# Rational Numbers – Definition with Examples
Kids are introduced to rational numbers early in their math education, forming a fundamental basis for understanding fractions, decimals, and percentages. While learning to solve problems on rational in math, they deal with addition, subtraction, multiplication, and division.
But first, what are rational numbers? Here, we’ll define rational number in math, cover the main types of rational numbers, provide examples, and share practice test problems.
## What is Rational Number?
Rational Numbers are like fractions written as p/q, where p and q are whole numbers, and q is not zero. Simply put, if a fraction has a number on top and a number on the bottom that is not zero, it’s called a rational number.
### Rational Numbers Definition
There are two major types of numbers: rational and irrational. We’ll cover the rational numbers definition and how they can be written as fractions. As you may have guessed, irrational numbers, on the other hand, cannot be written as simple fractions.
Rational numbers comprise two whole numbers: a numerator and a non-zero denominator. Everything in arithmetic can be related to rational numbers. For example, the number ½ is rational because it comprises two whole numbers, 1 and 2.
## Set of Rational Numbers
For rational numbers, the top number (numerator) is a whole number, and the bottom number (denominator) is a whole number that isn’t zero.
Rational numbers can be positive, negative, or zero. For example, ⅓, -⅔, and 5/1 are rational numbers. Rational numbers are important in math because we use them a lot in everyday life.
## Rational Numbers: Examples
Following what you now know of rational numbers, some examples include:
• ½
• -0.6667 or -⅔
• 10 or 10/1
• -6 or -6/1
• 5.0
• 0.8 or -8/10
• -0.2
## Types of Rational Numbers
The different types of rational numbers are:
• Integers: Integers are just whole numbers. They can be positive rational numbers, like 4, or negative, like -5, and they can also be zero. You can think of them as numbers with no fractions or decimals. For example, 4 is the same as 4/1, and -5 is the same as -5/1. So, they are all rational numbers because you can write them as fractions.
• Fractions: Fractions are numbers like ⅖ and ⅓. The top part is called the numerator, and the bottom is called the denominator. Fractions are rational numbers as long as the bottom part (denominator) is not zero. They show how one number (the top) is divided by another (the bottom).
• Terminating Decimals: Terminating decimals are numbers that end after a few digits. For example, 0.12 stops after two digits. You can turn these numbers into fractions, too. For instance, 0.12 is the same as 12/100, which can be simplified to 3/25.
• Non-Terminating Repeating Decimals: These decimals go on forever but have a repeating pattern. For example, 0.333… keeps repeating 3 forever. These can also be written as fractions. For example, 0.333… is the same as ⅓.
## How to Identify Rational Numbers?
The first thing to know about rational numbers is that:
• You can represent them as fractions and have their denominators be 1 (some examples are -4/1 and 9/1).
• As long as you have a fraction with a denominator greater than 0 and both denominator and numerator are whole numbers, you have a rational number. ⅝ is an example that fits this description.
• Numbers with digits after the decimal point, like 4.75, are rational and can be converted into fractions.
• Decimals that repeat a pattern forever, like 1.272727…, are also rational. They can be written as fractions.
• Decimals that go on forever without repeating are not rational. For example, √7 (about 2.645751…) is irrational because it never ends and doesn’t repeat.
## Rational Numbers in Decimal Form
Rational numbers can appear as decimals in different ways: terminating and non-terminating decimals. Terminating decimals are numbers that end after a few digits. For example, 0.1 is a terminating decimal and can be written as the fraction 2/10. Non-terminating repeating decimals are numbers that go on forever but have a repeating pattern. For example, 0.333… repeats the 3 and can be written as ⅓.
## List of Rational Numbers
There are so many rational numbers that we can’t list them all. For example, we can name a few, like 5, 2.5, ¾, -1, and 0.2. This list shows that rational numbers include natural numbers (like 1, 2, 3), whole numbers (like 0, 4), integers (like -3), fractions (like ⅖), and decimals that either end or repeat (like 0.666…). So, rational numbers are all the numbers you can write as fractions.
## Examples of Rational Numbers
Here are some rational number examples:
1. 4/9
2. -⅝
### Example 1: 3
The number 3 is rational because it can be written as a fraction. Think of it as 3 divided by 1. So, 3 is the same as 3/1. Since 3 can be written this way, it fits the rational number meaning, which are simply numbers that can be written as fractions with whole numbers on top and bottom.
### Example 2: 4/9
The fraction 4/9 is a rational number because it is already a fraction. It means 4 divided by 9. Any number that can be written this way, with one whole number on top and another one on the bottom, is rational. So, 4/9 fits the definition of a rational number perfectly.
### Example 3: -⅝
The fraction -⅝ is rational because it can be written as a fraction. Any number written this way is rational, with a whole number on top (the numerator) and another on the bottom (the denominator). So, -⅝ fits the definition of a rational number perfectly.
## Adding and Subtracting Rational Numbers
When we add or subtract rational numbers, we use similar rules to regular numbers. Here are the steps for the subtraction of rational numbers:
Example: Solve ¾ – (-½)
Solution:
• Step 1: Addition is the easiest operation on rational numbers. So, change the subtraction to addition and flip the sign of the second fraction. So, ¾ – (-½) becomes ¾ + ½ .
• Step 2: Add these fractions. To do that, find a common denominator. Here, the common denominator for 4 and 2 is 4. Get the equivalent fractions of the numerators. This equation becomes ¾ + ¼.
• Step 3: Now add ¾ and 2/4. That gives 5/4 or 1¼.
## Multiplying and Dividing Rational Numbers
To multiply rational numbers, just multiply the top numbers (numerators) together and the bottom numbers (denominators) together.
Example: Multiply ⅔ by 4/5
Solution:
• Step 1: Multiply the top numbers: 2 x 4 = 8
• Step 2: Multiply the bottom numbers: 3 x 5 = 15
• Step 3: So, ⅔ x ⅘ is 8/15.
To divide rational numbers, change the division to multiplication by flipping the second fraction (finding the reciprocal).
Example: Divide ¾ by ⅖.
Solution:
• Step 1: Flip the second fraction: ⅖ becomes 5/2.
• Step 2: Multiply ¾ by 5/2.
• Step 3: Multiply the numerators, so 3 x 5 = 15.
• Step 4: Multiply the denominators, so 4 x 2 = 8.
• Step 5: ¾ ÷ ⅖ = 15/8.
## Rational and Irrational Numbers
Rational numbers can be written as fractions, where both the top number (numerator) and the bottom number (denominator) are integers, and the denominator is not zero. For example, ⅔ is a rational number because it fits this rule. Rational numbers include positive numbers, negative numbers, and zero.
The difference between rational numbers and irrational numbers is that the latter cannot be written as simple fractions, expressed as a ratio of two integers, or have decimal expansions that go on forever without repeating. For example, √2 is irrational because it cannot be expressed as a fraction, and its decimal form does not terminate or repeat.
Example 1: Express 18/30 in standard form
Solution:
• Step 1: Find the greatest number to divide 18 and 30 without leaving a remainder. This number is 6.
• Step 2: Divide the numerator (18) and the denominator (30) by 6. 18 ÷ 6/ 30 ÷ 6 = ⅗.
So, 18/30 in standard form is ⅗.
Example 2: Add ⅔ and 4/9.
Solution:
• Step 1: Find a common denominator for both fractions. 9 is the least common multiple of 3 and 9.
• Step 2: When converting a rational number, both fractions after having 9 as the denominator. So, we have 9 (the common denominator) divided by 3 (the denominator) multiplied by 2 (the numerator). The outcome is 6. Thus, the new fraction is 6/9.
• Step 3: Add the fractions. 6/9 + 4/9 = 10/9.
So, ⅔ + 4/9 is 10/9 or 1 1/9.
Example 3: Subtract ⅖ from 7/8.
Solution:
• Step 1: Find a common denominator for both fractions. The least common multiple of 5 and 8 is 40.
• Step 2: Convert both fractions after having 40 as the denominator. So, we have 40 (the common denominator) divided by 5 (the denominator) multiplied by 2 (the numerator). The outcome is 16. Thus, the new fraction is 16/40.
• Step 3: Do something similar for the other fraction. So, we have 40 (the common denominator) divided by 8 (the denominator) multiplied by 5 (the numerator). The outcome is 35. Thus, the new fraction is 35/40.
• Step 4: Subtract the fractions. 35/40 – 19/40. The denominator is the same, so you pick one. The answer becomes 19/40.
So, when you subtract ⅖ from ⅞, you get 19/40.
## Rational Numbers Worksheets
Now that you know the definition of rational numbers in math, you should check out these Brighterly’s worksheets to improve your children’s interest in math and their ability to coast through common math problems. |
write 2 1/3 as a improper fraction
# write 2 1/3 as a improper fraction
Improper and Mixed Fractions Visual. Write each amount as a mixed number.5). Write each amount as an improper fraction. 6). First thing to know - remember that any number over itself as a fraction equals " 1". Now you have to create an improper fraction from the mixed number 4 1/ 2. Let me know if you have difficulties with this. Writing Mixed Numbers as Fractions. This mixed number can also be expressed as a fraction.Proper and Improper Fractions. A fraction in which the numerator is smaller than the denominator, like 1/3 or 2/5 is called a proper fraction. Write improper fractions as mixed numbers or whole numbers.Example 4: (Writing fractions) Represent the shaded part of each gure group as both an improper fraction and a mixed number. a) b). Unit Rates and Ratios of Fractions - Guided Lesson Explanation Explanation 1 Step 1) Divide the total minutes (1/2) by the amount of eaten banana (1/4). We will write 1/4 as an improper fraction. Solution: 3 14.
Write an improper fraction for the shaded parts. Then write each as a mixed number or a whole number. 1. 2. 3. 4. 5. 6. Mixed numbers and improper fractions. In the previous lesson, you learned about mixed numbers.Another way to write this would be 5/3, or five-thirds. These two numbers look different, but theyre actually the same. 5/ 3 is an improper fraction. Write 3 1/6 as an improper fraction. 19/6. 13 terms.
AmyTew. mixed fractions to improper fractions.a fraction that is less than 1, with the numerator less than t 12 terms. abagwell0325. Improper fractions. 1 2/3 is known as a mixed number, because it is made up of a whole number and a fraction. Improper fractions.Converting from a mixed number to an improper fraction. You can write the whole number part as a fraction, then add the fractions together. A mixed number can also be written as an improper fraction. A mixed number is the sum of its parts, the whole number and the fraction.Consider the same mixed number 5frac12. To convert from an improper fraction to a mixed number a. Divide the numerator by the denominator. b. Write down the whole number. c. Place any remainder in the quotient over the denominator. d. Simplify the fraction if needed. Improper Fractions or Write down the whole number answer Then write downHow do you write 3 1/4 as an improper fraction? Algebra. Examples: 1/3, 3/4, 2/7.What about when the numerator is equal to the denominator? For example 4/4 ? Well it is the same as a whole, but it is written as a fraction, so most people agree it is a type of improper fraction. In decimal numbers greater than 1 (such as 3.75), the fractional part of the number is expressed by the digits to the right of the decimal (with a value of 0.75 in this case). 3.75 can be written either as an improper fraction, 375/100, or as a mixed number You can only write 3 1/2 as an improper fraction one way. The improper fraction for 3 1/2 would be 3 16/15. Writing Mixed Numbers as Fractions. This mixed number can also be expressed as a fraction.Proper and Improper Fractions. A fraction in which the numerator is smaller than the denominator, like 1/3 or 2/5 is called a proper fraction. They are called "improper", because fractions are always supposed to be written with a numerator (number on top) that is less than the denominator (number on the bottom).The equivalent of two is 2/1 or 4/2. If you are unsure, check using division. Rewrite the mixed number as an addition problem. To convert the decimal 1.3 to a fraction follow these steps: Step 1: Write down the number as a fraction of one: 1.3 1.3/1.As the numerator is greater than the denominator, we have an IMPROPER fraction, so we can also express it as a MIXED NUMBER, thus 13/10 is also equal to 1 The improper fraction 8/3 may be considered as the division of 8. by 3. This division is carried out as follows: The truth of this can be verified another wayStep 1: Write as a whole number plus a fraction, 2 1/5. Change the mixed number 2(3/4) to an improper fraction and show how it happens.2 and 3/4 or 2 2/3 can be written like: 2/1 3/4. We make the denominator common , that is, instead of two different denominators, the common denominator 4. An equivalent form of 2/1 with denominator 4 is 8/4. To write an improper fraction as a whole number means to write the improper fraction as a mixed number, Indianapolis native Meghan Smith has been writing How do you write 2 1/3 as improper fraction? which has a specific gravity of 1.18? PRACTICE QUIZ - SECOND (version 2). Operations on Fractions. A. Write 3 equivalent fractions for the given fraction. |
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Math_112_Fall_2010.jkovalsky.Lesson_11_-_Lines_and_Linear_Equations
Math_112_Fall_2010.jkovalsky.Lesson_11_-_Lines_and_Linear_Equations
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JORDAN KOVALSKY Math - 112 Spring 2010 WeBWorK Assignment Lesson 11 - Lines and Linear Equations Due Date: 10/01/2010 at 11:59pm CDT. The primary purpose of WeBWorK is to let you know that you are getting the correct answer or to alert you if you are making some kind of mistake. You can use the Feedback button on each problem page to send e-mail to the course instructors. 1. (1 pt) Compute the slope of the line passing through the given points. The points ( 4 , 4 ) and ( 4 , 4 ) . Slope = The points ( 0 , 5 ) and ( 4 , 5 ) . Slope = 2. (1 pt) Here is a graph of a line: (Click on the image to enlarge it.) Compute the slope of the using each pair of points indicated. The points are A ( 3 , 2 ) , B ( 2 , 0 ) , and C ( 1 , 6 ) . Using A and B : Slope = Using B and C : Slope = Using A and C : Slope = The principle involved here is that no matter which pair of points you choose, the slope is the same. 3. (1 pt) Find an equation for the line having the slope and passing through the given point. Write your answer in the form y = mx + b . With m = 5; through ( 4 , 3 ) : y = x + With m = 4 5 ; through 5 , 4 5 : y = x + 4. (1 pt) Find the equation of a vertical line passing through the point ( 9 , 4 ) . Write your answer in the form Ax + By + C = 0. Line is : = 0 Find the equation of a horizontal line passing through the point ( 9 , 4 ) . Write your answer in the form Ax + By + C = 0. Line is : = 0 5. (1 pt) Find the equation of the line with slope 0 and intercept 11 and write it in slope-intercept form.
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# Factors of 99 | with Easy division and prime factorization
Contents
## Factors of 99
Factors of 99 are the natural numbers that divide an original number completely.
Therefore, if any factor divides 99, there is no remainder, and a quotient is a whole number. Take the number 8, whose factors are 1, 2, 4, and 8. As a result, dividing 8 by 4 gives us 2.
8 ÷ 4 = 2
As a result, there is no remainder, and a quotient is a whole number.
Similarly, we can find here in this article the factors of 99 as well as all pair factor and prime factors. When multiplied together, pair factors result in the original number, and prime factors are the prime number that divides the original number evenly.
## What are the Factors of -99?
When we multiply two numbers, and we get -99 as a product, those numbers are the factors of -99.
Here is the explanation:
-1 × 99 = -99
1 × -99 = -99
-3 × 33 = -99
3 × -33 = -99
-9 × 11 = -99
9 × -11 = -99
We can get -99 by having (-1, 99), (-3, 33), and(-9, 11), as a pair of factors. Similarly, we can also get -99 by having (1, -99), (3, -33), and(9, -11), as a pair of factors.
## How to Find Factors of 99?
The factor of 99 is a real number/integer that divides the original number evenly, without any remainder. It has more than two factors because 99 is a composite number. The smallest natural number is 1, so we’ll divide 99 by 1.
99 ÷ 1 = 99
99 ÷ 3 = 33
99 ÷ 9 = 11
99 ÷ 11 = 9
99 ÷ 33 = 3
99 ÷ 99 = 1
Therefore, the factors of 99 are 1, 3, 9, 11, 33, and 99.
Remember: 99 is an odd number, so it cannot be divided by an even number.
## What are the Factors of 99?
To get a complete set of the factors of 99, we will learn the pair factors of 99.
## Pair Factors of 99
If you multiply pair factors of 99, you will get the original number, which is as follows:
1 × 99 = 99
3 × 33 = 99
9 × 11 = 99
Thus, the pair factors are (1, 99), (3, 33), and (9, 11).
In this case, the pair factors were positive. When we consider negative pairs of factors, we can multiply the two negative numbers to get the original number.
-1 × -99 = 99
-3 × -33 = 99
-9 × -11 = 99
Hence, (-1, -99), (-3, 33), and (-9, 11) are the negative pair factors.
## Prime Factorization of 99
A prime factor of 99 is the prime number that divides this number evenly. Alternatively, we can say that 99 is divisible by its prime factors. To find the prime factors, we will use the prime factorization method.
### Step 1:
When we divide 99 by the smallest prime factor, which is 3, we get;
99/3 = 33
### Step 2:
Divide 33 again by the smallest prime factor, 3, to get;
33/33 = 11
### Step 3:
Since 11 is a prime factor, it can only be divided by 11.
11/11 = 1
Thus,
## Solved Examples
### Q.1: 99 friends have to go to a conference. There are 33 seats available on the bus. How many students can be seated in one seat?
Solution
Given,
Number of friends = 99
Number of seats in a bus = 33
Number of students sitting in a seat = 99/33 = 3
### Q.2: What is the sum of all 99 factors and 50?
Solution:
Factors of 99= 1, 3, 9, 11, 33 and 99.
Sum = 1+3+9+11+33+99 = 156
Thus, 156 is the required sum.
Factors of 50 = 1, 2, 5, 10, 25, and 50.
Sum=1+2+5+10+25+50=93
Thus, 41 is the required sum.
### Q.3: What is the common factor between 100 and 99?
Solution
Both numbers have factors, so let’s write them down.
Factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, and 100.
99 → 1, 3, 9, 11, 33, and 99 (99 is a composite number)
Hence, we can see the common factor is 1.
### Calculate the common factors of 99 and 24.
Solution:
99 → 1, 3, 9, 11, 33, and 99 (99 is a composite number)
Find the prime factors of 24 = 1, 2, 3, 4, 6, 8, 12, and 24.
The common factor of 99 and 24 is 1, and 3 only.
## FAQ`s
### 1. How many factors are there for number 99?
There are six factors of 99 in total. These are 1, 3, 9, 11, 33, and 99.
### 2. Is 99 a prime number?
There are more than two factors in the number 99, so it isn’t a prime number.
### 3. What are the multiples of 99?
These are the first 10 multiples of 99: 99, 198, 297, 396, 495, 594, 693, 792, 891 and 990.
### 4. What is 99 as a product of primes?
The prime factorization of 99 = 3 × 3 × 11 = 32 × 11
### 5. What is the GCF of 99 and 100?
Factors of 99 = 1, 3, 9, 11, 33, 99
Factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, 100
As a result, 99 and 100 share one greatest common factor (GCF). |
## Algebra wouldn't be fun if it were this easy to find x!
It's all about using step-by-step logic to figure out missing numbers.
## One Always Equals The Other
Equation comes from the word equal, so one side of an equation will always be equal to the other side. If you can find the value of one side, you'll know the value of the other.
## Work it out!
### 2x = 12 - 8
• What's the value of the right side?
• What number, if you multiply it by 2, will give you the same value of the right side?
What's x?
## Opposites Attract
Since both sides of an equation are equal, they'll balance each other out. If you look at one side's symbol and apply the opposite symbol to the other side, you can find the balance to work out the equation.
Think of it like a seesaw. If the person on one side of the seesaw is on the ground, the other person has to push down on the other side for the seesaw to even out.
So if one side has multiplication? Use division on the other side to work out the balance.
## Work it out!
### 5x - 3 = 12
You need to isolate x on the left side, so you need to get 5 and 3 to the right side.
### Follow the steps below to solve this equation:
• Move - 3 to the right side, make it + 3, and add it to 12
• The value of the right side: 12 + 3 = 15
• Move 5 to the right side and divide it into 15
What's x?
## Who knows what you'll be able to solve next!
Once you're comfortable with these go-to strategies for solving x, you'll be able to do more and more with algebra. |
Divide Four-Digit Numbers With Remainder
Start Practice
## How to Divide Four-Digit Numbers With a Remainder
Division is splitting a number into equal groups.
What if we can't split a number exactly?
That's when we get a remainder.
A remainder is a number that's left over from division.
### Dividing Four-Digit Numbers
In an earlier lesson, you learned how to divide 4-digit numbers without remainders.
Now, let's learn how to divide 4-digit numbers with remainders. It's mostly the same.
Let's learn with an example:
9317 ÷ 5 = ?
The first step is to write this in long division form.
We write the dividend, or the big number we want to split, inside the division bar.
We write the divisor on the outside.
Tip: the answer to our division equation is called the quotient.
We're all set! 🙂
👉 Start dividing from the first digit of the dividend.
What's the first digit in the dividend?
Yes! It's 9.
Let's divide 9 by 5.
To get the answer, we think about how many 5's can fit in 9. 🤔
You're right! Only 1.
We write that on top of 9.
Now let's multiply 1 × 5.
What product do we get?
Good! It's 5.
Let's write that below 9. Then we subtract.
Now, we bring down 3, the second digit in the dividend.
We have 43.
How many 5's are in 43?
Correct! There are 8.
Let's write that above 3.
Now we multiply 8 by 5.
We get 40. We write that below 43.
Then we subtract.
Now, we bring down the third digit in the dividend.
Now we divide 31 by 5.
How many 5's can fit in 31?
That's right! 6.
We write 6 on top of 1.
Now let's multiply 6 with our divisor.
What do we get when we multiply 6 x 5?
You got it! 30
We write 30 below 31. Then we subtract.
What's the next step? 🤔
Correct! We bring down 7, the fourth digit in the dividend.
Now we have 17.
How many 5's can fit in 17? 🤔
You're right! 3
We write 3 on top of 7.
Now we multiply 3 by 5, our divisor.
3 x 5 is equal to 15.
We write 15 below 17.
Then we subtract.
The difference is 2.
Are there any digits left to bring down?
There are none left! 🤷♂️
That means 2 is our remainder.
Let's complete our division by writing 2 as our remainder.
So this is our final answer:
9317 ÷ 5 = 1863 R2
Great work! 🌟
Did you get all those steps?
Try practice to find out! 💪
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# What is 1/52 as a decimal?
## Solution and how to convert 1 / 52 into a decimal
1 / 52 = 0.019
Convert 1/52 to 0.019 decimal form by understanding when to use each form of the number. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. So let’s dive into how and why you can convert 1/52 into a decimal.
## 1/52 is 1 divided by 52
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! To solve the equation, we must divide the numerator (1) by the denominator (52). Here's how our equation is set up:
### Numerator: 1
• Numerators sit at the top of the fraction, representing the parts of the whole. Small values like 1 means there are less parts to divide into the denominator. The bad news is that it's an odd number which makes it harder to covert in your head. Ultimately, having a small value may not make your fraction easier to convert. Let's take a look at the denominator of our fraction.
### Denominator: 52
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. Larger values over fifty like 52 makes conversion to decimals tougher. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. Now it's time to learn how to convert 1/52 to a decimal.
## Converting 1/52 to 0.019
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 52 \enclose{longdiv}{ 1 }$$
To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 52 \enclose{longdiv}{ 1.0 }$$
We've hit our first challenge. 1 cannot be divided into 52! Place a decimal point in your answer and add a zero. Now 52 will be able to divide into 10.
### Step 3: Solve for how many whole groups you can divide 52 into 10
$$\require{enclose} 00.0 \\ 52 \enclose{longdiv}{ 1.0 }$$
We can now pull 0 whole groups from the equation. Multiple this number by our furthest left number, 52, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 52 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If your remainder is zero, that's it! If you still have numbers left over, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. This is also true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 1/52 and 0.019 bring clarity and value to numbers in every day life. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 1/52 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 1/52 MPH. The radar will read: 90.1 MPH. This simplifies the value.
### When to convert 0.019 to 1/52 as a fraction
Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves.
### Practice Decimal Conversion with your Classroom
• If 1/52 = 0.019 what would it be as a percentage?
• What is 1 + 1/52 in decimal form?
• What is 1 - 1/52 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.019 + 1/2? |
## Approximation and Estimation Part 1
Practice Unlimited Questions
#### What is approximation?
Approximation means rounding off. We round off numbers to simplify them so they are easy to use and easy to remember.
Examples:
1. You bought a pair of shoes for \$99. If someone asks you the price of the shoes you are more likely to say that the price is \$100.
1. This is called rounding up where the new number is bigger than the given number.
2. \$99 ≈ \$100 (You say it as: \$99 is approximately equal to \$100)
2. A dress costs \$81. You will more likely to say that the cost of the dress was \$80.
1. This is called rounding down where the new number is smaller than the given number.
2. \$81 ≈ \$80 (You say it as: \$81 is approximately equal to \$80)
#### 1. a) Round off 77 250 to the nearest 1000 b) Round off 77 500 to the nearest 1000 c) Round off 77 876 to the nearest 1000
Rounding off to the nearest thousand
When rounding off to the nearest 1000, you only care about the part of the number up to the thousands place. So,
1. Check the digit in the hundreds place.
1. If it is less than 5, keep the digit in the thousands place unchanged (i.e. round down)
2. If it is equal to or more than 5, add 1 to the digit in the thousands place (i.e. round up)
2. Change the digits in the hundreds place, tens place and ones place to 0s.
1. 77 250 ≈ 77 000 (when rounded off to the nearest 1000, it is rounded down because it is closer to 77 000 than to 78 000)
2. 77 500 ≈ 78 000 (when rounded off to the nearest 1000, it is rounded up because it is closer to 78 000 than to 77 000)
3. 77 876 ≈ 78 000 (when rounded off to the nearest 1000, it is rounded up because it is closer to 78 000 than to 77 000)
Notes:
1. All numbers to the left of the midpoint on the number line will be rounded down.
2. All numbers to the right of the midpoint including the midpoint itself will be rounded up.
#### 2. a) Round off 1425 to the nearest 100 b) Round off 1450 to the nearest 100 c) Round off 1483 to the nearest 100
Rounding off to the nearest hundred
When rounding off to the nearest 100, you only care about the part of the number up to the hundreds place. So,
1. Check the digit in the tens place.
1. If it is less than 5, keep the digit in the hundreds place unchanged (i.e. round down)
2. If it is equal to or more than 5, add 1 to the digit in the hundreds place (i.e. round up)
2. Change the digits in the tens place and ones place to 0s.
1. 1425 ≈ 1400 (when rounded off to the nearest 100, it is rounded down because it is closer to 1400 than to 1500)
2. 1450 ≈ 1500 (when rounded off to the nearest 100, it is rounded up because it is closer to 1500 than to 1400)
3. 1483 ≈ 1500 (when rounded off to the nearest 100, it is rounded up because it is closer to 1500 than to 1400)
#### 3. a) Round off 1402 to the nearest 10 b) Round off 1405 to the nearest 10 c) Round off 1408 to the nearest 10
Rounding off to the nearest ten
When rounding off to the nearest 10, you only care about the part of the number up to the tens place. So,
1. Check the digit in the ones place.
1. If it is less than 5, keep the digit in the tens place unchanged (i.e. round down)
2. If it is equal to or more than 5, add 1 to the digit in the tens place (i.e. round up)
2. Change the digits in the ones place to 0.
1. 1402 ≈ 1400 (when rounded off to the nearest 10, it is rounded down because it is closer to 1400 than to 1410)
2. 1405 ≈ 1410 (when rounded off to the nearest 10, it is rounded up because it is closer to 1410 than to 1400)
3. 1408 ≈ 1410 (when rounded off to the nearest 10, it is rounded up because it is closer to 1410 than to 1400)
#### 4. If Jack has an income of \$6000 when rounded off to the nearest thousand. What could his actual income be?
1. When a number given to you has been round off to the nearest thousand, then there are 1000 possible options that the actual number could be.
1. Half of those numbers (500 of them) would be less than the given number, and
2. Half of those numbers (500 of them) would be equal to or greater than the given number.
2. When rounding off to the nearest 1000, all numbers from \$5500 to \$5999 will be rounded up to give \$6000.
3. When rounding off to the nearest 1000, all numbers from \$6001 to \$6499 will be rounded down to give \$6000.
So, Jack's actual income could be anywhere between \$5500 and \$6499.
#### 5. If Jack has an income of \$5920 and Harry has an income of \$4120, estimate the sum of money they earn together.
Jack's income = \$5920 ≈ \$6000
Harry's income = \$4120 ≈ \$4000
Total = Jack's income + Harry's income
≈ \$6000 + \$4000
= \$10 000
Jack and Harry earn approximately \$10 000 together. |
# Proof
(related to Proposition: Addition Of Rational Numbers)
Let $$x$$ and $$y$$ be rational numbers. By definition, it means that they are represented by some integers $$x=\frac ab$$, $$y=\frac cd$$, $$a,c\in \mathbb Z$$, $$b,d\in \mathbb Z\setminus\{0\}$$.
Note that $$ad$$, $$cb$$ and $$bd$$ are all integers, as they are the products of the respective integers $$a,d$$, $$c,d$$ and $$b,d$$. Also note that $$ad + cb$$ is an integer, as it is the sum of the integers $$ad$$ and $$cb$$. Moreover, we have $$bd\neq 0$$, since both $$b\neq 0$$ and $$d\neq 0$$ and their product cannot equal $$0$$, since integers form an integral domain. Therefore, the sum $\begin{array}{rcl} x+y=\frac {ad + cb}{bd}. \end{array}$ exists, because it denotes some new rational number, as it is represented by the integers $$ad + cb$$ and $$bd\neq 0$$.
It remains to be shown that the addition of rational numbers does not depend on the specific representatives of the numbers $$x$$ and $$y$$. Suppose, we have different representatives $\begin{array}{rcl} x=\frac{a_1}{b_1}=\frac{a_2}{b_2},~y=\frac{c_1}{d_1}=\frac{c_2}{d_2}.&&(*) \end{array}$ It follows from the definition of rational numbers that $$a_1=\frac{a_2b_1}{b_2}$$ and $$c_1=\frac{c_2d_1}{d_2}$$. We have to show that $x+y=\frac {a_1d_1 + c_1b_1}{b_1d_1}=\frac {a_2d_2 + c_2b_2}{b_2d_2}.$ In the following, we will use the following mathematical definitions and concepts: * definition of rational numbers, * definition of adding rational numbers (hypothesis), * commutativity law of multiplying integers , * integer one is neutral with respect to the multiplication of integers, * distributivity law for integers, and * multiplication of rational numbers:
$\begin{array}{rcll} x+y&=&\frac{a_1}{b_1}+\frac{c_1}{d_1}&\text{by definition of rational numbers}\\ &=&\frac{a_1d_1+c_1b_1}{b_1d_1}&\text{by hypothesis}\\ &=&\frac{a_1d_1\cdot 1+c_1b_1\cdot 1}{1\cdot b_1d_1}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\frac{(a_1d_1+c_1b_1)\cdot 1}{1\cdot b_1d_1}&\text{by distributivity law for integers}\\ &=&\frac{a_1d_1+c_1b_1}{1}\cdot \frac 1{b_1d_1}&\text{by definition of multiplying rational numbers}\\ &=&(a_1d_1+c_1b_1)\cdot \frac 1{b_1d_1}&\text{by definition of rational numbers}\\ &=&\left(\frac{a_2b_1}{b_2}d_1+\frac{c_2d_1}{d_2}b_1\right)\cdot \frac 1{b_1d_1}&\text{according to }(*)\\ &=&\left(\frac{a_2b_1\cdot d_1}{b_2\cdot 1}+\frac{c_2d_1\cdot b_1}{d_2\cdot 1}\right)\cdot \frac 1{b_1d_1}&\text{by definition rational numbers and of multiplying them}\\ &=&\left(\frac{a_2b_1d_1}{b_2}+\frac{c_2d_1b_1}{d_2}\right)\cdot \frac 1{b_1d_1}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\left(\frac{a_2b_1d_1d_2+c_2d_1b_1b_2}{b_2d_2}\right)\cdot \frac 1{b_1d_1}&\text{by hypothesis}\\ &=&\left(\frac{a_2d_2b_1d_1+c_2b_2b_1d_1}{b_2d_2}\right)\cdot \frac 1{b_1d_1}&\text{by commutativity law for multiplying integers}\\ &=&\left(\frac{(a_2d_2+c_2b_2)b_1d_1}{b_2d_2}\right)\cdot \frac 1{b_1d_1}&\text{by distributivity law for integers}\\ &=&\frac{a_2d_2+c_2b_2}{b_2d_2}\cdot \frac {b_1d_1\cdot 1}{b_1d_1}&\text{by definition of multiplying rational numbers}\\ &=&\frac{a_2d_2+c_2b_2}{b_2d_2}\cdot \frac {b_1d_1}{b_1d_1}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\frac{a_2d_2+c_2b_2}{b_2d_2}\cdot \frac {1}{1}&\text{by definition of rational numbers}\\ &=&\frac{(a_2d_2+c_2b_2)\cdot 1}{b_2d_2\cdot 1}&\text{by definition of multiplying rational numbers}\\ &=&\frac{a_2d_2+c_2b_2}{b_2d_2}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\frac{a_2}{b_2}+\frac{c_2}{d_2}&\text{by hypothesis}\\ \end{array}$
Thank you to the contributors under CC BY-SA 4.0!
Github:
### References
#### Bibliography
1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013 |
# Rd Sharma Solutions Class 6 Chapter 4
## Rd Sharma Solutions Class 6 Chapter 4 Operation on Whole Numbers
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Rd Sharma Class 6 Mathematics, Chapter 4, Operation on Whole Numbers. Here students can easily find Exercise wise solution for chapter 4, Operation on Whole Numbers. Students will find proper solutions for Exercise 4.1, 4.2, 4.3, 4.4 and 4.5. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.
Operation on Whole Numbers Exercise 4.1 Solution
Question no – (1)
Solution :
(i) 359 + 476 = 476 + 359 …(Commutative law)
(ii) 2008 + 1952 = 1952 + 2008 …(Commutative)
(iii) 90758 + 0 = 9078 …(Additive identity)
(iv) 54321 + (489 + 699) 489 + (54321 + 699) …(Associative)
Question no – (2)
Solution :
(i) 5628 + 39784
= 45412
Now, by reversing,
= 39784+ 5628
= 45412
(ii) 923584 + 178
= 923762
Now, by reversing,
178 + 923584
= 923762
(iii) 15409 + 112
= 15521
Now, by reversing,
= 112 + 15409
= 15521
(iv) 2359 + 641
= 3000
Now, by reversing,
641 + 2359
= 3000
Question no – (4)
Solution :
(i) Given statement is False.
(ii) Given statement is True.
(ii) Given statement is True.
(iv) Given statement is False.
(v) Given statement is True.
(vi) Given statement is False.
(vii) Given statement is False.
(viii) Given statement is True.
(ix) Given statement is True.
(x) Given statement is False.
(xi) Given statement is False.
(xii) Given statement is True.
Operation on Whole Numbers Exercise 4.2 Solution
Question no – (1)
Solution :
(i)
15 8 13 10 12 14 11 16 9
Diagonal values,
= 13 + 12 + 11
= 36
(ii)
22 29 6 13 20 28 10 12 19 21 9 11 18 25 27 15 17 24 26 8 16 23 30 7 14
Consider the diagonal values,
= 20 + 15 + 18 + 17 + 16
= 90
Question no – (2)
Solution :
(i) 57839 – 2983 = 54856
= 54856 + 2983
= 57839
(ii) 92507- 10879 = 81628
= 81628 + 10879
= 92507
(iii) 400000 – 98798 = 301202
301202 + 98798
= 400000
(iv) 5050501 – 969696 = 4080805
= 4080805 + 969696
= 5050501
(v) 200000 – 97531 = 1024469
= 102469 + 97351
= 200000
Question no – (4)
Solution :
We know that,
Largest no of 5 digits = 99999
Largest no of 6 digits = 100000
Difference is,
= (100000 – 99999)
= 1
Therefore, the difference will be 1.
Question no – (5)
Solution :
As we know that,
Largest number of 4 digits = 9999
Smallest number of 7 digits = 1000000
Difference is,
= (1000000 – 9999)
= 990001
Therefore, the difference will be 990001.
Question no – (6)
Solution :
In the given question,
Rohit deposited = Rs 125000
He withdrew = Rs 35425
Money left = ?
Money left in his account,
= (125000 – 25425)
= 89575 Rs
Therefore, Rs 89575 was left in Rohit’s account.
Question no – (7)
Solution :
According to the question,
Population of a town is 96209
Men = 29642
Women = 29167
Children = ?
Total no of number + women,
= (29642 + 29167)
= 58809
No of children,
= (96209 – 58809)
= 37400
Thus, the the number of children will be 37400.
Question no – (8)
Solution :
Original no = 39460
After interchanging 6 and 9
= 36490 (new number)
Difference is,
= (39460 – 36490)
= 2790
Therefore, the difference will be 2790.
Question no – (9)
Solution :
In the given question,
population of a town = 59000.
Increased due to new births = 4536
persons died or left the town = 9218
Population at the end of the year,
= 59000 + 4536 – 9218
= 54318
Therefore, the population at the end of the year will be 54318.
Operation on Whole Numbers Exercise 4.3 Solution
Question no – (1)
Solution :
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567
(iii) 475 × 129 = 129 × 475
(iv) 1243 × 8975 × 1243
(v) 10 × 100 × = 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × 5
(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
Question no – (2)
Solution :
(i) 2 × 1497 × 50
= (2 × 50) × 1497
= 100 × 1497
= 149700
(ii) 4 × 358 × 25
= (4 × 25) × 358
= 100 × 358
= 35800
(iii) 495 × 625 × 16
= (625 × 16) × 495
= 10000 × 495 ‘
= 4950000
(iv) 625 × 20 × 8 × 50
= 625 × 20 × 8 × 50
= (625 × 8) (20 × 500
= 5000000
Question no – (3)
Solution :
(i) 736 × 103
= 736 (100 + 3)
= (736 × 100) + (736 × 3)
= 73600 + (736 × 3)
= 788808
(ii) 258 × 1008
= 258 (1000 + 8)
= (258 × 1000) × (258 × 8)
= 258000 + 2064
= 260064
Question no – (4)
Solution :
(i) 736 × 93
= 736 × 93 = 736 (100 – 7)
= (736 × 100) – (736 × 7)
= 73600 – 5752
= 68448
So, the product will be 68448
(ii) 816 × 745
= 816 × 745 = 816 × (750 – 6)
= (816 × 750) – (816 × 5)
= 612000 – 4080
= 607920
So, the product will be 607920
(iii) 2032 × 613
= 2032 × 613 = 2032 (600 + 13)
= (2032 × 600) + (2032 × 13)
= 1219200 + 26416
= 1245616
So, the product will be 1245616
Question no – (8)
Solution :
In the question,
Dealer purchased = 125 colour television
Cost of each set = Rs 19820
Determine cost of all sets together = ?
Cost of 125 television sets,
= (19820 × 125) Rs
= 2477500 Rs
Therefore, the cost of all sets together will be 2477500 Rs.
Question no – (9)
Solution :
In the given question,
The annual fee of class VI = Rs 8880.
Students in class VI = 235
Annual fee changed for 235 students,
= (8880 × 235)
= 208600 Rs
Therefore, the total collection is 208600 Rs.
Question no – (10)
Solution :
In the question we get,
Constructed flats= 350
Cost of construction for each flat = Rs 993570
Cost of Constriction,
= (993570 × 350) Rs
= 347749500 Rs
Hence, total cost of construction of all the flats will be 347749500 Rs.
Question no – (11)
Solution :
I conclude, The product of two whole numbers is Zero when at least one number or both now are Zero,
Question no – (12)
Solution :
Two numbers when multiplied with gives the same number 0 × 0 = 0, 1 × 1 = 1
Question no – (13)
Solution :
Total apartment in 1 large building,
= 10 × 2
= 20
1 small building, no of apartment,
= 12 × 3
= 36
Total apartment,
= (22 × 20) + (15 × 36)
= 440 + 540
= 980
Hence, there will be total 980 apartments.
Operation on Whole Numbers Exercise 4.4 Solution
Question no – (1)
Solution :
Yes, there exist a whole number we know, the whole number is 1,
Where 1 ÷ 1 = 1.
Question no – (2)
Solution :
(i) 23457 ÷ 1
= 23457
(ii) 0 ÷ 97
= 0
(iii) 476 + (840 ÷ 84)
= 476 + 10
= 486
(iv) 964 – (425 ÷ 425)
= 964 – 1
= 963
(v) (2758 ÷ 2758) – (2758 ÷ 2758)
= 1 – 1
= 0
Question no – (3)
Solution :
(i) 10 ÷ (5×2) = (10÷5) × (10÷2)
= L.H.S 10 ÷ (5×2)
= 10 ÷ 10
= 1
R.H.S (10÷5) × (10÷2)
= 2 × 5
= 10
Therefore, L.H.S ≠ R.H.S
(ii) (53–14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7
= L.H.S (35–14) ÷ 7
= 21 ÷ 7
= 3
R.H.S, 35 ÷ 7 – 14 ÷ 7
= 5 – 2
= 3
Therefore, L.H.S = R.H.S (Proved)
(iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7
= L.H.S, 25 – 14 ÷ 7
= 35 – 2
= 33
R.H.S, 35 ÷ 7 – 14 ÷ 7
= 5 ÷ 2
= 3
Therefore, L.H.S ≠ R.H.S
(iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5
= L.H.S (20–5) ÷ 5
= 15 ÷ 5
= 3
R.H.S, 20 ÷ 5 – 5
= 4 – 5
= – 1
Therefore, L.H.S ≠ R.H.S
(v) 12×(14 ÷ 7) = (12×14) ÷ (12×7)
= L.H.S, 12 (14÷7)
= 12×2
= 24
R.H.S, (12×14) ÷ (12÷7)
= 168 ÷ 84
= 2
Therefore, L.H.S ≠ R.H.S
Question no – (4)
Solution :
(i) Given, 7772 ÷ 58
∴ Quotient = 134,
Remainder = 0
For check, We, know,
Divided = Divisor × Quotient + Remainder.
7772 = 58 × 134 + 0
= 7772
L.H.S = R.H.S (proved)
(ii) Given, 6906 ÷ 35
Hence,
Quotient = 197
Remainder = 11
(iii) Given, 16135 ÷ 875
Here,
Quotient = 18,
Remainder = 385
Question no – (5)
Solution :
As we know that,
Divided = Divisor × Quotient + Remainder
35 × 20 + 18
= 700 + 18
= 718
Therefore, the required number is 718.
Question no – (6)
Solution :
As we know,
Divided = Divisor × Quotient + Remainder
Divided = 58 × 40 + 31
= 2320 + 31
= 2351
Therefore, the required number is 2351.
Question no – (7)
Solution :
In the question,
Product of two numbers = 504347
One number = 1591
Other number = ?
Other number is,
= 504347/1591
= 317
Hence, the other number will be 317.
Question no – (8)
Solution :
As we know,
Divided = Divisor × Quotient + Remainder
59761 = Divisor × 189 + 37
= 59761 – 37 = 189 Divisor
Now the required divisor,
= 59724/189
= 316
Therefore, the Divisor will be 316.
Question no – (9)
Solution :
As we know that,
Divided = Divisor × Quotient + Remainder
55390 = 299 × Quotient + 75
= 299× Quotient = 55390 – 75
Now the required quotient,
= 55315/299
= 185
Hence, the quotient will be 185.
Operation on Whole Numbers Exercise 4.5 Solution
Question no – (1)
Solution :
(i) 10th square number
= 10th square number (10×10)
= (10)2
= 100
Therefore, 10th square number is 100.
