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# What is the Fourth Root of 81? | 4th root of 81
Let m be a number such that m4=81. In this case, m is called a fourth root of 81. The fourth roots of 81 are the solutions of x4=81. It has four solutions since it has degree four. In this article, we will learn about the fourth roots of eighty-one.
$\sqrt[4]{81}$ is the radical form of the fourth root of 81 with index 4.
Fourth root of 81 can be written as 811/4 according to the rule of indices. That is,
$\sqrt[4]{81}=81^{1/4}$
## What is the fourth root of 81?
Note that 81 can be expressed as a product of four numbers of 3‘s, that is,
81=3×3×3×3
81=34
Taking fourth root on both sides, we get that
$\sqrt[4]{81}=\sqrt[4]{3^4}$
As the fourth root is the same as the power 1/4, we have
$\sqrt[4]{81}=(3^4)^{1/4}$
$\Rightarrow \sqrt[4]{81}=3^{4 \times 1/4}$ $[\because (a^m)^n=a^{m \times n}]$
$\Rightarrow \sqrt[4]{81}=3^{1}=3$
So the value of the fourth root of 81 is 3. This is one of the real fourth roots of 81.
## Is Fourth Root of 81 Rational?
Since the fourth root of 81 is equal to 3 and we know that 3 is a rational number, then we conclude that the fourth root of 81 is a rational number; in fact, it is a whole number (or an integer).
For the above same reason, 81 is a perfect fourth number.
## How to Find Fourth Root of 81?
Now, we will find the fourth root of 81 by the prime factorization method. Observe that
81=3×27, 27=3×9 and 9=3×3
Thus, 81 = 3×27 = 3×3×9 = 3×3×3×3.
So finally we get that
81 = 3×3×3×3
$\therefore \sqrt[4]{81}=\sqrt[4]{3\times 3 \times 3 \times 3}=3$ as we know that $\sqrt[4]{a \times a \times a \times a}=a$
So the fourth root of 81 is 3.
Question: What is the fourth root of 81?
Video Solution:
## Complex Fourth Roots of 81
We will find the solutions of $x^4=81$ to find all the complex fourth roots of 81. We have:
$x^4=81 \cdot 1$
$\therefore x=81^{1/4} \cdot 1^{1/4}$ $\cdots (*)$
$\Rightarrow x=3 \cdot 1^{1/4}.$,
That is, $x$ is equal to $3$ times the fourth roots of $1.$
We know that there are $4$ fourth roots of $1$ and they are $\pm 1, \pm i$. Here $i$ is a complex imaginary number.
So from $(*)$, we obtain that the fourth roots of $81$ are $\pm 3$ and $\pm 3i.$ Here $\pm 3$ are the real roots and $\pm 3i$ are the complex roots of $x^4-81=0.$ Thus there are $2$ reals roots and $2$ complex roots of the equation $x^4=81.$ All these are the fourth roots of $81.$ |
We think you are located in South Africa. Is this correct?
# Inverse Proportion Patterns, Relationships And Graphs
## 2.4 Inverse proportion patterns, relationships and graphs (EMG3G)
### Inverse proportion relationships and graphs (EMG3H)
Some relationships between quantities give patterns that form inverse proportion graphs. How do we recognise an inverse proportion relationship?
Remember from Chapter 1: in an inverse proportion, as one quantity decreases, the other increases OR as one quantity increases, the other decreases.
## Worked example 5: A graph of an inverse proportion
A rectangle has a fixed area of $$\text{32}$$ square units, but the length $$l$$ and breadth $$b$$ can both change. If the length is smaller, the breadth gets bigger, because the area stays the same.
1. Complete the following table of possible values for the length and breadth of the rectangle.
length $$l$$ $$\text{1}$$ $$\text{2}$$ $$\text{4}$$ $$\text{8}$$ $$\text{16}$$ $$\text{32}$$ breadth $$b$$ $$\text{32}$$
2. Draw a graph to show all the possible values of the length and breadth.
3. Is the graph continuous or discrete? Explain.
4. Why does the curve not touch the axes?
1. length $$l$$ $$\text{1}$$ $$\text{2}$$ $$\text{4}$$ $$\text{8}$$ $$\text{16}$$ $$\text{32}$$ breadth $$b$$ $$\text{32}$$ $$\text{16}$$ $$\text{8}$$ $$\text{4}$$ $$\text{2}$$ $$\text{1}$$
2. The graph is continuous, because measurement values are continuous. The curve is solid and not dotted because there are an infinite number of values between the points.
3. The graph doesn't touch the axes because the length and breadth can never be $$\text{0}$$.
The important things to note about the graph of an inverse proportion is that it is a smooth curve, and that the curve never touches the axes. In most cases, we will deal with a positive inverse proportion, because most real life values are positive. But it is possible for them to be negative too.
## Inverse proportion patterns
Exercise 2.3
Lerato decides to get a group of friends together to play a lucky draw game. The bigger the group, the more tickets they can buy, but if they win, they will have to share the prize among more people. The total amount of money is $$\text{R}\,\text{2 000}$$.
If they win the prize money, how much will they have to share?
$$\text{R}\,\text{2 000}$$
How will the number in the group affect the amount each person receives?
The larger the number of people in the group, the smaller the shared amount that each person will receive.
What kind of relationship is this?
An inverse proportional relationship.
Copy and complete the table below including the first column which is the headings for the independent and dependent variables.
Number of people $$\text{1}$$ $$\text{2}$$ $$\text{3}$$ $$\text{4}$$ $$\text{8}$$ $$\text{10}$$ $$\text{50}$$ Share of the prize money $$\text{2 000}$$ $$\text{1 000}$$
Number of people $$\text{1}$$ $$\text{2}$$ $$\text{3}$$ $$\text{4}$$ $$\text{8}$$ $$\text{10}$$ $$\text{50}$$ Share of the prize money $$\text{2 000}$$ $$\text{1 000}$$ $$\text{666,67}$$ $$\text{500}$$ $$\text{250}$$ $$\text{200}$$ $$\text{100}$$
Plot a graph of these points to show the relationship.
Meryl wants to make a garden bed with an area of $$\text{16}$$ $$\text{m^{2}}$$.
Draw up a table to show a few possible length and breadth measurements of the garden bed.
Breadth (m) $$\text{1}$$ $$\text{2}$$ $$\text{4}$$ $$\text{8}$$ $$\text{16}$$ Length (m) $$\text{16}$$ $$\text{8}$$ $$\text{4}$$ $$\text{2}$$ $$\text{1}$$
Do the measurements have to be whole numbers? Explain.
No. Any two numbers whose product is $$\text{16}$$ can be used, because length and breadth are continuous variables. Whole numbers will be easier to calculate and plot on a graph, however.
Draw a graph to show the relationship between the length and the breadth of the garden bed. |
# SOLVING ABSOLUTE VALUE EQUATIONS
The general form of an absolute value equation is
|ax + b| = k
In the above absolute value equation, we can notice that there is only absolute part on the left side.
(Here 'a' and 'k' are real numbers and k 0)
Let us consider the absolute value equation |2x + 3| = 5.
We can solve the absolute value equation |2x + 3| = 5 as shown below.
The following steps will be useful to solve absolute value equations.
Step 1 :
Get rid of absolute sign and divide it into two branches.
Step 2 :
For the first branch, take the sign as it is on the right side.
Step 3 :
For the second branch, change the sign on the right side.
Step 4 :
Then solve both the branches.
Solve the following absolute value equations :
Example 1 :
Solve the absolute value equation :
|3x + 5| = 7
Solution :
|3x + 5| = 7
3x + 5 = 7 or 3x + 5 = -7
3x = 2 or 3x = -12
x = 2/3 or x = -4
Example 2 :
Solve the absolute value equation :
|7x| = 21
Solution :
|7x| = 21
7x = 21 or 7x = -21
x = 3 or x = -3
Example 3 :
Solve the absolute value equation :
|2x + 5| + 6 = 7
Solution :
|2x + 5| + 6 = 7
Subtract 6 from each side.
|2x + 5| = 1
2x + 5 = 1 or 2x + 5 = 1
2x = -4 or 2x = -6
x = -2 or x = -3
Example 4 :
Solve the absolute value equation :
|x - 3| + 6 = 6
Solution :
|x - 3| + 6 = 6
Subtract 6 from each side.
|x - 3| = 0
x - 3 = 0
x = 3
Example 5 :
Solve the absolute value equation :
2|3x +4| = 7
Solution :
2|3x +4| = 7
Divide each side by 2.
|3x + 4| = 7/2
3x + 4 = 7/2 or 3x + 4 = -7/2
3x = 7/2 - 4 or 3x = -7/2 - 4
3x = -1/2 or 3x = -15/2
x = -1/6 or x = -15/6
x = -1/6 or x = -5/2
Example 6 :
Solve the absolute value equation :
3|5x - 6| - 4 = 5
Solution :
3|5x - 6| - 4 = 5
3|5x - 6| = 9
Divide each side by 3.
|5x - 6| = 3
5x - 6 = 3 or 5x - 6 = -3
5x = 9 or 5x = 3
x = 9/5 or x = 3/5
Example 7 :
Solve the absolute value equation :
|x² - 4x - 5| = 7
Solution :
|x2 - 4x - 5| = 7
x2 - 4x - 5 = 7 or x2 - 4x - 5 = -7
x2 - 4x - 12 = 0 or x2 - 4x + 2 = 0
Solve the first quadratic equation x2 - 4x - 12 = 0.
x2 - 4x - 12 = 0
(x + 2)(x - 6) = 0
x + 2 = 0 or x - 6 = 0
x = -2 or x = 6
Solve the second quadratic equation x2 - 4x + 2 = 0.
This quadratic equation can not be solved by factoring. So, we can use quadratic formula and solve the equation as shown below.
Comparing ax2 + bx + c = 0 and x2 - 4x + 2 = 0, we get
a = 1, b = -4, c = 2
Substitute a = 1, b = -4 and c = 2.
So, the solution is x = -2, 7, 2 ± √2.
Example 8 :
Solve the absolute value equation :
0.5|0.5x| - 0.5 = 2.5
Solution :
0.5|0.5x| - 0.5 = 2.5
0.5|0.5x| = 3
Divide each side by 0.5
|0.5x| = 6
0.5x = 6 or 0.5x = -6
x = 12 or x = -12
Example 9 :
If the absolute value equation |2x + k| = 3 has the solution x = -2, find the value of k.
Solution :
Because x = -2 is a solution, substitute x = -2 in the given absolute value equation.
|2(-2) + k| = 3
|-4 + k| = 3
Solve for k :
-4 + k = 3 or -4 + k = -3
k = 7 or k = 1
Example 10 :
If the absolute value equation |x - 3| - k = 0 has the solution x = -5, find the value of k.
Solution :
Because x = -2 is a solution, substitute x = -2 in the given absolute value equation.
|-5 - 3| - k = 0
|-8| - k = 0
8 - k = 0
8 = k
Kindly mail your feedback to v4formath@gmail.com
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# 4.7 Solve Equations with Fractions
The topics covered in this section are:
## 4.7.1 Determine Whether a Fraction is a Solution of an Equation
As we saw in Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers: The Division Property of Equality, a solution of an equation is a value that makes a true statement when substituted for the variable in the equation. In those sections, we found whole number and integer solutions to equations. Now that we have worked with fractions, we are ready to find fraction solutions to equations.
The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, or a fraction.
## HOW TO: Determine whether a number is a solution to an equation.
1. Substitute the number for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.
#### Example 1
Determine whether each of the following is a solution of $x – \frac{3}{10} = \frac{1}{2}$.
1. $x=1$
2. $x= \frac{4}{5}$
3. $x= – \frac{4}{5}$
Solution
Since $x=1$ does not result in a true equation, $1$ is not a solution to the equation.
Since $x= \frac{4}{5}$ results in a true equation, $\frac{4}{5}$ is a solution to the equation $x – \frac{3}{10} = \frac{1}{2}$.
Since $x= – \frac{4}{5}$ does not results in a true equation, $- \frac{4}{5}$ is not a solution to the equation.
## 4.7.2 Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality
In Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers: The Division Property of Equality, we solved equations using the Addition, Subtraction, and Division Properties of Equality. We will use these same properties to solve equations with fractions.
### ADDITION, SUBTRACTION, AND DIVISION PROPERTIES OF EQUALITY
For any numbers $a,b$, and $c$,
In other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.
#### Example 2
Solve: $y+ \frac{9}{16} = \frac{5}{16}$
Solution
Since $y=- \frac{1}{4}$ makes $y+ \frac{9}{16} = \frac{5}{16}$ a true statement, we know we have found the solution to this equation.
We used the Subtraction Property of Equality in Example 2. Now we’ll use the Addition Property of Equality.
#### Example 3
Solve: $a – \frac{5}{9} = – \frac{8}{9}$.
Solution
Since $a= – \frac{1}{3}$ makes the equation true, we know that $a= – \frac{1}{3}$ is the solution to the equation.
The next example may not seem to have a fraction, but let’s see what happens when we solve it.
#### Example 4
Solve: $10q = 44$.
Solution
The solution to the equation was the fraction $\frac{22}{5}$. We leave it as an improper fraction.
## 4.7.3 Solve Equations with Fractions Using the Multiplication Property of Equality
Consider the equation $\frac{x}{4} = 3$. We want to know what number divided by $4$ gives $3$. So to “undo” the division, we will need to multiply by $4$. The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.
### THE MULTIPLICATION PROPERTY OF EQUALITY
For any numbers $a,b$, and $c$,
if $a=b$, then $ac=bc$.
If you multiply both sides of an equation by the same quantity, you still have equality.
Let’s use the Multiplication Property of Equality to solve the equation $\frac{x}{7} = -9$.
#### Example 5
Solve: $\frac{x}{7} = -9$.
Solution
#### Example 6
Solve: $\frac{p}{-8} = -40$.
Solution
Here, $p$ is divided by $-8$. We must multiply by $-8$ to isolate $p$.
#### Solve Equations with a Coefficient of $-1$
Look at the equation $-y=15$. Does it look as if $y$ is already isolated? But there is a negative sign in front of $y$, so it is not isolated.
There are three different ways to isolate the variable in this type of equation. We will show all three ways in Example 7.
#### Example 7
Solve: $-y=15$.
Solution
One way to solve the equation is to rewrite $-y$ as $-1y$, and then use the Division Property of Equality to isolate $y$.
Another way to solve this equation is to multiply both sides of the equation by $-1$.
The third way to solve the equation is to read $-y$ as “the opposite of $y$.” What number has $15$ as its opposite? The opposite of $15$ is $-15$. So $y=-15$.
For all three methods, we isolated $y$ and solve the equation.
Check:
#### Solve Equations with a Fraction Coefficient
When we have an equation with a fraction coefficient we can use the Multiplication Property of Equality to make the coefficient equal to $1$.
For example, in the equation:
$\large \frac{3}{4} x = 24$
The coefficient of $x$ is $\frac{3}{4}$. To solve for$x$, we need its coefficient to be $1$. Since the product of a number and its reciprocal is $1$, our strategy here will be to isolate $x$ by multiplying by the reciprocal of $\frac{3}{4}$. We will do this in Example 8.
#### Example 8
Solve: $\frac{3}{4} x = 24$.
Solution
Notice that in the equation $\frac{3}{4} x =24$, we could have divided both sides by $\frac{3}{4}$ to get $x$ by itself. Dividing is the same as multiplying by the reciprocal, so we would get the same result. But most people agree that multiplying by the reciprocal is easier.
#### Example 9
Solve: $- \frac{3}{8} w = 72$.
Solution
The coefficient is a negative fraction. Remember that a number and its reciprocal have the same sign, so the reciprocal of the coefficient must also be negative.
## 4.7.4 Translate Sentences to Equations and Solve
Now we have covered all four properties of equality—subtraction, addition, division, and multiplication. We’ll list them all together here for easy reference.
When you add, subtract, multiply or divide the same quantity from both sides of an equation, you still have equality.
In the next few examples, we’ll translate sentences into equations and then solve the equations. It might be helpful to review the translation table in Evaluate, Simplify, and Translate Expressions.
#### Example 10
Translate and solve: $n$ divided by $6$ is $-24$.
Solution
#### Example 11
Translate and solve: The quotient of $q$ and $-5$ is $70$.
Solution
#### Example 12
Translate and solve: Two-thirds of $f$ is $18$.
Solution
#### Example 13
Translate and solve: The quotient of $m$ and $\frac{5}{6}$ is $\frac{3}{4}$.
Solution
Our solution checks.
#### Example 14
Translate and solve: The sum of three-eighths and $x$ is three and one-half.
Solution
We write the answer as a mixed number because the original problem used a mixed number.
Check:
Is the sum of three-eighths and $3 \frac{1}{8}$ equal to three and one-half?
The solution checks.
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# Solve ((x+1)/(x+2))+((x+2)/(x+3))+7/2=2 ?
Sep 17, 2016
$x \cong - 1.63841$ or $\cong - 2.79016$
#### Explanation:
$\left(\frac{x + 1}{x + 2}\right) + \left(\frac{x + 2}{x + 3}\right) + \frac{7}{2} = 2$
$\frac{\left(x + 1\right) \left(x + 3\right) + {\left(x + 2\right)}^{2}}{\left(x + 2\right) \left(x + 3\right)} = - \frac{3}{2}$
$2 \left(x + 1\right) \left(x + 3\right) + 2 {\left(x + 2\right)}^{2} = - 3 \left(x + 2\right) \left(x + 3\right)$
$2 \left({x}^{2} + 4 x + 3\right) + 2 \left({x}^{2} + 4 x + 4\right) = - 3 \left({x}^{2} + 5 x + 6\right)$
$2 {x}^{2} + 8 x + 6 + 2 {x}^{2} + 8 x + 8 = - 3 {x}^{2} - 15 x - 18$
$7 {x}^{2} + 31 x + 32 = 0$
Using the quadratic formula:
$x = \frac{- 31 \pm \sqrt{{31}^{2} - 4 \cdot 7 \cdot 32}}{2 \cdot 7}$
$x = \frac{- 31 \pm \sqrt{65}}{14}$
$x \cong \frac{- 31 \pm 8.06226}{14}$
$x \cong - 1.63841$ or $- 2.79016$ |
# Completing the Square (Solving for $x$ with fractions)
Im working through some completing the square questions but this question I'm currently working on has a few fractions which seem to be tripping me up. Some advice on where I have gone wrong would be greatly appreciated.
Q) Complete the Square to solve for $x$: $\qquad x^2 - 5x - 6 = 0$
So far I have:
\begin{align} x^2 - 5x &= 6 \\ x^2 - 5x + \frac{25}4 &= \frac{49}4 \\ \left(x - \frac52\right)\left(x - \frac52\right) &= \frac{49}4 \\ \left(x - \frac52\right)^2 = \frac{49}4 \end{align}
Take the square root:
\begin{align} x - \frac52 &= \pm \frac72 \\ x_1 &= \frac72 + \frac52 = \frac{12}2 = 6 \\ x_2 &= -\frac72 + \frac52 = -\frac22 = -1 \end{align}
What you have done is correct, $$0= x^2-5x-6 = \left( x-\frac{5}{2} \right)^2 - \frac{5^2}{2^2}-6 \\ = \left( x-\frac{5}{2} \right)^2 - \frac{49}{4},$$ so $$\left( x-\frac{5}{2} \right) = \pm \frac{7}{2},$$ so $x=-1$ or $x=6$. You can check these in the original equation: $$(-1)^2-5(-1)-6 = 0\\ 6^2-5 \times 6 -6 = 0.$$ Also, it is possible to spot the factorisation as $$(x+1)(x-6).$$ |
# What is 1/155 as a decimal?
## Solution and how to convert 1 / 155 into a decimal
1 / 155 = 0.006
Convert 1/155 to 0.006 decimal form by understanding when to use each form of the number. Both represent numbers between integers, in some cases defining portions of whole numbers Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 1/155 is 1 divided by 155
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We must divide 1 into 155 to find out how many whole parts it will have plus representing the remainder in decimal form. Here's 1/155 as our equation:
### 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 below the vinculum at 155.
### Denominator: 155
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. Larger values over fifty like 155 makes conversion to decimals tougher. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. Now let's dive into how we convert into decimal format.
## Converting 1/155 to 0.006
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 155 \enclose{longdiv}{ 1 }$$
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
$$\require{enclose} 00. \\ 155 \enclose{longdiv}{ 1.0 }$$
We've hit our first challenge. 1 cannot be divided into 155! Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 155 into 10.
### Step 3: Solve for how many whole groups you can divide 155 into 10
$$\require{enclose} 00.0 \\ 155 \enclose{longdiv}{ 1.0 }$$
We can now pull 0 whole groups from the equation. Multiple this number by our furthest left number, 155, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 155 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If your remainder is zero, that's it! If you still have a remainder, 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. And the same is 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 each represent values in everyday life! Here are examples of when we should use each.
### When you should convert 1/155 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 1/155 MPH. The radar will read: 90.0 MPH. This simplifies the value.
### When to convert 0.006 to 1/155 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 1/155 = 0.006 what would it be as a percentage?
• What is 1 + 1/155 in decimal form?
• What is 1 - 1/155 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.006 + 1/2? |
# How do I solve this following inequality? Solve the following inequality. `(3x-5)/(x+3)lt=2` (Write the answer in interval notation. Simplify the answer. Use integers or fractions for any numbers...
How do I solve this following inequality?
Solve the following inequality.
`(3x-5)/(x+3)lt=2`
(Write the answer in interval notation. Simplify the answer. Use integers or fractions for any numbers in the expression.)
Asked on by amber756
lemjay | High School Teacher | (Level 3) Senior Educator
Posted on
`(3x-5)/(x+3)lt=2`
To start, change the inequality sign to "`=`" .Then, make the right side zero by subtracting both sides by 2.
`(3x-5)/(x+3)-2=0`
Simplify left side.
`(3x-5)/(x+3)- (2(x+3))/(x+3)=0`
`(3x-5)/(x+3)-(2x+6)/(x+3)=0`
`(x-11)/(x+3)=0`
Then, set the numerator and denominator equal to zero. And solve for x.
`x-11=0` and `x+3=0`
`x=11` `x=-3`
The two values of x above are referred as critical numbers and it divides the number line into three intervals. In each interval, assign a test value and substitute it to the original inequality equation to be able to determine if it satisfy the condition. The interval that satisfy or result to a true condition is the solution.
For interval `xlt-3` , test value is x=-4.
`(3(-4)-5)/(-4+3)lt=2`
`17lt=2 ` (False)
For interval `-3ltxlt11` , test value is x=0.
`(3*0-5)/(0+3)lt=2`
`-1 2/3lt=2` (True)
For interval, `xgt11` , test value is x=12
`(3*12-5)/(12+3)lt=2`
`2 1/15 lt= 2` (False)
Furthermore, substitute the critical numbers to check if they are included in the solution.
`x=-3`, `(3(-3)-5)/(-3+3)lt=2`
`-14/0lt=2`
Note that in fractions, zero denominator is not allowed. So, x=-3 is not a solution.
`x=11` , `(3*11-5)/(11+3)lt=2`
`2lt=2 ` (True)
Hence, the solution set of the inequality equation `(3x-5)/(x+3)lt=2` is `-3ltxlt=11` .
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Q:
# How do you subtract hexadecimal numbers?
A:
Subtracting hexadecimal numbers is not much different than subtracting decimal numbers once you understand what a hexadecimal number is. The hexadecimal system is based on the number 16 instead of the number 10 like the familiar decimal system.
## Keep Learning
1. Understand how to count with hexadecimal numbers
Just as the decimal system has 10 digits, 0 through 9, the hexadecimal system has 16 digits. The 16 digits of hexadecimal are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F.
2. Convert hexadecimal letters into decimal equivalents
When doing operations such as subtraction with hexadecimal numbers, think of the letters as equal to numbers, so A equals 10, B equals 11, C equals 12, D equals 13, E equals 14 and F equals 15.
3. Line up numbers with more than one place, and subtract from right to left
For a problem with more than one digit, line the two numbers up just as in decimal subtraction, aligning the placeholder positions. For example, FF minus D1 is aligned into F minus 1 and F minus D. As with decimal subtraction problems, begin at the right. F minus 1 is the same as 15 minus 1 which equals 14, or E in the hexadecimal system. F minus D is the same as 15 minus 13, which equals 2. Therefore, FF minus D1 equals 2E.
4. Borrow 16 from the next higher place when necessary
In decimal subtraction, when the number to be subtracted is greater than the number to be subtracted from, you solve by borrowing a 10 from the next higher place. In hexadecimal subtraction, you borrow a 16 instead. So for 26 minus 18, because 8 is greater than 6, you must borrow a 16 from the 2. Add the 16 to the 6. This gives you 22, from which the 8 is subtracted to equal 14, or E. Since you borrowed from the higher position, only 1 is left from which to subtract 1. In other words, 26 minus 18 is equal to E in the hexadecimal system, not 8 as it is in decimal.
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# Limits at Infinity
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Definition of Limits at infinity
Let 'L' be any real number.
1)$\lim_{x \rightarrow \infty}f(x) = L$ means that for each $\epsilon$ > 0 there exists M > 0 such that
|f(x) - L| < $\epsilon$ whenever x > M.
2) $\lim_{x \rightarrow - \infty}f(x) = L$ means that for each $\epsilon$ > 0 there exists N < 0 such that |f(x) - L| < $\epsilon$ whenever x < N.
Horizontal asymptotes
The line is a horizontal asymptote of the graph of when
$\lim_{x \rightarrow -\infty}f(x) = L$ OR $\lim_{x \rightarrow \infty}f(x)$ = L
Vertical asymptotes at x = a
The line is a vertical asymptote of the graph of when
$\lim_{x \rightarrow a}f(x) = \infty$ OR $\lim_{x \rightarrow \infty}f(x) = -\infty$
STEPS FOR FINDING LIMITS AT $\pm\infty$ OF RATIONAL FUNCTIONS
1) If the degree of the numerator < the degree of the denominator, then the limit of the rational function is 0.
2) If the degree of the numerator = the degree of the denominator, then the limit of the rational function is the ratio of the leading coefficients.
3) If the degree of the numerator > the degree of the denominator, then the limit of the rational function does not exist.
## Examples on Limits at Infinity
Example 1 : Find the limit of $\lim_{x \rightarrow \infty}\frac{2x-1}{3x +5}$
Solution : $\lim_{x \rightarrow \infty}\frac{2x-1}{3x +5}$
Take x as a common factor from numerator and denominator
= $\lim_{x \rightarrow \infty}\frac{x(2-\frac{1}{x})}{x(3 +\frac{5}{x})}$
= $\lim_{x \rightarrow \infty}\frac{2-\frac{1}{x}}{3 +\frac{5}{x}}$
( $\lim_{x \rightarrow \infty}\frac{1}{x} = 0)$
$\lim_{x \rightarrow \infty}\frac{2x-1}{3x +5} = \frac{2}{3}$
Example 2 : Find the limit of $\lim_{x \rightarrow \infty}\frac{2x^{2}+4 }{x^{2} -5x -1}$
Solution : $\lim_{x \rightarrow \infty}\frac{2x^{2}+4 }{x^{2} -5x -1}$
Take $x^{2}$ as a common factor from numerator and denominator
= $\lim_{x \rightarrow \infty}\frac{x^{2}(2+\frac{4}{x^{2}})}{x^{2}(1 -\frac{5}{x}-\frac{1}{x^{2}})}$
$\lim_{x \rightarrow \infty}\frac{(2+\frac{4}{x^{2}})}{(1 -\frac{5}{x}-\frac{1}{x^{2}})}$ ( $\lim_{x \rightarrow \infty}\frac{1}{x} = 0$ and $\lim_{x \rightarrow \infty}\frac{1}{x^{2}} = 0$)
$\lim_{x \rightarrow \infty}\frac{2x^{2}+4 }{x^{2} -5x -1}$ =2 |
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# NCERT solutions for Class 10 Mathematics chapter 8 - Introduction to Trigonometry
## Chapter 8: Introduction to Trigonometry
Ex. 8.10Ex. 8.20Ex. 8.30Ex. 8.40
#### Chapter 8: Introduction to Trigonometry Exercise 8.10 solutions [Page 181]
Ex. 8.10 | Q 1.1 | Page 181
In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine
sin A, cos A
Ex. 8.10 | Q 1.2 | Page 181
In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine sin C, cos C
Ex. 8.10 | Q 2 | Page 181
In Given Figure, find tan P – cot R.
Ex. 8.10 | Q 3 | Page 181
If sin A =3/4 , calculate cos A and tan A.
Ex. 8.10 | Q 4 | Page 181
Given 15 cot A = 8. Find sin A and sec A
Ex. 8.10 | Q 5 | Page 181
Given sec θ = 13/12 , calculate all other trigonometric ratios.
Ex. 8.10 | Q 6 | Page 181
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Ex. 8.10 | Q 7.1 | Page 181
If cot θ = 7/8 evaluate ((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ))
Ex. 8.10 | Q 7.2 | Page 181
If cot θ = 7/8, evaluate cot2 θ
Ex. 8.10 | Q 8 | Page 181
If 3 cot A = 4, Check whether ((1-tan^2 A)/(1+tan^2 A)) = cos^2 A - sin^2 A or not
Ex. 8.10 | Q 9 | Page 181
In ΔABC, right angled at B. If tan A = 1/sqrt3 , find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C − sin A sin C
Ex. 8.10 | Q 10 | Page 181
In ΔPQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Ex. 8.10 | Q 11.1 | Page 181
The value of tan A is always less than 1.
Ex. 8.10 | Q 11.2 | Page 181
sec A = 12/5 for some value of angle A.
Ex. 8.10 | Q 11.3 | Page 181
cos A is the abbreviation used for the cosecant of angle A.
Ex. 8.10 | Q 11.4 | Page 181
State whether the following are true or false. Justify your answer. cot A is the product of cot and A
Ex. 8.10 | Q 11.5 | Page 181
sin θ =4/3, for some angle θ
#### Chapter 8: Introduction to Trigonometry Exercise 8.20 solutions [Page 187]
Ex. 8.20 | Q 1.1 | Page 187
Evaluate the following in the simplest form: sin 60º cos 30º + cos 60º sin 30º
Ex. 8.20 | Q 1.2 | Page 187
Evaluate the following : 2tan245° + cos230° − sin260°
Ex. 8.20 | Q 1.3 | Page 187
Evaluate the following : (cos 45°)/(sec 30° + cosec 30°)
Ex. 8.20 | Q 1.4 | Page 187
Evaluate the following
(sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
Ex. 8.20 | Q 1.5 | Page 187
Evaluate the following
(5cos^2 60° + 4sec^2 30° - tan^2 45°)/(sin^2 30°+cos^2 30°)
Ex. 8.20 | Q 2.1 | Page 187
Choose the correct option and justify your choice
(2 tan 30°)/(1+tan^2 30°)
• sin 60°
• cos 60°
• tan 60°
• sin 30°
Ex. 8.20 | Q 2.2 | Page 187
Choose the correct option and justify your choice.
(1- tan^2 45°)/(1+tan^2 45°)
• tan 90°
• 1
• sin 45°
• 0
Ex. 8.20 | Q 2.3 | Page 187
Choose the correct option and justify your choice :
sin 2A = 2 sin A is true when A =
• 30°
• 45°
• 60°
Ex. 8.20 | Q 2.4 | Page 187
Choose the correct option and justify your choice :
(2 tan 30°)/(1-tan^2 30°)
• cos 60°
• sin 60°
• tan 60°
• sin 30°
Ex. 8.20 | Q 3 | Page 187
If tan (A + B) = sqrt3 and tan (A – B) = 1/sqrt3 ; 0° < A + B ≤ 90° ; A > B, find A and B.
Ex. 8.20 | Q 4.1 | Page 187
sin (A + B) = sin A + sin B
• True
• False
Ex. 8.20 | Q 4.2 | Page 187
The value of sinθ increases as θ increases
• True
• False
Ex. 8.20 | Q 4.3 | Page 187
The value of cos θ increases as θ increases
• True
• False
Ex. 8.20 | Q 4.4 | Page 187
sinθ = cos θ for all values of θ
• True
• False
Ex. 8.20 | Q 4.5 | Page 187
cot A is not defined for A = 0°
• True
• False
#### Chapter 8: Introduction to Trigonometry Exercise 8.30 solutions [Pages 189 - 190]
Ex. 8.30 | Q 1.1 | Page 189
Evaluate (sin 18^@)/(cos 72^@)
Ex. 8.30 | Q 1.1 | Page 189
Evaluate (sin 18^@)/(cos 72^@)
Ex. 8.30 | Q 1.2 | Page 189
Evaluate (tan 26^@)/(cot 64^@)
Ex. 8.30 | Q 1.2 | Page 189
Evaluate (tan 26^@)/(cot 64^@)
Ex. 8.30 | Q 1.3 | Page 189
Evaluate cos 48° − sin 42°
Ex. 8.30 | Q 1.4 | Page 189
Evaluate cosec 31° − sec 59°
Ex. 8.30 | Q 1.4 | Page 189
Evaluate cosec 31° − sec 59°
Ex. 8.30 | Q 2.1 | Page 189
Show that tan 48° tan 23° tan 42° tan 67° = 1
Ex. 8.30 | Q 2.2 | Page 189
Show that cos 38° cos 52° − sin 38° sin 52° = 0
Ex. 8.30 | Q 2.2 | Page 189
Show that cos 38° cos 52° − sin 38° sin 52° = 0
Ex. 8.30 | Q 3 | Page 189
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A
Ex. 8.30 | Q 3 | Page 189
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A
Ex. 8.30 | Q 4 | Page 189
If tan A = cot B, prove that A + B = 90
Ex. 8.30 | Q 4 | Page 189
If tan A = cot B, prove that A + B = 90
Ex. 8.30 | Q 5 | Page 189
If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Ex. 8.30 | Q 5 | Page 189
If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Ex. 8.30 | Q 6 | Page 190
If A, B and C are interior angles of a triangle ABC, then show that \sin( \frac{B+C}{2} )=\cos \frac{A}{2}
Ex. 8.30 | Q 7 | Page 190
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
Ex. 8.30 | Q 7 | Page 190
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
#### Chapter 8: Introduction to Trigonometry Exercise 8.40 solutions [Pages 193 - 194]
Ex. 8.40 | Q 1 | Page 193
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Ex. 8.40 | Q 1 | Page 193
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Ex. 8.40 | Q 2 | Page 193
Write all the other trigonometric ratios of ∠A in terms of sec A.
Ex. 8.40 | Q 2 | Page 193
Write all the other trigonometric ratios of ∠A in terms of sec A.
Ex. 8.40 | Q 3.1 | Page 193
Evaluate
(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)
Ex. 8.40 | Q 3.1 | Page 193
Evaluate
(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)
Ex. 8.40 | Q 3.2 | Page 193
Evaluate sin25° cos65° + cos25° sin65°
Ex. 8.40 | Q 3.2 | Page 193
Evaluate sin25° cos65° + cos25° sin65°
Ex. 8.40 | Q 4.1 | Page 193
Choose the correct option. Justify your choice.
9 sec2 A − 9 tan2 A =
• 1
• 9
• 8
• 0
Ex. 8.40 | Q 4.1 | Page 193
Choose the correct option. Justify your choice.
9 sec2 A − 9 tan2 A =
• 1
• 9
• 8
• 0
Ex. 8.40 | Q 4.2 | Page 193
Choose the correct option. Justify your choice.
(1 + tan θ + sec θ) (1 + cot θ − cosec θ)
• 0
• 1
• 2
• -1
Ex. 8.40 | Q 4.2 | Page 193
Choose the correct option. Justify your choice.
(1 + tan θ + sec θ) (1 + cot θ − cosec θ)
• 0
• 1
• 2
• -1
Ex. 8.40 | Q 4.3 | Page 193
Choose the correct option. Justify your choice.
(secA + tanA) (1 − sinA) =
• secA
• sinA
• cosecA
• cosA
Ex. 8.40 | Q 4.3 | Page 193
Choose the correct option. Justify your choice.
(secA + tanA) (1 − sinA) =
• secA
• sinA
• cosecA
• cosA
Ex. 8.40 | Q 4.4 | Page 193
Choose the correct option. Justify your choice.
(1+tan^2A)/(1+cot^2A)
• secA
• −1
• cotA
• tanA
Ex. 8.40 | Q 4.4 | Page 193
Choose the correct option. Justify your choice.
(1+tan^2A)/(1+cot^2A)
• secA
• −1
• cotA
• tanA
Ex. 8.40 | Q 5.01 | Page 193
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)
Ex. 8.40 | Q 5.01 | Page 193
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)
Ex. 8.40 | Q 5.02 | Page 193
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
Ex. 8.40 | Q 5.02 | Page 193
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
Ex. 8.40 | Q 5.03 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(tantheta)/(1-cottheta) + (cottheta)/(1-tantheta) = 1+secthetacosectheta
Ex. 8.40 | Q 5.03 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(tantheta)/(1-cottheta) + (cottheta)/(1-tantheta) = 1+secthetacosectheta
Ex. 8.40 | Q 5.04 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(1+ secA)/sec A = (sin^2A)/(1-cosA)
[Hint : Simplify LHS and RHS separately]
Ex. 8.40 | Q 5.04 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(1+ secA)/sec A = (sin^2A)/(1-cosA)
[Hint : Simplify LHS and RHS separately]
Ex. 8.40 | Q 5.05 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA
Ex. 8.40 | Q 5.05 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA
Ex. 8.40 | Q 5.06 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
sqrt((1+sinA)/(1-sinA)) = secA + tanA
Ex. 8.40 | Q 5.06 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
sqrt((1+sinA)/(1-sinA)) = secA + tanA
Ex. 8.40 | Q 5.07 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta
Ex. 8.40 | Q 5.07 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta
Ex. 8.40 | Q 5.08 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Ex. 8.40 | Q 5.09 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(cosec A – sin A) (sec A – cos A)=1/(tanA+cotA)
[Hint : Simplify LHS and RHS separately]
Ex. 8.40 | Q 5.1 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
((1+tan^2A)/(1+cot^2A))=((1-tanA)/(1-cotA))^2=tan^2A
## Chapter 8: Introduction to Trigonometry
Ex. 8.10Ex. 8.20Ex. 8.30Ex. 8.40
## NCERT solutions for Class 10 Mathematics chapter 8 - Introduction to Trigonometry
NCERT solutions for Class 10 Maths chapter 8 (Introduction to Trigonometry) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Textbook for Class 10 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 10 Mathematics chapter 8 Introduction to Trigonometry are Introduction to Trigonometry, Introduction to Trigonometry Examples and Solutions, Trigonometric Ratios, Trigonometric Ratios of an Acute Angle of a Right-angled Triangle, Trigonometric Ratios of Some Specific Angles, Trigonometric Ratios of Complementary Angles, Trigonometric Identities, Proof of Existence, Relationships Between the Ratios, Trigonometric Ratios of Complementary Angles, Trigonometric Identities.
Using NCERT Class 10 solutions Introduction to Trigonometry exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 8 Introduction to Trigonometry Class 10 extra questions for Maths and can use Shaalaa.com to keep it handy for your exam preparation
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# Card Turnover Math Game
Continuing on from yesterday’s post on other ways to help your students recall the times tables, here I share with you a fun times tables card game.
## Card Turnover
This game is for 2 players. You will need a deck of cards with the picture cards removed (King, Queen, Jack, Joker). There are 2 ways this game can be played, I’ll explain both ways below.
### All times tables facts
• Shuffle the cards and divide them in to two equal piles, face down on the playing surface.
• Players take it in turns to turn over the top card on each pile.
• Players multiply the two numbers, and the first player to call out the correct answer is the winner. That player gets to keep both cards. For example, if the cards turned up are a 7 and a 4, then the problem becomes: 7 x 4. The player that calls out the correct answer of 28 keeps the cards.
• If both players say the answer at the same time, then they keep one card each.
• If the answer is incorrect, then both cards are placed back at the bottom of the piles.
• The winner is the player with the most cards at the end.
### Individual times tables
This game can also be adapted to focus on an individual times table, for example the 7 times tables.
• Shuffle the cards and place a card with a 7 on the playing surface face up. Place all the other cards face down next to the 7.
• Players take it in turns to turn over the top card from the pile.
• Players multiply the two numbers, and the first player to call out the correct answer is the winner. That player gets to keep the card turned over. For example, if the card turned over is a 6, then the problem becomes: 6 x 7. The player that calls out the correct answer of 42 keeps the card.
• If both players say the answer at the same time, then the card is placed back at the bottom of the pile.
• If the answer is incorrect, then the card is placed back at the bottom of the pile.
• The winner is the player with the most cards at the end.
### Tips and extension
• Have a calculator handy for students to settle any disputes.
• The Ace card can be used as the number 11 or the number 1.
• Include more sets of cards and have up to 4 students playing.
• Use jumbo playing cards and place a magnetic strip on the back. Stick two cards on the whiteboard and get your students to call out the answer.
• The rules can be changed so that players can add the numbers, or subtract the lower number from the higher number.
• To make the game more challenging get your students to make their own sets of cards to include numbers up to 100. For example: 20, 30, 15, 37. They can then use these with an individual times tables, so when these cards are turned over they’re multiplied by the selected multiplier, e.g. 20 x 7.
• Alternatively change the multiplier to be 70, rather than 7. So for the example above the problem would become, 6 x 70.
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### 5 Responses to Card Turnover Math Game
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Can you type this in an easy to print overview format?
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Thanks so much for being here and your comment. Wow, that must be interesting, home schooling your granddaughter 🙂 I’m glad you like the game and I’m sure you’ll find many more ideas and activities here that will help you.
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# Geometry
## Spherical Coordinates and Trigonometric Functions
In addition to understanding points, vectors, normals, and matrices, mastering the concept of expressing vectors in spherical coordinates proves immensely valuable in image rendering (and CG in general). While it's possible to render images without this knowledge, incorporating spherical coordinates often simplifies complex shading challenges. This chapter also serves as a prime opportunity to revisit trigonometric functions, crucial for navigating geometric problems in computer graphics.
## Trigonometric Functions
Trigonometry and the Pythagorean theorem are foundational to creating computer-generated imagery, as rendering is fundamentally a geometric endeavor. Starting with the sine and cosine functions, we'll explore how to determine angles from 2D coordinates. These functions are typically defined with respect to the unit circle, a circle with a radius of one. For a point $$P$$ on this circle, the $$x$$-coordinate is found using the cosine of the angle ($$\theta$$) formed by the $$x$$-axis and a line from the origin to $$P$$. Similarly, the $$y$$-coordinate is obtained using the sine of $$\theta$$. It's important to note that $$\theta$$ is measured in radians, and although angles may be easier to conceptualize in degrees, a conversion to radians is necessary for computation in C++:
$$\theta_{radians} = {\pi \over 180} \theta_{degrees}$$
Where a full circle's rotation (360 degrees) equals $$2\pi$$ radians.
Trigonometric functions also stem from the relationships between the sides of a right triangle. The tangent of an angle, for example, is the ratio of the opposite side to the adjacent side of the triangle. The arctangent function, or the inverse of tangent, is particularly useful in graphics programming. While the atan function calculates the arctangent, it doesn't account for the quadrant of the angle, potentially leading to inaccuracies. The atan2 function, however, considers the signs of both $$x$$ and $$y$$ coordinates, accurately determining the angle $$\theta$$. For a point with coordinates (0.707, 0.707), $$\theta$$ is $$\pi / 4$$, whereas for (-0.707, -0.707), $$\theta$$ should logically be $$3\pi / 4$$, a distinction properly handled by atan2.
Summarizing the trigonometric functions discussed:
$$\sin(\theta) = \frac{\text{opposite side}}{\text{hypothenuse}}, \quad \cos(\theta) = \frac{\text{adjacent side}}{\text{hypothenuse}}, \quad \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}$$
And their inverse functions used to find angles given a point $$P$$:
$$\theta = \text{acos}(P_x), \quad \theta = \text{asin}(P_y), \quad \theta = \text{atan2}(P_y, P_x)$$
The atan2 function uniquely returns angles in the range $$[-\pi, \pi]$$, with positive values for counter-clockwise angles and negative for clockwise, providing a comprehensive understanding of angle orientation.
Finally, the Pythagorean theorem, essential for various calculations such as ray-sphere intersections, states:
$$\text{hypothenuse}^2 = \text{adjacent}^2 + \text{opposite}^2$$
This principle asserts that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the triangle's other two sides.
## Representing Vectors with Spherical Coordinates
Vectors, commonly represented in Cartesian coordinates by three values corresponding to each axis, can alternatively be described using spherical coordinates, which utilize just two angles. These are the vertical angle ($$\theta$$), shown in red, and the horizontal angle ($$\phi$$), depicted in green, as illustrated in the accompanying figures. The vertical angle, $$\theta$$, measures the vector's deviation from the vertical axis, while the horizontal angle, $$\phi$$, gauges the angle between the vector's projection onto the horizontal plane and a predefined right vector of the Cartesian system.
Consistently within the computer graphics (CG) community, $$\theta$$ denotes the vertical angle, and $$\phi$$ represents the horizontal angle, a convention that aids in maintaining clarity across different texts and applications. These angles are measured in radians, with $$\theta$$ spanning from 0 to $$\pi$$ and $$\phi$$ from 0 to $$2\pi$$, encompassing a full circle around the axis. This method of vector representation, referred to as spherical coordinates, offers a compact way to encode direction, reducing the information needed from three values to two, provided the vector's length is not of concern.
Spherical coordinates not only simplify the representation but also play a crucial role in shading techniques, where understanding the direction relative to light sources and surfaces is key. The transition from Cartesian to spherical coordinates involves recognizing the vector in terms of its orientation within a 3D space, defined against the right ($$V_r$$), up ($$Vu$$), and forward ($$V_f$$) axes, rather than the traditional $$x$$, $$y$$, and $$z$$. This choice avoids confusion with the Cartesian system's axis names and highlights the orientation's role in defining direction.
While vectors in spherical coordinates are often normalized to unit length for simplicity, the system also accommodates vectors of any length by including a radial distance component ($$r$$). This addition, representing the vector's magnitude, complements the angular measurements of $$\theta$$ (polar angle) and $$\phi$$ (azimuthal angle), offering a full description of direction and magnitude in three-dimensional space.
The conversion process from Cartesian coordinates to spherical involves calculating the angles $$\theta$$ and $$\phi$$ based on the vector's orientation relative to the vertical axis and its projection on the horizontal plane, respectively. This approach not only conserves memory in digital applications by minimizing data storage requirements but also enhances algorithms related to shading and light interaction, where angles play a significant role in determining visual outcomes.