(ii) 6th triangular number,
= 6 × (6 + 1)/2
= 6 × 7/2
= 42/2
= 21
Hence, the 6th triangular number is 21.
Question no – (2)
Solution :
(i) Yes, a rectangular number also be a square number.
(ii) Yes, a triangular number also be a square number.
Question no – (3)
Solution :
Here, we know that,
1 × 5 = 5
2 × 6 = 12
3 × 7 = 21
4 × 8 = 32
First four products of two numbers with difference 4.
= 5 – 1 = 4; 6 – 2 = 4
= 7 – 3 = 4; 8 – 4 = 4
Question no – (4)
Solution :
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 5 = 88888
98765 × 9 + 3 = 8888888
987654 × 9 + 2 = 8888888
9876543 × 9 + 1 = 88888888
Question no – (5)
Solution :
6 × 2 – 5 = 7
7 × 3 – 12 = 9
8 × 4 – 21 = 11
9 × 5 – 32 = 13
10 × 6 – 45 = 15
11 × 7 – 60 = 17
12 × 8 – 77 = 19
Question no – (6)
Solution :
By observing the above pattern,
(i) 1 + 3 + 5 + 7 + 9 + 11
= 6 × 6
= 36
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
= 8 × 8
= 64
(iii) 21 + 23 + 25 + __ + 51
= 262 – 102
= 676 – 100
= 576
Question no – (7)
Solution :
By observing the above pattern,
(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
= (10×11)/2
= 110/2
= 55
(ii) 50 + 51 + 52 + __ + 100
= 100 × 100/2 – 49 × 50/2
= 5050 – 1225
= 3825
(iii) 2 + 4 + 6 + 8+ 10 + __ + 100
= 2 (1 + 2 + 3 + 4 + 5 + ……)
= 2 × (50 ×57/2)
= 2550
Previous Chapter Solution :
Updated: June 7, 2023 — 2:04 pm |
## Formula For Finding The Hypotenuse Of A Right Triangle What Is Angle of Elevation?
You are searching about Formula For Finding The Hypotenuse Of A Right Triangle, today we will share with you article about Formula For Finding The Hypotenuse Of A Right Triangle was compiled and edited by our team from many sources on the internet. Hope this article on the topic Formula For Finding The Hypotenuse Of A Right Triangle is useful to you.
## What Is Angle of Elevation?
Angle of Elevation Definition: Let us first define Angle of Elevation. Let O and P be two points such that the point P is at higher level. Let OA and PB be horizontal lines through O and P respectively. If an observer is at O and the point P is the object under consideration, then the line OP is called the line of sight of the point P and the angle AOP, between the line of sight and the horizontal line OA, is known as the angle of elevation of point P as seen from O. If an observer is at P and the object under consideration is at O, then the angle BPO is known as the angle of depression of O as seen from P.
Angle of elevation formula: The formula we use for angle elevation is also known as altitude angle. We can measure the angle of the sun in relation to a right angle using angle elevation.Horizon Line drawn from measurement angle to the sun in right angle is elevation.Using opposite, hypotenuse, and adjacent in a right triangle we can find finding the angle elevation. From right triangle sin is opposite divided by hypotenuse; cosine is adjacent divided by hypotenuse; tangent is opposite divided by adjacent. To understand angle of the elevation we will take some
Angle of elevation problems. Suppose if a tower height is 100 sqrt(3) metres given. And we have to find angle elevation if its top from a point 100 metres away from its foot. So let us first collect information, we know height of tower given is 100sqrt3, and distance from the foot of tower is 100 m. Let us take (theta) be the angle elevation of the top of the tower…we will use the trigonometric ratio containing base and perpendicular. Such a ratio is tangent. Using tangent in right triangle we have,
tan (theta) = perpendicular / adjacent
tan (theta) = 100sqrt(3)/100 = sqrt(3).
tan (theta) = tan 60
theta = 60 degree.
Hence, the angle elevation will be 60 degree
Example: The elevation angle of the top of the tower from a point on the ground, which is 30 metre away from the foot of the tower, is 30 degree. Find the height of the tower.
Solution: Let AB be the top A of tower height h metres and C be a point on ground such that the angle elevation from the top A of tower AB is of 30 degree.
In triangle ABC we are given angle C = 30 degree and base BC = 30 m and we have to find perpendicular AB. So, we use those trigonometrically ratios which contain base and perpendicular. Clearly, such ratio is tangent. So, we take tangent of angle C.
In triangle ABC, taking tangent of angle C, we have
tan C = AB/AC
tan 30 = AB/AC
1/sqrt(3) = h/30
h = 30/sqrt(3) metres = 10 sqrt(3) metres.
Hence, the height of the tower is 10 sqrt(3) metres.
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## Formula For Finding The Hypotenuse Of A Right Triangle An Introduction to the Pythagorean Theorem
You are searching about Formula For Finding The Hypotenuse Of A Right Triangle, today we will share with you article about Formula For Finding The Hypotenuse Of A Right Triangle was compiled and edited by our team from many sources on the internet. Hope this article on the topic Formula For Finding The Hypotenuse Of A Right Triangle is useful to you.
## An Introduction to the Pythagorean Theorem
Mathematicians have labored for ages to discover relationships within triangles along with other polygons. One of the famous and beneficial relationships ended up being uncovered by the Greek mathematician called Pythagorus. He found the sides of a right triangle are related in the following way:
When the lengths of each of the shorter two sides (the legs) of the right triangle are squared and the squares added together, the total is the same as the length of the 3rd side (called the hypotenuse) squared. So should you notice a right triangle, keep in mind that the lengths of the 2 smaller sides are related to the length of the longest side.
If an individual had time to form three external squares from each side of any right triangle, you will discover the smaller squares result in an interesting relationship as compared to the big square.
A triangle whose edges measure 3 units, 4 units, and then five units is one of the most well-known triangles in just about all of mathematics. Squaring each of the 2 smaller sides yields nine sixteen Equals 25 sq units. The longer side is five units, which means its square has an area of twenty-five square units. This arrangement holds true for each and every right triangle.
Many problems that deal with right triangles yield decimal answers. However, there are many examples of whole numbers combinations that are possible in right triangles. Some of these are:
three, four, five
six, eight, ten
five, twelve, thirteen
seven, twenty-four, twenty-five
eight, fifteen, seventeen
Each of the above combinations represent the three lengths of a right triangle. In class, teachers often begin to teach the concept of the Pythagorean Theorem using whole number examples. Later in class, many of the answers may contain one or more sides whose lengths are not whole numbers.
In a right triangle, the 2 smaller sides are legs and the longest side is known as a hypotenuse. Usually a stands out as the shorter of the two legs and b is usually the longer of the legs. In some cases, a is the identical length as b. All right triangles contain a lengthiest side that is directly across from the right angle. This longest side is represented by the variable “c” and it is referred to as the hypotenuse.
Frequently, major findings in science and math obtain distinctive names. Since this special relationship within right triangles was unearthed by Pythagorus, it has been referred to as Pythagorean Theorem in his honor.
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## Formula For Finding The Area Of A Triangular Prism Geometry for Beginners – How To Find the Surface Area and Volume of Prisms
You are searching about Formula For Finding The Area Of A Triangular Prism, today we will share with you article about Formula For Finding The Area Of A Triangular Prism was compiled and edited by our team from many sources on the internet. Hope this article on the topic Formula For Finding The Area Of A Triangular Prism is useful to you.
## Geometry for Beginners – How To Find the Surface Area and Volume of Prisms
Welcome to Geometry for Beginners. In this article, we are going to start looking at the 3-dimensional equivalents of the 2-dimensional relationships of perimeter and area. Perimeter of polygons refers to “the distance around” while the area of polygons refers to “the space inside.” For 3-dimensional figures, like prisms, “the distance around” becomes surface area; and “the space inside” becomes volume. We will be introducing the formulas for finding both surface area and volume of prisms and discussing applications of these concepts.
Before we discuss the formulas, we need to make sure we are all picturing the same kind of figure. Let’s use a cereal box for our mental image. Our cereal box is an example of a right rectangular prism. RIGHT because the sides (lateral faces) are perpendicular to the top and bottom (bases). RECTANGULAR because the bases (top and bottom) are rectangles. PRISM because the figure has two identical polygonal bases that are parallel to each other and 4 lateral faces that are rectangles.
Note: If the sides were not perpendicular to the bases, the figure would be OBLIQUE–not right, and the lateral faces would be parallelograms rather than rectangles. An oblique triangular prism would have sides NOT perpendicular to the bases, the bases would be triangles, and the 3 lateral faces would be parallelograms.
Again, picturing our cereal box, surface area would refer to the packaging or the box itself. For package manufacturers, the amount of material needed for each box is extremely important. The cereal inside the box would represent the volume of the package, assuming the box is full. This is an equally important business concern.
The formulas for surface area and volume will look unusual because they use symbols we haven’t used before, but only the symbols are new. You already have the skills to use these formulas!
Formula for the Surface Area of Prisms: SA = 2B + LA, where SA is surface area, B is AREA of the base, and LA is lateral area.
Thus, to calculate the surface area of a prism, we must first calculate B, the area of whatever polygon forms the base, using the appropriate formula for the shape. Then, this value must be multiplied by two since there are 2 bases.
Next, we must calculate the lateral area which is the sum of the areas of the sides. Since the sides are either rectangles or parallelograms, we will use the formula A = bh. Then add the sides together for the lateral area. Caution! Be certain you are using the actual height and not a side if the sides are parallelograms.
The final step is to add your values of 2B and LA.
The best way to memorize and read the formula SA = 2B + LA is “The surface area of a prism is equal to two times the area of the base plus the sum of the areas of all the lateral sides.” Remember that area is always labeled as square units.
Formula for the Volume of Prisms: V = Bh, where, again, B is the AREA of the base and h is the height of the prism.
Caution! Remember that the edge of the prism is the height only if the figure is a right prism. If the prism is oblique, the height will have to be calculated just as is necessary in non-right triangles.
The formula V = Bh should be read as “The volume of a prism is equal to the area of the base times the height of the prism.” Remember that volume is labeled as cubic units.
I think you can tell that these formulas are actually rather simple to calculate IF you have mastered the terminology and the area formulas for 2-dimensional polygons. The secret to success in Geometry: Memorize! Memorize! Memorize!
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## Formula For Finding The Area Of A Triangular Prism Famous Skyscrapers
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## Famous Skyscrapers
Architectural design has advanced to a phenomenal stage where every modern city would like to boast having one of the tallest buildings. While criteria defining the tallest buildings or skyscrapers differ widely, listed below are the ten most popular architectural marvels of our time.
Empire State Building
A world famous New York City landmark and sky scaper, it rises above the island of Manhattan, about a quarter of a mile into the sky. The observatory is located on the 86th floor, 1050 feet above the ground and offers the most breathtaking and panoramic view of Manhattan and beyond from within a glass enclosed pavilion. Besides the observatory there are several tourist attractions, including, restaurants, shops and banks. It also has a New York SKYRIDE an independently owned and operated simulated helicopter ride and virtual-reality movie theatre. William Lamb, an architect at Shreve, Lamb and Harmon was chosen to design the Empire State Building in 1930
Petronas Towers
Petronas Twin Towers in Kuala Lampur lay claim to being the tallest twin towers of the 20th century, standing at a height of 1483 feet. They were designed by architect Cesar Pelli and completed in 1998. The 88-floor towers are constructed largely of reinforced concrete, with a steel and glass facade designed to resemble motifs found in Islamic art, a reflection of Malaysia’s Muslim religion. The towers feature a sky-bridge between the towers on the 41st and 42nd floors which is 170m high and 58m long.
Sears Tower
Architect Bruce Graham designed the Sears Tower of Chicago in 1974. At 1450 feet, and 110 stories, it is the tallest building in the United States of America. The construction system consists of steel fram with bronze tinted glass curtain wall. The Sears Tower Skydeck observation deck and tourist attraction is on the 103rd floor, 1353 feet above the ground.
Bank of China Tower
Located in Honk Kong, in addition to being one of the famous skyscrapers in the world, it is one of the most outstanding achievements of modern architecture. The construction was started in 1985 and completed in five years by the architects I. M. Pei & Partners and Sherman Kung & Associates. The building standing at 1205 feet, is a grouping of four triangular glass and aluminium towers of different heights, all emerging from a single granite podium. The changes rising from a square base to a single spire results in a magnificent façade of angles and profiles that reflect the light and seem almost crystalline in composition. On the 42nd floor is a sky-deck providing a panoramic view of the northwest Hong Kong.
Chrysler Building
Rising at 1046 feet, it was considered to be an engineering marvel and the tallest building in 1931. However, it still remains the tallest brick building in the world. The tower is a beautifully tapered stainless steel crown supporting the famous spire at its peak. A quintessence of skyscraper design, the Chrysler Building is a perfect example of Art Deco and has a lobby clad in different marble, onyx and amber.
Taipei 101
At a height of 1671 feet, this high-rise building has surpassed all to become the tallest skyscraper today. Taipei 101 holds the world record in three of the Council on Tall Buildings and Urban Habitat’s height categories: tallest to the structural top, tallest to the roof, and highest occupied floor. The 89th floor has an indoor observation area while the 91st floor has an outdoor observation deck, known as the highest in the world. The pagoda shaped design of this building is inspired by traditional Chinese architecture. The sectioned tower is also symbolic of the bamboo plant characterizing strength, resilience and elegance. The tower’s design specifications are based on the number 8, considered lucky in the Chinese culture.
Jin Mao Building
This building in Shanghai symbolizes the progress and advancement made by the Chinese. It boasts of being the first tallest sky scrape in the country and the third tallest in the world. A great blend of East-West architecture it denotes aptly the emergence of Shanghai as a modern global city. It follows the versatility model by offering retail at it base, offices above and the Grand Hyatt’s World’s highest Hotel occupying the upper 38 floors. The magnificently designed building combines the elements of traditional Chinese architecture and a vastly Gothic influence.
Burj Al Arab
The Burj Al Arab rises to a level of 1053 feet, and is known as one of the world’s tallest structure with a membrane façade, 24-meter wide helipad. This is the tallest operating hotel building in the world and the design is influenced by the profile of an Arabian sailing ship. The Al Muntaha restaurant is located 200 meters above the Persian Gulf offering a panoramic view of Dubai while the atrium is situated at a height of 180 meters.
CN Tower
At a height of 1815 feet, the CN Tower in Toronto is the tallest building and freestanding structure in the world. It is considered to be the signature icon of the city and hosts almost two million visitors a year. The view from the exterior glass floored observation deck located 342 meters above ground is breathtaking as it is exciting. It also has the Space Deck at 447 meters, the world’s tallest observation deck with a 160 km view and the revolving 360 Degree Restaurant.
Hancock Place
A reflective obelisk sky scrape at Boston is an architectural marvel. It is regarded as ‘icily magnificent’ wherein the surface changes as the day changes, each side reflecting the color of the sky it faces. Moreover, this dramatic effect is highlighted by the parallelogram shape of the prism, which provides uniquely differing reflections on adjoining surfaces.
The architectural buildings mentioned above are famous creations and literally considered architectural works of art today.
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## Formula 1 Drive To Survive Season 3 Release Date Analytics in Football – A Double Edged Sword
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## Analytics in Football – A Double Edged Sword
Sports as we know it today has come a long way. There were times when watching sports on television was considered a massive step forward in terms of technology. Fast forward 60 years, watching sports on television has become the most basic thing. Today we watch sports on the go on our mobile phones or any device with a screen and internet connectivity. Proud of how far we’ve come, aren’t we? Hopefully I can change your opinion on that by the end of this article.
What is sports all about? Sports is a bunch of people getting together to play a game with pre defined rules and a referee to ensure that these rules are adhered to during the passage of play. I am a sport lover and play sports all time. My love for tennis and soccer in particular cannot be defined. My issue when it came to technology and advanced analytics was with the game of soccer in particular. Soccer is such a beautiful game. The strategies that the coaching staff come up with and the way it is executed on field by the players, it actually is a thing of beauty. I was a soccer player myself (just an average one at that) and have been part of various teams. I know firsthand how strategies are built, how much thought goes into one single run of play.
Enter -> Advanced Analytics
Most of you would’ve seen the movie Moneyball. The movie was based on the book Michael Lewis wrote in 2003. It talks about how a jock turned luminary uses advanced statistics to gain a competitive edge over his better funded opponents. This book brought about a revolution is sports. Fans and boards of soccer clubs didn’t want to settle for subpar statistics or analytics anymore. What Moneyball did is, it took an old cliché – “sports are businesses” and made us move on to the next logical question – “how do we do things smarter?”
Now let’s talk about advanced analytics. Advanced analytics in today’s world plays a massive role in every business sector. Advanced analytics has been a boon for us. Moving from descriptive analytics to prescriptive analytics, we actually have come a long way. In various businesses, where the requirement is demanding, advanced analytics are of utmost importance.
When we look at soccer, its a game that does not require too much machine intelligence, it is a game that needs the human element. When you bring in analytics and technology and try to reduce the human element in the sports, it simply just crushes the spirit of the game.
Relying on analytics heavily killed the Premier Leagues long ball game and brought in the pressing, continual passing tiki-taka. Each league for that matter had its own style of play. The Premier League had the brash and brazen style of football that was termed “The way real men play football”. There were beautiful long balls, harsh tackles but all the players just sucked it up, walked it off and it was all up to the referee on the pitch to penalize the offender or not. There were arguments and fights, the passion from the fans was crazy, that was the football that screamed of passion, when players got in the face of other players not fearing punishment. The Eric Cantona’s, the Ivan Genaro Gattuso’s, the Jaap Stam’s of the football world went missing soon enough and the diving and the biting began. Then there was the tiki-taka style of football that was played in the Spanish La Liga, the silky style of play that caught everyone off guard. The legendary Pep Guardiola and his army at Barcelona were the masters of the tiki-taka. There was Real Madrid who were always a star studded line-up with excessive parts of their play relying on lightning quick counters which most often than not left the opponents stunned. There was Manchester United who had their own brand of football being managed by the legendary Sir Alex Ferguson. That United team was a team of sheer grit and character. Each of these leagues had their own beauty and the teams had their own style of play.
When you bring in excessive technology and analytics, there emerge sorry technologies like VAR (Video Assistant Referees).
There are 3 stages as to how the VAR works:
Step 1
Incident occurs
The referee informs the VAR, or the VAR recommends to the referee that a decision/incident should be reviewed.
Step 2
Review and advice by the VAR
The video footage is reviewed by the VAR, who advises the referee via headset what the video shows.
Step 3
Decision or action is taken
The referee decides to review the video footage on the side of the field of play before taking the appropriate action/decision, or the referee accepts the information from the VAR and takes the appropriate action/decision.
Now the referee can consult with VAR for basically any doubts he wants clarified. What does this do?
• Removes the human element from the game.
• Takes up excess time and brings too many stoppages within the game, a game that was previously free flowing and continuous.
This makes it similar to Formula 1 racing. The analytics which brought about the fuel weight management systems and the numerous pit stops took the continuity out of the race and viewership reduced with the increase in technology. A pretty similar trend might occur in football if this implementation becomes mandatory.
The Positive Side of Advanced Analytics in Soccer:
Analytics are not all that bad in football. Let’s take the case of when Simon Wilson joined Manchester City in 2006. Simon Wilson was a consultant for an analytics startup called Prozone initially. He joined City to start a department of analytics and hired the best data analysts under him. He wanted to change the way how data was used by football teams. He saw that, after a defeat there was no introspection as to why they had lost and what needed to be done next time. City were a mid table club at that time. In September 2008, when the club was acquired by the Abu Dhabi United Group for Development and Investment, a private-equity outfit owned by a member of the Abu Dhabi royal family, the team suddenly found itself with the resources necessary to mount a challenge for the Premier League. Today, Wilson is Manchester City’s manager of strategic performance analysis. He has five departments under him, including the team of performance analysis, which is now led by a sports scientist named Ed Sulley.
After each match, the team’s performance data would be examined. The list is extensive. Line breaks (a rugby term), ball possession, pass success rates, ball win/loss time ratio were what used to be analyzed. “Instead of looking at a list of 50 variables we want to find five, say, that really matter for our style of play,” says Pedro Marques, a match analyst at Manchester City.
“With the right data-feeds, the algorithms will output the statistics that have a strong relationship with winning and losing.” Wilson recalls one particular period when Manchester City hadn’t scored from corners in over 22 games, so his team decided to analyze over 400 goals that were scored from corners. It was noticed that about 75 percent resulted from in-swinging corners, the type where the ball curves towards the goal. The next 12 games of the next season saw City score nine goals from corner.
Teams are investing heavily in analytics today and it is working in their favor. Look at where Manchester City are today, sitting atop the Premier League table and not being threatened at all. Look at Manchester United this season, their game has been such where their possession percentages are low but their goal conversions are high. The Manchester Derby on 7th April 2018 saw United have only 35% of the possession but they managed to trump City 3-2. Each team has their set of analysts who provide inputs as per the strength of the team.
Advanced analytics is like the coin Two Face in Batman has, “Heads you die, Tails you survive!”
It can reap crazy rewards from a team’s point of view but at the same time can disrupt the lovely game by bringing in unnecessary stoppages, replays and by taking the human element out of it. The numerous replays and the different angles, show the fans if the referee has made an error or not. Let the error happen, after all to err is human. Refereeing in soccer is not an exact science and it’s all real time. Let there be arguments about a decision, let the passion in the argument come through. Do you want to watch a football match like the El Classico or the Manchester Derby and sit with your bunch of friends and say “it was a very clean game, the best team won!” Hell NO! Don’t drive the passion out of soccer with technology and analytics. Let soccer be soccer and let technology stay away!
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## Find The Formula For The Exponential Function With Points Understanding Marketing – An Overview of Strategies, Costs, Dangers and Risks
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## Understanding Marketing – An Overview of Strategies, Costs, Dangers and Risks
What is Marketing?
Marketing is a business discipline through which the targeted consumer is influenced to react positively to an offer. This can relate to the purchase of a product or a service, the joining of an organization, the endorsement of a candidate or ideology, the contribution or investment in a cause or company, or a variety of other choices of response.
The marketer can use a number of techniques to reach the consumer which can be based on artistic or scientific strategies, or a combination of the two.
Usually, the consumer is identified as a member of a particular segment of the populace, known as a market. For example, markets can be defined by age, income, area of residence, home value, interest, buying habits, industry or profession, etc., which facilitates and simplifies the marketing process. Knowing to whom the marketing effort is appealing greatly assists the marketer in developing appropriate language, reasoning and incentives to find success in its marketing efforts.
Choosing to target a particular market as opposed to the entire universe also greatly controls marketing expenditures but also may limit response. If anyone anywhere can be a customer, sales expectations may be higher but marketing costs will certainly also need to be higher as well with such a huge target as its goal.
To address this dilemma, more creative means of marketing are sometimes utilized to assist with marketing message delivery. If what is being marketed is considered newsworthy and of public interest, editorial coverage in the media can greatly assist marketing efforts. Since this usually is not reliant on major marketing funds other than what is needed to support the development, distribution, and yes, marketing of press releases to editors and publishers, the advantages of such publicity can be priceless, albeit usually miraculous on such a large scale.
Marketing is everywhere!
Everywhere we turn, everything we do is somehow connected to marketing, whether we have been induced to participate in some activity because of it or develop an interest in some idea as a result of it. Whether we realize it or not, there are personal, political or commercial agendas cloaked as news we read in the paper, behind the books, movies and music we experience as part of our culture, and within the confines of our stores and supermarkets where we shop. Of course, we easily recognize the blatant marketing efforts that reach us through direct mail, media advertising, and all over the Internet including the spam we receive ad nauseum. Marketing has become one of the most all-pervasive elements of life and we are fools if we do not question the validity or innocence of everything we read, see and hear.
Marketing is communication and education!
In order to be successful in business marketing, the customer must be reached in a variety of ways. First of all, not every customer gets the daily paper or listens to local radio. We have limited knowledge of which TV station they may watch, where they shop, what roads they travel or where they dine. Depending on what we are marketing, we may have to utilize a whole assortment of avenues of marketing to get their attention. And, if we reach them just once, that is hardly enough to make a lasting impression. Marketing is necessary on a repeated basis in a diverse number of ways in an ever-changing presentation to assure that every customer can relate to it in some way, learn what we are offering and understand how it can benefit them. To achieve long-term customer loyalty, the targeted consumer needs to be coddled into familiarity with what we are selling so they feel it is something they truly want as opposed to having it forced upon them as something they desperately need, only to find out later they were tricked!
Marketing Sounds Expensive!
Yes, marketing can get pricey particularly if it is done on a consistent basis. But in today’s world, we have marketing options we never had even twenty or thirty years ago. Now, instead of paying for expensive printing and postage to mail a brochure or postcard to a targeted consumer, we can utilize email marketing, website presentations or online banner ads to reach the same market, usually at a fraction of the cost. Today, instead of buying expensive print advertising, we can work on improving our website’s SEO (search engine optimization) – (something we can do for free, if we are so inclined) so that people in need of what we offer can find us through Internet searches, rather than our trying to find them at an astronomical expense.
What About Social Media Marketing?
In addition to alternative marketing options already mentioned, there is the latest craze for Facebook, Twitter, LinkedIn, and other incredibly popular social media where people, young and old, spend hours developing relationships with “friends” they may never have met or ever will meet. Yet they share intense secrets of their deepest thoughts and desires as well as actual photographic representations of the same which sometimes land people in trouble with the law, or at the very least, their employer, school or parents.
Whether social media marketing is a worthwhile endeavor for businesses remains to be seen since businesses rarely accumulate millions of followers the way celebrities do. But as a way for customers to interact with a business for which they may have developed a fondness cannot be disputed. Can this translate into more sales for the business? We’ll have to wait and see, while continuing to devote precious time to composing meaningful 140-character tweets and building a Facebook “persona” for the business. From this writer’s standpoint, the only worthwhile social medium for business is that of LinkedIn since it provides a serious platform on which to create a business résumé where anyone interested in your professional stature can quickly summarize your capabilities, experience and accomplishments.
Marketing Can Be Intuitive
Much of what becomes marketing strategy is based more on common sense than on some mysterious scientific formula. As we see on a daily basis in stock market gyrations as well as political leanings, the herd mentality rules. On any particular day, if the Japanese or European stock or bond markets are selling off for one reason or another, you can safely bet that the U.S. markets will follow suit. And in any political race, as we are witnessing in the U.S. presidential primaries, the more one candidate gains ground, baby step by baby step, the more likely that candidate will become the Party nominee. Today’s world is governed by a minute-by-minute opinion survey measured by the endlessly publicized polls where people see what other people are thinking and use those results to form their own opinions. Monkey see, monkey do. The same holds true for marketing.
If we are told that a certain brand of coffee is the leading brand in America, we will probably believe what we are told, assume it tastes best, perhaps buy it ourselves regardless of cost, and perhaps adopt it as our own favorite. All because we were told everyone else was doing it. Safety in numbers, as they say.
It is ironic that those who become successful marketers usually dwell on the outskirts of the herd, have a more astute grasp of mass psychology, and approach business and life in a more innovative, creative and unique way, a mindset they use to formulate the next marketing phenomenon. The world is made up of leaders and followers: a few choice leaders and a glut of followers. It takes a lot more gumption to become a leader than it does to join the herd. That’s why marketing is a profession based in psychological control by a choice few over the mindless masses who have no initiative or courage to decide for themselves.
What is the difference between marketing and selling?
Selling is one aspect of the greater process of marketing. Marketing begins long before the product or service is even ready to sell. Marketing encompasses the concept, naming, branding and promoting of the offer while selling is the much more individualized effort to convince a lead who has possibly responded to the marketing offer to make the purchase. You can’t have one without the other, at least not easily. Marketing is a process by which we strive to reach the final goal of making the sale. Without marketing, the sales process is extremely difficult because the entire onus of educating the consumer about the offer is on the shoulders of the sales representative. On the other hand, if marketing has been successful, the sales rep can waltz in knowing the consumer is well apprised of the offer and can work his magic to convert the prospect into a satisfied customer.
What are some of the instruments of marketing?
There are many ways to market an offer, some of which are expensive, and others of which can be free. The methods we use that cost us dearly may not work as well as some of those we receive as a gift. Among the costly ways are media advertising, direct mail, conference presentations, distribution of printed literature, online advertising, email marketing, etc. Of those that are free are efforts referred to as guerrilla marketing, which are things we do ourselves to spread the word, network and publicize what we are offering. This can include posting flyers on bulletin boards in supermarkets, libraries, delis, small shops, and government offices, etc. Every time we add a tag to our emails where people can click to go to our website, we are using guerrilla marketing at no cost. Making sure we are easily found in Internet searches through search engine optimization of our website or other online presence, is an excellent way to achieve free marketing. One way to do this is to register your company or organization on every possible free online directory in your industry, region or interest group which translates into exponential growth as time passes.
What is viral marketing?
Viral marketing (as it relates to the word “virus,” meaning contagious and capable of spreading) is another means of free promotion facilitated by shrewd decisions we can make to further our cause. The easiest way to define viral marketing is that which is communicated via “word-of-mouth.” Related to the herd mentality discussed above, if a friend or business acquaintance mentions a product or service in a favorable light, we will be much more inclined to remember it and check it out. This can happen in a business meeting, at a mall, at a soccer game or over lunch. However, since most of us spend so much time on the Internet, it can happen practically everywhere we turn by clicking on the “like” buttons on Facebook or the “1” button on Google, among others. These are our personal endorsements where we give a “thumbs up” to something we have experienced and want to share with our friends so they too can enjoy it. Getting your offerings out with such buttons attached can result in viral marketing in your favor.
Viral marketing can have powerful repercussions as experienced by one client with an online auto accessories store. Many of his customers frequent online special interest forums related to the model of car they drive where members discuss products they have installed and the source of their purchase, followed by a link to his referenced website. Such referrals are repeated in other ensuing discussions, multiplying the number of links back to his site, increasing the power of his SEO and catapulting him to the tops of Internet searches for what he sells. He paid nothing for this phenomenon of parlayed good fortune except the daily effort he consistently expends to offer top quality merchandise and equally excellent customer service.
Do you need marketing?
If you are in business, of course you do. While you can attempt to do as much of it as you can on your own, it is advised that you begin with a reliable base of professional name, logo, website and search engine optimization to get started on the right foot. From there, you can work on promotion via guerrilla marketing and seek professional marketing services as needed for special needs, like a strong, effective ad to run, the development of professional sales literature to distribute at an upcoming show, or a direct mail promotion to your list of repeat customers, for example. Some business people choose to handle their own taxes to save on the cost of using an accountant for such critical functions at the risk of getting audited. Likewise, you can certainly attempt to produce marketing tools yourself but for long-term branding purposes and best return on investment, it is advisable to leave marketing development to the professionals.
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## Find The Formula For The Exponential Function With Points Alcohol, Aging, and Curing Cancer
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## Alcohol, Aging, and Curing Cancer
Recent media coverage has been devoted to the ability of alcohol to extend human life span. A rather insolent statement considering the historical barrage of disparaging findings regarding alcohol and its secondary ill health effects. However, if we isolate this one finding without examining the broader context of alcohol’s adverse health effects, in the long run, we probably do more harm than good.
Using Albert Einstein’s famous equation,
E=MC2
I have formulated the Time Longevity equation. By replacing the velocity of light coefficient with its equivalent velocity variables (distance/time), I have generated the Time or Longevity equation. Time equals the square root of distance squared times mass times the inverse of energy. I have proclaimed this formula, The Rainer Time Longevity Equation.
Rainer’s Time Longevity Equation
T=√D2x M x 1/E
Rainer’s Time Longevity Equation states that an organism’s life span is interdependent on three variables. Those three variables are Distance (D), Mass (M), and Energy (E). Animals of various species, including man, may alter these three variables in order to prolong life. Since time is relative to movement through space, animals may slow the aging process by forward movement through space. The greater the distance covered, the greater the positive impact on longevity. Animals that have the capacity to fly have an increased (D) variable compared to their non flying counterparts. Likewise, animals that typically fly or swim large distances will have a similarly increased life span. In short, caging migratory animals equates to a death sentence. The Galapagos Tortoise for example, has capitalized on all three variables of Distance (D), Mass (M), and Energy (E) in order to maximize its average life span of 150 years. The tortoise shuttles through the ocean’s currents (Increased D); It possesses a large inert shell that requires no metabolic maintenance (Increased M); Tortoises possess exceedingly slow movement with a similarly decreased metabolic rate (Decreased E).
Man has achieved greater longevity through modernization which includes amongst other things, more rapid travel modalities (Increased D). Air travel has helped push man’s life span to higher limits. Space travel shows even greater promise in extending man’s life span. Increased Mass allows larger mammals to extend their life span by increasing the (M) variable. Man may also alter time by manipulating the Mass (M) variable. Ancient Egyptians brilliantly utilized this fact by spending time residing in pyramids. Bear in mind, that a pyramid by virtue of its construction, is virtually entirely mass. Modern French researchers were astounded when they placed a dead cat and raw fresh meat in the interior of the Great pyramid, only to find both items exquisitely preserved days later. Likewise, Dinosaurs improved their life span by virtue of their enormous mass.
Lastly, this brings us to Energy (E), the final variable of the Rainer Time Longevity Equation. Given that time or longevity has an inverse relationship to energy (E), all things that decrease Energy will invariably increase time. However, I would like to point out that items that decrease the (E) variable and therefore increase longevity, may not necessarily improve the quality of life. Items that decrease the Energy (E) component and cause a decrease in Basal Metabolic Rate and secondarily increase longevity, include the following: love, marriage, hibernation, sleep, coma, stupor, caloric reduction, prayer, meditation, medicinal substances that suppress the central nervous system.
This brings us to the discussion of alcohol. Alcohol is a central nervous system depressant. Furthermore, it decreases brain endorphin production. Alcohol may increase longevity by decreasing the Energy (E) variable thus reducing Basal Metabolic Rate. However, we must not overlook the obvious ill effects that alcohol has on our overall health. Alcohol causes brain atrophy, liver cirrhosis, and promotes the formation of various types of cancer. Alcohol in a broader context has damaging effects on the body. Additionally, by reducing endogenous brain endorphin levels, alcohol robs our subconscious mind of its creativity and severs our link with God. Alcohol steals our ambition which, many may argue, is our chief purpose for existence.
Invariably, future studies will likely support findings that opiates such as heroin and codeine increase longevity by inducing a comatose state. But few would claim that periodic heroin use is beneficial in improving one’s life. Let’s not be seduced into accepting isolated research findings as virtuous advances. Embracing a morally sound lifestyle shows the most promise in extending and improving the quality of life. Enhancing endogenous brain endorphin production by routine exercise and meditation, facilitates a quality lifestyle. Naturally, man desires a longer life span. Likewise, an associated increase in life span will occur as man adopts the following lifestyle and behavior modifications: swifter travel modalities (increased D); decrease in obesity(decreased E); acquire more sleep (decreased E); prayer (decrease E);universal love (decrease E); consumption of minimal amounts of wine (decrease E); reside in structures composed of denser building materials (increase M) such as granite high rises.
I would like to add a caveat while on the topic of Energy (E). Contrary to current popular belief and aside from acute injury, a lower bone density contributes to a longer life span. High bone masses require increased metabolic rates to maintain. Likewise, an increased Basal Metabolic Rate shortens life span. As Man evolves, he sacrifices the development of high bone density for superior brain development. Asian ethnic groups, known to have the lowest bone density, likewise have the longest life span. Hence, lower bone density promotes longer life. This may ultimately encourage the use of medicinal agents targeted to enhance osteoclast activity (cells that reabsorb bone tissue) in order to improve life span. Additionally, this may discourage the use of aggressive medicinal treatments used to build bone in order to prevent osteoporosis.
The human body, given its fine tuning due to an extensive evolutionary process, capitalizes on the energy (E) component of the Time Longevity equation. The body utilizes fever as a weapon to combat bacterial and viral offenders. By raising the body’s core temperature when faced with a crisis, effectively increasing Energy (E), the body’s immune system selectively shortens the life cycle of the offending bacterial or viral agent. Hence, fever can assist the body in its effort to destroy infecting organisms. However, the body must precipitously drop the temperature once the offending threat has been thwarted in order to circumvent its own demise. Fever, although an effective defensive tool, can also prove extremely hazardous if left unchecked. Severe systemic infections may increase the basal metabolic rate (E) beyond limits compatible with life.
We must also avoid the temptation to implement detrimental behavior patterns in an effort to extend life span. This is particularly true if newly adapted practices jeopardize the quality of life or blunt the creativity of the human mind. Our minds are our cherished tools that serve as the link to connect us to our higher power. Our minds resonate with one another allowing us to share our creativity and inspiration which may then develop into a material equivalent. No substance, compound, or medicinal agent should ever compromise this higher cerebral function. Our minds empower our capacity to connect to one another, our environment, and with God. We live to promote our thought capacity for now and beyond. Our bodies merely serve as our mind’s temporary housing. Alcohol excess can evoke the brain’s capacity for addiction, a biological disease. In all fairness, two glasses of red wine consumed on a weekly basis may have some positive health benefits. Red wine with its antioxidant properties coupled with relaxants, may enable us to cope with life’s rigors. But stronger types of alcohol such as whiskey, bourbon, rum, and vodka often tax the capacity of our liver’s detoxification capabilities and therefore cause systemic illness.
Time travel is also implicit to this discussion. As we photograph a tennis player strike a ball without the aid of a freeze frame lens, we see that motion is composed of a sequence of independent movements separated by distance and time. Likewise, photographing our solar system gives us the same effect. Photos of seemingly independent planets appear separated only by distance. But, however, this is effectively one of the same. Our solar system is the evolution of one planet only– Earth. Mars is earth’s future and Venus is earth’s past. The volcanoes on Mar’s surface are merely eroded Great Pyramids of Egypt. The eerie face peering at us from Mar’s surface is our own Great Sphinx looking towards the heavens. However, the dimension of time is well into our future. Time travel is before our eyes. What appears as a seemingly array of independent objects separated by various distances is merely one of the same. Given the temperature variation of Mars, global warming then becomes a moot point.
Let’s turn our sights to one of our most challenging issues, cancer. Is there a cure in sight? We race to find a cure but we must force ourselves to look beyond the narrow scope of our lens. Increasing distance (D) and mass (M) while decreasing energy (E) will not only increase life span but also may hold the definitive key to cancer’s cure. If we focus on energy (E), from a metabolic standpoint, we make reference to Basal Metabolic Rate. Whatever variable decreases basal metabolic rate likewise decreases the incidence of cancer. Current medical standards must be revised. At best, our attempts to preserve bone mass by medicinal means is in fact an absolute route to promoting the formation of cancer. Raising bone density increases metabolic rates. An increase in Basal Metabolic rate increases energy (E) which in turn shortens longevity (T). Hence, increasing bone density increases cancer. An increase in metabolic rate is an increase in energy (E) which causes a decrease in time longevity (T). Medicinal agents that increase bone density in post menopausal years directly shorten life span. God’s naturally occurring hormonal transition has been correct all along. The production of progesterone in women’s twilight years with the corresponding decline in bone mass potentiates the duration of life. Cancer is merely the byproduct of stimulating basal metabolic rates to an extreme thus creating a cellular replicating, self perpetuating vortex. We cannot eliminate this vortex by applying more energy such as radiation. In fact, we must do the very opposite. Hence, the promise for an absolute cure for cancer lies in the basal metabolic reductive capacity of cryogenics and frigid cold baths.
The following items below decrease energy (E) or Basal Metabolic Rate and therefore help PREVENT CANCER and increase Life Span:
– Decrease in Basal Metabolic Rate
– Decrease in Bone Density
– Osteoblast inhibitor
– Antibiotics
– Sleep
– Hibernation
– Estrogen Blocker (Tamoxifen)
– Osteoclast activator
– Meditation
– Coma
– Prayer
– Exercise
– Sex
– Love
– Classical and Jazz Music
– Relaxation
– Laughter
– Caloric Reduction
– Fasting
– Progesterone
– Depo Provera
– Excessive Cold
– Moderate Wine
– Lupron
– Decreased Body Mass Index
The items listed below increase Mass (M) and therefore help PREVENT CANCER and increase Life Span:
– Residing in the Great Pyramid of Egypt
– Residing in Stone House
– Residing in a Cave
– Burrowing and residing underground
The items listed below increase Distance (D) and therefore help PREVENT CANCER and increase Life Span:
– Space travel
– Flying
– Migration
– Forward Movement
– Swimming
The items listed below increase energy (E) or Basal Metabolic Rate and therefore CAUSE CANCER and decrease Life Span:
– Increase in Basal Metabolic Rate
– Increase in Bone Density
– Osteoclast Inhibitor
– Cigarettes
– Infection
– Fever
– Carcinogenic exposure
– Excessive Heat
– Genetic Predisposition
– Bacterial exposure
– Viral exposure
– Obesity
– Osteoblast Activator
– Sleep Deprivation
– Hate
– Anger
– Stress
– Fosomax
– Androgenic Steroids
– Unopposed Estrogen
– Cocaine
– Excessive Sun
– Mifepristone (RU486)
The items listed below decrease Mass (M) and therefore CAUSE CANCER and decrease Life Span:
– Residing in a trailer
– Residing in a tent
The items listed below decrease Distance (D) and therefore CAUSE CANCER and decrease Life Span:
– Sedentary lifestyle
– Stasis or staying still
*Some items listed above are currently theoretical and require case-control study to be regarded as scientific fact
Optimizing the physical laws of nature will assist us as we make progress in increasing longevity and eradicating cancer as a health menace. We must carefully observe and analyze many of nature’s obvious examples of perfection and harmony. These examples in nature all have purpose and intent and have required exponential years of evolutionary trial and error. As man universally harmonizes in the spirit of love, all things will ultimately be achieved.
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## Các vòi cảm ứng giá rẻ TPPRO cho chậu rửa lavabo uy tín nhất
Vòi nước tự động là một thiết bị vòi chậu thông minh, có dùng thêm cả pin và điện mang đến nhiều thuận tiện cho người sử dụng. Nếu bạn đang muốn sắm một chiếc vòi nước cảm biến giá rẻ, chất lượng cho gia đình mình thì đừng bỏ dở bài viết này của chúng tôi nhé! Top 5 Vòi cảm ưng giá rẻ cho chậu rửa lavabo bên dưới chắc chắn sẽ là những gợi ý lý tưởng dành cho bạn.
## Vòi cảm ứng giá rẻ TPPRO TP-20934
Vòi cảm biến TPPRO TP-2093 là vòi rửa tay tự động cho chậu rửa (lavabo), thay thế cho các loại vòi xả thủ công. Bạn cũng có thể lắp đặt trong nhà vệ sinh công cộng, nhà ở cao cấp.
Sản phẩm này không chỉ mang đến sự tiện lợi, sang trọng mà còn đảm bảo dọn dẹp vệ sinh, hiệu quả và tiết kiệm cho gia đình bạn. Vòi sẽ tự ngắt nguồn nước sau 1 giây nếu không nhận thấy vật thể dưới mắt hồng ngoại.
Sau đây là những đặc điểm nổi bật của sản phẩm này:
• TPPRO TP-2093 có giấy chứng nhận xuất xưởng (CO) được cấp cho từng lô hàng. Sản phẩm cũng được kiểm định chất lượng bởi Trung tâm Kỹ thuật Tiêu chuẩn Đo lường Chất lượng 3 – Quatest 3 (CQ) theo giấy chứng nhận số 022/N5.17/ĐG cấp ngày 22/02/2017.
• Thân vòi được làm bằng đồng tiêu chuẩn mạ chrome bền chắc, sáng bóng, không bị gỉ sét đi theo thời gian.
• Khi xử dụng, bạn chỉ cần đưa tay vào dưới vòi trong khoảng cách 12 – 18cm thì nước sẽ tự xả mà không cần thực hiện bất kỳ thao tác nào khác.
• Vòi nước thông minh tự ngắt TPPRO TP-2093 dùng 4 pin AA cho áp lực nước 0,05 – 0,6 MPa giúp bạn làm sạch tay nhanh hơn.
• Tuổi thọ của pin lâu dài lên tới 100.000 lần xử dụng. Bạn cần cung cấp nguồn nước sạch không cặn bẩn cho vòi, có thể xử dụng nguồn nước cấp nóng lạnh.
## Vòi cảm ứng giá rẻ TPPRO TP-20935
Vòi tự ngắt TP-20935 sở hữu thiết kế nhỏ gọn, màu sắc sáng bóng sang trọng, đồng thời nó có khả năng chống nước theo chuẩn IP56, rất an toàn cho người sử dụng khi sử dụng.
Vòi tự động TPPRO TP-20935 là một giải pháp thông minh giúp bạn tiết kiệm được lượng nước khá lớn hằng tháng cho gia đình. Sản phẩm có trang bị cảm biến hồng ngoại, khi bạn đưa tay vào vùng cảm ứng thì vòi sẽ chảy nước ra, khi đưa tay ra vòi sẽ tự động ngắt nước, tránh lãng phí nước.
Bên cạnh đó, thiết kế này giúp bạn không phải thao tác bật/tắt vòi nước theo kiểu thủ công, đảm bảo dọn dẹp sạch sẽ, tránh lây nhiễm bệnh.
TPPRO TP-20935 được thiết kế với kiểu dáng hơi uốn cong nhìn rất mềm mại. Toàn bộ thân vòi được làm bằng đồng mạ chrome khá ưa nhìn, đồng thời có khả năng chống gỉ sét cao. Sản phẩm phù hợp để lắp đặt với tất cả các loại chậu rửa mặt lavabo thông dụng trên thị trường. Thời gian lắp đặt mỗi vòi chỉ mất từ 10 – 15 phút nên cực kì tiện lợi..
## Vòi cảm ứng giá rẻ TPPRO TP-20948
Nhờ những tính năng đặc biệt mà TPPRO TP-20948 đang được dùng khá phổ biến tại những nơi công cộng như trung tâm thương mại, sân bay, khu vui chơi, gia đình, nhà hàng, văn phòng, trường học, bệnh viện…
TPPRO TP-20948 được ứng dụng công nghệ cảm ứng hồng ngoại. Khi người sử dụng đưa tay vào vùng cảm ứng thì vòi sẽ tự động xả nước để rửa sạch tay & tự động ngắt nước khi người sử dụng đưa tay ra giúp tiết kiệm nước và đảm bảo vệ sinh, tránh lây nhiễm bệnh tật.
Sản phẩm được làm từ chất liệu bằng đồng mạ chrome cho độ sáng bóng cao, bền hơn. Vòi tự ngắt TPPRO được ứng dụng công nghệ Hybrid dùng cả điện 220V AC & pin giúp vòi hoạt động được liên tục, ổn định. Độ nhạy của cảm ứng là 0,5s, thời gian ngắt trễ là 1s, khoảng cách cảm ứng từ 15 – 20 cm.
Bên cạnh đó, sản phẩm còn có khả năng chống nước đi theo hạt tiêu chuẩn IP56, giúp sử dụng an toàn về điện. Áp lực nước khỏe cung cấp dòng chảy mạnh. Vòi rửa tay cảm ứng có thể dùng được nguồn nước cấp nằm trong dải nhiệt độ từ 1 cho đến 45 độ C.
Hiện tại, sản phẩm vòi tự ngắt TPPRO TP-20948 đang được bán với giá 1.950.000 đồng và được bảo hành 12 tháng.
## Vòi cảm ứng giá rẻ thông minh TPPRO TP-20949
Vòi nước cảm ứng TPPRO TP-20949 là vòi rửa tay tự động cho chậu rửa (lavabo) giúp thay thế cho các loại vòi xả thủ công, có thể lắp đặt trong nhà dọn dẹp vệ sinh công cộng, nhà hàng, bệnh viện, trường học… mang cho đến sự tiện nghi, hiện đại và tiết kiệm.
Bên dưới là những đặc điểm nổi bật của sản phẩm này:
• Thân vòi được làm từ chất liệu đồng hạt tiêu chuẩn mạ chrome cho độ cứng cáp, bóng đẹp lâu dài và bảo vệ hệ thống mạch điện không bị hư hỏng trong môi trường ẩm ướt.
• Sản phẩm sử dụng 4 viên pin AA alkaline với tuổi thọ dùng 100.000 lần giúp tiết kiệm chi phí đáng kể cho người sử dụng.
• Đặc biệt, vòi tự động thông minh TPPRO TP-20949 được trang bị cảm biến hồng ngoại, có thể tự động điều chỉnh khoảng cách cảm ứng từ 12 – 18cm tùy vào môi trường. Bên cạnh đó, vòi cũng được tích hợp cơ chế tự động ngắt nước sau 1 giây nhằm hạn chế tình trạng thất thoát gây lãng phí nước.
• Bên cạnh đó, sản phẩm này còn có thể được lắp đặt dễ dàng với mọi thứ các loại chậu rửa (lavabo) thông dụng trên thị trường chỉ với thời gian lắp đặt từ 10 – 15 phút mang lại nhiều tiện ích cho khách hàng.
## Vòi cảm ứng giá rẻ TPPRO TP-20918
Nhờ những ưu điểm vượt trội mà vòi rửa tay cảm biến TPPRO TP-20918 đang được sử dụng khá phổ biến tại nhiều trung tâm thương mại, siêu thị, nhà hàng, gia đình… nhằm thay thế cho những chiếc vòi nước truyền thống thông thường.
Sản phẩm được trang bị hệ thống cảm biến thông minh, có thể tự động ngắt sau 1 giây khi người xử dụng đưa tay ra khỏi vùng cảm ứng. Điều này sẽ giúp cho người sử dụng có thể tiết kiệm nước tối đa, đồng thời hạn chế nguy cơ lây nhiễm bệnh tật.
TPPRO TP-20918 có kiểu dáng đơn giản với phần vỏ làm bằng đồng tiêu chuẩn mạ chrome cứng cáp, có thể vệ sinh, lau chùi dễ dàng. Vòi sử dụng nguồn điện áp 220V/50Hz kết hợp với 4 pin AA alkaline cho tớy 100.000 lần sử dụng, giúp tiết kiệm chi phí đáng kể.
Trên đây là top 5 vòicảm ứng giá rẻ cho chậu rửa lavabo. Nếu có nhu cầu trang bị những chiếc vòi nước cảm ứng đặc biệt này, bạn vui lòng tham khảo & chọn mua tại trang web TPPRO.VN
## Mỗi món ăn Thượng Hải đều mang trong mình những nét đặc trưng quyến rũ vô cùng khác lạ đối với thực khách.
Thượng Hải là một trong những đô thị phát triển nhất ở Trung Quốc và là trung tâm kinh tế, tài chính, thương mại của quốc gia này. Nơi đây có khí hậu dễ chịu, sản vật phong phú, đây cũng là thành phố nổi tiếng với những món ăn ngon bậc nhất Trung Quốc. Sau đây là 7 món ăn nổi tiếng của Thượng Hải nhất định phải nếm thử khi tới nơi đây.
Vịt hầm bát bảo
Vịt hầm bát bảo là một món ăn truyền thống đặc biệt ở Thượng Hải. Để làm được món ăn này, người đầu bếp phải loại bỏ toàn bộ nội tạng trong khi vẫn đảm bảo được hình dáng và bộ da vịt nguyên vẹn. Sau đó, người ta sẽ nhồi vào bụng vịt 8 loại nguyên liệu tạo nên sự đặc biệt của món vịt hầm bát bảo bao gồm: măng non, sò điệp khô, gạo nếp, thịt lợn băm, nấm Quan thoại, hạt trộn, hạt dẻ và mề vịt.
Sau khi nhồi 8 nguyện liệu đặc biệt vào bụng vịt, người đầu bếp sẽ khâu kín bụng rồi đem đi chiên giòn. Tiếp đó thịt vịt sẽ được đem hầm trong nước dùng đặc biệt gồm xương vịt, măng và gừng trong khoảng 90 phút cho đến khi thịt vịt mềm và ngấm gia vị. Món vịt hầm bát bảo sau khi hoàn thành có màu hồng hào, hương thơm đậm đà, mùi vị vô cùng đặc biệt.
Tôm pha lê
Tôm pha lê là món ăn được đánh giá là “Món ngon số 1 ở Thượng Hải”. Món ăn tưởng chừng như đơn giản này không cần có quá nhiều gia vị cầu kỳ nhưng lại đòi hỏi người đầu bếp phải có tay nghề cao, biết cách kiểm soát nhiệt độ để cho những con tôm vừa chín tới, mềm ngọt đúng độ, màu sắc tươi sáng.
Sở dĩ được gọi là tôm pha lê bởi món ăn sau khi hoàn thành sẽ tươi sáng, trong suốt như pha lê. Bề ngoài của chúng như được phủ lên một lớp thuỷ tinh phản chiếu ánh sáng lấp lánh, trong vắt cùng với đó là hương thơm ngập tràn, hương vị tươi ngon, tôm ngọt mềm, giòn như trân châu.
Gà trắng chặt Thượng Hải
Gà trắng hay gà trắng chặt là món đặc sản rất nổi tiếng ở Thượng Hải, vị béo, mềm, thơm ngon, da gà vàng ruộm, thịt trắng. Những con gà được dùng để chế biến gà trắng chặt Thượng Hải là giống gà trống nhỏ của địa phương. Những con gà sau khi được làm thịt sẽ được luộc với những gia vị đặc trưng để mang đến hương vị đậm đà, sau đó toàn bộ con gà được phết lên một lớp dầu mè để mang đến độ sáng bóng cho da gà. Thịt gà được chế biến theo cách này có màu trắng và thịt mềm và căng mọng. Đây là món ăn được nhiều người dân Thượng Hải ưa thích.
Thanh ngư ngốc phế (phổi cá trắm đen)
Đây là một món ăn truyền thống của người dân Thượng Hải nhưng thường rất khó để tìm thấy chúng trong thực đơn của các nhà hàng. Tuy tên món ăn là phổi cá nhưng thực chất nó lại được chế biến từ gan của cá trắm đen cùng với măng, đậu tằm và một số nguyên liệu khác. Chỉ cần cắn một miếng, thực khách có thể cảm nhận được miếng gan cá mềm, dẻo thơm, không hề có mùi tanh mà dư vị đậm đà. Tuy ngon nhưng món ăn này rất khó kiếm, bởi để chế biến ra được một đĩa Thanh ngư ngốc phế cần gan của ít nhất chục con cá trắm đen. Do đó đây là món ăn vừa đắt đỏ vừa khó tìm nhưng vô cùng gây thương nhớ của Thượng Hải.
Thịt kho Thượng Hải
Mặc dù món thịt kho có ở nhiều nơi, nhưng thịt kho kiểu Thượng Hải lại có một số đặc trưng rất riêng biệt. Món ăn này được làm từ thịt lợn ba chỉ, nước tương, rượu, đường và một số gia vị khác. Đặc điểm lớn nhất của ẩm thực Thượng Hải chính là nằm ở nước sốt đỏ đậm đà và món thịt kho Thượng Hải là đại diện tiêu biểu nhất cho đặc điểm này. Món ăn này có nước sốt đặc đặc sệt màu đỏ, thịt béo ngậy nhưng không ngấy, vị ngọt nhưng không gắt, đậm đà mà không mặn.
Cá cháy chưng dầu
Cá cháy chưng dầu là một món ăn truyền thống nổi tiếng của Thượng Hải. Ngoài hương vị thơm ngon, nó còn giúp bổ tỳ vị, bồi bổ cơ thể suy yếu, điều hoà dinh dưỡng. Thành phần chính của món ăn bao gồm cá cháy, mỡ lợn, giăm bông, nấm, măng, … Cá và mỡ lợn sẽ được hấp cùng những lát giăm bông và măng thái mỏng cùng các gia vị đặc biệt để hương vị hoà quyện với nhau, từ đó tạo nên một món ăn béo ngậy, thơm ngon được nhiều thực khách ưa chuộng.
Zaobotou
Zaobotou là một món ăn truyền thống đặc sắc của Thượng Hải, tương truyền nó được một người phụ nữ nông dân sáng tạo nên. Cách làm món ăn này khá cầu kỳ, những nguyên liệu bao gồm lưỡi lợn, phổi, tim, bao tử, ruột già lợn, chân giò, … sẽ được trần sơ qua, thái miếng nhỏ và cho vào một chiếc chén nhỏ bằng gốm (zaobotou có thể hiểu là món ăn được nấu trong chén gốm). Sau đó nước dùng được chế biến đặc biệt cùng các loại gia vị sẽ được cho thêm vào. Kế đó, miệng chén sẽ được bịt kín rồi đem đi hấp cách thuỷ trong khoảng nửa tiếng đồng hồ. Món ăn này có vị bùi, thơm, nước dùng trong vắt, mùi hương thoang thoảng và đặc biệt không hề có mùi hôi của nội tạng, thơm ngon vô cùng. Đây cũng là món ăn thường được phục vụ trong các nhà hàng lớn ở Thượng Hải. |
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## Algebra (all content)
### Course: Algebra (all content)>Unit 12
Lesson 5: Graphs of radical functions
# Transforming the square-root function
Sal shows various examples of functions and their graphs that are a result of shifting and/or flipping y=√x. Created by Sal Khan.
## Want to join the conversation?
• What if it 's like 2x+3? with that radical sign over it ? How would you graph that?
• If you have the function y = sqrt(2x+3), you can rewrite the right hand side of this as:
y = sqrt(2(x+3/2)).
Then, using the property that sqrt(ab) = sqrt(a)*sqrt(b), you can rewrite this again as:
y = sqrt(2) * sqrt(x+3/2).
Now, notice that sqrt(2) is no more than a constant, you all you've done is stretched the graph vertically byu a factor of sqrt(2). Then, notice that under the second radical sign, you've got a shift to the left by 3/2. To show how this process makes sense, try graphing both y = sqrt(2x+3) and y = sqrt(2) * sqrt(x+3/2). You should get the same thing.
To graph it, know what the graph of y = sqrt(x) looks like first (its a parabola on its side with only the top half). Then, notice that you've shifted the graph to the left by 3/2 and stretched the entire graph by sqrt(2). All done!
• Sal writes sqrt - (x + 3)
I can see the principle of reflection this radical function along line of x = -3. However will this not yield complex numbers? Thank you
• It will not yield imaginary numbers as long as "x" is chosen carefully. We can find exactly for which values of x no complex numbers result. We do this by finding the domain of the function:
ƒ(x) = √[-(x + 3)]
The radicand must be greater than or equal to 0 in order for the function to yield only real numbers:
-(x + 3) ≥ 0
-x - 3 ≥ 0
-x ≥ 3
x ≤ -3
Therefore, all values of "x" that produce real numbers in the aforementioned equation are all numbers lesser than or equal to -3. This is also why the graph of the function starts at -3 on the x-axis and keeps going in the negative direction, but does not go in the positive direction at all. Comment if you have questions.
• Why am I not getting any of this.
• lol so it's not just me
(1 vote)
• Another way to say why x is positive instead of negative, you could say that y=√x -> or x=y^2. Because you want to move x three spaces to the left, you would do x=y^2 -3. Then, to convert it back, it would become y^2=x+3, or y=√x+3. idk just another way to understand it I guess.
• Your logic is good up to here: x=y^2 -3
y^2 = x+3
Then take the square root of both sides (the entire side)
y = √(x+3)
You can't just do the square root of "x"
• You said you have to make sure that whatever is under the radical will equal 0 at the origin like if we want to shift 2 to left we +2 to make sure that the number under radical is zero,but i want to ask why is that necessary to make it 0.?
• Radicals are tricky things. To evaluate points on a radical function, you want to think about values for x that make the radicand (value under the radical) equal to a perfect square, like 0, but also 1, 4, 9, 16, etc.
• how do you compute √-(x+3) ? I thought you couldn't have negatives under the radical
• does anyone know a video or link that describes how to find the x intercept of a transformation of the function y = cube root of x^2?
(1 vote)
• I don't know of a video or link, but if you express the cube root of x^2 as x^(2/3) power, I'm sure you can find the x-intercepts by substituting 0 for y and solving for x with exponent rules.
• What would the graph of y = 3* sqrt(x) or y = sqrt(3x) or y = -3*sqrt(x), etc.?
(1 vote)
• y=3*sqrt(x) would be the a slightly steeper version of y=sqrt(x). Think of it this way: each value for sqrt(x) is multiplied by three, making it it much steeper.
y=-3*sqrt(x) would be the exact mirror along the y axis of y=3*sqrt(x).
Hope this helped.
• At , what is y when sqrt(x+3) is negative? I don't understand!
(1 vote)
• sqrt(x) means positive (principal) sqrt always. If someone wants the negative sqrt they must write -sqrt(x). This is a rule.
When -(x+3) is negative (that is, to the right of -3), y = sqrt(-(x+3)) is undefined. At -3, -(x+3) is 0. As x moves to the left of -3, -(x+3) increases from 0, and so its sqrt, y, increases.
(1 vote)
• When you graph a radical function how do you tell whether the x-value is negative or positive? I get that the y-value maintains it's negativity/positivity from the equation but the x seems to switch at random.
(1 vote)
• All the possible x values come from the domain of the function. It's not random.
(1 vote) |
2.9 The Problems Page
Chapter
Chapter 2
Section
2.9
11mathHarcourt2.9 12 Videos
Recall the problem from Chapter 1 in which you New Tab ed to move blue and green discs so that the positioning of the discs was reversed. Here is a variation of the game. What's different? Two things. First, we have three blue and three green discs, but also a one-disc space between them. Second, there are two allowable moves; you can move a disc one position if there is a blank space beside it, and you can jump one disc over another if there is an empty space to jump into. What is the minimum number of moves required to get the green discs to the three positions on the left and the blue discs to the three positions on the right? Can you give a general solution for \displaystyle k discs of each colour?
Q1
This is a set of questions about prime numbers. Your task is to answer the question posed and to draw conclusions where you can.
The numbers 2 and 3 differ by one. Can you find another such pair of primes?
Q2a
This is a set of questions about prime numbers. Your task is to answer the question posed and to draw conclusions where you can.
The numbers 3 and 5 differ by two. Can you find another such pair of primes? How many?
Q2b
This is a set of questions about prime numbers. Your task is to answer the question posed and to draw conclusions where you can. The numbers 3,5, and 7 are a triple set of primes differing by two. Can you find another such triple?
Q2c
This is a set of questions about prime numbers. Your task is to answer the question posed and to draw conclusions where you can. Can you find a prime that is one less than a perfect square? Can you find more than one such prime?
Q2d
This is a set of questions about prime numbers. Your task is to answer the question posed and to draw conclusions where you can. Can you find a prime that is one more than a perfect square? Can you find more than one such prime?
Q2e
When \displaystyle f(x) is divided by \displaystyle 3 x+1 , a quotient of \displaystyle 2 x-3 and a remainder of 5 are obtained. What is \displaystyle f(x) ?
Q3
Bus A and bus B leave the same terminal at 09:00. Bus A requires 45 min to complete its route while bus \displaystyle \mathrm{B} requires 54 min for its route. If the buses continue running, what is the next time the two leave the terminal together?
Q4
If two poles \displaystyle 10 \mathrm{~m} and \displaystyle 15 \mathrm{~m} high are \displaystyle 25 \mathrm{~m} apart, what is the height of the point of intersection of the lines that run from the top of each pole to the foot of the other pole?
We define a sequence of numbers by using the symbol \displaystyle t_{k} where \displaystyle k=1,2,3,4, \ldots This means that \displaystyle t_{1} represents the first number in the sequence, \displaystyle t_{2} the second, and so on. If \displaystyle t_{1}=1, t_{2}=2 , and every number thereafter is obtained from the formula \displaystyle t_{k+1}=\frac{t_{k}+1}{t_{k-1}}\left(\right. for example, \displaystyle t_{3}=\frac{t_{2}+1}{t_{1}}=\frac{2+1}{1}=3 ), what is \displaystyle t_{63} ? Can you determine the sum of the first 100 numbers in the sequence?
Show that for the set of numbers in Problem \displaystyle 6, t_{6}=t_{1} and \displaystyle t_{7}=t_{2} . |
# Lesson 8
Equations and Graphs
• Let’s write an equation for a parabola.
### 8.1: Focus on Distance
The image shows a parabola with focus $$(\text-2,2)$$ and directrix $$y=0$$ (the $$x$$-axis). Points $$A$$, $$B$$, and $$C$$ are on the parabola.
Without using the Pythagorean Theorem, find the distance from each plotted point to the parabola’s focus. Explain your reasoning.
### 8.2: Building an Equation for a Parabola
The image shows a parabola with focus $$(3,2)$$ and directrix $$y=0$$ (the $$x$$-axis).
1. Write an equation that would allow you to test whether a particular point $$(x,y)$$ is on the parabola.
2. The equation you wrote defines the parabola, but it’s not in a very easy-to-read form. Rewrite the equation to be in vertex form: $$y=a(x-h)^2+k$$, where $$(h,k)$$ is the vertex.
### 8.3: Card Sort: Parabolas
Your teacher will give you a set of cards with graphs and equations of parabolas. Match each graph with the equation that represents it.
In this section, you have examined points that are equidistant from a given point and a given line. Now consider a set of points that are half as far from a point as they are from a line.
1. Write an equation that describes the set of all points that are $$\frac{1}{2}$$ as far from the point $$(5,3)$$ as they are from the $$x$$-axis.
2. Use technology to graph your equation. Sketch the graph and describe what it looks like.
### Summary
The parabola in the image consists of all the points that are the same distance from the point $$(1,4)$$ as they are from the line $$y=0$$. Suppose we want to write an equation for the parabola—that is, an equation that says a given point $$(x,y)$$ is on the curve. We can draw a right triangle whose hypotenuse is the distance between point $$(x,y)$$ and the focus, $$(1,4)$$.
The distance from $$(x,y)$$ to the directrix, or the line $$y=0$$, is $$y$$ units. By definition, the distance from $$(x,y)$$ to the focus must be equal to the distance from the point to the directrix. So, the distance from $$(x,y)$$ to the focus can be labeled with $$y$$. To find the lengths of the legs of the right triangle, subtract the corresponding coordinates of the point $$(x,y)$$ and the focus, $$(1,4)$$. Substitute the expressions for the side lengths into the Pythagorean Theorem to get an equation defining the parabola.
$$(x-1)^2+(y-4)^2=y^2$$
To get the equation looking more familiar, rewrite it in vertex form, or $$y=a(x-h)^2+k$$ where $$(h,k)$$ is the vertex.
$$(x-1)^2+(y-4)^2=y^2$$
$$(x-1)^2+y^2-8y+16=y^2$$
$$(x-1)^2-8y+16=0$$
$$\text-8y=\text-(x-1)^2-16$$
$$y=\frac18 (x-1)^2+2$$
### Glossary Entries
• directrix
The line that, together with a point called the focus, defines a parabola, which is the set of points equidistant from the focus and directrix.
• focus
The point that, together with a line called the directrix, defines a parabola, which is the set of points equidistant from the focus and directrix.
• parabola
A parabola is the set of points that are equidistant from a given point, called the focus, and a given line, called the directrix. |
# 8.2 Angular acceleration (Page 3/6)
Page 3 / 6
So far, we have defined three rotational quantities— , and $\alpha$ . These quantities are analogous to the translational quantities , and $a$ . [link] displays rotational quantities, the analogous translational quantities, and the relationships between them.
Rotational and translational quantities
Rotational Translational Relationship
$\theta$ $x$ $\theta =\frac{x}{r}$
$\omega$ $v$ $\omega =\frac{v}{r}$
$\alpha$ $a$ $\alpha =\frac{{a}_{t}}{r}$
## Making connections: take-home experiment
Sit down with your feet on the ground on a chair that rotates. Lift one of your legs such that it is unbent (straightened out). Using the other leg, begin to rotate yourself by pushing on the ground. Stop using your leg to push the ground but allow the chair to rotate. From the origin where you began, sketch the angle, angular velocity, and angular acceleration of your leg as a function of time in the form of three separate graphs. Estimate the magnitudes of these quantities.
Angular acceleration is a vector, having both magnitude and direction. How do we denote its magnitude and direction? Illustrate with an example.
The magnitude of angular acceleration is $\alpha$ and its most common units are ${\text{rad/s}}^{2}$ . The direction of angular acceleration along a fixed axis is denoted by a + or a – sign, just as the direction of linear acceleration in one dimension is denoted by a + or a – sign. For example, consider a gymnast doing a forward flip. Her angular momentum would be parallel to the mat and to her left. The magnitude of her angular acceleration would be proportional to her angular velocity (spin rate) and her moment of inertia about her spin axis.
Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs.
## Section summary
• Uniform circular motion is the motion with a constant angular velocity $\omega =\frac{\Delta \theta }{\Delta t}$ .
• In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is $\alpha =\frac{\Delta \omega }{\Delta t}$ .
• Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as ${a}_{\text{t}}=\frac{\Delta v}{\Delta t}$ .
• For circular motion, note that $v=\mathrm{r\omega }$ , so that
${a}_{\mathrm{\text{t}}}=\frac{\text{Δ}\left(\mathrm{r\omega }\right)}{\Delta t}.$
• The radius r is constant for circular motion, and so $\mathrm{\text{Δ}}\left(\mathrm{r\omega }\right)=r\Delta \omega$ . Thus,
${a}_{\text{t}}=r\frac{\Delta \omega }{\Delta t}.$
• By definition, $\Delta \omega /\Delta t=\alpha$ . Thus,
${a}_{\text{t}}=\mathrm{r\alpha }$
or
$\alpha =\frac{{a}_{\text{t}}}{r}.$
## Conceptual questions
Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.
Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude.
In circular motion, a tangential acceleration can change the magnitude of the velocity but not its direction. Explain your answer.
Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, centripetal acceleration, or both when: (a) The plate starts to spin? (b) The plate rotates at constant angular velocity? (c) The plate slows to a halt?
## Problems&Exercises
At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?
$\omega =0\text{.}\text{737 rev/s}$
Integrated Concepts
An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is its angular acceleration in ${\text{rad/s}}^{2}$ ? (b) What is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) What is the radial acceleration in ${\text{m/s}}^{2}$ and multiples of $g$ of this point at full rpm?
Integrated Concepts
You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest?
(a) $-0\text{.}{\text{26 rad/s}}^{2}$
(b) $\text{27}\phantom{\rule{0.25em}{0ex}}\text{rev}$
Unreasonable Results
You are told that a basketball player spins the ball with an angular acceleration of . (a) What is the ball’s final angular velocity if the ball starts from rest and the acceleration lasts 2.00 s? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
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MANUFACTURING
# A math moment: growth rates
Staff Writer
Telegram & Gazette
The bad news: Studying companies well involves a little math. The good news: It’s not that hard.
Here’s a short lesson on calculating growth rates. Imagine Meteorite Insurance (ticker: HEDSUP), with sales of \$12 million in 2007 and \$48 million in 2010. If you have a slightly fancy calculator, it might have a feature that calculates growth rates for you. If not, here’s what to do:
Divide \$48 million by \$12 million and you’ll get 4. This means that sales quadrupled, or increased by a growth multiple of 4. That doesn’t translate to a gain of 400 percent, though. (After all, if something doubles, it increases by 100 percent, not 200 percent.) To get the percentage, you need to take the growth multiple, subtract 1, multiply by 100, and then tack on a percentage sign. So 4 minus 1 equals 3. And 3 times 100 is — bingo! — 300 percent. Between 2007 and 2010, sales grew by a total of 300 percent.
Another way to approach it is to take the \$48 million and subtract the \$12 million, getting \$36 million, which represents the growth. Divide \$36 million by \$12 million and you’ll get 3. Multiply that by 100 and you’ve got 300 percent. Same answer.
Next, you can annualize the growth rate, to see how much Meteorite sales are growing by each year, on average. You first need the time period involved. From 2007 to 2010 is three years, so we’ll be taking the third, or cube, root of the growth multiple. To do that, we’ll raise it to the 1/3 power.
You’ll need either a computer with a spreadsheet program or a calculator that can raise numbers to various powers. Raise the growth multiple of 4 to the 1/3 power, and you’ll get 1.59. (4 to the power of 1/3 equals 1.59. The calculator symbol is ^.) Now subtract 1, multiply by 100, and you’ve got 59 percent as the average annual growth rate. (If the time period had been 7.5 years, you’d raise the multiple to the 1/7.5 power.)
This math can prove profitable — after a little practice. |
2002 AMC 12A Problems/Problem 23
Problem
In triangle $ABC$, side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and $BD$ bisects $\angle ABC$. If $AD=9$ and $DC=7$, what is the area of triangle $ABD$?
$\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5$
Solution 1
$[asy] unitsize(0.25 cm); pair A, B, C, D, M; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; draw(A--B--C--cycle); draw(B--D--M); label("A", A, SW); label("B", B, N); label("C", C, SE); label("D", D, S); [/asy]$ Looking at the triangle $BCD$, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let $x = \angle C$, so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x, \angle ADB=2x$ because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means $\triangle ABD$ and $\triangle ACB$ are similar, so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$.
Then by using Heron's Formula on $ABD$ (with sides $12,7,9$), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$.
Solution 2
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, $BD = DC = 7$ and $BM = MC$. Also, by the angle bisector theorem, $\frac {AB}{BC} = \frac{9}{7}$. Thus, let $AB = 9x$ and $BC = 7x$. In addition, $BM = 3.5x$.
Thus, $\cos\angle CBD = \frac {3.5x}{7} = \frac {x}{2}$. Additionally, using the Law of Cosines and the fact that $\angle CBD = \angle ABD$, $81 = 49 + 81x^2 - 2(9x)(7)\cos\angle CBD$
Substituting and simplifying, we get $x = 4/3$
Thus, $AB = 12$. We now know all sides of $\triangle ABD$. Using Heron's Formula on $\triangle ABD$, $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$
"Note:-you could also drop a perpendicular from D to AB at point let say,F then BF = 3.5x by pyathgoras theorem we can find DF and (AB ×DF )÷2 is our answer"
Solution 3
Note that because the perpendicular bisector and angle bisector meet at side $AC$ and $CD = BD$ as triangle $BDC$ is isosceles, so $BD = 7$. By the angle bisector theorem, we can express $AB$ and $BC$ as $9x$ and $7x$ respectively. We try to find $x$ through Stewart's Theorem. So
$16(7^2+9\cdot7) = (7x)^2 \cdot 9 + (9x)^2 \cdot 7$
$16(49+63) = (49 \cdot 9 + 81 \cdot 7)x^2$
$16(49+63) = 9(49+63) \cdot x^2$
$x^2 = \frac{16}{9}$
$x=\frac{4}{3}$
We plug this to find that the sides of $\triangle ABD$ are $12,7,9$. By Heron's formula, the area is $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$. ~skyscraper
Solution 4
$[asy] size(12cm, 12cm); pair A, B, C, D, M, N; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; N = (6,4.27); draw(A--B--C--cycle); draw(B--D--M); draw(D--N--B); label("A", A, SW); label("B", B, N); label("C", C, SE); label("D", D, S); label("M", M, NE); label("N", N, NW); draw(rightanglemark(B, M, D), linewidth(.5)); draw(rightanglemark(A, N, D), linewidth(.5)); [/asy]$
Draw $DN$ such that $DN \bot AB$, $\triangle BND \cong \triangle BMD$
$\angle ACB = \angle DBC = \angle ABD$, $\triangle ABD \sim \triangle ACB$ by $AA$
$\frac{AB}{AD} = \frac{AC}{AB}$, $AB^2 = AD \cdot AC = 9 \cdot 16$, $AB = 12$
By the Angle Bisector Theorem, $\frac{BC}{AB} = \frac{CD}{AD}$
$\frac{BC}{12} = \frac{7}{9}$
$BC = \frac{28}{3}$, $CM = \frac{14}{3}$, $DN = DM = \sqrt{CD^2 - CM^2} = \frac{7 \sqrt{5} }{3}$
$[ABD] = \frac12 \cdot AB \cdot DN = \frac12 \cdot 12 \cdot \frac{7 \sqrt{5} }{3} = \boxed{\textbf{(D) } 14 \sqrt{5} }$
Solution 5 (Trigonometry)
Let $\angle ACB = \theta$, $\angle DBC = \theta$, $\angle ABD = \theta$, $\angle ADB = 2 \theta$, $\angle BAC = 180^\circ - 3 \theta$
$[ABD] = \frac12 \cdot AD \cdot BD \cdot \sin \angle ADB = \frac12 \cdot 9 \cdot 7 \cdot \sin 2\theta$
By the Law of Sines we have $\frac{ \sin \angle BAC }{BD} = \frac{ \sin \angle ABD }{AD}$
$\frac{ \sin(180^\circ - 3 \theta) }{7} = \frac{ \sin \theta }{ 9 }$
$\frac{ \sin 3 \theta }{7} = \frac{ \sin \theta }{ 9 }$
By the Triple-angle Identities, $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$
$3 - 4 \sin^2 \theta = \frac79$, $36 \sin^2 \theta = 20$, $\sin^2 \theta = \frac59$
$\sin \theta = \frac{\sqrt{5}}{3}$, $\cos \theta = \frac{2}{3}$
By the Double Angle Identity $\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{\sqrt{5}}{3} \cdot \frac{2}{3} = \frac{ 4 \sqrt{5} }{9}$
$[ABD] = \frac12 \cdot 9 \cdot 7 \cdot \frac{ 4 \sqrt{5} }{9} = \boxed{\textbf{(D) } 14 \sqrt{5} }$ |
# Place Values in Words and Numbers
Here we will learn about the place values in words and numbers. We have seen that a digit written in a particular place has its value according to the place.
1 written in one’s place stand for 1
1 written in ten’s place stand for 10 (1 ten)
1 written in hundred’s place stand for 100 (1 hundred)
2 written in one’s place stand for 2
2 written in ten’s place stand for 20 (2 tens)
2 written in hundred’s place stand for 200 (2 hundreds)
3 written in one’s place stand for 3
3 written in ten’s place stand for 30 (3 tens)
3 written in hundred’s place stand for 300 (3 hundreds)
4 written in one’s place stand for 4
4 written in ten’s place stand for 40 (4 tens)
4 written in hundred’s place stand for 400 (4 hundreds)
5 written in one’s place stand for 5
5 written in ten’s place stand for 50 (5 tens)
5 written in hundred’s place stand for 500 (5 hundreds)
6 written in one’s place stand for 6
6 written in ten’s place stand for 60 (6 tens)
6 written in hundred’s place stand for 600 (6 hundreds)
7 written in one’s place stand for 7
7 written in ten’s place stand for 70 (7 tens)
7 written in hundred’s place stand for 700 (7 hundreds)
8 written in one’s place stand for 8
8 written in ten’s place stand for 80 (8 tens)
8 written in hundred’s place stand for 800 (8 hundreds)
9 written in one’s place stand for 9
9 written in ten’s place stand for 90 (9 tens)
9 written in hundred’s place stand for 900 (9 hundreds)
Now to write the place values in words and numbers, such as 45, we see the number has two digits. Each digit is a different place value.
The first digit i.e. 4 is called the tens' place. Here, there are 4 tens.
The last or right digit i.e. 5 is called the ones' place.
Therefore, in the number 45 = 4 tens plus 5 ones = 40 + 5 = 45.
Questions and Answers on Place Values in Words and Numbers:
1. Count and write the number of hundreds, tens and ones. One has done for you.
(i) 2 hundreds20 tens200 ones (ii) ___ hundreds___ tens___ ones (iii) ___ hundreds___ tens___ ones (iv) ___ hundreds___ tens___ ones (v) ___ hundreds___ tens___ ones (vi) ___ hundreds___ tens___ ones (vii) ___ hundreds___ tens___ ones
1. (ii) 3 hundreds
30 tens
300 ones
(iii) 4 hundreds
40 tens
400 ones
(iv) 5 hundreds
50 tens
500 ones
(v) 6 hundreds
60 tens
600 ones
(vi) 7 hundreds
70 tens
700 ones
(vii) 8 hundreds
80 tens
800 ones
II. Count and write the number of hundreds, tens and ones. Then, write the number. One has done for you.
(i) H T O3 3 6 336 (ii) H T O__ __ __ ___ (iii) H T O__ __ __ ___ (iv) H T O__ __ __ ___ (v) H T O__ __ __ ___ (vi) H T O__ __ __ ___
II.
(ii) H T O 2 4 2 242 (iii) H T O 4 4 2 442 (iv) H T O 5 6 1 561 (v) H T O 7 1 3 713 (vi) H T O 8 8 8 888
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# 3.OA.3 MULTIPLICATION AND DIVISION STRATEGIES ARRAYS AND DRAWINGS.
## Presentation on theme: "3.OA.3 MULTIPLICATION AND DIVISION STRATEGIES ARRAYS AND DRAWINGS."— Presentation transcript:
3.OA.3 MULTIPLICATION AND DIVISION STRATEGIES ARRAYS AND DRAWINGS
GROUP SIZE UNKNOWN - ARRAY If 18 apples are arranged into 3 equal groups. How many apples will be in each group?
NUMBER OF GROUPS UNKNOWN - ARRAY I have 28 pieces of gum for my friends. Each friend gets 4 pieces. How many friends do I have?
UNKNOWN PRODUCT - ARRAY What multiplication problem is represented by the drawing below?
PRODUCT UNKNOWN– AREA MODEL ARRAY I have 18 packs of markers. Each pack has 4 markers in it. How many markers do I have in total? 10 + 8 4 72 total markers 4032
TRUE OR FALSE My friend multiplied these two numbers. Below is their reasoning. Are they correct? If not, what advice do you have for this student? 18 x 6 = 116 10 + 8 6 6056
WRITE A WORD PROBLEM REPRESENTED BY THE DRAWING
DIVISION - AREA MODEL ARRAY GROUP SIZE UNKNOWN If you have 48 inches of ribbon cut into 6 pieces, how long will each piece measure? 48 inches of Ribbon 6 pieces 3 X 6 = 18 5 X 6 = 30
DIVISION – ARRAY NUMBER OF GROUPS UNKNOWN If 56 pieces of fruit are to be packed 4 to a bag. How many bags will be needed? Use a drawing or array to solve. Be prepared to share your answer and explain.
DIVISION – ARRAY NUMBER OF GROUPS UNKNOWN Same Problem - If 56 pieces of fruit are to be packed 4 to a bag. How many bags will be needed? Can you explain this drawing? Is it correct? Is there anything missing?
DIVISION – ARRAY NUMBER OF GROUPS UNKNOWN Same Problem - If 56 pieces of fruit are to be packed 4 to a bag. How many bags will be needed? Can you explain this array? Is it correct? Is there anything missing?
GIVEN THE ARRAY WRITE A WORD PROBLEM 72 buttons 8 buttons
GIVEN THE ARRAY WRITE A WORD PROBLEM How many shirts can I make if a shirt requires 8 buttons and I have a total of 72 buttons? 9 shirts 72 buttons 4 x 8 buttons 5 x 8 buttons 8 buttons 40 buttons 32 buttons
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# Worksheet on 18 Times Table | Printable 18 Times Table Worksheet with Answers
Worksheet on Multiplication Table of 18 is provided here. Check out the important questions involved in the 18 Times Table Multiplication Chart. Have a look at the various methods, rules, formulas that helps to solve the questions of the 18 Times Table. After reading this page, you will be able to understand and importance of the 18 tables along with their applications. Solved example questions on the Worksheet on 18 Times Table will help you to get a piece of detailed information and also helps you to score good marks in the exam.
### 18 Times Multiplication Table Chart up to 25
Check out the multiplication table of 18 and remember the output to make your math-solving problems easy.
18 x 0 = 0 18 x 1 = 18 18 x 2 = 36 18 x 3 = 54 18 x 4 = 72 18 x 5 = 90 18 x 6 = 108 18 x 7 = 126 18 x 8 = 144 18 x 9 = 162 18 x 10 = 180 18 x 11 = 198 18 x 12 = 216 18 x 13 = 234 18 x 14 = 252 18 x 15 = 270 18 x 16 = 288 18 x 17 = 306 18 x 18 = 324 18 x 19 = 342 18 x 20 = 360 18 x 21 = 378 18 x 22 = 396 18 x 23 = 414 18 x 24 = 432 18 x 25 = 450
Problem 1:
Find each product using the multiplication table of 18
(i) 16 x 18
(ii) 21 x 18
(iii) 8 x 18
(iv) 5 x 18
(v) 9 x 18
Solution:
(i) The given two numbers are 16, 18
16 x 18 = 288
The product of 16 and 18 is 288.
(ii) The given two numbers are 21, 18
21 x 18 = 378
The product of 21 and 18 is 378.
(iii) The given two numbers are 8, 18
8 x 18 = 144
The product of 8 and 18 is 144.
(iv) The given two numbers are 5, 18
5 x 18 = 90
The product of 5 and 18 is 90.
(v) The given two numbers are 9, 18
9 x 18 = 162
The product of 9 and 18 is 162.
Problem 2:
Name the number just after:
(i) 7 x 18
(ii) 18 x 18
(iii) 15 x 18
(iv) 10 x 18
Solution:
(i) The given numbers are 7, 18
7 x 18 = 126
The number just after is 126 + 1 = 127.
(ii) The given numbers are 18, 18
18 x 18 = 324
The number just after is 324 + 1 = 325.
(iii) The given numbers are 15, 18
15 x 18 = 270
The number just after is 270 + 1 = 271.
(iv) The given numbers are 10, 18
10 x 18 = 180
The number just after is 180 + 1 = 181.
Problem 3:
Name the number just before:
(i) 18 x 12
(ii) 18 x 24
(iii) 18 x 17
(iv) 18 x 4
Solution:
(i) The given numbers are 18, 12
18 x 12 = 216
The number just before 216 is 216 – 1 = 215.
(ii) The given numbers are 18, 24
18 x 24 = 432
The number just before 432 is 432 – 1 = 431.
(iii) The given numbers are 18, 17
18 x 17 = 306
The number just before 306 is 306 – 1 = 305.
(iv) The given numbers are 18, 4
18 x 4 = 72
he number just before 72 is 72 – 1 = 71.
Problem 4:
The cost of a cricket bat is $18. How much will 50 such cricket bats cost? Solution: Given that, The cost of one cricket bat =$18
The cost of 50 such cricket bats is obtained by multiplying one cricket bat by 50.
So, the cost of 50 cricket bats = 18 x 50 = 900
Therefore, the cost of 50 cricket bats is $900. Problem 5: A doll costs$15. How much will 18 such dolls cost?
Solution:
Given that,
The cost of a doll = $15 The cost for 18 dolls = 18 x 15 = 270 Therefore,$270 is the cost for 18 dolls.
Problem 6:
Complete the chart by multiplying the numbers by 18.
(i) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 (ii) 52, 85, 96, 152, 638, 452
Solution:
The 18 table multiplication chart is along the lines.
(i) 18 x 1 = 18
18 x 2 = 36
18 x 3 = 54
18 x 4 = 72
18 x 5 = 90
18 x 6 = 108
18 x 7 = 126
18 x 8 = 144
18 x 9 = 162
18 x 10 = 180
(ii) 18 x 52 = 936
18 x 85 = 1530
18 x 96 = 1728
18 x 152 = 2736
18 x 638 = 11,484
16 x 452 = 8136
Problem 7:
(i) What are 18 elevens?