## Conventions Again: Z is Up!
In the realms of mathematics and physics, a conventional approach is to represent spherical coordinates within a Cartesian coordinate system where the z-axis is designated as the up vector. This differs from some computer graphics contexts where the y-axis might serve as the up vector. The standard convention, as illustrated in Figure 5, employs a left-hand coordinate system for spherical coordinates, with the z-axis as the up vector and the x- and y-axes as the right and forward vectors, respectively.
This convention is widely accepted in scholarly articles and educational resources on spherical coordinates. Although this might differ from the y-up axis convention familiar to some graphics applications, it's important to adapt to this norm for consistency in discussions and code related to spherical coordinates. The significance of this convention becomes apparent when we delve into shading, where a common technique involves transforming vectors from world space to a local coordinate system. In this local system, the surface normal at the point being shaded acts as the up vector.
In the forthcoming chapter on Creating an Orientation Matrix or Local Coordinate System, we'll explore constructing a transformation matrix that aligns with this convention. Unlike the typical approach of aligning the tangent (x-axis), normal (y-axis), and bitangent (z-axis) with the matrix's rows, we'll arrange them as follows, swapping the positions of the normal and bitangent vectors:
$$\begin{bmatrix}T_x&T_y&T_z&0\\B_x&B_y&B_z&0\\N_x&N_y&N_z&0\\0&0&0&1\end{bmatrix}$$
Here, $$T$$, $$B$$, and $$N$$ symbolize the tangent, bitangent, and normal vectors, respectively. Consider a scenario where the normal vector points directly upwards in world space, with coordinates (0, 1, 0), and we construct a matrix as described:
$$\begin{bmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$$
For a vector $$v$$ with coordinates (0, 1, 0), paralleling the world space y-axis, matrix-vector multiplication yields:
$$\begin{array}{l} x = Vx \cdot M_{00} + Vy \cdot M_{10} + Vz \cdot M_{20} = 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 0\\ y = Vx \cdot M_{01} + Vy \cdot M_{11} + Vz \cdot M_{21} = 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 1 = 0\\ z = Vx \cdot M_{02} + Vy \cdot M_{12} + Vz \cdot M_{22} = 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 \end{array}$$
Following transformation, the vector aligns with the z-axis, now representing the up vector in this local coordinate system. This adjustment might seem counterintuitive, especially when visualizing the result in a 3D application where the y-axis traditionally represents up. However, this approach effectively swaps the y- and z-coordinates, aligning the vector with the z-up convention used in mathematics and physics.
## Translating Cartesian Coordinates into Spherical Coordinates
Assuming a vector is normalized simplifies its conversion to spherical coordinates. The left-hand image in Figure 6 mirrors the upper image from Figure 4, with a notable difference: the z-axis now serves as the vertical reference. Rotating this diagram by 45 degrees, as shown on the right side of Figure 6, aligns it with the scenario depicted in Figure 1, where x-coordinates were derived using $$\cos(\theta)$$. This parallel allows us to deduce that Vz is similarly determined by $$\cos(\theta)$$ in this context. Consequently, $$\theta$$ can be accurately calculated through the arccosine of Vz:
$$V_z = \cos(\theta) \implies \theta = \text{acos}(V_z)$$
In practical terms, within a C++ environment, this translates to:
float theta = acos(Vz);
Next, we delve into determining $$\phi$$, the horizontal angle. Observing Figure 7, which reflects the lower illustration of Figure 4, the axes are relabeled to x (right axis) and y (forward axis) in red and green, respectively. Recalling our brief overview of trigonometric functions, the tangent of an angle is identified by the ratio of Vy (opposite side) to Vx (adjacent side) within a right-angled triangle. Although one might consider computing $$\phi$$ using arccosine similar to $$\theta$$, it's crucial to remember $$\phi$$'s range extends from 0 to $$2\pi$$. Employing the tangent and specifically the atan2 function in C++ offers an advantage by factoring in the signs of Vy and Vx. This method yields an angle ranging between 0 to $$\pi$$ for vectors on the unit circle's right and between 0 to $$-\pi$$ for those on the left. Programmers may need to adjust this outcome to fit within the $$[0:2\pi]$$ interval, ensuring universal applicability:
$$\tan(\phi) = \frac{V_y}{V_x} \implies \phi = \text{atan2}(V_y, V_x)$$
This computation is encapsulated in C++ as:
float phi = atan2(Vy, Vx);
This methodology facilitates the translation of vectors from Cartesian to spherical coordinates, enhancing both the theoretical understanding and practical application of vectors in 3D spaces.
## Converting Spherical Coordinates to Cartesian Coordinates
Transitioning from spherical to Cartesian coordinates involves a direct and elegant formula:
$$x = \cos(\phi) \sin(\theta), \quad y = \sin(\phi) \sin(\theta), \quad z = \cos(\theta)$$
While memorizing this conversion might seem daunting, logical reasoning simplifies its recall. The $$z$$ component's dependence on $$\theta$$ alone is straightforward, with $$V_z = \cos(\theta)$$ demonstrating this relationship. For the $$x$$ component, consider a scenario where $$V$$ aligns with the $$x$$-axis, specifically at coordinates (1, 0, 0), achievable when $$\theta = \pi / 2$$ and $$\phi = 0$$. Given that $$\sin(\pi / 2) = 1$$ and $$\cos(0) = 1$$, it follows that $$x = \sin(\theta) \cos(\phi)$$, a deduction also applicable for determining the $$y$$ component.
Here's a snippet of C++ code to perform the conversion from spherical angles to Cartesian coordinates:
template<typename T>
Vec3<T> sphericalToCartesian(const T &theta, const T &phi) {
return Vec3<T>(cos(phi) * sin(theta), sin(phi) * sin(theta), cos(theta));
};
This straightforward approach enables the efficient computation of Cartesian coordinates from spherical ones, enhancing the flexibility and utility of geometric transformations in computational applications.
## Enhanced Techniques with Trigonometric Functions
Following the explanation of converting between Cartesian and spherical coordinates and their inverse process, we introduce a few practical functions. These functions are invaluable in rendering applications for manipulating vectors across both coordinate systems.
### Calculating $$\theta$$ from Cartesian Coordinates
In our discussions, we adopt a left-hand coordinate system where the z-axis signifies the vertical direction for spherical coordinates. As previously discussed, the calculation of $$\theta$$ is given by:
template<typename T>
inline T sphericalTheta(const Vec3<T> &v) {
return acos(clamp<T>(v[2], -1, 1));
}
It's essential to ensure the input vector is normalized, keeping the z-coordinate within the [-1, 1] range. Employing clamping enhances the robustness of this function.
### Deriving $$\phi$$ from Cartesian Coordinates
The computation of $$\phi$$ presents a different challenge, as the atan function spans the range $$[-\pi, \pi]$$. To accommodate our needs, we adjust the range to $$[0, 2\pi]$$ as follows:
template<typename T>
inline T sphericalPhi(const Vec3<T> &v) {
T p = atan2(v[1], v[0]);
return (p < 0) ? p + 2 * M_PI : p;
}
### Simplifying Calculations for Trigonometric Ratios
Beyond calculating angular values, obtaining direct trigonometric ratios such as $$\cos(\theta)$$, $$\sin(\theta)$$, $$\cos(\phi)$$, and $$\sin(\phi)$$ can be more straightforward:
template<typename T> inline T cosTheta(const Vec3<T> &w) { return w[2]; }
However, determining $$\sin(\theta)$$ necessitates a deeper understanding. Given a unit-length vector, the Pythagorean theorem allows us to assert $$V_x^2 + V_y^2 = 1$$. With $$V_x = \cos(\theta)$$ and $$V_y = \sin(\theta)$$, it follows that:
$$\cos(\theta)^2 + \sin(\theta)^2 = 1 \implies \sin(\theta)^2 = 1 - \cos(\theta)^2$$
To facilitate this calculation, we introduce two functions: one for $$\sin(\theta)^2$$ and another for $$\sin(\theta)$$, leveraging the square root of the former's output.
### Projecting Vectors onto the XY Plane
The computation of $$\cos(\phi)$$ and $$\sin(\phi)$$ becomes nuanced when considering the vector's shadow—or its projection—onto the xy plane. As depicted in Figure 8, a unit-length vector in 3D doesn't necessarily translate to a unit-length projection unless $$\theta = \pi/2$$. This discrepancy arises from the vector's projection, $$v_p$$, varying in length depending on $$\theta$$.
The length of the projected vector $$v_p$$ is directly tied to $$\sin(\theta)$$, as shown in Figure 9. Normalizing $$v_p$$ by dividing its components by $$\sin(\theta)$$ enables the accurate calculation of $$\cos(\phi)$$ and $$\sin(\phi)$$ through the normalized components:
template<typename T>
inline T cosPhi(const Vec3<T> &w) {
T sintheta = sinTheta(w);
if (sintheta == 0) return 1;
return clamp<T>(w[0] / sintheta, -1, 1);
}
template<typename T>
inline T sinPhi(const Vec3<T> &w) {
T sintheta = sinTheta(w);
if (sintheta == 0) return 0;
return clamp<T>(w[1] / sintheta, -1, 1);
}
This advanced approach to vector manipulation using trigonometric functions and projections enriches the toolbox for rendering, allowing for sophisticated control over vector orientations and interactions within a 3D environment. |
# 3.5 Addition of velocities (Page 2/12)
Page 2 / 12
## Take-home experiment: relative velocity of a boat
Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the boat from one side of the tub to the other and perpendicular to the flow of water. Which way do you need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat.
## Adding velocities: a boat on a river
Refer to [link] , which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat’s velocity relative to an observer on the shore, ${\mathbf{\text{v}}}_{\text{tot}}$ . The velocity of the boat, ${\mathbf{\text{v}}}_{\text{boat}}$ , is 0.75 m/s in the $y$ -direction relative to the river and the velocity of the river, ${\mathbf{\text{v}}}_{\text{river}}$ , is 1.20 m/s to the right.
Strategy
We start by choosing a coordinate system with its $x$ -axis parallel to the velocity of the river, as shown in [link] . Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the $y$ -axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations ${v}_{\text{tot}}=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}$ and $\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)$ directly.
Solution
The magnitude of the total velocity is
${v}_{\text{tot}}=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}\text{,}$
where
${v}_{x}={v}_{\text{river}}=1\text{.}\text{20 m/s}$
and
${v}_{y}={v}_{\text{boat}}=0\text{.}\text{750 m/s.}$
Thus,
${v}_{\text{tot}}=\sqrt{\left(1\text{.}\text{20 m/s}{\right)}^{2}+\left(0\text{.}\text{750 m/s}{\right)}^{2}}$
yielding
${v}_{\text{tot}}=1\text{.}\text{42 m/s.}$
The direction of the total velocity $\theta$ is given by:
$\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)={\text{tan}}^{-1}\left(0\text{.}\text{750}/1\text{.}\text{20}\right)\text{.}$
This equation gives
$\theta =\text{32}\text{.}0º\text{.}$
Discussion
Both the magnitude $v$ and the direction $\theta$ of the total velocity are consistent with [link] . Note that because the velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only $32.0º$ ) the total velocity has relative to the riverbank.
## Calculating velocity: wind velocity causes an airplane to drift
Calculate the wind velocity for the situation shown in [link] . The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction $\text{20}\text{.0º}$ west of north.
Strategy
In this problem, somewhat different from the previous example, we know the total velocity ${\mathbf{\text{v}}}_{\text{tot}}$ and that it is the sum of two other velocities, ${\mathbf{\text{v}}}_{\text{w}}$ (the wind) and ${\mathbf{\text{v}}}_{\text{p}}$ (the plane relative to the air mass). The quantity ${\mathbf{\text{v}}}_{\text{p}}$ is known, and we are asked to find ${\mathbf{\text{v}}}_{\text{w}}$ . None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of ${\mathbf{\text{v}}}_{\text{w}}$ , then we can combine them to solve for its magnitude and direction. As shown in [link] , we choose a coordinate system with its x -axis due east and its y -axis due north (parallel to ${\mathbf{\text{v}}}_{\text{p}}$ ). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Vector Addition and Subtraction: Analytical Methods .)
#### Questions & Answers
Is the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?
what's electromagnetic induction
electromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.
Lukman
wow great
Salaudeen
what is mutual induction?
je
mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
how to undergo polarization
show that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin
show that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin
No idea.... Are you even sure this question exist?
Mavis
I can't even understand the question
yes it was an assignment question "^"represent raise to power pls
Gabriel
Gabriel
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .
no ideas
Augstine
if u at an angle of 12 degrees their period will be same so as their velocity, that means they both move simultaneously since both both hovers at same length meaning they have the same length
Modern cars are made of materials that make them collapsible upon collision. Explain using physics concept (Force and impulse), how these car designs help with the safety of passengers.
calculate the force due to surface tension required to support a column liquid in a capillary tube 5mm. If the capillary tube is dipped into a beaker of water
find the time required for a train Half a Kilometre long to cross a bridge almost kilometre long racing at 100km/h
method of polarization
Ajayi
What is atomic number?
The number of protons in the nucleus of an atom
Deborah
type of thermodynamics
oxygen gas contained in a ccylinder of volume has a temp of 300k and pressure 2.5×10Nm
why the satellite does not drop to the earth explain
what is a matter
Yinka
what is matter
Yinka
what is matter
Yinka
what is a matter
Yinka
I want the nuclear physics conversation
Mohamed
because space is a vacuum and anything outside the earth 🌎 can not come back without an act of force applied to it to leave the vacuum and fall down to the earth with a maximum force length of 30kcm per second
Clara
at t=0second,aparticles moving in x-y plain with aconstant acceleration has avelocity of initial velocity =(3i-2j)m/s and is at the origion.at t=3second the particle's velocity is final velocity=(9i+7j)then how to find the acceleration?
how about the formula like v^2=u^2+2as
Bayuo
a=v-u/t
Doreen
what is physics
Yinka
why is there a maximum distance at which the image can exist behind a convex mirror
The ball of a simple pendulum take 0.255 to swing from its equilibrium position to one extreme. Calculate it period.
The Ball of a simple pendulum take 0.255 to swing from its equilibrium position to one extreme. calculate its period
Abubakr
why is there a maximum distance at which the image can exist behind a convex mirror
Alfred
amplitude=0 .255s period=4×.255=1.02 sec period is one complete cycle
MUKHTAR |
# 2.10: Chapter 2 Formula Review
## 2.2 Measures of the Location of the Data
$$i=\left(\frac{k}{100}\right)(n+1)$$
where $$i$$ = the ranking or position of a data value,
$$k$$ = the $$k$$th percentile,
$$n$$ = total number of data.
Expression for finding the percentile of a data value: $$\left(\frac{x+0.5 y}{n}\right)(100)$$
where $$x$$ = the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile,
$$y$$ = the number of data values equal to the data value for which you want to find the percentile,
$$n$$ = total number of data
## 2.3 Measures of the Center of the Data
$$\mu=\frac{\sum f m}{\sum f}$$ Where $$f$$ = interval frequencies and $$m$$ = interval midpoints.
The arithmetic mean for a sample (denoted by $$\overline{x}$$) is $$\overline{x}=\frac{\text { Sum of all values in the sample }}{\text { Number of values in the sample }}$$
The arithmetic mean for a population (denoted by μ) is $$\boldsymbol{\mu}=\frac{\text { Sum of all values in the population }}{\text { Number of values in the population }}$$
## 2.5 Geometric Mean
The Geometric Mean: $$\overline{x}=\left(\prod_{i=1}^{n} x_{i}\right)^{\frac{1}{n}}=\sqrt[n]{x_{1} \cdot x_{2} \cdots x_{n}}=\left(x_{1} \cdot x_{2} \cdots x_{n}\right)^{\frac{1}{n}}$$
## 2.6 Skewness and the Mean, Median, and Mode
Formula for skewness: $$a_{3}=\sum \frac{\left(x_{i}-\overline{x}\right)^{3}}{n s^{2}}$$
Formula for Coefficient of Variation:$$C V=\frac{s}{\overline{x}} \cdot 100 \text { conditioned upon } \overline{x} \neq 0$$
## 2.7 Measures of the Spread of the Data
$$s_{x}=\sqrt{\frac{\sum f m^{2}}{n}-\overline{x}^{2}} \text { where }$$ $$\begin{array}{l}{s_{x}=\text { sample standard deviation }} \\ {\overline{x}=\text { sample mean }}\end{array}$$
Formulas for Sample Standard Deviation $$s=\sqrt{\frac{\Sigma(x-\overline{x})^{2}}{n-1}} \text { or } s=\sqrt{\frac{\Sigma f(x-\overline{x})^{2}}{n-1}} \text { or } s=\sqrt{\frac{\left(\sum_{t=1}^{n} x^{2}\right)-n x^{2}}{n-1}}$$ For the sample standard deviation, the denominator is n - 1, that is the sample size - 1.
Formulas for Population Standard Deviation $$\sigma=\sqrt{\frac{\Sigma(x-\mu)^{2}}{N}} \text { or } \sigma=\sqrt{\frac{\Sigma f(x \mu)^{2}}{N}} \text { or } \sigma=\sqrt{\frac{\sum_{i=1}^{N} x_{i}^{2}}{N}-\mu^{2} F}$$ For the population standard deviation, the denominator is N, the number of items in the population. |
Question Video: Finding the Coefficient of a Certain Term in a Binomial Expansion | Nagwa Question Video: Finding the Coefficient of a Certain Term in a Binomial Expansion | Nagwa
# Question Video: Finding the Coefficient of a Certain Term in a Binomial Expansion Mathematics • Third Year of Secondary School
## Join Nagwa Classes
Determine the coefficient of π₯β»βΆ in the expansion of (π₯ + (1/π₯Β²))βΆ.
03:21
### Video Transcript
Determine the coefficient of π₯ to the power of negative six in the expansion of π₯ plus one over π₯ squared to the power of six.
In this question, we have a binomial expansion written in the form π plus π to the πth power. We know that the general term denoted π sub π plus one is equal to π choose π multiplied by π to the power of π minus π multiplied by π to the power of π. We will begin by rewriting our expression as π₯ plus π₯ to the power of negative two raised to the sixth power. We can do this as one over π₯ to the power of π is equal to π₯ to the power of negative π. The general term of this expansion will therefore be equal to six choose π multiplied by π₯ to the power of six minus π multiplied by π₯ to the power of negative two to the power of π. This can be rewritten as six choose π multiplied by π₯ to the power of six minus π multiplied by π₯ to the power of negative two π.
We know from our laws of exponents that when the base is the same, we can add the exponents or powers. Our expression becomes six choose π multiplied by π₯ to the power of six minus π plus negative two π. Simplifying this again gives us six choose π multiplied by π₯ to the power of six minus three π. In this question, weβre interested in the coefficient when the exponent of π₯ is negative six. We need to calculate the value of π when six minus three π equals negative six. We can subtract six from both sides of this equation so that negative three π is equal to negative 12. Dividing both sides by negative three gives us π is equal to four. We can now substitute this value back into the expression for our term.
This gives us six choose four multiplied by π₯ to the power of negative six. The coefficient of this term is therefore equal to six choose four. We know that π choose π or πCπ is equal to π factorial divided by π minus π factorial multiplied by π factorial. Six choose four is therefore equal to six factorial divided by four factorial multiplied by two factorial. Six factorial is the same as six multiplied by five multiplied by four factorial. Canceling the four factorial gives us six multiplied by five divided by two factorial. As two factorial is equal to two and 30 divided by two is 15, six choose four is equal to 15. The coefficient of π₯ to the power of negative six in the expansion is 15.
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# 20 Times Table
Learn here the 20 times table. Twenty times table 20 x 1 = 20, 20 x 2 = 40, 20 x 3 = 60, 20 x 4 = 80, 20 x 5 = 100, 20 x 6 = 120, 20 x 7 = 140, 20 x 8 = 160, 20 x 9 = 180 and 20 x 10 = 200.
Hello, math enthusiasts! Are you ready to dive into the realm of the 20 times table? In this blog, we will explore the wonders and intricacies of multiplying by 20.
Prepare to be amazed as we unravel patterns, discover tricks, and delve into the fascinating world of the 20 times table.
Let's embark on this multiplication adventure together and unlock the power of the mighty 20!
## Discovering the Marvels
The 20 times table holds some incredible surprises and patterns that make multiplication a breeze. Let's explore a few captivating aspects of the 20 times table:
### The Power of Zero
Multiplying any number by 20 always results in a product with a zero at the end. This fascinating property makes it easy to identify the product when multiplying by 20.
Embrace the power of zero and let it guide you through the multiplication journey!
## Mastering the 20 Times Table
Now, let's explore some strategies and techniques to master the 20 times table and become multiplication masters:
### Visualize and Practice
Visual aids can be incredibly helpful in mastering the 20 times table. Create colorful charts, multiplication grids, or use manipulatives to help visualize and understand the products.
Practice regularly with flashcards or online quizzes to reinforce your knowledge and improve your speed.
### Real-Life Applications
Apply the 20 times table to real-life scenarios to make it more relevant and engaging. Calculate the total cost of items priced at \$20 each or determine the length of a room in feet when each tile measures 20 inches.
By connecting multiplication to everyday situations, you'll deepen your understanding and see the practical value of the 20 times table.
### Find Patterns
Explore patterns within the 20 times table. Notice that the units digit alternates between 0 and the number being multiplied (e.g., 20 times 4 is 80, 20 times 5 is 100, and so on).
Discovering these patterns can make multiplication by 20 feel like unraveling a mathematical puzzle.
## Twenty Multiplication Table
Read, Repeat and Learn Twenty times table and Check yourself by giving a test below
## Table of 20
20 Times table Test
## How much is 20 multiplied by other numbers?
@2024 PrintableMultiplicationTable.net |
# Surface Area of a Cube
In these lessons, we will learn
• how to calculate the surface area of a cube.
• how to find the length of a cube given the surface area.
• how to use the net of a cube to find its surface area.
• how to find the surface area of a cube in terms of its volume.
Related Topics: More Geometry Lessons
### Surface Area of a Cube
A cube is a three-dimensional figure with six equal square faces.
The surface area of a cube is the sum of the area of the six squares that cover it.
The figure above shows a cube. The dotted lines indicate edges hidden from your view.
If s is the length of one of its sides, then the area of one face of the cube is s2.
Since a cube has six faces the surface area of a cube is six times the area of one face.
Surface area of a cube = 6s2
Example
Find the surface area of a cube with a side of length 3 cm
Solution:
Given that s = 3
Surface area of a cube = 6s2 = 6(3)2 = 54 cm2
How to find the surface area of a cube? Example:
Given a side of 3 cm, find the surface area of the cube. How to find the surface area of a cube?
Step 1: Find the length of a side
Step 2: Substitute into the formula: side × side × 6 and evaluate
Step 3: Write the units How to find the length of a cube given the surface area?
This video shows how to find the length of a cube given the surface area.
Example: The total surface area of a cube is 216 in2. What is the length of each side of the cube?
How to use the net of a cube to find its surface area?
Another way to look at the surface area of a cube is to consider a net of the cube. The net is a 2-dimensional figure that can be folded to form a 3-dimensional object.
Imagine making cuts along some edges of a cube and opening it up to form a plane figure. The plane figure is called the net of the cube.
The following net can be folded along the dotted lines to form a cube.
We can then calculate the area of each square in the net and then multiply the area by 6 to get the surface area of the cube.
There are altogether 11 possible nets for a cube as shown in the following figures. Notice that the surface area of each of the net is the same.
How to use nets and 3-dimensional figures to find surface area of cubes and prisms? Surface Area of a Cube in terms of its Volume
How to find the surface area of a cube in terms of volume
S = 6V2/3
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# How do you solve -7+ 6x = - ( - 5x + 5)?
Sep 8, 2017
$x = 2$
#### Explanation:
Begin by distributing the negative on the left side to $\left(- 5 x + 5\right)$
$- 7 + 6 x = \left[- - 5 x\right] - \left[+ 5\right]$
$- 7 + 6 x = 5 x - 5$
Add $7$ to both sides:
cancel(-7)cancelcolor(red)(+7)+6x=5x-5color(red)(+7
$6 x = 5 x + 2$
Subtract $5 x$ from both sides:
$6 x \textcolor{red}{- 5 x} = \cancel{5 x} \cancel{\textcolor{red}{- 5 x}} + 2$
$x = 2$ |
# Proof for Radian is a Constant Angle | Circular System | 1 radian = 2/π Right Angles
Radian is a measure of the trigonometric angle. The angle formed by an arc in a circle is exactly one radian and it is equal to the radius of the circle. We use radian as a unit of measuring angles in a circular system. Here we are giving the proof for the statement radian is a constant angle. Also, check example questions on expressing radian in the units of the sexagesimal system.
## Prove that Radian is a Constant Angle
A radian is defined as an angle subtended at the center of a circle by an arc whose length is equal to the radius of that circle. Here is the proof for the statement i.e radian is a constant angle.
Let us take a circle with center O and radius r. Also, take any two points on the circle A, B and AB = OA = r, then by the definition ∠AOB = 1 radian.
Produce AO to intersect the circle at point C so that the length of arc ABC is equal to half of the circumference and ∠AOC, the angle at the center subtended by the arc = arc straight angle = two right angles.
$$\frac { arc AB }{ arc ABC }$$ = $$\frac { r }{ ½ x 2πr }$$ = $$\frac { 1 }{ π }$$
$$\frac { ∠AOB }{ ∠AOC }$$ = $$\frac { 1 radian }{ 2 right angles }$$
The arc of a circle is proportional to the angle it subtended at the center of the circle.
Therefore, $$\frac { ∠AOB }{ ∠AOC }$$ = $$\frac { arc AB }{ arc ABC }$$
$$\frac { 1 radian }{ 2 right angles }$$ = $$\frac { 1 }{ π }$$
So, 1 radian = $$\frac { 2 }{ π }$$ right angles
Both 2, π are constants.
Hence, proved.
Also, Check
Question 1:
Express one radian in the units of the sexagesimal system.
= (180 x 7)°/22
= 57° 16′ 22″
Question 2:
Express 3 radians in the units of the sexagesimal system.
= (540 x 7)°/22
= 171°49’5″ |
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## Introduction to Quadratic Type Expressions
An expression with three terms where the power of the first term is twice that of the second and the third term is a constant is called a quadratic type expression. They factor in the same way as quadratic polynomials. The power on x in the factorization will be the power on x in the middle term. To see the effect of changing the exponents, let us look at x 2 – 2 x – 3 = ( x – 3)( x + 1).
x 4 – 2 x 2 – 3 = ( x 2 – 3)( x 2 + 1)
x 6 – 2 x 3 – 3 = ( x 3 – 3)( x 3 + 1)
x 10 – 2 x 5 – 3 = ( x 5 – 3)( x 5 + 1)
x –4 – 2 x –2 – 3 = ( x –2 – 3)( x –2 + 1)
x 2/3 – 2 x 1/3 – 3 = ( x 1/3 – 3)( x 1/3 + 1)
x 1 – 2 x 1/2 – 3 = ( x 1/2 – 3)( x 1/2 + 1)
## Quadratic Type Expressions Practice Problems
#### Practice
1. x 4 – 3 x 2 + 2 =
2. x 10 – 3 x 5 + 2 =
3. x 2/5 – 3 x 1/5 + 2 =
4. x –6 – 3 x –3 + 2 =
5. x 1/2 – 3 x 1/4 + 2 =
6. x 4 + 10 x 2 + 9 =
7. x 6 – 4 x 3 – 21 =
8. 4 x 6 + 4 x 3 – 35 =
9. 10 x 10 + 23 x 5 + 6 =
10. 9 x 4 – 6 x 2 + 1 =
11. x 2/7 – 3 x 1/7 – 18 =
12. 6 x 2/3 – 7 x 1/3 – 3 =
13. x 1/3 + 11 x 1/6 + 10 =
14. 15 x 1/2 – 8 x 1/4 + 1 =
16. 25 x 6 + 20 x 3 + 4 =
#### Solutions
1. x 4 – 3 x 2 + 2 = ( x 2 – 2)( x 2 – 1)
2. x 10 – 3 x 5 + 2 = ( x 5 – 2)( x 5 – 1)
3. x 2/5 – 3 x 1/5 + 2 = ( x 1/5 – 2)( x 1/5 – 1)
4. x –6 – 3 x –3 + 2 = ( x –3 – 2)( x –3 – 1)
5. x 1/2 – 3 x 1/4 + 2 = ( x 1/4 – 2)( x 1/4 –1)
6. x 4 + 10 x 2 + 9 = ( x 2 + 9)( x 2 + 1)
7. x 6 – 4 x 3 – 21 = ( x 3 – 7)( x 3 + 3)
8. 4 x 6 + 4 x 3 – 35 = (2 x 3 + 7)(2 x 3 – 5)
9. 10 x 10 + 23 x 5 + 6 = (10 x 5 + 3)( x 5 + 2)
10. 9 x 4 – 6 x 2 + 1 = (3 x 2 – 1)(3 x 2 – 1) = (3 x 2 – 1) 2
11. x 2/7 – 3 x 1/7 – 18 = ( x 1/7 – 6)( x 1/7 + 3)
12. 6 x 2/3 – 7 x 1/3 – 3 = (2 x 1/3 – 3)(3 x 1/3 + 1)
13. x 1/3 + 11 x 1/6 + 10 = ( x 1/6 + 10)( x 1/6 + 1)
14. 15 x 1/2 – 8 x 1/4 + 1 = (3 x 1/4 – 1)(5 x 1/4 – 1)
16. 25 x 6 + 20 x 3 + 4 = (5 x 3 + 2)(5 x 3 + 2) = (5 x 3 + 2) 2
Practice problems for this concept can be found at: Algebra Factoring Practice Test.
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Goal:
## Theory:
Part 1
When there is an ordered pair that satisfies two or more equations, then this ordered pair is a simultaneous solution of the equations.
For example, consider these two equations:
y = 2x – 1
y = -x + 5
When ‘x’ is 2, then ‘y’ is 3 for both:
y = 2×2 – 1 = 3
y = -2 + 5 = 3
The ordered pair (x,y) = (2,3) is said to satisfy both equations, and is called a simultaneous solution of the equations.
If we graph both equations together, the graph lines will coincide at the point (2,3). This means that simultaneous solutions of equations can be found by graphing the equations and finding where they intersect.
Graphing with pencil and paper is inaccurate, but the Maths Helper Plus software allows you to use this approach with high accuracy.
## Method:
Part 2
IMPORTANT: This topic assumes that you know how to graph equations in Maths Helper Plus. Find out how by completing the ‘Easy Start’ tutorial in Maths Helper Plus help. To view the tutorial, select the ‘Tutorial’ option from the ‘Help’ menu in Maths Helper Plus.
We will use the example described in the ‘theory’ section above to demonstrate the steps for using Maths Helper Plus to solve simultaneous equations graphically.
If you have just launched the software then you already have an empty document, otherwise, hold down ‘Ctrl’ while you briefly press the ‘N’ key.
##### Step 2 Enter the two equations
NOTE: You can type powers by using the ‘^’ character. Type x2 as ‘x^2’, or x3 as ‘x^3’.
• Type your first equation: y = 2x – 1
• Press the Enter key
• Type the second equation: y = -x + 5
• Press the Enter key
For the example equations, the graph now looks like this:
• ##### Step 3 Adjust the graph scale
Depending on the equations you entered, important parts of the graphs may lie outside of the default graph scale. You can check on this with a simple zoom operation:
• Zoom out. Briefly press the F10 key one or more times until you are sure that all of the intersection points between the two graphs are visible.
• Zoom in. If you have zoomed out too far, hold down ‘Shift’ while you press F10 to zoom back in.
For the example graphs, there is no need to adjust the graph scale settings.
##### Step 4 Find the intersection points of the graph lines
• Select the intersection tool by clicking this button: on the ‘math tools’ toolbar.
The intersection tool dialog box will appear (see below) and the mouse cursor will have this shape: when the mouse is moved over the graph.
NOTE: If the intersection tool dialog box covers up part of your graph, you can move it. Point with the mouse to the title: ‘Intersection Tool’ at the top of the dialog box. Now click and drag to a new location.
• Move the cursor close to the point where the graph lines intersect. (See diagram below) Click the left mouse button. If the intersection point was found, a black dot will appear on the graph, and the dialog box will display its coordinates.
Repeat for each intersection point between the two graphs.
Cancel the intersection tool by clicking the ‘Cancel’ button on the intersection tool dialog box.
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# Trigonometry, Part 2
Now let’s use the unit circle to see some of the common trig identities. These identities (rules) will be used in future posts.
Let’s assume we have an acute angle π. An acute angle is one that is between 0 and π/2 (or 0 to 90Β°). The following identities are valid for any angle, not just acute ones – it is just easier to see the logic in the diagram if we assume this.
The following picture shows the relationship between an angle π in the first quadrant, and an angle in the second quadrant which is symmetric with π:
You can see that to measure this symmetric angle from the postive x-axis, you just subtract it from π. The coordinates of the intersected point on the unit circle are negative for the x coordinate but the same y coordinate as the original angle π. So the following identities are evident from this picture:
cos(π – π) = -cosπ
sin(π – π) = sinπ
tan(π – π) = -cosπ/sinπ = -tanπ
Again, these are true for any angle, not just acute ones.
As an example, let π = π/3, (60Β°). The following is true for π/3:
cos(π/3) = 1/2
sin(π/3) = β3Μ
/2
tan(π/3) = β3Μ
Now π – π/3 = 2π/3. So using these identities, we know that
cos(2π/3) = -1/2
sin(2π/3) = β3Μ
/2
tan(2π/3) = -β3Μ
Now let’s look at a symmetric angle in the third quadrant. To measure this angle from the positive x-axis, you add it to π. The corresponding coordinates of the intersected point on the unit circle are both the negative of the coordinates for π. So the following identities are shown in this picture:
So these identities are
cos(π + π) = -cosπ
sin(π + π) = -sinπ
tan(π + π) = -cosπ/-sinπ = tanπ
Using our same example, π + π/3 = 4π/3. Using these identities:
cos(4π/3) = -1/2
sin(4π/3) = -β3Μ
/2
tan(4π/3) = β3Μ
As was mentioned before, angles measured clockwise from the positive x-axis are negative. So the following trig identities are shown in the figure above:
cos(-π) = cosπ
sin(-π) = -sinπ
tan(-π) = cosπ/-sinπ = -tanπ
So,
cos(-π/3) = 1/2
sin(-π/3) = -β3Μ
/2
tan(-π/3) = -β3Μ
There are a couple more identities I would like to show but I’ll save that for next time. |
# Points A and B are at (9 ,4 ) and (7 ,2 ), respectively. Point A is rotated counterclockwise about the origin by (3pi)/2 and dilated about point C by a factor of 2 . If point A is now at point B, what are the coordinates of point C?
May 26, 2018
$C = \left(1 , - 20\right)$
#### Explanation:
$\text{under a counterclockwise rotation about the origin of } \frac{3 \pi}{2}$
• " a point "(x,y)to(y,-x)
$A \left(9 , 4\right) \to A ' \left(4 , - 9\right) \text{ where A' is the image of A}$
$\vec{C B} = \textcolor{red}{2} \vec{C A '}$
$\underline{b} - \underline{c} = 2 \left(\underline{a} ' - \underline{c}\right)$
$\underline{b} - \underline{c} = 2 \underline{a} ' - 2 \underline{c}$
$\underline{c} = 2 \underline{a} ' - \underline{b}$
$\textcolor{w h i t e}{\underline{c}} = 2 \left(\begin{matrix}4 \\ - 9\end{matrix}\right) - \left(\begin{matrix}7 \\ 2\end{matrix}\right)$
$\textcolor{w h i t e}{\underline{c}} = \left(\begin{matrix}8 \\ - 18\end{matrix}\right) - \left(\begin{matrix}7 \\ 2\end{matrix}\right) = \left(\begin{matrix}1 \\ - 20\end{matrix}\right)$
$C = \left(1 , - 20\right)$ |
1. ## Derivatives and Tangents
Can someone describe how to solve this problem to me?
Determine a, b, and c so that the curve y=ax^2+bx+c shall be tangent to the line 3x-y=5 at (2,1) and shall also pass through (3,-1).
Thank you!
2. You need three conditions to uniquely determine a, b and c. We have:
y(x) = ax˛+bx+c, tangent line y = 3x-5 at (2,1) and point (3,-1).
So, the conditions are:
1) y(2) = 1
2) y'(2) = 3
3) y(3) = -1
Do you understand why? If so: solve the system.
3. Hello, BubblegumPops!
Determine $a,\,b,\text{ and }c$ so that the curve $y\:=\:ax^2+bx+c$ shall be tangent
to the line $3x-y\:=\:5$ at (2,1) and shall also pass through (3,-1).
Since (2,1) is on the curve: . $a\!\cdot2^2 + b\!\cdot\!2 + c \:=\:1\quad\Rightarrow\quad{\color{blue}4a +2b + c \:=\:-1}$
Since (3,-1) is on the curve: . $a\!\cdot\!3^2 + b\!\cdot\!3 + c\:=\:-1\quad\Rightarrow\quad{\color{blue}9a + 3b + c \:=\:-1}$
The line $y \:=\:3x-5$ has slope 3.
The slope of the tangent to the curve: . $y' \:=\:2ax + b$
At $x=3\!:\;y'\;=\;2a\!\cdot\!3 + b \:=\:3\quad\Rightarrow\quad{\color{blue}6a + b \:=\:3}$
Solve the system of equations for $a,\,b,\text{ and }c.$
4. hmm...i tried to solve the problem, but I don't think I got it right. I got a=15, b= -87, and c=125.
5. Originally Posted by BubblegumPops
hmm...i tried to solve the problem, but I don't think I got it right. I got a=15, b= -87, and c=125.
$4a+2b+c=-1$ ...[1]
$9a+3b+c=-1$ ...[2]
$6a+b=3$ ...[3]
[2] - [1]:
$5a+b=0$ ...[4]
[3] - [4]:
$a=3$
$\implies b=-15$
$\implies c = 17$
6. Thank you! I think i have it, but I don't understand how you got ...[1]. I thought that should equal 1, not -1.
7. Sorry, you're right. It should be $4a+2b+c=1$. Solving them similarly will yield $(a,b,c)=(5,-27,35)$
8. Thank you so much for your help! |
# 12.4: Finding Outcomes
Difficulty Level: At Grade Created by: CK-12
## Introduction
The Talent Show Outfit
Alicia is going to sing for the Talent Show. She is very excited and has selected a wonderful song to sing. She has been practicing with her singing teacher for weeks and is feeling very confident about her ability to do a wonderful job.
Her performance outfit is another matter. Alicia has selected a few different skirts and a few different shirts and shoes to wear. Here are her options for shirts
Striped shirt
Solid shirt
Here are her options for skirts.
Blue skirt
Red skirt
Brown skirt
Here are her options for shoes
Dance shoes
Black dress shoes
How many different outfits can Alicia create given these options?
This is best done using a tree diagram. Alicia needs to organize her clothing options using a tree diagram. This lesson will show you all about tree diagrams. When finished, you will know how many possible outfits Alicia can create.
What You Will Learn
In this lesson you will learn how to correctly apply the following skills.
• Use tree diagrams to list all possible outcomes of a series of events involving two or more choices or results.
• Recognize all possible outcomes of an experiment as the sample space.
• Find probability of specified outcomes using tree diagrams.
Teaching Time
I. Use Tree Diagrams to List all Possible Outcomes of a Series of Events Involving Two or More Choices or Results
Nadia’s soccer team has 2 games to play this weekend. How many outcomes are there for Nadia’s team?
A good way to find the total number of outcomes for events is to make a tree diagram. A tree diagram is a branching diagram that shows all possible outcomes for an event.
To make a tree diagram, split the different events into either-or choices. You can list the choices in any order. Here is a tree diagram for game 1 and game 2.
As you can see, there are four different outcomes for the two games:
\begin{align*}\text{win-win} \quad \quad \text{win-lose}\!\\ \text{lose-win} \quad \quad \text{lose-lose}\end{align*}
What happens when you increase the number of games to three? Just add another section to your tree diagram.
In all, there are 8 total outcomes.
A tree diagram is a great way to visually see all of the possible options. It can also help you to organize your ideas so that you don’t miss any possibilities.
Example
To remodel her kitchen, Gretchen has the following choices: Floor: tile or wood; Counter: Granite or formica; Sink: white, steel, stone. How many different choices can Gretchen make?
First, let’s create a tree diagram that shows all of the possible options.
Step 1: List the choices.
Choice 1 Floor: tile, wood Choice 2 Counter: granite, formica Choice 3 Sink: white, steel, stone
Step 2: Start the tree diagram by listing any of the choices for Choice 1. Then have Choice 1 branch off to Choice 2. Make sure Choice 2 repeats for each branch of Choice 1. Can you identify the missing labels in the tree below?
Step 3: Fill in the third choice. We have left some of the spaces for you to fill in.
Step 4: Fill in the outcomes. Again some of the spaces are left for you to fill in.
You can see that there are 12 possible outcomes for the kitchen design.
II. Recognize all Possible Outcomes of an Experiment as the Sample Space
When you conduct an experiment, there may be few or many possible outcomes. For example, if you are doing an experiment with a coin, there are two possible outcomes because there are two sides of the coin. You can either have heads or tails. If you have an experiment with a number cube, there are six possible outcomes, because there are six sides of the number cube and the sides are numbered one to six. We can think of all of these possible outcomes as the sample space.
A sample space is the set of all possible outcomes for a probability experiment or activity. For example, on the spinner there are 5 different colors on which the arrow can land. The sample space, \begin{align*}S\end{align*}, for one spin of the spinner is then:
\begin{align*}S = \text{red, yellow, pink, green, blue}\end{align*}
These are the only outcomes that result from a single spin of the spinner.
Changing the spinner changes the sample space. This second spinner still has 5 equal-sized sections. But its sample space now has only 3 outcomes:
\begin{align*}S = \text{red, yellow, blue}\end{align*}
Let’s look at an example having to do with sample spaces.
Example
A small jar contains 1 white, 1 black, and 1 red marble. If one marble is randomly chosen, how many possible outcomes are there in the sample space?
Since only a single marble is being chosen, the total number of possible outcomes, the sample space, matches the marble colors.
\begin{align*}S = \text{white, black, red}\end{align*}
Sometimes, the sample space can change if an experiment is performed more than once. If a marble is selected from a jar and then replaced and if the experiment is conducted again, then the sample space can change. The number of outcomes is altered. When this happens, we can use tree diagrams to help us figure out the number of outcomes in the sample space.
Example
A jar contains 1 white and 1 black marble. If one marble is randomly chosen, returned to the jar, then a second marble is chosen, how many possible outcomes are there?
This is an example where a tree diagram is very useful. Consider the marbles one at a time. After the first marble is chosen, it is returned to the jar so now there are again two choices for the second marble. Use a tree diagram to list the outcomes.
From the tree diagram, you can see that the sample space is:
\begin{align*}S = \text{white-white, white-black, black-white, black-black}\end{align*}
12G. Lesson Exercises
What is the sample space in each example?
1. A spinner with red, blue, yellow and green.
2. A number cube numbered 1 – 6.
3. A bag with a blue and a red marble. One marble is drawn and then replaced.
Take a few minutes to check your work with a friend.
III. Find Probabilities of Specified Outcomes Using Tree Diagrams
In the last section, you started to see how tree diagrams could be very helpful when looking for a sample space. Tree diagrams can also be helpful when finding probability.
Finding the probability of an event is a matter of finding the ratio of favorable outcomes to total outcomes. For example, the sample space for a single coin flip has two outcomes: heads and tails. So the probability of getting heads on any single coin flip is:
\begin{align*}P (\text{heads}) = \frac{favorable \ outcomes}{total \ outcomes} =\frac{1}{2}\end{align*}
You can see that the sample space is represented by a number in the total outcomes. For example, if you had a spinner with four colors, the colors by name would be the sample space and the number four would be the total possible outcomes.
What about if we flipped a coin more than one time?
To find the probability of a single outcome for more than one coin flip, use a tree diagram to find all possible outcomes in the sample space.
Then count the number of favorable outcomes within that sample space to find the probability.
For example, to find the probability of tossing a single coin twice and getting heads both times, make a tree diagram to find all possible outcomes.
The diagram shows there are 8 total outcomes and they are paired with first toss option and second toss option.
Then pick out the favorable outcome–in this case, the outcome “heads-heads” is shown in red. You could have selected any of the favorable outcomes for the probability to be accurate.
Now write the ratio of favorable outcomes to total outcomes in the sample space.
\begin{align*}P (\text{heads-heads}) = \frac{favorable \ outcomes}{total \ outcomes}=\frac{1}{4}\end{align*}
You can see that since 1 of 4 outcomes is a favorable outcome, the probability of the coin landing on heads 2 times in a row is \begin{align*}\frac{1}{4}\end{align*}.
Let’s look at another example.
Example
What is the probability of flipping a coin two times and getting two matching results–that is, either two heads or two tails?
First, let’s create a tree diagram to see our options.
Once again, just pick out the favorable outcomes on the same tree diagram. They are shown in red.
You can see that 2 of 4 total outcomes match.
\begin{align*}P (2 \ \text{heads or 2 tails}) = \frac{favorable \ outcomes}{total \ outcomes}=\frac{2}{4}=\frac{1}{2}\end{align*}
You can see that the probability of flipping two heads or two tails is 1:2.
## Real–Life Example Completed
The Talent Show Outfit
Here is the original problem once again. Reread it and then look at the tree diagram created.
Alicia is going to sing for the Talent Show. She is very excited and has selected a wonderful song to sing. She has been practicing with her singing teacher for weeks and is feeling very confident about her ability to do a wonderful job.
Her performance outfit is another matter. Alicia has selected a few different skirts and a few different shirts and shoes to wear. Here are her options for shirts
Striped shirt
Solid shirt
Here are her options for skirts.
Blue skirt
Red skirt
Brown skirt
Here are her options for shoes
Dance shoes
Black dress shoes
How many different outfits can Alicia create given these options?
This is best done using a tree diagram. Alicia needs to organize her clothing options using a tree diagram. To do this, we can take each option and create a diagram to show all of the options.
Based on this tree diagram, you can see that Alicia has twelve possible outfits to choose from.
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Tree Diagram
a visual way of showing all of the possible outcomes of an experiment. Called a tree diagram because each option is drawn as a branch of a tree
Sample Space
The possible outcomes of an experiment
Favorable Outcome
the outcome that you are looking for in an experiment
Total Outcome
the number of options in the sample space
## Time to Practice
Directions: Use Tree Diagrams for each of the following problems.