(ii) 18 times 9?
(iii) 12 eighteen’s?
(iv) Eighteen times 9?
(v) What are 6 times 18?
(vi) Eighteen’s 4?
(vii) 18 times 11?
(viii) Eighteen’s 7?
(ix) What is 18 multiplied by 8?
(x) 5 multiplied by 18?
Solution:
(i) 18 elevens mean the product of 18 and 11
18 x 11 = 198
So, 18 elevens is 198.
(ii) 18 times 9 means the product of 18 and 9
18 x 9 = 162
So, 18 times 9 is 162.
(iii) 12 eighteen’s means the product of 12 and 18
12 x 18 = 216
So, 12 eighteen’s is 216.
(iv) Eighteen times 9 means the product of 18 and 9
18 x 9 = 162
So, Eighteen times 9 is 162.
(v) 6 times 18 means the product of
6 x 18 = 108
So, 6 times 18 is 108
(vi) Eighteen’s 4 means the product of 18 and 4
18 x 4 = 72
So, Eighteen’s 4 is 72.
(vii) 18 times 11 means the product of 18 and 11
18 x 11 = 198
So, 18 times 11 is 198.
(viii) Eighteen’s 7 means the product of 18 and 7
18 x 7 = 126
So, eighteen’s 7 is 126.
(ix) 18 multiplied by 8 means the product of 18 and 8
18 x 8 = 144
So, 18 multiplied by 8 is 144.
(x) 5 multiplied by 18 means the product of 5 and 18
5 x 18 = 90
So, 5 multiplied by 18 is 90.
Problem 8:
The cost of a pocket Radio set is 18 dollars. What will be the cost of 7 such Radio sets?
Solution:
Given that,
The cost of a pocket Radio set = 18 dollars
The cost of 7 such Radio sets = The cost of one pocket radio set x 7
= 18 x 7
= 126
Therefore, the cost of 7 such Radio sets is 126 dollars.
Problem 9:
What does 18 × 82 mean? What number is it equal to?
Solution:
18 x 82 means the product of 18 and 82
18 x 82 = 1476
So, 18 x 82 is equal to 1476 and 18 x 82 means 1476.
Problem 10:
Solve the following using the 18 times table.
(i) How many eighteen’s in 198?
(ii) 18 times 6 minus 4
(iii) How many eighteen’s in 972?
(iv) How many eighteen’s in 1728?
(v) 18 times 16 minus 45
(vi) How many eighteen’s in 1134?
(vii) 18 times 2 plus 6
(viii) How many eighteen’s in 108?
(ix) How many eighteen’s in 1170?
(x) How many eighteen’s in 1332?
Solution:
(i) How many eighteen’s in 198 means divide 198 by 18.
198 ÷ 18 = 11
There are 11 eighteen’s in 198.
(ii) 18 times 6 minus 4
18 times 6 means the product of 18 and 6
18 x 6 = 108
108 – 4 = 104
18 times 6 minus 4 is 104.
(iii) How many eighteen’s in 972 means divide 972 by 18
972 ÷ 18 = 54
So, there are 54 eighteen’s in 972.
(iv) How many eighteen’s in 1728 means divide 1728 by 18
1728 ÷ 18 = 96
So, there are 96 eighteen’s in 1728.
(v) 18 times 16 minues 45
18 times 16 means the product of 18 and 16
18 x 16 = 288
18 times 16 minues 45 = 288 + 45 = 333
So, 18 times 16 minues 45 is 333.
(vi) How many eighteen’s in 1134 means divide 1134 by 18
1134 ÷ 18 = 63
So, there are 63 eighteen’s in 1134
(vii) 18 times 2 plus 6
18 times 2 means the product of 18 and 2
18 x 2 = 36
18 times 2 plus 6 = 36 + 6 = 42
So, 18 times 2 plus 6 is 42
(viii) How many eighteen’s in 108 means divide 108 by 18
108 ÷ 18 = 6
So, there are 6 eighteen’s in 108.
(ix) How many eighteen’s in 1170 means divide 1170 by 18
1170 ÷ 18 = 65
So, there are 65 eighteen’s in 1170
(x) How many eighteen’s in 1332 means divide1332 by 18
1332 ÷ 18 = 74
So, there are 74 eighteen’s in 1332
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Say we have a hungry caterpillar that wants to eat a leaf off of a tree. There are ten trees it can choose from. So, as far as we know, the probability of it choosing a specific tree is ${1 \over 10}=10%=0.10$. Once it chooses a tree, and that tree has 100 branches, then the probability of choosing a specific branch is ${1 \over 100}=1%=0.01$. Finally, once it chooses a branch, and that branch has 25 leaves, then the probability of it choosing a specific leaf is ${1 \over 25}=4%=0.04$. So, before it has chosen a tree, the probability of it choosing a specific leaf is: $0.10×0.01×0.04=0.00004=0.004%={1 \over 25000}$ This makes sense because there are $10×100×25=25000$ leaves total to choose from.
When you multiply probabilities together, you will end up with a smaller probability unless one probability is $1.00=100%$. For instance, let’s say all the probabilities are $0.90=90%$ and that there are ten probabilities. Then, the resulting probability is:
$$0.90×0.90×0.90×0.90×0.90×0.90×0.90×0.90×0.90×0.90=0.90^{10}≈0.3486784401≈34.86%$$
Also, probabilities are typically only useful if you don’t have all the information. If you had all the information, the probability would either be $0%$ or $100%$. It is typically impossible to know all the information – especially if that information only exists in the future. Additional information can also change what the probability is. Take our caterpillar. Let’s say he’s a maple-loving specie of caterpillar, and half our trees are oak, and half our trees are maple. Now his probability of choosing a specific oak leaf is $0%$, while his probability of choosing a specific maple leaf is: $0.20×0.01×0.04=0.00008=0.008%={1 \over 12500}$ Also, if we don’t know what type of tree he likes, but somehow we determine there is a $50%$ of him being eaten by a bird before he gets to eat a leaf, then the probability of him eating a specific leaf is: $0.50 ×0.10×0.01×0.04=0.00002=0.002%={1 \over 50000}$ Also – probability isn’t really like something might or might not happen. Just because we don’t know what kind of tree he prefers, doesn’t mean he doesn’t have a preference. Just because we don’t know a bird is going to eat him, doesn’t mean the bird won’t.
Something else that might affect probability is scope. With all the previous caterpillar examples, I always said “a specific leaf.” Before the bird came into the equation, the probability that he was going to eat a leaf, any leaf, was $100%$. After the bird, it was $50%$. |
# Lesson 14
Solving Systems by Elimination (Part 1)
### Lesson Narrative
This is the first of three lessons that develop the idea of solving systems of linear equations in two variables by elimination.
Students warm up to the idea of adding equations visually. They examine a diagram of three hangers where the third hanger contains the combined contents of the first two hangers and all three hangers are balanced. Then, they analyze the result of adding two linear equations in standard form and notice that doing so eliminates one of the variables, enabling them to solve for the other variable and, consequently, to solve the system. In studying and testing a new strategy of adding equations and then offering their analyses, students construct viable arguments and critique the reasoning of others (MP3).
Next, students connect the solution they found using this method to the graphs of the equations in a system and the graph of the third equation (that results from adding or subtracting the original equations). They observe that the solution they found is the solution to the system, and that the graph of the third equation intersects the other two graphs at the exact same point—at the intersection of the first two.
The foundational idea is that adding or subtracting equations in a system creates a new equation whose solutions coincide with those of the original system. Students begin using this insight to solve systems, but they are not yet expected to construct an argument as to why this approach works.
### Learning Goals
Teacher Facing
• Recognize that adding or subtracting equations in a system creates a new equation with a solution that coincides with that of the original system, so the new equation can be used to solve the original system.
• Solve systems of equations by adding or subtracting the equations strategically to eliminate a variable.
• Use graphing technology to graph the sums and differences of the equations in a system, and analyze and describe (orally and in writing) the behaviors of the graphs.
### Student Facing
• Let’s investigate how adding or subtracting equations can help us solve systems of linear equations.
### Required Preparation
Acquire devices that can run Desmos (recommended) or other graphing technology. It is ideal if each student has their own device. (If students typically access the digital version of the materials, Desmos is always available under Math Tools.)
### Student Facing
• I can solve systems of equations by adding or subtracting them to eliminate a variable.
• I know that adding or subtracting equations in a system creates a new equation, where one of the solutions to this equation is the solution to the system.
Building Towards
### Glossary Entries
• elimination
A method of solving a system of two equations in two variables where you add or subtract a multiple of one equation to another in order to get an equation with only one of the variables (thus eliminating the other variable). |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 3.23: Stem-and-Leaf Plots, Mean, Median and Mode
Difficulty Level: At Grade Created by: CK-12
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Practice Stem and Leaf Plots, Mean, Median, and Mode
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Do you remember calculating the mean, median and mode of a data set? You can also use a stem - and - leaf plot to find these measures. Here is Julie's stem - and - leaf plot once again.
This Concept will show you how to use Julie's stem - and - leaf plot to find the mean, median and mode of the data set.
### Guidance
Remember back to when you learned about data?
We worked with data sets and found the mean, median and mode of each set of data.
The mean is the average of a set of data.
The median is the middle number of a set of data.
The mode is the number that occurs the most in a set of data.
We can use a stem-and-leaf plot to find the mean, median and mode of a set of data.
Here we have a data set with numbers that range from 35 to 59. The largest interval is from 55 to 59. The smallest interval is from 35 to 38.
What is the mean for this set of data?
To find the mean, we add up all of the numbers in the set and divide by the number of values that we added.
35 + 36 + 37 + 38 + 40 + 40 + 41 + 42 + 43 + 55 + 55 + 55 + 56 + 57 + 58 + 59 = 747
We divide by the number of values, which is 16.
74716=46.68\begin{align*}\frac{747}{16} = 46.68\end{align*}
After rounding, our answer is 47.
What is the median for this set of data?
Well, remember that the median is the middle score. We just wrote all of the scores in order from the smallest to the greatest. We can find the middle score by counting to the middle two scores.
42 + 43 These are the two middle scores.
We can find the mean of these two scores and that will give us the median.
42 + 43 = 42.5
The median score is 42.5 for this data set.
What is the mode for this data set?
The mode is the value that appears the most.
In this set of data, 55 is the number that appears the most.
The mode is 55 for this data set.
Use this stem - and - leaf plot to answer the following questions.
#### Example A
What is the mean?
Solution: 30
#### Example B
What is the median?
Solution: 28.5 or round up to 29
#### Example C
What is the mode?
Solution: There isn't a mode for this data set.
Now back to Julie.
Here is Julie's stem - and - leaf plot once again.
What is the mean?
What is the median?
What is the mode?
To find the mean we add up all of the values in the data set and divide by the number of values in the set.
52+67+70+75+78+78+86=506\begin{align*}52 + 67 + 70 + 75 + 78 + 78 + 86 = 506\end{align*}
506÷7=72.2\begin{align*}506 \div 7 = 72.2\end{align*}
The mean is 72.
To find the median, we look for the middle score.
The median value is 75.
To find the mode, we look for the value that appears the most.
The mode of this set is 78.
### Vocabulary
Here are the vocabulary words in this Concept.
Stem-and-leaf plot
a way of organizing numbers in a data set from least to greatest using place value to organize.
Data
information that has been collected to represent real life information
Mean
the average of a data set.
Median
the middle value in a data set.
Mode
the value that occurs the most in a data set.
### Guided Practice
Here is one for you to try on your own.
What is the mean, median and mode of this data set?
Weight of Trout Caught
Stem Leaf
2 9
3 1
4 0 5
5 2
6 2
7 6
8 3
9 2 2
To find the mean, we add up the weights and divide by the total number of weights.
29+31+40+45+52+62+76+83+92+92=602\begin{align*}29 + 31 + 40 + 45 + 52 + 62 + 76 + 83 + 92 + 92 = 602\end{align*}
602÷10=60.2\begin{align*}602 \div 10 = 60.2\end{align*}
The mean is 60.
To find the median, we look for the middle value of the data set.
The median is between 52 and 62. So we find the average of those two numbers.
52+62=114÷2=57\begin{align*}52 + 62 = 114 \div 2 = 57\end{align*}
The median value is 57.
To find the mode, we look for the value that appears the most.
The mode of the data set is 92.
### Video Review
Here are videos for review.
Great video on organizing, building and interpreting a stem and leaf plot.
### Practice
1. Define the mean.
2. Define the median.
3. Define the mode.
Directions: Use the stem and leaf plot to answer the following questions.
Stem Leaf
6 8
7 5 7 9
8 0 2
9 2 6 6 7
4. What is the lowest value in the plot?
5. What is the greatest value in the plot?
6. Which interval has the most values in it?
7. What is the mean of the data set?
8. What is the median of the data set?
9. What is the mode of the data set?
10. What is the range of the data set?
Stem Leaf
0 8
1 2 7 8 9
2 2 3
3 1 5
4 0
11. What is the smallest value in the plot?
12. What is the greatest value in the plot?
13. What is the mean of the data set?
14. What is the median of the data set?
15. What is the mode of the data set?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Data Data is information that has been collected to represent real life situations, usually in number form.
Mean The mean of a data set is the average of the data set. The mean is found by calculating the sum of the values in the data set and then dividing by the number of values in the data set.
Median The median of a data set is the middle value of an organized data set.
Mode The mode of a data set is the value or values with greatest frequency in the data set.
Stem-and-leaf plot A stem-and-leaf plot is a way of organizing data values from least to greatest using place value. Usually, the last digit of each data value becomes the "leaf" and the other digits become the "stem".
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# How to write the slope-intercept form of the equation of a line described (– 2, – 1) parallel to $y=\dfrac{-3}{2}x-1?$
Last updated date: 26th Feb 2024
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Hint: We are asked to find the slope-intercept form of the line which is parallel to $y=\dfrac{-3}{2}x-1$ and it also passes by (– 2, – 1). To solve this we will first understand what slope-intercept form of the line is and then how we can use the line parallel to the given line to solve our problem. We will use the slope-intercept form of a line as y = mx + c where m is the slope and c is the intercept.
We are asked to find the slope-intercept form of a line that is parallel to $y=\dfrac{-3}{2}x-1$ and is passing through (– 2, – 1). Before we find the slope solution, we will see what slope-intercept is. The slope-intercept form of a line means the equation of the line which is formed using the slope and the intercept. It is given as y = mx + c where m is the slope and c is the intercept.
Now, remember we know that when two lines are parallel to each other then it means their slopes are always equal. So, it means the slope of our required line is the same as the slope of the line $y=\dfrac{-3}{2}x-1.$
Now, we will reduce the line in the slope-intercept form to get the slope. If we look closely at $y=\dfrac{-3}{2}x-1$ we can see it is already in slope-intercept form. So, comparing it with y = mx + c, we get, $m=\dfrac{-3}{2}$ and hence it means its slope is $\dfrac{-3}{2}.$ So, the slope of our equation is $\dfrac{-3}{2}.$ Now our equation becomes $y=\dfrac{-3}{2}x+c.$ Now as our line passes by (– 2, – 1), so we put x = – 2 and y = – 1 and solve for c. So, putting the above values, we get,
$\Rightarrow -1=\dfrac{-3}{2}\times \left( -2 \right)+c$
On simplifying, we get,
$\Rightarrow -1=3+c$
Subtracting 3 on both the sides, we get,
$\Rightarrow c=-1-3$
$\Rightarrow c=-4$
So, c = – 4.
Hence, we get our equation as $y=\dfrac{-3}{2}x-4.$
Note: While we solve the equation we should be careful on solving the term that includes brackets. When we have the term inside the bracket and we have to multiply that with some term, so error happens like a (b + c) = ab + c, we may miss calculating the product with each term. Also, y – intercept is defined as the point on the y – axis where the given line cuts the y – axis. If y – intercept is 4, it means the line cuts y – axis at 4 and x – intercept means the point where the graph cut at x – axis. |
## The Number Zero: Activities for Teaching Preschoolers
written by: Sharon Dominica • edited by: Elizabeth Wistrom • updated: 12/29/2014
Struggling to teach the number zero? The preschool activities offered here will help children learn the concept in a fun way!
• slide 1 of 6
The concept of zero is usually harder than counting and other early number concepts. Thus, we usually introduce it only after a child has understood the value of numbers to some extent. The difference between 0 and other numbers is that all of the other numbers have a tangible visual form, whereas 0 does not.
Some children really struggle to understand that 0 means nothing. Here are a few number zero preschool activities that you can use to teach and reinforce the zero concept. We first associate the number 0 with empty and nothing, and slowly help them understand the relation of 0 with other numbers.
• slide 2 of 6
### Bowls and Number Cards
Make a few cards with 0 written on them. Take a set of 8-10 bowls and arrange them in a line. In a few of the bowls place a piece of candy, or any other object. The child needs to place the number 0 card in front of the bowls that are empty.
• slide 3 of 6
### Paper Cups and Branches of Leaves
For this activity, fill paper cups with sand and place 8- 10 of them in a line. Place a twig with leaves in each cup to make miniature trees. On some of the trees, remove all the leaves. The child has to place the number 0 card in front of the trees that have no leaves.
• slide 4 of 6
### Trees with Apples Pictures
This activity can be done as a bulletin board activity, a white board activity or even a worksheet. Draw or make a set of trees. Inside some of the trees, draw some apples. The child is to write the number 0 under the trees that have no apples.
• slide 5 of 6
### Counters
This activity is very easy to make and hardly requires any space. Keep a lot of pennies, or even checkers counters, available for this activity. On a sheet of paper, draw a grid of squares. Inside every square write a number between 0 and 5. The child has to see the number and place the right number of coins on it. On the squares that have 0 written on them, the child must not place any coins.
• slide 6 of 6
### Stair Stepping
Here is another game to help children understand number 0. The children must start at the top of the stairs. When you call out "one," the child must jump one step down. If you call out "zero," the child must stay where he is. If a child steps down when you call out 0, he is out of the game. In this manner children must try to stay in the game until they reach the bottom of the stairs.
Thus, these are some number zero preschool activities that you can use. Here are some more ideas for teaching math through activities. If you have ideas to add, please leave them in the comments section below. |
# How can we use fractions in everyday life?
0
0
1. Slide
60 seconds
Fractions: A Fun Way to Learn Division
Fractions are like pieces of a whole. They can represent parts of a larger number or a whole. Fractions can be added, subtracted, multiplied and divided. This allows you to solve complex equations. Fractions can help you understand the world around you, from measuring ingredients in baking to understanding fractions of an hour.
2. Slide
60 seconds
Did you know?
A fraction can be expressed as a decimal by dividing the numerator by the denominator. In mathematics, fractions are used to describe how many parts of a whole there are. Fractions can also be written in scientific notation, for example 1/4 can be written as 0.25.
3. Slide
60 seconds
Did you know?
A fraction can be expressed as a decimal by dividing the numerator by the denominator. In mathematics, fractions are used to describe how many parts of a whole there are. Fractions can also be written in scientific notation, for example 1/4 can be written as 0.25.
4. Open question
300 seconds
How can we use fractions in everyday life?
5. Drawings
450 seconds
Brain break: Draw a hamburger with legs and arms, dancing on a hotdog bun stage
6. Drawings
1260 seconds
Clues
Question: You are a baker and you need to divide a cake into 8 equal slices. How do you do it?
Remember that a fraction is a part of a whole - 8 is a fraction of the whole cake. To find 8 equal parts you can divide the cake into halves, then quarters, then eighths. Each time you divide the cake, it becomes easier to cut the slices to be equal.
A: Draw the cake and the 8 equal slices. B: Explain to your partner how you divided the cake into 8 equal slices.
7. Poll
60 seconds
What is the top number in a fraction called?
• Denominator
• Divisor
• Numerator
• Whole number
8. Poll
60 seconds
What is the bottom number in a fraction called?
• Equal sign
• Denominator
• Fraction bar
• Numerator
9. Poll
60 seconds
Which fraction is bigger: 1/4 or 3/8?
• 3/8
• 1/4 |
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# If x = y (log xy), then find $\dfrac{{dy}}{{dx}}$.
Last updated date: 13th Jun 2024
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Hint: We are given x = y (log xy). For calculating the derivative of y with respect to x, we will first use the logarithmic relation log (xy) = log x + log y and then we will differentiate with respect to x. on further simplification, we will get the value of $\dfrac{{dy}}{{dx}}$.
We are given that x = y (log xy).
We need to calculate $\dfrac{{dy}}{{dx}}$.
First of all, we will use the logarithmic identity log (xy) = log x + log y in the given equation. On substituting, we get
$\Rightarrow x = y\left( {\log x + \log y} \right)$
Now, we will differentiate this equation with respect to x and we will use the product rule for differentiating y (log x + log y) which is defined as $\dfrac{{d\left( {mn} \right)}}{{dx}} = n\dfrac{{dm}}{{dx}} + m\dfrac{{dn}}{{dx}}$ . On differentiating w. r. t. x, we get
$\Rightarrow 1 = \dfrac{{dy}}{{dx}}\left( {\log x + \log y} \right) + y\left( {\dfrac{1}{x} + \dfrac{1}{y}\dfrac{{dy}}{{dx}}} \right)$
$\Rightarrow 1 = \dfrac{{dy}}{{dx}}\left( {\log x + \log y} \right) + \dfrac{y}{x} + \dfrac{{dy}}{{dx}} \\ \Rightarrow 1 - \dfrac{y}{x} = \dfrac{{dy}}{{dx}}\left( {1 + \log x + \log y} \right) \\$
By using the logarithmic identity log (xy) = log x + log y again
$\Rightarrow \dfrac{{x - y}}{x} = \dfrac{{dy}}{{dx}}\left( {1 + \log xy} \right) \\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x - y}}{{x\left( {1 + \log xy} \right)}} \\$
Therefore, $\dfrac{{dy}}{{dx}}$ is found to be $\dfrac{{x - y}}{{x\left( {1 + \log xy} \right)}}$.
Note: In such problems you can get confused while using various identities and also when you will differentiate a product of two functions. You should be handy of all the identities.
Additional Information: In mathematics, many logarithmic identities are there with whom we solve the various problems based on logarithmic functions.
In mathematics, the derivative $\dfrac{{dy}}{{dx}}$ is a function that characterizes the rate of change of the function y with respect to x. The process of finding the derivatives is called differentiation.
Derivatives are fundamentals to the solution of problems on equations in calculus. |
### Numerical Expressions
• Mathematicians have agreed upon a convention for performing arithmetic operations, called the order of operations, so that any mathematical expression will always have the same value.
• Order of Operations
1. Parentheses (work inside grouping symbols)
2. Exponents
3. Multiply and Divide in order from left to right
4. Add and Subtract in order from left to right
## 18 ×3 − 49 ÷7 ×3 =
• 54 − 49 ÷7 ×3 =
• 54 − 7 ×3 =
• 54 − 21 =
## 44 ÷(16 − 12) + 7 ×5 =
• 44 ÷4 + 7 ×5 =
• 11 + 35 =
## (14 − 7) + 72 ÷9 − (8 + 3) =
• 7 + 72 ÷9 − (8 + 3) =
• 7 + 8 − (8 + 3) =
• 7 + 8 − 11 =
• 15 − 11 =
## 7.4 + ((57.7 − 41.5) − 9.1) =
• 7.4 + (16.2 − 9.1) =
• 7.4 + 7.1 =
## 2(5 + (27.4 − 6 − (15.4 − 2.5))) =
• 2(5 + (27.4 − 6 − 12.9)) =
• 2(5 + (21.4 − 12.9)) =
• 2(5 + 8.5) =
• 2(13.5) =
## ((11 ×2 + 3) ÷5 ×2 + 10) ÷4 =
• ((22 + 3) ÷5 ×2 + 10) ÷4
• (25 ÷5 ×2 + 10) ÷4 =
• (5 ×2 + 10) ÷4 =
• (10 + 10) ÷4 =
• 20 ÷4 =
## ((17 + 5 ×2) ÷3 − 1) ÷4 =
• ((17 + 10) ÷3 − 1) ÷4 =
• (27 ÷3 − 1) ÷4 =
• (9 − 1) ÷4 =
• 8 ÷4 =
## While shopping for school, you buy two notebooks for \$ 4.25 each, three pens for \$ 1.50 each, and 5 pencils for \$ 0.75 each. How much money did you spend total?
• 2(\$ 4.25) + 3(\$ 1.50) + 5(\$ 0.75) =
• \$ 8.50 + \$ 4.50 + \$ 3.75 =
• \$ 13.00 + \$ 3.75 =
## A family with 3 children, 2 adults, and 1 senior citizen purchase tickets for a movie. A child ticket costs \$ 7, an adult ticket costs \$ 10.50, and a senior ticket costs \$ 8. What is the total cost of their tickets?
• 3(\$ 7) + 2(\$ 10.50) + \$ 8 =
• \$ 21 + \$ 21 + \$ 8 =
• \$ 42 + \$ 8 =
## Michelle is shopping for groceries. She buys 2 apples for \$ 1.50 each, 6 bananas for \$ 0.75 each, 10 strawberries for \$ 0.25 each, and 1 pineapple for \$ 5. How much did Michelle spend on groceries?
• 2(\$ 1.50) + 6(\$ 0.75) + 10(\$ 0.25) + \$ 5 =
• \$ 3 + \$ 4.50 + \$ 2.50 + \$ 5 =
• \$ 7.50 + \$ 2.50 + \$ 5 =
• \$ 10 + \$ 5 =
### \$ 15
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer. |
## The Tiny Seed
In this lesson, children will count the seeds by ones and twos and then put the packets in order from smallest to largest and vice versa.
### Math Lesson for:
Toddlers/Preschoolers
(See Step 5: Adapt lesson for toddlers or preschoolers.)
Algebra
Measurement
### Learning Goals:
This lesson will help toddlers and preschoolers meet the following educational standards:
• Understand numbers, ways of representing numbers, relationships among numbers and number systems
• Represent and analyze mathematical situations and structures using algebraic symbols
• Apply appropriate tools and formulas to determine measurements
• Understand patterns, relations and functions
### Learning Targets:
After this lesson, toddlers and preschoolers should be more proficient at:
• Developing understanding of the relative position and magnitude of whole numbers and of ordinal and cardinal numbers and their connection
• Sorting, classifying and ordering objects by size, number and other properties
• Using concrete, pictorial and verbal representations to develop an understanding of invented and conventional symbolic notations
• Counting with understanding and recognizing “how many” in sets of objects
• Developing common referents for measures to make comparisons and estimates
## The Tiny Seed
### Lesson plan for toddlers/preschoolers
#### Step 1: Gather materials.
• The book, The Tiny Seed by Eric Carle
• 12×18 pieces of construction paper
• Glue
• Packets of seeds (Make sure to create seed packets that contain a mixture of small and large seeds of different colors. Make a packet for each child, as well as a packet for demonstration purposes.)
Note: Small parts pose a choking hazard and are not appropriate for children age five or under. Be sure to choose lesson materials that meet safety requirements.
#### Step 2: Introduce activity.
1. Show the children a seed packet. Ask the children to guess/estimate how many seeds are in the packet.
2. Field and record answers. Open the packet and count the seeds. Ask: “Were the estimates more or less than the actual number of seeds?” (Extension: For your older students, ask: “What is the difference between the estimate and the actual number of seeds?)”
3. Explain that, after we read the book, The Tiny Seed, the children will receive their own seed packets , which they will use to estimate, count and record the number of seeds.
#### Step 3: Engage children in lesson activities.
1. Read the book, The Tiny Seed by Eric Carle.
2. Ask the children to count how many seeds they have by ones. Then have them put the seeds in a row and count them by twos. Ask: “Can you count them by fives?” Make sure that, no matter how they count the seeds, they always come up with the same number.
3. Now have the students put the seeds in the mathematical order of their choice (such as ordering the seeds from smallest to largest or vice versa).
4. Once the children decide on a way to organize and display their seeds, have them glue their seeds and seed packets onto construction paper. They should also write their mathematical thinking next to the seeds. For example, if they are grouping their seeds into rows of twos, their mathematical thinking should exhibit skip counting by twos. They should also write down the total number of seeds found in the packet.
• Ask: “How much more?” or “What is the difference?” between the estimate and the actual number of seeds.
• Create simple addition and subtraction word problems. Use the seeds and what happens to them throughout the book. You can use some of the animals in your word problems as well.
• The children can group the seeds in a variety of ways. The children can group the seeds according to any criteria they want to use (color, size, texture, etc.), but they will have to explain the criteria to you.
#### Step 4: Math vocabulary.
• Group: Equal sets (e.g.,”Group the seeds by twos.”)
• Estimate: To form an approximate judgment or opinion regarding the amount, worth, size, weight, etc., of; calculate approximately (e.g.,”Estimate how many seeds are in your packets.”)
• Sort: Separating the items according to a given attribute (e.g.,”Sort the seeds according to color.”)
• Count: To identify the amount of something by number (e.g.,”Count the number of seeds in your packet.”)
#### Step 5: Adapt lesson for toddlers or preschoolers.
##### Adapt Lesson for Toddlers
###### Toddlers may:
• Not have mastered one-to-one correspondence
###### Home child care providers may:
• Help the children count their seeds
• Have the children count their seeds by ones
• When the children glue their seeds onto construction paper, help them write the corresponding numbers under each seed
• Have the children group the seeds into a named attribute, such as color or shape
##### Adapt Lesson for Preschoolers
###### Preschoolers may:
• Have a grasp of one-to-one correspondence and can easily count by and group by twos
• Understand attributes
• Be beginning to develop number sense and add and subtract single-digit numbers
###### Home child care providers may:
• Have the children count and group the seeds by a higher number
• Allow the children to group the seeds according to any criteria they want to use (color, size, texture, etc.) and then explain the criteria
• Create simple addition and subtraction word problems using the seeds and what happens to them throughout the book (some of the animals can be used in the word problems as well)
### Suggested Books
• The Tiny Seed by Eric Carle (New York: Little Simon, 2009)
### Outdoor Connections
• Plant seeds and measure their growth. Group the seeds that you plant (vegetables, flowers, plants, fruits). A wonderful activity for all small children.
• Grow the seeds indoors. Take time to observe how the seeds grow. A simple way to plant seeds is to use cotton balls, water, plastic baggies, a bowl of water and seeds. Before introducing the activity to the children, soak the seeds overnight to speed up the process. Have the children dip their cotton balls into water and place them in their baggies. They should use enough cotton balls to fill the bottoms of their baggies. Next, the children will add seeds to their baggies. Remind them that the seeds need space to grow and four or five days to sprout. Once the children have finished adding the cotton balls and seeds to the baggies, close up the baggies and tape them to a window (make sure the baggies are sealed tightly to keep the seeds moist). Wait and see what happens. Talk about the growth of the seeds as they start to shoot out some sprouts. |
# Operating With Rational Numbers Worksheet
A Rational Numbers Worksheet will help your child become more informed about the principles powering this percentage of integers. In this particular worksheet, students are able to resolve 12 different issues relevant to realistic expressions. They will figure out how to increase 2 or more numbers, group of people them in couples, and figure out their products and services. They may also process simplifying logical expressions. After they have mastered these methods, this worksheet will certainly be a important resource for furthering their reports. Operating With Rational Numbers Worksheet.
## Reasonable Figures certainly are a ratio of integers
The two main forms of phone numbers: rational and irrational. Reasonable figures are understood to be whole figures, whereas irrational numbers will not repeat, and possess an endless amount of numbers. Irrational numbers are no-no, non-terminating decimals, and sq . roots which are not best squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To outline a logical variety, you need to realize exactly what a rational number is. An integer is a whole variety, as well as a reasonable variety is really a rate of two integers. The rate of two integers will be the number on the top split by the variety on the bottom. For example, if two integers are two and five, this would be an integer. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be created right into a small fraction
A reasonable quantity includes a numerator and denominator which are not zero. Consequently they may be indicated as being a small fraction. In addition to their integer numerators and denominators, logical amounts can also have a negative benefit. The adverse value should be placed to the left of as well as its absolute worth is its range from zero. To make simpler this example, we are going to claim that .0333333 is actually a small fraction that can be created as a 1/3.
Along with adverse integers, a rational quantity can also be created in a small fraction. As an example, /18,572 is really a reasonable number, while -1/ is just not. Any portion made up of integers is realistic, as long as the denominator fails to consist of a and will be created as an integer. Also, a decimal that leads to a level is also a realistic amount.
## They create sensation
Despite their title, reasonable phone numbers don’t make very much feeling. In mathematics, they can be single organizations having a unique length in the amount collection. Which means that once we add up one thing, we are able to buy the size and style by its rate to the unique number. This retains correct even if you can find endless reasonable figures among two certain phone numbers. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To obtain the length of a pearl, as an example, we could add up its thickness. A single pearl weighs in at twenty kilograms, which is a realistic number. Furthermore, a pound’s bodyweight equals ten kilos. As a result, we must be able to split a lb by twenty, with out be concerned about the duration of just one pearl.
## They are often expressed being a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal quantity might be created as being a numerous of two integers, so four times 5 various is the same as 8. An identical problem necessitates the frequent portion 2/1, and either side ought to be divided up by 99 to obtain the proper respond to. But how can you make your conversion process? Here are a few good examples.
A realistic variety will also be written in great shape, including fractions plus a decimal. One method to represent a logical quantity inside a decimal is usually to separate it into its fractional equivalent. There are actually three ways to divide a reasonable amount, and every one of these approaches produces its decimal counterpart. One of these simple approaches is usually to divide it into its fractional comparable, and that’s what’s known as the terminating decimal. |
constant multiple rule | derivatives,ab calculus,12th grade
# Constant Multiple Rule
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The constant multiple rule : If 'f' is a differentiable function and k is any real number, then kf is also differentiable and
$\frac{\text{d}[kf(x)]}{\text{d}x}=k\frac{\text{d}f(x)}{\text{d}x}= kf'(x)$
Prove that: $\frac{\text{d}[kf(x)]}{\text{d}x}= kf'(x)$
Proof : $\frac{\text{d}[kf(x)]}{\text{d}x} = \lim_{\triangle x \rightarrow 0}\frac{k f(x+\triangle x)-kf(x)}{\triangle x}$
= $\lim_{\triangle x \rightarrow 0}k[\frac{ f(x+\triangle x)-f(x)}{\triangle x}]$
$\frac{\text{d}[kf(x)]}{\text{d}x}=k[\lim_{\triangle x \rightarrow 0}\frac{ f(x+\triangle x)-f(x)}{\triangle x}]$
According to the definition of derivative
$\frac{\text{d}[f(x)]}{\text{d}x} = \lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$
So, $\frac{\text{d}[kf(x)]}{\text{d}x} = k\frac{\text{d}[f(x)]}{\text{d}x}$
1) $\frac{\text{d}[kf(x)]}{\text{d}x} = k\frac{\text{d}[f(x)]}{\text{d}x}$
2) ${\frac{\text{d}[\frac{f(x)}{k}]}{\text{d}x} = \frac{\text{d}[\frac{1}{k}f(x)]}{\text{d}x}}=\frac{1}{k}\frac{\text{d}[f(x)]}{\text{d}x}$
## Examples of constant multiple rule
1) Find : $\frac{\text{d}[5x^{4}]}{\text{d}x}$
Solution : $\frac{\text{d}[5x^{4}]}{\text{d}x}$
=$5\frac{\text{d}x^{4}}{\text{d}x}$
=$5x^{4-1}$
$\frac{\text{d}[5x^{4}]}{\text{d}x}=5x^{3}$
2)
Find : $\frac{\text{d}[\frac{3}{x}]}{\text{d}x}$
Solution : $\frac{\text{d}[\frac{3}{x}]}{\text{d}x}$
= 3$\frac{\text{d}[\frac{1}{x}]}{\text{d}x}$
=3$\frac{\text{d}[x^{-1}]}{\text{d}x}$
=3$x^{-1-1}$
=3$x^{-2}$
=3$\frac{1}{x^{2}}$
$\frac{\text{d}[\frac{3}{x}]}{\text{d}x} =\frac{3}{x^{2}}$
3)
Find: $\frac{\text{d}[\frac{5}{\sqrt[3]{x^{2}}}]}{\text{d}x}$
=$\frac{\text{d}[\frac{5}{\sqrt[3]{x^{2}}}]}{\text{d}x}$
=5$\frac{\text{d}[\frac{1}{\sqrt[3]{x^{2}}}]}{\text{d}x}$
=5$\frac{\text{d}[\frac{1}{{x^{\frac{2}{3}}}}]}{\text{d}x}$
=5$\frac{\text{d}[{{x^{\frac{-2}{3}}}}]}{\text{d}x}$
=-5$x^{(\frac{-2}{3}-1)}$
=-5$x^{\frac{-5}{3}}$
= -5$\frac{1}{x^{\frac{5}{3}}}$
$\frac{\text{d}[\frac{5}{\sqrt[3]{x^{2}}}]}{\text{d}x}=-\frac{5}{x^{\frac{5}{3}}}$
The constant multiple rule and the power Rule can be combined into one rule. The combination rule is given by
$\color{red}{\frac{\text{d}(kx^{n})}{\text{d}x}=knx^{n-1}}$
Practice questions :
1) Find the derivative of $\frac{-3x^{2}}{2}$
2) Find the derivative of $\frac{7}{2x^{3}}$
3) Find the derivative of $16x^{5}$
4) Find the derivative of $6x^{-3}$
5) Find the derivative of $\frac{1}{2x^{-2}}$ |
# AP Calculus AB : Derivatives
## Example Questions
### Example Question #11 : Ap Calculus Ab
Find the equation of the line tangent to at .
Explanation:
The first step is to find the derivative of the function given, which is .
Next, find the slope at by plugging in and solving for , which is the slope. You should get . This means the slope of the new line is also 3 because at the point where a slope and a line are tangent they have the same slope. Use the equation to express your line. Y and x are variables and m is the slope, so the only thing you need to find is b. Plug in the point and slope into to get . Now you can express the general equation of the line as .
### Example Question #12 : Ap Calculus Ab
Find the equation of the line perpendicular to the line tangent to the following function at x=1, and passing through (0, 6):
Explanation:
To find the equation of the line perpendicular to the tangent line to the function at a certain point, we must find the slope of the tangent line to the function, which is the derivative of the function at that point:
The derivative was found using the following rules:
Now, we evaluate the derivative at the given point:
We now know the slope of the tangent line, but because the line we are solving for is perpendicular to this line, its slope is the negative reciprocal, .
With a slope and a point, we can now find the equation of the line:
### Example Question #11 : Tangent Line To A Curve At A Point And Local Linear Approximation
Find the equation of the line that is tangent to the graph of when .
Explanation:
First, evaluate .
Then, to find the slope of the tangent line, find .
, so .
Therefore, the equation of the tangent line is
.
### Example Question #11 : Derivatives
Find the minimum value of
Explanation:
In order to find the extreme value we need to take the derivative of the function.
After setting it equal to 0, we see that the only candidate is for . After setting into , we get the coordinate as an extreme value. To confirm it is a minimum we can plot the function.
### Example Question #15 : Tangent Line To A Curve At A Point And Local Linear Approximation
Let . Find the equation of the line tangent to at the point .
Explanation:
To find the equation of a tangent line, we need two things: The tangent point, which is given as , and the slope of the tangent line at that point, which is the derivative at that point.
To find the derivative at the point, we will find , using derivative rules. Then we will plug the given point's x-coordinate into and that will give us the slope we need.
Finding the derivative of will require the power rule for each term. Recall that the power rule is . For the , the power rule effectively removes the . Also, the derivative of a constant is , so the will drop when we get the derivative.
Applying these rules, we get
Now that we have the derivative, we effectively have a formula to find the slope of a tangent line at any point we choose. The question asks for the tangent line at . So we plug the x-coordinate of the point into the derivative function to find the slope.
The last step is to use the point-slope equation of a line to construct the equation we need. The point-slope equation is , where is the slope, and is the given point.
Plugging our slope into , and our original point in for and , we get
Now we can solve for to compare to the answer choices.
This is the equation of the tangent line at the given point. We have our answer.
### Example Question #16 : Tangent Line To A Curve At A Point And Local Linear Approximation
Let . Find the equation of the line tangent to at the point .
Explanation:
To find the equation of a tangent line at a point, we will need the slope of the function at that point. To find this, we find the derivative of the function.
Finding this derivative will use trigonometric function derivative rules, and the product rule.
Recall that the derivative of is . This takes care of the first term.
To find the derivative of the next term, we need to be careful with our signs. We will use the product rule which results in two terms. The negative sign can mess us up if we aren't careful. The way we will handle this is to associate the minus sign with the term when doing the product rule. So we will find the derivative of . The product rule is as follows, where the red and blue are the two factors of the term we are differentiating. In our case, and .
by the special case of the power rule. The drops.
by the trigonometric derivative rules.
Putting these together we get the derivative of to be
Simplifying, we get
Assembling all the pieces of the derivative together, we have found
Combining like terms gives
Now we have a formula for the tangent slope of at any point. The point we care about is . To find this point's tangent slope, we will plug its x-coordinate into our derivative.
Simplifying gives
So we have found the slope of the tangent line at our point, , is .
The last step is the point-slope equation of a line, , where is the slope and is the given point. Plugging in for , for , and for , we get
Solving for , we get
This is the equation of the tangent line at the given point. We have our answer.
### Example Question #1 : Derivatives Of Functions
The Second Fundamental Theorem of Calculus (FTOC)
Consider the function equation (1)
(1)
The Second FTOC states that if:
1. is continuous on an open interval
2. is in .
3. and is the anti derivative of
then we must have,
(2)
Differentiate,
Explanation:
Differentiate:
Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.
Let,
Therefore,
Now we can write the derivative using the chain rule as:
Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to
Therefore we have,
### Example Question #2 : Derivatives Of Functions
Find the derivative.
Explanation:
Use the power rule to find the derivative.
### Example Question #3 : Derivatives Of Functions
Find the derivative of the following:
None of the above
Explanation:
To take the derivative you need to bring the power down to the front of the equation, multiplying it by the coefficient and then drop the power.
So:
becomes
because the degree of "x" is just one, and once you multiply 3 by 1 you get 3 and drop the power of "x" to 0. The second term is just a constant and the derivative of any term is just 0.
Find : |
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Exercises with Z-Scores and Percentiles
Not what you're looking for?
Six steps of hypothesis testing are:
Identify populations, comparison distribution, and assumption; state null and research hypothesis; determine characteristics of the comparisons distribution; determine critical values, or cutoffs; calculate test statistic; and make a decision.
Question 7.18: Calculate the following percentages for a z- score using z table: -0.08
(-0.08 = 93.62 (46.81+ 46.81) (above) 6.38 (3.19 + 3.19) (below), 0.08 (at least as extreme as this z score).
Question 7.20: Rewrite percentages as probabilities, or p levels: a. 5% (0.05) b. 83% (0.83) c. 51% (0.51).
Question 7.22: If the critical values for a hypothesis test occur where 2.5% of the distribution is in each tail, what are the cutoffs for z?
Question 7.28: If cutoffs for a z test are -2.58 and 2.58, determine whether you would reject or fail to reject the null hypothesis in each of the following cases: a. z = -0.94 b. z = 2.12
Question 7.32: z distribution and height. 15 year old girl is 58 inches tall, average height for girls is at this age is 63.80 inches, with a standard deviation of 2.66 inches; a. Calculate her z score, b. What percentage of girls are taller than her? c. What percentage of girls are shorter? d. How much would she have to grow to be perfectly average?
Question 7.34: The z statistic and height. An average class of thirty three 15 yr. old girls with an average height of 62.6 inches. Remember, u = 63.8 inches and σ = 2.66 inches; a. Calculate z score, b. How does this sample of girls compare to the distribution of sample means?