1. The Triplex Theater has 3 different movies tonight: Bucket of Fun, Bozo the Great, and Pickle Man. Each movie has an early and late show. How many different movie choices are there?
2. Raccoon Stadium offers the following seating plans for football games:
• lower deck, middle loge, or upper bleachers
• center, side, end-zone
How many different kinds of seats can you buy?
3. Cell-Gel cell phone company offers the following choices:
• Free internet plan or Pay internet plan
• 1200, 2000, or 3000 minutes
How many different kinds of plans can you get?
4. Jen’s soccer team is playing 4 games next week. How many different outcomes are there for the four games?
5. The e-Box laptop computer offers the following options.
• Screen: small, medium, or large
• Memory: standard 1 GB, extra 2 GB
• Colors: pearl, blue, black
List the number of different activity choices a camper can make. Use a tree diagram to list them all. Double click to check your answers.
6. What is the sample space for a single toss of a number cube?
7. What is the sample space for a single flip of a coin?
8. A coin is flipped two times. List all possible outcomes for the two flips.
9. A coin is flipped three times in a row. List all possible outcomes for the three flips.
10. A bag contains 3 ping pong balls: 1 red, 1 blue, and 1 green. What is the sample space for drawing a single ball from the bag?
11. A bag contains 3 ping pong balls: 1 red, 1 blue, and 1 green. What is the sample space for drawing a single ball, returning the ball to the bag, then drawing a second ball?
12. What is the sample space for a single spin of the spinner with red, blue, yellow and green sections?
13. What is the sample space for 2 spins of the first spinner?
14. A box contains 3 socks: 1 black, 1 white, and 1 brown. What is the sample space for drawing a single sock, NOT returning the sock to the box, then drawing a second sock?
15. A box contains 3 socks: 1 black, 1 white, and 1 brown. What is the sample space for drawing all 3 socks from the box, one at a time, without returning any of the socks to the box?
16. A box contains 3 black socks. What is the sample space for drawing all 3 socks from the box, one at a time, without returning any of the socks to the box?
17. A box contains 2 black socks and 1 white sock. What is the sample space for drawing all 3 socks from the box, one at a time, without returning any of the socks to the box?
Directions: Answer each question. Use tree diagrams when necessary.
18. What is the probability that the arrow of the spinner will land on red on a single spin?
19. If the spinner is spun two times in a row, what is the probability that the arrow will land on red both times?
20. If the spinner is spun two times in a row, what is the probability that the spinner will land on the same color twice?
21. If the spinner is spun two times in a row, what is the probability that the arrow will land on red at least one time?
22. If the spinner is spun two times in a row, what is the probability that the spinner will land on a different color both times?
23. If the spinner is spun two times in a row, what is the probability that the arrow will land on blue or green at least one time?
24. Two cards, the Ace and King of hearts, are taken from a deck, shuffled, and placed face down. What is the probability that a single card chosen at random will be an Ace?
25. If one card is chosen from the 2-card stack above, then returned to the stack and a second card is chosen, what is the probability that both cards will be Kings?
26. If one card is chosen from the 2-card stack above, then returned to the stack and a second card is chosen, what is the probability that both cards will match?
27. If one card is chosen from the 2-card stack above, then returned to the stack and a second card is chosen, what is the probability that both cards NOT match?
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Thursday , July 7 2022
# NCERT 5th Class (CBSE) Mathematics: Fractions
(b) 5×3/8 = ?
5×3/8 = 5/1×3/8
5/1×3/8 = 15/8
(c) 5/6×1 = ?
5/6×1 = 5
(d) 0×3/7 = ?
0×3/7 = 0
(e) In a class of 40 children, 3/4 use the school bus to come to school. How many children use the school bus?
To find 3/4 of 40, you can multiply.
3/4×40 ?
3/4×40 = 3/4×40/1 = 120/4 = 30
Answer: 30 students in the class come to school by the school bus.
(f) What is 2/3 of 4 meters
### B. Multiplying Fractions by Fractions
(a) This is Venkatesh’s plot land. He wants to build a house on it.
He wants to keep 1/5 of the land as a garden in front of the house.
Of the portion meant for the house, he keeps 1/4 as garage space for his car. The rest of the space is for building the house.
What part of the total land does Venkatesh use to build the house?
From the diagram we can see that Venkatesh uses 12/20 part of the total land to build his house.
To be able to find this answer without a diagram, we must solve it as follows.
Venkatesh uses 3/4 of 4/5 of the total land to build his house.
3/4 of 4/5 = ?
3/4 of 4/5 is the same as 3/4×4/5
3/4×4/5 = ?
Step 1. Multiply the numerators: 3/4×4/5 = 12
Step 2. Multiply the denominators: 3/4×4/5 = 12/20
Step 3. Reduce to the lowest term: 3/4×4/5 = 12/20 = 3/5
Venkatesh uses 3/5 of the land to build his house.
## चुनौती हिमालय की 5th NCERT CBSE Hindi Rimjhim Ch 18
चुनौती हिमालय की 5th Class NCERT CBSE Hindi Book Rimjhim Chapter 18 प्रश्न: लद्दाख जम्मू-कश्मीर राज्य में …
### One comment
1. I need worksheet for Class 5th Mathematics. |
# Perpendicular bisectors
##### Intros
###### Lessons
1. How to use a protractor?
• What is a perpendicular bisector?
• How to draw it?
##### Examples
###### Lessons
1. Draw the perpendicular bisector of the line below.
2. Line AB below represents the bottom edge of a sign. The sign needs a perpendicular bisector support so that it can be put up on the corner of North Avenue. Draw the support.
###### Topic Notes
In previous section, we were introduced to parallel and perpendicular line segments. In this section, we will expand our knowledge on this topic by learning how to draw perpendicular bisectors. Perpendicular bisectors are found everywhere in the world around us. For example, engineers use perpendicular bisectors when designing support beams for structures, including buildings and bridges.
## Introduction to Perpendicular Bisectors
Welcome to our exploration of perpendicular bisectors in geometry! These fascinating lines are essential in understanding shape properties and solving various geometric problems. A perpendicular bisector is a line that cuts another line segment exactly in half at a right angle. It's like a magical line that divides and conquers! To kick off our journey, I've prepared an introduction video that will visually demonstrate this concept. This video is crucial as it lays the foundation for more complex applications we'll encounter later. You'll see how perpendicular bisectors create equal distances and form the basis for constructing triangles and circles. As we progress, you'll discover their role in finding the center of a circle and solving real-world problems. So, grab your geometry tools and let's dive into the world of perpendicular bisectors together! Remember, understanding this concept opens doors to many exciting geometric principles.
## What is a Perpendicular Bisector?
In geometry, a perpendicular bisector is a fundamental concept that plays a crucial role in various mathematical problems and constructions. To understand what a perpendicular bisector is, let's break down the term into its two components: 'perpendicular' and 'bisector'.
First, let's define 'perpendicular'. When two lines or line segments intersect at a 90-degree angle, they are said to be perpendicular to each other. Imagine the corner of a square or a rectangle - the two sides meeting at that corner form a perpendicular angle.
Now, let's look at 'bisector'. A bisector is a line or line segment that divides something into two equal parts. For example, if you cut a sandwich exactly in half, the line where you cut it would be the bisector of the sandwich.
Combining these two concepts, a perpendicular bisector is a line that cuts another line segment into two equal parts (bisects it) and forms right angles (90 degrees) with that line segment. In other words, it's a line that passes through the midpoint of another line segment and is perpendicular to it.
So, what does a perpendicular bisector look like? Imagine a straight line segment. Now, picture another line crossing through the exact middle of this segment, forming a perfect 'T' shape. This crossing line is the perpendicular bisector. It divides the original line segment into two equal parts and forms right angles where it intersects.
Here's a simple example to help visualize a perpendicular bisector: Think of a seesaw on a playground. The long plank of the seesaw represents a line segment. The support in the middle, which holds up the plank and allows it to balance, represents the perpendicular bisector. It's exactly in the middle of the plank (bisecting it) and forms right angles with the plank (perpendicular to it).
Perpendicular bisectors have many practical applications in geometry. They are used to find the center of a circle, construct equal segments, and solve problems involving equidistant points. For instance, any point on the perpendicular bisector of a line segment is equidistant from both endpoints of that segment. This property is often used in solving real-world problems, such as finding the best location for a facility that needs to be equally accessible from two different points.
In construction and drawing, perpendicular bisectors can be created using a compass and straightedge. By drawing two intersecting arcs from the endpoints of a line segment, then connecting the points where these arcs intersect, you can create a perpendicular bisector without measuring angles or distances.
Understanding perpendicular bisectors is essential for students learning geometry, as this concept forms the basis for more advanced topics like the properties of triangles, circle theorems, and coordinate geometry. It's a fundamental tool in the geometer's toolkit, helping to solve a wide range of problems and construct various geometric figures with precision.
## Properties and Importance of Perpendicular Bisectors
Perpendicular bisectors are fundamental elements in geometry, possessing unique properties that make them invaluable in various mathematical and real-world applications. A perpendicular bisector is a line that intersects the midpoint of a line segment at a right angle, effectively dividing it into two equal parts. This simple definition belies the significant role these geometric constructs play in both theoretical and practical contexts.
One of the key perpendicular bisector properties is that every point on the perpendicular bisector is equidistant from the endpoints of the original line segment. This property forms the basis for many geometric proofs and constructions. It also means that the perpendicular bisector of a line segment is the set of all points that are equidistant from the segment's endpoints, making it a locus of points.
Another crucial property is that the perpendicular bisector of a chord always passes through the center of the circle. This relationship between perpendicular bisectors and circles is fundamental in circular geometry and has practical applications in fields such as engineering and architecture. Furthermore, the intersection of the perpendicular bisectors of any two sides of a triangle determines the circumcenter of that triangle the point where the circumcircle (the circle that passes through all three vertices of the triangle) is centered.
The importance of perpendicular bisectors extends beyond their geometric properties. They play a vital role in solving various geometric problems and constructing complex shapes. For instance, they are essential in creating Voronoi diagrams, which partition a plane into regions based on distance to points in a specific subset of the plane. These diagrams have applications in diverse fields, from computer graphics to urban planning.
Perpendicular bisectors also relate closely to other geometric concepts. They are intrinsically linked to the concept of symmetry, as they define lines of reflection symmetry in many shapes. In the study of triangles, perpendicular bisectors are one of the four types of special line segments (along with medians, angle bisectors, and altitudes) that provide crucial information about the triangle's properties and center points.
In real life, perpendicular bisectors find numerous applications. In cartography, they are used to determine the boundaries between regions based on equidistance from certain points. In telecommunications, the concept is applied in cellular network planning to optimize coverage areas. Architects and engineers use perpendicular bisectors in designing structures with balanced load distribution. In sports, the principle of perpendicular bisectors is utilized in field layouts, such as determining the exact center of a soccer field or the positioning of a tennis net.
The concept also appears in nature. The growth patterns of certain crystals follow principles related to perpendicular bisectors, creating symmetrical structures. In biology, the arrangement of leaves on some plants follows patterns that can be described using perpendicular bisectors, optimizing sunlight exposure.
Understanding and applying the properties of perpendicular bisectors is crucial in many STEM fields. From solving complex geometric problems to designing efficient structures and systems, these fundamental geometric constructs continue to play a vital role in shaping our understanding of space and form. Their importance in geometry and their wide-ranging applications make perpendicular bisectors an essential topic for students, professionals, and anyone interested in the mathematical foundations of our world.
## How to Construct a Perpendicular Bisector
A perpendicular bisector is a line that divides another line segment into two equal parts at a right angle. Learning how to draw a perpendicular bisector is an essential skill in geometry. In this guide, we'll explore two methods to construct a perpendicular bisector: the ruler and protractor method and the compass method.
### Method 1: Ruler and Protractor Method
Follow these steps to construct a perpendicular bisector using a ruler and protractor:
1. Draw a line segment AB of any length on your paper.
2. Use your ruler to measure the length of AB.
3. Find the midpoint of AB by dividing its length by 2. Mark this point as M.
4. Place your protractor at point M, aligning it with the line segment.
5. Measure and mark a 90-degree angle from point M.
6. Draw a line from M through the 90-degree mark, extending it beyond AB.
7. This line is the perpendicular bisector of AB.
Common mistakes to avoid:
• Inaccurately measuring the midpoint of AB.
• Not aligning the protractor correctly with the line segment.
• Failing to draw the perpendicular line long enough.
### Method 2: Compass Method
The compass method is more precise and doesn't require a protractor. Here's how to draw a perpendicular bisector using a compass:
1. Draw a line segment AB on your paper.
2. Set your compass to a radius greater than half the length of AB.
3. Place the compass point on A and draw an arc above and below the line segment.
4. Without changing the compass width, repeat step 3 from point B.
5. The arcs should intersect at two points. Label these points C and D.
6. Use your ruler to draw a line connecting points C and D.
7. This line CD is the perpendicular bisector of AB.
Common mistakes to avoid:
• Using a compass radius that's too small, resulting in arcs that don't intersect.
• Changing the compass width between steps.
• Not drawing the arcs large enough to clearly see the intersection points.
To ensure you've correctly constructed your perpendicular bisector, you can:
• Measure the distances from the intersection point to A and B. They should be equal.
• Check that the angle between the original line segment and the bisector is 90 degrees using a protractor.
### Applications of Perpendicular Bisectors
Understanding how to draw perpendicular bisectors is crucial for various geometric constructions and real-world applications, including:
• Finding the center of a circle
• Constructing triangles and other polygons
• Solving problems in architecture and engineering
• Creating symmetrical designs in art and graphic design
Mastering the technique of constructing perpendicular bisectors opens up a world of geometric possibilities. Whether you prefer the ruler and protractor method or the more precise compass method, practice is key to perfecting your skills. Remember to double-check your work and avoid common pitfalls to ensure accuracy in your constructions. With these step-by-step instructions, you'll be able to confidently draw perpendicular bisectors for various applications in geometry and beyond.
## Finding the Equation of a Perpendicular Bisector
Understanding how to find the equation of a perpendicular bisector is a crucial skill in geometry and coordinate algebra. A perpendicular bisector is a line that passes through the midpoint of a line segment at a right angle. This concept is widely used in various mathematical applications, from constructing geometric shapes to solving real-world problems. In this section, we'll explore the process of finding the perpendicular bisector equation and provide examples to illustrate its practical use.
To find the equation of a perpendicular bisector, we follow a step-by-step process:
1. Identify the coordinates of the two points that define the line segment.
2. Calculate the midpoint of the line segment.
3. Determine the slope of the original line segment.
4. Find the negative reciprocal of the slope to get the perpendicular slope.
5. Use the point-slope form of a line equation to write the perpendicular bisector equation.
The perpendicular bisector equation is derived from these steps and can be expressed as:
(x - x) = -m(y - y)
Where (x, y) is the midpoint of the original line segment, and m is the negative reciprocal of the original line's slope.
Let's walk through an example to illustrate how to find the perpendicular bisector of two points:
Example: Find the equation of the perpendicular bisector for the line segment with endpoints (2, 3) and (6, 7).
1. We have the points (2, 3) and (6, 7).
2. Calculate the midpoint: ((2+6)/2, (3+7)/2) = (4, 5)
3. Find the slope of the original line: m = (7-3)/(6-2) = 1
4. The perpendicular slope is the negative reciprocal: -1/1 = -1
5. Use the point-slope form with the midpoint (4, 5) and slope -1: y - 5 = -1(x - 4)
Simplifying this equation gives us the perpendicular bisector equation: y = x + 1
Another method to find the equation of a perpendicular bisector is using the general formula:
(x - x)(x - x) + (y - y)(y - y) = 0
Where (x, y) and (x, y) are the endpoints of the original line segment.
Using this formula for our example:
(x - 2)(6 - 2) + (y - 3)(7 - 3) = 0
Simplifying: 4(x - 2) + 4(y - 3) = 0
x - 2 + y - 3 = 0
x + y - 5 = 0
y = -x + 5
This equation is equivalent to our previous result, just in a different form.
Understanding how to find the equation of a perpendicular bisector is essential in various geometric applications. It's used in constructing triangles, finding the center of circles, and solving problems related to distances and symmetry. By mastering this concept, you'll be better equipped to tackle more complex geometric challenges and apply these skills in real-world scenarios.
Remember, the key to finding the perpendicular bisector equation lies in understanding the relationship between the original line segment and its perpendicular bisector. Practice with different sets of points to reinforce your understanding.
## Solving Perpendicular Bisector Problems
Perpendicular bisectors are fundamental concepts in geometry that play a crucial role in various problem-solving scenarios. In this section, we'll explore different types of problems involving perpendicular bisectors and demonstrate both geometric and algebraic approaches to solve them. Let's dive into some examples and walk through step-by-step solutions.
### Example 1: Finding the Center of a Circle
Problem: Given two points on a circle's circumference, find the center of the circle.
Solution:
1. Draw a line segment connecting the two given points.
2. Construct the perpendicular bisector of this line segment.
3. The point where the perpendicular bisector intersects the line segment is the center of the circle.
Reasoning: The perpendicular bisector of a chord always passes through the center of a circle.
### Example 2: Locating Equidistant Points
Problem: Find a point that is equidistant from three given points.
Solution:
1. Draw line segments connecting any two pairs of points.
2. Construct perpendicular bisectors for these two line segments.
3. The point where the two perpendicular bisectors intersect is equidistant from all three points.
Reasoning: The perpendicular bisector of a line segment is the set of all points equidistant from the segment's endpoints.
### Example 3: Algebraic Approach to Perpendicular Bisectors
Problem: Find the equation of the perpendicular bisector of a line segment with endpoints (2, 3) and (6, 7).
Solution:
1. Find the midpoint: ((2+6)/2, (3+7)/2) = (4, 5)
2. Calculate the slope of the original line: m = (7-3)/(6-2) = 1
3. The perpendicular slope is the negative reciprocal: m_perpendicular = -1
4. Use the point-slope form with the midpoint: y - 5 = -1(x - 4)
5. Simplify to get the equation: y = -x + 9
Reasoning: The perpendicular bisector passes through the midpoint and has a slope perpendicular to the original line segment.
### Example 4: Constructing a Perpendicular Bisector
Problem: Construct a perpendicular bisector of a line segment using only a compass and straightedge.
Solution:
1. Set the compass to more than half the length of the line segment.
2. Place the compass point at one endpoint and draw an arc above and below the line.
3. Repeat from the other endpoint, creating intersecting arcs.
4. Draw a line through the two intersection points of the arcs.
Reasoning: This construction ensures that every point on the bisector is equidistant from both endpoints.
### Example 5: Application in Triangles
Problem: Prove that the perpendicular bisectors of a triangle's sides intersect at a single point.
Solution:
1. Construct perpendicular bisectors for two sides of the triangle.
2. Mark their intersection point.
3. Prove that this point is equidistant from all three vertices.
4. Conclude that it must lie on the perpendicular bisector of the third side as well
## Applications of Perpendicular Bisectors
Perpendicular bisectors, a fundamental concept in geometry, have numerous practical applications across various fields, including engineering, architecture, and computer graphics. Understanding these applications can help solve real-world problems efficiently and accurately. In engineering, perpendicular bisectors play a crucial role in structural design and analysis. For instance, when designing bridges, engineers use perpendicular bisectors to determine the optimal placement of support structures, ensuring equal distribution of weight and stress. This application of perpendicular bisectors in engineering contributes to the stability and longevity of structures.
In architecture, perpendicular bisectors are essential for creating balanced and aesthetically pleasing designs. Architects utilize these geometric constructs to determine the center points of circular elements, such as domes or arches, ensuring symmetry and structural integrity. Moreover, in landscape architecture, perpendicular bisectors help in planning garden layouts, positioning fountains, and creating visually appealing pathways that intersect at precise angles.
The field of computer graphics heavily relies on perpendicular bisectors for various applications. In 3D modeling and animation, these geometric principles are used to create accurate reflections and symmetrical objects. Game developers employ perpendicular bisectors to design realistic collision detection systems, ensuring that objects interact naturally within virtual environments. Additionally, in computer-aided design (CAD) software, perpendicular bisectors are fundamental for creating precise technical drawings and blueprints.
Real-world use of perpendicular bisectors extends to surveying and cartography. Land surveyors use these principles to accurately measure and map terrains, determining property boundaries and creating topographical maps. In GPS technology, perpendicular bisectors help in triangulating positions, improving the accuracy of location services. The concept is also applied in robotics for path planning and obstacle avoidance, allowing autonomous machines to navigate complex environments efficiently.
Understanding perpendicular bisectors is crucial in solving various real-world problems. For example, in urban planning, these geometric principles help in optimizing the placement of public facilities like hospitals or fire stations, ensuring equidistant access for the maximum number of residents. In telecommunications, perpendicular bisectors aid in the optimal positioning of cell towers to provide the best coverage. Even in sports, coaches and analysts use these concepts to analyze player movements and develop strategic plays.
In conclusion, the applications of perpendicular bisectors are vast and varied, demonstrating their significance in solving complex real-world problems across multiple disciplines. From enhancing structural integrity in engineering to creating visually appealing designs in architecture and enabling precise calculations in computer graphics, perpendicular bisectors continue to be an indispensable tool in modern problem-solving and design processes.
## Conclusion
In summary, perpendicular bisectors are fundamental geometric concepts with crucial properties. They divide line segments into two equal parts at right angles, creating a line of symmetry. The key points to remember include their equidistance property, intersection at the circumcenter of triangles, and applications in various geometric constructions. The introduction video serves as an essential foundation for understanding these concepts, providing visual demonstrations and clear explanations. It's crucial to grasp these basics before advancing to more complex topics. To solidify your understanding, practice solving problems involving perpendicular bisectors and explore their real-world applications. Consider how they relate to other geometric concepts like circumcircles and equidistant points. By mastering perpendicular bisectors, you'll enhance your overall geometric reasoning skills and problem-solving abilities. Remember, a strong grasp of this topic will prove invaluable in more advanced mathematical studies and practical applications in fields like engineering and design.
Understanding the line of symmetry in geometric figures can also aid in comprehending more complex structures. Additionally, regularly solving problems related to these concepts will further reinforce your knowledge and application skills.
### Example:
Draw the perpendicular bisector of the line below.
#### Step 1: Measure the Length of the Line Segment
To begin drawing the perpendicular bisector of the line segment AB, the first step is to measure the length of the line. You can do this using a ruler or the edge of a protractor. Align your measuring tool along the line segment AB to get an accurate measurement. In this example, the length of the line segment AB is 14 centimeters.
#### Step 2: Find the Midpoint of the Line Segment
Next, you need to find the midpoint of the line segment AB. The midpoint is the point that divides the line segment into two equal parts. To find the midpoint, divide the total length of the line segment by 2. Since the length of AB is 14 centimeters, the midpoint will be at 7 centimeters (14 divided by 2 equals 7). Mark this midpoint on the line segment AB.
#### Step 3: Draw a 90-Degree Angle at the Midpoint
Now that you have the midpoint, the next step is to draw a line at a 90-degree angle to the line segment AB at this midpoint. To do this, use a protractor. Place the midpoint at the center of the protractor and align the baseline of the protractor with the line segment AB. Locate the 90-degree mark on the protractor and draw a small mark at this point.
#### Step 4: Draw the Perpendicular Bisector
Using a ruler or the straight edge of your protractor, draw a straight line through the midpoint and the 90-degree mark you made in the previous step. This line should be perpendicular to the line segment AB and should intersect it at the midpoint. This newly drawn line is the perpendicular bisector of the line segment AB.
#### Step 5: Verify the Perpendicular Bisector
To ensure accuracy, verify that the angle between the line segment AB and the perpendicular bisector is indeed 90 degrees. You can do this by placing the protractor at the intersection point and checking the angle. If the angle is 90 degrees, you have successfully drawn the perpendicular bisector.
#### Conclusion
By following these steps, you can accurately draw the perpendicular bisector of any line segment. This process involves measuring the length of the line segment, finding the midpoint, drawing a 90-degree angle at the midpoint, and verifying the perpendicular bisector. Practice these steps to become proficient in constructing perpendicular bisectors.
### FAQs
#### 1. What is a perpendicular bisector?
A perpendicular bisector is a line that passes through the midpoint of another line segment at a 90-degree angle, dividing it into two equal parts. It is perpendicular to the original line segment and bisects it, hence the name.
#### 2. How do you construct a perpendicular bisector?
To construct a perpendicular bisector:
1. Draw a line segment AB.
2. Set your compass to more than half the length of AB.
3. Place the compass point at A and draw an arc above and below the line.
4. Repeat from point B without changing the compass width.
5. Connect the points where the arcs intersect to form the perpendicular bisector.
#### 3. What is the equation of a perpendicular bisector?
The general equation of a perpendicular bisector for a line segment with endpoints (x, y) and (x, y) is:
(x - x)(x - x) + (y - y)(y - y) = 0
This equation represents all points equidistant from the two endpoints.
#### 4. What are the properties of perpendicular bisectors?
Key properties include:
• It passes through the midpoint of the line segment it bisects.
• It forms right angles (90 degrees) with the original line segment.
• Any point on the perpendicular bisector is equidistant from the endpoints of the original line segment.
• Perpendicular bisectors of a triangle's sides intersect at the circumcenter.
#### 5. How are perpendicular bisectors used in real life?
Perpendicular bisectors have various practical applications:
• In construction for ensuring symmetry and balance in structures.
• In cartography for determining equidistant points on maps.
• In engineering for designing load-bearing structures.
• In computer graphics for creating symmetrical designs and reflections.
• In urban planning for optimal placement of public facilities.
### Prerequisite Topics
Understanding perpendicular bisectors is a crucial concept in geometry, but to fully grasp this topic, it's essential to have a solid foundation in several prerequisite areas. These fundamental concepts not only provide the necessary background knowledge but also help in developing a deeper understanding of perpendicular bisectors and their applications.
One of the key prerequisites is the understanding of right angles. Right angles are fundamental to perpendicular bisectors, as these bisectors always intersect the line they're bisecting at a 90-degree angle. Knowing how to work with right angles, including calculating their measurements using trigonometric ratios, is crucial when dealing with perpendicular bisectors in various geometric problems.
Another important prerequisite is the ability to work with point-slope form of a line equation. This algebraic concept is vital when you need to determine the equation of a perpendicular bisector. Understanding how to manipulate linear equations and graph them will greatly enhance your ability to work with perpendicular bisectors, especially in coordinate geometry.
Additionally, familiarity with constructing perpendicular lines is essential. This skill directly relates to creating perpendicular bisectors, as you'll need to know how to draw a line perpendicular to another at its midpoint. Understanding the properties of parallel and perpendicular lines in linear functions will help you recognize and analyze perpendicular bisectors in various geometric contexts.
These prerequisite topics form the foundation upon which your understanding of perpendicular bisectors will be built. By mastering these concepts, you'll be better equipped to tackle more complex problems involving perpendicular bisectors. For instance, when you're working on problems related to the circumcenter of a triangle (the point where all three perpendicular bisectors of a triangle's sides intersect), you'll draw upon your knowledge of right angles, linear equations, and perpendicular line construction.
Moreover, these prerequisites are not isolated concepts but interconnected ideas that support each other. Your understanding of right angles enhances your ability to work with perpendicular lines, which in turn helps you grasp the nature of perpendicular bisectors. Similarly, your skills in graphing linear functions will aid in visualizing and analyzing perpendicular bisectors in the coordinate plane.
As you delve deeper into the study of perpendicular bisectors, you'll find that these prerequisite topics continually resurface, reinforcing their importance. Whether you're solving problems related to the equidistance property of points on a perpendicular bisector or using perpendicular bisectors to locate the center of a circle, your foundational knowledge will prove invaluable.
In conclusion, taking the time to thoroughly understand these prerequisite topics will significantly enhance your ability to work with perpendicular bisectors. It will not only make the learning process smoother but also enable you to approach more advanced geometric concepts with confidence and a deeper understanding. |
Related Articles
# How to solve time complexity Recurrence Relations using Recursion Tree method?
• Difficulty Level : Easy
• Last Updated : 19 Jun, 2021
The Recursion Tree Method is a way of solving recurrence relations. In this method, a recurrence relation is converted into recursive trees. Each node represents the cost incurred at various levels of recursion. To find the total cost, costs of all levels are summed up.
Steps to solve recurrence relation using recursion tree method:
1. Draw a recursive tree for given recurrence relation
2. Calculate the cost at each level and count the total no of levels in the recursion tree.
3. Count the total number of nodes in the last level and calculate the cost of the last level
4. Sum up the cost of all the levels in the recursive tree
Let us see how to solve these recurrence relations with the help of some examples:
Question 1: T(n) = 2T(n/2) + c
Solution:
• Step 1: Draw a recursive tree
Recursion Tree
• Step 2: Calculate the work done or cost at each level and count total no of levels in recursion tree
Recursive Tree with each level cost
Count the total number of levels –
Choose the longest path from root node to leaf node
n/20 -→ n/21 -→ n/22 -→ ……… -→ n/2k
Size of problem at last level = n/2k
At last level size of problem becomes 1
n/2k = 1
2k = n
k = log2(n)
Total no of levels in recursive tree = k +1 = log2(n) + 1
• Step 3: Count total number of nodes in the last level and calculate cost of last level
No. of nodes at level 0 = 20 = 1
No. of nodes at level 1 = 21 = 2
………………………………………………………
No. of nodes at level log2(n) = 2log2(n) = nlog2(2) = n
Cost of sub problems at level log2(n) (last level) = nxT(1) = nx1 = n
• Step 4: Sum up the cost all the levels in recursive tree
T(n) = c + 2c + 4c + —- + (no. of levels-1) times + last level cost
= c + 2c + 4c + —- + log2(n) times + Θ(n)
= c(1 + 2 + 4 + —- + log2(n) times) + Θ(n)
1 + 2 + 4 + —– + log2(n) times –> 20 + 21 + 22 + —– + log2(n) times –> Geometric Progression(G.P.)
= c(n) + Θ(n)
Thus, T(n) = Θ(n)
Question 2: T(n) = T(n/10) + T(9n/10) + n
Solution:
• Step 1: Draw a recursive tree
Recursive Tree
• Step 2: Calculate the work done or cost at each level and count total no of levels in recursion tree
Recursion Tree with each level cost
Count the total number of levels –
Choose the longest path from root node to leaf node
(9/10)0n –> (9/10)1n –> (9/10)2n –> ……… –> (9/10)kn
Size of problem at last level = (9/10)kn
At last level size of problem becomes 1
(9/10)kn = 1
(9/10)k = 1/n
k = log10/9(n)
Total no of levels in recursive tree = k +1 = log10/9(n) + 1
• Step 3: Count total number of nodes in the last level and calculate cost of last level
No. of nodes at level 0 = 20 = 1
No. of nodes at level 1 = 21 = 2
………………………………………………………
No. of nodes at level log10/9(n) = 2log10/9(n) = nlog10/9(2)
Cost of sub problems at level log2(n) (last level) = nlog10/9(2) x T(1) = nlog10/9(2) x 1 = nlog10/9(2)
• Step 4: Sum up the cost all the levels in recursive tree
T(n) = n + n + n + —- + (no. of levels – 1) times + last level cost
= n + n + n + —- + log10/9(n) times + Θ(nlog10/9(2))
= nlog10/9(n) + Θ(nlog10/9(2))
Thus, T(n) = Θ(nlog10/9(n))
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# The probability of getting a girl student selected for IAS is $0.4$ and that of a boy candidate selected is $0.6$. The probability that at least one of them will be selected for IAS is ?A. $0.76$B. $0.24$C. $1.0$D. $0.8$
Last updated date: 20th Jul 2024
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Hint: We first assume the events for the given probabilities. We use the formula of $p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right)$. Then we use events A and B are independent and therefore we can write $p\left( A\cap B \right)=p\left( A \right)p\left( B \right)$.
The probability of getting a girl student selected for IAS is $0.4$ and that of a boy candidate selected is $0.6$. We assume them as events where we take event A for probability of getting a girl student selected for IAS and event B for probability of getting a boy student selected for IAS.So,
$p\left( A \right)=0.4$ and $p\left( B \right)=0.6$
The probability that at least one of them will be selected for IAS can be expressed as $p\left( A\cup B \right)$. The events A and B are independent and therefore we can write,
$p\left( A\cap B \right)=p\left( A \right)p\left( B \right)$
$p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right)$
$p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right) \\ \Rightarrow p\left( A\cup B \right)=0.4+0.6-0.4\times 0.6 \\ \therefore p\left( A\cup B \right)=0.76$
Note:We use the independent event to find the dependency for the given probabilities. In case of exclusiveness, we could have written it in the form of $p\left( A\cap B \right)=0$.Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Maths to predict how likely events are to happen. |
# Two lines passing through the point (2, 3) intersects each other at an angle of 60°.
Question:
Two lines passing through the point $(2,3)$ intersects each other at an angle of $60^{\circ}$. If slope of one line is 2 , find equation of the other line.
Solution:
It is given that the slope of the first line, $m_{1}=2$.
Let the slope of the other line be $m_{2}$.
The angle between the two lines is $60^{\circ}$.
$\therefore \tan 60^{\circ}=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
$\Rightarrow \sqrt{3}=\left|\frac{2-m_{2}}{1+2 m_{2}}\right|$
$\Rightarrow \sqrt{3}=\pm\left(\frac{2-m_{2}}{1+2 m_{2}}\right)$
$\Rightarrow \sqrt{3}=\frac{2-m_{2}}{1+2 m_{2}}$ or $\sqrt{3}=-\left(\frac{2-m_{2}}{1+2 m_{2}}\right)$
$\Rightarrow \sqrt{3}\left(1+2 m_{2}\right)=2-m_{2}$ or $\sqrt{3}\left(1+2 m_{2}\right)=-\left(2-m_{2}\right)$
$\Rightarrow \sqrt{3}+2 \sqrt{3} m_{2}+m_{2}=2$ or $\sqrt{3}+2 \sqrt{3} m_{2}-m_{2}=-2$
$\Rightarrow \sqrt{3}+(2 \sqrt{3}+1) m_{2}=2$ or $\sqrt{3}+(2 \sqrt{3}-1) m_{2}=-2$
$\Rightarrow m_{2}=\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}$ or $m_{2}=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$
Case I:
$m_{2}=\left(\frac{2-\sqrt{3}}{2 \sqrt{3}+1}\right)$
The equation of the line passing through point $(2,3)$ and having a slope of $\frac{(2-\sqrt{3})}{(2 \sqrt{3}+1)}$ is
$(y-3)=\frac{2-\sqrt{3}}{2 \sqrt{3}+1}(x-2)$
$(2 \sqrt{3}+1) y-3(2 \sqrt{3}+1)=(2-\sqrt{3}) x-2(2-\sqrt{3})$
$(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-4+2 \sqrt{3}+6 \sqrt{3}+3$
$(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$
In this case, the equation of the other line is $(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$.
Case II :
$m_{2}=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$
The equation of the line passing through point $(2,3)$ and having a slope of $\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$ is
$(y-3)=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}(x-2)$
$(2 \sqrt{3}-1) y-3(2 \sqrt{3}-1)=-(2+\sqrt{3}) x+2(2+\sqrt{3})$
$(2 \sqrt{3}-1) y+(2+\sqrt{3}) x=4+2 \sqrt{3}+6 \sqrt{3}-3$
$(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}$
In this case, the equation of the other line is $(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}$
Thus, the required equation of the other line is $(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$ or $(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}$. |
# Search by Topic
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### There are 42 results
Broad Topics > Fractions, Decimals, Percentages, Ratio and Proportion > Fractions
### Rod Fractions
##### Stage: 3 Challenge Level:
Pick two rods of different colours. Given an unlimited supply of rods of each of the two colours, how can we work out what fraction the shorter rod is of the longer one?
### The Greedy Algorithm
##### Stage: 3 Challenge Level:
The Egyptians expressed all fractions as the sum of different unit fractions. The Greedy Algorithm might provide us with an efficient way of doing this.
### Couples
##### Stage: 3 Challenge Level:
In a certain community two thirds of the adult men are married to three quarters of the adult women. How many adults would there be in the smallest community of this type?
### History of Fractions
##### Stage: 2 and 3
Who first used fractions? Were they always written in the same way? How did fractions reach us here? These are the sorts of questions which this article will answer for you.
### How Long Is the Cantor Set?
##### Stage: 3 Challenge Level:
Take a line segment of length 1. Remove the middle third. Remove the middle thirds of what you have left. Repeat infinitely many times, and you have the Cantor Set. Can you find its length?
### Egyptian Fractions
##### Stage: 3 Challenge Level:
The Egyptians expressed all fractions as the sum of different unit fractions. Here is a chance to explore how they could have written different fractions.
### Farey Sequences
##### Stage: 3 Challenge Level:
There are lots of ideas to explore in these sequences of ordered fractions.
### Teaching Fractions with Understanding: Part-whole Concept
##### Stage: 1, 2 and 3
Written for teachers, this article describes four basic approaches children use in understanding fractions as equal parts of a whole.
### Do Unto Caesar
##### Stage: 3 Challenge Level:
At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won \$1 200. What were the. . . .
##### Stage: 3 Challenge Level:
Can you work out which drink has the stronger flavour?
### Keep it Simple
##### Stage: 3 Challenge Level:
Can all unit fractions be written as the sum of two unit fractions?
### Unit Fractions
##### Stage: 3 Challenge Level:
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation.
### Twisting and Turning
##### Stage: 3 Challenge Level:
Take a look at the video and try to find a sequence of moves that will take you back to zero.
### Chocolate
##### Stage: 2 and 3 Challenge Level:
There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time?
### Matching Fractions, Decimals and Percentages
##### Stage: 3 Challenge Level:
An activity based on the game 'Pelmanism'. Set your own level of challenge and beat your own previous best score.
### More Twisting and Turning
##### Stage: 3 Challenge Level:
It would be nice to have a strategy for disentangling any tangled ropes...
### Peaches Today, Peaches Tomorrow....
##### Stage: 3 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### All Tangled Up
##### Stage: 3 Challenge Level:
Can you tangle yourself up and reach any fraction?
### 3388
##### Stage: 3 Challenge Level:
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### Six Notes All Nice Ratios
##### Stage: 4 Challenge Level:
The Pythagoreans noticed that nice simple ratios of string length made nice sounds together.
### Ratio Sudoku 3
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios or fractions.
### Mathematical Swimmer
##### Stage: 3 Challenge Level:
Twice a week I go swimming and swim the same number of lengths of the pool each time. As I swim, I count the lengths I've done so far, and make it into a fraction of the whole number of lengths I. . . .
### Diminishing Returns
##### Stage: 3 Challenge Level:
In this problem, we have created a pattern from smaller and smaller squares. If we carried on the pattern forever, what proportion of the image would be coloured blue?
### Racing Odds
##### Stage: 3 Challenge Level:
In a race the odds are: 2 to 1 against the rhinoceros winning and 3 to 2 against the hippopotamus winning. What are the odds against the elephant winning if the race is fair?
### Water Lilies
##### Stage: 3 Challenge Level:
There are some water lilies in a lake. The area that they cover doubles in size every day. After 17 days the whole lake is covered. How long did it take them to cover half the lake?
### Harmonic Triangle
##### Stage: 4 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
##### Stage: 3 Challenge Level:
What is the total area of the first two triangles as a fraction of the original A4 rectangle? What is the total area of the first three triangles as a fraction of the original A4 rectangle? If. . . .
### Stretching Fractions
##### Stage: 4 Challenge Level:
Imagine a strip with a mark somewhere along it. Fold it in the middle so that the bottom reaches back to the top. Stetch it out to match the original length. Now where's the mark?
### Stretching Fractions: A Discussion
##### Stage: 4 and 5
This article extends and investigates the ideas in the problem "Stretching Fractions".
### Bull's Eye
##### Stage: 3 Challenge Level:
What fractions of the largest circle are the two shaded regions?
### The Cantor Set
##### Stage: 3 Challenge Level:
Take a line segment of length 1. Remove the middle third. Remove the middle thirds of what you have left. Repeat infinitely many times, and you have the Cantor Set. Can you picture it?
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### Equal Temperament
##### Stage: 4 Challenge Level:
The scale on a piano does something clever : the ratio (interval) between any adjacent points on the scale is equal. If you play any note, twelve points higher will be exactly an octave on.
### Tangles
##### Stage: 3 and 4
A personal investigation of Conway's Rational Tangles. What were the interesting questions that needed to be asked, and where did they lead?
### Fracmax
##### Stage: 4 Challenge Level:
Find the maximum value of 1/p + 1/q + 1/r where this sum is less than 1 and p, q, and r are positive integers.
### Plutarch's Boxes
##### Stage: 3 Challenge Level:
According to Plutarch, the Greeks found all the rectangles with integer sides, whose areas are equal to their perimeters. Can you find them? What rectangular boxes, with integer sides, have. . . .
### What a Joke
##### Stage: 4 Challenge Level:
Each letter represents a different positive digit AHHAAH / JOKE = HA What are the values of each of the letters?
### Circuit Training
##### Stage: 4 Challenge Level:
Mike and Monisha meet at the race track, which is 400m round. Just to make a point, Mike runs anticlockwise whilst Monisha runs clockwise. Where will they meet on their way around and will they ever. . . .
### Pythagoras’ Comma
##### Stage: 4 Challenge Level:
Using an understanding that 1:2 and 2:3 were good ratios, start with a length and keep reducing it to 2/3 of itself. Each time that took the length under 1/2 they doubled it to get back within range.
### Rationals Between...
##### Stage: 4 Challenge Level:
What fractions can you find between the square roots of 65 and 67?
### Counting Fish
##### Stage: 4 Challenge Level:
I need a figure for the fish population in a lake, how does it help to catch and mark 40 fish ?
### The Genes of Gilgamesh
##### Stage: 4 Challenge Level:
Can you work out the parentage of the ancient hero Gilgamesh? |
Depending on your AS-level maths exam board, you might encounter the equation of a circle in C1 (OCR) or C2 (everyone else). It’s really just a restatement of Pythagoras’ Theorem: saying $(x-a)^2 + (y-b)^2 = r^2$ is the same as saying “the square of the horizontal distance between $(a,b)$ and $(x,y)$ plus the square of the vertical distance is the same as the square of the direct distance” - which is, of course, the hypotenuse.
A typical question gives you an expanded form - maybe $x^2 + y^2 + 8x + 6y = 0$ - and asks you for the centre and the radius.
Of course, there’s more than one way to do it.
### Complete the square
This is the traditional method: you would say $x^2 + 8x \equiv (x + 4)^2 - 16$ and $y^2 + 6y = (x+3)^2 - 9$, so the left hand side works out to be $(x+4)^2 + (y+3)^2 - 25$ - meaning the circle has centre $(-4,-3)$ and radius $5$.
Which is all well and good, as long as you’re comfortable completing the square. However, you can apply some different tools, too.
### Partial differentiation
I’m going to skip the details of what’s going on with the partial differentiation method and cut straight to the chase: to find the $x$-coordinate of the centre, ignore the $y$s and differentiate: $2x + 8 = 0$, so $x = -4$. Now there’s a coincidence. Surely it can’t work for $y$ as well?
Of course it works for $y$ as well. Ignore the $x$ and differentiate: $2y + 6 = 0$, so $y=-3$. Brilliant. Now, what about the radius?
Ah, that’s the clever bit: you rearrange, if you need to, so that you have an expression that’s equal to $0$, and put your values of $x$ and $y$ for the centre into that expression. Here, we don’t need to rearrange (it already equals $0$), so we work out $(-4)^2 + (-3)^2 + 8(-4) + 6(-3)$ to get $16 + 9 -32 - 18 = -25$. Quick, look over there while I take the negative of it and then take a square root! The radius is 5.
#### Wait a minute… I saw that!
Aw, you spotted it. Right: when you rearrange the circle equation, you get the expression $(x-a)^2 + (y-b)^2 - r^2$ as the left hand side. If you put $x=a$ and $y=b$ in, the brackets vanish and you’re left with $-r^2$. Happy now?
### Implicit differentiation
Another option - very similar, in fact, to partial differentiation - is to notice that $\frac{dy}{dx}=0$ directly above and below the centre; similarly, $\frac{dx}{dy}=0$ to the left and right.
If you differentiate $x^2 + y^2 + 8x + 6y =0$ with respect to $x$, you get $2x + 2y \frac{dy}{dx} + 8 + 6 \frac{dy}{dx} = 0$; substituting in $\frac{dy}{dx}=0$ gives you $2x + 8 = 0$, giving the $x$-coordinate of the centre as $-4$
With respect to $y$, you get $2y + 6 = 0$ after you set the derivatives equal to $0$. Those two equations give you the centre of the circle.
### Implicit differentiation for smartarses
Pick a point, any point.
Evaluate $\frac{dy}{dx}$ at that point, even if it’s not on the circle.
Find the line perpendicular to that gradient through the point.
Repeat those three steps with a point that’s not on the line you just found.
These lines cross at the centre of the circle. Why? |
# Find the total number of ways in which a beggar can be given at least one rupee from four 25paise coins,three 50paise coins & 2 one rupee coins ? (1rupee=100paise)
98
#### Explanation:
There are nine coins in total. I can choose to give one, some, or all of the coins to the beggar. Since each choice for each coin is a yes/no, I can use ${2}^{x}$ to figure out the total number of combinations.
${2}^{7} = 128$
~~~~~
To check this, let's say we have 3 coins called A, B, C. We should have ${2}^{3} = 8$ ways to give the coins:
A
B
C
AB
BC
AC
ABC
none
~~~~~
How many ways out of the 128 makes a rupee?
One way to do this is to count the number directly and the other is to figure out how many ways we don't make a rupee and subtract that from 128. I think I'll approach this by not going over 1 rupee.
What are the different coin combinations that will keep us under? Each of these, being a combination, will fall under this general equation:
C_(n,k)=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$
1 X 25 paise
2 X 25 paise
3 X 25 paise
1 X 50 paise
1 X 50 paise + 1 X 25 paise
We can evaluate each of these:
1 X 25 paise
C_(4,1)=(4!)/((1!)(4-1)!)=(4!)/(3!)=24/6=4
2 X 25 paise
C_(4,2)=(4!)/((2!)(4-2)!)=(4!)/((2!)(2!))=24/4=6
3 X 25 paise
C_(4,3)=(4!)/((3!)(4-3)!)=(4!)/((3!)(1!))=24/6=4
1 X 50 paise
C_(3,1)=(3!)/((1!)(3-1)!)=(3!)/((1!)(2!))=6/2=3
1 X 50 paise + 1 X 25 paise
${C}_{3 , 1} {C}_{4 , 1} = 3 \times 4 = 12$
$4 + 6 + 4 + 3 + 12 = 29$
$128 - 1 - 29 = 98$ |
• Fifth Grade Mathematics Curriculum Modules
Summarized from A Story Of Units
Module 1: Place Value and Decimal Fractions
Whole number patterns with number disks are generalized to decimal numbers. Students solve word problems with measurements in the metric system using these same patterns. Students will apply their work with place value to adding, subtracting, multiplying, and dividing decimal numbers with tenths and hundredths.