Question 7.39: Directional versus no directional hypotheses; for each example, identify whether the research has expressed a directional or a no directional hypotheses; a. A researcher is interested in studying the relation between the use of antibacterial products and the dryness of people's skin. He thinks these products might alter the moisture in skin compared to other products that are not antibacterial; b. A student wonders if grades in a class are in any way related to where a student sits in the classroom. In particular, do students who sit in the front row get better grades, on average, than the general population of students?
1. Fill in the blank with the best word or words.
a. Values of a test statistic beyond which you reject the null hypothesis are called critical values.
b. Critical region is the area in the tails in which the null can be rejected.
c. If your data differ from what you would expect if chance were the only thing operating, you would call your finding statistically significant.
d. A hypothesis test in which the research hypothesis is directional is a (n) one- tailed test.
e. A hypothesis test in which the research hypothesis specifies that there will be a difference but does not specify the direction of that difference is a (n) two- tailed test.
f. If your z-statistic exceeds the critical cutoff, you can reject the null hypothesis.
The police department of a major city has found that the average height of their 1,250 officers is 71 inches (in.) with = 2.3 inches.
a. How many officers are at least 75 inches tall? 513 (1250 - 71/2.3).
b. What percent of officers are between 65 and 72 inches tall? 5.2% and 5.76%
c. The top 10% of the officers in terms of height also make higher salaries than the shorter officers. How tall does an officer have to be to get in that top 10% group? (I.e. what is the height marking off the top 10 percent?) 6 ft. tall.
2. The verbal part of the Graduate Record Exam (GRE) has a of 500 and = 100. Use the normal distribution to answer the following questions:
a. If you wanted to select only students at or above the 90th percentile, what verbal GRE score would you use as a cutoff score?
b. What verbal GRE score corresponds to a percentile rank of 15%?
c. What's the percentile rank of a GRE score of 628?
3. The police department of a major city has found that the average height of their 1,250 officers is 71 inches (in.) with = 2.3 inches.
d. How many officers are at least 75 inches tall? 513 (1250 - 71/2.3).
e. What percent of officers are between 65 and 72 inches tall? 5.2% and 5.76%
f. The top 10% of the officers in terms of height also make higher salaries than the shorter officers. How tall does an officer have to be to get in that top 10% group? (I.e. what is the height marking off the top 10 percent?) 6 ft. tall.
4. The verbal part of the Graduate Record Exam (GRE) has a of 500 and = 100. Use the normal distribution to answer the following questions:
a. If you wanted to select only students at or above the 90th percentile, what verbal GRE score would you use as a cutoff score?
b. What verbal GRE score corresponds to a percentile rank of 15%?
c. What's the percentile rank of a GRE score of 628?
Solution Summary
This solutions deals with how you work with z-scores and percentiles in a data set and involves a little work with the Central Limit Theorem.
Measures of Central Tendency
Tests knowledge of the three main measures of central tendency, including some simple calculation questions.
Measures of Central Tendency
This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research.
Terms and Definitions for Statistics
This quiz covers basic terms and definitions of statistics. |
# ( ) FACTORING. x In this polynomial the only variable in common to all is x.
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## Transcription
1 FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5*. The multiples are 5 and. In a polynomial it is the same way, however, the procedure is somewhat more complicated since variables, not just numbers, are involved. There are different ways of factoring an equation depending on the complexity of the polynomial. Factoring out the greatest common factor The first thing to do when factoring is to look at all terms and break up each term into its multiples: Examples: Factor: 8x + 4x + 10x ( 8x + 4x + 10) x In this polynomial the only variable in common to all is x. x(4x+x+5) The two is also common to all terms. Therefore, this is as far as the polynomial can be factored. Factor: 6x + 8x + 16x 6x + 8x + 16x The common variable is x, and the smallest exponent is 1. The common multiple is. x x + 4x + 8 Therefore, the greatest common factor is x. ( ) x x = 6x You can also look at it this way: ( x)( 4x) = 8x ( x)( 8) = 16x
2 Factoring by grouping There will be occasions when there are no common factors for all terms, but there will be terms that have a variable in common while other terms have a different variable in common. In this case, we can factor by grouping together these common terms. Examples: Factor: ax + ay + bx + by ax + ay + bx + by Notice how there are two variables, x and y. Group ax + bx + ay + by together the x s and the y s and then x a + b + y a + b factor out the common factor in each group. ( ) ( ) Factor: + 4 x x y y + There are two variables present, x and y. 4 x x y y Start by grouping two terms at a time. ( x ) + y ( x ) Notice how the two polynomials inside the parentheses are similar. To make them the same change the sign in the second polynomial. ( x ) y ( ) x Now that they are the same, factor out the common term. x y Now it is completely factored out. Factoring a difference of squares For a difference of squares, a b, the factors will be ( a b)( a + b). Examples: x y = ( x y)( x + y) ( ) ( ) ( ) 1 x = 1 x = 1 x 1+ x = 1 x 1+ x 1+ x 4
3 Factoring the sum and difference of two cubes For the addition or subtraction of two cubes, the following formulas apply: a + b = a + b a ab + b a b = a b a + a + bb Examples: x + 8 ( x) ( ) ( x ) x ( x)( ) = + Manipulate to be in a + b form. ( ) ( x )( x x 4) = + + follow formula. = + + 8y 7 Manipulate to be in a b = y follow formula. ( ) ( ) ( y ) ( y) ( y)( ) ( ) = + + ( y )( 4y 6y 9) = + + form. Factoring trinomials Factoring trinomials is based on finding the two integers whose sum and product meet the given requirements. Example: Find two integers whose sum is 11 and whose product is 0. Step 1 The product is positive, the sum is negative, therefore, both integers must be =+ and + = negative. ( ) ( ) Step Possible pairs of factors that will give a positive answer: 10 = 0 1. ( 1)( 0) = 0.. ( )( 15) = = 0 Step The pair whose sum is 11 is 5 and 6. Therefore, the critical integers are = 11. and 6, ( )
4 Factoring trinomials of the form x + bx + c To factor trinomials of the form x + bx + c follow the same principle described above. Find two integers whose sum equals the middle term and whose product equals the last term. Examples: Factor: x x 6 Step 1 Find two integers whose sum is equal to 1 and whose product is equal to 6. The product is negative; therefore, we need integers that have different sign. The sum is negative, so the bigger integer is negative. Step Possible pairs of factors that will give us 6: 1. ( )( ) = 6. ( )( ) = 6. ( 1)( 6) = 6 4. ( 1)( 6) = 6 Step The pair whose sum is 1 is and, ( + = 1) The critical integers are and. Step 4 The factors for x x 6 are ( x )( x + ).. Factor: x + 16x + 60 Step 1 Find two integers whose sum is 16 and whose product is 60. Both integers must be positive since their sum and product are positive. Step Possible pairs of factors: 1. ( 1)( 60). ( )( 0) 5.. ( )( 0) 4. ( 4)( 15) Step The pair whose sum is 16 is (6, 10). Therefore, the critical integers are 6, 10. Step 4 The factors for x + 16x+ 60 are ( x + 6)( x + 10 ).
5 Factoring trinomials of the form ax + bx + c In this type of trinomials the coefficient of x is not 1, therefore we must consider the product of Examples: Factor: x 7x 4 Step 1 Find two integers whose sum is 7 and whose product is ( )( ) 4 = 8. The a c. integers must be of different signs since their product is a negative number. The larger integer is negative since their sum is negative. Step Possible pairs of factors: 1. ( 1)( 8) = 8.. ( 1)( 8) = = 8 4 = 8 Step The pair whose sum is 7 is 1 and 8. The critical integers are 1 and 8. Step 4 Use the critical integers to break the first degree term into two parts: x 7x 4 = x 8x + x 4 Step 5 Factor by grouping the first two and last two terms: x 8x + x 4 = x( x 4) + 1( x 4) There are now two terms in the expression. The binomial factor in each of these two terms should be the same; otherwise, there is an error. In this case, the common binomial factor is( x 4). Step 6 Factor out the common binomial factor: x( x 4) + 1( x 4) = ( x 4)( x + 1) Factor: 6x x + 0 Step 1 Find two integers whose sum is and whose product is 6(0) = 10. Since the product is positive and the sum is negative, the two integers are negative. Step Possible pairs of factors: 1. ( 1)( 10) = ( 4)( 0) = ( )( 60) = ( 5)( 4) = ( )( 40) = ( 6)( 0) 8 15 = = 10 Step The sum of 8 and 15 is. Therefore, the critical integers are 8 and 15.
6 Step 4 Use the critical integers to break the first degree term into two parts: 6x x + 0 = 6x 15x 8x + 0 Step 5 Factor, separately, the first two and last two terms: 6x 15x 8x + 0 = x( x 5) 4( x 5) Step 6 Factor out the common binomial factor. x( x 5) 4( x 5) = ( x 5)( x 4)
7 FACTORING EXERCISES 1. x 10x 8. x ( x ) 5x( x ) ( x ). x + x 9x x 6 x 10x + 5 y y 1 a ( x + 5) + 4( x + 5) t + t 4 x 18x + 7x x 100x x x x + xy + y 1. x ( x + ) + 1x( x + ) + 15( x + ) 14. x + x x y + y + x a x + 8x x 0x 50x 19. 9x + x x + x
8 FACTORING ANSWER TO EXERCISES x. x ( x ) 5x( x ) ( x ) 1. 10x 8 x 1x + x 8 ( x )(x 5x ) x( x 4) ( x 4) ( x )( x 4) + ( x )(x 6x + x ) + ( x )(x( x ) + 1( x )) ( x )( x+ 1)( x ) = ( x ) ( x+ 1). x + x 9x x x ( x + ) 9( x + ) ( x 9)( x + ) 10x + 5 y ( x 10x + 5) y ( x 5) y x 9 = ( x + )( x ) Difference of squares: ( a b ) = ( a b)( a + b) ( )( x + )( x + ) x where ( ) ( x 5 y)( x 5 + y) a = x 5 and b= y x y 6. ( ) ( ) 1 a x y = x y Difference of cubes: Difference of squares: ( a b ) = ( a b)( a + b) where x y x y x y 6 6 a b = ( a b)( a + a + bb ) a x b y = and = a 1 a = and b = = a 1 = a a = + a a ( a) Sum & Difference of Cubes: a + b = a + b a ab + b a b = a b a + a + bb = ( x + y)( x xy + y )( x y)( x + xy + y ) = ( x + y)( x y)( x + xy + y )( x xy + y )
9 7. ( x+ 5) + 4( x+ 5) + 4= ( x+ 7)( x+ 7) = ( x+ 7) 8. 6 ( )( ) ( ) ( ) 1 t + t = t 1 t 1 = t 1 t + t ( ) ( ) x 18x + 7x = x x 6x+ 9 = x x x 100x 600 = 100( x )( x+ ) 11. 8x x 15 = ( 4x+ 5)( x ) 1. x + xy + y = ( x + y)( x + y) 1. x ( x+ ) + 1x( x+ ) + 15( x+ ) = ( x+ )( x+ )( x+ 5) 14. x + x 1 = ( x+ 4)( x ) 15. xy + y+ x + 6= ( x + )( y+ ) a 50 = ( a+ 5)( a 5) 17. x + 8x+ 15 = ( x+ 5)( x+ ) x 0x 50x= x( x 5)( 4x+ 5) 19. 9x + x+ 0 = ( x+ )( x+ 5) 0. 1x + x = ( x 1)( x+ 1)
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# McGraw Hill My Math Grade 1 Chapter 6 Lesson 2 Answer Key Count On Tens and Ones
All the solutions provided in McGraw Hill My Math Grade 1 Answer Key PDF Chapter 6 Lesson 2 Count On Tens and Ones will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 1 Answer Key Chapter 6 Lesson 2 Count On Tens and Ones
Explore and Explain
Teacher Directions: Use and to model. There are 35 people at a baseball game. 3 more people come to the game. How many people are at the game in all? Draw the rods and units that you used. Explain to a classmate how you found the answer.
Explanation:
Start at 35
Count on 36, 37, 38
The sum is 38.
See and Show
Find 26 + 3.
Count on by ones.
Find 26 + 30.
Count on by tens.
Count on to add. Use and . Write the sum.
Question 1.
47 + 2 = ______________
49
Explanation:
Count by 1’s
Start at 47
Count on 48, 49
The sum is 49.
Question 2.
47 + 20 = ______________
67
Explanation:
Count by 10’s
Start at 47
Count on 57, 67
The sum is 67.
Question 3.
13 + 3 = _______________
16
Explanation:
Count by 1’s
Start at 13
Count on 14, 15, 16
The sum is 16.
Question 4.
13 + 30 = _______________
43
Explanation:
Count by 10’s
Start at 13
Count on 23, 33, 43
The sum is 43.
Talk Math How many tens do you count on to add 32 + 40? Explain.
4 tens
Explanation:
In 40 there are 4 tens
40 = 4 tens
So, i will count on 4 tens to add 40.
On My Own
Count on to add. Use and . Write the sum.
Question 5.
50 + 14 = _____________
64
Explanation:
Count by 10’s
Start at 50
Count on 60
Count by 1’s
Start at 60
Count on 61, 62, 63, 64
The sum is 64.
Question 6.
25 + 3 = _______________
28
Explanation:
Count by 1’s
Start at 28
Count on 26, 27, 28
The sum is 28.
Question 7.
30 + 22 = ______________
Explanation:
Count by 10’s
Start at 30
Count on 40, 50
Count by 1’s
Start at 60
Count on 51, 52
The sum is 52.
Question 8.
66 + 2 = _______________
Explanation:
Count by 1’s
Start at 66
Count on 67, 68
The sum is 68.
Question 9.
53 + 20 = ______________
Explanation:
Count by 10’s
Start at 53
Count on 63, 73
The sum is 73.
Question 10.
14 + 3 = _______________
Explanation:
Count by 1’s
Start at 14
Count on 15, 16, 17
The sum is 17.
Question 11.
51 + 3 = _______________
Explanation:
Count by 1’s
Start at 51
Count on 52, 53, 54
The sum is 54.
Question 12.
20 + 76 = ________________
Explanation:
Count by 10’s
Start at 20
Count on 30, 40, 50, 60, 70, 80, 90
Count by 1’s
Start at 90
Count on 91, 92, 93, 94, 95, 96
The sum is 96.
Question 13.
Explanation:
The sum of 44 and 3 is 47.
Question 14.
Explanation:
The sum of 10 and 88 is 98.
Question 15.
Explanation:
The sum of 88 and 1 is 89.
Question 16.
Explanation:
The sum of 33 and 3 is 36.
Question 17.
Explanation:
The sum of 12 and 2 is 14.
Question 18.
Explanation:
The sum of 79 and 20 is 99.
Problem Solving
Question 19.
Percy sees 3 ants in his ant farm. He sees 12 more ants. How many ants does he see in all?
_____________ ants
15 ants
Explanation:
Percy sees 3 ants in his ant farm
He sees 12 more ants
Count by 1’s
Start at 12
Count on, 13, 14, 15
The sum is 15
So, Percy saw 15 ants in all.
Question 20.
Kevin’s team and Alonso’s team each have 25 points. Kevin’s team scores 30 more points.
Kevin’s team has _____________ points.
Alonso’s team scores 30 more points.
Alonso’s team has _____________ points.
Kevin’s team has 25 points.
Alonso’s team scores 30 more points.
Alonso’s team has 55 points.
Explanation:
Kevin’s team and Alonso’s team each have 25 points
Kevin’s team scores 30 more points
Count by 10’s
Start at 25
Count on 35, 45, 55
The sum is 55
So, Alonso’s team has 55 points.
Write Math Find 54 + 3. Explain how you add the ones.
54 + 3 = 57
Explanation:
54 + 3
Count by 1’s
Start at 54
Count on 55, 56, 57
The sum is 57
So, 54 + 3 = 57.
### McGraw Hill My Math Grade 1 Chapter 6 Lesson 2 My Homework Answer Key
Practice
Question 1.
27 + 30 = _____________
57
Explanation:
Count by 10’s
Start at 27
Count on 37, 47, 57
The sum is 57.
Question 2.
74 + 3 = ______________
77
Explanation:
Count by 1’s
Start at 74
Count on 75, 76, 77
The sum is 77.
Question 3.
66 + 3 = _______________
Explanation:
Count by 1’s
Start at 66
Count on 67, 68, 69
The sum is 69.
Question 4.
12 + 70 = _______________
Explanation:
Count by 10’s
Start at 70
Count on 80
Count by 1’s
Start at 80
Count on 81, 82
The sum is 82.
Question 5.
51 + 3 = _______________
Explanation:
Count by 1’s
Start at 51
Count on 52, 53, 54
The sum is
Question 6.
87 + 1 = _____________
Explanation:
Count by 1’s
Start at 87
Count on 88
The sum is 88.
Question 7.
Explanation:
The sum of 32 and 20 is 52.
Question 8.
Explanation:
The sum of 46 and 10 is 56.
Question 9.
Explanation:
The sum of 97 and 2 is 99.
Question 10.
Explanation:
The sum of 26 and 10 is 36.
Question 11.
Explanation:
The sum of 40 and 2 is 42.
Question 12.
Explanation:
The sum of 64 and 20 is 84.
Question 13.
The children’s choir sang 17 songs on Wednesday and 2 songs on Thursday. How many songs did they sing in all?
_____________ songs
19 songs
Explanation:
The children’s choir sang 17 songs on Wednesday
2 songs on Thursday
Count by 1’s
Strat at 17
Count on 18, 19
The sum is 19
So, the children’s choir sang 19 songs in all.
Test Practice
Question 14.
70 + 20 = ____________
(A) 78
(B) 85
(C) 88
(D) 90 |
Polynomials and Polynomial Functions
### Learning Objectives
By the end of this section, you will be able to:
• Determine the degree of polynomials
• Evaluate a polynomial function for a given value
• Add and subtract polynomial functions
Before you get started, take this readiness quiz.
1. Simplify:
If you missed this problem, review (Figure).
2. Subtract:
If you missed this problem, review (Figure).
3. Evaluate: when and
If you missed this problem, review (Figure).
### Determine the Degree of Polynomials
We have learned that a term is a constant or the product of a constant and one or more variables. A monomial is an algebraic expression with one term. When it is of the form where a is a constant and m is a whole number, it is called a monomial in one variable. Some examples of monomial in one variable are. Monomials can also have more than one variable such as and
Monomial
A monomial is an algebraic expression with one term.
A monomial in one variable is a term of the form where a is a constant and m is a whole number.
A monomial, or two or more monomials combined by addition or subtraction, is a polynomial. Some polynomials have special names, based on the number of terms. A monomial is a polynomial with exactly one term. A binomial has exactly two terms, and a trinomial has exactly three terms. There are no special names for polynomials with more than three terms.
Polynomials
polynomial—A monomial, or two or more algebraic terms combined by addition or subtraction is a polynomial.
monomial—A polynomial with exactly one term is called a monomial.
binomial—A polynomial with exactly two terms is called a binomial.
trinomial—A polynomial with exactly three terms is called a trinomial.
Here are some examples of polynomials.
Polynomial Monomial 14 Binomial Trinomial
Notice that every monomial, binomial, and trinomial is also a polynomial. They are just special members of the “family” of polynomials and so they have special names. We use the words monomial, binomial, and trinomial when referring to these special polynomials and just call all the rest polynomials.
The degree of a polynomial and the degree of its terms are determined by the exponents of the variable.
A monomial that has no variable, just a constant, is a special case. The degree of a constant is 0.
Degree of a Polynomial
The degree of a term is the sum of the exponents of its variables.
The degree of a constant is 0.
The degree of a polynomial is the highest degree of all its terms.
Let’s see how this works by looking at several polynomials. We’ll take it step by step, starting with monomials, and then progressing to polynomials with more terms.
Let’s start by looking at a monomial. The monomial has two variables a and b. To find the degree we need to find the sum of the exponents. The variable a doesn’t have an exponent written, but remember that means the exponent is 1. The exponent of b is 2. The sum of the exponents, is 3 so the degree is 3.
Working with polynomials is easier when you list the terms in descending order of degrees. When a polynomial is written this way, it is said to be in standard form of a polynomial. Get in the habit of writing the term with the highest degree first.
Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. Then, find the degree of each polynomial.
15
Polynomial Number of terms Type Degree of terms Degree of polynomial
3 Trinomial 2, 1, 0 2
1 Monomial 4, 2 6
5 Polynomial 5, 3, 2, 1, 0 5
2 Binomial 1, 4 4
15 1 Monomial 0 0
Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. Then, find the degree of each polynomial.
monomial, 0
polynomial, 3 trinomial, 3
binomial, 2 monomial, 10
Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. Then, find the degree of each polynomial.
binomial, 3 trinomial, 3 monomial, 0 polynomial, 4 monomial, 7
We have learned how to simplify expressions by combining like terms. Remember, like terms must have the same variables with the same exponent. Since monomials are terms, adding and subtracting monomials is the same as combining like terms. If the monomials are like terms, we just combine them by adding or subtracting the coefficients.
Remember that like terms must have the same variables with the same exponents.
Simplify:
We can think of adding and subtracting polynomials as just adding and subtracting a series of monomials. Look for the like terms—those with the same variables and the same exponent. The Commutative Property allows us to rearrange the terms to put like terms together.
Find the sum:
Find the sum:
Find the sum:
Be careful with the signs as you distribute while subtracting the polynomials in the next example.
Find the difference:
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Find the difference:
Find the difference:
To subtract from we write it as placing the first.
Subtract from
Subtract from
Subtract from
Find the sum:
Find the sum:
Find the sum:
When we add and subtract more than two polynomials, the process is the same.
Simplify:
Simplify:
Simplify:
### Evaluate a Polynomial Function for a Given Value
A polynomial function is a function defined by a polynomial. For example, and are polynomial functions, because and are polynomials.
Polynomial Function
A polynomial function is a function whose range values are defined by a polynomial.
In Graphs and Functions, where we first introduced functions, we learned that evaluating a function means to find the value of for a given value of x. To evaluate a polynomial function, we will substitute the given value for the variable and then simplify using the order of operations.
For the function find:
Simplify the exponents. Multiply. Simplify.
Simplify the exponents. Multiply. Simplify.
Simplify the exponents. Multiply. Simplify.
For the function find
18 50
For the function find
20 2
The polynomial functions similar to the one in the next example are used in many fields to determine the height of an object at some time after it is projected into the air. The polynomial in the next function is used specifically for dropping something from 250 ft.
The polynomial function gives the height of a ball t seconds after it is dropped from a 250-foot tall building. Find the height after seconds.
The polynomial function gives the height of a stone t seconds after it is dropped from a 150-foot tall cliff. Find the height after seconds (the initial height of the object).
The height is feet.
The polynomial function gives the height of a ball t seconds after it is dropped from a 175-foot tall bridge. Find the height after seconds.
The height is 31 feet.
### Add and Subtract Polynomial Functions
Just as polynomials can be added and subtracted, polynomial functions can also be added and subtracted.
Addition and Subtraction of Polynomial Functions
For functions and
For functions and find:
Rewrite without the parentheses. Put like terms together. Combine like terms.
In part (a) we found and now are asked to find
Notice that we could have found by first finding the values of and separately and then adding the results.
Find Find Find
Rewrite without the parentheses. Put like terms together. Combine like terms.
For functions and find:
For functions and find
Access this online resource for additional instruction and practice with adding and subtracting polynomials.
### Key Concepts
• Monomial
• A monomial is an algebraic expression with one term.
• A monomial in one variable is a term of the form where a is a constant and m is a whole number.
• Polynomials
• Polynomial—A monomial, or two or more algebraic terms combined by addition or subtraction is a polynomial.
• monomial —A polynomial with exactly one term is called a monomial.
• binomial — A polynomial with exactly two terms is called a binomial.
• trinomial —A polynomial with exactly three terms is called a trinomial.
• Degree of a Polynomial
• The degree of a term is the sum of the exponents of its variables.
• The degree of a constant is 0.
• The degree of a polynomial is the highest degree of all its terms.
#### Practice Makes Perfect
Determine the Type of Polynomials
In the following exercises, determine if the polynomial is a monomial, binomial, trinomial, or other polynomial.
4
trinomial, 5 polynomial, 3 binomial, 1 monomial, 1
binomial, 1
19
binomial trinomial
polynomial trinomial
monomial
17
15
In the following exercises, add or subtract the monomials.
Subtract from
Subtract from
In the following exercises, add the polynomials.
In the following exercises, subtract the polynomials.
In the following exercises, subtract the polynomials.
Subtract from
Subtract from
Subtract from
Subtract from
In the following exercises, find the difference of the polynomials.
Find the difference of and
Find the difference of and
In the following exercises, add the polynomials.
In the following exercises, add or subtract the polynomials.
Evaluate a Polynomial Function for a Given Value
In the following exercises, find the function values for each polynomial function.
For the function find:
187 40 2
For the function find:
For the function find:
4 40
For the function find:
In the following exercises, find the height for each polynomial function.
A painter drops a brush from a platform 75 feet high. The polynomial function gives the height of the brush t seconds after it was dropped. Find the height after seconds.
The height is 11 feet.
A girl drops a ball off the cliff into the ocean. The polynomial gives the height of a ball t seconds after it is dropped. Find the height after seconds.
A manufacturer of stereo sound speakers has found that the revenue received from selling the speakers at a cost of p dollars each is given by the polynomial function Find the revenue received when dollars.
The revenue is ?10,800.
A manufacturer of the latest basketball shoes has found that the revenue received from selling the shoes at a cost of p dollars each is given by the polynomial Find the revenue received when dollars.
The polynomial gives the cost, in dollars, of producing a rectangular container whose top and bottom are squares with side x feet and height 6 feet. Find the cost of producing a box with feet.
The cost is ?456.
The polynomial gives the cost, in dollars, of producing a rectangular container whose top and bottom are squares with side x feet and height 4 feet. Find the cost of producing a box with feet.
In each example, find (f + g)(x) (f + g)(2) (fg)(x) (fg)(−3).
and
and
and
and
#### Writing Exercises
Using your own words, explain the difference between a monomial, a binomial, and a trinomial.
Using your own words, explain the difference between a polynomial with five terms and a polynomial with a degree of 5.
Ariana thinks the sum is What is wrong with her reasoning?
Is every trinomial a second degree polynomial? If not, give an example.
#### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no – I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
### Glossary
binomial
A binomial is a polynomial with exactly two terms.
degree of a constant
The degree of any constant is 0.
degree of a polynomial
The degree of a polynomial is the highest degree of all its terms.
degree of a term
The degree of a term is the sum of the exponents of its variables.
monomial
A monomial is an algebraic expression with one term. A monomial in one variable is a term of the form where a is a constant and m is a whole number.
polynomial
A monomial or two or more monomials combined by addition or subtraction is a polynomial.
standard form of a polynomial
A polynomial is in standard form when the terms of a polynomial are written in descending order of degrees.
trinomial
A trinomial is a polynomial with exactly three terms.
polynomial function
A polynomial function is a function whose range values are defined by a polynomial. |
In this short video I show you what we mean by factorising.
The first step in factorising is:
ALWAYS check to see if the expression contains any common factors.
In this video, I demonstrate examples on factorising where there is a highest common factor (HCF).
Sometimes when there is no common factor, by grouping several of the terms together it is possible to factorise an expression.
Quadratic expressions have a distinct look about them. They have the form: ax2 + bx + c where a ≠ 0
When it comes to factorising a quadratic expression, there are particular methods depending on the form of the expression
In the tutorials which follow, I will show you how to factorise the above expressions.
## HCF types
Your first priority in factorising is to check to see if the quadratic expression has a common factor. In this tutorial I introduce you to what a quadratic expression is and then look at factorising quadratic expressions where there is a highest common factor (HCF).
Difference of two squares type (D.O.T.S)
Some quadratic expressions do not have a common factor but consist of two terms separated by a minus sign and each term is the square of something. This is known as the difference of 2 squares type. It has the form a^2 – b^2
## Trinomials
Some quadratic expressions have 3 terms (trinomials) and no common factor yet can still be factorised.
There are two methods of factorising such expressions. 1) by grouping 2) by inspection. Which method you choose is up to you but I generally prefer method 2 as it can be quicker but requires a bit more practice.
### Method 1
Trinomials Grouping Method
### Method 2
Trinomials Inspection Method
I now extend the work on factorising quadratic trinomials (3 terms) to ones where the first term is not x2 only. I would encourage you to look at all the videos as the thinking changes depending on the signs in the trinomial. This is not an easy topic and requires patience.
In the first of these tutorials I show you how to factorise a trinomial with all terms positive.
In the next tutorial I show you how to factorise a trinomial where the 2nd term is negative and remind you of an important stage in the 2nd example.
In the next tutorial I show you how to factorise a trinomial where the 3rd term is negative. An important one to try.
In the next tutorial I show you how to factorise a trinomial where the last 2 terms are negative.
In the last tutorial in this section I show you how to factorise a trinomial where the x2 term is not a prime number and so can be split into factors. This can often be a lot harder as there are more combinations to try.
Sometimes a quadratic trinomial can be disguised as in these two examples below. Check out the tutorial if you cannot factorise them. |
# Types and terms
## Sets and their elements
In mathematics you have seen many examples of sets and their elements. For example the real numbers `ℝ` is a set, and it has elements such as `37` and `12345`. It is difficult to give a formal definition of a set. Typically a student thinks of a set as a “collection of things”, and the elements of the set are the things.
Lean uses something called type theory as a foundation of mathematics rather than set theory. We will not be launching into a deep study of type theory in this course; the idea of this section is to give you a working knowledge of the key differences between type theory and set theory.
In Lean the type plays the role of the “collection of things”, and the things in the type are called terms. For example, in Lean the real numbers `ℝ` are a type, not a set, and specific real numbers like `37` and `12345` are called terms of this type.
The notation used is also different to what you have usually seen. In set theory, we write `37 ∈ ℝ` to mean that `37` is a real number. More formally we might say “`37` is an element of the set of real numbers”. In type theory the notation is different. In type theory we express the idea that `37` is a real number by writing `37 : ℝ`, and more formally we would say “`37` is a term of the type of real numbers”. Basically the colon `:` in type theory plays the role of the “is an element of” symbol `∈` in set theory.
## The universe of all types
Some of you might know that whilst it’s unproblematic to talk about the set of real numbers in set theory, it is problematic to talk about the set of all sets. Russell’s Paradox is the observation that if `X` is the set of all sets which are not elements of themselves, then `X ∈ X` if and only if `X ∉ X`, a contradiction. For similar reasons one cannot expect there to be a type of all types – this type of “self-referentiality” can lead to logical problems. In Lean, there is a universe of all types, and this universe is called `Type`. The statement that the real numbers are a type can be expressed as `ℝ : Type`.
Here is another example. If `G` is a group, and `g` is an element of this group, then in set theory one might say that `G` is a set and `g ∈ G` is an element of the set. In type theory one says instead that `G : Type` and that `g : G`.
The mental model which you should have in your mind is that in Lean, every object exists at one of three “levels”. There are universes, such as `Type`, there are types such as `ℝ` or `G` and there are terms such as `37` and `g`. Every mathematical object you know fits neatly into this heirarchy. For example rings, fields and topological spaces are all types in Lean, and their elements are terms.
## Function types
Say `X` and `Y` are types. The standard notation which you have seen for a function from `X` to `Y` is `f : X → Y`. This is also the notation used in Lean. A mathematician might write `Hom(X,Y)` for the set of all functions from `X` to `Y`. In Lean this set is of course a type, and the notation for this type is `X → Y`. So `f : X → Y` says that `f` is a term of type `X → Y`, i.e., the type theory version of the idea that `f` is an element of the set `Hom(X,Y)`, or equivalently that `f` is a function from `X` to `Y`.
## The universe Prop
Mathematicians define objects such as the real numbers and groups, but they also prove theorems about these objects. What is a theorem? It has two parts, a statement and a proof. Lean needs to be able to manipulate theorem statements and proofs as well as being able to manipulate objects such as the real numbers and groups. How do theorem statements and theorem proofs fit into the picture?
The answer to this question is beautifully simple. Lean regards a theorem statement as a type, not living in the `Type` universe, but in another universe called `Prop` – the universe of true-false statements. A true-false statement, otberwise known as a proposition in this course, is a statement such as `2+2=4` or `2+2=5` or the Riemann hypothesis. Note in particular that we are reclaiming the word “proposition” from its traditional usage in other mathematics courses. You might have seen the word “proposition” being used to mean the same thing as “lemma” or “theorem” or “corollary” or “sublemma” or… . We don’t need so many words to express the same idea, so in this course we will use the word “proposition” to mean the same thing as the logicians and the computer scientists: propositions, unlike theorems, can be false! A proposition is the same thing as a true-false statement. The notation `P : Prop` means that `P` is a proposition. For example `2 + 2 = 4 : Prop` and `2 + 2 = 5 : Prop`.
The idea that a proposition can be thought of as a type means in particular that a proposition has somehow got “elements”. This is not the way that true-false statements are usually thought of by mathematicians, but it is a key idea in Lean’s type theory. The “elements” (or, to use Lean’s language, the terms) of a proposition are its proofs! Every proposition in Lean has at most one term. The true propositions have one term, and the false propositions have no terms. To give a concrete example, we have `2 + 2 = 4 : Prop`, because 2+2=4 is a true-false statement. we will learn in this course how to make a term `h : 2 + 2 = 4`; this term `h` should be thought of as a proof that 2+2=4. You could read it as the hypothesis that `2 + 2 = 4` or however you like, but under the hood what is happening is that `h` is a term of the type `2 + 2 = 4`.
We also have the proposition `2 + 2 = 5 : Prop`. It is however impossible to make a term whose type is `2 + 2 = 5`, because 2+2=5 is a false proposition. If you like, you can think of `2 + 2 = 4` as a set with one element, and `2 + 2 = 5` as a set with no elements. This is initially a rather bizarre way of thinking about true-false statements, however you will soon get used to it.
The reason it is important to start thinking of elements of sets and proofs of propositions as “the same sort of thing”, is that when formalising mathematics one frequently runs into things like the type of non-negative real numbers. To give a term of this type is to give a pair `(x,h)` consisting of a real number `x` and a proof `h` that `x ≥ 0`, or, to put it in Lean’s language, a term `x : ℝ` and a term `h : x ≥ 0`. When doing mathematics like this in Lean, one just gets used to the fact that some variables are representing elements of sets and others are representing proofs of propositions.
## `P ⇒ Q` is `P → Q`
Here’s an interesting analogy.
In the usual set-theoretic language which mathematicians use, we might say the following: If `X` and `Y` are sets, then we can consider the set `Hom(X,Y)` of functions from `X` to `Y`, and an element `f ∈ Hom(X,Y)` is a function from `X` to `Y`.
In Lean’s type theory we say it like this: if `X` and `Y` are types in the `Type` universe, then we can consider the type `X → Y` of functions from `X` to `Y`, and a term `f : X → Y` of this type is a function from `X` to `Y`.
In usual mathematical logic, we might say the following: If `P` and `Q` are true-false statements, then `P ⇒ Q` is also a true-false statement (for example if `P` is true and `Q` is false, then `P ⇒ Q` is false). If we have a hypothesis `h` that says that `P ⇒ Q` is true, we might write `h : P ⇒ Q`.
In Lean’s type theory we say it like this. If `P` and `Q` are types in the `Prop` universe, i.e., propositions, then we can consider the type `P → Q` of functions from proofs of `P` to proofs of `Q`. If we have such a function `h`, which takes as input a proof of `P` and spits out a proof of `Q`, then `h` can be thought of as a proof that `P ⇒ Q`. In Lean the function type `P → Q` lives in the `Prop` universe – it’s also a true-false statement.
What Lean’s type theory is suggesting here is that an interesting model for a true/false statement is a set with at most one element. If the set has an element, it correponds to a true statement, and if it has no elements then it corresponds to a false statement.
As an exercise, imagine that `P` and `Q` are sets with either 0 or 1 element, and try and work out in each of the four cases the size of the set `Hom(P,Q)`, which in Lean we would write as `P → Q`. The answer you should get is that the size of `Hom(P,Q)` should be either 0 or 1, and it is 0 if `P ⇒ Q` is false, and 1 if `P ⇒ Q` is true. |
How do you solve equations with variables?
If the equation is in the form, ax + b = c, where x is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division. Solve 3y + 2 = 11. Subtract 2 from both sides of the equation to get the term with the variable by itself.
How do you solve an equation with two variables?
In a two-variable problem rewrite the equations so that when the equations are added, one of the variables is eliminated, and then solve for the remaining variable. Step 1: Multiply equation (1) by -5 and add it to equation (2) to form equation (3) with just one variable.
What are the 4 steps to solving an equation?
We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal.
What is the golden rule for solving equations?
Do unto one side of the equation, what you do to the other! When solving math equations, we must always keep the ‘scale’ (or equation) balanced so that both sides are ALWAYS equal.
How do you solve 3 variable equations?
Pick any two pairs of equations from the system. Eliminate the same variable from each pair using the Addition/Subtraction method. Solve the system of the two new equations using the Addition/Subtraction method. Substitute the solution back into one of the original equations and solve for the third variable.
What are 3 types of variables?
A variable is any factor, trait, or condition that can exist in differing amounts or types. An experiment usually has three kinds of variables: independent, dependent, and controlled. The independent variable is the one that is changed by the scientist.
What is the formula for a linear equation?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
What is the unknown variable?
An unknown is a variable in an equation which has to be solved for. An indeterminate is a symbol, commonly called variable, that appears in a polynomial or a formal power series. Formally speaking, an indeterminate is not a variable, but a constant in the polynomial ring or the ring of formal power series.
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# Advice 1: How to find the bisector of isosceles triangle
Have an isosceles triangle two sides equal, the angles at its base will also be equal. Therefore the bisectors drawn to the lateral sides are equal to each other. Bisector drawn to the base of an isosceles trianglewill be both a median and a height of this triangle.
Instruction
1
Let the bisector of AE is drawn to the base BC of isosceles triangle ABC. Triangle AEB will be rectangular, since the angle bisector AE will be its height. Side AB is the hypotenuse of this triangle, and BE and AE for his legs.By the Pythagorean theorem AB^2) = (YZ^2)+(AE^2). Then (BE^2) = sqrt (AB^2)-(AE^2)). Since AE and median of the triangle ABC, BE = BC/2. Therefore, (BE^2) = sqrt (AB^2)-((BC^2)/4)).If you specify the base angle ABC, then from rectangular triangle the angle bisector AE equal AE = AB/sin(ABC). Angle BAE = BAC/2, since AE is the bisector. Hence, AE = AB/cos(BAC/2).
2
Suppose now that the height of BK is conducted to the side AC. This height is neither a median nor a bisector of the triangle. To calculate its length there is a formula Stewart.The perimeter of a triangle is the sum of the lengths of all its sides is P = AB+BC+AC. And properiter equal to half the sum of the lengths of all its sides: P = (AB+BC+AC)/2 = (a+b+c)/2 where BC = a, AC = b, AB = c.Stuart the formula for the length of the bisector drawn to the side c (i.e., AB), will be of the form: l = sqrt(4abp(p-c))/(a+b).
3
From the formula of Stuart can be seen that the angle bisector drawn to the side b (AC), will have the same length as b = c.
# Advice 2: The angle bisector and its properties
The angle bisector has a number of properties. If properly harnessed, can solve tasks of different level of complexity. But even having data on all three bisectors, you can't build a triangle.
## What is a bisector
The study of the properties of triangles and solution of problems associated with them is an interesting process. It allows you to develop both the logic and spatial thinking. One of the important components of a triangle is a bisector. The bisector is a segment that goes from the corner of the triangle and divides it into equal parts.
In many geometry problems in conditions provides information about the bisector, thus it is required to find the angle or length of the opposite side and so on. In other tasks it is necessary to find the parameters of the bisecting line. To determine the correct response to any of the tasks associated with the bisector, you need to know its properties.
## The properties of bisectors
First, the bisector is the locus of points that are removed at equal distances from the sides adjacent to the corner.
Second, the angle bisector divides the opposite corner side into segments that are proportional to the adjacent sides. For example, there is a triangle ABS, it out of the corner B enters the bisector that connects the vertex of the angle with the point M on the adjacent side AC. After the analysis, have the formula: AM/MS=AB/BS.
Thirdly, the point which is the intersection of the bisectors of all the angles of a triangle, acts as the center of the circle inscribed in this triangle.
Fourth, if two of the bisectors of one triangle are equal, then the triangle is isosceles.
Fifth, if there are data about all three bisectors, it is impossible to build a triangle, even if you use a compass.
Often to solve the problem bisector is unknown, it is necessary to find its length. To solve the problem, you need to know the angle from which it comes, and the lengths of the sides adjacent to it. In this case, the length of the bisector is equal to twice the product of the adjacent sides to the cosine of the angle divided by half the sum of the lengths of adjacent sides.
## Right triangle
In a right triangle the angle bisector has the same properties as in the normal. But adds the additional property that the bisector of the right angle forms a crossing angle of 45 degrees. Moreover, an isosceles right triangle, the bisector, which is lowered onto the base, will also act as altitude and median.
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# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18
In the following lectures, we will be concerned with integration theory, i.e. we want to study and to compute the area of a surface which is bounded by the graph of a function (the integrand)
${\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }$
and the ${\displaystyle {}x}$-axis. At the same time, there is a direct relation with finding a primitive function of ${\displaystyle {}f}$, these are functions such that their derivative equals ${\displaystyle {}f}$. The concept of the area of a surface itself is problematic, which is understood thoroughly within measure theory. However, this concept can be understood from an intuitive perspective, and we will only use some basic facts. These are only used for motivation, and not as arguments. The starting point is that the area of a rectangle is just the product of the side lengths, and that the area of a surface, which one can exhaust with rectangles, equals the sum of the areas of these rectangles. We will work with the Riemann integral, which provides a satisfactory theory for continuous functions. Here, all rectangles are parallel to the coordinate system, their width (on the ${\displaystyle {}x}$-axis) can vary and their height (length) is in relation to the value of the function over the base. By this method, the functions are approximated by so-called step functions.
Step functions
## Definition
Let ${\displaystyle {}I}$ be a real interval with endpoints ${\displaystyle {}a,b\in \mathbb {R} }$. Then a function
${\displaystyle t\colon I\longrightarrow \mathbb {R} }$
is called a step function, if there exists a partition
${\displaystyle {}a=a_{0}
of ${\displaystyle {}I}$ such that ${\displaystyle {}t}$ is constant
on every open interval ${\displaystyle {}]a_{i-1},a_{i}[}$.
This definition does not require a certain value at the partition points. We call the interval ${\displaystyle {}]a_{i-1},a_{i}[}$ the ${\displaystyle {}i}$-th interval of the partition, and ${\displaystyle {}a_{i}-a_{i-1}}$ is called the length of this interval. If the lengths of all intervals are constant, then the partition is called an equidistant partition.
## Definition
Let ${\displaystyle {}I}$ be a real interval with endpoints ${\displaystyle {}a,b\in \mathbb {R} }$, and let
${\displaystyle t\colon I\longrightarrow \mathbb {R} }$
denote a step function for the partition ${\displaystyle {}a=a_{0}, with the values ${\displaystyle {}t_{i}}$, ${\displaystyle {}i=1,\ldots ,n}$. Then
${\displaystyle {}T:=\sum _{i=1}^{n}t_{i}(a_{i}-a_{i-1})\,}$
is called the step integral of ${\displaystyle {}t}$ on ${\displaystyle {}I}$.
We denote the step integral also by ${\displaystyle {}\int _{a}^{b}t(x)\,dx}$. If we have an equidistant partition of interval length ${\displaystyle {}{\frac {b-a}{n}}}$, then the step integral equals $\displaystyle {{}} \frac{b-a}{n} { \left( \sum_{i [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18]] __NOINDEX__ 1}^n t_i \right) }$ . The step integral does not depend on the partition chosen. As long as we have a step function with respect to the partition, one can pass to a refinement of the partition.
## Definition
Let ${\displaystyle {}I}$ denote a bounded interval, and let
${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
denote a function. Then a step function
${\displaystyle t\colon I\longrightarrow \mathbb {R} }$
is called a step function from above for ${\displaystyle {}f}$, if ${\displaystyle {}t(x)\geq f(x)}$ holds for all ${\displaystyle {}x\in I}$. A step function
${\displaystyle s\colon I\longrightarrow \mathbb {R} }$
is called a step function from below for ${\displaystyle {}f}$, if ${\displaystyle {}s(x)\leq f(x)}$ holds for all
${\displaystyle {}x\in I}$.
A step function from above (below) for ${\displaystyle {}f}$ exists if and only if ${\displaystyle {}f}$ is bounded from above (from below).
## Definition
Let ${\displaystyle {}I}$ denote a bounded interval, and let
${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
denote a function. For a step function from above
${\displaystyle t\colon I\longrightarrow \mathbb {R} }$
of ${\displaystyle {}f}$, with respect to the partition ${\displaystyle {}a_{i}}$, ${\displaystyle {}i=0,\ldots ,n}$, and values ${\displaystyle {}t_{i}}$, ${\displaystyle {}i=1,\ldots ,n}$, the step integral
${\displaystyle {}T:=\sum _{i=1}^{n}t_{i}{\left(a_{i}-a_{i-1}\right)}\,}$
is called a step integral from above for ${\displaystyle {}f}$ on ${\displaystyle {}I}$.
## Definition
Let ${\displaystyle {}I}$ denote a bounded interval, and let
${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
denote a function. For a step function from below
${\displaystyle t\colon I\longrightarrow \mathbb {R} }$
of ${\displaystyle {}f}$, with respect to the partition ${\displaystyle {}a_{i}}$, ${\displaystyle {}i=0,\ldots ,n}$, and values ${\displaystyle {}s_{i}}$, ${\displaystyle {}i=1,\ldots ,n}$, the step integral
${\displaystyle {}S:=\sum _{i=1}^{n}s_{i}{\left(a_{i}-a_{i-1}\right)}\,}$
is called a step integral from below for ${\displaystyle {}f}$ on ${\displaystyle {}I}$.
Different step functions from above yield different step integrals from above.
For further integration concepts, we need the following definitions which refer to arbitrary subsets of the real numbers.
## Definition
For a nonempty subset ${\displaystyle {}M\subseteq \mathbb {R} }$, an upper bound ${\displaystyle {}T}$ of ${\displaystyle {}M}$ is called the supremum of ${\displaystyle {}M}$, if ${\displaystyle {}T\leq S}$
holds for all upper bounds ${\displaystyle {}S}$ of ${\displaystyle {}M}$.
## Definition
For a nonempty subset ${\displaystyle {}M\subseteq \mathbb {R} }$, a lower bound ${\displaystyle {}t}$ of ${\displaystyle {}M}$ is called the infimum of ${\displaystyle {}M}$, if ${\displaystyle {}t\geq s}$
holds for all lower bounds ${\displaystyle {}s}$ of ${\displaystyle {}M}$.
The existence of infimum and supremum follows from the completeness of the real numbers.
## Theorem
Every nonempty subset of the real numbers, which is bounded from above, has a supremum in ${\displaystyle {}\mathbb {R} }$.
### Proof
This proof was not presented in the lecture.
${\displaystyle \Box }$
## Definition
Let ${\displaystyle {}I}$ denote a bounded interval, and let
${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
denote a function, which is bounded from above. Then the infimum of all step integrals of step functions from above
of ${\displaystyle {}f}$ is called the upper integral of ${\displaystyle {}f}$.
## Definition
Let ${\displaystyle {}I}$ denote a bounded interval, and let
${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
denote a function, which is bounded from below. Then the supremum of all step integrals of step functions from below
of ${\displaystyle {}f}$ is called the lower integral of ${\displaystyle {}f}$.
The boundedness from below makes sure that there exists at all a step function from below, so that the set of step integrals from below is not empty. This condition alone does not guarantee that a supremum exists. However, if the function is bounded from both sides, then the upper integral and the lower integral exist. If a partition is given, then there exists a smallest step function from above (a largest from below) which is given by the suprema (infima) of the function on the intervals of the partition. For a continuous function on a closed interval, these are maxima and minima. To compute the integral, we have to look at all step functions for all partitions.
Riemann-integrable functions
In the following, we will talk about compact interval, which is just a bounded and closed interval, hence of the form ${\displaystyle {}I=[a,b]}$ with ${\displaystyle {}a,b\in \mathbb {R} }$.
## Definition
Let ${\displaystyle {}I}$ denote a compact interval and let
${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
denote a function. Then ${\displaystyle {}f}$ is called Riemann-integrable if the upper integral and the lower integral
of ${\displaystyle {}f}$ exist and coincide.
It might by historically more adequate to call this Darboux-integrable.
## Definition
Let ${\displaystyle {}I=[a,b]}$ denote a compact interval. For a Riemann-integrable function
$\displaystyle f \colon I [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18]] __NOINDEX__ [a,b] \longrightarrow \R , t \longmapsto f(t) ,$
we call the upper integral of ${\displaystyle {}f}$ (which by definition coincides with the lower integral) the definite integral of ${\displaystyle {}f}$ over ${\displaystyle {}I}$. It is denoted by
${\displaystyle \int _{a}^{b}f(t)\,dt{\text{ or by }}\int _{I}^{}f(t)\,dt.}$
The computation of such integrals is called to integrate. Don't think too much about the symbol ${\displaystyle {}dt}$. It expresses that we want to integrate with respect to this variable. The name of the variable is not relevant, we have
${\displaystyle {}\int _{a}^{b}f(t)\,dt=\int _{a}^{b}f(x)\,dx\,.}$
## Lemma
Let ${\displaystyle {}I}$ denote a compact interval, and let
${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
denote a function. Suppose that there exists a sequence of lower step functions ${\displaystyle {}{\left(s_{n}\right)}_{n\in \mathbb {N} }}$ with ${\displaystyle {}s_{n}\leq f}$ and a sequence of upper step functions ${\displaystyle {}{\left(t_{n}\right)}_{n\in \mathbb {N} }}$ with ${\displaystyle {}t_{n}\geq f}$. Suppose furthermore that the corresponding sequences of step integrals converge to the same real number. Then ${\displaystyle {}f}$ is Riemann-integrable, and the definite integral equals this limit, so
${\displaystyle {}\lim _{n\rightarrow \infty }\int _{a}^{b}s_{n}(x)\,dx=\int _{a}^{b}f(x)\,dx=\lim _{n\rightarrow \infty }\int _{a}^{b}t_{n}(x)\,dx\,.}$
### Proof
${\displaystyle \Box }$
## Example
We consider the function
${\displaystyle f\colon [0,1]\longrightarrow \mathbb {R} ,t\longmapsto t^{2},}$
which is strictly increasing in this interval. Hence, for a subinterval ${\displaystyle {}[a,b]\subseteq [0,1]}$, the value ${\displaystyle {}f(a)}$ is the minimum, and ${\displaystyle {}f(b)}$ is the maximum of the function on this subinterval. Let ${\displaystyle {}n}$ be a positive natural number. We partition the interval ${\displaystyle {}[0,1]}$ into the ${\displaystyle {}n}$ subintervals ${\displaystyle {}\left[i{\frac {1}{n}},(i+1){\frac {1}{n}}\right]}$, ${\displaystyle {}i=0,\ldots ,n-1}$, of length ${\displaystyle {}{\frac {1}{n}}}$. The step integral for the corresponding lower step function is
$\displaystyle {{}} \sum_{i [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18]] __NOINDEX__ 0}^{n-1} \frac{1}{n} { \left(i \frac{1}{n}\right) }^2 = \frac{1}{n^3} \sum_{i [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18]] __NOINDEX__ 0}^{n-1} i^2 = \frac{1}{n^3} { \left( \frac{1}{3} n^3 - \frac{1}{2}n^2 + \frac{1}{6} n\right) } = \frac{1}{3} - \frac{1}{2n} + \frac{1}{6n^2} \,$
(see Exercise 2.10 for the formula for the sum of the squares). Since the sequences ${\displaystyle {}{\left(1/2n\right)}_{n\in \mathbb {N} }}$ and ${\displaystyle {}{\left(1/6n^{2}\right)}_{n\in \mathbb {N} }}$ converge to ${\displaystyle {}0}$, the limit for ${\displaystyle {}n\rightarrow \infty }$ of these step integrals equals ${\displaystyle {}{\frac {1}{3}}}$. The step integral for the corresponding step function from above is
$\displaystyle {{}} \sum_{i [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18]] __NOINDEX__ 0}^{n-1} \frac{1}{n} { \left((i+1) \frac{1}{n}\right) }^2 = \frac{1}{n^3} \sum_{i [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18]] __NOINDEX__ 0}^{n-1} (i+1)^2 = \frac{1}{n^3} \sum_{j [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18]] __NOINDEX__ 1}^{n} j^2 = \frac{1}{n^3} { \left(\frac{1}{3} n^3 + \frac{1}{2}n^2 + \frac{1}{6} n\right) } = \frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2} \, .$
The limit of this sequence is again ${\displaystyle {}{\frac {1}{3}}}$. By Lemma 18.13 , the upper integral and the lower integral coincide, hence the function is Riemann-integrable, and for the definite integral we get
${\displaystyle {}\int _{0}^{1}t^{2}\,dt={\frac {1}{3}}\,.}$
## Lemma
Let ${\displaystyle {}I=[a,b]}$ be a compact interval, and let
${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
be a function. Then the following statements are equivalent.
1. The function ${\displaystyle {}f}$ is Riemann-integrable.
2. There exists a partition ${\displaystyle {}a=a_{0}, such that the restrictions ${\displaystyle {}f_{i}:=f|_{[a_{i-1},a_{i}]}}$ are Riemann-integrable.
3. For every partition ${\displaystyle {}a=a_{0}, the restrictions ${\displaystyle {}f_{i}:=f|_{[a_{i-1},a_{i}]}}$ are Riemann-integrable.
In this situation, the equation
$\displaystyle {{}} \int_{ a }^{ b } f ( t) \, d t = \sum_{i [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18]] __NOINDEX__ 1}^n \int_{ a_{i-1} }^{ a_i } f_i ( t) \, d t \,$
holds.
### Proof
${\displaystyle \Box }$
## Definition
Let ${\displaystyle {}f\colon I\rightarrow \mathbb {R} }$ be a function on a real interval. Then ${\displaystyle {}f}$ is called Riemann-integrable, if the restriction of ${\displaystyle {}f}$ to every compact interval ${\displaystyle {}[a,b]\subseteq I}$ is
Riemann-integrable.
Due to this lemma, both definitions coincide for a compact interval ${\displaystyle {}[a,b]}$. The integrability of a function ${\displaystyle {}f\colon \mathbb {R} \rightarrow \mathbb {R} }$ does not mean that ${\displaystyle {}\int _{\mathbb {R} }f(x)dx}$ has a meaning or exists.
Riemann-integrability of continuous functions
## Theorem
Let ${\displaystyle {}f\colon I\rightarrow \mathbb {R} }$ denote a continuous function. Then ${\displaystyle {}f}$ is Riemann-integrable.
### Proof
This proof was not presented in the lecture.
${\displaystyle \Box }$
## Lemma
Let ${\displaystyle {}I=[a,b]}$ denote a compact interval, and let ${\displaystyle {}f,g\colon I\rightarrow \mathbb {R} }$ denote Riemann-integrable
functions. Then the following statements hold.
1. If ${\displaystyle {}m\leq f(x)\leq M}$ holds for all ${\displaystyle {}x\in I}$, then ${\displaystyle {}m(b-a)\leq \int _{a}^{b}f(t)\,dt\leq M(b-a)}$ holds.
2. If ${\displaystyle {}f(x)\leq g(x)}$ holds for all ${\displaystyle {}x\in I}$, then ${\displaystyle {}\int _{a}^{b}f(t)\,dt\leq \int _{a}^{b}g(t)\,dt}$ holds.
3. The sum ${\displaystyle {}f+g}$ is Riemann-integrable, and the identity ${\displaystyle {}\int _{a}^{b}(f+g)(t)\,dt=\int _{a}^{b}f(t)\,dt+\int _{a}^{b}g(t)\,dt}$ holds.
4. For ${\displaystyle {}c\in \mathbb {R} }$ we have ${\displaystyle {}\int _{a}^{b}(cf)(t)\,dt=c\int _{a}^{b}f(t)\,dt}$.
5. The functions ${\displaystyle {}{\max {\left(f,g\right)}}}$ and ${\displaystyle {}{\min {\left(f,g\right)}}}$ are Riemann-integrable.
6. The function ${\displaystyle {}\vert {f}\vert }$ is Riemann-integrable.
7. The product ${\displaystyle {}fg}$ is Riemann-integrable.
### Proof
For (1) to (4) see Exercise 18.14 . For (5) see Exercise 18.17 . (6) follows directly from (5), because of ${\displaystyle {}\vert {f}\vert ={\max {\left(f,-f,\right)}}}$. For (7), see Exercise 18.18 .
${\displaystyle \Box }$
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# Converting decimal to binary
Problem:
Is it possible to partition the numbers 1-100 into two sets, such that no number in either set will be double any other number in the same group?
Solution:
There are many solutions. Here's one: Try to put every number in the first set, and only put a number in the second set if you have to. You'll find that all the odd numbers will appear in the first set. So we've placed every odd number. Now we have to place all the even numbers. Every even number can be expressed as even * odd. But we also know that odd * odd = odd, so we only need to figure out how to distribute the even multiples of odd numbers. All the doubles of the odd numbers must go in the second set. The quadruples of the odd numbers, therefore must go in the first set. The sextuples in the first set, and so on, alternating in this way. We can express every whole number $$m,$$ in the form $$m = 2^kd,$$ where $$d$$ is some odd number. When $$k$$ is even, we put $$m$$ in the first set, and when $$k$$ is odd, we put $$m$$ in the second set. The number $$2^k,$$ in binary, is a 1, followed by $$k$$ zeros. Thus, $$2^kd$$ is $$d$$ followed by $$k$$ zeros. Thus, we have another way of deciding the set to which $$m$$ belongs. If the binary representation of $$m$$ has an even number of trailing zeros, put $$m$$ in the first set. If it has an odd number of trailing zeros, put $$m$$ in the second set.
Going further:
What if, instead of avoiding doubles, we wanted to avoid triples? quadruples? What's the general strategy, if any? |
# JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3
Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.
## JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.3
Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the media, mean and mode of the data and compare them.
Monthly consumption (in units) No. of consumers 65 – 85 4 85 – 105 5 105 – 125 13 125 – 145 20 145 – 165 14 165 – 185 8 185 – 205 4
Solution:
Mean is 137 units.
Median is 137 units.
Mode is 135.76 units.
The three measures are approximately the same.
Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
Class interval Frequency 0 – 10 5 10 – 20 x 20 – 30 20 30 – 40 15 40 – 50 y 50 – 60 5 Total 60
Solution:
Class interval Frequency Cumulative frequency 0 – 10 5 5 10 – 20 x 5 + x 20 – 30 20 25 + x 30 – 40 15 40 + x 40 – 50 y 40 + x + y 50 – 60 5 45 + x + y 60
n = 60, 45 + x + y = 60
x + y = 60 – 45
x + y = 15
The median is 28.5. It lies in the class interval 20 – 30.
∴ l = 20, f = 20, cf = 5 + x, h = 10
Question 3.
A life insurance agent found the following data for distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.
Age (in years) No. of policyholders Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100
Solution:
Class interval No. of policyholders c.f. Below 20 2 2 20 – 25 4 6 25 – 30 18 24 30 – 35 21 45 35 – 40 33 78 40 – 45 11 89 45 – 50 3 92 50 – 55 6 98 55 – 60 2 100 n = 100 $$\frac{n}{2}$$ = 50
l = 35, $$\frac{n}{2}$$ = 50, cf = 45, f = 33, h = 5
Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
Length (in mm) Number of leaves 118 – 126 3 127 – 135 5 136 – 144 9 145 – 153 12 154 – 162 5 163 – 171 4 172 – 180 2
Find the median length of the leaves.
(Hint: The data need to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, ….., 171.5 – 180.5).
Solution:
The data have to be converted to continuous classes for finding the median since the formula. assumes continuous classes.
Class interval No. of leaves Cumulative frequency (cf) 117.5 – 126.5 3 3 126.5 – 135.5 5 8 135.5 – 144.5 9 17 144.5 – 153.5 12 29 153.5 – 162.5 5 34 162.5 – 171.5 4 38 171.5 – 180.5 2 40
n = 40, $$\frac{n}{2}$$ = 20
The median lies in the class interval 144.5 – 153.5.
l = 144.5, $$\frac{\mathrm{n}}{2}=\frac{40}{2}$$ = 20, cf = 17, f = 12, h = 9.
Median = l + $$\left[\frac{\frac{n}{2}-c f}{f}\right]$$ × h
= 144.5 + $$\left[\frac{20-17}{12} \times 9\right]$$
= 144.5 + $$\left[\frac{27}{12}\right]$$
= 144.5 + 2.25
= 146.75 mm.
Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:
Lifetime (in hours) Number of lamps 1500 – 2000 14 2000 – 2500 56 2500 – 3000 60 3000 – 3500 86 3500 – 4000 74 4000 – 4500 62 4500 – 5000 48
Find the median lifetime of a lamp.
Solution:
Lifetime in hours (CI) No. of lamps (l) Cumulative frequency 1500 – 2000 14 14 2000 – 2500 56 70 2500 – 3000 60 130 3000 – 3500 86 216 3500 – 4000 74 290 4000 – 4500 62 352 4500 – 5000 48 400
The median lies in the class interval 3000 – 3500.
$$\frac{\mathrm{n}}{2}=\frac{400}{2}=200$$
l = 3000, $$\frac{n}{2}$$ = 200, cf = 130, f = 86, h = 500.
Median = l + $$\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]$$ × h
= 3000 + $$\left[\frac{200-130}{86}\right]$$ × 500
= 3000 + $$\left[\frac{70}{86}\right]$$ × 500
= 3000 + $$\frac{35000}{86}$$
= 3000 + 406.976
Median life of a lamp is 3406.98 hours.
Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:
No. of letters No. of surnames 1 – 4 6 4 – 7 30 7 – 10 40 10 – 13 16 13 – 16 4 16 – 19 4
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:
Hence the modal size of the surnames is 7.88.
Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Solution:
Weight in kg. No. of students Cumulative frequency (c.f.) 40 – 45 2 2 45 – 50 3 5 50 – 55 8 13 55 – 60 6 19 60 – 65 6 25 65 – 70 3 28 70 – 75 2 30
$$\frac{n}{2}$$ = 15
The median lies in the class 55 – 60.
l = 55, $$\frac{n}{2}$$ = 15, c.f. = 13, f = 6, h = 5.
Median = l + $$\frac{1}{2}$$ × h
= 55 + $$\left[\frac{15-13}{6}\right]$$ × 5
= 55 + $$\frac{2}{6}$$ × 5
= 55 + $$\frac{5}{3}$$
= 55 + 1.666 = 56.666
∴ Median = 56.67 kg.
Hence, the median weight of the students is 56.67 kg. |
Area of a trapezoid. The area of triangle formula is given as. Radius = r = 3 Area = Ï × r 2 = Ï × 3 2 = Ï × (3 × 3) = 3.14159... × 9 = 28.27 (to 2 decimal places) Example: What is the area of this triangle? What are called as vertices of a triangle? So our area of our original triangle is one half base times height. Area of a rhombus. The area of a triangle depends upon the type of triangle. Lesson Quiz: Part I. The formula is: Area = w × h w = width h = height. Area of a regular polygon. There are many types of quadrilaterals, each having its own properties and formula of area. Now we will learn this formula. Enter the values of A, B, C, or drag the vertices of the triangle and see how the area changes for different values. We are providing you with the quadrilateral area, formula, types ⦠Construction of triangles - I Construction of triangles - II. The area of Triangle B is 4.5 square units because $$\frac{1}{2}\cdot 3\cdot 3=4.5$$. The meeting point of any two line segments, we call it as a vertex of the triangle. For a triangle with base b b b and height h h h, the area A A A is given by. Area of a rhombus. Knowing Two Sides and the Included Angle. Then the formula for area of scalene triangle is given as \sqrt { s\times \left( s-a \right) \times \left( s-b \right) \times \left( s-c \right) } square units. $\endgroup$ â user137731 Apr 30 '15 at 13:30 Area of a triangle (Heron's formula) Area of a triangle given base and angles. Area = ½ × b × h = ½ × 20 × 12 = 120 . Area of a cyclic quadrilateral. An easy to use, free area calculator you can use to calculate the area of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, octagon, and sector of a circle. Related Topics: More Geometry Lessons In these lessons, we have compiled. To find the area of the triangle on the left, substitute the base and the height into the formula for area. There's also a formula to find the area of any triangle when we know the lengths of all three of its sides. How to Calculate the Area of a Triangle Using Heron's Formula? Definitions and formulas for the area of a triangle, the sum of the angles of a triangle, the Pythagorean theorem, Pythagorean triples and special triangles (the 30-60-90 triangle and the 45-45-90 triangle) The Sum of any two sides of a triangle is always greater than the third side. GEOMETRY. He also extended it to the area of quadrilaterals and higher-order polygons. Formulas, explanations, and graphs for each calculation. If you're seeing this message, it means we're having trouble loading external resources on our website. Use the calculator on below to calculate the area of a triangle given 3 sides using Heron's formula. Home List of all formulas of the site; Geometry. Quad means four and hence the name quadrilateral is given to the closed shapes with four sides. We know w = 5 and h = 3, so: Area = 5 × 3 = 15. Area of a rectangle. And the area of the parallelogram was equal to its base times its height. Area of a cyclic quadrilateral. A triangle may be regarded as a quadrilateral with one side of length zero. Area of a square. Types of angles Types of triangles. Area of a parallelogram given base and height. AB = c, BC = a, CA = b are the sides of the ÎABC ( triangle ABC). Height = h = 12. Heron was one of the great mathematicians of antiquity and came up with this formula sometime in the first century BC, although it may have been known earlier. Area and perimeter. Developing Formulas for . Area of a triangle calculation using all different rules, side and height, SSS, ASA, SAS, SSA, etc. Terminology and Formulas of the Triangles: Triangle Sides : The three lines segments that form the triangle area called sides of triangle. Find the area of the following triangle. So, an equilateral triangleâs area can be calculated if the length of its side is known. The step-by-step explanation of using Heron's formula is shown using an example below. An extension of Heron's triangle area formula to quadrilaterals was discovered by the Hindu geometer Brahmagupta (620 ⦠Using coordinate geometry, it is possible to find the distance between two points, dividing lines in a ratio, finding the mid-point of a line, calculating the area of a triangle in the Cartesian plane, etc. Area of Triangle Formula. The Geometry of Triangles. So hopefully that makes you feel pretty good about this formula that you will see in geometry, that area of a triangle is one half base times height, while the area of a rectangle or a paralleogram is ⦠But there was no general formula for the quadrilateral's area. Area of a quadrilateral. i.e a + b > c , b + c > a , c + a > b. Vertex : The area of an equilateral triangle is the amount of space that it occupies in a 2-dimensional plane. This can be found on the Heron's Formula page. Let's derive the formula for the area of a triangle when the coordinates of its vertices are given. A = 182 x 2 mm 2. h = 9.1 x mm. For a triangle, its area can be calculated using the formula: How to find the area of triangles, rectangles, parallelograms, rhombuses and trapeziums. If we know the length of all sides of triangle & quadrilateral & also known the length of a diagonal of the quadrilateral. I am not even sure if making 2 triangles is the best way to get the area of the quadrilateral. The quadrilateral area formulas are as follows: Note: The median of [â¦] Holt Geometry. Example: What is the area of this circle? In this section, you will learn how to find area of triangles and quadrilaterals. Coordinate geometry is defined as the study of geometry using the coordinate points on the plane with any dimension. The formula for area of triangle with vertices comes in coordinate geometry. Base = b = 20. Solution. Develop and apply the formulas for the areas of triangles and special quadrilaterals. General formulas are given for all types of triangles, special cases for equilateral, isosceles and right triangles. Area of a quadrilateral Area of a rectangle. Triangles and Quadrilaterals . Quadrilateral Area, Formula, Types, & Examples. When we know two sides and the included angle (SAS), there is another formula (in fact three equivalent formulas) we can use. Properties of parallelogram. or . Let's derive the formula for the area of a triangle when the coordinates of its vertices are given. Simplify. Area of a square. This formula generalizes Heron's formula for the area of a triangle. The area of Triangle C is 24 square units because $$\frac{1}{2}\cdot 12\cdot 4=24$$. The area of Triangle A is 15 square units because $$\frac{1}{2}\cdot 5\cdot 6=15$$. We just substitute the values of the sides of the triangle $$a,b$$, and $$c$$ in the Heron's formula to find its area. Heron's formula is used to find the area of any type of triangles or quadrilaterals. The area of this quadrilateral can be calculated by summing up the area of its four triangles. If A is the area of the quadrilateral and T is the area of triangle PBC. Description of Task Multiple Choice Questions based on Herons formula. a more detailed explanation (in text and video) of each area formula. Area of a parallelogram given sides and angle. $\endgroup$ â Rubab Apr 30 '15 at 13:06 $\begingroup$ You can definitely break it into two triangles. A = 1 2 b × h. A = \frac{1}{2} b \times h.\ _\square A = 2 1 b × h. Observe that this is exactly half the area of a rectangle which has the same base and height. But in the case of equilateral triangles, where all three sides are the same length, there is a simpler formula: where s is the length of any side of the triangle. Please check the visualization of the area of a triangle in coordinate geometry. Properties of triangle. Example. Depending on which sides and angles we know, the formula ⦠Totally for a triangle there exist 3 ⦠A quadrilateral is a 2-dimensional shape containing 4 sides. Home » Derivation of Formulas » Formulas in Plane Geometry Derivation of Formula for Area of Cyclic Quadrilateral For a cyclic quadrilateral with given sides a, b, c, and d, the formula for the area ⦠Substitute. Area of a triangle (Heron's formula) Area of a triangle given base and angles. Find each measurement. Construction of triangles - III. 1. the height of the parallelogram, in which. ... Mensuration formulas. These formulas we will learn in geometry. Calculator. Area of a triangle; Area of a right triangle; Heron's formula for area; Area of an isosceles triangle ; Area of an equilateral triangle; Area of a triangle - "side angle side" (SAS) method; Area of a triangle - "side and two angles" (AAS or ASA) method; Area of a square; Area of a rectangle ; Area of a parallelogram ; Area ⦠Area = \frac{1}{2} (base \cdot height) \\ =\frac{1}{2} (12 ⦠To ï¬nd the area of.dont need to memorize the formula in junior high. To recall, an equilateral triangle is a triangle in which all the sides are equal and the measure of all the internal angles is 60°. Area of a triangle given sides and angle. Heron's formula can be used to express the area of triangle PBC 2. Area of Triangles: The following triangle area formula was developed nearly.To ï¬nd the area of quadrilateral by dividing them into two triangle. a table of area formulas and perimeter formulas used to calculate the area and perimeter of two-dimensional geometrical shapes: square, rectangle. This is the most common formula used and is likely the first one that you have seen. Above are formulas and a calculator that will help calculate the area of a triangle or check already performed calculations. Area of Triangle = A = ½ (b × h) square units. Now, 1. Area of a trapezoid. Formula for area of scalene triangle as follows: We know that for a scalene triangle, all the three sides will be different. If you want to do it that way I'd suggest using Heron's formula to find the area of each triangle. The similarity of triangles PBC and PAD can be use to effect various (but tedious) substitutions. Let us assume the sides of the triangle to be: \[a=5\\[0.1cm] ⦠Area of a parallelogram given sides and angle. Area of plane shapes. Volume. 9-1. Formula for area of a rhombus . parallelogram, trapezoid (trapezium), triangle, rhombus, kite, regular polygon, circle, and ellipse. ⦠where b is the base and h is the height of the triangle. Let the three sides of scalene triangle be âaâ, âbâ and âcâ units. 2. the perimeter of a rectangle in ⦠Note that is constant that has the value of approximately 0.433, so the formula simplifies a little to Methods for finding triangle area Various formulas are used to calculate the area of a triangle, depending on the known input data. Area of a parallelogram given base and height. Sum of the angle in a triangle is 180 degree. The formula for the area of a triangle is $$\dfrac{1}{2}\times\text{base}\times\text{altitude}$$. We know that triangle consists of 3 line segments. There are five formulas that you can use to calculate the area of the seven special quadrilaterals. With appropriate perseverance and and algebraic ⦠From this perspective, as d approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula. 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Regular polygon, circle, and ellipse shapes with four sides a 2-dimensional shape 4! To express the area of triangle of Task Multiple Choice Questions based Herons... Bc = a = ½ ( b × h = 3, so: area = w × =. Heron 's formula is shown using an example below in these Lessons, we compiled... Greater than the third side height h h h, the area of triangle 's derive the formula used! Is defined as the study of geometry using the coordinate points on Heron! You can use to calculate the area of a triangle given base and angles, +... & quadrilateral & also known the length of its side is known it that way I 'd suggest using 's! } { 2 } \cdot 3\cdot 3=4.5\ ) use to calculate the and! Third side plane with any dimension ( triangle ABC ) may be regarded as a vertex of the triangle angles! Having its own properties and formula of area, trapezoid ( trapezium ), triangle, the..., circle, and ellipse suggest using Heron 's formula is shown using example! Have compiled explanation of using Heron 's formula can be found on the Heron 's formula be. Different rules, side and height, SSS, ASA, SAS SSA. Bc = a, c + a > b. vertex for area of triangle c is 24 units... Calculator on below to calculate the area of this circle: the three sides will be.. To ï¬nd the area of triangle c is 24 area of quadrilateral triangle formula units because \ ( \frac { 1 {! Third side you 're seeing this message, it means we 're having trouble loading resources. B > c, BC = a, CA = b are the of. 5 and h is the base and angles 20 × 12 = 120 calculate! What is the base and angles triangle b is 4.5 square units calculation all... = b are the sides of triangle = a = 182 x 2 mm 2. h = ½ × ×... Any dimension: More geometry Lessons in these Lessons, we call it as a quadrilateral with one of. I.E a + b > c, BC = a, CA = b are the of... Base b b and height, SSS, ASA, SAS, SSA, etc area of quadrilateral triangle formula formula... Called sides of triangle & quadrilateral & also known the length of a diagonal of the,... 'Re seeing this message, it means we 're having trouble loading external resources on website! Three sides of the angle in a triangle with vertices comes in coordinate.! |
# 1504 and Level 3
### Today’s Puzzle:
Since this is a level 3 puzzle the clues are given in a logical order from top to bottom. Write the factors 1 to 10 in the first column and again in the top row.
Usually, you only have to consider the previous clues when finding the factors in a level 3 puzzle, but when you consider if 4 = 2 × 2 or 1 × 4, you will also have to look at a clue below it. You can do this!
### Factors of 1504:
• 1504 is a composite number.
• Prime factorization: 1504 = 2 × 2 × 2 × 2 × 2 × 47, which can be written 1504 = 2⁵ × 47.
• 1504 has at least one exponent greater than 1 in its prime factorization so √1504 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1504 = (√16)(√94) = 4√94.
• The exponents in the prime factorization are 5 and 1. Adding one to each exponent and multiplying we get (5 + 1)(1 + 1) = 6 × 2 = 12. Therefore 1504 has exactly 12 factors.
• The factors of 1504 are outlined with their factor pair partners in the graphic below.
### More about the Number 1504:
1504 is the difference of two squares in four different ways:
377² – 375² = 1504
190² – 186² = 1504
98² – 90² = 1504
55² – 39² = 1504 |
DIRECTIONS FOR THE TRUSTEE AND PERSONAL REPRESENTATIVE
Document Sample
``` Section 4: Annuities
Definition 4.1: An annuity is a sequence of equal payments made at regular intervals of time.
Remark: The regular intervals are usually months.
Example 4.1: Suppose that someone deposits \$100 a month into a savings account paying an annual
rate of interest of 4.25% (compounded monthly) for 20 years.
Let’s examine the mathematics of an annuity via Example 4.1. For the sake of terminology, the
monthly payment of an annuity is called the rent and we will denote the rent by R . In Example 4.1,
R 100 .
The amount in the account after 20 years of monthly payments are made is called the future
value of the annuity and we will denote this quantity by F .
We’d like to find a formula for F . For convenience, let R be the rent, r be the annual rate of
interest, and n the compound period of an annuity. Define the compound interest rate as i ,
r
where i .
n
Note, after the first period, F R because no interest is paid in the first period. After two
periods, F R R1 i , because interest is paid on the first rent payment. After three periods,
F R R1 i R1 i , etc., so that, after k rent payments,
2
F R R1 i R1 i R1 i 1
2 k 1
We’d like to come up with a compact form for 1 . Note first that we can factor R from every term in
1 to obtain
F R 1 1 i 1 i 1 i
2 k 1
The trick is to let x 1 i . The expression in brackets then becomes the k 1 degree polynomial
th
1 x x 2 x k 1 . It is well-known in algebra that
xk 1
1 x x x
2 k 1
.
x 1
Hence, 1 1 i 1 i 1 i
2 k 1 1 i k 1 1 i k 1 , so that
1 i 1 i
1 i 1
k
F R .
i
We can now solve Example 4.1: we want to know the future value of an annuity when R 100 ,
n 12 , and k 240 . Hence,
.0425 240
1 1
F 100
12 \$37,726.20.
.0425
12
```
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Multiplying Monomials
Multiplying monomials involves two steps: 1) Multiply the numbers (coefficients) 2) Multiply the variables. Remember that when you multiply powers, you add the exponents. Examples: 1) 3x.5x=15x2 First you multiply 3.5=15. Then x.x=x2 2) 4y3.2y2=8y5 4.2=8 y3.y2=y5 because when you multiply powers you add the exponents (because if you were to write it out you have y.y.y.y.y=y5). 3) 2x(-8x2 )=-16x3 2(-8)=-16. x.x2=x3 . Remember that any variable with no exponent has an understood exponent of 1, so this is like x1.x2=x3. To multiply powers, add the exponents. 1+2=3. 4) -4x2 y(7x4 y3 )=-28x6 y4 Notice that first the numbers are multiplied, then the x's (2+4=6), then the y's (understood 1+3=4) 5) 3x(4y)=12xy. Notice the numbers were multiplied then the variables were both included in the answer. Practice: Multiply the monomials 1) 3y2 (5y) 2) -2x5 (-9x) 3) 7x2 y(4x3 y4) 4) 3x(4x)(2x) 5) 5x2 (-8y) Answers: 1) 15y3 2) 18x6 3) 28x5 y5 4) 24x3 5) -40x2 y
Related Links: Math Algebra Factors Polynomials Algebra Topics Classifying Polynomials Writing Polynomials in Standard Form Simplifying Polynomials Adding Polynomials Subtracting Polynomials Distributive Property (Multiplying a monomial by a polynomial) Multiplying binomials Multiplying trinomials and polynomials Dividing monomials Dividing polynomials by monomials Dividing polynomials by binomials Factoring Polynomials: Common Factors Factoring Polynomials: The difference of two squares |
# A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall
Question:
A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize.
Solution:
In the given problem,
Total amount of money (Sn) = Rs 700
There are a total of 7 prizes and each prize is Rs 20 less than the previous prize. So let us take the first prize as Rs a.
So, the second prize will be Rs, third prize will be Rs.
Therefore, the prize money will form an A.P. with first term a and common difference −20.
So, using the formula for the sum of n terms,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
We get,
$700=\frac{7}{2}[2(a)+(7-1)(-20)]$
$700=\frac{7}{2}[2 a+(6)(-20)]$
$700=\frac{7}{2}(2 a-120)$
$700=7(a-60)$
On further simplification, we get,
$\frac{700}{7}=a-60$
$100+60=a$
$a=160$
Therefore, the value of first prize is Rs 160.
Second prize = Rs 140
Third prize = Rs 120
Fourth prize = Rs 100
Fifth prize = Rs 80
Sixth prize = Rs 60
Seventh prize= Rs 40
So the values of prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, Rs 40
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# A tale of two triangles: Heron triangles and elliptic curves
Originating author is William Mc Callum.
If two triangles have the same area and the same perimeter, are they necessarily congruent? It turns out that the answer is no. For example, the triangle with sides and has the same area and perimeter as the triangle with sides , , and .
Both triangles have perimeter :
Amazingly, the two triangles also have the same area. The right triangle has area . To find the area of the other triangle, we use Heron’s formula, which states that the area of a triangle with side lengths , , and , is given by
where is the semiperimeter of the triangle. A quick calculation using this formula shows that the area of the second triangle is also 6.
The space of triangles
How do we find examples like this? The secret is to find the right way of representing the space of all triangles. There are many possible ways to do this. One way is to represent a triangle by the triple consisting of its three side lengths in some order. In this way we represent a triangle by a point in space. Not every point corresponds to a triangle; for example, all the coordinates must be positive. Can you think of other restrictions?
There’s another way of putting coordinates on the space of triangles using angles instead of lengths. Every triangle has an inscribed circle, and the radius of the circle has a simple relationship with the area and semiperimeter , namely
To see why this is true, drop perpendiculars from the center of the circle to the sides of the triangle, as in the left diagram in Figure 2. These perpendiculars form the altitudes of 3 smaller triangles with bases on the sides of the big triangle and vertices at the center of the inscribed circle. Adding up the areas of these triangles we get .
Figure 2: Parameterizing the space of triangles
This equation tells us that if two triangles have the same area and same semiperimeter, then the radii of their inscribed circles are also the same. So if we are looking for two such triangles we will find them amongst all triangles inscribed around a fixed circle. Instead of using lengths to describe these triangles, we will use the angles formed by the three radii at the center of incircle, as in the right diagram in Figure 2.
Parameterizing triangles with constant area and perimeter
Inside the space of triangles we can find curves corresponding to a whole family of triangles with the same values of and .
First, we express in terms of the angles , , and and the radius of the incircle, as follows. The radii and the lines from the vertices to the incenter break the triangle into six right triangles. Because the lines from the vertices to the center bisect the angles of the big triangle, these right triangles occur in congruent pairs. Taking one base length from each pair and adding, we get
This equation and together tell us that if the area and semiperimeter are constant, then so is the sum of the tangents:
(1)
Second, we translate this condition into an equation defining a curve in the plane. Let , , and . Since , we have
so
Then, if is the constant , equation (1) becomes for fixed , the equation
(2)
which we rewrite as
(3)
Every triangle with area and semiperimeter determines a point on this curve, and every point on the curve in a certain region of the plane corresponds to a triangle. The region corresponds to angles that actually work in Figure 2, namely angles satisfying and , which corresponds to the region , and (since ).
The following figure shows this curve for , the value corresponding to the triangle with sides 3, 4, and 5. Every point on the component of this curve in the positive quadrant corresponds to a triangle; the side lengths of the triangle are , , and . In particular, the points , , , , , and all correspond to the triangle with sides , , and , with the sides taken in different orders. This figure is interactive: see what happens for some other points on the curve or some other values for the area and perimeter!
Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.5 (or later) is installed and active in your browser (Click here to install Java now)
Finding points on the curve
Figure 3: Curve parameterizing triangles of perimeter 12 and area 6.
Because the curve in Figure 3 is defined by an equation of degree 3, we can find points on it using the method of tangents and secants. Two points on the curve determine a secant which cuts the curve in one more point; finding the point amounts to solving a cubic equation in , two of whose roots are already known. Since we already have 6 points on the curve, there are lots of possibilities for secants, and generating more points generates more possibilities. In fact, the curve has infinitely many points with rational coordinates. The two-secant procedure illustrated in Figure 3 leads to the point (marked with a circle), which corresponds to the triangle with sides , , and .
The secant procedure works for any cubic curve in the plane; such curves are called elliptic curves (not because the curves are themselves ellipses, but because they arise in the study of a certain class of complex functions called elliptic functions). The secant procedure allows one to define a group structure on the set of rational points on a elliptic curves (that is, points whose coordinates are rational numbers).
The study of elliptic curves is a central area of research in number theory, with applications to the cryptographic schemes behind secure financial transactions on the web. Elliptic curves played a central role in the proof of Fermat’s Last Theorem.
The story described in this article shows the remarkable unity of mathematics, starting as it does in high school and ending in research. Along the way we encountered a fundamental idea in modern mathematics: the idea of solving a problem about a particular type of object (triangles with area 6 and perimeter 12, for example) by situating the object in a more general space (the space of all triangles) and finding the right way of parameterizing that space.
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1. Beautiful stuff. A related article is Rosenberg, Spillane, Wulf, “Heron Triangles and Moduli Spaces,” The Mathematics Teacher 101 (2008), 656-663. The article contains insightful Geometer’s Sketchpad pictures created by HS teacher extraordinaire Dan Wulf. If you contact Dan, he’ll send you the Sketchpad files, which include great animations of all triangles of a fixed area and perimeter.
• Bill McCallum says:
This article was in fact the inspiration for this piece, thanks Steve for mentioning it. I’m still trying to figure out how to embed Sketchpad or Geogebra animations in a WordPress blog.
2. PAK SO KRETH says:
dear sir!
i am a mathematical student at Royal University of Phnom Penh!i am interested in klein project serously.i think that this program is a guideline for us to be a good mathematical teacher. i know,if we want to be good at mathematics,it is very hard. However,i want to tell u that i really like and love math so much. But , sometimes i don’t have enough mathematical document to search for,for example math exercise book and i also don’t have enough time to reading book too because my lifestyle thannks for your presentation at my university.
sincerely from me
Pak So Kreth
3. Awesome detail about using Hero’s formula and cool explanations on the Elliptic Curves. Very rare to find such blogs online.
4. David Fowler says:
Excellent article, in the same spirit as “Roots to Research,” by Judith and Paul Sally. There should be more material like this in the courses that prepare secondary teachers. |
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# squiz1 - teams we get the same game so the total number of...
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Solutions to quiz # 1 (January 12) 1. a) We can choose a committee of 3 from a group of 10 people in ± 10 3 ² = 10! 3!7! ways and then we can choose a committee of 4 from the group of remaining 7 people in ± 7 4 ² = 7! 4!3! ways. Altogether, there are 10! 3!7! · 7! 4!3! = 10! 4!3!3! = 4200 ways to choose the committees. Answer: the two committees can be chosen in 10! 4!3!3! ways. b) We can choose the first team in ± 10 4 ² = 10! 4!6! ways, after which we can choose the second team in ± 6 4 ² = 6! 4!2! ways. Hence the total number of ways to choose first the first team and then the second team is 10! 4!6! · 6! 4!2! = 10! 4!4!2! ways. However, the order in which the teams are chosen does not matter (if we switch the
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Unformatted text preview: teams we get the same game), so the total number of diļ¬erent games is 1 2 Ā· 10! 4!4!2! = 1575 . Answer: there are 10! 4!4!2!2! diļ¬erent games. 2. The sample space is { AB,AC,BC } . We have P ( { AB,AC } ) = P ( AB ) + P ( AC ) = 5 8 , P ( { AB,BC } ) = P ( AB ) + P ( BC ) = 5 8 and P ( AB ) + P ( AC ) + P ( BC ) =1 . Therefore, P ( { AC,BC } ) = P ( AC ) + P ( BC ) =2 Ā³ P ( AB ) + P ( AC ) + P ( BC ) Ā“-Ā³ P ( AB ) + P ( AC ) Ā“-Ā³ P ( AB ) + P ( BC ) Ā“ =2-5 8-5 8 = 3 4 . Answer: The probability that one of the chosen courses is Communications is 3 / 4. 1...