Module 2: Multi-Digit Whole Number and Decimal Fraction Operations
This unit uses place value patterns and the distributive and associative properties to multiply multi-digit numbers by multiples of 10 and leads to fluency with multi-digit whole number multiplication. Division with two-digit divisors is also covered.
Module 3: Addition and Subtraction of Fractions
Students will learn to add and subtract fractions with unlike denominators. Fraction models, area models, and tape diagrams will be some of the tools used to help with the understanding of this concept. Students will learn to form units in tape diagrams to aid in solving word problems.
Module 4: Multiplication and Division of Fractions and Decimal Fraction
Students will extend knowledge from the previous unit to apply to multiplying fractions and dividing by a unit fraction. Continued use of hands-on methods will be used. Students will also explore multi-digit decimal multiplication and division.
Module 5: Addition and Multiplication with Volume and Area
Measuring volume and area are the emphasis of the unit. Work done in the fraction modules will be reinforced here as students will answer questions about how the area of a figure changes when scaled by a whole or fractional scale. Students may be asked to find missing fractional sides. Students will also find the volume of simple shapes composed of rectangular prisms.
Module 6: Problem Solving with the Coordinate Plane
Scaling is revisited in this final module. Line plots will be presented on the coordinate plane and students are asked about the scaling relationship between the increase in the units of the vertical axis for 1 unit of increase in the horizontal axis. This unit will frame the work to be done in middle school with ratios and proportions. |
How do you graph x-y<=4?
May 31, 2017
Refer to the explanation for the process to follow.
Explanation:
Solve and Graph:
$x - y \le 4$
Turn the inequality into slope intercept form: $y = m x + b$, where $m$ is the slope and $b$ is the y-intercept (the value of $y$ when $x = 0$.
Subtract $x$ from both sides.
$- y \le - x + 4$
Multiply both sides by $- 1$. This will change the direction of the inequality and make $x$ and $y$ positive.
color(blue)(y>=x-4
In order to graph this inequality, you will need to determine some points on the line.
Points
$x = - 4 ,$$y = - 8$
$x = 0 ,$$y = - 4$
$x = 4 ,$$y = 0$
Plot the points and draw a solid straight line through them. Then shade in the area above the line to represent the inequality. straight line through them. Then shade in the area above the line to represent the inequality. graph{y>=x-4 [-15.95, 16.07, -10, 6.02]} |
# Piecewise Function - Domain & Range
• Sep 21st 2009, 07:09 AM
BeSweeet
Piecewise Function - Domain & Range
For the function
_____{x+3 if -5≤x<2
f(x)={x^2 if 2≤x≤4
I have to sketch the graph, determine the domain and range of f(x), and determine f(2). Help please? (Headbang)
• Sep 21st 2009, 07:22 AM
red_dog
$\displaystyle f(x)=\left\{\begin{array}{ll}x+3 & ,-5\leq x<2\\x^2 & ,2\leq x\leq 4\end{array}\right.$
The domain is $\displaystyle [-5,4]$.
To find the range we can use the following statement abou function:
If $\displaystyle f:A\to B$ and $\displaystyle X, \ Y$ are two subsets of A, then
$\displaystyle f(X\cup Y)=f(X)\cup f(Y)$
In this case let $\displaystyle X=[-5,2), \ Y=[2,4]$
We have $\displaystyle f(X)=[-2,5), \ f(Y)=[4,16]$.
Then the range is $\displaystyle [-2,5)\cup[4,16]=[-2,16]$
• Sep 21st 2009, 07:34 AM
BeSweeet
Quote:
Originally Posted by red_dog
$\displaystyle f(x)=\left\{\begin{array}{ll}x+3 & ,-5\leq x<2\\x^2 & ,2\leq x\leq 4\end{array}\right.$
The domain is $\displaystyle [-5,4]$.
To find the range we can use the following statement abou function:
If $\displaystyle f:A\to B$ and $\displaystyle X, \ Y$ are two subsets of A, then
$\displaystyle f(X\cup Y)=f(X)\cup f(Y)$
In this case let $\displaystyle X=[-5,2), \ Y=[2,4]$
We have $\displaystyle f(X)=[-2,5), \ f(Y)=[4,16]$.
Then the range is $\displaystyle [-2,5)\cup[4,16]=[-2,16]$
Wow... That is completely confusing. Never seen anything that looks anything like that stuff.
• Sep 21st 2009, 07:47 AM
red_dog
Quote:
Originally Posted by BeSweeet
Wow... That is completely confusing. Never seen anything that looks anything like that stuff.
Then let's do it in another way.
$\displaystyle -5\leq x<2$
Add 3 to all members: $\displaystyle -2\leq x+3<5$
Then, if $\displaystyle x\in[-5,2)$ then $\displaystyle f(x)\in[-2,5)$
$\displaystyle 2\leq x\leq 4$
Square all members: $\displaystyle 4\leq x^2\leq 16$
Then, if $\displaystyle x\in[2,4]$ then $\displaystyle f(x)\in[4,16]$.
So, the range is $\displaystyle [-2,5)\cup[4,16]=[-2,16]$
Is it better now?
• Sep 21st 2009, 08:08 AM
stapel
Quote:
Originally Posted by BeSweeet
For the function
_____{x+3 if -5≤x<2
f(x)={x^2 if 2≤x≤4
I have to sketch the graph...
Sketch y = x + 3, and then erase everything before x = -5 and after x = 2. Make sure to draw a filled-in circle for the left-hand endpoint and an "open" circle for the right-hand endpoint.
Then sketch y = x^2, and erase everything before x = 2 and after x = 4. Make sure to draw filled-in circles for each of the endpoints.
Quote:
Originally Posted by BeSweeet
...determine the domain and range of f(x)...
The domain is given: it's the x-values for which the function is defined.
To find the range, look at your graph. Which y-values are covered by this graph? (If you "collapsed the graph sideways onto the y-axis, which portions would be covered?)
Quote:
Originally Posted by BeSweeet
...and determine f(2).
The function is defined for x = 2. So look at the function rule, find the half which is defined for x = 2, and plug 2 in for x in that half's rule.
(Wink)
• Sep 21st 2009, 02:11 PM
BeSweeet
Quote:
Originally Posted by stapel
Sketch y = x + 3, and then erase everything before x = -5 and after x = 2. Make sure to draw a filled-in circle for the left-hand endpoint and an "open" circle for the right-hand endpoint.
Then sketch y = x^2, and erase everything before x = 2 and after x = 4. Make sure to draw filled-in circles for each of the endpoints.
The domain is given: it's the x-values for which the function is defined.
To find the range, look at your graph. Which y-values are covered by this graph? (If you "collapsed the graph sideways onto the y-axis, which portions would be covered?)
The function is defined for x = 2. So look at the function rule, find the half which is defined for x = 2, and plug 2 in for x in that half's rule.
(Wink)
I'm still unsure at how you are supposed to graph this thing. I don't understand the domain & range either. I do understand the f(2) thing. The answer for that part is 4, right?
• Sep 21st 2009, 02:35 PM
stapel
Quote:
Originally Posted by BeSweeet
I'm still unsure at how you are supposed to graph this thing.
To learn how to graph linear equations, such as y = x + 3, try here.
to learn how to graph quadratics, such as y = x^2, try here.
They were supposed to have covered this material way before moving on to piecewise functions, etc. (Wondering)
• Sep 21st 2009, 02:53 PM
BeSweeet
Quote:
Originally Posted by stapel
To learn how to graph linear equations, such as y = x + 3, try here.
to learn how to graph quadratics, such as y = x^2, try here.
They were supposed to have covered this material way before moving on to piecewise functions, etc. (Wondering)
I get it!
a. Figured out how to sketch it.
bA. For the domain, is it (-, -5)∪(-5, 2)∪(2, 4)∪(4, )... Something like that?
bB. For the range, I'm still not getting that part.
c. Determining $\displaystyle f(2)$ is simple. Should be 4, right? |
By accessing our 180 Days of Math for Sixth Grade Answers Key Day 32 regularly, students can get better problem-solving skills.
Directions Solve each problem.
Question 1.
(-12) ÷ (-14) = ___
(-12) ÷ (-14) = 0.85.
Explanation:
The division of -12 and -14 is (-12) ÷ (-14) = 0.85.
Question 2.
6 • 9 = _____
6×9 = 54.
Explanation:
By multiplying 6 and 9 we will get 6×9 = 54.
Question 3.
540 ÷ 6 = ___
540 ÷ 6 = 90.
Explanation:
The division of 540 by 6 we will get 540 ÷ 6 = 90.
Question 4.
Write 573 in expanded notation.
__________________
________________
500+70+3.
Explanation:
The expanded form of 573 is 500+70+3.
Question 5.
$$\frac{1}{2}$$ • $$\frac{1}{3}$$ = ____
$$\frac{1}{6}$$.
Explanation:
Given $$\frac{1}{2}$$ • $$\frac{1}{3}$$ which is
$$\frac{1}{6}$$.
Question 6.
32 – 8 • 2 = ___
32 – 8 • 2 = 16.
Explanation:
Given the expression is 32 – 8 • 2 which is
32 – 8 • 2 = 32 – 16
= 16.
Question 7.
Solve for x. $$\frac{x}{7}$$ = 7
x = ___
x = 49.
Explanation:
Given the expression is $$\frac{x}{7}$$ = 7,
x = 7×7
= 49.
Question 8.
30 10 = 300
30×10 = 300.
Explanation:
By multiplying 30 and 10 we will get 30×10 = 300.
Question 9.
Calculate the volume of a cube in which each side measures 3 cm.
_______________________
The volume of the cube is 9 cubic cm.
Explanation:
The volume of a cube in which each side measures 3 cm is
= 3×3×3
= 9 cubic cm.
Question 10.
What is the sum of the angles in a triangle?
_______________________
180 degrees.
Explanation:
The sum of the angles in a triangle is 180 degrees.
Question 11.
What is the probability you will spin red or green?
_________________________
$$\frac{2}{3}$$.
Explanation:
The probability of spinning red or green will be $$\frac{2}{3}$$.
Question 12.
Complete the subtraction table. |
# Difference between revisions of "2007 AMC 12A Problems/Problem 13"
## Problem
A piece of cheese is located at $(12,10)$ in a coordinate plane. A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$. At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$?
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$
## Solution
We are trying to find the point where distance between the mouse and $(12, 10)$ is minimized. This point is where the line that passes through $(12, 10)$ and is perpendicular to $y=-5x+18$ intersects $y=-5x+18$. By basic knowledge of perpendicular lines, this line is $y=\frac{x}{5}+\frac{38}{5}$. This line intersects $y=-5x+18$ at $(2,8)$. So $a+b=\boxed{10}$. - MegaLucario1001
## Solution 2
If the mouse is at $(x, y) = (x, 18 - 5x)$, then the square of the distance from the mouse to the cheese is$(x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).$The value of this expression is smallest when $x = 2$, so the mouse is closest to the cheese at the point $(2, 8)$, and $a+b=2+8 = \boxed{10}$. -Paixiao
## See also
2007 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# How to solve 1^3+2^3+3^3+4^3+5^3 to 85^3 formula calculator
We always provide the solution of Math question to our student in easy words to my article How to solve 1^3+2^3+3^3+4^3+5^3 to 85^3 formula calculator. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve 1^3+2^3+3^3+4^3+5^3 to 85^3 formula calculator, read and understand it carefully till the end.
Let us know how to solve this question How to solve 1^3+2^3+3^3+4^3+5^3 to 85^3 formula calculator.
First write the question on the page of the notebook.
## How to solve 1^3+2^3+3^3+4^3+5^3 to 85^3 formula calculator
To know the solution of this article, first of all, according to your understanding, we will write this question in such a way.
\displaystyle {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+……+{{85}^{3}}
By understanding this question, we come to know that it is going on according to a chain. Therefore, we will solve this question with the help of the following formula.
FORMULA :
\displaystyle \mathop{S}_{n}={{\left( {\frac{{n(n+1)}}{2}} \right)}^{2}}
here ,
n = 85
putting the value of n in this formula-
\displaystyle \mathop{S}_{{85}}={{\left( {\frac{{85(85+1)}}{2}} \right)}^{2}}
\displaystyle \mathop{S}_{{85}}={{\left( {\frac{{85(86)}}{2}} \right)}^{2}}
\displaystyle \mathop{S}_{{85}}={{\left( {\frac{{\cancel{2}\times 85(43)}}{{\cancel{2}}}} \right)}^{2}}
\displaystyle \mathop{S}_{{85}}={{\left( {85(43)} \right)}^{2}}
\displaystyle \mathop{S}_{{85}}={{\left( {3655} \right)}^{2}}
\displaystyle \mathop{S}_{{85}}=13359025 |
# How do you find the domain and range of sqrt(x+5)?
May 18, 2017
Domain$: \left\{x \in \mathbb{R} | x \ge q - 5\right\}$
Range$: \left\{f \left(x\right) \in \mathbb{R} | f \left(x\right) \ge q 0\right\}$
#### Explanation:
We have: $f \left(x\right) = \sqrt{x + 5}$
The domain of any square root function is dependent on its argument.
The argument must be greater than or equal to zero:
$R i g h t a r r o w x + 5 \ge q 0$
$R i g h t a r r o w x \ge q - 5$
The square root function never produces a negative result.
So the range will be greater than or equal to zero as well:
$R i g h t a r r o w f \left(x\right) \ge q 0$
Therefore, for the function $f \left(x\right) = \sqrt{x + 5}$, the domain is $\left\{x \in \mathbb{R} | x \ge q - 5\right\}$ and the range is $\left\{f \left(x\right) \in \mathbb{R} | f \left(x\right) \ge q 0\right\}$. |
# Interest Over Several Periods
In this section, you’ll learn what happens when $n$ has a value greater than $1$. In other words, there is more than one time period over which interest will compound.
Over time, this can lead to significant increases in the balance, because compounding interest means that interest is being calculated based on prior interest.
When saving, compound interest is quite appealing, because it gives you more money. But with credit cards and loans, compound interest means you end up paying interest charges on interest you’ve already been charged!
Formula
### Interestcalculations
${K}_{n}={K}_{0}\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{n}$
where ${K}_{0}$ is the initial balance, and $p$ is the interest rate. After $n$ periods (or number of times the interest is added) the capital is ${K}_{n}$.
Example 1
### Interest over time
You borrow $\text{}\text{}5000\text{}$ from the bank to finance your graduation party. The bank decides that you have to pay $\text{}40\text{}\phantom{\rule{0.17em}{0ex}}\text{%}$ in interest per year. You ask to pay the total amount in five years. How much do you have to pay back?
The initial balance is ${K}_{0}=5000$. ${K}_{n}$ is unknown, $p=40$ and $n=5$. You can put these figures straight into the formula and solve the equation: $\begin{array}{llll}\hfill {K}_{n}& =5000\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{40}{100}\right)}^{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5000\cdot 1.{4}^{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =26\phantom{\rule{0.17em}{0ex}}891.2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Example 3
### Interest over time
Your grandfather was always a cool guy. In his will, he has left his classic car to you. He bought the car 55 years ago, and it cost $\text{}\text{}5000\text{}$ at the time. The annual increase in the car’s value is $\text{}5\text{}\phantom{\rule{0.17em}{0ex}}\text{%}$. How much is the car worth today?
Again, begin by finding out where in the formula the different numbers fit. In this case, ${K}_{0}=5000$, $p=5$, $n=55$, and ${K}_{n}$ is the unknown. Plug these straight into the formula and solve the equation to find the answer: $\begin{array}{llll}\hfill {K}_{n}& ={K}_{0}\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{n}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5000\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{5}{100}\right)}^{55}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5000{\left(1+0.05\right)}^{55}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5000\cdot 1.0{5}^{55}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 73\phantom{\rule{0.17em}{0ex}}178\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
The car is worth \$$73\phantom{\rule{0.17em}{0ex}}178$ today.
Example 4
### Find the interest rate
Ola bought a car for $\text{}\text{}25\phantom{\rule{0.17em}{0ex}}000\text{}$ 15 years ago. The present value of the car is $\text{}\text{}2200\text{}$. What was the yearly percentage decrease in value for Ola’s car?
Again you can begin by finding where the numbers go in the formula. In this case we know that ${K}_{0}=25\phantom{\rule{0.17em}{0ex}}000$, ${K}_{n}=2200$, $n=15$, and $p$ is the unknown we want to find. Insert this information into the formula to get the answer. As $p$ is unknown inside the parenthesis raised to the power of 15, you need to take the 15th root to solve the equation:
$\begin{array}{llll}\hfill 25\phantom{\rule{0.17em}{0ex}}000& =2200\phantom{\rule{-0.17em}{0ex}}{\left(1-\frac{p}{100}\right)}^{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{25\phantom{\rule{0.17em}{0ex}}000}{2200}& =\phantom{\rule{-0.17em}{0ex}}{\left(1-\frac{p}{100}\right)}^{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 11.36& =\phantom{\rule{-0.17em}{0ex}}{\left(1-\frac{p}{100}\right)}^{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt[15]{0.088}& =\sqrt[15]{\phantom{\rule{-0.17em}{0ex}}{\left(1-\frac{p}{100}\right)}^{15}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 0.85& =1-\frac{p}{100}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -0.15& =-\frac{p}{100}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill p& =15\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
$\begin{array}{llllllll}\hfill 25\phantom{\rule{0.17em}{0ex}}000& =2200\phantom{\rule{-0.17em}{0ex}}{\left(1-\frac{p}{100}\right)}^{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{1em}{0ex}}|÷22\phantom{\rule{0.17em}{0ex}}000\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 11.36& =\phantom{\rule{-0.17em}{0ex}}{\left(1-\frac{p}{100}\right)}^{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt[15]{0.088}& =\sqrt[15]{\phantom{\rule{-0.17em}{0ex}}{\left(1-\frac{p}{100}\right)}^{15}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 0.85& =1-\frac{p}{100}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -0.15& =-\frac{p}{100}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{1em}{0ex}}|\cdot -100\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill p& =15\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
The yearly decrease in value was $15$ %. |
# Range of a Set of Data in Math and Statistics
## Definition of a Range (Statistics)
In statistics, the range is a measure of spread: it’s the difference between the highest value and the lowest value in a data set.
Note: In some areas of math, the range can also mean the entire range of numbers — for example, the range of cell phone prices might be \$40 to \$550. In calculus, the range is defined differently. It is all of the output values of a function. See: How to Find the Domain and Range of a Function.
## How to Find a Range in Statistics
Watch the video, or read the article below:
The same two steps are used whether you are dealing with positive numbers, negative numbers, or time (e.g. seconds or minutes).
## How to Find a Range
Example question 1: What is the range for the following set of numbers? 10, 99, 87, 45, 67, 43, 45, 33, 21, 7, 65, 98?
Step 1: Sort the numbers in order, from smallest to largest:
7, 10, 21, 33, 43, 45, 45, 65, 67, 87, 98, 99
Step 2: Subtract the smallest number in the set from the largest number in the set:
99 – 7 = 92
The range is 92
That’s it!
Example question 2: What is the range of these integers?
14, -12, 7, 0, -5, -8, 17, -11, 19
Step 1: Sort the numbers in order, from smallest to largest:
-12, -11, -8, -5, 0, 7, 14, 17, 19
Step 2: Subtract the smallest number in the set from the largest number in the set:
19 – -12 = 19 + 12 = 31
The range is 31.
That’s it!
Example question 3: What is the range of the following times?
2.7 hrs, 8.3 hrs, 3.5 hrs, 5.1 hrs, 4.9 hrs
Step 1: Sort the numbers in order, from smallest to largest:
2.7, 3.5, 4.9, 5.1, 8.3
Step 2: Subtract the smallest number in the set from the largest number in the set:
8.3 hr – 2.7 hr = 5.6 hr
The range is 5.6 hr.
That’s how to find a range!
### Another Example.
Problem: You take 7 statistics tests over the course of a semester. You score 94, 88, 73, 84, 91, 87, and 79. What is the range of your scores?
Solution:
Step 1: Order your scores from smallest to largest:
73, 79, 84, 87, 88, 91, 94.
Step 2: Subtract the smallest number from the highest = 94 – 73 = 21.
## When it Might be Misleading
The range only uses the smallest and the largest number in a set; The rest of the values are ignored. That could lead to a misleading result. Take the above test scores. Let’s say you had the flu one test day and scored a 10. Assuming your highest score on another test was 94, then:
94 – 10 = 84!
That’s not a good reflection of your overall test performance at all.
The score of 10 in the example above is what we call an outlier. It’s an extremely high or low value that can throw off stats. That’s why other measures of spread are sometimes preferred, like the mean.
## Rule of Thumb
The rule of thumb says that the range is about four times the standard deviation. The standard deviation is another measure of spread in statistics. It tells you how your data is clustered around the mean. What the rule of thumb tells you in most cases is that the bulk of the data can be found pretty close to the mean (within a couple of standard deviations); The result is that those erroneous “outliers” should have very little effect on your final statistic.
Procedure for finding a standard deviation using the rule of thumb:
Step 1: Find the range.
Step 2: Divide Step 1 by four.
The rule of thumb doesn’t work that well for small data sets. And it doesn’t work at all if you don’t have data that fits a normal distribution. That’s why you’ll rarely see it used in statistics.
## Range in Excel 2013-2016
Watch the video or read the steps below:
To find a range in Excel, you have two options: you can use the MAX and MIN functions to find the largest and smallest numbers in a data set and then you can subtract the two. For example, if you had a data set in cells A1 to A10, you’d need three formulas in three blank cells. Lastly the format (assuming you put these formulas into cells B1:B3) would be:
B1 = MAX(A1:A10)
B2 = MIN(A1:A10)
B3 =(B1-B2)
A much easier way is to use Data Analysis, where in just a couple of clicks (with no entering formulas) you can display a variety of summary statistics, including the range (How to load the Data Analysis Toolpak).
### Range in Excel: Data Analysis Steps
1st: Click the “Data” tab and then click “Data Analysis.”
2nd: Click “Descriptive Statistics” and then click “OK.”
3rd: Click the Input Range box and then type the location for your data. For example, if you typed your data into cells A1 to A10, type “A1:A10” into that box
4th: Click the radio button for Rows or Columns, depending on how your data is laid out.
5th: Click the “Labels in first row” box if your data has column headers.
6th: Click the “Descriptive Statistics” check box.
7th: Select a location for your output. For example, click the “New Worksheet” radio button.
8th: Click “OK.”
## Origins
The origin of the word “Range” in mathematics is unknown, but a few early uses of the word as it’s used in statistics can be found as far back as 1848, in H. Lloyd, “On Certain Questions Connected with the Reduction of Magnetical and Meteorological Observations,” Proceedings of the Royal Irish Academy, 4, 180-183 (David, 1995). Following this, the word was later used in a book on Calculus in 1865: The Differential Calculus by John Spare mentions: “…in respect to the range of values which the function and its variable may sustain, and to their mutual dependence” [University of Michigan Digital Library]. Although technically not statistics, the range in calculus has practically the same meaning (the spread from the smallest value to the largest).
Check out our YouTube channel for more stats help and tips!
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If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track. |
# Why Do we Count by 5s?
5 teachers like this lesson
Print Lesson
## Objective
SWBAT make connections between counting by 5s and the minute hand.
#### Big Idea
Students learn the WHY behind counting by 5s in this lesson, perfect for extension in 1st grade or an introduction in 2nd.
## Setting Up the Learning
5 minutes
CCSS Context:
A key aspect of the CCSS shift is the push towards students developing a conceptual understanding of mathematics. In this lesson, students understand why we count by 5s, instead of just memorizing the idea of half hour. This also allows students to connect to their understanding of why we group-count in general, which gets at number sense concepts.
Review:
Yesterday we talked about how the minute hand says "00" when it points to the 12. We use the word "o'clock" to describe this time. Today we will talk about WHY the minute hand says 00 there.
Connect
We use the clock to tell the time and measure the time everyday. In football, they have to measure how long each quarter of the game is in minutes. In school, we have to check the clock to see what time it is.
Objective :
Your thinking job today is: How does the minute hand count?
## Opening Discussion
10 minutes
Counting Around the Clock:
• The minute hand looks at the little lines between the big numbers.
• We are only going to say the numbers loud when we get to the big numbers. (Whisper count 1-4 and say 5 loud)
• Let’s do that together and say the numbers loud and I’ll write them down. (I'll puts a post it over the big numbers with the minute hand info)
• Let me say the numbers that we wrote down. How are we counting?
• Discuss the WHY: We count by 5s because each part of the clock has 5 minutes in it (model checking each section to make sure there are 5 minutes between the big numbers).
Practice figuring out what the minute hand is saying on teacher clock.
“When we tell time, we have to follow these steps to figure out and write the time”
Steps:
2. Is the hour hand pointing to a number or in the middle?
3. Look at the minute hand.
4. Count by 5s.
Partner Talk: Why do we have to count by 5s?
## Student Share
15 minutes
Students add to paper plate clocks: We started making these clocks in this lesson!
Students write the numbers they say around the outside rim of the paper plate.
• On your clocks today, you are going to add in what the minute hand sees and says on the outside of the plate.
• How will you figure out what the minute hand says on each part of the clock?
• Do 00 together at the 12, then do 1, 2, and 3 together. (05, 10, 15)
• Compare with your partner’s. Does your plate have the same numbers going around it?
Student Practice:
• Put your minute hand on the 25 mark. What number is it pointing to?
• Follow this routine with other minute hand times. If students need a challenge, add in hour information. (The time is 2:40. Where will the hour hand go? Where will the minute hand go?)
## Independent Practice
15 minutes
Group A:
Students show how to count by 5s by labeling the clock. Students draw the minute hand to point at the number that represents the minute.
Group B:
Students order clocks from 1:00 to 2:00. Students cut out the clock to represent the time.
Group C Extension: See Extension video for more on the Group C work.
Students use the exercise schedule clues to figure out what time each exercise starts. Students use critical thinking skills to interpret clues and order the exercises.
## Closing
5 minutes
Pass out post its/cards that say 00 all the way to 60 by 5s. Have students figure out where to put them to label the classroom clock. |
Find the inverse of the matrix $$\begin{bmatrix} 3& -3 \\ 4& -2 \end{bmatrix}$$
1. $$\frac{-1}{6}\begin{bmatrix} -2& 3 \\ -4& -3 \end{bmatrix}$$
2. $$\frac{1}{6}\begin{bmatrix} -2& 3 \\ -4& 3 \end{bmatrix}$$
3. $$\frac{1}{6}\begin{bmatrix} -3& 3 \\ 4& -2 \end{bmatrix}$$
duplicate options found. English Question 1 options 1,2
4. None of these
Option 2 : $$\frac{1}{6}\begin{bmatrix} -2& 3 \\ -4& 3 \end{bmatrix}$$
Detailed Solution
Concept:
For any matrix $$A =\begin{bmatrix} a& b \\ c& d \end{bmatrix}$$, If matrix A = [aij]
A-1 = $$\rm \text{(adj A)}\over|A|$$
The adj A matrix is the transpose of the matrix [Aij] is the cofactor of the element aij.
Note: To calculate adjoint of 2 × 2 matrix [A], interchange the a and d and reverse the sign of b and c.
Calculation:
Given that,
A =$$\begin{bmatrix} 3& -3 \\ 4& -2 \end{bmatrix}$$
⇒ adj(A) = $$\begin{bmatrix} -2& 3 \\ -4& 3 \end{bmatrix}$$
|A| = 6 - (-12) = 6
We knowe that
A-1 = $$\rm \text{(adj A)}\over|A|$$
⇒ A-1 = $$\frac{1}{6}\begin{bmatrix} -2& 3 \\ -4& 3 \end{bmatrix}$$
Alternate Method
A = $$\begin{bmatrix} 3& -3 \\ 4& -2 \end{bmatrix}$$ = [aij]
⇒ |A| = 3 × (-2) - (-3) × 4
⇒ |A| = 6
Now, finding cofactor of matrix
a11 = 3, cof(a11) = -2
a12 = -3, cof(a12) = -4
a21 = 4, cof(a21) = 3
a22 = -2, cof(a22) = 3
Cof(A) = $$\begin{bmatrix}- 2& -4 \\ 3& 3 \end{bmatrix}$$
adj(A) = $$\begin{bmatrix} -2& 3 \\ -4& 3 \end{bmatrix}$$
A-1 = $$\rm \text{(adj A)}\over|A|$$
⇒ A-1 = $$\frac{1}{6}\begin{bmatrix} -2& 3 \\ -4& 3 \end{bmatrix}$$ |
# 3 Ways to Find the Perimeter of a Triangle
Finding the perimeter of a triangle consists of finding the distance of the line that passes through its edges. The simplest way to do this is to add up the length of all sides, but if you don't know them yet, you need to calculate them first. This article will first teach you how to find the perimeter of a triangle when all three side lengths are known; this is the simplest and most common way. It will then teach you how to find the perimeter of a right triangle when only two of the side lengths are known. Finally, we'll teach you how to find the perimeter of any triangle of which you know two sides and the angle between them (a “CAC triangle”), with the Law of Cosines.
## Steps
### Method 1 of 3: Finding the Perimeter When Three Sides Are Known
#### Step 1. Recall the formula for finding the perimeter of a triangle
For given triangle with sides The, B and ç, the perimeter FOR is defined as: P = a + b + c.
### What this formula means, in simple terms, is that to find the perimeter of a triangle you only need to join the lengths of each of its three sides
#### Step 2. Look at your triangle and determine the lengths of the three sides
In this example, the length of the side a = 5, the one on the side b = 5 and the one on the side c = 5.
### This particular example is called an equilateral triangle because all three sides have equal measurements. Remember, however, that the formula for the perimeter is the same for any type of triangle
#### Step 3. Add up the lengths of the three sides to find the perimeter
In the present example, 5 + 5 + 5 = 15. Soon, P = 15.
• In another example, in which a = 4, b = 3 and c = 5, the perimeter would be: P = 3 + 4 + 5, or
Step 12..
If the sides of the triangle are measured in centimeters, the answer must also be given in centimeters. If they are given in terms of a variable like x, your answer must also be defined in terms of x.
### Method 2 of 3: Finding the Perimeter of a Right Triangle When Two Sides Are Known
#### Step 1. Recall what a right triangle is
A right triangle is one that has a right angle (90 degrees). The side of the triangle opposite the right angle will always be the largest, being called the hypotenuse. Right triangles often appear on math tests, and luckily, there's a very useful formula for figuring out the value of unknown sides!
#### Step 2. Recall the Pythagorean Theorem
The Pythagorean Theorem tells us that for every right triangle with sides of size a and b, and hypotenuse of size c, The2 + b2 = c2.
#### Step 3. Look at your triangle and label sides “a”, “b” and “c”
Remember that the longest side is called the hypotenuse. It will be opposite the right angle and should be named ç. Name the two smallest sides as The and B. It doesn't really matter which one is represented by which letter - the result will be the same!
#### Step 4. Enter the known side lengths in the Pythagorean Theorem
remember that The2 + b2 = c2. Replace the side lengths with the corresponding letters in the equation.
• If, for example, you know that the side a = 3 and that side b = 4, enter these values into the formula as follows: 32 + 42 = c2.
• If you know the lengths of one side a = 6 and the hypotenuse c = 10, you need to describe the equation as follows: 62 + b2 = 102.
#### Step 5. Solve the equation to find the length of the unknown side
You must first square the known side lengths, that is, multiply each value by itself (for example: 32 = 3 × 3 = 9). If you're looking for the hypotenuse, simply add the two values together and find the square root of that number to find the length. If it's an unknown side length, you'll need to do some simple subtractions and then extract the square root to get the desired side length.
• In the first example, square the values present in 32 + 42 = c2 and find out that 25 = c2. Then calculate the square root of 25 to find that c = 25.
• In the second example, square the values in 62 + b2 = 102 to find that 36+b2 = 100. Subtract 36 from each side to find that B2 = 64 and then extract the square root of 64 to get the result b = 8.
#### Step 6. Add up the lengths of the three sides to find the perimeter
Remember the perimeter formula P = a + b + c. Now, knowing the value of the sides The, B and ç, you simply add up the lengths and figure out the perimeter.
• In our first example, P = 3 + 4 + 5 = 12.
• In our second example, P = 6 + 8 + 10 = 24.
### Method 3 of 3: Finding the Perimeter of a CAC Triangle Using Cosine Law
#### Step 1. Learn the Cosine Law
The Cosine Law allows you to unravel any triangle if you know the lengths of two sides and the measure of the angle between them. It works on any triangle and is a very useful formula. The Cosine Law states that for any triangle with sides The, B and ç, with opposite angles THE, B and Ç: ç2 = the2 + b2 - 2b cos (C).
#### Step 2. Look at your triangle and assign variable letters to its components
The first known side should be called the The and the angle opposite to it, of THE. The second known side must be named B; the opposite angle to it, B. The known angle must be defined by Ç, and the third side, for which the problem must be solved in order to find the perimeter of the triangle, will be the ç.
• For example, imagine a triangle with side lengths equal to 10 and 12, and an angle between them of 97°. We will define the variables as follows: a = 10, b = 12 and C = 97°.
#### Step 3. Enter the known information into the equation and solve the problem to find side c
You must first find the squares of a and b, adding them up next. Then find the cosine of C with the cos function on your calculator or on an online cosine calculator. Multiply cos (C) per 2b and subtract the product from the sum of The2 + b2. The result will be equal to ç2. Find the square root of this value, and you get the size of the side ç. Using our triangle as an example:
• ç2 = 102 + 122 - 2 × 10 × 12 × cos (97)
• ç2 = 100 + 144 - (240 × -0, 12187)
### Round the cosine to 5 places
• ç2 = 244 - (-29, 25)
• ç2 = 244 + 29, 25
### When cos (C) is negative, remember the sign
• ç2 = 273, 25
• c = 16.53
#### Step 4. Use a length of side c to find the perimeter of the triangle
Remember that the perimeter P = a + b + c, so all that has to be done is to add the newly calculated length for the side ç to the values already known for The and B. Easy!
• In our example: 10 + 12 + 16, 53 = 38, 53, the perimeter of our triangle! |
New Hampshire - Grade 1 - Math - Geometry - Shape Attributes - 1.G.1
Description
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
• State - New Hampshire
• Standard ID - 1.G.1
• Subjects - Math Common Core
• Math
• Geometry
More New Hampshire Topics
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). |
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# LESSON PLAN IN MATHEMATICS 4
SECOND QUARTER
September 28 29, 2017
Objectives:
Divide two-to-four-digit numbers by one-to-two digit numbers without remainder;
Divide two-to-four digit numbers by one-to-two digit numbers with remainder;
Solve word problems involving division.
## Topic: Operations on Whole Numbers
Lesson Proper Learning Activities Learning Materials
Day 1: Concept of Review:
Division Multiplication
Lesson: Dividing Properties of Multiplication Show-Me-Board
Whole Numbers Multiples and LCM
Whiteboard Marker
Core Idea: Seatwork: Random problem set.
Multiplication and Paper
Division are two of the Engage:
four fundamental Post this problem: Pen
operations on numbers.
Knowing how to The 1, 152 student participants of the Worksheet for Review
perform these two Reduce-Reuse-Recycle Waste
operations on whole Management workshop were divided into
numbers allows you to 8 groups for the different tasks and
work with numbers as presentations. How many students were
used in daily life, such in each group?
as applications on
money and Building the Understanding
as solve problems
involving numbers. Step 1: Divide 11 hundreds by 8.
## Step 2: Divide 35 tens by 8.
Remember: 3 tens = 30
Step 3: Divide 32 tens by 8.
## Therefore, there were 144 student
participants each group
## Changing the Routinize Problem to a
Non-Routinize Problem.
(Using pictures)
## Four pupils equally divided 24 paper clips
among themselves. How many paper
clips did each pupil get?
24 4 = 6
## Give examples from Our World of Math 4
pp. 48-49
Lets Try 1-3.
Board Activity
See pp. 50-51, Our World of Math 4
3 persons competing, solving under time
limit.
Activities:
Group Work (Quiz Bee)
Group the class into 4 (application of
Division)
Prepare the Show Me Board per group,
paper (for scratch/computation), pen and
white board marker.
a. 700 f. 903
b. 242 g. 472
c. 105 h. 1000
d. 468 i. 2501
e. 301 j. 3802
## B. Find the quotient:
a. 2412 6 e. 2993 7
b. 7212 3 f. 8052 11
c. 8544 4 g. 5475 33
d. 4555 2
Individual Work
Find the quotient.
1. 2412 7
2. 9217 8
3. 2254 49
4. 6648 11
Solve:
Day 2: Part 1:
Part 1: Short Review about Division Worksheets
Part 2: Problem
Solving using Division For Slow and Average Learners:
See page 77 and 84, , Math for All 4
## For Fast Learners:
Our World of Math 4
See page 51, Lets Try This B
page 53, Lets Extend Your
Understanding
## Part 2: Problem Solving using Division
and the 4-steps of Problem Solving
## Step 1: Understand the Plan
Step 2: Devise a Plan
Step 3: Carry Out the Plan
Step 4: Look back and Check
For Examples,
See
See Math for All 4 pp. 101-102
LESSON PLAN IN MATHEMATICS 4
SECOND QUARTER
October 5 - 6, 2017
Objectives:
Follow the correct order of mathematical operations; and
Solve multi-step problems.
## Topic: Operations on Whole Numbers
Lesson Proper Learning Activities Learning Materials
Day 1: Concept of Review:
Problem Solving Division
Show-Me-Board
Seatwork: Random problem set.
Core Idea: Whiteboard Marker
Multiplication and Engage:
Division are two of the Post this problem: Paper
four fundamental To raise awareness for clean air and to
operations on numbers. promote biking as a mode of Pen
Knowing how to transportation, bikers hold the Tour of the
perform these two Fireflies. 200 bottles of water were at the Worksheet for Review
operations on whole water station at the start of the biking
numbers allows you to event. After 173 riders have each gotten a
work with numbers as bottle, the organizers brought out 48 more
used in daily life, such bottles of water. How many bottles of
as applications on water were at the water station at this
money and time?
measurements, as well
as solve problems
involving numbers. |
Credit:pitelCC-BY-SA 2.0
Q:
# How do you convert a percentage to a ratio?
A:
To convert a percentage to a ratio, write out the percentage number as a fraction, reduce the fraction to its simplest form and convert the new fraction to a ratio by replacing the slash mark with a colon. Converting a percentage to a ratio takes only a few minutes and requires paper and a pencil.
Know More
1. Write the percentage down as a fraction
Start by writing out the percentage as a fraction over 100. So, if the percentage is 46 percent, write it as the following: 46/100.
2. Reduce the fraction
Take the fraction down to its simplest form. The example, 46/100, can be reduced by dividing both numbers by 2, or 2/2. This yields the fraction 23/50, which cannot be reduced further.
3. Express the fraction as a ratio
Expressing a fraction as a ratio simply means replacing the slash mark with a colon. So, in the example, 23/50 becomes the ratio 23:50.
4. Convert a ratio into a fraction
To reverse the process, turn the ratio back into a fraction. To use a new example, convert the ratio 5:7 into 5/7. Divide the top number by the bottom number. In the example, this yields 0.71. Convert the decimal into a percentage by moving the period two places to the right and adding the percent symbol. In this case, the answer is 71 percent, or 71 percent.
Sources:
## Related Questions
• A:
You can easily write a number in percentage form as a fraction in its simplest form by converting your numbers from one form to the other. A percentage can be directly converted to a fraction, or a percentage can be converted to a decimal first, then to a fraction. In converting percentages to simplified fractions, it is important to keep in mind that percent means "out of 100."
• A:
A percentage is solved by transferring a percentage into a decimal or a fraction, or by transferring a decimal or fraction into a percentage. To transfer a percentage into a decimal, divide the percentage by 100. To transfer it into a fraction, multiply the fraction by 100. |
# Class 6 Mathematics | Decimals Practice Problems
CLASS 6 Mathematics | DECIMALS – Lesson 8 – Practice Problems
1. Write each of the following as decimals:
a) Two ones and three- tenths
b) Twenty and two-tenths
2. Write each of the following as decimals:
a) 20 + 5 + 3/10
b) 300 + 4+7/10
3. Express the following as cm using decimals:
a) 4 mm
b) 112 mm
c) 4 cm 3 mm
4. Write as fractions in lowest terms.
a) 0.07
b) 2.56
c) 0.432
5. Write each of the following as a decimal.
a) 500+40+3+ 1/10 +7/100
b) 23 + 4/10 + 2/100
6. Write each of the following as decimal.
a) Two hundred seven and three- hundredths
b) Twelve point three four two.
7. Which is greater?
a) 0.2 or 0.02
b) 1.3 or 1.30
8. Express as rupees using decimals.
a) 50 paise
b) 20 rupees 25 paise
9. Express as kg using decimals.
a) 3g
b) 2550 g
10. Express as km using decimals.
a) 4 m
b) 4321 m
11. Sona spent Rs 20.50 for buying a chocolate and Rs 6.50 for one pen. How much money did she spend?
12. Anu travelled 2 km 25 m by bus, 3 km 450 m by car and the rest 1 km 25 m he walked. How much distance did he travel in all?
13. Manu bought 3 kg 80 g of oranges, 4 kg 60 g of bananas and 5 kg 450 g of apples. Find the total weight of all the fruits he bought.
14. Janvi had Rs 100. She bought chocolates for Rs 55. 50. Find the balance amount left with her.
15. Hisham bought fruits weighing 10 kg. Out of this, 3 kg 750 g is grapes, 2 kg 50 g is apples and the rest is bananas. What is the weight of the bananas?
1. a) 2 + 3/10 = 2.3
b) 20 + 2/10 = 20.2
2. a) 25.3
b) 304.7
3.a) We know that 1 cm = 10 mm
So 4 mm = 4/10 = 0.4 cm
b) 112 mm = 112/10 = 11.2 cm
c) 4 cm 3 mm = 4 + 3/10 = 4.3 cm
4. a) 0.07 = 7 /100
b) 2.56 = 256/100 = 128/50 = 64/25.
c) 0.432 = 432/ 1000 = 216/500 = 108/250 = 54/125
5. a) 543.17
b) 23.42
6. a) 207 + 3/100 = 207.03
b) 12.342
7. a) 0.2
b) Both are equal.
8. a) We know that 1 rupee = 100 paise
So 50 paise = 50/100 = 0.50 rupees
b) 20 rupees 25 paise = 20 + 25/100 = 20.25 rupees
9. We know that 1 kg = 1000g
a) 3g = 3/1000 = 0.003 kg
b) 2550 g = 2550/1000 = 2.550 kg
10. a) We know that 1 km = 1000m
So 4m = 4/1000 = 0.004 km
b) 4321m = 4321/1000 = 4.321 km
11. Money spent for chocolate = 20.50 rupees
Money spent for one pen = 6.50 rupees
Total money spent = 20.50 + 6.50 = Rs 27
12. Distance travelled by bus = 2 km 25 m = 2.025 km
Distance travelled by car = 3 km 450m = 3.450 km
Distance travelled by walking = 1 km 25m = 1.025 km
Total distance travelled = 2.025 + 3.450 + 1.025 = 6.500 km
13. Weight of Oranges = 3 kg 80 g = 3. 080 kg
Weight of bananas = 4 kg 60g = 4. 060 kg
Weight of apples = 5 kg 450g = 5.450 kg
Total weight of fruits = 3.080 + 4.060 + 5.450 = 12.590 kg = 12 kg 590g
14. Total money Janvi had = Rs 100
Money spent for chocolate = 55.50 rupees
Balance amount left with her = 100 – 55.50 = 44.50 rupees.
15. Total weight of fruits = 10 kg
Weight of grapes = 3 kg 750g = 3.750g
Weight of apples = 2kg 50g = 2.050 g
Total weight of grapes and apples = 3.750 + 2.050 = 5.800 kg
Weight of bananas = 10 – 5.800 = 4.200 kg = 4kg 200g |
# Closure Property
If a and b are two whole numbers and their sum is c, i.e. a + b = c, then c is will always a whole number.
For any two whole numbers a and b, (a + b) is also a whole number. This is called the Closure-Property of Addition for the set of W.
Whole number + whole number = Whole number
Some solved examples :
1) 3 + 4 = 7
Here 3 and 4 are whole numbers.
The addition of 3 and 4 which is 7 is also a whole number.
So, property of closure is true for addition.
2) 4 - 3 = 1
Here, 4 and 3 are whole numbers and 1 is also a whole number.
So the property is true.
But 3 - 4 = -1
Here 3 and 4 are whole numbers.
The subtraction of 3 and 4 is -1 which is not a whole number.
So the property of closure for subtraction is not always true.
3) 12 + 0 = 12
Here, 12 and 0 both are whole numbers.
The addition of them which is 12 again is also a whole number.
So the property of closure is true.
Closure property for multiplication :
If a and b are whole numbers then their multiplication is also a whole number.
For any two whole numbers a and b, (a x b) is also a whole number.This is called the property of closure for Multiplication for the set of W.
Whole number x whole number = whole number
Some solved examples :
1) 30 x 7 = 210
Here 30 and 7 are whole numbers.
The multiplication of 30 and 7 which is 210 is also a whole number.
So property of closure for multiplication is true.
2) 40 x 0 = 0
Here 40 and 0 both are whole numbers.
Their multiplication 0 which is the smallest whole number.
Note : Property of closure is not always true for division.
Example : 45 ÷ 0 = not defined
As division with zero is not possible.
Whole Number
Closure property
Commutative property
Associative property |
# New Pattern Reasoning Questions for SBI PO 2021 | SBI PO 2021 New Pattern Alphanumeric Questions PDF at Smartkeeda
Directions: Study the following numbers carefully and answer the questions given below:
5836 7469 8251 6293 4172
Important for :
1
If the digits of all the numbers are to be arranged in ascending order within the number from right to left then the numbers thus formed are to be arranged in descending order from left to right then what would be the sum of second digit of third number from left end and third digit of fourth number from right end?