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### CHAPTER 1 STRAIGHT LINES
LEARNING OBJECTIVES
Upon completion of this chapter, you should be able to do the following:
1. Calculate the distance between two points.
2. Locate a point by dividing a line segment.
3. Define the inclination of a line and determine the line's slope.
4. Solve for the slopes of parallel and perpendicular lines.
5. Compute the angle between two lines.
6. Determine the equation of a straight line using the point-slope form, the slope-intercept form, and the normal form.
7. Determine the equations of parallel and perpendicular lines.
8. Calculate the distance from a point to a line.
INTRODUCTION
The study of straight lines provides an excellent introduction to analytic geometry. As its name implies, this branch of mathematics is concerned with geometrical relationships. However, in contrast to plane and solid geometry, the study of these relationships in analytic geometry is accomplished by algebraic analysis.
The invention of the rectangular coordinate system made algebraic analysis of geometrical relationships possible. Rene Descartes, a French mathematician, is credited with this invention; the coordinate system is often designated as the Cartesian coordinate system in his honor.
You should recall our study of the rectangular coordinate system in Mathematics, Volume l, NAVEDTRA 10069-D1, in which we reviewed the following definitions and terms:
1. The values of x along, or parallel to, the X axis are abscissas. They are positive if measured to the right of the origin; they are negative if measured to the left of the origin. (See fig. 1-1.)
2. The values of y along, or parallel to, the Y axis are ordinates. They are positive if measured above the origin; they are negative if measured below the origin.
3. The abscissa and ordinate of a point are its coordinates.
Any point on the coordinate system is designated by naming its abscissa and ordinate. For example, the abscissa of point P (fig. 1-1) is 3 and the ordinate is - 2. Therefore, the symbolic notation for P is
P(3, -2)
Figure 1-1.-Rectangular coordinate system.
In using this symbol to designate a point, the abscissa is always written first, followed by a comma. The ordinate is written last. The general form of the symbol is
P(x,y) |
# Decay Formula
## Decay Formula
Students should first go over the definition of exponential decay before learning the exponential Decay Formula. In exponential decay, a quantity initially drops gradually before decreasing quickly. In addition to using the exponential Decay Formula to calculate population decline (depreciation), students can also use it to calculate half-life (the amount of time for the population to become half of its size).
The exponential decline, which is a rapid reduction over time, can be calculated with the use of the exponential Decay Formula. The exponential Decay Formula is used to determine population decay, half-life, radioactivity decay, and other phenomena. f(x) = a (1 – r)x is the general form.
Where
a = initial amount
1-r = decay factor
x= time period
### Decay Formula
A variable exponent, a positive base, and a base that is not equal to one are all characteristics of an exponential function. Since the exponent in F (x) = 4x is fixed rather than changeable, it is an example of an exponential function. f (x) = x3 is a basic polynomial function rather than an exponential function. Exponential function graphs never reach a horizontal asymptote and are continually curved. Exponential or logarithmic functions can be used to explain a variety of real-world occurrences.
In Mathematics, the concept of “exponential decay” describes the process of a constant percentage rate decline in an amount over time. It can be written as y=a(1-b)x, where x is the amount of time that has passed, and is the initial amount, b is the decay factor, and y is the final amount.
The exponential Decay Formula is helpful in many real-world situations, most notably for keeping track of inventory that is consumed consistently in the same amount (such as food for a school cafeteria). It is also particularly helpful in that it can be used to quickly calculate the overall cost of using a product over time.
What are Exponential Decay Formulas?
The quantity drops gradually, followed by a dramatic decline in the rate of change and growth over time. The exponential Decay Formula is used to determine this decline in growth. One of the following formats can be used to express the exponential Decay Formula:
f(x) = abx
f(x) = a (1 – r)x
P = P
0
e-k t
Where,
a (or) P
0
= Initial amount
b = decay factor
r = Rate of decay (for exponential decay)
x (or) t = time intervals (time can be in years, days, (or) months, whatever students are using should be consistent throughout the problem).
k = constant of proportionality
e- Euler’s constant
### Exponential Decay Formula
Understanding the Exponential Decay Formula and being able to name each of its components is crucial:
y = a (1-b)x
The term “decay factor” (represented by the letter b in the exponential Decay Formula), which is a percentage by which the initial amount will decline each time, is crucial to comprehend in order to fully appreciate the utility of the Decay Formula.
If students are thinking about this practice, the original amount would be the number of apples a bakery purchases, and the exponential factor would be the proportion of apples used each hour to make pies. The original amount, denoted by the letter and in the formula, is the amount before the decay occurs.
The exponent indicates how frequently the decay occurs and is typically stated in seconds, minutes, hours, days, or years. In the case of exponential decay, the exponent is always time and is represented by the letter x.
In exponential decay, a quantity first decreases gradually before quickly doing so. The exponential Decay Formula can be used to calculate half-life as well as population decline (depreciation) (the amount of time for the population to become half of its size).
### Examples Using Exponential Decay Formulas
• The half-life of carbon-14 is 5,730 years. Find the exponential decay model for carbon-14. To the nearest decimal point, please round the response.
Solution: Use the formula of exponential decay
P = P0 e– k t
P0 = initial amount of carbon
The half-life of carbon-14 is 5,730 years,
P = P0 / 2 = Half of the initial amount of carbon when t = 5, 730.
P0 / 2 =P0 e– k (5730)
Divide both sides by P0
0.5 = e– k (5730)
Take “ln” on both sides,
ln 0.5 = -5730k
Divide both sides by -5730,
k = ln 0.5 / (-5730) ≈ 1.2097
The exponential decay model of carbon-14 is P = P0 e– 1.2097k
• A new couch cost $350,000 from Andrew. The sofa’s value is decreasing exponentially at a rate of 5% yearly. What is the sofa worth now, two years later? To the nearest decimal point, please round the response. Solution: Initial value of Sofa=$350,000
Rate of decay r = 5% = 0.05
Time t = 2 years
Use the exponential Decay Formula,
A = P (1 – r)t
A = 350000 x (1 – 0.05)
2
A = 315,875
The value of the sofa after 2 years = \$315,875 |
### Prompt Cards
These two group activities use mathematical reasoning - one is numerical, one geometric.
### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
# Street Sequences
## Street Sequences
Let's look at a street!
Most of the streets around here have their numbers going in order from one side of the street to the other - so that odds end up on one side and evens on the other.
So here's what is could be like:
Looking from above the numbers could appear as:
Well you can imagine walking down this street and adding the house numbers in various ways.
and carry on as far as you can.
Explore the answers you get for these additions. What do you notice?
Can you explain why?
If you want to go a bit further with this you could change the grouping of houses in different ways. How about:
A.
Or:
B.
But maybe the houses are already grouped in some way - like all semi-detached:
Then one side of the street the totals are: 4, 12, 20, etc
and the other side are: 6, 14, 22 etc.
You could continue this and explore it further.
Some streets have terraces of three houses together:
and some have fours:
You can explore the addition of these groups too!
### Why do this problem?
This problem is very useful when you want to give pupils opportunities to explore number patterns in their own way. There are various ways of exploring this activity with no particular 'right' way.
### Possible approach
The way in which you introduce this activity will depend on the children you are working with. It would be helpful to show the arrangement of numbers in the problem on a board (if a big group) or on paper (if a small group).
### Key questions
Some more experienced pupils may be asked "Why are these sequences like they are?"
Less experienced younger pupils can be asked simply "What do you notice about the answers we've got?"
Generally you can ask "What could you do now?"
### Possible extension
An extension may be suitable for some pupils that comes from comparing the results from the sequences that we get when the houses are grouped.
### Possible support
Some pupils may need number cards in front of them and calculators to free them up to think creatively. |
# Multivariate Function, Chain Rule / Multivariable Calculus
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## 1. Multivariate Function Definition
A multivariate function has several different independent variables. While “classroom” calculus usually deals with one variable, you’ll deal with their multivariate counterparts in applied sciences. While calculus commonly deals with functions with just one independent and dependent variable, multivariate functions are far more useful in everyday life. This is because most real-world processes depend on more than one parameter.
The functions we typically deal with have a single input and single output. “Input” and “output” refers to independent and dependent variables, typically denoted with x and y
## Notation
In calculus we write multivariate functions as having a dependent variable z and independent variables x and y:
z = f(x, y)
The inputs of a multivariate function the domain. The set of all possible outputs is the range
## 2. What is Multivariable Calculus?
Multivariable calculus (also called multivariate calculus) deals with functions of several variables. Many techniques can be transferred from single variable calculus, including finding derivatives and integrals.
Multivariable calculus is a huge field that usually covers an entire semester, usually after at least one full year of single variable calculus.
## Comparison of Single and Multivariable Calculus
Functions of one variable (left) are graphed on an x-y axis; The graph on the right is multivariate and is graphed on a x, y, z axis.
In single variable (univariate) calculus, the focus is on a function with a single input and a single output, like y = f(x). With multivariable calculus, more than one input is involved, such as z = x2 + y2.
You can also have multiple outputs, like:
• x = f(t),
• y = g(t),
• z = h(t).
While derivatives in single variable calculus give you a tangent line, in multivariate calculus the result is a tangent vector. The tangent vector is just your usual vector, extended to a 3-dimensional plane.
Another interesting difference is limits. In single variable calculus, you have a limit from the left and a limit from the right. When both of those limits exist and agree with each other, then we say there’s a limit. With multivariable calculus, this is a lot more challenging, because discontinuities don’t happen on a single line graph: they happen to 3D objects, which you can approach from multiple sides. Discontinuities are no longer simple broken lines, they often behave more like black holes in space, with the function sucked into (or blown out of) the point of discontinuity.
Hue-luminance plot of exp(1/z), centered on the essential singularity at zero. The function behaves differently depending on which direction you approach the function from. Credit: Functor Salad | Wikimedia Commons.
## 4. 3. The Multivariate Chain Rule
In the multivariate chain rule (or multivariable chain rule) one variable is dependent on two or more variables. The chain rule consists of partial derivatives
For the function f(x,y) where x and y are functions of variable t, we first differentiate the function partially with respect to one variable and then that variable is differentiated with respect to t. The chain rule is written as:
## Example
Let’s take a look at an example that shows how the chain rule works.
Example: The number of footballs a sporting goods store sells over time is x(t) t4 + 1; y(t) = 3t2 + 6 describes the price of footballs over time. Let z(t) = xy represent the revenue the store earns from the sale of footballs. How does revenue change with respect to time?
While there are several ways you could approach the solution, the easiest way to solve is to use the multivariate chain rule.
First, note that as a function of t, z is a function of a single variable. In this example, we have:
• Δ z / Δ x = y
• Δ z / Δ y = x
• dx/dt = 4t3
• dy/dt = 6t
Apply the chain rule formula from above to get the change in revenue over time:
dz/dt = (y)(4t3) + (x)(6t)
## References
Evans, M.; Hastings, N.; and Peacock, B. Statistical Distributions, 3rd ed. New York: Wiley, p. 5, 2000.
CITE THIS AS:
Stephanie Glen. "Multivariate Function, Chain Rule / Multivariable Calculus" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/multivariate-function/
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### Linear functions
1. 1. Linear FunctionsA linear function in two variables is any equation of that may be written in the formy = mx + b where m and b are real number coefficients and x and y represent any realnumbers that make up a solution. Furthermore, we observe that• The point (0, b) will always be the y-intercept.• The slope of the line will always equal m.• The slope is defined as m = (y2 – y1)/(x2 – x1) for any two points on the line.We call y = mx + b the Slope-Intercept Form of the linear equation.SlopeThe slope of a line is defined descriptively as the ratio of how far up you move divided byhow far to the right you move, as you move from one point to any other point on the line.Example: The graph of y = 2/3x – 1 has a slope of 2/3 and a y-intercept of (0,-1) asshown below.Negative SlopeIf we must move down instead of up, a negative sign is part of the slope. Also, if wemust move left instead of right, a negative sign is part of the slope. So if slope isnegative, we can interpret this in one of two ways:• m = -(A/B) = (-A)/B which would indicate that you move down A units and then right B units.• m = -(A/B) = A/(-B) which would indicate that you move up A units and then left B units.Note that if we must move both down and left when moving from point to point, twonegative signs are incorporated into the slope and the result is a positive ratio. From MathMotivation.com – Permission Granted For Use and Modification For Non-Profit Purposes
2. 2. Example: The graph of y = -2x +1 indicates a slope of –2. We may write m = -2 asm = (-2)/1. So we move down 2 and right 1 when moving from point to point as shownbelow. Alternatively, we could write this slope as m = 2/(-1) and move up 2 and left 1when moving from point to point.Example: Calculate the slope of the line passing through (4,-1) and (2, -2). Then usethis slope to help graph this line.The slope is given by m = (y2 – y1)/(x2 – x1). So we substitute in our values to getm = -2 – (-1) = -2 + 1 = -1 = 1 2–4 -2 -2 2The graph is given by From MathMotivation.com – Permission Granted For Use and Modification For Non-Profit Purposes
3. 3. Finding The Equation of a Line: The Point-Slope FormThe equation of a line that passes through (x1, y1) and has slope m is given bym(x – x1) = y – y1where (x, y) is any point that lies on the line and (x1, y1) is a specific point given.Example: Find the equation of the line with slope m = 3 that passes through (4, -1).Then, write this equation in slope-intercept form.We substitute m = 3 and x1=4, y1=-1 into m(x – x1) = y – y1 to get3(x – 4) = y - (-1) which simplifies to3(x – 4) = y + 1Now, to write this in slope-intercept form, we must solve this equation for y.Begin by multiplying out with the Distributive Property to get3x – 12 = y + 1Use the Addition Property of Equality to add –1 to both sides to get3x – 13 = y or y = 3x –13.Note: You can quickly check your answer by verifying that the slope of this equation(m=3) matches the given slope (m=3) and also by substituting in the given point (4,-1)into y = 3x - 13 to get –1= 3•4 – 13 and verify that it is a solution.Example: Find the equation of the line that passes through the points (4,-1) and (2, -2).In this case, we must first calculate the slope. Since we are given two points, we can dothis using the slope formula.The slope is given by m = (y2 – y1)/(x2 – x1). So we substitute in our values to getm = -2 – (-1) = -2 + 1 = -1 = 1 2–4 -2 -2 2Now, substitute in m=1/2 and either point (4,-1) or (2,-2) into the point-slope formm(x – x1) = y – y1.I will choose (4,-1) to be (x1, y1). Why? No reason. Either point will produce the sameequation so I simply chose this first point because I had to choose one.The equation becomesApplying the Distributive Property and simplifying results in(1/2)x – 2 = y +1 Continued. . . From MathMotivation.com – Permission Granted For Use and Modification For Non-Profit Purposes
4. 4. Now, apply the Addition Property of Equality to add –1 to both sides to getAs a check, both points (4,-1) and (2,-2) should be solutions for this equation. If theyare, you have the correct equation since only one linear equation will contain these twopoints.Where Does The Point-Slope Form Come From?The Point-Slope form is a generalization of the slope formula, which states:m = (y2 – y1)/(x2 – x1)If we replace the specific point (x2, y2) with a general solution point (x, y), we getm = (y – y1)/(x – x1)Applying the Multiplication Property of Equality to multiply both sides of the equation by(x – x1) results in m(x – x1) = y – y1 .Parallel Lines and Perpendicular LinesParallel lines are lines with the same slope. If the equations of the lines are in slope-intercept form, we can quickly determine whether or not the lines are parallel. Forexample, the lines given by y = 3x – 2 and y = 3x + 100 are parallel since both have aslope of 3.Perpendicular lines have slopes that are negative reciprocals of each other. Forexample, y = (3/2)x and y = (-2/3)x are perpendicular since the slopes, 3/2 and –2/3 arenegative reciprocals of each other. The graphs of these two functions are shown below. From MathMotivation.com – Permission Granted For Use and Modification For Non-Profit Purposes
5. 5. Horizontal and Vertical LinesHorizontal lines have slope m = 0 and form y = k, where k is the y-value of everypoint on the horizontal line. The slope is zero because if you move from point to point onany horizontal line, you will move up zero units and right or left some non-zero numberof units. This results in a slope of 0 since the numerator of the slope quotient is zero.Vertical lines have slope m = undefined and form x = k, where k is the x-value ofevery point on the vertical line. The slope is undefined because if you move from point topoint on any vertical line, you will move up some non-zero number of units and right orleft zero units. This results in an undefined slope since the denominator of the slopequotient is zero.Example: Graph the line through the points (2,1), (2, 3), find its slope, and find itsequation.The slope is undefined, since if you use the slope formula, you get m = (3-1)/(2-2) = 2/0.This is a vertical line with equation x = 2, since all points have an x-value of 2.Example: Graph the line through the points (2,1), (0, 1), find its slope, and find itsequation.The slope is m = (1-1)/(0-2) = 0/(-2) = 0. Or, if you see that it is a horizontal line, youknow that the slope is zero. This is a horizontal line with equation y = 1, since all pointshave a y-value of 1. From MathMotivation.com – Permission Granted For Use and Modification For Non-Profit Purposes
6. 6. Putting It All TogetherThe following 2-part problem requires you to use most of the concepts covered in thissection: A. Find the equation of a line that has graph parallel to the graph of 3x + y = 6 and passes through the point (2,-1). B. Also, find the equation of a line that has graph perpendicular to the graph of 3x + y = 6 and passes through the point (2,-1).First, we need to realize that two quantities are needed in order to find the equation of aline. We need:• The slope, or two points so we can calculate the slope.• One point in addition to the slope.We can then substitute these values into the point slope form m(x – x1) = y – y1.In this problem, we are given the same point (2,-1) for both parts A and B. So all weneed is the slope of the line in each part.In Part A, our line must be parallel to the line given by 3x + y = 6. This means it musthave the same slope as 3x + y = 6. To easily find the slope of 3x + y = 6, rewrite it inslope-intercept form. We gety = -3x + 6 by applying the Addition Property of Equality.This means the slope is m=-3. Our parallel line in Part A must also have slope m=- 3.Substituting m= - 3 and the point (2,-1) into m(x – x1) = y – y1 results in- 3(x – 2) = y – (-1) or -3(x – 2) = y + 1.We may use the Distributive Property to multiply out to –3x + 6 = y + 1 and then use theAddition Property of Equality to add –1 to both sides to get y = -3x + 5 for Part A.In Part B, our line must be perpendicular to the line given by 3x + y = 6. This means itmust have a slope that is the negative reciprocal of that of 3x + y = 6. We rewrote thegiven line equation asy = -3x + 6This means the slope of our perpendicular line is m =- 1/(-3) = 1/3.Substituting m= 1/3 and the point (2,-1) into m(x – x1) = y – y1 results in(1/3)(x – 2) = y – (-1) or 1/3(x – 2) = y + 1.We may use the Distributive Property to multiply out to 1/3x –2/3 = y + 1 and then usethe Addition Property of Equality to add –1 to both sides to get y = 1/3x –5/3 for Part B.To check both answers, substitute in x=2, y=-1 into both equations. You should get asolution. From MathMotivation.com – Permission Granted For Use and Modification For Non-Profit Purposes |
# Solve 11x + 15y + 23 = 0, 7x – 2y – 20 = 0 by elimination method.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Solving 11x + 15y + 23 = 0, 7x – 2y – 20 = 0 by elimination method we get x = 2 and y = -3. When one or both equations are multiplied by factors, only one variable remains after adding or subtracting the resulting two equations. One method for resolving a system of linear equations is the elimination method. In this approach, the equation in one variable is obtained by either adding or subtracting the equations. We can add the equation to delete a variable if its coefficients are the same and have the opposite sign from the other variables. Similarly to this, we can subtract the equation to get the equation in one variable if the coefficients of one of the variables are the same and their signs are the same.
You can start by multiplying one or both equations by a constant value on both sides of an equation to obtain the equivalent linear system of equations, and then eliminate the variable by simply adding or subtracting equations if we do not have the equation to directly add or subtract the equations to do so.
### Steps to Solve the Equation by Elimination Method
Given: 11x + 15y + 23 = 0, 7x – 2y – 20 = 0
When the first equation is multiplied by 2 and the second equation by 15, we obtain
2 x (11x + 15y + 23) = 0
22x + 30y + 46 = 0 …… (1)
15 x (7x – 2y – 20) = 0
105x – 30y – 300 = 0 …. (2)
22x + 30y + 46 + 105x – 30y – 300 = 0
127x – 254 = 0
127x = 254
On simplifying we get
x = 254/127
x = 2
Put the value of x in equation (1) we get
22x + 30y + 46 = 0
22 (2) + 30y + 46 = 0
44 + 30y + 46 = 0
30y + 90 = 0
30y = -90
On simplifying we get
y = -90/30
y = -3
Consequently, x and Y have values of 2 and -3.
Summary:
## Solve 11x + 15y + 23 = 0, 7x – 2y – 20 = 0 by elimination method.
Solving 11x + 15y + 23 = 0, 7x – 2y – 20 = 0 by elimination method we get x = 2 and y = -3. In this method, the equation present in one variable is got by either addition or subtraction of the equations.
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Math 115 Spring 2014 Written Homework 10-SOLUTIONS Due Friday, April 25
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1 Math 115 Spring 014 Written Homework 10-SOLUTIONS Due Friday, April 5 1. Use the following graph of y = g(x to answer the questions below (this is NOT the graph of a rational function: (a State the domain and range of g in interval notation. Solution: There is a point on the graph corresponding to every x value except x =. Hence, the domain of g is (, (, 5]. The range is (, 1 (1, ] (b What is g(? What is g(? What is g(4? Solution: We are looking for the y-values corresponding to the points where x =, x = and x = 4. We see that g( =, g( is undefined (as x = is not in the domain of g, and g(4 =. (c For what values of x does (i g(x =?, (ii g(x =?, (iii g(x = 1? Solution: We start by looking at the set of points on the graph of g that intersect the horizontal lines, y =, y = and y = 1, respectively. For y = we see that y = intersects the graph at the points (1,, (, and the line interval from (, to (0,. Hence, the solutions to the equation g(x = are the set {x R x 0, x = 1 or x = }. For y =, there is only one point of intersection with the graph; (,. g(x = when x =. Hence, For y = 1, there is no point of intersection with the graph. Hence, there are no solutions to the equation g(x = 1.
2 (d Determine all of the following limits from the graph of y = g(x: x g(x x g(x x + g(x x 0 g(x x 4 g(x x 4 + g(x x g(x Solution: g(x = 1 x g(x = x x + g(x = x 0 g(x does not exist (since the limit from the right of x = 0 is 1 and the limit from the left of x = 0 is x 4 g(x = 1 g(x = x 4 + x g(x =
3 . Express the lengths a and b in the figure below in terms of θ. Solution: Use the right-triangle definitions of the trig ratios and then solve for a and b: sin θ = a 4 a = 4 sin θ and cos θ = b 4 b = 4 cos θ. A boat approaches a 0-ft lighthouse whose base is at sea level. Let b be the distance between the boat and the base of the lighthouse. Let L be the distance between the boat and the top of the lighthouse. Let θ be the angle of elevation between the boat and the top of the lighthouse. (a Express b as a function of θ. Solution: To model this situation, we can use a right triangle: Using the right-triangle definitions, we know tan θ = 0 0. Thus, b = b tan θ.
4 (b Express L as a function of θ. Solution: Using the same right triangle from above, sin θ = 0 0. Thus, L = L sin θ. 4. (a In a circle of radius r, an arc of length 10 is swept out by an angle of radians. What is the exact radius of the circle? Solution: The radian was defined in such a way that the arc length, s, is equal to the radius times the angle, when the angle is in radians. i.e. Here we have that s = 10 and θ =. Thus, s = rθ. 10 = r r = 10. (b In a circle of radius r =, what is the exact length of the arc swept by an angle of 10? Solution: Before we can apply the arc length formula, we must have an angle in radians. Thus, here we must convert the angle from degree measure into radian measure. 10 ( π 180 = 10π 180 = 1π 18 = 7π. Now we have r = and θ = 7π. s = rθ = ( 7π = 7π. The length of the are subtended by the angle 10 on the circle is 7π. 5. Suppose θ is an angle in standard position (meaning it starts on the positive x-axis whose radian measure is an integer value between 0 and π. (a If the angle falls in the third quadrant, what is the radian measure θ? Solution: The angles in the third quadrant are between the angles π = and = Thus, the only integer value between these is 4. So, θ = 4. π (b If the terminal side of the angle had fallen in any other quadrant, could you still answer the question with a single value? Explain. Solution: If the terminal side had fallen in quadrant I, we would be able to answer this question because those angles fall between 0 and π = Hence, in quadrant I, θ would be equal to 1. In quadrants II and IV, there would be two possible values for θ; and in II and 5 and in IV.
5 . For each angle θ given below, when sketched in standard position, determine the quadrant in which it lies. (Hint: in some cases, it might help to determine a coterminal angle that is between 0 and π. You should justify your answer in some way - either with a brief explanation or a picture. (a θ = 7π Solution: θ = 7π = π + π. π is full revolutions and then π is less than π terminal side of θ will end up in the first quadrant. so the (b θ = 11π 4 Solution: θ = 11π 4 = π π 4 clockwise π which is more than π 4 in the third quadrant.. π is 1 full revolution clockwise an then we continue but less than π. Thus, the terminal side of θ will end up (c θ = 5.5 Solution: θ = 5.5 is between the quadrantal angles π = and π = Thus, the terminal side of θ will end up in the fourth quadrant. 7. (a Suppose θ is an angle in standard position which intersects the unit circle at the point (x, y. If y = 1, what are the possible values of cos θ? If you know that cos θ is positive, in 8 which quadrant does the angle lie? Solution: The equation of the unit circle is x + y = 1. If we plug y = 1 8 we get two possible values of x and thus two possible values of cos θ: into the equation, x + y = 1 x + ( 1 8 = 1 x = 1 x = 4 x = ± 8 The possible values are cos θ = and cos θ =. If we know that cos θ is positive, then 8 8 the angle must lie in quadrant I because both the x and the y coordinate of the corresponding point on the unit circle are positive.
6 (b Suppose sin α = 4 5 and π < α < π. Find cos α, tan α, cot α, sec α, and csc α. Solution: Since π < α < π, we know that the angle lies in quadrant III. One method would be to draw a right triangle in quadrant III with the opposite side length 4 and the hypotenuse 5. This is a -4-5 triangle so the adjacent side is. Then we can use the right triangle trig definitions to determine the other trig values - remembering that the angle is in the third quadrant. Another method would be to use the equation for the unit circle: x + y = 1 or the equivalent identity cos (x + sin (x = 1. By plugging in sin α = y = 4, we can then solve for 5 cos α = y = 5. Either way, we get the following solutions: cos α = 5, tan α = 4, cot α = 4, sec α = 5, csc α = 5 4 (c Suppose cos β = and β lies in quadrant II. Find sin β, tan β, cot β, sec β, and csc β. Solution: One method would be to draw a right triangle in quadrant II with the adjacent side length and the hypotenuse. This is a triangle so the opposite side is 1. Then we can use the right triangle trig definitions to determine the other trig values - remembering that the angle is in the second quadrant. Another method would be to recognize the point on the Unit Circle here. The point is in ( the second quadrant with an x-coordinate of. Thus, the point is, 1. Either way, we get the following solutions: sin β = 1, tan β = 1 = 1, cot β =, sec β =, csc β = 8. Find( the exact value of each of the following. You must justify your answer in some way. 9π (a cos Solution: 9π = 8π + π = 4π + π. Thus the angle is coterminal to π. Thus, cos ( 9π ( π = cos = 0
7 (b sin ( 4π Solution: The terminal side of 4π ends up in quadrant so it s sine value will be negative. The reference angle here is π. Using either a special right triangle or the unit circle, we find that ( 4π ( π sin = sin = ( 11π (c tan Solution: The terminal side of 11π ends up in quadrant 4 so it s tangent value will be negative. The reference angle is π. Using either a special right triangle or the unit circle, we find that (d ( csc 5π 4 ( 11π tan ( π = tan = sin π cos π = 1 = 1 Solution: The angle 5π 4 will fall in quadrant with a reference angle of θ r = π 4. Using either a special right triangle or the unit circle, we find that ( csc 5π ( π = csc = sin π = 1 1 = 4 (e ( 48π sec Solution: The angle 48π = 1π is a quadrantal angle that is coterminal to π. Thus, ( 48π 1 sec = sec(1π = cos 1π = 1 cos π = 1 1 = 1 ( (f cos 11π Solution: The angle 11π = 9π π = π π will fall in quadrant 1 with a reference angle of θ r = π. Using either a special right triangle or the unit circle, we find that ( cos 11π ( π = cos = 1
8 9. As we did with the sine function in lecture, state the properties of the cosine function. (a State the domain and range of f(x := cos x. Solution: The domain is x R. The range is f(x [ 1, 1]. (b Determine all values of x where f(x is zero (i.e. the x-intercepts. Solution: The zeros occur when x = π + kπ where k is any integer. (c Determine the y-intercept for the graph y = cos x. Solution: The y-intercept occurs at (0, Use your knowledge of transformations to sketch the graph of at least two periods/cycles of each function. ( (a f(x := sin x + π 4 Solution: We begin with the graph of y = sin x: Next we reflect the graph over the x axis to get the graph of y = sin x: Finally, we shift the previous graph to the left π ( 4 to get the graph of y = sin x + π : 4
9 (b g(x := cos(x + 4 Solution: We begin with a graph of y = cos x: Then we shift the graph up 4 to get the graph of y = cos(x + 4:
10 (c ( h(x := cos x π + 4 Solution: We begin with a graph of y = cos x. Then we shift the graph to the right π 4 to get the graph of y = cos(x π 4 Now applying the vertical stretch cause by A = yields the graph y = cos ( x π 4. Lastly, we apply the vertical shift up. This results in the final graph y = cos ( x π 4 +.
11 Remember - there is more than correct answer for each of these graphing questions. ANY complete cycles is correct. Some of you may have used a different method on part (c and ended up with the following graph:
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# How do you divide (-4+5i) / (2+i) in trigonometric form?
$\frac{- 4 + 5 i}{2 + i}$
in trigonometric form is
$\frac{r 1}{r 2} c i s \left(\theta 1 - \theta 2\right)$
where,
$r 1 = \sqrt{\left({\left(- 4\right)}^{2} + {5}^{2}\right)}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{5}{- 4}\right)$
and
$r 2 = \sqrt{\left({2}^{2} + {1}^{2}\right)}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{1}{2}\right)$
#### Explanation:
Divide $\frac{- 4 + 5 i}{2 + i}$
in trigonometric form
Expressing numerator and denominator as
$z = r c i s \theta$
For $z 1 = - 4 + 5 i$
$r 1 = \sqrt{\left({\left(- 4\right)}^{2} + {5}^{2}\right)}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{5}{- 4}\right)$
For $z 2 = 2 + i$
$r 2 = \sqrt{\left({2}^{2} + {1}^{2}\right)}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{1}{2}\right)$
Thus,
$\frac{- 4 + 5 i}{2 + i} = \frac{r 1 c i s \theta 1}{r 2 c i s \theta 2}$
$= \frac{r 1}{r 2} \frac{c i s \theta 1}{c i s \theta 2}$
By De-Moivre's theorem,
$\frac{c i s \theta 1}{c i s \theta 2} = c i s \left(\theta 1 - \theta 2\right)$
Now,
$\frac{- 4 + 5 i}{2 + i} = \frac{r 1}{r 2} c i s \left(\theta 1 - \theta 2\right)$ |
# Math in Focus Grade 2 Chapter 17 Practice 1 Answer Key Reading Picture Graphs with Scales
This handy Math in Focus Grade 2 Workbook Answer Key Chapter 17 Practice 1 Reading Picture Graphs with Scales detailed solutions for the textbook questions.
## Math in Focus Grade 2 Chapter 17 Practice 1 Answer Key Reading Picture Graphs with Scales
The picture graph shows the food a team ate after a softball game.
Food Eaten After the Game
Example
They had 6 helpings of hot dogs.
Question 1.
They had the same number of helpings of ________ as hot dogs.
They has the same number of helpings of carrots as hot dogs
Explanation:
There are 3 boxes for hot dogs = 6 helpings
There are 3 boxes for carrots = 6 helpings
Both are equal.
Question 2.
They has 4 more helpings of salad than apples
Explanation:
There are 4 boxes for salad = 8 helpings
There are 2 boxes for apples = 4 helpings
8 – 4 = 4
So, they has 4 more helpings of salad than apples.
Question 3.
They have 12 helpings of salad and apples in all.
Explanation:
There are 4 boxes for salad = 8 helpings
There are 2 boxes for apples = 4 helpings
8 + 4 = 12
So, they have 12 helpings of salad and apples in all.
Jane and her classmates chose their favorite fairy tale character. This picture graph shows their choices.
Favorite Fairy Tale Characters
Fill in the blanks. Use the picture graph on page 182 to help you.
Example
How many characters are shown?
4
Question 4.
Which is the most common favorite character?
Prince is the most common favorite character.
Explanation:
There are 7 books for prince
Each book stands for 2 children
So, 7 x 2 = 14
14 children chose their favorite character as Prince
It is the most common character chosen by children.
Question 5.
Which is the least common favorite character?
King is the least common favorite character
Explanation:
There are 3 books for king
Each book stands for 2 children
3 x 2 = 6
So, 6 children chose their favorite character as King
It is the least common character chosen by children.
Question 6.
8 children like the Queen.
What does each stand for?
Each stand for 2 children
Explanation:
There are 4 books for Queen
8 children like Queen
So, 8 / 4 = 2
Therefore, each book stand for 2 children.
Question 7.
How many children chose the Prince?
___________ children
14 children chose prince
Explanation:
There are 7 books for prince
Each book stands for 2 children
So, 7 x 2 = 14
14 children chose their favorite character as Prince.
Question 8.
How many more children chose the Fairy than the Queen as their favorite character?
____________ more children
4 more children
Explanation:
12 children chose Fairy
8 children chose Queen
12 – 8 = 4
So, 4 more children chose the Fairy than the Queen as their favorite character.
Question 9.
The total number of children who chose ____________ and _____________ as their favorite character is the same as the number of children who chose the Prince.
The total number of children who chose Queen and King as their favorite character is the same as the number of children who chose the Prince.
Explanation:
14 children chose Prince
8 children chose Queen
6 children chose King
8 + 6 = 14
So, The total of children who chose Queen and King as their favorite character is the same as the number of children who chose the Prince.
Randy’s home is near a School, a Bus Stop, a Store, and a Post Office. He draws a picture graph to show how far his home is from these places.
Number of Steps from Home
Example
The School is 100 steps from Randy’s home.
Each stands for 10 steps.
Question 10.
The Store is _________ steps from Randy’s home.
The Store is 90 steps from Randy’s home.
Explanation:
There are 9 pictures for store
Each stands for 10 steps
9 x 10 = 90
So, the store is 90 steps from Randy’s home.
Question 11.
Randy’s home is 50 steps from the Bus Stop.
He will draw ________ on the graph.
Explanation:
Each stands for 10 steps
Randy’s home is 50 steps from the Bus Stop
50 / 10 = 5
So, i drew 5 pictures on the graph for bus stop.
Question 12.
The Post Office is 80 steps from Randy’s home.
He will draw _________ more on the graph. |
# Lesson 6 The Locker Problem
• Let’s figure out what’s happening in a game about lockers.
## Warm-up Choral Count: Twos and Fours
What patterns do you see?
## Activity 1 Questionable Lockers
The picture shows lockers in a school hallway.
The 20 students in Tyler’s fourth-grade class are playing a game in a hallway that is lined with 20 lockers in a row.
• The first student starts with the first locker and goes down the hallway and opens all the lockers.
• The second student starts with the second locker and goes down the hallway and shuts every other locker.
• The third student stops at every third locker and opens the locker if it is closed or closes the locker if it is open.
This process continues until all 20 students in the class have touched the lockers.
• How does your representation show lockers?
• How does your representation keep track of students who touch lockers?
• How does your representation show which lockers are open or closed?
## Activity 2 An Open and Shut Case
The 20 students in Tyler’s fourth-grade class are playing a game in a hallway with 20 lockers in a row.
Your goal is to find out which lockers will be touched as all 20 students take their turn touching lockers.
1. Which locker numbers does the 3rd student touch?
2. Which locker numbers does the 5th student touch?
3. How many students touch locker 17? Explain or show how you know.
4. Which lockers are only touched by 2 students? Explain or show how you know.
5. Which lockers are touched by only 3 students? Explain or show how you know.
6. Which lockers are touched the most? Explain or show how you know.
7. If you have time… Which lockers are still open at the end of the game? Explain or show how you know.
## Problem 1
1. Find the factor pairs of 36.
2. How many factors does 36 have?
3. List the factors of 15. |
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# Scale Factor of Similar Polygons
## Compare matching side lengths using ratios and proportions.
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Scale Factor of Similar Polygons
Have you ever wondered about the relationship between side lengths of a figure?
Jessie looked at a drawing with two quadrilaterals in it. The side lengths of the quadrilaterals were listed as follows:
MN = 3 inches QT = 6 inches
NO = 2 inches QR = 4 inches
MP = 4 inches TS = 8 inches
OP = 2 inches RS = 4 inches
Given the relationship between these sides, can you figure out the scale factor?
This Concept will teach you how to do exactly that. We'll revisit this problem at the end of the Concept.
### Guidance
Similar figures are shapes that exist in proportion to each other. They have congruent angles, but their sides are different lengths. Squares, for example, are similar to each other because they always have four $90^{\circ}$ angles and four equal sides, even if the lengths of their sides differ. Other figures can be similar too, if their angles are equal. Let’s look at some pairs of similar figures.
Notice that in each pair the figures look the same, but one is smaller than the other. As you can see, similar figures have congruent angles but sides of different lengths.
Even though similar figures have sides of different lengths, corresponding sides still have a relationship with each other. Each pair of corresponding sides has the same relationship as every other pair of corresponding sides, so that, altogether, the pairs of sides exist in proportion to each other. For instance, if a side in one figure is twice as long as its corresponding side in a similar figure, all of the other sides will be twice as long too.
In this Concept, we are going to use these relationships to find the measures of unknown sides. This method is called indirect measurement.
Let’s take a step back to similar figures to understand how indirect measurement works.
First, let’s make sure we can recognize corresponding parts of similar figures. Similar figures have exactly the same angles. Therefore each angle in one figure corresponds to an angle in the other.
These triangles are similar because their angles have the same measures. Which corresponds to which? Angle $B$ is $100^{\circ}$ . Its corresponding angle will also measure $100^{\circ}$ : that makes angle $Q$ its corresponding angle. Angles $A$ and $P$ correspond, and angles $C$ and $R$ correspond.
Similar figures also have corresponding sides, even though the sides are not congruent. Corresponding sides are not always easy to spot. We can think of corresponding sides as those which are in the same place in relation to corresponding angles. For instance, side $AB$ , between angles $A$ and $B$ , must correspond to side $PQ$ , because $A$ corresponds to $P$ and $B$ corresponds to $Q$ .
Corresponding sides also have lengths that are related, even though they are not congruent. Specifically, the side lengths are proportional. In other words, each pair of corresponding sides has the same ratio as every other pair of corresponding sides. Take a look at the rectangles below.
These rectangles are similar because the sides of one are proportional to the other. We can see this if we set up proportions for each pair of corresponding sides. Let’s put the sides of the large rectangle on the top and the corresponding sides of the small rectangle on the bottom.
$\frac{LM}{WX} &= \frac{8}{4}\\\frac{MN}{XY} &= \frac{6}{3}\\\frac{ON}{ZY} &= \frac{8}{4}\\\frac{LO}{WZ} &= \frac{6}{3}$
Now you can clearly see each relationship. To figure out if the pairs do indeed form a proportion, we have to divide the numerator by the denominator. If the quotient is the same, then the ratios each form the same proportion and the figures are similar.
$\frac{LM}{WX} &= \frac{8}{4} = 2\\\frac{MN}{XY} &= \frac{6}{3} = 2\\\frac{ON}{ZY} &= \frac{8}{4} = 2\\\frac{LO}{WZ} &= \frac{6}{3} = 2$
Each quotient is the same so these ratios are proportional. The side lengths are proportional and the figures are similar.
We call this the scale factor. The scale factor is the ratio that determines the proportional relationship between the sides of similar figures. For the pairs of sides to be proportional to each other, they must have the same scale factor. In other words, similar figures have congruent angles and sides with the same scale factor. A scale factor of 2 means that each side of the larger figure is twice as long as its corresponding side is in the smaller figure.
Now it's time for you to try a few on your own. Determine each scale factor.
#### Example A
$\frac{18}{6}$ and $\frac{24}{8}$
Solution: $3$
#### Example B
$\frac{12}{6}$ and $\frac{8}{4}$
Solution: $2$
#### Example C
$\frac{25}{5}, \frac{45}{9}, \frac{15}{3}$
Solution: $5$
Here is the original problem once again.
Jessie looked at a drawing with two quadrilaterals in it. The side lengths of the quadrilaterals were listed as follows:
MN = 3 inches QT = 6 inches
NO = 2 inches QR = 4 inches
MP = 4 inches TS = 8 inches
OP = 2 inches RS = 4 inches
Given the relationship between these sides, can you figure out the scale factor?
To figure out the scale factor, we can write each corresponding side as a ratio comparing side lengths.
$\frac{3}{6}$
$\frac{2}{4}$
$\frac{4}{8}$
$\frac{2}{4}$
The scale factor of these figures is $\frac{1}{2}$ .
### Vocabulary
Similar Figures
figures that have the same angle measures but not the same side lengths.
Scale Factor
the proportional relationship between two side lengths.
Proportions
two equal ratios.
### Guided Practice
Here is one for you to try on your own.
What is the scale factor of the figures below?
We need to find the scale factor, so let’s set up the proportions of the sides. Let’s put all the sides from the large figure on top and the sides from the small figure on the bottom. It doesn’t matter which we put on top, as long as we keep all the sides from one figure in the same place.
$\frac{QR}{HI} &= \frac{15}{5}\\\frac{TS}{KJ} &= \frac{21}{7}\\\frac{RS}{IJ} &= \frac{6}{3}\\\frac{QT}{HK} &= \frac{15}{5}$
Now all we have to do is divide to find the scale factor.
$\frac{QR}{HI} &= \frac{15}{5} = 3\\\frac{TS}{KJ} &= \frac{21}{7} = 3\\\frac{RS}{IJ} &= \frac{6}{3} = 3\\\frac{QT}{HK} &= \frac{15}{5} = 3$
The scale factor for these similar figures is 3. This means that the first quadrilateral is exactly three times bigger than the second.
What happens if we had written the numbers the other way around?
If we had put the measurements of the smaller figure on top and the larger figure on the bottom, we would have found a scale factor of $\frac{1}{3}$ . You can write them the other way too as long as you understand how to read the scale factor. This is another way of saying the scale factor is three. When we say that the larger figure is three times bigger than the small one, it’s the same as saying that the small figure is one-third the size of the larger one.
### Practice
Directions: Find the scale factor of the pairs of similar figures below.
Directions: Use each ratio to determine scale factor.
5. $\frac{3}{1}$
6. $\frac{8}{2}$
7. $\frac{2}{8}$
8. $\frac{10}{5}$
9. $\frac{12}{4}$
10. $\frac{16}{2}$
11. $\frac{15}{3}$
12. $\frac{24}{4}$
13. $\frac{4}{2}$
14. $\frac{6}{2}$
15. $\frac{3}{9}$
### Vocabulary Language: English
Perimeter
Perimeter
Perimeter is the distance around a two-dimensional figure.