» Explain it
D
We have,
The given sequence = 5836 7469 8251 6293 4172
After arranging the digits of all the numbers in ascending order within the number from right to left, we get:
8653 9764 8521 9632 7421
After, arranging the newly formed in descending order from left to right, we get:
9764 9632 8653 8521 7421
Here, third number from left end is ‘8653’ and second digit of ‘8653’ is ‘6’.
And, fourth number from right end is ‘9632’ and third digit of ‘9632’ is ‘3’.
Required Sum = 6 + 3 = 9
Hence, the correct answer is option D.
2
If in each number first digit is interchanged with second digit and third digit is interchanged with fourth digit after that first digit is interchanged with fourth digit then what would be the difference of highest and second lowest numbers thus formed?
» Explain it
B
We have,
The given sequence = 5836 7469 8251 6293 4172
After interchanging first digit with second digit and third digit with fourth digit, we get:
8563 4796 2815 2639 1427
Now, interchanging first and fourth digit of each of the above number, we get:
3568 6794 5812 9632 7421
Here, the highest and second lowest numbers are ‘9632’ and ‘5812’
Required Difference = 9632 – 5812 = 3820
Hence, the correct answer is option B.
3
If all the odd digits of each number are decreased by 1 and all the even digits of each number are divided by 2 then the even numbers thus formed are arranged in descending from left to right and on the right of these all odd numbers are arranged in ascending order then which of the following numbers will be in the middle of the sequence?
» Explain it
C
We have,
The given sequence = 5836 7469 8251 6293 4172
After subtracting 1 from all the odd digits of each number and dividing all the even digits of each number by 2, we get:
4423 6238 4140 3182 2061
Now, arranging all the even numbers in descending order from left to right, we get:
6238 4140 3182
After arranging all the odd number in ascending order on the right of above numbers, we get:
6238 4140 3182 2061 4423
The, number in the middle of the sequence is ‘3182’.
Hence, the correct answer is option C.
4
If all the numbers are arranged in descending order then what will the product of second digit of the third number from the right end and third digit of the forth number from the left end?
» Explain it
C
We have,
The given sequence = 5836 7469 8251 6293 4172
After changing the numbers in descending order, we get:
8251 7469 6293 5836 4172
Now, we have:
The Third number from the right end - 6293
The second digit of the third number -2
The forth number from the left end - 5836
The third digit of the forth number – 3
The product of the numbers -2*3 = 6
the product of second digit of the third number from the right end and third digit of the forth number from the left end is 6.
Hence, the correct answer is option C.
5
Find the number the sum of all the digits of which is equal to the sum of all the smallest digits taken from all the numbers.
» Explain it
E
We have,
The given sequence = 5836 7469 8251 6293 4172
The smallest digit of each number – 3, 4, 1, 2, 1
The sum of all smallest digits of each number,
3+4+1+2+1=11
The sum of all digits of each number,
5+8+3+6 = 22
7+4+6+9 = 26
8+2+5+1 = 16
6+2+9+3 = 20
4+1+7+2 = 14
the sum of all the digits of each number is not equal to the sum of all the smallest digits taken from all the numbers.
Hence, the correct answer is option E.
Not Now Open App |
## Yearly interest rate to quarterly
14 Sep 2019 Multiply the principal amount by one plus the annual interest rate to the compounding or quarterly compounding, etc), the formula changes.
r = annual rate of interest (as a decimal) An amount of $1,500.00 is deposited in a bank paying an annual interest rate of 4.3%, compounded quarterly. What is Our third account is compounded quarterly and receives eight interest deposits— one at the end of each three-month period. If we view the annual interest rate of The annual percentage rate (APR) of an account, also called the nominal rate,$3,000 in an investment account paying 3% interest compounded quarterly, d) compounded quarterly, n = 4: A = 5000(1 + 0.06/4)(4)(4) = 5000(1.015)(16) = If the interest rate is compounded n times per year, the compounded amount
## 2 Feb 2019 For whatever reason, students often freak out when they see things like “quarterly ” or “semi-annually” on GMAT interest rate problems. Fear not!
If you have a nominal interest rate of 10% compounded quarterly, then the Annual Equivalent rate is the same as 10.38%. If you have a nominal interest rate of 10% compounded monthly, then the Annual Equivalent rate is the same as 10.47%. Effective annual interest rate calculation. The effective annual interest rate is equal to 1 plus the nominal interest rate in percent divided by the number of compounding persiods per year n, to the power of n, minus 1. Effective Rate = (1 + Nominal Rate / n) n - 1. Example. What is the effective annual interest rate for nominal annual interest rate of 5% compounded monthly? Solution: Effective Rate = (1 + 5% / 12) 12 - 1 = (1 + 0.05 / 12) 12 - 1 = 0.05116 = 5.116% . Effective interest rate Effective Period Rate = Nominal Annual Rate / n Effective annual interest rate calculation The effective interest rate is equal to 1 plus the nominal interest rate in percent divided by the number of compounding persiods per year n, to the power of n, minus 1. For example, if you need to compare an interest rate of 12% p.a., payable monthly with an interest rate of 12.50% p.a., payable annually to find which one is expensive in terms of effective cost, convert the former into annual one or the latter into monthly one using this tool - to check out which one is more (or less) expensive than the other. To convert an annual interest rate to monthly, use the formula "i" divided by "n," or interest divided by payment periods. For example, to determine the monthly rate on a $1,200 loan with one year of payments and a 10 percent APR, divide by 12, or 10 ÷ 12, to arrive at 0.0083 percent as the monthly rate. In order to determine your mortgage loan's APR, these fees are added to the original loan amount to create a new loan amount of$205,000. The 6% interest rate is then used to calculate a new annual payment of $12,300. Divide the annual payment of$12,300 by the original loan amount of $200,000 to get an APR The effective annual rate is the rate that actually gets paid after all of the compounding. When compounding of interest takes place, the effective annual rate becomes higher than the overall interest rate. The more times the interest is compounded within the year, the higher the effective annual rate will be. ### Dec 10, 2018 When you are using monthly or quarterly interest rates instead of annual, you can find the appropriate rate by dividing the annual interest rate by 10 Nov 2015 r = annual interest rate (divide the number by 100) coupled with higher frequency of compounding (quarterly, half-yearly), can work magic. Interest is quoted in terms of an annual rate, but frequently is compounded over shorter intervals. For example, an 8% interest rate when compounded quarterly when compounding of interest is done on a Monthly, Quarterly, Half Yearly or Fixed Deposits are a great way to invest for those who rate safety higher than 23 May 2019 Annual interest rates allow you to quickly compare how much interest you'll earn or pay on different types of accounts. However, the annual ### Calculate Principal, Interest Rate, Time or Interest. at a$\color{blue}{12\%}$nominal annual interest rate compounded$\color{blue}{\text{quarterly}}$. Interest is quoted in terms of an annual rate, but frequently is compounded over shorter intervals. For example, an 8% interest rate when compounded quarterly when compounding of interest is done on a Monthly, Quarterly, Half Yearly or Fixed Deposits are a great way to invest for those who rate safety higher than 23 May 2019 Annual interest rates allow you to quickly compare how much interest you'll earn or pay on different types of accounts. However, the annual After one year, you would earn$10 of interest ($100 * 10%) and still have the original$100 in the bank. After the second and third to simple interest? The loan is $10,000 at an annual rate of 8.7% for 3 years. Assume quarterly compounding. ## To convert an annual interest rate to monthly, use the formula "i" divided by "n," or interest divided by payment periods. For example, to determine the monthly rate on a$1,200 loan with one year of payments and a 10 percent APR, divide by 12, or 10 ÷ 12, to arrive at 0.0083 percent as the monthly rate.
Converts the nominal annual interest rate to the effective one and vice versa. effective (R). Compounded (k); annually semiannually quarterly monthly daily. A bank offers an account that yields a nominal rate of return of. 3.3% per year, compounded quarterly. What is the annual effective rate of return? How many years
Feb 21, 2020 Below is a breakdown of the results of these different compound periods with a 10% nominal interest rate: Semi-annual = 10.250%; Quarterly = Jun 7, 2006 Quarterly rate = (1 + annual rate )(1/4) – 1 I need to calculate the effective interest rate, using compounding base on the formula below? You'll often see interest rates quoted as an annual percentage—either an annual To calculate a monthly interest rate, divide the annual rate by 12 to account for the 12 months in the year. For a quarterly rate, divide the annual rate by four. Very often, we are presented with a rate of interest expressed as monthly, annual, or as quarterly, and need to be able to compare it with another rate denominated For example, you have a loan at an annual rate of 4% that compounds monthly ( m=12) however your payments are made quarterly (q=4) so your interest will be For instance, you can convert interest rate from annual to semi annual or monthly to annual, quarterly etc. Interest Rate, % p.a.. Payment frequency. Daily, Monthly |
# NCERT Solutions For Class 3 Maths Chapter 1 Where to look from?
### Where to look from ?
Question 1:
Have you looked at things from different sides? Do they look the same or different?
Yes, I have looked at things from different side. Most of them do not look the same from different sides.
Question 2:
Look at the pictures drawn here. How does the table look from the side? Which picture is from the top?
From the side, the table looks like this-
Picture from the top is-
PRACTICE TIME
Question A:
A cat is peeping into a classroom. Can you help her find out where the teacher is?
The teacher is standing in front of the students, near her table.
Question B:
Here are some pictures. Find out from where you have to look to see the things this way.
Question C:
Draw top views of a few things and ask your friends to guess what they are.
Top view of a train
Top view of a sofa
Top view of a bottle
MAKE THE OTHER PATTERNS YOURSELF
Question 1:
Copy these shapes on the dot grid. Note that some lines in the shapes are straight and some are not.
Question 2:
Use the dot grid given below to draw your own designs and shapes.
Question 3.
Complete these figures to make squares and rectangles.
Question 4:
On the dot grid given below, draw the following:
(a) a kite (b) a leaf (c) a flower (d) a boat (e) a star (f ) a pot
TIT FOR TAT
Question 1:
The painter had made many such pictures in which he drew only one half of the things. Draw the other half of these pictures and find out what these things are. Try doing it with a mirror.
Question 2:
If you ask the painter to draw things which cannot be divided into two similar mirror halves, then he cannot play the trick. Draw three more such things which do not have similar mirror halves.
MIRROR HALVES
Question 1:
Look at the pictures given below. Does the dotted line divide each picture into two similar mirror halves?
Question 2:
Give some more examples.
Question 3:
Using a dotted line, can you divide the following pictures into two similar halves?
Question 4:
Can you guess these letters from their halves?
A C H U B K
Question 5:
Guess the words by looking at their halves. |
What Does Difference Mean In Math?
What Does Difference Mean In Math?
In mathematics, the difference between two numbers results from subtracting one number from the other. For example, if you remove five from 10, the difference is 5.
The symbol for the difference is Δ (delta), and it is typically used when working with sequences or series. In a row, the difference between two consecutive terms is called a “term differential.” In a series, the sum of the differences between successive terms is called the “difference series.”
There are a few different ways to calculate the difference between two numbers. One way is to use subtraction:
Another way to calculate the difference between two numbers is to use division
The difference between two numbers can also be harmful, depending on what numbers you subtract. For example, if you remove five from -10, the difference is -5.
The absolute value of a number is the distance of that number from zero on the number line. So, in this case, the total value of -5 is five because it is five units away from zero. The absolute value of 10 is ten because it is ten teams away from zero. And the total value of -10 is also ten because it is ten units away from zero.
When working with negatives, it’s important to remember that the sign of the difference (positive or negative) depends on which numbers you are subtracting. This is because subtraction is only defined with two positive numbers.
For example, if you wanted to find the difference between -2 and -3, you would first have to convert each negative number into a positive number by changing its sign. When that is done, then you can subtract them.
So, for this problem, it would work like this:
The difference between two numbers can also be represented using a Venn diagram. This is a diagram that shows how different sets of data overlap. In the case of difference, it would show how the set of numbers {1, 2, 3, 4} minus the set of numbers {2, 3, 4, 5} would look.
As you can see in the diagram, the difference between these two sets is {1, 1}. This means that there are two numbers in set A that are not in set B (1 and 4) and two numbers in set B that are not in set A (2 and 5).
The mathematical concept of difference has many real-world applications. For example, when you’re shopping and see two prices for the same item, the difference is the amount you would save by buying the item at the lower price.
In business, the difference between two products sets them apart from one another. And in physics, the difference between kinetic and potential energy determines how an object will move.
No matter what field of mathematics you’re studying, the understanding difference is essential to mastering basic operations like addition, subtraction, multiplication, and division. With a strong foundation in these concepts, you’ll be able to tackle more complex problems with ease.
What does twice the difference mean in math?
In mathematics, “twice the difference” usually means to find the difference between two numbers and then multiply that result by 2. For example, if you are asked to see twice the difference between 4 and 7, you would first find the difference (3) and then multiply it by 2 to get 6. Alternatively, some people might think of “twice the difference” as taking the average of two numbers and then doubling it. So in our example, you would take the average of 4 and 7 (5) and then double it to get 10. Whichever method you use, make sure you are consistent with your way throughout the problem.
What does a difference of squares mean in math?
In mathematics, the difference of squares is a mathematical operation that finds the square root of the difference between two squares. To see the difference of squares, you first need to find the squares of each number involved in the problem.
Then, you subtract the squares and take the square root of the result. For example, if you are asked to find the difference of squares between 16 and 25, you would first see 16 and 25 (256 and 625). Then, you would subtract 256 from 625 to get 369. Finally, you would take the square root of 369 to get 13.5.
As another example, if you are asked to find the difference of squares between -16 and 25, you would first see the squares of -16 and 25 (4 and 625). Then, you would subtract four from 625 to get 621. Finally, you would take the square root of 621 to get 13.1875. No matter which method you use, make sure that it is consistent throughout the problem so that all your computations are accurate. |
# Telangana SCERT Class 9 Math Solution Chapter 2 Polynomials and Factorisation Exercise 2.4
## Telangana SCERT Solution Class IX (9) Math Chapter 2 Polynomials and Factorisation Exercise 2.4
(1) (i) given, p(x) = x3 – x2 – x – 1 ; g(x) = (x + 1)
∴ using remainder theorem we get
P(-1) = (-1)3 – (-1)2 – (-1) + 1
= -1 – 1 + 1 + 1
= 0
Hence,
∴ (x + 1) is a factor of p(x) since the remainder is 0.
(ii) x4 – x3 + x2 – x + 1 = p(x) ; g(x) = x + 1
∴ using remainder theorem
P(-1_ = (-1)4 – (-1)3 + (-1)2 – (-1) + 1
= 1 + 1 + 1 + 1 + 1
= 5
Hence, x + 1 is not a factor of p(x) since the remainder is not equal to 0.
(iii) given, p(x) = x4 + 2x3 + 2x2 + x + 1 ; g(x) = x + 1
∴ by remainder theorem,
P – 1 = (-1)4 + 2(-1)3 + 2(-1)2 + (-1) + 1
= 1 – 2 + 2 -1 + 1
= 1
Hence, x + 1 is not a factor of p(x) since the remainder is not equal to 0.
(iv) x3 – x2 – (3 – √3)x + √3
∴ by remainder theorem,
P(-1) = (-1)3 – (-1)2 – (3 – 3) (-1) + 3
= -1 – 1 + 3 – 3 + 3
= 1
Hence, x + 1 is not a factor of p(x) since the remainder is not equal to 0.
(2) (i) given f(x) = 5x3 + x2 – 5x – 1, g(x) = x + 1
∴ zero of g(x) = -1
Now, f(-1) = 5(-1)3 + (-1)2 – 5(-1) – 1
= -5 + 1 + 5 – 1
= 0
∴ By factor theorem g(x) is a factor of f(x)
(ii) given, f(x) = x3 + 3x2 + 3x + 1 , g(x) = x + 1
Zero of g(x) = -1
Now, f(-1) = (-1)3 + 3(-1)2 + 3(-1) + 1
= -1 + 3 – 3 + 1
= 0
∴ By factor theorem g(x) is a factor of f(x)
(iii) Given, f(x) = x3 – 4x2 + x + 6, g(x) = x – 2
∴ Zero of g(x) = +2
Now,
f(x) = 23 – 4(2)2 + 2 + 6
= 8 – 16 + 2 + 6
∴ By factor theorem g(x) is a factor of f(x).
(3) Given, f(x) = x3 – 3x2 – 10x + 24;
Let, g(x) = x – 2
L(x) = x + 3
P(x) = x – 4
Now, zero of g(x) = +2
Then, f(2) = 23 – 3x(2)2 – 10×2 + 24
= 8 – 12 – 20 + 24
= 32 – 32
= 0
∴ By factor theorem we can say that g(x) is a factor of f(x).
Now, zero of L(x) = -3
Then, f(-3) = (-3)3 – 3 x (-3)2 – 10 x -3 + 24
= -27 – 27 + 30 + 24
= 54 – 54
= 0
Hence, by factor theorem L(x) is a factor of f(x).
Now, the zero of p(x) = + 4
Then f(4) = 43 – 3x(4)2 – 10 x 4 + 24
= 64 – 48 – 40 + 24
= 88 – 88
= 0
Hence, by factor theorem p(x) is the factor of p(x).
(4) Let, f(x) = x3 – 6x2 – 19x + 84; P(x) = x + 4, q(x) = x – 3, R(x) = x – 7
Now, zero of p(x) = -4
Then, f(-4) = (-4)3 – 6x(-4)2 – 19 x -4 + 84
= -64 – 96 + 76 + 84
= 160 – 160
= 0
Hence by factorisation, p(x) is the factor of f(x).
Now, zero of f(x) = 3
Then, f(3) = 33 – 6 x 32 – 19 x 3 + 84
= 27 – 54 – 57 + 84
= 111 – 111
= 0
Hence by factor theorem, q(x) is a factor of f(x)
Now, zero of R(x) = 7
Then, f(7) = 73 – 6 x 72 – 19 x 7 + 84
= 343 – 294 – 133 + 84
= 427 – 427
= 0
Hence by factor theorem, R(x) is a factor of f(x)
(5) Let, f(x) = Px2 + 5x + r , g(x) = x – 2, p(x) = x – ½
Given, that both g(x) and p(x) are factors of f(x).
∴ By factor theorem,
Zero of g(x) = 2
F(2) = p x 22 + 5 x 2 + r
Or, 0 = p x 4 + 10 + r
Or, 4p + r = -10 ….(i)
(6) Let, f(x) = ax4 + bx3 + cx2 + d + e ;
g(x) = x2 -1
∴ zero of g(x)
X2 – 1 = 0
Or, x2 = 1
Or, x = √1
= 1
∴ by factor theorem,
f(1) = ax14 + bx13 + cx12 + dx1 + e
or, 0 = a + b + c + d + e
or, a + b+ c + d + e = 0 (Hence proved)
(7) (i) x3 – 2x2 – x + 2
= x3 – x2 – x2 + x – 2x + 2 [divide the two middle terms in this case -2x2 and –x into two parts such that each pair out of the six terms when divided by power of x or a number gives the same value for all three pairs]
= x2 (x – 1) – x(x – 1) -2(x – 1)
= (x-1) (x2 –x – 2)
Further,
X2 –x -2
= x2 + x -2x – 2
= x(x+1) -2(x+1)
= (x+1) (x-2)
∴ The factors of x3 -2x2 –x+ 2 are (x – 1), (X+1), (X-2).
∴ x2 – 4x – 5 is another factor of f(x)
Now, x2 – 4x – 5
= x2 + x – 5x – 5
= x(x + 1) – 5(x+1)
= (x + 1) (x – 5)
∴ (x+1) and (x+5) are the factors of (x).
(iii) x3 + 13x2 + 32x + 20
= x3 + x2 + 12x2 + 12x + 20x + 20
= x2(x+1) + 12x(x+1) + 20(x+1)
= (x+1) (x2 + 12x + 20)
Now,
X2 + 12x + 20
= x2 + 10x + 2x + 20
= x(x + 10) + 2 (x + 10)
= (x + 10) (x + 2)
∴ (x +1), (x + 10) and (x + 2) are the factors of x3 + 13x2 + 32x + 20.
(iv) y3 + y2 – y – 1
= y3 y2 + 2y2 – 2y + y – 1
= y2 (y – 1) + 2y (y – 1) + 1(y – 1)
= (y – 1) (y2 + 2y + 1)
Now, y2 + 2y + 1
= y2 + y + y + 1
= y(y + 1) + 1(y + 1)
= (y + 1)2
∴ The factors of y3 + y2 – y + 1 are (y – 1), (y + 1)2
(8) If ax2 + bx + c and bx2 + ax + c have a common factor x + 1 then show that c = 0 and a = b
Let, f(x) = ax2 + bx + c
P(x) = bx2 + ax + c
g(x) = x + 1
∴ zero of g(x) = -1
Now, given that g(x) is a common factor of both f(x) and p(x).
∴ Using factor theorem we can say that
f(-1) = a(-1)2 + b x -1 + c
or, 0 = a – b – c
or, a – b + c = 0….(i)
similarly,
p(-1) = b x (-1)2 + a x -1 + c
or, 0 = b – a + c
or, -a + b + c = 0 ….(ii)
Now, adding both equation (i) and (ii) we get
a – b + c = 0
– a + b + c = 0
____________
2c = 0
or, c = 0 (proved)
Putting the value of c in equation (i) we get
a – b + c = 0
or, a – b + 0 = 0
or, a = b (proved).
(9.) If x2 − x − 6 and x2 + 3x − 18 have a common factor (x − a) then find the value of a.
Let, f(x) = x2 – x – 6
P(x) = x2 + 3x – 18
g(x) = x – a
zero of g(x) = a
given that g(x) is the common factor of f(x) and p(x)
∴ By factor theorem
f(a) = a2 – a – 6
or, 0 = a2 – a – 6
or, a2 – a – 6 = 0
or, a2 + 2a – 3a – 6
or, a (a + 2) – 3 (a + 2)
or, (a+2) (a – 3) = 0
∴ either a = -2 or, a = 3.
Now also p(x) = a2 + 3 x a – 18
Or, 0 = a2 + 3a – 18
Or, a2 – 3a + 6a – 18 = 0
Or, a(a – 3) + 6(a – 3) = 0
Or, (a – 3) (a+6) = 0
∴ either a = 3 or, a = -6
∴ The values are -2, 3 , -6.
(10.) If (y − 3) is a factor of y3 − 2y2 − 9y + 18 then find the other two factors
Updated: September 18, 2021 — 4:44 pm |
# 2013 AIME I Problems/Problem 7
## Problem 7
A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$.
## Solution 1
Let the height of the box be $x$.
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\sqrt{\left(\frac{x}{2}\right)^2 + 64}$, and $\sqrt{\left(\frac{x}{2}\right)^2 + 36}$. Since the area of the triangle is $30$, the altitude of the triangle from the base with length $10$ is $6$.
Considering the two triangles created by the altitude, we use the Pythagorean theorem twice to find the lengths of the two line segments that make up the base of $10$.
We find: $$10 = \sqrt{\left(28+x^2/4\right)}+x/2$$
Solving for $x$ gives us $x=\frac{36}{5}$. Since this fraction is simplified: $$m+n=\boxed{041}$$
## Solution 2
We may use vectors. Let the height of the box be $2h$. Without loss of generality, let the front bottom left corner of the box be $(0,0,0)$. Let the center point of the bottom face be $P_1$, the center of the left face be $P_2$ and the center of the front face be $P_3$.
We are given that the area of the triangle $\triangle P_1 P_2 P_3$ is $30$. Thus, by a well known formula, we note that $\frac{1}{2}|\vec{P_1P_2} \text{x} \vec{P_1P_3}|=30$ We quickly attain that $\vec{P_1P_2}=<-6,0,h>$ and $\vec{P_1P_3}=<0,-8,h>$ (We can arbitrarily assign the long and short ends due to symmetry)
Computing the cross product, we find: $$\vec{P_1P_2} x \vec{P_1P_3}=-<6h,8h,48>$$
Thus: $$\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60$$ $$h=3.6$$ $$2h=7.2$$
$$2h=36/5$$
$$m+n=\boxed{041}$$
## Solution 3
Let the height of the box be $x$.
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\sqrt{(x/2)^2 + 64}$, and $\sqrt{(x/2)^2 + 36}$. Therefore, we can use Heron's formula to set up an equation for the area of the triangle.
The semiperimeter is (10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2
900 = $\frac{1}{2}$((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2)((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2 - 10)((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2 - $\sqrt{(x/2)^2 + 64}$)((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2 - $\sqrt{(x/2)^2 + 36}$).
Solving, we get $\boxed{041}$. |
# Algebra 1 Notes Warm Up 1. If y = 16, y – 3.6 = ____ 3. = ____
## Presentation on theme: "Algebra 1 Notes Warm Up 1. If y = 16, y – 3.6 = ____ 3. = ____"— Presentation transcript:
Algebra 1 Notes Warm Up 1. If y = 16, y – 3.6 = ____ 3. = ____
– (- 14) = ___ = ____ 12.4 14 16 – 3.6 59 14 GOAL USE THE DIST. PROP. TO EVALUATE/SIMPLIFY EXPRESSIONS; RECOGNIZE AND USE THE COMMUTATIVE AND ASSOCIATIVE PROP.; FIND SQUARE ROOTS; CLASSIFY AND ORDER REAL NUMBERS. KEY WORDS & PROPERTIES COEFFICIENT, EQUIVALENT EXPRESSIONS, IRRATIONAL NUMBERS, LIKE TERMS, PERFECT SQUARE, PRINCIPAL SQUARE ROOT, RADICAL SIGN, RATIONAL APPROXIMATION, REAL NUMBERS, SIMPLEST FORM, SQUARE ROOT, TERM,
Properties Commutative Associative Distributive Identity
Order of the terms does not change your answer 3 + 5 = a + b = b + a 4 * 3 = 3 * 4 x * y = y * x Moving grouping symbols does not change your answer 3 + (5 + 7) = (3 + 5) (6 * 5) * 9 = 6 * (5 * 9) a + (b + c) = (a + b) + c a * (b * c) = (a * b) * c Multiply your outside term times all terms inside the parentheses a(b + c) = ab + ac (7 + 9) = 5(7) + 5(9) Of Addition: = x + 0 = x Of Multiplication: 5 * 1 = x * 1 = x
Example 1 Use the Distributive Property to simplify the expressions. 3(2x + 6) 5(6m + 4n – 3n)
Example 2 Use the distributive property to write this product in another way.
Simplify by collecting like terms.
Example 3 Simplify by collecting like terms.
Example 4 Simplify. 3.2(x + y) + 2.3(x + y) + 4x
Example 5 Evaluate.
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# Plus and Minus (Same Variables)
When you add variables together, you’re actually using the definition of multiplication. In other words: $x+x=2x$ and
The same applies to subtraction, $-x-x-x=-3x$. Remember that $x-x=0$. Below is an example of how you can add and subtract the same variables.
Example 1
Calculate and simplify
$x+x-x-x-x+x+x$
There are several ways to do this, and you’ll look at two separate methods. The first is to add the positive and negative $x$s separately:
$\begin{array}{llll}\hfill & \phantom{=}x+x-x-x-x+x+x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x+x+x+x-x-x-x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{4-3=1}{\underbrace{4x-3x}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Note! You write $1x=x$.
Another method is to rearrange the terms so you have minus against plus, so you can cancel them with each other:
$\begin{array}{llll}\hfill & \phantom{=}x+x-x-x-x+x+x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x-x+x-x+x-x+x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\text{x}-\text{x}+\text{x}-\text{x}+\text{x}-\text{x}+x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
It’s not always the case that the variables in an expression are the same. When there are different variables in an expression, you can’t perform the calculation like you did in the example above. |
# Area of Integer Heronian Triangle is Multiple of 6
## Theorem
Let $\triangle {ABC}$ be an integer Heronian triangle.
Then the area of $\triangle {ABC}$ is a multiple of $6$.
## Proof
Heron's Formula gives us that:
$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where:
$\AA$ denotes the area of the triangle
$a$, $b$ and $c$ denote the lengths of the sides of the triangle
$s = \dfrac {a + b + c} 2$ denotes the semiperimeter of the triangle.
We set out to eliminate $s$ and simplify as best possible:
$\ds \AA$ $=$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ $\ds \AA^2$ $=$ $\ds \dfrac {a + b + c} 2 \paren {\dfrac {a + b + c} 2 - a} \paren {\dfrac {a + b + c} 2 - b} \paren {\dfrac {a + b + c} 2 - c}$ substituting for $s$ and squaring $\ds \leadsto \ \$ $\ds 16 \AA^2$ $=$ $\ds \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}$ multiplying through by $16$ and simplifying $\ds$ $=$ $\ds 2 a^2 b^2 + 2 b^2 c^2 + 2 c^2 a^2 - a^4 - b^4 - c^4$ multiplying out and simplifying $\ds \leadsto \ \$ $\ds \paren {4 \AA}^2 + \paren {b^2 + c^2 - a^2}$ $=$ $\ds \paren {2 b c}^2$ factorising
This is now in the form $p^2 + q^2 = r^2$.
From Solutions of Pythagorean Equation, $\tuple {p, q, r}$ has the parametric solution:
$\tuple {m^2 - n^2, 2 m n, m^2 + n^2}$
There are two steps to showing $6 \divides \AA$:
### Step $1$: $2 \divides \AA$
By Euclid's Lemma for Prime Divisors, we just need to show:
$2 \divides \AA^2 = s \paren {s - a} \paren {s - b} \paren {s - b}$
By Semiperimeter of Integer Heronian Triangle is Composite, $s \in \N$.
There are $3$ cases:
#### Case $1$: There are sides with odd and even lengths
Under this condition, one of $s - x$ will be even.
Then $\AA^2$ is also even.
$\Box$
#### Case $2$: All sides are of odd length
Suppose all sides are of odd length.
Then the perimeter is also odd.
But then the semiperimeter cannot be an integer.
$\Box$
#### Case $3$: All sides are of even length
Suppose all sides are of even length.
If the semiperimeter is even, the result follows.
Therefore we consider the case where $s$ is odd.
Then $x := s - a$, $y := s - b$ and $z := s - c$ are also odd.
Note that $x + y + z = 3 s - a - b - c = s$.
Hence $\AA^2 = x y z \paren {x + y + z}$.
Note that each of $x, y, z$ must be equivalent to $\pm 1 \pmod 4$.
If all of $x, y, z \equiv 1 \pmod 4$:
$x + y + z \equiv -1 \pmod 4$
If two of $x, y, z \equiv 1 \pmod 4$:
$x + y + z \equiv 1 \pmod 4$
If one of $x, y, z \equiv 1 \pmod 4$:
$x + y + z \equiv -1 \pmod 4$
If none of $x, y, z \equiv 1 \pmod 4$:
$x + y + z \equiv 1 \pmod 4$
In any of the above cases, we have:
$\AA^2 \equiv -1 \pmod 4$
which is impossible by Square Modulo 4.
$\Box$
In each valid case, we see that $2 \divides \AA^2$.
$\Box$
### Step $2$: $3 \divides \AA$
By Euclid's Lemma for Prime Divisors, we just need to show:
$3 \divides 16 \AA^2 = \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}$
We split the problem into $4$ cases:
#### Case $1$: None of $a, b, c$ are divisible by $3$
If $a \equiv b \equiv c \pmod 3$:
$a + b + c \equiv 3 a \equiv 0 \pmod 3$
Hence:
$3 \divides \paren {a + b + c}$
If two of the lengths have the same remainder when divided by $3$, say $a \equiv b \not \equiv c \pmod 3$:
$a \equiv b \equiv -c \pmod 3$
Hence:
$a + b - c \equiv 3 a \equiv 0 \pmod 3$
Thus:
$3 \divides \paren {a + b - c}$
$\Box$
#### Case $2$: One of $a, b, c$ is divisible by $3$
Without loss of generality suppose that number is $a$.
If $b \equiv c \pmod 3$:
$a + b - c \equiv a \equiv 0 \pmod 3$
Hence
$3 \divides \paren {a + b - c}$
If $b \not \equiv c \pmod 3$:
$b \equiv -c \pmod 3$
Hence:
$a + b + c \equiv -c + c \equiv 0 \pmod 3$
Thus:
$3 \divides \paren {a + b + c}$
$\Box$
#### Case $3$: Two of $a, b, c$ is divisible by $3$
Without loss of generality, suppose $3 \divides a, b$.
Then:
$\ds 16 \AA^2$ $=$ $\ds \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}$ $\ds$ $\equiv$ $\ds c \cdot c \cdot c \cdot \paren {-c}$ $\ds \pmod 3$ $\ds$ $\equiv$ $\ds -c^4$ $\ds \pmod 3$ $\ds$ $\equiv$ $\ds -1$ $\ds \pmod 3$ Square Modulo 3
By Square Modulo 3, $16 \AA^2 = \paren {4 \AA}^2$ is not a square.
Therefore this case is not valid.
$\Box$
#### Case $3$: $a, b, c$ are all divisible by $3$
We have:
$3 \divides \paren {a + b + c}$
$\Box$
We see that in every valid case, $3 \divides 16 \AA^2$.
Hence $3 \divides \AA$, and thus $6 \divides \AA$.
$\blacksquare$ |
## Series Convergence: What Does It Mean?
1. So we discussed what it means for a sequence to converge, and we are really interested in infinite series.
It makes sense to first ask “Does this series equal a finite number?” BEFORE we ask “What number is this series equal to?” So we construct a test for a series to see if it converges to a finite number.
2. Consider the series $\sum 2^{-n}$. We look at the partial sums, so, $S_{1}=1/2$, $S_{2} = 3/4$, $S_{3}=7/8$, $S_{4}=15/16$, and in general $S_{n}= (2^{n}-1)/2^{n} = 1-2^{-n}$.
We see we have a sequence $\{S_{n}\}$ and the sequence has its limit be
(1)$\displaystyle \lim_{n\to\infty}S_{n} = \lim_{n\to\infty} 1 - \frac{1}{2^{n}} = 1-\lim_{n\to\infty}2^{-n}= 1-0=1$.
We then define the series to converge to
(2)$\displaystyle \sum^{\infty}_{n=1}2^{-n}=1$.
This is our convention and definition, a series converges if and only if its sequence of partial sums converge.
3. Definition. Let $\displaystyle\sum^{\infty}_{n=1}a_{n}$ be given. Let $S_{n}=\sum^{n}_{k=1}a_{k}$ be the sequence of partial sums.
If $\displaystyle \lim_{n\to\infty}S_{n}=L$, we define $\displaystyle\sum^{\infty}_{n=1}a_{n}=L$ and say the series “Converges”.
If the sequence $\{S_{n}\}$ diverges, then the series $\sum a_{n}$ “Diverges”.
Example 1. Consider the series $\displaystyle\sum^{\infty}_{n=1}\cos(n\pi/2) = 0-1+0+1+\dots$. The sequence of partial sums does not “settle”: it diverges. So the series diverges.
Example 2. Consider the series
(3)$\displaystyle \sum^{\infty}_{n=1}\frac{6}{(2n+1)(2n-1)}=\sum^{\infty}_{n=1}\frac{6}{4n^{2}-1}=\sum^{\infty}_{n=1}\frac{3}{2n-1}-\frac{3}{2n+1}$.
We see that the partial sum is $S_{n} = 3 - 3/(2n+1)$. So
(4)$\displaystyle \lim_{n\to\infty}3-\frac{3}{2n+1}=3$.
Thus the series in Equation (3) converges to 3.
Theorem 1. If $\displaystyle\sum^{\infty}_{n=1}a_{n}$ converges, then $\displaystyle\lim_{n\to\infty}a_{n}=0$.
Proof. Let $\sum^{\infty}_{n=1}a_{n}=L$. Then $\lim S_{n}=L$ and $S_{n-1}=L$. We see $S_{n}-S_{n-1}=a_{n}$, so
$\displaystyle \lim_{n\to\infty}a_{n}=\lim_{n\to\infty}S_{n}-S_{n-1}=L-L=0$.
That concludes the proof.
4. Aside on Logic. A proposition of the form “If A, then B” can give us two different propositions.
(1) The Contrapositive “If not B, then not A.” It is logically equivalent to “If A, then B.”
(2) The Converse “If B, then A.” This is not logically equivalent to “If A, then B.”
The contrapositive of Theorem 1 is “If $\lim a_{n}\not=0$, then $\sum a_{n}$ diverges.”
Whenever you are handed a proposition, you should check its converse to see if its true or not. |
Associated Topics || Dr. Math Home || Search Dr. Math
### Multiple Step Equations
```
Date: Wed, 7 Dec 1994 16:31:42 -0800 (PST)
From: Phong Huynh
Subject: Math Probs
I am having problems with math problems like:
29 + q = -2(q - 13)
and -2(c + 7) - 30 = 9c
These are multiple step equations that I am having problems with.
```
```
Date: Wed, 7 Dec 1994 19:58:16 -0500 (EST)
From: Dr. Sydney
Subject: Re: Math Probs
Hello there! Thanks for writing Dr. Math! We'd love to help you with
your math problems. These can get a little tricky because it seems like
there are lots of steps involved, but if we take it one step at a time,
I think we can make sense of it.
Okay, you are given an equation with an unknown, and you are asked to
solve for the unknown. The general approach to problems like this is to
get all the terms with the unknown on one side of the equals sign with the
constants on the other side of the equals sign. Then you work with the
equation until you have only the unknown on one side of the equals sign.
Let's look at your problems, and try this strategy:
First, what is q if 29 + q = -2(q - 13)?
Our first step is to get all the terms with q's on one side of the equals
sign with constants on the other side. But before we can do that, we have
to simplify the right hand side of the equation. We need to distribute the
-2. So simplifying the right hand side of the equation, we get:
-2(q - 13) = -2q + 26
So, the equation becomes:
29 + q = -2q + 26
Now, we want all the terms that have a q in them on one side of the equals
sign and all the constants on the other side. So, let's add 2q to both
sides and subtract 29 from both sides. Then we have:
2q - 29 + 29 + q = -2q + 26 + 2q - 29
This reduces to:
3q = -3
Now we are almost done. We want to know what q is, so if we divide both
sides of the equation by 3, we will get our answer:
3q/3 = -3/3
So, we have:
q = -1
You can always check your answer by plugging in -1 to the original
equation. So, check: is 29 - 1 equal to -2( -1 - 13)? Yes, it is, so we
did everything right.
Did that make sense to you? I guess the steps you should take are:
1) Simplify (distribute if necessary)
2) Put like terms with like terms (Put all the terms containing the unknown
on one side of the equals sign with the constants on the other side)
3) Divide or do what you must to simplify
I hope this helps. Maybe you could do the second example and write back
--Sydney
```
Associated Topics:
Middle School Equations
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# Lesson 6Taking SidesPractice Understanding
## Learning Focus
Understand similarities and differences in solving equations and inequalities.
Learn to avoid common errors and misunderstandings about inequalities.
What are some of the common misconceptions of inequalities?
How does having a deep understanding of what inequalities mean help to avoid errors?
Can an inequality have no solutions?
## Open Up the Math: Launch, Explore, Discuss
Joaquin and Serena work together productively in their math class. They both contribute their thinking and when they disagree, they both give their reasons and decide together who is right. In their math class right now, they are working on inequalities. Recently, they had a discussion that went something like this:
Joaquin: The problem says that “ less than a number is greater than .” I think that we should just follow the words and write: .
Serena: I don’t think that works because if is and you do less than that you get . I think we should write .
Joaquin: Oh, you’re right. Then it makes sense that the solution will be , which means we can choose any number greater than .
The situations below are a few more of the disagreements and questions that Joaquin and Serena have. Your job is to decide how to answer their questions, decide who is right, and give a mathematical explanation of your reasoning.
### 1.
Joaquin and Serena are assigned to graph the inequality .
Joaquin thinks the graph should have an open dot at .
Serena thinks the graph should have a closed dot at .
Explain who is correct and why.
### 2.
Joaquin and Serena are looking at the problem .
Serena says that the inequality is always true because multiplying a number by and then adding to it makes the number greater than .
Is she right? Explain why or why not.
### 3.
The word problem that Joaquin and Serena are working on says, “ greater than .
Joaquin says that they should write:
Serena says they should write:
Explain who is correct and why.
### 4.
Joaquin is thinking hard about equations and inequalities and comes up with this idea:
If , then .
So, if , then .
Is he right? Explain why or why not.
### 5.
Joaquin’s question in problem 4 made Serena think about other similarities and differences in equations and inequalities. Serena wonders about the equation and the inequality . Explain to Serena ways that solving these two problems are alike and ways that they are different. How are the solutions to the problems alike and different?
### 6.
#### a.
Joaquin solved by adding to each side of the inequality. Serena said that he was wrong. Who do you think is right and why?
#### b.
Joaquin’s solution was . He checked his work by substituting for in the original inequality. Does this prove that Joaquin is right? Explain why or why not.
#### c.
Joaquin is still skeptical and believes he is right. Find a number that satisfies his solution but does not satisfy the original inequality.
### 7.
Serena is checking her work with Joaquin and finds that they disagree on a problem. Here’s what Serena wrote:
Is she right? Explain why or why not.
### 8.
Joaquin and Serena are having trouble solving .
Explain how they should solve the inequality, showing all the necessary steps and identifying the properties you would use.
### 9.
Joaquin and Serena know that some equations are true for any value of the variable and some equations are never true, no matter what value is chosen for the variable. They are wondering about inequalities. What could you tell them about the following inequalities? Do they have solutions? What are they? How would you graph their solutions on a number line?
### 10.
The partners are given the literal inequality to solve for . Joaquin says that he will solve it just like an equation. Serena says that he needs to be careful because if is a negative number, the solution will be different. What do you say? What are the solutions for the inequality?
## Ready for More?
Use your reasoning skills to solve the following inequalities. Write your solutions in both interval and set notation and graph on the number line.
## Takeaways
Similarities in Solving Inequalities and Equations:
Differences in Solving Inequalities and Equations:
## Lesson Summary
In this lesson, we examined common mistakes and misconceptions about inequalities. We analyzed the similarities and differences in solving equations and inequalities, determining which properties applied to both and which properties were different for inequalities.
## Retrieval
### 1.
Graph each set of equations and find the point where they intersect.
Point:
Solve each literal equation for the indicated variable.
Solve for .
Solve for . |
# 6th Grade Math: Graphing Integers
Up until 6th grade, most students work with positive whole numbers. Starting in 6th grade, you'll learn about integers, which include negative numbers. You'll learn to locate positive and negative integers on a number line and also plot points on the coordinate plane.
## How to Find Integers and Points
In 6th grade, you'll use both number lines and coordinate planes for graphing. Number lines are used for locating integers, which include all of the negative and positive numbers as well as zero. Number lines usually have zero at the center and list the positive numbers to the right of zero and the negative numbers to the left, like this: …-3, -2, -1, 0, 1, 2, 3… Some number lines are written vertically, with the positive numbers counted upward from zero and the negative numbers counted downward. A thermometer is a good example, since it shows both positive and negative temperatures.
The integers on a number line are in order, and they're usually labeled to make them easy to find. For instance, negative 12 (-12) will be in between -13 and -11. You may also be asked to locate rational numbers, like decimals, on a number line. These are usually not labeled, but you can find them if you know which integers they fall in between. For instance, 15.5 would be in between 15 and 16.
### Graphing Points on the Coordinate Plane
A coordinate plane is composed of two perpendicular number lines. They cross at zero, with one positioned vertically and the other positioned horizontally. The vertical number line is called the y-axis, and the horizontal number line is the x-axis. Coordinate points are points on the graph that are represented by ordered pairs.
#### Ordered Pairs
An ordered pair is a simple set of directions to the coordinate point, and they're written like this: (2, 3). The first number in the ordered pair tells you where the coordinate point is in relation to the x-axis, and the second number describes the coordinate point's position relative to the y-axis.
#### Finding Coordinate Points
To find a coordinate point that is represented by an ordered pair, place your pencil at zero on your graph. Since the first number in the ordered pair represents the point's relation to the x-axis, move your pencil right or left on that axis to the point specified in the ordered pair. For instance, if the ordered pair is (2, 3), you would move your pencil to the right to positive two; if your ordered pair is (-4, -5), you would move it to the left to negative four.
Now, it's time to locate the coordinate point relative to the y-axis, which is the second number in your ordered pair. You start at whatever point the first number in the ordered pair specified, and the move up or down relative to the y-axis.
For the ordered pair (2, 3), you would move your pencil up from two on the x-axis until it was parallel with three on the y-axis. This is the location of the coordinate point (2, 3).
For (-4, -5), you would move your pencil down from negative four on the x-axis until it was parallel with negative five on the y-axis. This is where the point (-4, -5) is located.
### How to Practice
To practice graphing integers on number lines and coordinate points on the coordinate plane, it's helpful to have a pad of graph paper. You can draw number lines and coordinate planes on it, and then practice finding points. You can make up the points by drawing numbers out of a hat or by asking a friend or adult to make them up for you.
Did you find this useful? If so, please let others know!
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# Construct a triangle with sides 5 cm,
Question:
Construct a triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$ and then another triangle whose sides are$\frac{\mathbf{7}}{\mathbf{5}}$ of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
1. Construct a $\triangle \mathrm{ABC}$ such that $\mathrm{AB}=5 \mathrm{~cm}$,
$\mathrm{BC}=7 \mathrm{~cm}$ and $\mathrm{AC}=6 \mathrm{~cm}$
2. Draw a ray $\mathrm{BX}$ such that $\angle \mathrm{CBX}$ is an acute angle.
3. Mark 7 points $X_{1}, X_{2}, X_{3}, X_{4}, X_{5}, X_{6}$ and $X_{7}$ on $B X$ such that $B X_{1}=X_{1} X_{2}=X_{2} X_{3}=X_{3} X_{4}$ $=X_{4} X_{5}=X_{5} X_{6}=X_{6} X_{7}$
4. Join $X_{5}$ to $C$.
5. Draw a line through $X_{7}$ intersecting $B C$ (produced) at $C^{\prime}$ such that $X_{5} C \| X_{7} C^{\prime}$
6. Draw a line through $\mathrm{C}^{\prime}$ parallel to $\mathrm{CA}$ to intersect $\mathrm{BA}$ (produced) at $\mathrm{A}^{\prime}$.
Thus, $\triangle \mathrm{A}^{\prime} \mathrm{BC}^{\prime}$ is the required triangle. |
Take a picture of your homework and get answers is a mathematical tool that helps to solve math equations. We can solve math problems for you.
## The Best Take a picture of your homework and get answers
Take a picture of your homework and get answers can be found online or in mathematical textbooks. In order to solve a quadratic equation, we first of all need to understand what a quadratic equation is. This can be done by first reviewing the basic properties of a quadratic equation, such as: The solution is always a linear function It always contains at least one real root (a real number) At least one root must be negative (This is the only way that a cubic equation can have an absolute value solution.) If this is the case, then the solution will also be negative. It can be shown that if the function has two real roots, then it is always possible to find at least one absolute value solution. If there are more than 2 real roots, then there will always be at least one solution. This can be either positive or negative.
Elimination equations are one of the most common types of algebra problems. They involve solving an equation that has two variables in it (x and y). The goal of this type of problem is to determine which one of the two factors (x or y) can be eliminated from the equation. The elimination process involves moving the factor with the smaller value to the left side of the equation, while leaving the value of that factor on the right side. In math terms, you are subtracting from both sides of the equation (right side minus left side) to get a smaller value on one side. Since any factor with a smaller value will always cancel out with a larger value, only one variable needs to be eliminated in order to solve an elimination equation. This typeable is why elimination equations are so common in math. If you have two variables in an equation and only need one to be solved, then you can move that variable to the left side and eliminate it from further consideration. For example, if you have x = 5 and y = 10, then you could take away 5 from both sides of the equation and get x = 3 and y = 7. This would indicate that y could be eliminated from further consideration based on its smaller value -3 compared to 10. Once you know which factor can be eliminated from one side of the equation, you can substitute that value for one of
In right triangle ABC, angle BAC is the right angle. The length of the hypotenuse AC is equal to the sum of the lengths of the other two sides, so angle BAC is equal to 90 degrees. Because 90 degrees is a right angle, it means that angle BAC is a right angle. It follows that: To solve for angle in right triangle ,> you first determine the length of side AB>. Then you can use trigonometry to calculate AC>. This can be done using one of three methods: Trigonometry Method - The Trigonometry method is by far the easiest and most common way to determine angles in right triangle ,>. It involves only simple addition and subtraction formulas. For example, if we know that side AB> = 4 units long, then we can simply subtract 4 from both sides of our equation to get AC> = 6> units long. The Trigonometry method has many benefits including its ability to simplify calculations and provide more accurate results (especially in cases where exact values are critical). Measuring Tool Method - Another way to solve for angle in right triangle ,>, is by using a measuring tool. A measuring tool consists of a set of straight-edge rulers or protractor which can be used to measure angles on any object. There are many different measuring tools available
The sine function is used to find the angle between two lines. It takes the form of sin(x) where x is in radians, and is used to calculate the angle between two distinct lines, or theta. To solve for the angle, we use the cosine function (see below). The sine function can be used to find the values for other trigonometric functions as well as other angles. For example, if you know the value of one of these functions, you can use the sine function to determine the value of other trigonometric functions. This technique is known as triangulation. The following equation shows how this works: sin(A) = Acos(B) + Bsin(A) In this equation, sin(A) represents the value of one trigonometric function (e.g., tan, arc tangent), while A and B represent a pair of distinct lines (e.g., x-axis and y-axis). To solve for another trigonometric function in terms of sin(A), you simply plug in that value for sin(A). For example, if you know that tan(60°) = 1.5, you can use this equation to determine that 1.5 = cos(60°) + sin(60°). You can also use equations like this one to determine
Good app with lot of math problems. It helps me to solve all the sums which I can't understand too along with the steps. I like it 97%. That 3% is for some time it doesn't scan the sums properly and rarely it doesn't have a solution for problems which I expected. Thanks for giving such kind of app.
Flor Taylor
The app is great a very useful. 2 stars (the app is worth 5 but because of that now after editing you get 4) because of that stupid way you implemented the premium feature, I mean adding a premium mode to your app is perfectly okay if you add new features and then asking for money but taking something old and asking for money for that is complete laziness and lack of interest from the developers!
Xandra Bryant |
# How do you solve the following linear system: -3x + 2y = 1 , x + 1/2y = 7 ?
Apr 1, 2017
Find the x value using one off the equations.
$x + \frac{1}{2} y = 7$
$x = 7 - \frac{1}{2} y$
Plug the x value into the other equation.
$- 3 \left(7 - \frac{1}{2} y\right) + 2 y = 1$
Simplify
$- 21 + \frac{3}{2} y + 2 y = 1$
$\frac{3}{2} y + 2 y = 22$
$\frac{3}{2} y + \frac{4}{2} y = 22$
$\frac{7}{2} y = 22$
$y = \frac{22 \left(2\right)}{7}$
$y = \frac{44}{7}$
Plug in the y value to find the x value.
$x + \frac{1}{2} \left(\frac{44}{7}\right) = 7$
$x + \frac{22}{7} = 7$
$x = \frac{49}{7} - \frac{22}{7}$
$x = \frac{27}{7}$
ANSWER: $\left(\frac{27}{7} , \frac{44}{7}\right)$
Apr 1, 2017
$\text{answer:" x=27/7" , } y = \frac{44}{7}$
#### Explanation:
$- 3 x + 2 y = 1 \text{ } \left(1\right)$
$x + \frac{1}{2} y = 7 \text{ } \left(2\right)$
$\text{let us multiply both sides of equation (2) by 4}$
$\textcolor{red}{4} \cdot \left(x + \frac{1}{2} y\right) = \textcolor{red}{4} \cdot 7$
$\text{We get ;}$
$4 x + 2 y = 28 \text{ } \left(3\right)$
$\text{the coefficients of term 'y' in both equation (1) and (3) is equal.}$
$\text{let subtract (3) from (1) to eliminate 'y'}$
$- 3 x + 2 y - \left(4 x + 2 y\right) = 1 - 28$
$- 3 x + 2 y - 4 x - 2 y = - 27$
$- 3 x + \cancel{2 y} - 4 x - \cancel{2 y} = - 27$
$- 7 x = - 27$
$x = \frac{27}{7}$
$\text{now ,let us use (1) or (2)}$
$x + \frac{1}{2} y = 7$
$\frac{27}{7} + \frac{y}{2} = 7$
$\frac{y}{2} = 7 - \frac{27}{7}$
$\frac{y}{2} = \frac{49 - 27}{7}$
$\frac{y}{2} = \frac{22}{7}$
$y = \frac{44}{7}$ |
1.2 Prime factorization
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. This chapter contains many examples of arithmetic techniques that are used directly or indirectly in algebra. Since the chapter is intended as a review, the problem-solving techniques are presented without being developed. Therefore, no work space is provided, nor does the chapter contain all of the pedagogical features of the text. As a review, this chapter can be assigned at the discretion of the instructor and can also be a valuable reference tool for the student.
Overview
• Prime And Composite Numbers
• The Fundamental Principle Of Arithmetic
• The Prime Factorization Of A Whole Number
Prime and composite numbers
Notice that the only factors of 7 are 1 and 7 itself, and that the only factors of 23 are 1 and 23 itself.
Prime number
A whole number greater than 1 whose only whole number factors are itself and 1 is called a prime number.
The first seven prime numbers are
2, 3, 5, 7, 11, 13, and 17
The number 1 is not considered to be a prime number, and the number 2 is the first and only even prime number.
Many numbers have factors other than themselves and 1. For example, the factors of 28 are 1, 2, 4, 7, 14, and 28 (since each of these whole numbers and only these whole numbers divide into 28 without a remainder).
Composite numbers
A whole number that is composed of factors other than itself and 1 is called a composite number. Composite numbers are not prime numbers.
Some composite numbers are 4, 6, 8, 10, 12, and 15.
The fundamental principle of arithmetic
Prime numbers are very important in the study of mathematics. We will use them soon in our study of fractions. We will now, however, be introduced to an important mathematical principle.
The fundamental principle of arithmetic
Except for the order of the factors, every whole number, other than 1, can be factored in one and only one way as a product of prime numbers.
Prime factorization
When a number is factored so that all its factors are prime numbers, the factorization is called the prime factorization of the number.
Sample set a
Find the prime factorization of 10.
$10=2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5$
Both 2 and 5 are prime numbers. Thus, $2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5$ is the prime factorization of 10.
Find the prime factorization of 60.
$\begin{array}{lllll}60\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}30\hfill & \hfill & \text{30\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}prime}.\text{\hspace{0.17em}}30=2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15\hfill \\ \hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15\hfill & \hfill & 15\text{\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}prime}.\text{\hspace{0.17em}}15=3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill \\ \hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & \hfill & \text{We'll\hspace{0.17em}use\hspace{0.17em}exponents}\text{.\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2={2}^{2}\text{\hspace{0.17em}}\hfill \\ \hfill & =\hfill & {2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & \hfill & \hfill \end{array}$
The numbers 2, 3, and 5 are all primes. Thus, $2^{2}·3·5$ is the prime factorization of 60.
Find the prime factorization of 11.
11 is a prime number. Prime factorization applies only to composite numbers.
The prime factorization of a whole number
The following method provides a way of finding the prime factorization of a whole number. The examples that follow will use the method and make it more clear.
1. Divide the number repeatedly by the smallest prime number that will divide into the number without a remainder.
2. When the prime number used in step 1 no longer divides into the given number without a remainder, repeat the process with the next largest prime number.
3. Continue this process until the quotient is 1.
4. The prime factorization of the given number is the product of all these prime divisors.
Sample set b
Find the prime factorization of 60.
Since 60 is an even number, it is divisible by 2. We will repeatedly divide by 2 until we no longer can (when we start getting a remainder). We shall divide in the following way.
$\begin{array}{l}\text{30\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}2\hspace{0.17em}again}\text{.}\hfill \\ \text{15\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}2,\hspace{0.17em}but\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3,\hspace{0.17em}the\hspace{0.17em}next\hspace{0.17em}largest\hspace{0.17em}prime}\text{.}\hfill \\ \text{5\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3,\hspace{0.17em}but\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}5,\hspace{0.17em}the\hspace{0.17em}next\hspace{0.17em}largest\hspace{0.17em}prime}\text{.}\hfill \\ \text{The\hspace{0.17em}quotient\hspace{0.17em}is\hspace{0.17em}1\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}stop\hspace{0.17em}the\hspace{0.17em}division\hspace{0.17em}process}\text{.}\hfill \end{array}$
The prime factorization of 60 is the product of all these divisors.
$\begin{array}{lllll}60\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & \hfill & \text{We\hspace{0.17em}will\hspace{0.17em}use\hspace{0.17em}exponents\hspace{0.17em}when\hspace{0.17em}possible}.\hfill \\ 60\hfill & =\hfill & {2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & \hfill & \hfill \end{array}$
Find the prime factorization of 441.
Since 441 is an odd number, it is not divisible by 2. We’ll try 3, the next largest prime.
$\begin{array}{l}\text{147\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3}.\hfill \\ \text{49\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3\hspace{0.17em}nor\hspace{0.17em}by\hspace{0.17em}5},\text{\hspace{0.17em}but\hspace{0.17em}by\hspace{0.17em}7}.\hfill \\ \text{7\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}7}\text{.}\hfill \\ \text{The\hspace{0.17em}quotient\hspace{0.17em}is\hspace{0.17em}1\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}stop\hspace{0.17em}the\hspace{0.17em}division\hspace{0.17em}process}.\hfill \end{array}$
The prime factorization of 441 is the product of all the divisors.
$\begin{array}{lllll}441\hfill & =\hfill & 3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\hfill & \hfill & \text{We\hspace{0.17em}will\hspace{0.17em}use\hspace{0.17em}exponents\hspace{0.17em}when\hspace{0.17em}possible}.\hfill \\ 441\hfill & =\hfill & {3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{7}^{2}\hfill & \hfill & \hfill \end{array}$
Exercises
For the following problems, determine which whole numbers are prime and which are composite.
23
prime
25
27
composite
2
3
prime
5
7
prime
9
11
prime
34
55
composite
63
1044
composite
339
209
composite
For the following problems, find the prime factorization of each whole number. Use exponents on repeated factors.
26
38
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}19$
54
62
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}31$
56
176
${2}^{4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}11$
480
819
${3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}13$
2025
148,225
${5}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{7}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{11}^{2}$
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
Do somebody tell me a best nano engineering book for beginners?
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
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
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Difference between revisions of "2018 AMC 10A Problems/Problem 12"
How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? $$x+3y=3$$ $$\big||x|-|y|\big|=1$$ $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
## Solutions
### Solution 1
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this: $[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy]$ Now, it becomes clear that there are $\boxed{\textbf{(C) } 3}$ intersection points. (pinetree1)
### Solution 2
$x+3y=3$ can be rewritten to $x=3-3y$. Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$. Splitting this question into casework for the ranges of $y$ will give us the total number of solutions.
$\textbf{Case 1:}$ $y>1$: $3-3y$ will be negative so $|3-3y| = 3y-3.$ $|3y-3-y| = |2y-3| = 1$
Subcase 1: $y>\frac{3}{2}$
$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$
Subcase 2: $1
$2y-3$ is negative so $|2y-3| = 3-2y = 1$. $2y = 2$ and so there are no solutions ($y$ can't equal to $1$)
$\textbf{Case 2:}$ $y = 1$: It is fairly clear that $x = 0.$
$\textbf{Case 3:}$ $y<1$: $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$
Subcase 1: $y>\frac{4}{3}$
$3-4y$ will be negative so $4y-3 = 1$ $\rightarrow$ $4y = 4$. There are no solutions (again, $y$ can't equal to $1$)
Subcase 2: $y<\frac{4}{3}$
$3-4y$ will be positive so $3-4y = 1$ $\rightarrow$ $4y = 2$. $y = \frac{1}{2}$ and $x = \frac{3}{2}$. Thus, the solutions are: $(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)$, and the answer is $\boxed{\textbf{(C) } 3}$. $\text{\LaTeX}$ edit by pretzel, very minor $\text{\LaTeX}$ edits by Bryanli, very very minor $\text{\LaTeX}$ edit by ssb02
### Solution 3
Note that $||x| - |y||$ can take on either of four values: $x + y$, $x - y$, $-x + y$, $-x -y$. Solving the equations (by elimination, either adding the two equations or subtracting), we obtain the three solutions: $(0, 1)$, $(-3,2)$, $(1.5, 0.5)$ so the answer is $\boxed{\textbf{(C) } 3}$
~trumpeter, ccx09 |
## Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.3 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3
Question 1.
In each of the following examples verify that the given expression is a solution of the corresponding differential equation.
(i) xy = log y + c; $$\frac{d y}{d x}=\frac{y^{2}}{1-x y}$$
Solution:
xy = log y + c
Differentiating w.r.t. x, we get
Hence, xy = log y + c is a solution of the D.E.
$$\frac{d y}{d x}=\frac{y^{2}}{1-x y^{\prime}}, x y \neq 1$$
(ii) y = (sin-1x)2 + c; (1 – x2) $$\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2$$
Solution:
y = (sin-1 x)2 + c …….(1)
Differentiating w.r.t. x, we get
Differentiating again w.r.t. x, we get
(iii) y = e-x + Ax + B; $$e^{x} \frac{d^{2} y}{d x^{2}}=1$$
Solution:
y = e-x + Ax + B
Differentiating w.r.t. x, we get
∴ $$e^{x} \frac{d^{2} y}{d x^{2}}=1$$
Hence, y = e-x + Ax + B is a solution of the D.E.
$$e^{x} \frac{d^{2} y}{d x^{2}}=1$$
(iv) y = xm; $$x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0$$
Solution:
y = xm
Differentiating twice w.r.t. x, we get
This shows that y = xm is a solution of the D.E.
$$x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0$$
(v) y = a + $$\frac{b}{x}$$; $$x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0$$
Solution:
y = a + $$\frac{b}{x}$$
Differentiating w.r.t. x, we get
Differentiating again w.r.t. x, we get
Hence, y = a + $$\frac{b}{x}$$ is a solution of the D.E.
$$x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0$$
(vi) y = eax; x $$\frac{d y}{d x}$$ = y log y
Solution:
y = eax
log y = log eax = ax log e
log y = ax …….(1) ……..[∵ log e = 1]
Differentiating w.r.t. x, we get
$$\frac{1}{y} \cdot \frac{d y}{d x}$$ = a × 1
∴ $$\frac{d y}{d x}$$ = ay
∴ x $$\frac{d y}{d x}$$ = (ax)y
∴ x $$\frac{d y}{d x}$$ = y log y ………[By (1)]
Hence, y = eax is a solution of the D.E.
x $$\frac{d y}{d x}$$ = y log y.
Question 2.
Solve the following differential equations.
(i) $$\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$$
Solution:
(ii) log($$\frac{d y}{d x}$$) = 2x + 3y
Solution:
(iii) y – x $$\frac{d y}{d x}$$ = 0
Solution:
y – x $$\frac{d y}{d x}$$ = 0
∴ x $$\frac{d y}{d x}$$ = y
∴ $$\frac{1}{x} d x=\frac{1}{y} d y$$
Integrating both sides, we get
$$\int \frac{1}{x} d x=\int \frac{1}{y} d y$$
∴ log |x| = log |y| + log c
∴ log |x| = log |cy|
∴ x = cy
This is the general solution.
(iv) sec2x . tan y dx + sec2y . tan x dy = 0
Solution:
sec2x . tan y dx + sec2y . tan x dy = 0
∴ $$\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0$$
Integrating both sides, we get
$$\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y=c_{1}$$
Each of these integrals is of the type
$$\int \frac{f^{\prime}(x)}{f(x)} d x$$ = log |f(x)| + c
∴ the general solution is
∴ log|tan x| + log|tan y | = log c, where c1 = log c
∴ log |tan x . tan y| = log c
∴ tan x . tan y = c
This is the general solution.
(v) cos x . cos y dy – sin x . sin y dx = 0
Solution:
cos x . cos y dy – sin x . sin y dx = 0
$$\frac{\cos y}{\sin y} d y-\frac{\sin x}{\cos x} d x=0$$
Integrating both sides, we get
∫cot y dy – ∫tan x dx = c1
∴ log|sin y| – [-log|cos x|] = log c, where c1 = log c
∴ log |sin y| + log|cos x| = log c
∴ log|sin y . cos x| = log c
∴ sin y . cos x = c
This is the general solution.
(vi) $$\frac{d y}{d x}$$ = -k, where k is a constant.
Solution:
$$\frac{d y}{d x}$$ = -k
∴ dy = -k dx
Integrating both sides, we get
∫dy = -k∫dx
∴ y = -kx + c
This is the general solution.
(vii) $$\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0$$
Solution:
$$\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0$$
∴ y cos2y dy + x cos2x dx = 0
∴ $$x\left(\frac{1+\cos 2 x}{2}\right) d x+y\left(1+\frac{\cos 2 y}{2}\right) d y=0$$
∴ x(1 + cos 2x) dx + y(1 + cos 2y) dy = 0
∴ x dx + x cos 2x dx + y dy+ y cos 2y dy = 0
Integrating both sides, we get
∫x dx + ∫y dy + ∫x cos 2x dx + ∫y cos 2y dy = c1 ……..(1)
Using integration by parts
Multiplying throughout by 4, this becomes
2x2 + 2y2 + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c1
∴ 2(x2 + y2) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0, where c = -4c1
This is the general solution.
(viii) $$y^{3}-\frac{d y}{d x}=x^{2} \frac{d y}{d x}$$
Solution:
(ix) 2ex+2y dx – 3 dy = 0
Solution:
(x) $$\frac{d y}{d x}$$ = ex+y + x2 ey
Solution:
∴ 3ex + 3e-y + x3 = -3c1
∴ 3ex + 3e-y + x3 = c, where c = -3c1
This is the general solution.
Question 3.
For each of the following differential equations, find the particular solution satisfying the given condition:
(i) 3ex tan y dx + (1 + ex) sec2y dy = 0, when x = 0, y = π
Solution:
3ex tan y dx + (1 + ex) sec2y dy = 0
(ii) (x – y2x) dx – (y + x2y) dy = 0, when x = 2, y = 0
Solution:
(x – y2x) dx – (y + x2y) dy = 0
∴ x(1 – y2) dx – y(1 + x2) dy = 0
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x2)(1 – y2) = 5.
(iii) y(1 + log x) $$\frac{d x}{d y}$$ – x log x = 0, y = e2, when x = e
Solution:
y(1 + log x) $$\frac{d x}{d y}$$ – x log x = 0
(iv) (ey + 1) cos x + ey sin x $$\frac{d y}{d x}$$ = 0, when x = $$\frac{\pi}{6}$$, y = 0
Solution:
(ey + 1) cos x + ey sin x $$\frac{d y}{d x}$$ = 0
$$\int \frac{f^{\prime}(x)}{f(x)} d x$$ = log|f(x)| + c
∴ from (1), the general solution is
log|sin x| + log|ey + 1| = log c, where c1 = log c
∴ log|sin x . (ey + 1)| = log c
∴ sin x . (ey + 1) = c
When x = $$\frac{\pi}{4}$$, y = 0, we get
$$\left(\sin \frac{\pi}{4}\right)\left(e^{0}+1\right)=c$$
∴ c = $$\frac{1}{\sqrt{2}}$$(1 + 1) = √2
∴ the particular solution is sin x . (ey + 1) = √2
(v) (x + 1) $$\frac{d y}{d x}$$ – 1 = 2e-y, y = 0, when x = 1
Solution:
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e0 = c(1 + 1)
∴ 3 = 2c
∴ c = $$\frac{3}{2}$$
∴ the particular solution is 2 + ey = $$\frac{3}{2}$$ (x + 1)
∴ 2(2 + ey) = 3(x + 1).
(vi) cos($$\frac{d y}{d x}$$) = a, a ∈ R, y (0) = 2
Solution:
cos($$\frac{d y}{d x}$$) = a
∴ $$\frac{d y}{d x}$$ = cos-1 a
∴ dy = (cos-1 a) dx
Integrating both sides, we get
∫dy = (cos-1 a) ∫dx
∴ y = (cos-1 a) x + c
∴ y = x cos-1 a + c
This is the general solution.
Now, y(0) = 2, i.e. y = 2,
when x = 0, 2 = 0 + c
∴ c = 2
∴ the particular solution is
∴ y = x cos-1 a + 2
∴ y – 2 = x cos-1 a
∴ $$\frac{y-2}{x}$$ = cos-1a
∴ cos($$\frac{y-2}{x}$$) = a
Question 4.
Reduce each of the following differential equations to the variable separable form and hence solve:
(i) $$\frac{d y}{d x}$$ = cos(x + y)
Solution:
(ii) (x – y)2 $$\frac{d y}{d x}$$ = a2
Solution:
(iii) x + y $$\frac{d y}{d x}$$ = sec(x2 + y2)
Solution:
Integrating both sides, we get
∫cos u du = 2 ∫dx
∴ sin u = 2x + c
∴ sin(x2 + y2) = 2x + c
This is the general solution.
(iv) cos2(x – 2y) = 1 – 2 $$\frac{d y}{d x}$$
Solution:
Integrating both sides, we get
∫dx = ∫sec2u du
∴ x = tan u + c
∴ x = tan(x – 2y) + c
This is the general solution.
(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1
Solution:
(2x – 2y + 3) dx – (x – y + 1) dy = 0
∴ (x – y + 1) dy = (2x – 2y + 3) dx
∴ $$\frac{d y}{d x}=\frac{2(x-y)+3}{(x-y)+1}$$ ………(1)
Put x – y = u, Then $$1-\frac{d y}{d x}=\frac{d u}{d x}$$
∴ u – log|u + 2| = -x + c
∴ x – y – log|x – y + 2| = -x + c
∴ (2x – y) – log|x – y + 2| = c
This is the general solution.
Now, y = 1, when x = 0.
∴ (0 – 1) – log|0 – 1 + 2| = c
∴ -1 – o = c
∴ c = -1
∴ the particular solution is
(2x – y) – log|x – y + 2| = -1
∴ (2x – y) – log|x – y + 2| + 1 = 0
## Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.2 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2
Question 1.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) x3 + y3 = 4ax
Solution:
x3 + y3 = 4ax ……..(1)
Differentiating both sides w.r.t. x, we get
3x2 + 3y2 $$\frac{d y}{d x}$$ = 4a × 1
∴ 3x2 + 3y2 $$\frac{d y}{d x}$$ = 4a
Substituting the value of 4a in (1), we get
x3 + y3 = (3x2 + 3y2 $$\frac{d y}{d x}$$) x
∴ x3 + y3 = 3x3 + 3xy2 $$\frac{d y}{d x}$$
∴ 2x3 + 3xy2 $$\frac{d y}{d x}$$ – y3 = 0
This is the required D.E.
(ii) Ax2 + By2 = 1
Solution:
Ax2 + By2 = 1
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y $$\frac{d y}{d x}$$ = 0
∴ Ax + By $$\frac{d y}{d x}$$ = 0 ……..(1)
Differentiating again w.r.t. x, we get
Substituting the value of A in (1), we get
This is the required D.E.
Alternative Method:
Ax2 + By2 = 1 ……..(1)
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y $$\frac{d y}{d x}$$ = 0
∴ Ax + By $$\frac{d y}{d x}$$ = 0 ……….(2)
Differentiating again w.r.t. x, we get,
The equations (1), (2) and (3) are consistent in A and B.
∴ determinant of their consistency is zero.
This is the required D.E.
(iii) y = A cos(log x) + B sin(log x)
Solution:
y = A cos(log x) + B sin (log x) ……. (1)
Differentiating w.r.t. x, we get
(iv) y2 = (x + c)3
Solution:
y2 = (x + c)3
Differentiating w.r.t. x, we get
This is the required D.E.
(v) y = Ae5x + Be-5x
Solution:
y = Ae5x + Be-5x ……….(1)
Differentiating twice w.r.t. x, we get
This is the required D.E.
(vi) (y – a)2 = 4(x – b)
Solution:
(y – a)2 = 4(x – b)
Differentiating both sides w.r.t. x, we get
2(y – a) . $$\frac{d}{d x}$$(y – a) = 4 $$\frac{d}{d x}$$(x – b)
∴ 2(y – a) . ($$\frac{d y}{d x}$$ – 0) = 4(1 – 0)
∴ 2(y – a) $$\frac{d y}{d x}$$ = 4
∴ (y – a) $$\frac{d y}{d x}$$ = 2 ……..(1)
Differentiating w.r.t. x, we get
This is the required D.E.
(vii) y = a + $$\frac{a}{x}$$
Solution:
y = a + $$\frac{a}{x}$$
Differentiating w.r.t. x, we get
Substituting the value of a in (1), we get
This is the required D.E.
(viii) y = c1e2x + c2e5x
Solution:
y = c1e2x + c2e5x ………(1)
Differentiating twice w.r.t. x, we get
$$\frac{d y}{d x}$$ = c1e2x × 2 + c2e5x × 5
The equations (1), (2) and (3) are consistent in c1e2x and c2e5x
∴ determinant of their consistency is zero.
This is the required D.E.
Alternative Method:
y = c1e2x + c2e5x
Dividing both sides by e5x, we get
This is the required D.E.
(ix) c1x3 + c2y2 = 5.
Solution:
c1x3 + c2y2 = 5 ……….(1)
Differentiating w.r.t. x, we get
Differentiating again w.r.t. x, we get
The equations (1), (2) and (3) in c1, c2 are consistent.
∴ determinant of their consistency is zero.
This is the required D.E.
(x) y = e-2x(A cos x + B sin x)
Solution:
y = e-2x(A cos x + B sin x)
∴ e2x . y = A cos x + B sin x ………(1)
Differentiating w.r.t. x, we get
Differentiating again w.r.t. x, we get
This is the required D.E.
Question 2.
Form the differential equation of family of lines having intercepts a and b on the coordinate axes respectively.
Solution:
The equation of the line having intercepts a and b on the coordinate axes respectively, is
$$\frac{x}{a}+\frac{y}{b}=1$$ ……….(1)
where a and b are arbitrary constants.
[For different values of a and b, we get, different lines. Hence (1) is the equation of family of lines.]
Differentiating (1) w.r.t. x, we get
Differentiating again w.r.t. x, we get $$\frac{d^{2} y}{d x^{2}}=0$$
This is the required D.E.
Question 3.
Find the differential equation all parabolas having length of latus rectum 4a and axis is parallel to the X-axis.
Solution:
Let A(h, k) be the vertex of the parabola whose length of latus rectum is 4a.
Then the equation of the parabola is (y – k)2 = 4a (x – h), where h and k are arbitrary constants.
Differentiating w.r.t. x, we get
Differentiating again w.r.t. x, we get
This is the required D.E.
Question 4.
Find the differential equation of the ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$
i.e., $$\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1$$
∴ $$\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1$$
∴ x2 + 4y2 = 4b2
Differentiating w.r.t. x, we get
2x + 4 × 2y $$\frac{d y}{d x}$$ = 0
∴ x + 4y $$\frac{d y}{d x}$$ = 0
This is the required D.E.
Question 5.
Form the differential equation of family of lines parallel to the line 2x + 3y + 4 = 0.
Solution:
The equation of the line parallel to the line 2x + 3y + 4 = 0 is 2x + 3y + c = 0, where c is an arbitrary constant.
Differentiating w.r.t. x, we get
2 × 1 + 3 $$\frac{d y}{d x}$$ + 0 = 0
∴ 3 $$\frac{d y}{d x}$$ + 2 = 0
This is the required D.E.
Question 6.
Find the differential equation of all circles having radius 9 and centre at point (h, k).
Solution:
Equation of the circle having radius 9 and centre at point (h, k) is
(x – h)2 + (y – k)2 = 81 …… (1)
where h and k are arbitrary constant.
Differentiating (1) w.r.t. x, we get
Differentiating again w.r.t. x, we get
From (2), x – h = -(y – k) $$\frac{d y}{d x}$$
Substituting the value of (x – h) in (1), we get
This is the required D.E.
Question 7.
Form the differential equation of all parabolas whose axis is the X-axis.
Solution:
The equation of the parbola whose axis is the X-axis is
y2 = 4a(x – h) …… (1)
where a and h are arbitrary constants.
Differentiating (1) w.r.t. x, we get
2y $$\frac{d y}{d x}$$ = 4a(1 – 0)
∴ y $$\frac{d y}{d x}$$ = 2a
Differentiating again w.r.t. x, we get
This is the required D.E.
## Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.1 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1
1. Determine the order and degree of each of the following differential equations:
Question (i).
$$\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x$$
Solution:
The given D.E. is $$\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 1.
∴ the given D.E. is of order 2 and degree 1.
Question (ii).
$$\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}$$
Solution:
The given D.E. is $$\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}$$
On cubing both sides, we get
$$1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 3.
∴ the given D.E. is of order 2 and degree 3.
Question (iii).
$$\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}$$
Solution:
The given D.E. is $$\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}$$
∴ $$\left(\frac{d y}{d x}\right)^{2}$$ = 2 sin x + 3
This D.E. has highest order derivative $$\frac{d y}{d x}$$ with power 2.
∴ the given D.E. is of order 1 and degree 2.
Question (iv).
$$\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}$$
Solution:
The given D.E. is $$\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}$$
On squaring both sides, we get
$$\left(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x\right)^{2}=1+\frac{d^{3} y}{d x^{3}}$$
This D.E. has highest order derivative $$\frac{d^{3} y}{d x^{3}}$$ with power 1.
∴ the given D.E. has order 3 and degree 1.
Question (v).
$$\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0$$
Solution:
The given D.E. is $$\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 1.
∴ the given D.E. has order 2 and degree 1.
Question (vi).
(y”‘)2 + 3y” + 3xy’ + 5y = 0
Solution:
The given D.E. is (y”‘)2 + 3y” + 3xy’ + 5y = 0
This can be written as:
$$\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+3 \frac{d^{2} y}{d x^{2}}+3 x \frac{d y}{d x}+5 y=0$$
This D.E. has highest order derivative $$\frac{d^{3} y}{d x^{3}}$$ with power 2.
∴ The given D.E. has order 3 and degree 2.
Question (vii).
$$\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$$
Solution:
The given D.E. is $$\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$
∴ order = 2
Since this D.E. cannot be expressed as a polynomial in differential coefficients, the degree is not defined.
Question (viii).
$$\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}$$
Solution:
The given D.E. is $$\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}$$
On squaring both sides, we get
$$\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=8^{2} \cdot\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 2.
∴ the given D.E. has order 2 and degree 2.
Question (ix).
$$\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20$$
Solution:
The given D.E. is $$\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20$$
∴ $$\left(\frac{d^{3} y}{d x^{3}}\right)^{3} \cdot\left(\frac{d y}{d x}\right)^{2}=20^{6}$$
This D.E. has highest order derivative $$\frac{d^{3} y}{d x^{3}}$$ with power 3.
∴ the given D.E. has order 3 and degree 3.
Question (x).
$$x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}$$
Solution:
The given D.E. is $$x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}$$
On squaring both sides, we get
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 1.
∴ the given D.E. has order 2 and degree 1.
## Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5
I. Choose the correct option from the given alternatives:
Question 1.
The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by
(a) 12 sq units
(b) 8 sq units
(c) 25 sq units
(d) 32 sq units
(a) 12 sq units
Question 2.
The area of the region enclosed by the curve y = $$\frac{1}{x}$$, and the lines x = e, x = e2 is given by
(a) 1 sq unit
(b) $$\frac{1}{2}$$ sq units
(c) $$\frac{3}{2}$$ sq units
(d) $$\frac{5}{2}$$ sq units
(a) 1 sq unit
Question 3.
The area bounded by the curve y = x3, the X-axis and the lines x = -2 and x = 1 is
(a) -9 sq units
(b) $$-\frac{15}{4}$$ sq units
(c) $$\frac{15}{4}$$ sq units
(d) $$\frac{17}{4}$$ sq units
(c) $$\frac{15}{4}$$ sq units
Question 4.
The area enclosed between the parabola y2 = 4x and line y = 2x is
(a) $$\frac{2}{3}$$ sq units
(b) $$\frac{1}{3}$$ sq units
(c) $$\frac{1}{4}$$ sq units
(d) $$\frac{3}{4}$$ sq units
(b) $$\frac{1}{3}$$ sq units
Question 5.
The area of the region bounded between the line x = 4 and the parabola y2 = 16x is
(a) $$\frac{128}{3}$$ sq units
(b) $$\frac{108}{3}$$ sq units
(c) $$\frac{118}{3}$$ sq units
(d) $$\frac{218}{3}$$ sq units
(a) $$\frac{128}{3}$$ sq units
Question 6.
The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is
(a) 1 sq unit
(b) 2 sq units
(c) 3 sq units
(d) 4 sq units
(d) 4 sq units
Question 7.
The area bounded by the parabola y2 = 8x, the X-axis and the latus rectum is
(a) $$\frac{31}{3}$$ sq units
(b) $$\frac{32}{3}$$ sq units
(c) $$\frac{32 \sqrt{2}}{3}$$ sq units
(d) $$\frac{16}{3}$$ sq units
(b) $$\frac{32}{3}$$ sq units
Question 8.
The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is
(a) 4 sq units
(b) $$\frac{3}{4}$$ sq units
(c) $$\frac{2}{3}$$ sq units
(d) $$\frac{4}{3}$$ sq units
(d) $$\frac{4}{3}$$ sq units
Question 9.
The area of the circle x2 + y2 = 25 in first quadrant is
(a) $$\frac{25 \pi}{3}$$ sq units
(b) 5π sq units
(c) 5 sq units
(d) 3 sq units
(a) $$\frac{25 \pi}{3}$$ sq units
Question 10.
The area of the region bounded by the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ is
(a) ab sq units
(b) πab sq units
(c) $$\frac{\pi}{a b}$$ sq units ab
(d) πa2 sq units
(b) πab sq units
Question 11.
The area bounded by the parabola y2 = x and the line 2y = x is
(a) $$\frac{4}{3}$$ sq units
(b) 1 sq unit
(c) $$\frac{2}{3}$$ sq unit
(d) $$\frac{1}{3}$$ sq unit
(a) $$\frac{4}{3}$$ sq units
Question 12.
The area enclosed between the curve y = cos 3x, 0 ≤ x ≤ $$\frac{\pi}{6}$$ and the X-axis is
(a) $$\frac{1}{2}$$ sq unit
(b) 1 sq unit
(c) $$\frac{2}{3}$$ sq unit
(d) $$\frac{1}{3}$$ sq unit
(d) $$\frac{1}{3}$$ sq unit
Question 13.
The area bounded by y = √x and line x = 2y + 3, X-axis in first quadrant is
(a) 2√3 sq units
(b) 9 sq units
(c) $$\frac{34}{3}$$ sq units
(d) 18 sq units
(b) 9 sq units
Question 14.
The area bounded by the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ and the line $$\frac{x}{a}+\frac{y}{b}=1$$ is
(a) (πab – 2ab) sq units
(b) $$\frac{\pi a b}{4}-\frac{a b}{2}$$ sq units
(c) (πab – ab) sq units
(d) πab sq units
(b) $$\frac{\pi a b}{4}-\frac{a b}{2}$$ sq units
Question 15.
The area bounded by the parabola y = x2 and the line y = x is
(a) $$\frac{1}{2}$$ sq unit
(b) $$\frac{1}{3}$$ sq unit
(c) $$\frac{1}{6}$$ sq unit
(d) $$\frac{1}{12}$$ sq unit
(c) $$\frac{1}{6}$$ sq unit
Question 16.
The area enclosed between the two parabolas y2 = 4x and y = x is
(a) $$\frac{8}{3}$$ sq units
(b) $$\frac{32}{3}$$ sq units
(c) $$\frac{16}{3}$$ sq units
(d) $$\frac{4}{3}$$ sq units
(c) $$\frac{16}{3}$$ sq units
Question 17.
The area bounded by the curve y = tan x, X-axis and the line x = $$\frac{\pi}{4}$$ is
(a) $$\frac{1}{3}$$ log 2 sq units
(b) log 2 sq units
(c) 2 log 2 sq units
(d) 3 log 2 sq units
(a) $$\frac{1}{3}$$ log 2 sq units
Question 18.
The area of the region bounded by x2 = 16y, y = 1, y = 4 and x = 0 in the first quadrant, is
(a) $$\frac{7}{3}$$ sq units
(b) $$\frac{8}{3}$$ sq units
(c) $$\frac{64}{3}$$ sq units
(d) $$\frac{56}{3}$$ sq units
(d) $$\frac{56}{3}$$ sq units
Question 19.
The area of the region included between the parabolas y2 = 4ax and x2 = 4ay, (a > 0) is given by
(a) $$\frac{16 a^{2}}{3}$$ sq units
(b) $$\frac{8 a^{2}}{3}$$ sq units
(c) $$\frac{4 a^{2}}{3}$$ sq units
(d) $$\frac{32 a^{2}}{3}$$ sq units
(a) $$\frac{16 a^{2}}{3}$$ sq units
Question 20.
The area of the region included between the line x + y = 1 and the circle x2 + y2 = 1 is
(a) $$\frac{\pi}{2}-1$$ sq units
(b) π – 2 sq units
(c) $$\frac{\pi}{4}-\frac{1}{2}$$ sq units
(d) π – $$\frac{1}{2}$$ sq units
(c) $$\frac{\pi}{4}-\frac{1}{2}$$ sq units
(II) Solve the following:
Question 1.
Find the area of the region bounded by the following curve, the X-axis and the given lines:
(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
(ii) y = sin x, x = 0, x = π
(iii) y = sin x, x = 0, x = $$\frac{\pi}{3}$$
Solution:
(i) Required area = $$\int_{0}^{5} y d x$$, where y = 2
= $$\int_{0}^{5} 2 d x$$
= $$[2 x]_{0}^{5}$$
= 2 × 5 – 0
= 10 sq units.
(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.
Two bounded regions A1 and A2 are obtained. Both the regions have equal areas.
∴ required area = A1 + A2 = 2A1
(iii) Required area = $$\int_{0}^{\pi / 3} y d x$$, where y = sin x
Question 2.
Find the area of the circle x2 + y2 = 9, using integration.
Solution:
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 3.
From the equation of the circle, y2 = 9 – x2.
In the first quadrant, y > 0
∴ y = $$\sqrt{9-x^{2}}$$
∴ area of the circle = 4 (area of the region OABO)
Question 3.
Find the area of the ellipse $$\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$$ using integration.
Solution:
By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the ellipse
$$\frac{y^{2}}{16}=1-\frac{x^{2}}{25}=\frac{25-x^{2}}{25}$$
∴ y2 = $$\frac{16}{25}$$ (25 – x2)
In the first quadrant y > 0
∴ y = $$\frac{4}{5} \sqrt{25-x^{2}}$$
∴ area of the ellipse = 4(area of the region OABO)
Question 4.
Find the area of the region lying between the parabolas:
(i) y2 = 4x and x2 = 4y
(ii) 4y2 = 9x and 3x2 = 16y
(iii) y2 = x and x2 = y.
Solution:
(i)
For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations.
From the equation x2 = 4y, y = $$\frac{x^{2}}{4}$$
y = $$\frac{x^{4}}{16}$$
$$\frac{x^{4}}{16}$$ = 4x
∴ x4 – 64x = 0
∴ x(x3 – 64) = 0
∴ x = 0 or x3 = 64 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = $$\frac{4^{2}}{4}$$ = 4
∴ the points of intersection are 0(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x, i.e. y = 2√x between x = 0 and x = 4
(ii)
For finding the points of intersection of the two parabolas, we equate the values of 4y2 from their equations.
From the equation 3x2 = 16y, y = $$\frac{3 x^{2}}{16}$$
∴ y = $$\frac{3 x^{4}}{256}$$
∴ $$\frac{3 x^{4}}{256}$$ = 9x
∴ 3x4 – 2304x = 0
∴ x(x3 – 2304) = 0
∴ x = 0 or x3 = 2304 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = $$\frac{4^{2}}{4}$$
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x,
i.e. y = 2√x between x = 0 and x = 4
Area of the region ODABO = area under the rabola x2 = 4y,
i.e. y = $$\frac{x^{2}}{4}$$ between x = 0 and x = 4
(iii)
For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations.
From the equation x2 = y, y = $$\frac{x^{2}}{y}$$
∴ y = $$\frac{x^{2}}{y}$$
∴ $$\frac{x^{2}}{y}$$ = x
∴ x2 – y = 0
∴ x(x3 – y) = 0
∴ x = 0 or x3 = y
i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = $$\frac{4^{2}}{4}$$ = 4
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x,
i.e. y = 2√x between x = 0 and x = 4
Area ofthe region ODABO = area under the rabola x2 = 4y,
i.e. y = $$\frac{x^{2}}{3}$$ between x = 0 and x = 4
Question 5.
Find the area of the region in the first quadrant bounded by the circle x2 + y2 = 4 and the X-axis and the line x = y√3.
Solution:
For finding the points of intersection of the circle and the line, we solve
x2 + y2 = 4 ………(1)
and x = y√3 ……..(2)
From (2), x2 = 3y2
From (1), x2 = 4 – y2
3y2 = 4 – y2
4y2 = 4
y2 = 1
y = 1 in the first quadrant.
When y = 1, r = 1 × √3 = √3
∴ the circle and the line intersect at A(√3, 1) in the first quadrant
Required area = area of the region OCAEDO = area of the region OCADO + area of the region DAED
Now, area of the region OCADO = area under the line x = y√3, i.e. y = $$\frac{x}{\sqrt{3}}$$ between x = 0
and x = √3
Question 6.
Find the area of the region bounded by the parabola y2 = x and the line y = x in the first quadrant.
Solution:
To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1).
Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO
Now, area of the region OCADO = area under the parabola y2 = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1
Area of the region OBADO = area under the line y = x between x = 0 and x = 1
Question 7.
Find the area enclosed between the circle x2 + y2 = 1 and the line x + y = 1, lying in the first quadrant.
Solution:
Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)
Now, area of the region OACBO = area under the circle x2 + y2 = 1 between x = 0 and x = 1
Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1
∴ required area = $$\left(\frac{\pi}{4}-\frac{1}{2}\right)$$ sq units.
Question 8.
Find the area of the region bounded by the curve (y – 1)2 = 4(x + 1) and the line y = (x – 1).
Solution:
The equation of the curve is (y – 1)2 = 4(x + 1)
This is a parabola with vertex at A (-1, 1).
To find the points of intersection of the line y = x – 1 and the parabola.
Put y = x – 1 in the equation of the parabola, we get
(x – 1 – 1)2 = 4(x + 1)
∴ x2 – 4x + 4 = 4x + 4
∴ x2 – 8x = 0
∴ x(x – 8) = 0
∴ x = 0, x = 8
When x = 0, y = 0 – 1 = -1
When x = 8, y = 8 – 1 = 7
∴ the points of intersection are B (0, -1) and C (8, 7).
To find the points where the parabola (y – 1)2 = 4(x + 1) cuts the Y-axis.
Put x = 0 in the equation of the parabola, we get
(y – 1)2 = 4(0 + 1) = 4
∴ y – 1 = ±2
∴ y – 1 = 2 or y – 1 = -2
∴ y = 3 or y = -1
∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).
To find the point where the line y = x – 1 cuts the X-axis.
Put y = 0 in the equation of the line, we get
x – 1 = 0
∴ x = 1
∴ the line cuts the X-axis at the point G (1, 0).
Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO
Now, area of the region BFAB = area under the parabola (y – 1)2 = 4(x + 1), Y-axis from y = -1 to y = 3
Since, the area cannot be negative,
Area of the region BFAB = $$\left|-\frac{8}{3}\right|=\frac{8}{3}$$ sq units.
Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG
Since, area cannot be negative,
area of the region = $$\left|-\frac{1}{2}\right|=\frac{1}{2}$$ sq units.
∴ required area = $$\frac{8}{3}+\frac{109}{6}+\frac{1}{2}$$
= $$\frac{16+109+3}{6}$$
= $$\frac{128}{6}$$
= $$\frac{64}{3}$$ sq units.
Question 9.
Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.
Solution:
The equation of the line is
2y = 5x + 7, i.e., y = $$\frac{5}{2} x+\frac{7}{2}$$
Required area = area of the region ABCDA = area under the line y = $$\frac{5}{2} x+\frac{7}{2}$$ between x = 2 and x = 5
Question 10.
Find the area of the region bounded by the curve y = 4x2, Y-axis and the lines y = 1, y = 4.
Solution:
By symmetry of the parabola, the required area is 2 times the area of the region ABCD.
From the equation of the parabola, x2 = $$\frac{y}{4}$$
In the first quadrant, x > 0
∴ x = $$\frac{1}{2} \sqrt{y}$$
∴ required area = $$\int_{1}^{4} x d y$$
## Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Ex 5.1 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1
1. Find the area of the region bounded by the following curves, X-axis, and the given lines:
(i) y = 2x, x = 0, x = 5.
Solution:
Required area = $$\int_{0}^{5} y d x$$, where y = 2x
= $$\int_{0}^{5} 2x d x$$
= $$\left[\frac{2 x^{2}}{2}\right]_{0}^{5}$$
= 25 – 0
= 25 sq units.