Proportion
Proportion
A proportion is an equation that shows two equivalent ratios.
Scale Factor
Scale Factor
A scale factor is a ratio of the scale to the original or actual dimension written in simplest form.
Similar
Similar
Two figures are similar if they have the same shape, but not necessarily the same size. |
# NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6
These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.6
Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given: Two circles of centre O and O’ intersect each other at A and B.
To prove that: ∠OAO’ = ∠OBO’
Construction: Join OO’.
Proof: In ∆AOO’ and ∆BOO’
AO = BO (Radii of the same circle)
AO’ = BO’ (Radii of the same circle)
OO’ = OO’ (Common)
So, by S-S-S congruency condition
∆AOO’ ≅ ∆BOO’
∴ ∠OAO’ = ∠OBO’ (By CPCT)
Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Solution:
Let O be the centre of the given circle and let its radius be r cm.
Draw OM ⊥ AB and ON ⊥ CD.
Since OM ⊥ AB, ON ⊥ CD, and AB || CD.
Therefore point M, O and N are collinear. So MN = 6 cm.
Let OM = x cm, Then ON = 6 – x cm
Join OA and OC, then OA = OC = r.
Since the perpendicular from the centre to a chord of the circle bisect the chord.
∴ AM = MB = $$\frac {5}{2}$$ cm
and CN = ND = $$\frac {11}{2}$$ cm.
In right triangle OAM,
OA2 = OM2 + AM2
r2 = x2 + $$\left(\frac{5}{2}\right)^{2}$$ ……(i)
Now, in right triangle OCN
OC2 = ON2 + CN2
r2 = (6 – x)2 + $$\left(\frac{11}{2}\right)^{2}$$ …….(ii)
From equation (i) and (ii)
$$x^{2}+\left(\frac{5}{2}\right)^{2}=(6-x)^{2}+\left(\frac{11}{2}\right)^{2}$$
Hence, the radius of the circle is $$\frac{5 \sqrt{5}}{2}$$ cm.
Question 3.
The length of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance of 4 cm from the centre, what is the distance of the other chord from the centre.
Solution:
Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 8 cm.
Let the radius of the circle be r cm.
Draw OP ⊥ AB and OQ ⊥ CD.
The length of OP is 4 cm.
In right triangle OAP,
r2 = OP2 + AP2
⇒ r2 = (4)2 + (3)2 [∵ AP = $$\frac {1}{2}$$ AB = 3 cm]
⇒ r2 = 16 + 9 = 25
⇒ r = 5 cm ……(i)
Now, in right triangle OQC,
r2 = (OQ)2 + (CQ)2
⇒ (5)2 = (OQ)2 + (4)2
[∵ r = 5 cm prove above and CQ = $$\frac {1}{2}$$ CD = 4 cm]
⇒ 25 = (OQ)2 + 16
⇒ (OQ)2 = 25 – 16 = 9
⇒ OQ = 3 cm
Therefore the distance of the chord CD from the centre is 3 cm.
Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with a circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
Given: Vertex of an angle ABC be located outside a circle, and AD = CE.
Let ∠AOC = ∠x, ∠AOD = ∠z and ∠DOE = ∠y.
To prove that: ∠ABC = $$\frac {1}{2}$$ (x – y)
∴ ∠AOD = ∠COE (Angle made by equal chord at the centre are equal)
Therefore,
∠x + ∠y + ∠z + ∠z = 360°
⇒ ∠x + ∠y + 2∠z = 360°
⇒ 2∠z = 360° – ∠x – ∠y
⇒ ∠z = 180 – $$\frac {1}{2}$$ ∠x – $$\frac {1}{2}$$ ∠y ……(A)
Now, ∠ODB = ∠OAD + ∠DOA
(Exterior angle is equal to sum of opposite interior angles)
∠ODB = ∠OAD + ∠z ……(i)
Again, ∠OAD + ∠z + ∠ODA = 180°
(Sum of all three angles of triangles)
(Angle opposite to equal sides are equal and OA = OD radii of circle)
or, 2∠OAD = 180° – ∠z
or, ∠AOD = $$\frac {1}{2}$$ (180° – ∠z) = 90° – $$\frac {1}{2}$$ ∠z
Putting the value of ∠AOD in equation (i)
∴ ∠ODB = 90° – $$\frac {1}{2}$$ ∠z + ∠z
(∵ ∠OAD = 90° – $$\frac {1}{2}$$ ∠z)
or, ∠ODB = 90° + $$\frac {1}{2}$$ ∠z ……..(ii)
Similarly, ∠OEB = 90° + $$\frac {1}{2}$$ ∠z ……(iii)
Now, in ODBE
∠ODB + ∠B + ∠OEB + ∠y = 360°
(Sum of all angle of ∆ is 360°)
90 + $$\frac {1}{2}$$ ∠z + ∠B + 90 + $$\frac {1}{2}$$ ∠z + ∠y = 360°
(Use equation (ii) and (iii))
or, 180 + ∠z + ∠B + ∠y = 360
or, ∠B = 180 – ∠y – ∠z …..(iv)
⇒ ∠ABC = $$\frac {1}{2}$$ [(Angle subtended by the chorde DE at the centre) – (Angle subtended by the chord AC at the centre)]
⇒ ∠ABC = $$\frac {1}{2}$$ [(Difference of the angles subtended by the chorde DE and AC at the centre)]
Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Solution:
We have a rhombus ABCD such that its diagonals AC and BD intersect at O.
P, Q, R and S are the mid-points of DC, AB, AD and BC respectively
∴ $$\frac {1}{2}$$ AD = $$\frac {1}{2}$$ BC
⇒ RA = SB
⇒ RA = OQ ……(ii)
⇒ $$\frac {1}{2}$$ AB = $$\frac {1}{2}$$ AD
⇒ AQ = AR ……(iii)
From (i), (ii) and (iii), we have
AQ = QB = OQ
i.e., A circle drawn with Q as centre, will pass through A, B and O.
Thus, the circle passes through the intersection ‘O’ of the diagonals rhombus ABCD.
Question 6.
ABCD is a parallelogram. The circle through A, B, and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
We have a circle passing through A, B, and C is drawn such that it intersects CD at E.
∴ ∠AEC + ∠B = 180° ……(i)
[Opposite angles of a cyclic quadrilateral are supplementary]
But ABCD is a prallelogram (Given)
∴ ∠D = ∠B ……(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
∠AEC + ∠D = 180° …….(iii)
But ∠AEC + ∠AED = 180° (Linear pair) ……(iv)
From (iii) and (iv),
We have ∠D = ∠AED
i.e., The base angles of ∆ADE are euqal.
Question 7.
AC and BD are chords of a circle which bisects each other. Prove that
(i) AC and BD are diameters
(ii) ABCD is a rectangle.
Solution:
Given: A circle in which two chords AC and BD are such that they bisect each other.
Let their point of intersection be O.
To Prove: (i) AC and BD are diameters.
(ii) ABCD is a rectangle.
Construction: Join AB, BC, CD, and DA.
Proof: (i) In ∆AOB and ∆COD, we have
(∵ ∠ADC = ∠ABC, opposite angles of || gm ABCD)
From equation (i) and (ii)
∠AED + ∠ABC = ∠ADE + ∠ABC
Thus, in ∆AED, we have
∴ AD = AE. (Side opposite to equal angles are equal)
Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angle of the triangle DEF are 90° – $$\frac {1}{2}$$ ∠A, 90° – $$\frac {1}{2}$$ ∠B and 90° – $$\frac {1}{2}$$ ∠C.
Solution:
Given: In ∆ABC, AD, BE and CF is the angle bisector of ∠A, ∠B, and ∠C respectively.
Where D, E, and F lie on the circumcircle of ∆ABC.
To prove that:
(i) ∠FDE = 90° – $$\frac {1}{2}$$ ∠A
(ii) ∠DEF = 90° – $$\frac {1}{2}$$ ∠B
(iii) ∠EFD = 90° – $$\frac {1}{2}$$ ∠C
Proof:
(i) ∠ADF = ∠ACF ……(i) (Angles in the same segments of a circle are equal)
Again, ∠ADE = ∠ABE ……(ii) (Angles in the same segment of atcircle are equal)
∠APF + ∠ACF = ∠ACF + ∠ABE
or, ∠FDE = $$\frac {1}{2}$$ ∠C + $$\frac {1}{2}$$ ∠B ……(iii)
(∵ CF and BE are the bisector of ∠B and ∠C respectively)
Now, In ∆ABC
∠A + ∠B + ∠C = 180° (Sum of all angles of a ∆ is 180°)
or, $$\frac {1}{2}$$ (∠A + ∠B + ∠C) = 90°
or, $$\frac {1}{2}$$ ∠B + $$\frac {1}{2}$$ ∠C = 90 – $$\frac {1}{2}$$ ∠A ……(iv)
From equation (iii) and (iv)
∠FDE = 90° – $$\frac {1}{2}$$ ∠A
Similarly we can prove
(ii) ∠DEF = 90 – $$\frac {1}{2}$$ ∠B
(iii) ∠EFD = 90 – $$\frac {1}{2}$$ ∠C.
Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q, lie on the two circles. Prove that BP = BQ.
Solution:
Let C(0, r) and C(O’, r) be two congruent circles.
Since AB is a common chord of two congruent circles. Therefore,
∴ ∠BPA = ∠BQA
Thus, in ∆BPQ, we have ∠BPA = ∠BQA
∴ BP = BC (Sides opposite to equal angles of a triangle are equal)
Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Solution:
Given: ABC is a triangle in which the bisector of ∠A intersects the circumcircle of ∆ABC at D.
To prove that: Perpendicular bisector of side BC intersects the angle bisector of ∠A at D.
Construction: Join BD and CD. |
# Tricks to solve Number Series Questions for Competitive Exam
Number Series Complete Tricks for Bank PO, SSC and Other Competitive Exam. Number Series is commonly asked in all the Competitive Exams. Here in this Article we will Understand Tricks to Solve Number Series Questions for Bank and Other Competitive Exam.
Number series is an arrangement of numbers in a certain order, where some numbers are wrongly put into the series of numbers and some number is missing in that series, we need to observe and find the accurate number to the series of numbers.
In competitive exams number series are given and where you need to find missing numbers. The number series come in different types. At first, you have to decide what type of series are given in papers and then you have to use shortcut tricks as fast as you can.
Different types of Number Series. There are some format of series which are given in Exams.
1). Perfect Square Series:
This Types of Series are based on square of a number which is in same order and one square number is missing in that given series.
Example 1: 441, 484, 529, 576?
Answer: 441 = 21*2, 484 = 22*2, 529 = 23*2, 576 = 24*2 , 625 = 25*2.
2). Perfect Cube Series:
This Types of Series are based on cube of a number which is in same order and one cube number is missing in that given series
Example 2: 1331, 1728, 2197, ?
Answer : 11*3 , 12*3 , 13*3 , 14*3
3). Geometric Series:
This type of series are based on ascending or descending order of numbers and each successive number is obtain by multiplying or dividing the previous number with a fixed number.
Example 3: 5, 45, 405, 3645,?
Answer: 5 x 9 = 45, 45 x 9 = 405, 405 x 9 = 3645, 3645 x 9 = 32805.
4). Two stage Type Series:
A two tier Arithmetic series is one in which the differences of successive numbers themselves form an arithmetic series.
Example 4: i. 3, 9, 18, 35, 58,——
ii. 6, 9, 17, 23,———-
5). Mixed Series:
This type of series are more than one different order are given in a series which arranged in alternatively in a single series or created according to any non-conventional rule. This mixed series Examples are describes in separately.
Examples 5:
11, 24, 50, 102, 206, ?
11 x 2 = 22 +2 = 24,
24 x 2 = 48 + 2 = 50,
50 x 2 = 100 + 2 = 102,
102 x 2 = 204 + 2 = 206,
206 x 2 = 412 + 2 = 414.
So the missing number is 414.
Directions (1-10): What will come in place of the question marks (?) in the following Number series?
1. 0, 6, 24, 60, 120, 210, ?
A. 336
B. 349
C. 312
D. 337
E. None of these
2. 11, 14, 19, 22, 27, 30, ?
A. 39
B. 34
C. 36
D. 35
E. None of these
3. 6, 12, 21, ? , 48
A. 33
B. 39
C. 36
D. 31
E. None of these
4. 18, 22, 30, ? ,78, 142
A. 44
B. 35
C. 46
D. 48
E. None of these
5. 73205, 6655, 605, 55, ?
A. 9
B. 5
C. 13
D. 11
E. None of these
6. 25, 100, ?, 1600, 6400
A. 400
B. 300
C. 360
D. 420
E. None of these
7. 125, ?, 343, 512, 729, 1000
A. 216
B. 215
C. 256
D. 225
E. None of these
8. 1 , 9 , 125 , 343 , ? , 1331
A. 730
B. 729
C. 512
D. 772
E. None of these
9. 121, 144, 169, ?, 225
A. 180
B. 172
C. 186
D. 196
E. None of these
10. ?, 2116, 2209, 2304, 2401, 2500
A. 2124
B. 1972
C. 1521
D. 2025
E. None of these
Quantitative Aptitude Study Materials PDF
1000+ Number Series Asked in Previous Year Bank Exam pdf - Download
200+ High Level Missing Data Interpretation Questions pdf - Download
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# Video: KS1-M18 • Paper 1 • Question 23
98 − ＿ = 28.
04:06
### Video Transcript
98 take away what equals 28.
Let’s draw a bar model to help us understand what this question is asking us. This bar represents the first number in the subtraction. It’s the number that we take away something from. And it’s worth 98. This bar represents part of 98. It’s the part that we take away. But we don’t know what this bar is worth. This is the value we need to find out. Our final bar represents what’s left after we’ve worked out the subtraction. This bar is worth 28.
Another way to look at this problem is to start with 28 and work backwards. Because we’re working backwards, we need to use the inverse operation, which means we need to add instead of subtract. So instead of working out 98 take away something equals 28, we can start with 28 and think about what we add to it to make 98.
We’re working backwards and so we add instead of subtract. 28 plus what equals 98. Let’s look at how the number 28 changes as it becomes 98. We can model 28 as two tens and eight ones. And then, we can model the number 98 as nine tens and eight ones. How does this number change to get to this number? Let’s start by looking at how the ones change. 28 has eight ones. And 98 also has eight ones. We don’t need to add any ones. So we know that the number we add must end in a zero. It has no ones.
Now, let’s look at how the tens change. And they do change. We can see that they become a lot larger. 28 has two tens. And 98 has nine tens. How many tens do we add to get from two tens to nine tens? Well, two plus seven equals nine. So we must add seven tens or 70. 98 is 70 more than 28. And so we can say that 98 take away 70 equals 28.
Let’s use a place value grid and some counters just to check that our answer is correct. Here’s the number 98, nine tens and eight ones. Watch how the two digits behave as we take away our seven tens or 70. The tens digit is going to get smaller or decrease. But because we’re only taking away seven tens and no ones, watch how the ones digit stays the same.
Let’s take away 70 then, seven tens, and see whether we end up with 28. We’ll start with 98, 88, 78, 68. That’s three tens or 30 we’ve already taken away, 58, 48, 38. That’s now six tens or 60 that we’ve taken away. We’ve got to take away one more ten, 28.
And so 98 take away 70 equals 28. |
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# Multiplication and Division Facts
Last updated date: 04th Mar 2024
Total views: 138.3k
Views today: 3.38k
## An Introduction to Multiplication and Division
Multiplication and division are basic knowledge for school students. Multiplication is the process of calculating the product of two or more numbers. It is a basic arithmetic operation that is frequently used in everyday life. When we need to combine groups of equivalent size, we employ multiplication. Division is the separating of a large group into smaller groups so that each group has an equal number of things. In Mathematics, it is an operation used for equal grouping and equal sharing. In this article, we will introduce you to some well-known multiplication and division facts.
## What are Multiplication and Division Facts?
The numbers that are being multiplied are referred to as factors, and the outcome of the multiplication is referred to as the product.
In division, the dividend is the number being divided, the divisor is the number dividing it, and the quotient is the outcome of the division.
Multiplication and Division
## Division and Multiplication-related Facts
• You can write a multiplication fact simply by writing out a multiplication calculation. For example: 12 × 12 = 144, 3 × 6 = 18 and 8 × 5 = 40
• You can also write multiplication facts using different methods of multiplication. For example, repeated addition is a way of finding multiplication facts by repeatedly adding a number until you reach the answer. 2 × 5 is the same as 2 + 2 + 2 + 2 + 2 = 10. The multiplication fact in that case is 10.
• If the dividend is ‘zero’ then any number as a divisor will give the quotient as ‘zero’.
Example: If ‘zero’ sweets are to be distributed among 8 children, naturally no one will get any sweets.
• If the divisor is ‘1’ then any dividend will have the quotient equal to itself.
Example: There are 15 sweets; each child is to get 1 sweet. How many children get sweets?
Division of 15 by 1
## Multiplication and Division Drills
Q1. At a zoo, there are 45 deer. Find out the total number of legs of all the deer.
Ans: There are a total of 45 deer. We know that each deer will have 4 legs.
Therefore, the total number of legs deer will have will be equal to the:
$= 45 \times 4$
= 180.
Hence, 180 will be the total number of legs of all the deer.
Q2. Write two multiplication for each division:
(i) 15 ÷ 3 = 5
Ans: The two multiplication for it will be: 5 × 3 = 15 and 3 × 5 = 15
(ii) 3 ÷ 3 = 1
Ans: The two multiplication for it will be: 1 × 3 = 3 and 3 × 1 = 3
## Practice Questions
Q1. Write two multiplication for each division:
(i) 21 ÷ 3 = 7
(ii) 30 ÷ 3 = 10
(iii) 33 ÷ 3 = 11
Ans:
(i) 7 × 3 = 21 and 3 × 7 = 21
(ii) 10 × 3 = 30 and 3 × 10 = 30
(iii) 11 × 3 = 33 and 3 × 11 = 33
Q2. Write two divisions for each multiplication:
(i) 6 × 3 = 18 (Ans: 18 ÷ 3 = 6 and 18 ÷ 6 = 3)
(ii) 2 × 3 = 6 (Ans: 6 ÷ 2 = 3 and 6 ÷ 3 = 2)
(iii) 9 × 3 = 27 (Ans: 27 ÷ 9 = 3 and 27 ÷ 3 = 9)
(iv) 12 × 3 = 36 (Ans: 36 ÷ 3 = 12 and 36 ÷ 12 = 3)
(v) 8 × 3 = 24 (Ans: 24 ÷ 3 = 8 and 24 ÷ 8 = 3)
## Summary
In this article, we have learned about multiplication which is a mathematical operation that conveys the fundamental concept of repeated addition of the same integer. Later, we also learned about division sentences, as it is a basic mathematical operation in which a bigger number is divided into smaller groups with the same number of components. In the end, we have seen some practice problems based on the facts. We hope you have a clear picture of the facts on multiplication and division. Practise further to enhance your understanding of the topic.
## FAQs on Multiplication and Division Facts
1. What are the 3 types of division?
The methods of division are of three types according to the difficulty level. These are the chunking method or division by repeated subtraction, short division method or bus stop method, and long division method.
2. What are the parts of multiplication?
The different parts of multiplication are expressed as follows. Let us understand this with an example: 6 × 4 = 24.
• Multiplicand (Factor): Multiplicand is the first number. In this case, 6 is the multiplicand.
• Multiplier (Factor): Multiplier is the second number. In this case, 4 is the multiplier.
• Product: The final result after multiplying the multiplicand and multiplier. In this example, 24 is the product.
• Multiplication Symbol: '×' (which connects the entire expression)
3. How does multiplication work in real life?
In our daily lives, we frequently use multiplication. For example, we may calculate the price of things based on the rate per quantity, discover the correct quantity of an ingredient to use in cooking, calculate the value of numerous items when the value of one item is known, and so on. |
# Sets and Functions Online Test Aptitude Questions With Solutions, Online Test On Sets and Functions
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Sets and functions are basic mathematical concepts that find applications in different fields like computer science, engineering, economics, and physics. Evaluating your understanding of these concepts and sharpen your problem-solving abilities can be achieved by taking online tests.
Set:
A set is a collection of distinct objects called elements. In an online set test you are asked to identify the properties of sets, such as subsets, intersections, unions.
Function:
A function is a relation between a set of inputs and a set of possible outputs with each input related to exactly one output. In an online function test you may be asked to identify the properties of functions, such as domain, range, inverse, and composition.
The best way to improve your understanding of Sets and Functions and sharpen your skills by utilizing the Free Online Test platform, where you can access various practice questions and their solutions.
## Learn these tips and tricks to improve your Set theory calculations.
A – A = Ø
B – A = B∩A
B – A = B – (A∩B)
n(A∪B) = n(A – B) + n(B – A) + n(A∩B)
n(A – B) = n(A∪B) – n(B)
n(A – B) = n(A) – n(A∩B)
(A – B) = A if A∩B = Ø
(A – B) ∩ C = (A∩C) – (B∩C)
A Δ B = (A-B) ∪ (B- A)
n(A‘) = n(∪) – n(A)
### Some Important Set Formulas
Laws of Universal Set and Empty set: Ø’ = ∪ and ∪’ = Ø
Law of Double complementation: (A’)’ = A
Complement Law : A∪A’ = U, A⋂A’ = Ø and A’ = U – A
De Morgan’s Law: (A ∪B)’ = A’ ⋂B’ and (A⋂B)’ = A’ ∪ B’
## Some Question On Sets and Functions
Question-1)
If p = {a, b, c, d, e, f}, A = {a, b, c}, find (P ∪ A)′
P ∪ A = {a, b, c, d, e, f} ∪ {a, b, c}
= {a, b, c, d, e, f} = P
= Hence (P ∪ A)′ = Φ
Question-2)
If N = {a, b, c, d, e, f}, P = {a, b, c}, Q = {c, d, e, f}, and R = {c, d, e} find (P ∪ Q) ∪ R
P ∪ Q = { a, b, c } ∪ { c, d, e, f }
= { a, b, c, d, e, f }
∴ (P ∪ Q) ∪ R
= { a, b, c, d, e, f } ∪ { c, d, e }
= { a, b, c, d, e, f } = N
Hence All elements are in set U so Answer is N
Question-3)
If R = { 1, 2, 3 } and S = { 4 , 5 } then what is the value of R × S ?
The Cartesian product X × Y of sets X , Y is the set of all ordered pairs ( x , y ).
Two ordered pairs \$( x_1 , y_1 ), ( x_2 , y_2 ) \$
Given that R = { 1, 2, 3 }
and S = { 4 , 5 }
then, R × S = { 1, 2, 3 } × { 4 , 5 }
⇒ R × S = { 1 } × { 4 , 5 } ,
{ 2 } × { 4 , 5 } , { 3 } × { 4 , 5 }
⇒ R × S = { 1 , 4 } , { 1 , 5 } , { 2 , 4 } , { 2 , 5 } , { 3 , 4 }, { 3 , 5 }
⇒ R × S = { (1 , 4 ) , ( 1 , 5 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 3 , 4 ), ( 3 , 5) }
Question-4)
Lets keep A and B a two finite sets where n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, so find n(A ∩ B)
Using Set Theory Formula: n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Wherein n(A ∩B) = n(A) + n(B) – n(A ∪B)
= 20 + 28 – 36
= 48 – 36 = 12
Question-5)
If A and B are two sets such that A ∪ B has 18 elements, A has 8 elements, and B has 15 elements, find how many elements does A ∩ B have?
From question,
(A ∪ B) = 18,
(A) = 8, (B) = 15
Using the formula
(A ∩ B) = (A) + (B) – (A ∪ B),
So, (A ∩ B) = 8 + 15 – 18 = 5
Question-6)
In a class of 50 students, 35 opted for mathematics and 37 opted for Biology. Find how many students have opted for both Mathematics and Biology:
From question,
(M ∪ B) = 50, (here M = math students, B = biology students)
(M) = 35, (B) = 37, (M ∩ B) =?
By formula
(M ∪ B) = (M) + (B) – (M ∩ B)
So, 50 = 35 + 37 – (M ∩ B)
⇒ (M ∩ B) = 35 + 37 – 50
= 72 – 50 = 22
Online Test - 1 (Sets and Functions) TAKE TEST Number of questions : 20 | Time : 30 minutes
Online Test - 2 (Sets and Functions) TAKE TEST Number of questions : 20 | Time : 30 minutes |
Linear Inequations in One Variable
Linear Inequations in One Variable:
We know that equations are mathematical statements having two sides connected by a sign of equality. In an inequation, the two sides of the statement are connected by a sign of inequality. The signs of inequality are <, >, ≤, ≥, ≠.
Statements like 2x < 3, 12x – 7 ≤ 2, (x – 2)/3 > 2x – 1, 3x – 2 < x + 1 etc., are known as linear inequations in one variable. In general, a linear inequation in one variable may be written as, ax + b > 0, ax + b < 0, ax + b ≥ 0, ax + b ≤ 0, where a, b ∈ R.
Solving a linear inequation of one variable is more or less similar to solving a linear equation in one variable as most of the basic rules are applicable except one exception discussed below. The permissible rules are-
(i) If the same expression is added to or subtracted from both sides of an inequation, the solution of the inequation remains unchanged.
(ii) If both the sides of an inequation are multiplied or divided by the same positive number, the solution of the resulting inequation remains unaltered.
(iii) If both the sides of an inequation are multiplied or divided by the same negative number, the resulting inequation has the same solution provided the sign of inequality is reversed.
Thus, the basic difference, between solving a linear equation and solving an inequation concerns the multiplication or division of both sides by a negative number.
The set from which the values of the variable involved in the inequation are chosen is generally known as the replacement set. For example, if the replacement set is the set of natural numbers, then from the values obtained after solving the inequation, only the natural numbers are to be taken as the solution.
Steps to Solve a Linear Inequation in One Variable (Simple Inequation):
(i) Both sides are simplified by removing the group symbols and collecting the like terms.
(ii) Fractions and decimals are removed (if necessary) after multiplying both sides by an appropriate factor.
(iii) Terms containing the variable are transposed to the left side and the constant terms are kept on the right side.
(iv) Both the sides are divided by the coefficient of the variable to get the values of the unknown.
(v) The solution set is, then, chosen from the replacement set.
Graphs of Inequations Involving One Variable:
We know that the solution set of the inequality x < 3 consists of all numbers less than 3, i.e., all the numbers lying to the left of 3 on the number line. We draw a picture of this set of numbers by shading the number line to the left of 3. The statement that the set extends up to infinity to the left of 3 is indicated by an arrowhead. We use an open circle at 3 to indicate that the number 3 is not included in the solution set. This picture representing the solution set is called the graph of the inequation (Fig. a).
The graph of the inequality x ≥ -1/2 is drawn by shading the number line to the right of -1/2. In this case, a filled circle is drawn at x = -1/2 to indicate that x = -1/2 is included in the solution set (Fig. b). |
# Algebra Training for RMO and Pre-RMO: let’s continue: Fibonacci Problem and solution
Fibonacci Problem:
Leonardo of Pisa (famous as Fibonacci) (1173) wrote a book “Liber Abaci” (1202), wherein he introduced Hindu-Arabic numerals in Europe. In 1225, Frederick II declared him as the greatest mathematician in Europe when he posed the following problem to defeat his opponents.
Question:
Determine the rational numbers x, y and z to satisfy the following equations:
$x^{2}+5=y^{2}$ and $x^{2}-5=z^{2}$.
Solution:
Definition: Euler defined a congruent number to be a rational number that is the area of a right-angled triangle, which has rational sides. With p, q, and r as a Pythagorean triplet such that $r^{2}=p^{2}+q^{2}$, then $\frac{pq}{2}$ is a congruent number.
It can be shown that square of a rational number cannot be a congruent number. In other words, there is no right-angled triangle with rational sides, which has an area as 1, or 4, or $\frac{1}{4}$, and so on.
Characteristics of a congruent number: A positive rational number n is a congruent number, if and only if there exists a rational number u such that $u^{2}-n$ and $u^{2}+n$ are the squares of rational numbers. (Thus, the puzzle will be solved if we can show that 5 is a congruent number and we can determine the rational number $u(=x)$). First, let us prove the characterisitic mentioned above.
Necessity: Suppose n is a congruent number. Then, for some rational number p, q, and r, we have $r^{2}=p^{2}+q^{2}$ and $\frac{pq}{2}=n$. In that case,
$\frac{p+q}{2}, \frac{p-q}{2}$ and n are rational numbers and we have
$(\frac{p+q}{2})^{2}=\frac{p^{2}+q^{2}}{4}+\frac{pq}{2}=(\frac{r}{2})^{2}+n$ and similarly,
$(\frac{p-q}{2})^{2}=(\frac{r}{2})^{2}-n$.
Setting $u=\frac{r}{2}$, we get $u^{2}-n$ and $u^{2}+n$ are squares of rational numbers.
Sufficiency:
Suppose n and u are rational numbers such that $\sqrt{u^{2}-n}$ and $\sqrt{u^{2}+n}$ are rational, when
$p=\sqrt{u^{2}+n}+\sqrt{u^{2}-n}$ and $q=\sqrt{u^{2}+n}-\sqrt{u^{2}-n}$
and 2n are rational numbers satisfying $p^{2}+q^{2}=(2n)^{2}$ is a rational square and also $\frac{pq}{2}=n$, a rational number which is a congruent number.
So, we see that the Pythagorean triplets can lead our search for a congruent number. Sometimes a Pythagorean triplet can lead to more than one congruent number as can be seen with $(9,40,41)$. This set obviously gives 180 as a congruent number. But, as $180=5 \times 36=5 \times 6^{2}$, we can also consider a rational Pythagorean triplet $(\frac{9}{6}, \frac{40}{6}, \frac{41}{6})$, which gives a congruent number 5 (we were searching for this congruent number in this puzzle!). We also determine the corresponding $u=\frac{r}{2}=\frac{41}{12}$.
The puzzle/problem is now solved with $x=\frac{41}{12}$, which gives $y=\frac{49}{12}$, and $z=\frac{31}{12}$.
One can further show that if we take three rational squares in AP, $u^{2}-n$, and $u^{2}$, and $u^{2}+n$, with their product defined as a rational square $v^{2}$ and n as a congruent number, then $x=u^{2}$, $y=v$ is a rational point on the elliptic curve $y^{2}=x^{3}-n^{2}x$.
Reference:
1) Popular Problems and Puzzles in Mathematics: Asok Kumar Mallik, IISc Press, Foundation Books, Amazon India link:
https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_1?s=books&ie=UTF8&qid=1525099935&sr=1-1&keywords=popular+problems+and+puzzles+in+mathematics
2) Use the internet, or just Wikipedia to explore more information on Fibonacci Numbers, Golden Section, Golden Angle, Golden Rectangle and Golden spiral. You will be overjoyed to find relationships amongst all the mentioned “stuff”.
Cheers,
Nalin Pithwa
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# If the point P on the parabola ${y^2} = 4ax$ for which |PR - PQ| is maximum, where R =(- a,0), Q =(0,a) is represented as $(k_1a, k_2a)$, then find the value of $(k_1+k_2)$.
Last updated date: 15th Aug 2024
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Answer
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70.5k+ views
Hint: Use triangle inequality to find the general points of parabola in parametric form to find the sum of $K_1$ and $K_2$.
Complete step-by-step answer:
First draw the diagram of the parabola equation which is horizontally parabola opening right side.
P point is given as $({K_1}a,{K_2}a)$,Q point is given as $(0,a)$ and R point is given as $(- a,0)$
When you join these points it makes a triangle, So according to triangle inequality the modulus of difference of two sides is always less than or equal to the third side.
$\Rightarrow \left| {PR - PQ} \right| \leqslant \left| {QR} \right|$
Maximum value of $\left| {PR - PQ} \right|\; = QR$ because everything is less than or equal to QR.
$\Rightarrow \left| {PR - PQ} \right|\; = QR$, when P, Q, R are collinear
So, the equation of line joining R and Q is
$\Rightarrow y - 0 = \dfrac{{a - 0}}{{0 - \left( { - a} \right)}}\left( {x - \left( { - a} \right)} \right)$
$\Rightarrow y = x + a$
Equation of parabola is $y^2 = 4ax$
In parametric form the coordinates of parabola is $( {a{t^2},2at})$
Where t is a parameter and these parametric coordinates represents every point on parabola.
$\Rightarrow$ p point is $( {a{t^2},2at})$
Now P, Q, R is in straight line, then this p point satisfy the equation of straight line
$\Rightarrow 2at - a{{\text{t}}^2} = a$
$\Rightarrow {t^2} - 2t + 1 = 0$
$\Rightarrow {\left( {t - 1} \right)^2} = 0$
$\Rightarrow \left( {t - 1} \right) = 0$
$\Rightarrow t = 1$
So, put the value of t in P points
$\Rightarrow {\text{P}}\left( {a,2a} \right)$, So compare this points with given points $({K_1}a,{K_2}a)$
So, on comparing $K_1 = 1, K_2 = 2$
$\Rightarrow$ The value of $\left( {{{\text{K}}_1} + {{\text{K}}_2}} \right) = 1 + 2 = 3$
So, this is your answer.
NOTE: - In this type of problem always apply inequality identity, after that find out the general points of parabola in parametric form and satisfy these points in the equation which pass from this point. |
# Constructing Triangles Worksheet Grade 7
👤 will chen 🗓 July 29, 2021, 3:55 pm ( Last Modified )
Constructing Triangles Worksheet Grade 7. Worksheet 1. Worksheet 2. Grade 7 Worksheets on Data Organization and Central Values. Worksheet 1. Worksheet 2. Worksheets on Division of Algebraic Expressions. Worksheet 1. Worksheet 2. 7th Grade Math Powers and Exponents Worksheets. Worksheet 1..Modeling Linear Relationships - Lesson 7.1. Solve One Variable Equations - Lesson 7.2 (Part 1) Solve One Variable Equations - Lesson 7.2 (Part 2) Linear Inequalities in Two Variables - Lesson 7.3. Practice Test for Unit 3. Review for Unit 3 Test on Linear Functions and Equations.For each angle, we either have a measure or an equation. For that reason, let's add all of the angles together to equal 180 degrees. Doing so, we have 40 + 10x + 20 + 20 = 180, and by combining ..Scalene triangles are triangles with three sides of different lengths. The math term for sides of different triangles is noncongruent sides, so you may also see this phrase in your math book..
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# Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 35
Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 35 Textbook Exercise Important Questions and Answers.
## Maharashtra State Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 35
Determine whether the pairs of numbers given below are co-prime numbers.
(1) 22, 24
Common factors of 22 and 24 are 1 and 2. (Not only 1 common factor) So, 22, 24 are not co-prime numbers.
(2) 14, 21
Common factors of 14 and 21 are 1 and 7. So, this pair is not co-prime numbers.
(3) 10, 33
Common factors of 10 and 33 is only 1. So, 10 and 33 are co-prime numbers.
(4) 11, 30
Common factors of 11 and 30 is only 1. So, 11 and 30 are co-prime numbers.
(5) 5, 7
Common factor of 5 and 7 is only 1. So, 5 and 7 are co-prime numbers.
(6) 15, 16
Common factors of 15 and 16 is only 1. So, 15 and 16 are co-prime numbers.
(7) 50, 52
Common factors of 50 and 52 are 1 and 2. So, 50 and 52 are not co-prime numbers.
(8) 17, 18
Common factors of 17 and 18 is only 1. So, 17 and 18 are co-prime numbers.
Activity 1 :
• Write numbers from 1 to 60.
• Draw a blue circle around multiples of 2.
• Draw a red circle around multiples of 4.
• Do all numbers with a blue circle also have a red circle around them?
• Do all the numbers with a red circle have a blue circle around them?
• Are all multiples of 2 also multiples of 4?
• Are all multiples of 4 also multiples of 2?
Activity 2 :
• Write numbers from 1 to 60.
• Draw a triangle around multiples of 2.
• Draw a circle around multiples of 3.
• Now find numbers divisible by 6. Can you find a property that they share?
Eratosthenes’ method of finding prime numbers
Eratosthenes was a mathematician who lived in Greece about 250 BC. He discovered a method to find prime numbers. It is called Eratosthenes’ Sieve. Let us see how to find prime numbers between 1 and 100 with this method.
• 1 is neither a prime nor a composite number. Put a square [ ] around it
• 2 is a prime number, so put a circle around it.
• Next, strike out all the multiples of 2. This tells us that of these 100 numbers more than half of numbers are not prime numbers.
• The first number after 2 not yet struck off is 3. So, 3 is a prime number.
• Draw a circle around 3. Strike out all the multiples of 3.
• The next number after 3 not struck off yet is 5. So, 5 is a prime number.
• Draw a circle around 5. Put a line through all the multiples of 5.
• The next number after 5 without a line through it is 7. So, 7 is a prime number.
• Draw a circle around 7. Put a line through all the multiples of 7.
In this way, every number between 1 and 100 will have either a circle or a line through it. The circled numbers are prime numbers. The numbers with a line through them are composite numbers.
One more method to find prime numbers
See how numbers from 1 to 36 have been arranged in six columns in the table alongside.
Continue in the same way and write numbers up to 102 in these six columns.
You will see that, in the columns for 2, 3, 4, and 6, all the numbers are composite numbers except for the prime numbers 2 and 3. This means that all the remaining prime numbers will be in the columns for 1 and 5. Now isn’t it easier to find them? So, go ahead, find the prime numbers!
Something more
• Prime numbers with a difference of two are called twin prime numbers. Some twin prime number pairs are 3 and 5, 5 and 7, 29 and 31 and 71 and 73. 5347421 and 5347423 are also a pair of twin prime numbers.
• There are eight pairs of twin prime numbers between 1 and 100. Find them.
• Euclid the mathematician lived in Greece about 300 BC. He proved that if prime numbers, 2, 3, 5, 7, ……., are written in serial order, the list will never end, meaning that the number of prime numbers is infinite.
Determine whether the pairs of numbers given below are co-prime numbers.
(1) (12,18) |
# Finding Percent of Change - PowerPoint PPT Presentation
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670 – 632 = 38 Find the amount of change. 38 632. n 100. = Write a proportion. 632 n = 38 ( 100 ) Find the cross products. 632 n 632. 38(100) 632. = Divide each side by 632. n = 6.01265 Simplify.
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Finding Percent of Change
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#### Presentation Transcript
670 – 632 = 38Find the amount of change.
38
632
n
100
= Write a proportion.
632n = 38(100)Find the cross products.
632n
632
38(100)
632
= Divide each side by 632.
n = 6.01265Simplify.
COURSE 2 LESSON 6-8
### Finding Percent of Change
Last year, a school had 632 students. This year the school has 670 students. Find the percent of increase in the number of students.
The number of students increased by 6%.
6-8
13,775 – 10,590 = 3,185Find the amount of markup.
3,185
10,590
n
100
= Write a proportion.
10,590n = 3,185(100) Find the cross products.
10,590n
10,590
3,185(100)
10,590
= Divide both sides by 10,590.
n 30 Simplify.
COURSE 2 LESSON 6-8
### Finding Percent of Change
Find the percent markup for a car that a dealer buys for \$10,590 and sells for \$13,775.
The percent markup is 30%.
6-8 |
Variance and Standard Deviation
Standard deviation is the positive square root of the mean of the squared deviation from the arithmetic mean.The square of standard deviation is called variance.Three methods for calculating variance and standard deviation are direct method, short-cut method or assumed mean method and step deviation method.Variance and standard deviation from an ungrouped data can be computed using direct method short-cut methodVariance and standard deviation from grouped data can be computed using step deviation method.Variance and standard deviation from grouped data within class can be computed using step deviation method.The measure of variability which is independent of units is called coefficient of variation (denoted as C.V.).C.V. = (S. D./Mean) × 100, where mean is not equal to 0.For two series with equal means, the series with greater standard deviation (or variance) is more variable or dispersed than the other. Also, the series with lesser value of standard deviation (or variance) is said to be more consistent than the other.Combined mean is the collective arithmetic mean of several sets of data combined into a single arithmetic mean.Combined Standard deviation is the collective standard deviation of several sets of data combined into a single standard deviation.
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• Q1
The variance of the first ‘n’ natural numbers is
Marks:1
(n2 – 1)/12.
Explanation:
• Q2
Megha takes a test of 25 marks. The observations of marks of 5 students are 10, 9, 14, 17 and 20. Now, the standard deviation is
Marks:1
4.15.
Explanation:
• Q3
The standard deviation of the following series is
Marks:1
9.
Explanation:
Let us assume an arbitrary mean a = 25
Class interval h = 10
Construct the following table:
• Q4
If the standard deviation of a set of observation is 4 and if each observation is divided by 4, the standard deviation of the new set of observations will be
Marks:1
1.
Explanation:
We know that, if y = x/h, then Sy = Sx/|h|
Since each observation is divided by 4.
Therefore, the S.D of new set of observation will be 4/4 = 1.
• Q5
The standard deviation for the following frequency distribution is
Marks:1 |
# Roots of unity
The Roots of unity are a topic closely related to trigonometry. Roots of unity come up when we examine the complex roots of the polynomial $x^n=1$.
## Solving the Equation
First, we note that since we have an nth degree polynomial, there will be n complex roots.
Now, we can convert everything to polar form by letting $x = re^{i\theta}$, and noting that $1 = e^{2\pi ik}$ for $k\in \mathbb{Z}$, to get $r^ne^{ni\theta} = e^{2\pi ik}$. The magnitude of the RHS is 1, making $r^n=1\Rightarrow r=1$ (magnitude is always expressed as a positive real number). This leaves us with $e^{ni\theta} = e^{2\pi ik}$.
Taking the natural logarithm of both sides gives us $ni\theta = 2\pi ik$. Solving this gives $\theta=\frac{2\pi k}n$. Additionally, we note that for each of $k=0,1,2,\ldots,n-1$ we get a distinct value for $\theta$, but once we get to $k > n-1$, we start getting coterminal angles.
Thus, the solutions to $x^n=1$ are given by $x = e^{2\pi k i/n}$ for $k=0,1,2,\ldots,n-1$. We could also express this in trigonometric form as $x=\cos\left(\frac{2\pi k}n\right) + i\sin\left(\frac{2\pi k}n\right) = \mathrm{cis }\left(\frac{2\pi k}n\right).$
## Geometry
All of the roots of unity lie on the unit circle in the complex plane. This can be seen by considering the magnitudes of both sides of the equation $x^n = 1$. If we let $x = re^{i\theta}$, we see that $r^n = 1$, since the magnitude of the RHS of $x^n=1$ is 1, and for two complex numbers to be equal, both their magnitudes and arguments must be equivalent.
Additionally, we can see that when the nth roots of unity are connected in order (more technically, we would call this their convex hull), they form a regular n-sided polygon. This becomes even more evident when we look at the arguments of the roots of unity.
## Properties
Listed below is a quick summary of important properties of roots of unity.
• They occupy the vertices of a regular n-gon in the complex plane.
• For $n>1$, the sum of the nth roots of unity is 0. More generally, if $\zeta$ is a primitive nth root of unity (i.e. $\zeta^m\neq 1$ for $1\le m\le n-1$), then $\sum_{k=0}^{n-1} \zeta^{km}=\begin{cases} n & {m\mid n}, \\ 0 & \mathrm{otherwise.}\end{cases}$
• If $\zeta$ is a primitive nth root of unity, then the roots of unity can be expressed as $1, \zeta, \zeta^2,\ldots,\zeta^{n-1}$.
• Also, don't overlook the most obvious property of all! For each $n$th root of unity, $\zeta$, we have that $\zeta^n=1$
## Uses
Roots of unity show up in many surprising places. Here, we list a few: |
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