(ii) x = 2y, y = 0, y = 4.
Solution:
Required area = $$\int_{0}^{4} x d y$$, where x = 2y
= $$\int_{0}^{4} 2 y d y$$
= $$\left[\frac{2 y^{2}}{2}\right]_{0}^{4}$$
= 16 – 0
= 16 sq units.
(iii) x = 0, x = 5, y = 0, y = 4.
Solution:
Required area = $$\int_{0}^{5} y d x$$, where y = 4
= $$\int_{0}^{5} 4 d x$$
= $$[4 x]_{0}^{5}$$
= 20 – 0
= 20 sq units.
(iv) y = sin x, x = 0, x = $$\frac{\pi}{2}$$
Solution:
Required area = $$\int_{0}^{\pi / 2} y d x$$, where y = sin x
= $$\int_{0}^{\pi / 2} \sin x d x$$
= $$[-\cos x]_{0}^{\pi / 2}$$
= -cos $$\frac{\pi}{2}$$ + cos 0
= 0 + 1
= 1 sq unit.
(v) xy = 2, x = 1, x = 4.
Solution:
For xy = 2, y = $$\frac{2}{x}$$
Required area = $$\int_{1}^{4} y d x$$, where y = $$\frac{2}{x}$$
= $$\int_{1}^{4} \frac{2}{x} d x$$
= $$[2 \log |x|]_{1}^{4}$$
= 2 log 4 – 2 log 1
= 2 log 4 – 0
= 2 log 4 sq units.
(vi) y2 = x, x = 0, x = 4.
Solution:
The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y2 = x, y = √x
Required area = A1 + A2 = 2A1
(vii) y2 = 16x, x = 0, x = 4.
Solution:
The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y2 = x, y = √x
Required area = A1 + A2 = 2A1
2. Find the area of the region bounded by the parabola:
(i) y2 = 16x and its latus rectum.
Solution:
Comparing y2 = 16x with y2 = 4ax, we get
4a = 16
∴ a = 4
∴ focus is S(a, 0) = (4, 0)
For y2 = 16x, y = 4√x
Required area = area of the region OBSAO
= 2 [area of the region OSAO]
(ii) y = 4 – x2 and the X-axis.
Solution:
The equation of the parabola is y = 4 – x2
∴ x2 = 4 – y
i.e. (x – 0)2 = -(y – 4)
It has vertex at P(0, 4)
For points of intersection of the parabola with X-axis,
we put y = 0 in its equation.
∴ 0 = 4 – x2
∴ x2 = 4
∴ x = ± 2
∴ the parabola intersect the X-axis at A(-2, 0) and B(2, 0)
Required area = area of the region APBOA
= 2[area of the region OPBO]
3. Find the area of the region included between:
(i) y2 = 2x and y = 2x.
Solution:
The vertex of the parabola y2 = 2x is at the origin O = (0, 0).
To find the points of intersection of the line and the parabola, equaling the values of 2x from both the equations we get,
y2 = y
∴ y2 – y = 0
∴ y = 0 or y = 1
When y = 0, x = $$\frac{0}{2}$$ = 0
When y = 1, x = $$\frac{1}{2}$$
∴ the points of intersection are 0(0, 0) and B($$\frac{1}{2}$$, 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y2 = 2x between x = 0 and x = $$\frac{1}{2}$$
Area of the region OCBDO = area under the line y = 2x between x = 0 and x = $$\frac{1}{2}$$
(ii) y2 = 4x and y = x.
Solution:
The vertex of the parabola y2 = 4x is at the origin O = (0, 0).
To find the points of intersection of the line and the parabola, equaling the values of 4x from both the equations we get,
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = $$\frac{0}{2}$$ = 0
When y = 1, x = $$\frac{1}{2}$$
∴ the points of intersection are O(0, 0) and B($$\frac{1}{2}$$, 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y2 = 4x between x = 0 and x = $$\frac{1}{2}$$
Area of the region OCBDO = area under the line y = 2x between x = 0 and x = $$\frac{1}{2}$$
(iii) y = x2 and the line y = 4x.
Solution:
The vertex of the parabola y = x2 is at the origin 0(0, 0)
To find the points of the intersection of a line and the parabola.
Equating the values of y from the two equations, we get
x2 = 4x
∴ x2 – 4x = 0
∴ x(x – 4) = 0
∴ x = 0, x = 4
When x = 0, y = 4(0) = 0
When x = 4, y = 4(4) = 16
∴ the points of intersection are 0(0, 0) and B(4, 16)
Required area = area of the region OABCO = (area of the region ODBCO) – (area of the region ODBAO)
Now, area of the region ODBCO = area under the line y = 4x between x = 0 and x = 4
= $$\int_{0}^{4} y d x$$, where y = 4x
= $$\int_{0}^{4} 4 x d x$$
= 4$$\int_{0}^{4} x d x$$
= 4$$$\int_{0}^{4} x d x$$$
= 2(16 – 0)
= 32
Area of the region ODBAO = area under the parabola y = x2 between x = 0 and x = 4
= $$\int_{0}^{4} y d x$$, where y = x2
= $$\int_{0}^{4} x^{2} d x$$
= $$\left[\frac{x^{3}}{3}\right]_{0}^{4}$$
= $$\frac{1}{3}$$ (64 – 0)
= $$\frac{64}{3}$$
∴ required area = 32 – $$\frac{64}{3}$$ = $$\frac{32}{3}$$ sq units.
(iv) y2 = 4ax and y = x.
Solution:
The vertex of the parabola y2 = 4ax is at the origin O = (0, 0).
To find the points of intersection of the line and the parabola, equaling the values of 4ax from both the equations we get,
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = $$\frac{0}{2}$$ = 0
When y = 1, x = $$\frac{1}{2}$$
∴ the points of intersection are O(0, 0) and B($$\frac{1}{2}$$, 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO
= area under the parabola y2 = 4ax between x = 0 and x = $$\frac{1}{2}$$
Area of the region OCBDO
= area under the line y
= 4ax between x = 0 and x = $$\frac{1}{4 a x}$$
(v) y = x2 + 3 and y = x + 3.
Solution:
The given parabola is y = x2 + 3, i.e. (x – 0)2 = y – 3
∴ its vertex is P(0, 3).
To find the points of intersection of the line and the parabola.
Equating the values of y from both the equations, we get
x2 + 3 = x + 3
∴ x2 – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
∴ the points of intersection are P(0, 3) and B(1, 4)
Required area = area of the region PABCP = area of the region OPABDO – area of the region OPCBDO
Now, area of the region OPABDO
= area under the line y = x + 3 between x = 0 and x = 1
Area of the region OPCBDO = area under the parabola y = x2 + 3 between x = 0 and x = 1
## Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Miscellaneous Exercise 4 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4
I. Choose the correct option from the given alternatives:
Question 1.
$$\int_{2}^{3} \frac{d x}{x\left(x^{3}-1\right)}=$$
(a) $$\frac{1}{3} \log \left(\frac{208}{189}\right)$$
(b) $$\frac{1}{3} \log \left(\frac{189}{208}\right)$$
(c) $$\log \left(\frac{208}{189}\right)$$
(d) $$\log \left(\frac{189}{208}\right)$$
(a) $$\frac{1}{3} \log \left(\frac{208}{189}\right)$$
Question 2.
$$\int_{0}^{\pi / 2} \frac{\sin ^{2} x \cdot d x}{(1+\cos x)^{2}}=$$
(a) $$\frac{4-\pi}{2}$$
(b) $$\frac{\pi-4}{2}$$
(c) 4 – $$\frac{\pi}{2}$$
(d) $$\frac{4+\pi}{2}$$
(a) $$\frac{4-\pi}{2}$$
Question 3.
$$\int_{0}^{\log 5} \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3} \cdot d x=$$
(a) 3 + 2π
(b) 4 – π
(c) 2 + π
(d) 4 + π
(b) 4 – π
Question 4.
$$\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{2} x \cdot d x=$$
(a) $$\frac{7 \pi}{256}$$
(b) $$\frac{3 \pi}{256}$$
(c) $$\frac{5 \pi}{256}$$
(d) $$\frac{-5 \pi}{256}$$
(c) $$\frac{5 \pi}{256}$$
Question 5.
If $$\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{X}}=\frac{k}{3}$$, then k is equal to
(a) √2(2√2 – 2)
(b) $$\frac{\sqrt{2}}{3}$$(2 – 2√2)
(c) $$\frac{2 \sqrt{2}-2}{3}$$
(d) 4√2
(d) 4√2
Question 6.
$$\int_{1}^{2} \frac{1}{x^{2}} e^{\frac{1}{x}} \cdot d x=$$
(a) √e + 1
(b) √e − 1
(c) √e(√e − 1)
(d) $$\frac{\sqrt{e}-1}{e}$$
(c) √e(√e − 1)
Question 7.
If $$\int_{2}^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] \cdot d x=a+\frac{b}{\log 2}$$, then
(a) a = e, b = -2
(b) a = e, b = 2
(c) a = -e, b = 2
(d) a = -e, b = -2
(a) a = e, b = -2
Question 8.
Let $$\mathrm{I}_{1}=\int_{e}^{e^{2}} \frac{d x}{\log x}$$ and $$\mathrm{I}_{2}=\int_{1}^{2} \frac{e^{x}}{\boldsymbol{X}} \cdot d x$$, then
(a) I1 = $$\frac{1}{3}$$ I2
(b) I1 + I2 = 0
(c) I1 = 2I2
(d) I1 = I2
(d) I1 = I2
Question 9.
$$\int_{0}^{9} \frac{\sqrt{X}}{\sqrt{X}+\sqrt{9-X}} \cdot d x=$$
(a) 9
(b) $$\frac{9}{2}$$
(c) 0
(d) 1
(b) $$\frac{9}{2}$$
Question 10.
The value of $$\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right) \cdot d \theta$$ is
(a) 0
(b) 1
(c) 2
(d) π
(a) 0
II. Evaluate the following:
Question 1.
$$\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x$$
Solution:
Let I = $$\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x$$
Put Numerator = A(Denominator) + B[$$\frac{d}{d x}$$(Denominator)]
∴ cos x = A(3 cos x + sin x) + B[$$\frac{d}{d x}$$(3 cos x + sin x)]
= A(3 cos x + sin x) + B(-3 sin x + cos x)
∴ cos x + 0 . sin x = (3A + B) cos x + (A – 3B) sin x
Comparing the coefficients of sinx and cos x on both the sides, we get
3A + B = 1 ………. (1)
A – 3B = 0 ………. (2)
Multiplying equation (1) by 3, we get
9A + 3B = 3 ………(3)
Adding (2) and (3), we get
10A = 3
∴ A = $$\frac{3}{10}$$
Question 2.
$$\int_{\pi / 4}^{\pi / 2} \frac{\cos \theta}{\left[\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right]^{3}} d \theta$$
Solution:
Question 3.
$$\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x$$
Solution:
Let I = $$\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x$$
Put √x = t
∴ x = t2 and dx = 2t . dt
When x = 0, t = 0
When x = 1, t = 1
Question 4.
$$\int_{0}^{\pi / 4} \frac{\tan ^{3} x}{1+\cos 2 x} d x$$
Solution:
Question 5.
$$\int_{0}^{1} t^{5} \sqrt{1-t^{2}} d t$$
Solution:
Question 6.
$$\int_{0}^{1}\left(\cos ^{-1} x\right)^{2} d x$$
Solution:
Question 7.
$$\int_{-1}^{1} \frac{1+x^{3}}{9-x^{2}} d x$$
Solution:
Question 8.
$$\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{4} x d x$$
Solution:
Question 9.
$$\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x} d x$$
Solution:
Question 10.
$$\int_{1}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x$$
Solution:
III. Evaluate the following:
Question 1.
$$\int_{0}^{1}\left(\frac{1}{1+x^{2}}\right) \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$$
Solution:
Question 2.
$$\int_{0}^{\pi / 2} \frac{1}{6-\cos x} d x$$
Solution:
Question 3.
$$\int_{0}^{a} \frac{1}{a^{2}+a x-x^{2}} d x$$
Solution:
Question 4.
$$\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x$$
Solution:
Question 5.
$$\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$$
Solution:
Let I = $$\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$$
Question 6.
$$\int_{0}^{\pi / 4} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x$$
Solution:
Question 7.
$$\int_{0}^{\pi / 2}[2 \log (\sin x)-\log (\sin 2 x)] d x$$
Solution:
Question 8.
$$\int_{0}^{\pi}\left(\sin ^{-1} x+\cos ^{-1} x\right)^{3} \sin ^{3} x d x$$
Solution:
Question 9.
$$\int_{0}^{4}\left[\sqrt{x^{2}+2 x+3}\right]^{-1} d x$$
Solution:
Question 10.
$$\int_{-2}^{3}|x-2| d x$$
Solution:
|x – 2|= 2 – x, if x < 2
= x – 2, if x ≥ 2
IV. Evaluate the following:
Question 1.
If $$\int_{a}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x$$, find the value of $$\int_{a}^{a+1} x d x$$.
Solution:
Question 2.
If $$\int_{0}^{k} \frac{1}{2+8 x^{2}} \cdot d x=\frac{\pi}{16}$$, find k.
Solution:
Question 3.
If f(x) = a + bx + cx2, show that $$\int_{0}^{1} f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right]$$
Solution:
## Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.2 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2
I. Evaluate:
Question 1.
$$\int_{1}^{9} \frac{x+1}{\sqrt{x}} \cdot d x$$
Solution:
Question 2.
$$\int_{2}^{3} \frac{1}{x^{2}+5 x+6} \cdot d x$$
Solution:
Question 3.
$$\int_{0}^{\pi / 4} \cot ^{2} \cdot d x$$
Solution:
The integral does not exist since cot 0 is not defined.
Question 4.
$$\int_{-\pi / 4}^{\pi / 4} \frac{1}{1-\sin x} \cdot d x$$
Solution:
Question 5.
$$\int_{3}^{5} \frac{1}{\sqrt{2 x+3}-\sqrt{2 x-3}} \cdot d x$$
Solution:
Question 6.
$$\int_{0}^{1} \frac{x^{2}-2}{x^{2}+1} \cdot d x$$
Solution:
Question 7.
$$\int_{0}^{\pi / 4} \sin 4 x \sin 3 x \cdot d x$$
Solution:
Question 8.
$$\int_{0}^{\pi / 4} \sqrt{1+\sin 2 x} \cdot d x$$
Solution:
Question 9.
$$\int_{0}^{\pi / 4} \sin ^{4} x \cdot d x$$
Solution:
Question 10.
$$\int_{-4}^{2} \frac{1}{x^{2}+4 x+13} \cdot d x$$
Solution:
Question 11.
$$\int_{0}^{4} \frac{1}{\sqrt{4 x-x^{2}}} \cdot d x$$
Solution:
Question 12.
$$\int_{0}^{1} \frac{1}{\sqrt{3+2 x-x^{2}}} \cdot d x$$
Solution:
Question 13.
$$\int_{0}^{\pi / 2} x \cdot \sin x \cdot d x$$
Solution:
Question 14.
$$\int_{0}^{1} x \cdot \tan ^{-1} x \cdot d x$$
Solution:
Question 15.
$$\int_{0}^{\infty} x \cdot e^{-x} \cdot d x$$
Solution:
II. Evaluate:
Question 1.
$$\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} \cdot d x$$
Solution:
Question 2.
$$\int_{0}^{\pi / 4} \frac{\sec ^{2} x}{3 \tan ^{2} x+4 \tan x+1} \cdot d x$$
Solution:
Question 3.
$$\int_{0}^{4 \pi} \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} \cdot d x$$
Solution:
Question 4.
$$\int_{0}^{2 \pi} \sqrt{\cos x} \cdot \sin ^{3} x \cdot d x$$
Solution:
Question 5.
$$\int_{0}^{\pi / 2} \frac{1}{5+4 \cos x} \cdot d x$$
Solution:
Question 6.
$$\int_{0}^{\pi / 4} \frac{\cos x}{4-\sin ^{2} x} \cdot d x$$
Solution:
Question 7.
$$\int_{0}^{\pi / 2} \frac{\cos X}{(1+\sin x)(2+\sin x)} \cdot d x$$
Solution:
Question 8.
$$\int_{-1}^{1} \frac{1}{a^{2} e^{x}+b^{2} e^{-x}} \cdot d x$$
Solution:
Question 9.
$$\int_{0}^{\pi} \frac{1}{3+2 \sin x+\cos x} \cdot d x$$
Solution:
Question 10.
$$\int_{0}^{\pi / 4} \sec ^{4} x \cdot d x$$
Solution:
Question 11.
$$\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \cdot d x$$
Solution:
Question 12.
$$\int_{0}^{\pi} \sin ^{3} x(1+2 \cos x)(1+\cos x)^{2} \cdot d x$$
Solution:
Question 13.
$$\int_{0}^{\pi / 2} \sin 2 x \cdot \tan ^{-1}(\sin x) \cdot d x$$
Solution:
Question 14.
$$\int_{\frac{1}{\sqrt{2}}}^{1} \frac{\left(e^{\cos ^{-1} x}\right)\left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}} \cdot d x$$
Solution:
Question 15.
$$\int_{2}^{3} \frac{\cos (\log x)}{x} \cdot d x$$
Solution:
III. Evaluate:
Question 1.
$$\int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} \cdot d x$$
Solution:
Question 2.
$$\int_{0}^{\pi / 2} \log \tan x \cdot d x$$
Solution:
Question 3.
$$\int_{0}^{1} \log \left(\frac{1}{x}-1\right) \cdot d x$$
Solution:
Question 4.
$$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cdot \cos x} \cdot d x$$
Solution:
Question 5.
$$\int_{0}^{3} x^{2}(3-x)^{\frac{5}{2}} \cdot d x$$
Solution:
Question 6.
$$\int_{-3}^{3} \frac{x^{3}}{9-x^{2}} \cdot d x$$
Solution:
Question 7.
$$\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2+\sin x}{2-\sin x}\right) \cdot d x$$
Solution:
Question 8.
$$\int_{-\pi / 4}^{\pi / 4} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} \cdot d x$$
Solution:
Question 9.
$$\int_{-\pi / 4}^{\pi / 4} x^{3} \cdot \sin ^{4} x \cdot d x$$
Solution:
Question 10.
$$\int_{0}^{1} \frac{\log (x+1)}{x^{2}+1} \cdot d x$$
Solution:
Question 11.
$$\int_{-1}^{1} \frac{x^{3}+2}{\sqrt{x^{2}+4}} \cdot d x$$
Solution:
Question 12.
$$\int_{-a}^{a} \frac{x+x^{3}}{16-x^{2}} \cdot d x$$
Solution:
Question 13.
$$\int_{0}^{1} t^{2} \sqrt{1-t} \cdot d t$$
Solution:
Question 14.
$$\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{2} x \cdot d x$$
Solution:
Question 15.
$$\int_{0}^{1} \frac{\log x}{\sqrt{1-x^{2}}} \cdot d x$$
Solution:
## Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.1 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1
I. Evaluate the following integrals as a limit of a sum.
Question 1.
$$\int_{1}^{3}(3 x-4) \cdot d x$$
Solution:
Let f(x) = 3x – 4, for 1 ≤ x ≤ 3
Divide the closed interval [1, 3] into n subintervals each of length h at the points
1, 1 + h, 1 + 2h, 1 + rh, ….., 1 + nh = 3
∴ nh = 2
∴ h = $$\frac{2}{n}$$ and as n → ∞, h → 0
Here, a = 1
∴ f(a + rh) = f(1 + rh)
= 3(1 + rh) – 4
= 3rh – 1
Question 2.
$$\int_{0}^{4} x^{2} d x$$
Solution:
Let f(x) = x2, for 0 ≤ x ≤ 4
Divide the closed interval [0, 4] into n subintervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 4
i.e. 0, h, 2h, ….., rh, ….., nh = 4
∴ h = $$\frac{4}{n}$$ as n → ∞, h → 0
Here, a = 0
Question 3.
$$\int_{0}^{2} e^{x} d x$$
Solution:
Let f(x) = ex, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n equal subntervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = $$\frac{2}{n}$$ and as n → ∞, h → 0
Here, a = 0
Question 4.
$$\int_{0}^{2}\left(3 x^{2}-1\right) d x$$
Solution:
Let f(x) = 3x2 – 1, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n subintervals each of length h at the points.
0, 0 + h, 0 + 2h, ….., 0 + rh, ……, 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = $$\frac{2}{n}$$ and as n → ∞, h → 0
Here, a = 0
∴ f(a + rh) = f(0 + rh)
= f(rh)
= 3(rh)2 – 1
= 3r2h2 – 1
Question 5.
$$\int_{1}^{3} x^{3} d x$$
Solution:
Let f(x) = x3, for 1 ≤ x ≤ 3.
Divide the closed interval [1, 3] into n equal su bintervals each of length h at the points
1, 1 + h, 1 + 2h, ……, 1 + rh, ……, 1 + nh = 3
∴ nh = 2
∴ h = $$\frac{2}{n}$$ and as n → ∞, h → 0
Here a = 1
## Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Miscellaneous Exercise 1 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1
(I) Choose the correct option from the given alternatives:
Question 1.
Let f(1) = 3, f'(1) = $$-\frac{1}{3}$$, g(1) = -4 and g'(1) = $$-\frac{8}{3}$$. The derivative of $$\sqrt{[f(x)]^{2}+[g(x)]^{2}}$$ w.r.t. x at x = 1 is
(a) $$-\frac{29}{15}$$
(b) $$\frac{7}{3}$$
(c) $$\frac{31}{15}$$
(d) $$\frac{29}{15}$$
(d) $$\frac{29}{15}$$
Question 2.
If y = sec(tan-1 x), then $$\frac{d y}{d x}$$ at x = 1, is equal to
(a) $$\frac{1}{2}$$
(b) 1
(c) $$\frac{1}{\sqrt{2}}$$
(d) 2
(c) $$\frac{1}{\sqrt{2}}$$
Question 3.
If f(x) = $$\sin ^{-1}\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)$$, which of the following is not the derivative of f(x)?
(a) $$\frac{2 \cdot 4^{x} \log 4}{1+4^{2 x}}$$
(b) $$\frac{4^{x+1} \log 2}{1+4^{2 x}}$$
(c) $$\frac{4^{x+1} \log 4}{1+4^{4 x}}$$
(d) $$\frac{2^{2(x+1)} \log 2}{1+2^{4 x}}$$
(c) $$\frac{4^{x+1} \log 4}{1+4^{4 x}}$$
Question 4.
If xy = yx, then$$\frac{d y}{d x}$$ = _______
(a) $$\frac{x(x \log y-y)}{y(y \log x-x)}$$
(b) $$\frac{y(y \log x-x)}{x(x \log y-y)}$$
(c) $$\frac{y^{2}(1-\log x)}{x^{2}(1-\log y)}$$
(d) $$\frac{y(1-\log x)}{x(1-\log y)}$$
(b) $$\frac{y(y \log x-x)}{x(x \log y-y)}$$
Question 5.
If y = sin (2 sin-1 x), then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}$$
(b) $$\frac{2+4 x^{2}}{\sqrt{1-x^{2}}}$$
(c) $$\frac{4 x^{2}-1}{\sqrt{1-x^{2}}}$$
(d) $$\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}$$
(a) $$\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}$$
Question 6.
If y = $$\tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)+\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{x}{\sqrt{1-x^{2}}}$$
(b) $$\frac{1-2 x}{\sqrt{1-x^{2}}}$$
(c) $$\frac{1-2 x}{2 \sqrt{1-x^{2}}}$$
(d) $$\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}$$
(c) $$\frac{1-2 x}{2 \sqrt{1-x^{2}}}$$
Question 7.
If y is a function of x and log(x + y) = 2xy, then the value of y'(0) = _______
(a) 2
(b) 0
(c) -1
(d) 1
(d) 1
Question 8.
If g is the inverse of function f and f'(x) = $$\frac{1}{1+x^{7}}$$, then the value of g'(x) is equal to:
(a) 1 + x7
(b) $$\frac{1}{1+[g(x)]^{7}}$$
(c) 1 + [g(x)]7
(d) 7x6
(c) 1 + [g(x)]7
Question 9.
If $$x \sqrt{y+1}+y \sqrt{x+1}=0$$ and x ≠ y, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{1}{(1+x)^{2}}$$
(b) $$-\frac{1}{(1+x)^{2}}$$
(c) (1 + x)2
(d) $$-\frac{x}{x+1}$$
(b) $$-\frac{1}{(1+x)^{2}}$$
Question 10.
If y = $$\tan ^{-1}\left(\sqrt{\frac{a-x}{a+x}}\right)$$, where -a < x < a, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{x}{\sqrt{a^{2}-x^{2}}}$$
(b) $$\frac{a}{\sqrt{a^{2}-x^{2}}}$$
(c) $$-\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
(d) $$\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
(c) $$-\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
[Hint: Put x = a cos θ]
Question 11.
If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), then $$\left[\frac{d^{2} y}{d x^{2}}\right]_{\theta=\frac{\pi}{4}}$$ = _______
(a) $$\frac{8 \sqrt{2}}{a \pi}$$
(b) $$-\frac{8 \sqrt{2}}{a \pi}$$
(c) $$\frac{a \pi}{8 \sqrt{2}}$$
(d) $$\frac{4 \sqrt{2}}{a \pi}$$
(a) $$\frac{8 \sqrt{2}}{a \pi}$$
Question 12.
If y = a cos (log x) and $$A \frac{d^{2} y}{d x^{2}}+B \frac{d y}{d x}+C=0$$, then the values of A, B, C are _______
(a) x2, -x, -y
(b) x2, x, y
(c) x2, x, -y
(d) x2, -x, y
(b) x2, x, y
(II) Solve the following:
Question 1.
Let u(x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v'(1) and w'(1). if it doesn’t exist then explain why?
Solution:
Question 2.
The values of f(x), g(x), f'(x) and g'(x) are given in the following table:
Match the following:
Solution:
Question 3.
Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.
(i) The derivative of f[g(x)] w.r.t. x at x = 0 is _______
(ii) The derivative of g[f(x)] w.r.t. x at x = 0 is _______
(iii) The value of $$\left[\frac{d}{d x}\left[x^{10}+f(x)\right]^{-2}\right]_{x=1}$$ is _______
(iv) The derivative of f[(x+g(x))] w.r.t. x at x = 0 is _______
Solution:
Question 4.
Differentiate the following w.r.t. x:
(i) $$\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$
Solution:
Let y = $$\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$
Put x = cos θ, Then θ = cos-1 x and
(ii) $$\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]$$
Solution:
Let y = $$\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]$$
Put x = cos θ. Then θ = cos-1 x and
(iii) $$\tan ^{-1}\left[\frac{\sqrt{x}(3-x)}{1-3 x}\right]$$
Solution:
(iv) $$\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$$
Solution:
Let y = $$\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$$
Put x = cos θ. Then θ = cos-1 x and
(v) $$\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)+\cot ^{-1}\left(\frac{1-10 x^{2}}{7 x}\right)$$
Solution:
(vi) $$\tan ^{-1}\left[\sqrt{\frac{\sqrt{1+x^{2}+x}}{\sqrt{1+x^{2}}-x}}\right]$$
Solution:
Question 5.
(i) If $$\sqrt{y+x}+\sqrt{y-x}=c$$, show that $$\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$$
Solution:
$$\sqrt{y+x}+\sqrt{y-x}=c$$
Differentiating both sides w.r.t. x, we get
(ii) If $$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$$, then show that $$\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$$
Solution:
$$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$$
$$y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$$
Differentiating both sides w.r.t. x, we get
(iii) If x sin(a + y) + sin a cos(a + y) = 0, then show $$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$$
Solution:
x sin(a + y) + sin a . cos (a + y) = 0 ….. (1)
Differentiating w.r.t. x, we get
(iv) If sin y = x sin(a + y), then show that $$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$$
Solution:
(v) If x = $$e^{\frac{x}{y}}$$, then show that $$\frac{d y}{d x}=\frac{x-y}{x \log x}$$
Solution:
x = $$e^{\frac{x}{y}}$$
$$\frac{x}{y}$$ = log x …..(1)
y = $$\frac{x}{\log x}$$
(vi) If y = f(x) is a differentiable function of x, then show that $$\frac{d^{2} x}{d y^{2}}=-\left(\frac{d y}{d x}\right)^{-3} \cdot \frac{d^{2} y}{d x^{2}}$$
Solution:
If y = f(x) is a differentiable function of x such that inverse function x = f-1(y) exists,
Question 6.
(i) Differentiate $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$ w.r.t. $$\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$$
Solution:
(ii) Differentiate $$\log \left[\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}-x}\right]$$ w.r.t. cos(log x)
Solution:
(iii) Differentiate $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$ w.r.t. $$\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^{2}}}{2 \sqrt{1+x^{2}}}}\right)$$
Solution:
Question 7.
(i) If y2 = a2 cos2x + b2 sin2x, show that $$y+\frac{d^{2} y}{d x^{2}}=\frac{a^{2} b^{2}}{y^{3}}$$
Solution:
y2 = a2 cos2x + b2 sin2x …… (1)
Differentiating both sides w.r.t. x, we get
(ii) If log y = log(sin x) – x2, show that $$\frac{d^{2} y}{d x^{2}}+4 x \frac{d y}{d x}+\left(4 x^{2}+3\right) y=0$$
Solution:
(iii) If x = a cos θ, y = b sin θ, show that $$a^{2}\left[y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]+b^{2}=0$$
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get
(iv) If y = A cos(log x) + B sin(log x), show that x2y2 + xy1 + y = o.
Solution:
y = A cos (log x) + B sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get
(v) If y = A emx + B enx, show that y2 – (m + n) y1 + (mn) y = 0.
Solution:
y = A emx + B enx
Differentiating w.r.t. x, we get
## Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.4 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4
Question 1.
Maximize : z = 11x + 8y subject to x ≤ 4, y ≤ 6,
x + y ≤ 6, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and ED whose equations are x = 4, y = 6 and x + y = 6 respectively.
The feasible region is shaded portion OAPDO in the graph.
The vertices of the feasible region are O (0, 0), A (4, 0), P and D (0, 6)
P is point of intersection of lines x + y = 6 and x = 4.
Substituting x = 4 in x + y = 6, we get
4 + y = 6 ∴ y = 2 ∴ P is (4, 2).
∴ the corner points of feasible region are O (0, 0), A (4, 0), P(4, 2) and D(0 ,6).
The values of the objective function z = 11x + 8y at these vertices are
z (O) = 11(0) + 8(0) = 0 + 0 = 0
z(a) = 11(4) + 8(0) = 44 + 0 = 44
z (P) = 11(4) + 8(2) = 44 + 16 = 60
z (D) = 11(0) + 8(2) = 0 + 16 = 16
∴ z has maximum value 60, when x = 4 and y = 2.
Question 2.
Maximize : z = 4x + 6y subject to 3x + 2y ≤ 12,
x + y ≥ 4, x, y ≥ 0.
Solution:
First we draw the lines AB and AC whose equations are 3x + 2y = 12 and x + y = 4 respectively.
The feasible region is the ∆ABC which is shaded in the graph.
The vertices of the feasible region (i.e. corner points) are A (4, 0), B (0, 6) and C (0, 4).
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(4) + 6(0) = 16 + 0 = 16
z(B) = 4(0)+ 6(6) = 0 + 36 = 36
z(C) = 4(0) + 6(4) = 0 + 24 = 24
∴ has maximum value 36, when x = 0, y = 6.
Question 3.
Maximize : z = 7x + 11y subject to 3x + 5y ≤ 26
5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 3x + 5y = 26 and 5x + 3y = 30 respectively.
The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (6, 0), p and B(0, $$\frac{26}{5}$$)
The vertex P is the point of intersection of the lines
3x + 5y = 26 … (1)
and 5x + 3y = 30 … (2)
Multiplying equation (1) by 3 and equation (2) by 5, we get
9x + 15y = 78
and 25x + 15y = 150
On subtracting, we get
16x = 72 ∴ x = $$\frac{72}{16}=\frac{9}{2}$$ = 4.5
Substituting x = 4.5 in equation (2), we get
5(4.5) + 3y = 30
22.5 + 3y = 30
∴ 3y = 7.5 ∴ y = 2.5
∴ P is (4.5, 2.5)
The values of the objective function z = 7x + 11y at these corner points are
z (O) = 7(0) + 11(0) = 0 + 0 = 0
z (C) = 7(6) + 11(0) = 42 + 0 = 42
z (P) = 7(4.5) + 11 (2.5) = 31.5 + 27.5 = 59.0 = 59
z(B) = 7(0) + 11$$\left(\frac{26}{5}\right)=\frac{286}{5}$$ = 57.2
∴ z has maximum value 59, when x = 4.5 and y = 2.5.
Question 4.
Maximize : z = 10x + 25y subject to 0 ≤ x ≤ 3,
0 ≤ y ≤ 3, x + y ≤ 5 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.
The feasible region is OAPQDO which is shaded in the i graph.
The vertices of the feasible region are O (0, 0), A (3, 0), P, Q and D(0, 3).
t P is the point of intersection of the lines x + y = 5 and x = 3.
Substituting x = 3 in x + y = 5, we get
3 + y = 5 ∴ y = 2
∴ P is (3, 2)
Q is the point of intersection of the lines x + y = 5 and y = 3
Substituting y = 3 in x + y = 5, we get
x + 3 = 5 ∴ x = 2
∴ Q is (2, 3)
The values of the objective function z = 10x + 25y at these vertices are
z(O) = 10(0) + 25(0) = 0 + 0 = 0
z(a) = 10(3) + 25(0) = 30 + 0 = 30
z(P) = 10(3) + 25(2) = 30 + 50 = 80
z(Q) = 10(2) + 25(3) = 20 + 75 = 95
z(D) = 10(0)+ 25(3) = 0 + 75 = 75
∴ z has maximum value 95, when x = 2 and y = 3.
Question 5.
Maximize : z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21,
x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.
The feasible region is OCPQBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (7, 0), P, Q and B (0, 6).
P is the point of intersection of the lines
3x + y = 21 … (1)
and x + y = 9 … (2)
On subtracting, we get 2x = 12 ∴ x = 6
Substituting x = 6 in equation (2), we get
6 + y = 9 ∴ y = 3
∴ P = (6, 3)
Q is the point of intersection of the lines
x + 4y = 24 … (3)
and x + y = 9 … (2)
On subtracting, we get
3y = 15 ∴ y = 5
Substituting y = 5 in equation (2), we get
x + 5= 9 ∴ x = 4
∴ Q = (4, 5)
∴ the corner points of the feasible region are 0(0,0), C(7, 0), P (6, 3), Q (4, 5) and B (0, 6).
The values of the objective function 2 = 3x + 5y at these corner points are
z(O) = 3(0)+ 5(0) = 0 + 0 = 0
z(C) = 3(7) + 5(0) = 21 + 0 = 21
z(P) = 3(6) + 5(3) = 18 + 15 = 33
z(Q) = 3(4) + 5(5) = 12 + 25 = 37
z(B) = 3(0)+ 5(6) = 0 + 30 = 30
∴ z has maximum value 37, when x = 4 and y = 5.
Question 6.
Minimize : z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3,
x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 5x + y = 5 and x + y = 3 respectively.
The feasible region is XCPBY which is shaded in the graph.
The vertices of the feasible region are C (3, 0), P and B (0, 5).
P is the point of the intersection of the lines
5x + y = 5
and x + y = 3
On subtracting, we get
4x = 2 ∴ x = $$\frac{1}{2}$$
Substituting x = $$\frac{1}{2}$$ in x + y = 3, we get
$$\frac{1}{2}$$ + y = 3
∴ y = $$\frac{5}{2}$$ ∴ P = $$\left(\frac{1}{2}, \frac{5}{2}\right)$$
The values of the objective function z = 7x + y at these vertices are
z(C) = 7(3) + 0 = 21
z(B) = 7(0) + 5 = 5
∴ z has minimum value 5, when x = 0 and y = 5.
Question 7.
Minimize : z = 8x + 10y subject to 2x + y ≥ 7, 2x + 3y ≥ 15,
y ≥ 2, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.
The feasible region is EPQBY which is shaded in the graph. The vertices of the feasible region are P, Q and B(0,7). P is the point of intersection of the lines 2x + 3y = 15 and y = 2.
Substituting y – 2 in 2x + 3y = 15, we get 2x + 3(2) = 15
∴ 2x = 9 ∴ x = 4.5 ∴ P = (4.5, 2)
Q is the point of intersection of the lines
2x + 3y = 15 … (1)
and 2x + y = 7 … (2)
On subtracting, we get
2y = 8 ∴ y = 4
∴ from (2), 2x + 4 = 7
∴ 2x = 3 ∴ x = 1.5
∴ Q = (1.5, 4)
The values of the objective function z = 8x + 10y at these vertices are
z(P) = 8(4.5) + 10(2) = 36 + 20 = 56
z(Q) = 8(1.5) + 10(4) = 12 + 40 = 52
z(B) = 8(0) +10(7) = 70
∴ z has minimum value 52, when x = 1.5 and y = 4
Question 8.
Minimize : z = 6x + 21y subject to x + 2y ≥ 3, x + 4y ≥ 4,
3x + y ≥ 3, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 2y = 3, x + 4y = 4 and 3x + y = 3 respectively.
The feasible region is XCPQFY which is shaded in the graph.
The vertices of the feasible region are C (4, 0), P, Q and F(0, 3).
P is the point of intersection of the lines x + 4y = 4 and x + 2y = 3
On subtracting, we get
2y = 1 ∴ y = $$\frac{1}{2}$$
Substituting y = $$\frac{1}{2}$$ in x + 2y = 3, we get
x + 2$$\left(\frac{1}{2}\right)$$ = 3
∴ x = 2
∴ P = (2, $$\frac{1}{2}$$)
Q is the point of intersection of the lines
x + 2y = 3 … (1)
and 3x + y = 3 ….(2)
Multiplying equation (1) by 3, we get 3x + 6y = 9
Subtracting equation (2) from this equation, we get
5y = 6
∴ y = $$\frac{6}{5}$$
∴ from (1), x + 2$$\left(\frac{6}{5}\right)$$ = 3
∴ x = 3 – $$\frac{12}{5}=\frac{3}{5}$$
Q ≡ $$\left(\frac{3}{5}, \frac{6}{5}\right)$$
The values of the objective function z = 6x + 21y at these vertices are
z(C) = 6(4) + 21(0) = 24
z(P) = 6(2) + 21$$\left(\frac{1}{2}\right)$$
= 12 + 10.5 = 22.5
z(Q)= 6$$\left(\frac{3}{5}\right)$$ + 21$$\left(\frac{6}{5}\right)$$
= $$\frac{18}{5}+\frac{126}{5}=\frac{144}{5}$$ = 28.8
2 (F) = 6(0) + 21(3) = 63
∴ z has minimum value 22.5, when x = 2 and y = $$\frac{1}{2}$$. |
# Multiplication strategies — Equal groups
by C. Elkins, OK Math and Reading Lady
Thanks for checking in on another multiplication strategy! The focus for this post will be on the equal groups strategy — looking at how students can efficiently use this strategy to help learn basic multiplication facts. My angle will be at the conceptual level by using concrete and pictorial methods. Be sure to see the links at the end for books and my free equal groups story cards.
Basics:
• Instead of in array or area format, equal groups are separate groups.
• The “x” means “groups of.” So 3 x 4 means “3 groups of 4.”
What things normally come in equal groups? Conduct a brainstorming session. I love the book “What Comes in 2’s, 3’s, and 4’s” as a springboard. After reading the book, let students brainstorm other things that come in equal groups. See the pictures below for some more ideas. After some internet research, I also made this attached list to use (in case you or your students draw a blank): click here: Equal groups pictures and list template
Use these lists to help students generate stories about equal groups. When students can create (and maybe illustrate) their own stories, they are much better at solving problems they must read on their own. This also helps students think carefully about what in the story constitutes a “group” and what the “groups of” represents:
1. There were 5 bowling balls on the rack. If you count all of the holes (3 per ball), how many holes are there all together? (5 x 3). The bowling balls are the groups. The holes are what is being counted in each group.
2. How many numbers are shown on 3 clocks? (3 x 12). The clocks are the groups. The numbers are what is being counted in each group.
3. I bought 8 pair of earrings. How many earrings are there? (8 x 2). The pairs are the groups.
4. Seven ladybugs were crawling on the leaves. How many legs would there be? (7 x 6). The ladybugs are the groups. The legs are what is being counted in each group.
Ways to show equal groups with objects and drawings:
• Hula hoops (great to use these in PE class to emphasize multiplication)
• Embroidery hoops
• Circles of yarn
• Dishes: cup, bowl, plate, tray
• Baskets
• Shelves
Objects to use to show equal groups:
• people
• cubes
• tiles
• mini erasers
• teddy bear manipulatives
• base ten materials
• food: pinto beans, macaroni, cereal, candy
• practically anything you have an abundance of!!
Teaching concepts regarding equal groups:
• When students are placing objects or drawing inside, do they randomly place objects? Or do they organize them to enable ease in counting? Showing students how to organize the objects in each set contributes to their knowledge of equal groups — AND it’s a big help to you as the teacher as you check on students. If the dots are randomly placed, the teacher and student must count one at a time to check. If they are organized, teacher and student can tell at a glance if the amount in each group is correct. Notice the difference below: Which ones show a student’s understanding of 9? Which ones can a student or teacher check rapidly?
• When counting the objects or drawings to determine the product of these equal groups, are students counting one at a time? Or are they counting in equal groups (such as by 2’s, 5’s, 3’s, etc.)? If we allow students to just count by ones, then they are not practicing multiplication . . .just counting!!
Activities to practice equal groups strategy:
1. Circles and Stars: Roll a dice once. This is the number of circles to draw. Roll a dice again. This is the number of stars to draw inside. If played with a partner, students can keep track of their totals to determine a winner. Dice can be varied depending on the facts that need to be practiced. A spinner can also be used. (See picture at beginning of this post.)
2. Variation of above: Use other materials (such as those listed above).
• Dice roll #1 = # of cups. Dice roll #2 = number of cubes
• Dice roll #1 = # of hoops. Dice roll #2 = # of pinto beans
• Dice roll #1 = # of plates. Dice roll #2 = # of Cheerios
3. Write and illustrate stories: Provide a problem for students to illustrate (example: 6 x 3 or 3 x 6). Then each student can decide how to form the story and illustrate. I always tell students to choose items they like to draw to make their story. Here are some examples. See some examples from former students.
• There were 6 monsters in the cave. Each monster had 3 eyeballs. How many eyeballs all together?
• Six princesses lived in the castle. They each had 3 ponies. How many ponies in all?
• There are 3 plants in the garden. They each have 6 flowers. How many flowers are in my garden?
• I made 3 pizzas. Each pizza had 6 slices. How many slices of pizza did I make?
4. PE Class activities: If your PE teacher likes to help you with your learning objectives, let them know you are working on equal groups strategies. While I’ve not done this personally, I think having relay races related to this would work perfectly. For example, the teacher presents a problem and each team must use hula hoops and objects to show the problem (and the answer).
5. Try these story books about multiplication:
6. Equal groups story problems to solve: Here are some story problem task cards and templates for solving multiplication and division problems using the equal groups strategy. Click to see the blog post on equal groups story problems and get my FREE set of story problem cards: HERE
Enjoy!! Many of you are now off for a well-deserved summer break. Use your summer time to catch up on my postings from this past year, or email me for more information. Also, feel free to comment on any article about your experience or additional tips.
# Multiplication Concepts Part 3: Equal Groups
by C. Elkins, OK Math and Reading Lady
Thanks for checking in on part 3 of my multiplication posts. Focus will be on the equal groups strategy — looking at how students can efficiently use this strategy to help learn basic multiplication facts. My angle will be at the conceptual level by using concrete and pictorial methods.
Basics:
• Instead of in array or area format, equal groups are separate groups.
• The “x” means “groups of.” So 3 x 4 means “3 groups of 4.”
What things normally come in equal groups? Conduct a brainstorming session. I love the book “What Comes in 2’s, 3’s, and 4’s” as a springboard. After reading the book, let students brainstorm other things that come in equal groups. See the pictures below for some more ideas. After some internet research, I also made this attached list to use (in case you or your students draw a blank): click here: Equal groups pictures and list template
Use these lists to help students generate stories about equal groups. When students can create (and maybe illustrate) their own stories, they are much better at solving problems they must read on their own. This also helps students think carefully about what in the story constitutes a “group” and what the “groups of” represents:
1. There were 5 bowling balls on the rack. If you count all of the holes (3 per ball), how many holes are there all together? (5 x 3). The bowling balls are the groups. The holes are what is being counted in each group.
2. How many numbers are shown on 3 clocks? (3 x 12). The clocks are the groups. The numbers are what is being counted in each group.
3. I bought 8 pair of earrings. How many earrings are there? (8 x 2). The pairs are the groups.
4. Seven ladybugs were crawling on the leaves. How many legs would there be? (7 x 6). The ladybugs are the groups. The legs are what is being counted in each group.
Ways to show equal groups with objects and drawings:
• Hula hoops (great to use these in PE class to emphasize multiplication)
• Embroidery hoops
• Circles of yarn
• Dishes: cup, bowl, plate, tray
• Baskets
• Shelves
Objects to use to show equal groups:
• people
• cubes
• tiles
• mini erasers
• teddy bear manipulatives
• base ten materials
• food: pinto beans, macaroni, cereal, candy
• practically anything you have an abundance of!!
Teaching concepts regarding equal groups:
• When students are placing objects or drawing inside, do they randomly place objects? Or do they organize them to enable ease in counting? Showing students how to organize the objects in each set contributes to their knowledge of equal groups — AND it’s a big help to you as the teacher as you check on students. If the dots are randomly placed, the teacher and student must count one at a time to check. If they are organized, teacher and student can tell at a glance if the amount in each group is correct. Notice the difference below: Which ones show a student’s understanding of 9? Which ones can a student or teacher check rapidly?
• When counting the objects or drawings to determine the product of these equal groups, are students counting one at a time? Or are they counting in equal groups (such as by 2’s, 5’s, 3’s, etc.)? If we allow students to just count by ones, then they are not practicing multiplication, just counting!!
Activities to practice equal groups strategy:
1. Circles and Stars: Roll a dice once. This is the number of circles to draw. Roll a dice again. This is the number of stars to draw inside. If played with a partner, students can keep track of their totals to determine a winner. Dice can be varied depending on the facts that need to be practiced. A spinner can also be used. (See picture at beginning of this post.)
2. Variation of above: Use other materials (such as those listed above).
• Dice roll #1 = # of cups. Dice roll #2 = number of cubes
• Dice roll #1 = # of hoops. Dice roll #2 = # of pinto beans
• Dice roll #1 = # of plates. Dice roll #2 = # of Cheerios
3. Write and illustrate stories: Provide a problem for students to illustrate (example: 6 x 3 or 3 x 6). Then each student can decide how to form the story and illustrate. I always tell students to choose items they like to draw to make their story. Here are some examples. See some examples from former students.
• There were 6 monsters in the cave. Each monster had 3 eyeballs. How many eyeballs all together?
• Six princesses lived in the castle. They each had 3 ponies. How many ponies in all?
• There are 3 plants in the garden. They each have 6 flowers. How many flowers are in my garden?
• I made 3 pizzas. Each pizza had 6 slices. How many slices of pizza did I make?
4. PE Class activities: If your PE teacher likes to help you with your learning objectives, let them know you are working on equal groups strategies. While I’ve not done this personally, I think having relay races related to this would work perfectly. For example, the teacher presents a problem and each team must use hula hoops and objects to show the problem (and the answer).
5. Try these story books about multiplication:
6. Equal groups story problems to solve: See my previous post related to this. You will find some story problem task cards and templates for solving multiplication and division problems using the equal groups strategy. Click HERE
Enjoy!!
# Division Basics Part 2: Equal Groups
by C. Elkins, OK Math and Reading Lady
Last post featured division using arrays and the area model. This post will focus on helping children see division as equal groups. Most of us have used the “plates of cookies” analogy to help kids see how to represent equal groups in a drawing. I will just take that a few more steps to increase efficiency.
Much like multiplication, there are different aspects of division children should get familiar with:
• Arrays
• Equal Groups
• Repeated Subtraction
• Number lines
• Skip counting
In this post, I will break down the benefits of equal groups models to help children understand division (and how it is related to multiplication). Check out the freebies within this post.
If you haven’t utilized this book with your students, please try to find a copy! It’s called The Doorbell Rang by Pat Hutchins. In this story, Ma makes some cookies to be split between the kids. Then the doorbell rings and more kids come, so the problem has to be refigured. This scenario repeats. As a class, you can duplicate the story with a different # of cookies and children.
Another great story emphasizing equal groups (as well as arrays) is the story One Hundred Hungry Ants by Elinor Pinczes. In this story, 100 ants are on their way to raid a picnic. They start off in one straight line (1 x 100), but then rearrange into different equal groups to shorten the line (2 lines of 50, 4 lines of 25, etc.). A nice project after reading this book is to see how many ways a different given # of ants (or other animals / objects) can be divided into equal groups / rows.
By clicking on the links for each book above, you will be taken to Amazon for more details.
As I mentioned earlier, many children’s view of equal groups regarding division is to use manipulatives and/or draw circles / plates to match the divisor and then divide up the “cookies” equally in these groups. Let’s say you had this problem: “There are 12 cookies to be divided onto 3 plates equally. How many cookies would go on each plate?” As you observe the students:
• How are they dividing up the cookies? One at a time, two at a time, randomly, trial and error?
• Are the “cookies” scattered randomly on the plate / circle? Or, are they arranged in an easy-to-see pattern so they are easily counted (by the student and yourself as you walk around the room)?
• Are the students able to verbally tell you how they divided them?
• Are the students making the connection to multiplication by noting that 3 x 4 = 12?
• Can they solve similar problems using language other than plates / cookies?
• Try shelves / books; trays / brownies; buildings / windows; flowers / petals; students / rows of desks, stars / # of points; aquariums / fish; boxes / donuts; etc.
Use of manipulatives of various types (cubes, tiles, counters) is important for children to have their hands on the objects being divided. This is how they work out their thinking. Then work toward paper/pencil drawings before going to the abstract use of numbers only. This doesn’t have to be done in separate lessons, however. There is great value for children to see how the concrete, pictorial, and abstract representations all work together.
Also, help children list synonyms for the dividing process: distribute, share, split, separate, halve, quarter, partition
Here are a few strategies I believe help make the equal groups process more efficient: Continue reading
# Geometry Part 4: More Composing and Decomposing
by C. Elkins, OK Math and Reading Lady
There are so many good ways to help students compose and decompose shapes (2D and 3D), so I will focus on some more by using tangrams and 2D paper shapes. In case you missed it, my last post focused on ways to use 1″ color tiles and pattern blocks to compose and decompose shapes. Click HERE to link back to that.
1. Give students paper shapes of these polygons: rectangle, square, hexagon, trapezoid, rhombus. Click here for a FREE pdf copy: Decompose and Compose Polygons.
• Students should color each paper shape one solid color (a different color for each shape). My advice is to use light colors because they will be drawing lines on the shapes and light colors enable them to see the lines.
• Model how to draw 1 or 2 lines to decompose the shape into smaller shapes. For first and 2nd grade, I recommend you show them how to use at least one corner of the shape to connect to another corner or side using a straight edge or ruler. This way the newly created shapes will resemble ones they already know (triangle, trapezoid, rectangle). Older students can be given a little more leeway — their decomposing may result in other more irregular polygons. Here is one way to decompose.
• Cut apart on the lines. Have students put their initials or name on the back of each piece (in case it gets separated or ends up on the floor).
• Each student puts their cut-up pieces in a baggy for safe-keeping. Then the student can take them out and try to compose them back into their original shapes. This is where the color-coding comes in handy (all the yellow go together, all the green, etc.).
• Students can trade their baggies with others to compose their shapes.
• When students are done with the shape puzzles, they can glue them back together on background construction paper (or take them home for practice, or keep at school for ongoing work).
• Discuss together how many different ways these shapes were decomposed using 1 or 2 lines.
2. Use the book, “The Greedy Triangle” by Marilyn Burns as a springboard to compose other polygons using various numbers of triangles. In this book, the triangle keeps adding a shape to himself (after a visit to the “Shapeshifter”). There are many good pictures in this book illustrating common things with the named shape. This is also a great way to connect art to math. You can start with squares which the students must cut in half on the diagonal, or start with pre-cut triangles. Length of edges must match. Level 0 students can just try out different combinations. Level 1-2 students would analyze the properties more and name the new shapes. You can even emphasize symmetry (as I have shown with the bottom row). Here is the link to the full article about this wonderful activity. Math Art: The Greedy Triangle Activity
# Math Art Part 2: Decomposing and composing squares and triangles
by C. Elkins, OK Math and Reading Lady
I wanted to show you another example of math art, this time using squares and triangles. This project also falls under the standards dealing with decomposing and composing shapes. With this project, students can create some unique designs while learning about squares, triangles, symmetry, fractions, and elements of art such as color and design. It would be a great project for first grade (using 2 squares) or for higher grades using 3 to 4 squares.
A great literature connection to this project is the book “The Greedy Triangle” by Marilyn Burns. (Click link to connect to Amazon.) The triangle in this book isn’t content with being 3-sided and transforms himself into other shapes (with the help of the Shapeshifter). Lots of great pictures showing real objects in the shape of triangles, squares, pentagons, hexagons, and more.
Marilyn Burns is a great math educator to check out, if you haven’t already. She has a company called Math Solutions (check out MathSolutions.com). Marilyn and her consultants have wonderful resources and advocate for constructivist views regarding math education. She is also the author of Number Talks and many math and literature lesson ideas.
### The 4 Triangle Investigation
Materials needed:
• Pre-cut squares 3″, 4″ or 5″ (I used brightly colored cardstock.)
• Scissors and glue
• Background paper to glue shapes to
Directions
1. Model how to cut a square in half (diagonally) to make two right triangles. (I advocate folding it first so that the two resulting triangles are as equivalent as possible.)
2. Guide students into showing different ways to put two triangles together to form another shape. Rule: Sides touching each other must be the same length. Let students practice making these shapes on their desk top (no gluing needed).
3. Help students realize they may need to use these actions:
• Slide the shape into place
• Flip it over to get a mirror image
• Rotate it around in a circular motion to align the edges
4. Students are then given 2 squares (to be cut into 4 triangles) and investigate different shapes they can make following the above rule. Here are some possibilities:
5. As the teacher, you can decide how many creations you want each student to attempt.
6. These shapes can be glued onto construction paper (and cut out if desired).
7. As an extension, shapes can be sorted according to various attributes:
• # of sides
• symmetry
• # of angles
• regular polygons vs. irregular
# First Day Math & Literature Activity K- 5
by C. Elkins, OK Math and Reading Lady
The book, Chrysanthemum, by Kevin Henkes is one my my all time favorite first-day-of-school stories to share with my students – no matter what grade level. The main character is Chrysanthemum, who is all excited about her first day of school until the other students start making fun of her name because it is soooo long. This makes her reluctant to go to school until everyone finds out their favorite music teacher has a long name (Delphinium) and is planning to name her new baby Chrysanthemum. A poignant story to help children develop a sense of empathy and compassion and realize that everyone’s name is special – no matter what it is or how long or short it is!
Math Connection Grades K-2
• Letter and name recognition
• Counting letters in names
• Name graph with a variety of methods (paper graph, color tile or unifix cube graph, etc.)
• Name grid art activity (see below)
• Comparing name lengths
Math Connection Grades 3-5
• Name graph – can use first, middle, and/or last names. To start, just have students write their name on a post-it-note and stick it on the board. Then rearrange into columns or rows according to how you are collecting your data. Or make a frequency table, line plot, percentage pie chart, etc.
• Name grid art activity (see below). Review terms: row, column, grid, array.
• Use some type of strategy to determine total number of letters in first names in the class (repeated addition, multiplication). Using the example graph, students could add 3 + (4 x 5) + (5 x 8), and so on. Let students think of the strategy though!
• Determine most often and least often used letters.
• Determine the mean, median, mode, and range using length of names.
Name grid art activity Continue reading |
# Year 4
## Overview of progression in Year 4
### Number and place value
In Year 4, children use place value in four-digit numbers, such as 3742 is three thousands, seven hundreds, four tens and two ones.
They learn to count in 6s, 7s, 9s, 25s and 1000s, and say 1000 more or less than a specific number. They encounter negative numbers by counting back past zero on number lines, and continue work on rounding (to the nearest 10, 100 or 1000) and estimation. Children are introduced to Roman numerals to 100 and find out how the number system has changed over time.
Children extend previous years’ work by adding and subtracting numbers with up to four digits, using mental and written methods, including columnar addition and subtraction.
They keep practising mental methods of addition and subtraction as well as written methods, performing calculations increasingly quickly and confidently. They continue using estimation as well as inverse operations to help check answers.
### Multiplication and division
Children learn the remaining multiplication tables up to the 12 multiplication table, and use facts from the tables to solve increasingly complex multiplication and division problems.
They build on their work with mental methods of calculation in Year 3, using their knowledge of place value and number facts to multiply and divide confidently. They begin to use a formal written layout for multiplication when multiplying two-digit and three-digit numbers by one-digit numbers.
### Fractions (including decimals)
Developing ideas from Year 3, children confidently count up and down in hundredths.
They learn about and recognise equivalent fractions, simplifying them when necessary (for example, understanding that 1/3 = 2/6 = 4/12). They move on to understand and show families of equivalent fractions. They build on earlier work, practising adding and subtracting fractions with the same denominator (2/3 + 7/9 = 11/9). Children also work with decimal equivalents of tenths and hundredths and of 1/2, 1/4, 3/4, understanding that decimals and fractions are different ways of expressing numbers. They round numbers with one decimal place to the nearest whole number, and compare numbers with the same number of decimal places, up to two decimal places. They use fractions and decimals to solve straightforward money and measure problems.
### Measurement
In Year 3, children learned to measure the perimeter of 2D shapes; they now extend this, calculating the perimeter of rectilinear shapes including squares.
They work out the area of rectilinear shapes by counting. Children compare digital clocks and analogue clocks, reading, writing and converting time between the two systems. They begin using £ and p notation to record money.
### Geometry: properties of shapes
Children learn about a wider range of geometric shapes, including different types of triangles and quadrilaterals.
They develop work on acute and obtuse angles from Year 3, comparing and ordering angles up to two right angles. They work with lines of symmetry in 2D shapes.
### Geometry: position and direction
Children begin to work with a coordinate grid (first quadrant only), using coordinates to describe positions on a grid.
### Statistics
Children are introduced to the difference between discrete and continuous data, using bar charts for discrete data (numbers of children travelling to school by different methods) and line graphs for continuous data (children’s heights).
Children will build further on their work with line graphs in Year 5. |
In the previous section, we learned about several properties that distinguish
parallelograms
from other
. Most of the work we did was computation-based because
we were already given the fact that the figures were parallelograms. In this section,
we will use our reasoning skills to put together two-column geometric
proofs
for parallelograms. We can apply much of what we learned in the previous
section to help us throughout this lesson, but we will be much more formalized and
organized in our arguments.
## Using Definitions and Theorems in Proofs
The ways we start off our proofs are key steps toward arriving at a conclusion.
Therefore, comprehending the information that we are given by an exercise may be
the single most important part of proving a statement.
As we will see, there are different ways in which we can essentially say the same
statement. Recall, that many of our
angle theorems
had converses. The converses of the theorems essentially
gave the same information, but in a reversed order. We will have to approach problems
involving parallelograms in the same way. That is, we must be conscious of the arguments
we make based on whether we are given that a certain quadrilateral is a parallelogram,
or if we want to prove that the quadrilateral is a parallelogram. Let’s take
a look at these statements so that we understand how to use them properly in our
proofs.
### Given a Parallelogram
We can use the following statements in our proofs if we are given that a quadrilateral
is a parallelogram.
Definition: A parallelogram is a type of quadrilateral whose pairs of opposite
sides are parallel.
If a quadrilateral is a parallelogram, then…
Much of the information above was studied in the previous section. The purpose of
organizing it in the way that it has been laid out is to help us see the difference
in our statements depending on whether we are given a parallelogram, or if we are
trying to prove that a quadrilateral is a parallelogram.
Let’s look at the structure of our statements when we are trying to prove that a
### Proving a Parallelogram
Definition: A parallelogram is a type of quadrilateral whose pairs of opposite
sides are parallel.
If…
Let’s use these statements to help us prove the following exercise. We will need
to use both forms of the statements above, because we will be given one parallelogram,
and we will have to prove that another one exists. This will give us practice using
regular theorems and definitions, as well as their converses.
## Exercise
Solution:
As stated before this exercise, we need to be conscious of how to use theorems and
definitions, as well as their converses because we are given that NRSM
is a parallelogram, but we also want to prove that ERAM is
a parallelogram. We were also given that ?4??5, which will help us
prove our conclusion.
To begin, we know that ?R??M because they are the opposite angles
of parallelogram NRSM.
Knowing this allows us to claim that ?3??6 by the Angle Subtraction
Postulate
. We see that ?R is composed of two smaller angles
(?3 and ?4). Likewise, we see that ?M
is composed of ?5 and ?6. Since the whole of the angles
are congruent, and two of the smaller angles in them are congruent, then their remainders
are also congruent.
Now, we have proven that one pair of opposite angles are congruent. If we can show
that ?2 and ?7 are also congruent, we can prove that
Because NRSM is a parallelogram, we know that its opposite sides are
parallel. So, we have that segments NR and MS are parallel.
Considering these lines, we know that segments EM and RA
are transversals to the parallel lines, since they intersect both lines. Thus, we
can use the Alternate Interior Angles Theorem to prove that ?1??6
and ?3??8.
By transitivity, we can say that ?1 is congruent to ?8.
It is a bit difficult to imagine the chain of congruences that allows us to make
this claim, but it is as such:
Notice that ?3 and ?6 are congruent, opposite angles,
just as ?2 and ?7 are. Let’s look at our new illustration
to help us visualize what we’ve done.
We have proven that ERAM is a parallelogram because both pairs of
its opposite angles are congruent. The two-column proof for our argument is shown
below. |
# Video: Determining Whether Relations Are Functions
Which of the relations shown by the set of ordered pairs below does not represent a function? [A] {(−8, 4), (−9, 4), (10, 4), (11, 4)} [B] {(3, 4), (11, 8), (3, 12), (11, 11)} [C] {(3, 11), (11, 19), (27, 35), (43, 43)} [D] {(3, 11), (4, 11), (5, 11), (6, 11)}
09:53
### Video Transcript
Which of the relations shown by the set of ordered pairs below does not represent a function? So we have four options here, each of which is a set of ordered pairs, each of which is a relation, but only three of those four also represent a function.
Let’s go through the options one by one, starting with option A. So option A is the set of ordered pairs minus eight, four; minus nine, four; 10, four; and 11, four. And this defines a relation. Let’s first remind our self what relation is. A relation is defined on two sets, which I’m going to call 𝑥 and 𝑦. And let’s see how this works by filling in the relation defined by the set in option A.
The first element of the set is the ordered pair minus eight, four. And the fact that the first component of that ordered pair is minus eight tells us that minus eight has to go in the set 𝑥, and the fact that the second component of that ordered pair is four tells us that four has to go in the set 𝑦. But more than that, we’re told not just a minus eight and four in those two sets. We’re told that there’s a relation between these two elements. We represent that on the diagram using an arrow, like so.
Okay, so let’s move on to the second element of our set, which is the ordered pair minus nine, four. The first component that ordered pair is minus nine, so that goes in 𝑥. The second component, four, you already put in our set 𝑦; we don’t need to draw it again. We just need to represent the relation using an arrow. And I’ve done the same, the last two elements of the set, which were 10, four and 11, four.
Now perhaps 𝑥 contains some other things apart from minus eight, minus nine, 10, and 11. It could be the set of integers for example. And similarly, 𝑦 might be more than just a set of the number four. It could be the set of whole numbers. In fact, both of the sets could be the set of the real numbers, but we’ve just marked those numbers, those elements of those sets which are involved in this relation here. And that’s all we need to do. This is a perfectly good relation, but the question is is it also a function.
So let’s ignore the diagram for the moment and just have a look at the set. If this is a function, then each ordered pair represents a pair of input and output values of that function. So for example, the first ordered pair — minus eight, four — tells us that minus eight is an input whose output given by the function is four. Moving along, we see that the input nine has the output four as well. As does the input 10, it has the output four. And the final pair of input and output values tells us that the input 11 gives the output four.
Going back to the diagram, we can say that the set 𝑥 is now the set of inputs to that function, otherwise known as the domain of the function. And the set 𝑦 is the set of outputs of the function, otherwise known as the range of the function. And looking at the diagram, we see that every input in the domain is given an output in the range, and there’s no problem at all about that. So this is a function; A is a function. And so it’s not our answer. Let’s move on to the second option, B.
Option B is quite a similar looking set of ordered pairs to what we had before, and we’re going to go through the same process of representing this relation on a diagram. So the first element of this set is the ordered pair three, four. The first component of the pair, three, goes in the set 𝑥 the second component, four, goes in the set 𝑦. And of course, they are related. There’s a relation between them, so we draw an arrow. Next, we have 11, eight. We do the same, so 11 goes in the set 𝑥; eight goes in the set 𝑦; and we draw an arrow between them.
So far so good. How about three, 12? Well three goes in the set 𝑥. Oh! we’ve already got three in the set 𝑥; you don’t need to draw it again. We’ve got 12, the second component, has to go in the set 𝑦. And we draw an arrow between them. And finally, 11, 11. So 11 in the first component means that it’s 11 in the first set 𝑥, which we already have, and 11 as a second component of the ordered pair tells us that we need to put 11 into the set 𝑦 as well, so this is a different 11 because it’s in a different set. And we showed that these related like so.
I just like to stress this is a perfectly fine relation. In fact, any set of ordered pairs is a perfectly fine relation that you can establish like we have. But the question is is it also a function. So remember, a function can be defined as the set of ordered pairs of the input and output values of that function. So for example the first ordered pair — three, four — in that set tells us that the input three when given to that function gives the output four. So maybe if the function is called 𝑓, we’ll write 𝑓 of three is equal to four.
The next ordered pair tells us that 𝑓 of 11, the input, is eight, the output. Okay, what about the third ordered pair? Well we’ve got an input of three gives an output of 12, and so 𝑓 of three is 12. But hang on, we just said that 𝑓 of three was four. How can that make sense? How can 𝑓 of three be both four and 12? Well it can’t.
We can notice the same thing on the diagram. I’ve got three going to four and also to 12, so there’re two arrows coming out of the number three on the left. Again this is fine for a relation, but not fine for a function. We can’t have 𝑓 of three being both four and 12, so this set of ordered pairs does not define a function. As a result, B is our answer. This is the set of ordered pairs which defines a relation but it does not represent a function.
Just to conclude, let’s see a quick way of finding whether a set of ordered pairs represents a function. Look at the first components of those ordered pairs, so three, 11, three, and 11, and you’ll notice that there are some repeats there. So there’s some inputs which are given twice, for example three, with different outputs, four and 12, and 11 with eight and 11. So if you ever see repeats in the first components of the ordered pairs, you know that’s not going to be a function.
And if you have a diagram which represents relation, like we do below, you can ask whether that also represents a function by thinking about it as a mapping diagram. And as we saw before, three was mapped to both four and 12. There were two arrows coming out of three from the left going to the set of outputs as we like to think about it on the right, and of course it doesn’t make sense. So if there are two or more arrows coming out of any element on the left, then we know that’s not a function
You can go back to earlier in the video to see that these two problems didn’t occur with option A, so that was fine; that was a function. And you can check as well that options C and D do represent functions. You can check that in each of those options in each of those sets, the first component of those ordered pairs is never repeated. And you can check that when you draw a diagram from those relations, there’s never more than one arrow coming out of any input on the left.
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# What is the square root of 444*111?
Friday, October 1st, 2010
### Solution:
1. Notice that 444 * 111 = 4*111*111
2. The square root of 4 is 2
3. The square root of 111*111 is 111
4. So the square root of 4*111*111 = 2*111
5. The answer is therefore 222
# X has 5 base-ten blocks. She models a number that is less than 999. The number she models has a zero in the tens place and a zero in the ones place. What number is she modeling?
Friday, October 1st, 2010
### Solution:
Base ten blocks are a mathematical manipulative used to learn basic mathematical concepts including addition, subtraction, number sense, place value and counting. You can manipulate the blocks in different ways to express numbers and patterns. Generally, the 3-dimensional blocks are made of a solid material such as plastic or wood and come in four sizes to indicate their individual place value: Units (one’s place), Longs (ten’s place), Flats (hundred’s place) and Big Blocks (thousand’s place). There are also computer programs available that simulate base ten blocks.
Since the number that is being modeled has a zero in the tens place and a zero in the ones place, there are no Longs or Units. Since the number is less than 999, there are no Big Blocks. That leaves only Flats, and there must be five Flats. Since each Flat represents one hundred, the number must be 500.
# Find the minimum number of students needed to guarantee that 4 of them were born: hint use pigeonhole principle. 1. on the same day of the week; 2. in the same month. 3. Which counting principle applies to the questions above?
Friday, October 1st, 2010
### Solution:
1. There are seven days in a week
2. If there are 7 students, they could all have been born on different days of the week
3. If there are 8 students, we could guarantee that 2 of them were born on the same day
4. If there are 15 students, we could guarantee that 3 of them were born on the same day
5. F there are 22 students, we could guarantee that 4 of them were born on the same day
# Suppose a Word is a String of 8 letters of the Alphabet with Repeated Letters Allowed: (5 points, 1 point each) Show all work. Do not Answer with just a Number. 1. How many words are there? 2. How many words end with the letter N? 3. How many words begin with R and end with N? 4. How many words start with A or B? 5. How many words begin with A or end with B
Friday, October 1st, 2010
### Solution:
1. If you have 26 choices for each letter in a word, then there are 26 such one-letter words, 262 two-letter words, 263 three-letter words, etc. There would be 268 eight-letter words.
2. There would be 267 eight-letter words that end with the letter N, which is the same as the number of 7 letter words (just add “N” to the end of each seven-letter word.
3. There would be 266 eight-letter words that begin with R and end with N (just add R to the front and N to the end of all six-letter words.
4. There would be 2*267 eight-letter words that begin with an A or B (add A to the beginning of every seven-letter word, then add B to the beginning of every seven-letter word).
5. There would be 2*267 eight-letter words that begin with A or end with B (add A to the beginning of every seven-letter word, then add B to the end of every 7 letter word).
# If Utility Function is Expressed as U(x,y) = x^0.5 y^0.5 What is the Marginal Utility at Point (64,25) and (49,36)? Treat Y as a Constant. Would the Answer be MUx(x,y) = y. MUx(64,25) = 5 ? and MUx(49,36) = 6?
Tuesday, September 7th, 2010
### Solution:
MUx(x,y) = Derivative of U(x,y), treating y constant.
MUx(64,25) =
MUx(49,36) =
# Use Mathematical Induction to Prove That the Statements are true for Every Positive Integer n. 1*3+2*4+3*5+⋯+n(n+2)=n(n+1)(2n+7)/6
Monday, September 6th, 2010
### Solution:
First, let’s make sure we understand what we mean by “mathematical induction”: “Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers. It is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one.” (Wikipedia) We have pasted below another more detailed explanation of how to create a proof using mathematical induction (see “Appendix” below).
Let’s now take the equation you provided: 1*3+2*4+3*5+…+n(n+2)=
We can show that this is true for n=1: 1*3 = = = 3
Now let us assume that the statement is true for n = k. If it is, then we will prove that it has to be true for n=k+1:
WTS è = +
=
=
=
=
= QED
In the above proof, WTS means “want to show” and QED means “quod erat demonstratum” (“which was to be demonstrated”).
## Appendix
Here’s another more detailed explanation of how to create a proof using mathematical induction:
# A Jeweler Needs to mix an Alloy with 16% Gold Content & an Alloy with a 28% Gold Content to Obtain 32oz. of a new Alloy with a 25% Gold Content. How Many oz. of Each of the Original Alloys Must be Used?
Monday, September 6th, 2010
### Solution:
• A = number of ounces of the first alloy
• B = number of ounces of the second all
• A+B=32
• 0.16 A +0.28 B = 32 x 0.25
• Since A + B = 32, we know that A = 32 – B. You can then re-write the equation above as 0.16 (32-B) + 0.28 B = 32 x 0.25
• This can be simplified to 5.12- 0.16 B + 0.28 B =8
• Or, 0.12 B = 2.88
• Or B = 24
• So A = 8 |
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# How to Simplify Absolute Values
Absolute value is an expression of a number’s distance from 0. It is denoted by two vertical bars on either side of a number, variable, or expression. Anything inside the absolute value bars is called the “argument.” Absolute value bars do not function as parentheses or brackets do, so it is crucial that you use them appropriately.
### Method 1 Simplifying When the Argument Is a Number
1. 1
Determine your expression. Simplifying a numerical argument is an easy process: because absolute zero represents a distance between your number and 0, your answer will always be positive. Start by performing any operations within the absolute value bars to determine your expression.
• For example, say you are trying to simplify the absolute value of an expression, -6 + 3. Since the entire expression is within the absolute value bars, do the addition first. The problem is now to simplify the absolute value of -3.
2. 2
Simplify the absolute value. Once you have performed any operations inside the absolute value bars, you can simplify the absolute value. Whatever number you have as your argument, whether positive or negative, represents a distance from 0, so your answer is that number, and it is positive.
• In the example above, the simplified absolute value is 3. This is true because the distance between 0 and -3 is 3.
3. 3
• For the example above, your number line should look like this.
### Method 2 Simplifying When the Argument Includes a Variable
1. 1
Deal with an argument that is just a variable. If your argument is just a variable by itself, set equal to a number, then simplifying is very easy. Since absolute value represents a distance from 0, your variable could be either the positive number to which it is equal, or it could be the negative version of that number. There is no way to tell, so include both possibilities in your solution.
• For example, say you know that the absolute value of a variable, x, is equal to 3. You cannot tell if x is positive or negative; you are looking for any number whose distance is 3 from 0. Therefore, you solution is either 3 or -3.
• If this is the kind of argument you need to simplify, stop here. Your work is done. If, however, you have an inequality, continue on.
2. 2
Recognize absolute value inequalities. If, however, you are given an argument with a variable, expressed as an inequality, more steps are required. Interpret these inequalities as asking you to find all the possible numbers that could work.
• For example, say you have the following inequality.
This can be interpreted as “Show all numbers whose absolute value is less than 7.” In other words, find all numbers whose distance is 7 from 0, excluding 7 itself. Note that the inequality is constructed as “less than” rather than “less than or equal to.” If it were the later, then 7 itself would be included.
3. 3
Graph a number line. The first thing to do, when faced with an absolute value inequality, is to draw a number line. Tag points corresponding with the numbers with which you are working.
• In the example above, your number line would look like this.
The open circles indicate numbers excluded from your final result. Remember: if the inequality were stated as “greater than or equal to” or “less than or equal to,” then these numbers would be included instead. In that case, the circles would be solid.
4. 4
Consider the numbers on the left side of the number line. Since you don’t know whether your variable is positive or negative, you are really dealing with two possible ranges of numbers: those on the left side of the number line and those on the right. First, consider the numbers on the left. Make the variable negative, and convert your absolute value bars to parentheses. Solve.
• In the example above, you would convert the absolute value bars to parentheses to show that (-x) is less than 7. Multiply both sides of the inequality by -1. Note that when you multiply by a negative number, you must switch the inequality sign (from less than to greater than, or vice versa). Your inequality would look like this.
You now know that for the left side of the number line, x will be greater than -7. On a number line, that would look like this.
5. 5
Consider the numbers on the right side of the number line. Now you can look at the other range of numbers, those that are positive. This is even simpler: make the variable positive, convert your absolute value bars to parentheses.
• In the example above, you would convert the absolute value bars to parentheses to show that (x) is less than 7. No further work is necessary for this step. On a number line, that would look like this.
6. 6
Find the intersect of the two intervals. Once you have considered both sides, you need to determine where the solutions overlap. Draw both intervals on the same number line to get a final result.
• In the example above, you would highlight values greater than -7 and less than 7 (but excluding -7 and 7 themselves). These are your solutions.
## Community Q&A
• How do you simplify problems that have a negative sign outside the brackets?
## Tips
• Remember that absolute value bars are different than brackets or parentheses. You can convert them to parentheses at the appropriate step, but they do not necessarily mean the same thing.
## Article Info
Categories: Mathematics
In other languages:
Español: simplificar valores absolutos, Русский: упрощать (решать) абсолютные величины, Deutsch: Beträge vereinfachen, Italiano: Semplificare i Valori Assoluti, Português: Simplificar Valores Absolutos, Français: simplifier les valeurs absolues
Thanks to all authors for creating a page that has been read 3,743 times. |
In this section we will obtain the expressions for the differentials dH, dA, dG using the relation $dU=TdS-PdV$ from the definitions $H=U+PV$, $A=U-TS$ and $G=H- TS$.
We start by getting dH: $$dH=d(U+PV)=dU+PdV+VdP=TdS-PdV+PdV+VdP=TdS+VdP$$ Similarly we get dA and dG. $$dA=d(U-TS)=dU-SdT-TdS=\cancel{TdS}-PdV-SdT-\cancel{TdS}=-SdT-PdV$$ $$dG=d(H-TS)=dH-TdS-SdT=\cancel{TdS}+VdP-\cancel{TdS}-SdT=-SdT+VdP$$ Therefore the Gibbs equations are: \begin{eqnarray} dU & = & TdS - PdV\\ dH & = & TdS+VdP\\ dA & = & -SdT-PdV\\ dG & = & -SdT+VdP \end{eqnarray}
From equation (29) the following relations can be obtained: $$\left(\frac{\partial U}{\partial S}\right)_v=T\;\;\; \left(\frac{\partial U}{\partial V}\right)_s=-P$$ The first relation is obtained by making dV=0 in (29), and the second with dS=0.
From equation (30) we obtain: $$\left(\frac{\partial H}{\partial S}\right)_p = T\;\;\; \left(\frac{\partial H}{\partial P}\right)_s = V$$ From equation (31) we obtain: $$\left(\frac{\partial A} {\partial T}\right)_v = -S\;\;\; \left(\frac{\partial A}{\partial V}\right)_T = -P$$ From equation (32) we obtain: $$\left(\frac{\partial G }{\partial T}\right)_p = -S\;\;\; \left(\frac{\partial A}{\partial P}\right)_T = V$$ The objective of these equations is to relate thermodynamic properties that are difficult to obtain with others that are easily measurable, such as: $C_p$, $\alpha$ and $\kappa$. |
# How to Calculate the Number of Months to Pay Off a Loan
When you make monthly payments on a loan, it helps to know how long you have left to pay it off so you can better budget your money. By using a formula and some basic information about your loan, you can calculate the number of months until you're free of the debt. This formula works for a typical mortgage, auto loan or personal loan that is fully amortizing, which means its payments include both principal and interest and its balance reduces to zero over a fixed term.
#### Things You'll Need
• Most recent loan statement
• Scientific calculator
Video of the Day
## Step 1
Find your monthly principal and interest payment, outstanding balance and annual interest rate on your most recent loan statement. Exclude any property taxes, insurance or other charges from the payment.
Video of the Day
For example, assume you have a 30-year mortgage with a current balance of \$167,371.45, a monthly payment of \$1,199.10 and a 6 percent annual interest rate.
## Step 2
Divide your annual interest rate by 12 to calculate your monthly interest rate.
In the example, divide 0.06 by 12 to get a monthly interest rate of 0.005:
0.06 / 12 = 0.005
## Step 3
Substitute the loan balance, monthly payment, and monthly interest rate into the loan term formula:
N = –[ln(1 – [(PV * i) / PMT_] ) / ln(1 + _i)]
In the formula, "ln" stands for natural logarithm, a math function used to calculate exponents. The formula also contains four variables:
N = the number of months remaining
PV = present value, or outstanding loan balance
PMT = monthly payment
i = monthly interest rate
In the example, substitute \$167,371.45 for PV, \$1,199.10 for PMT and 0.005 for i:
N = –[ln(1 – [(\$167,371.45 * 0.005) / \$1,199.10] ) / ln(1 + 0.005)]
## Step 4
Multiply the balance by the monthly interest rate and divide the result by the monthly payment in the numerator.
In the example, multiply \$167,371.45 by 0.005 to get \$836.86. Divide \$836.86 by \$1,199.10 to get 0.6979.
N = –[ln(1 – 0.6979) / ln(1 + 0.005)]
## Step 5
Subtract the figures in parentheses in the numerator, and add the figures in the parentheses in the denominator.
In the example, subtract 0.6979 from 1 to get 0.3021 in the numerator. Add 1 and 0.005 to get 1.005 in the denominator:
N = –[ln(0.3021) / ln(1.005)]
## Step 6
Input the figure in parentheses in the numerator into the scientific calculator, and push the natural logarithm button, "ln," to calculate the natural logarithm in the numerator.
In the example, input "0.3021" in the calculator, and push "ln" to get –1.197:
N = –[–1.197 / ln(1.005)]
## Step 7
Input the figure in parentheses in the denominator into the calculator, and push the natural logarithm button to figure the natural logarithm in the denominator.
In the example, input "1.005" in the calculator, and push "ln" to get 0.00499:
N = –(–1.197 / 0.00499)
## Step 8
Divide the remaining figures in parentheses.
In the example, divide –1.197 by 0.00499 to get –239.9:
N = –(–239.9)
## Step 9
Apply the negative sign outside the parentheses to the number in parentheses to calculate the number of months remaining on your loan.
In the example, apply the negative sign to –239.9 to get positive 239.9, or approximately 240 months left on the loan:
N = 240
This means if you make all your payments on time, you will pay off the loan in 240 months, or 20 years, from the current month. |
# 2009 AMC 12A Problems/Problem 22
## Problem
A regular octahedron has side length $1$. A plane parallel to two of its opposite faces cuts the octahedron into the two congruent solids. The polygon formed by the intersection of the plane and the octahedron has area $\frac {a\sqrt {b}}{c}$, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $b$ is not divisible by the square of any prime. What is $a + b + c$?
$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 14$
## Solutions
### Solution 1
Firstly, note that the intersection of the plane must be a hexagon. Consider the net of the octahedron. Notice that the hexagon becomes a line on the net. Also, notice that, given the parallel to the faces conditions, the line must be parallel to precisely $\frac{1}{3}$ of the sides of the net. Now, notice that, through symmetry, 2 of the hexagon's vertexes lie on the midpoint of the side of the "square" in the octahedron. In the net, the condition gives you that one of the intersections of the line with the net have to be on the midpoint of the side. However, if one is on the midpoint, because of the parallel conditions, all of the vertices are on the midpoint of a side. Thus, we have a regular hexagon with a side length of the midline of an equilateral triangle with side length 1, which is $\frac{1}{2}$. Thus, the answer is$\frac {3\sqrt {3}}{8}$, and $a + b + c = 14\ \mathbf{(E)}$.
(Can somebody clarify this and provide a diagram?)
### Solution 2
The plane cuts the octahedron into two congruent solids, which allows us to consider only the cut through the top half (a square pyramid). Because the cut is parallel to one side of the pyramid and must create two congruent solids, we can see that it must take the shape of a trapezoid with right, left, and top sides being 0.5 and a bottom side of 1. This is equivalent to a unit equilateral triangle with the tip cut off at the midpoint. Accordingly, the area of the polygon is $2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{4}$ or $\frac{3\sqrt{3}}{8}$, and $a + b + c = 14$. |
# Change of Variables
The Idea: If you can't solve it here, then move somewhere else where you can solve it, and then move back to the original position.
Like this:
These are the steps:
• Replace an expression (like "2x-3") with a variable (like "u")
• Solve,
• Then put the expression (like "2x-3") back into the solution (where "u" is).
## Example
Here is a simple example: solving (x+1)2 - 4 = 0.
Replace "x+1" with "u" ... Solve ... Replace "u" with "x+1":
## More Examples
OK, you could have solved that without doing that "u=x+1" thing, but here is question where "changing variables" is very useful:
### Example: (x2+2)2 - 2(x2+2) - 15 = 0
It could be hard to solve, but let's try a change of variables:
Let u = x2+2, then our equation becomes:
u2 - 2u - 15 = 0
Which is a quadratic equation that factors nicely into:
(u-5)(u+3)
And the solutions are simply:
u = 5 or u = -3
But wait! We still need to turn "u" back into "x2+2":
First Solution Second Solution
u = 5 u = -3
x2+2 = 5 x2+2 = -3
x2 = 5-2 = 3 x2 = -3-2
x = ±√3 x2 = ±√(-5)
The second solution is imaginary (it has the square root of a negative number), so let us just use the First Solution:
Check: ((√3)2+2)2 - 2((√3)2+2) - 15 = = 52 - 2·5 - 15 = 25-10-15 = 0
Check: ((-√3)2+2)2 - 2((-√3)2+2) - 15 = = 52 - 2·5 - 15 = 25-10-15 = 0
### Example: 3x8 + 5x4 - 2 = 0
It sort of looks Quadratic, but it is degree 8 which could be impossible to solve.
But if we use:
u = x4
Then it becomes:
3u2 + 5u - 2 = 0
Which is Quadratic. And solving it gives:
u = 1/3 or u = -2
Now put the original back again:
First Solution Second Solution
u = 1/3 u = -2
x4 = 1/3 x4 = -2
x = (1/3)1/4 x = (-2)1/4
Answer: x = (1/3)1/4 and x = (-2)1/4
Check: You can check this answer! |
Set Theory - PowerPoint PPT Presentation
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Set Theory
Set Theory
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
1. Set Theory
2. Information
3. Sets and elements A set consists of objects or elements. The elements can be numbers, letters, etc. Elements are listed inside curly brackets. Sets can have a finite or infinite number of elements, or they can be empty. Empty sets, called the null set, have no elements and are denoted by the symbol ∅. A = {1, 2, 3, 4, 5, 6, 7, 8, 9} is a set of single digit positive numbers. B = {red, yellow, blue} is the set of primary colors C = {x|0 < x < 5} is the set of numbers between 0 and 5
4. Subsets What is the set of possible outcomes from rolling a die? A = {1, 2, 3, 4, 5, 6} What is the set of possible even outcomes? Ae = {2, 4, 6} The universal set is the largest possible set for a given scenario. For this example, Ais the universal set. Ae is a subset of all the possible outcomes, A, because is contains some of the elements of A. The null set is a subset of every set. Ae⊆A Subsets are denoted using the ⊆ symbol.
5. Complement The complement of a set is all of the elements in the universal set but not in the set of interest. Complements of sets are denoted using a prime symbol ′. What is the complement of the set of even outcomes when rolling a die? A = {1, 2, 3, 4, 5, 6} universal set: set of even outcomes: Ae = {2, 4, 6} complement of set: Ae′ = {1, 3, 5}, the set of possible odd numbers.
6. Unions The union of two or more sets contains all the elements in all the sets. The union of sets A and B is the elements in either set Aor B or both. Unions of sets are denoted by the symbol ∪. What is the union of set A, the outcomes of rolling a dice, and set B, all positive integers between 7 and 11? set A = {1, 2, 3, 4, 5, 6} set B = {7, 8, 9, 10, 11} A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
7. Intersections The intersection of two or more sets contains all the elements that are in all sets. For example, all the elements in set Aand B. Intersections of sets are denoted by the symbol ∩. What is the intersection of set A, the outcomes of rolling a dice, and set B, all positive integers between and inclusive of 4 and 9? set A = {1, 2, 3, 4, 5, 6} set B = {4, 5, 6, 7, 8, 9} A ∩B = {4, 5, 6,}
8. Disjoint sets Disjoint sets are two or more sets that have no elements in common, therefore the intersection is an empty set. For example: a set of all 3D shapes with a curved surfaces, and B, a set of all polyhedrons, are disjoint sets. The intersection of disjoints sets is the empty set: A∩B = ∅
9. Language of sets
10. Venn diagram
11. Unions and intersections Let A be the set of all letters in the alphabet. A = {A, B, C, …} Let B be the set of all vowels in the alphabet. B = { A, E, I, O, U} LetC be the set of all letters in your name. Draw a Venn diagram to represent these sets. What is the universal set and which are subsets? A is the universal set in this scenario. B is a subset of A. C is a subset of A. Describe B′. B′ is the set of all consonants in the English alphabet. List the elements in B ∩CandB ∪C?
12. Multiple sets
13. Sets in the animal kingdom |
# Tutorial
## Algebra Interaction
In this section, we will discuss the basics on the interaction type Algebra.
The three core principles of the algebra interaction are:
AlgebraKiT automatically simplifies an expression.
The default task when creating an algebra interaction is set to Simplify. With this task selected, AlgebraKiT automatically simplifies the given expressen.
##### Example
With a task $4+5$, AlgebraKiT will evaluate the sum, because $9$ is a simpler expression than $4+5$.
Try it yourself:
• $2x+4x+4$
AlgebraKiT will combine the like terms.
• $x(x+1)$
AlgebraKiT will do nothing. Expanding the brackets is not necessarily ‘simpler’.
• $x(x+1)-x^2$
Now AlgebraKiT will expand the brackets, because it can then cancel terms and reach a simple result.
• $\frac{2x+4}{x^2+2x}$
Using the mathematical knowledge that AlgebraKiT has, it decides to factor both numerator and denominator and cancel factors to reach a simpler result.
0 / 0
The interaction tasks indicate what the student should do.
An important principle in AlgebraKiT is that an author does not specify the answer, but the problem. This allows AlgebraKiT to evaluate intermediate steps and support the student with the derivation.
You can use the dropdown elements to choose from the available tasks.
Try it yourself:
• Solve $2(2x+5)=5-x$
• Expand $x(x+1)$
• Factor $4x^2+12x$
• Write $184300$ into scientific notation.
You can use audience and modes to influence how AlgebraKiT solves the problem.
#### Audience
The audience defines the type of student for which the exercise was meant. It influences what mathematical steps are available and how expressions are written.
##### Note
In a low-level audience such as Arithmetic, AlgebraKiT will not be able to solve more advanced problems, like equations.
#### Modes
##### Examples
AlgebraKiT will interpret $\sin{30}$ by default as the sine of 30 radians. To use degrees, you can set mode gonio_degrees.
To force AlgebraKiT (and the student) to write the fraction $\frac{10}{8}$ into the mixed fraction 1¼, set mode nf_rational.
AlgebraKiT will solve the equation $x^2+5x+6=0$ by default by factoring. To force use of the Quadratic Formula (‘abc-formula’), set mode solve_quadratic_abc. |
# Arithmetic and Geometric Progression Question 1
Problem:
The second, third and sixth terms of an arithmetic progression are consecutive terms of a geometric progression. Find the common ratio of the geometric progression.
My attempt:
I thought of rewriting the terms as $$a+d,a+2d,a+5d$$ where $$a$$ is the first term and $$d$$ is the common difference. Now, since these are consecutive terms of a GP, $$\Longrightarrow \dfrac{a+2d}{a+d}=\dfrac{a+5d}{a+2d}=r$$ where $$r$$ is the common ratio. However, when I cross-multiplied the terms$$\dfrac{a+2d}{a+d}=\dfrac{a+5d}{a+2d},$$ I got $$a^2+4ad+4d^2=a^2+6ad+5d^2$$ I'm unable to proceed any further. Any help would be really appreciated. Many thanks!
• Drop the first term and you have 'the first, second and fifth term'. That doesn't essentially change the problem, but simplifies calculations a little bit. Jul 5, 2015 at 18:25
$$a^2+4ad+4d^2=a^2+6ad+5d^2$$
$$2ad+d^2=0$$
$$d(2a+d)=0$$
$$d=0 \vee 2a+d=0$$
If $d$ is zero, then the ratio is clearly 1.
If $2a+d=0$ is zero, you have $d=-2a$. Substitute this in the first ratio we get $r=\frac{-3a}{-a}=3$.
• You are welcome. Jul 5, 2015 at 17:48
$$a^2+4ad+4d^2=a^2+6ad+5d^2$$
This gives you $$d(2a+d)=0$$ so that either $d=0$ or $d= -2a$.
Your common ratio is given by $$\frac{a+2d}{a+d}$$ Substituting $d=0$ into that gives you the common ratio as $1$. This would make both the geometric and the arithmetic series constant, so we discard that solution.
Substitute $d=-2a$ into the common ratio equation to get $$r = 3.$$
Note that $$a^2+4ad+4d^2=a^2+6ad+5d^2$$ $$\iff 2ad+d^2=0\iff d(2a+d)=0\iff d=0\ \ \text{or}\ \ 2a+d=0$$
More generally, if $i < j < k$, suppose the $i$-th, $j$-th, and $k$-th terms in an arithmetic progression are in a geometric progression with ratio $r$.
I will show that (1) if $ik \ne j^2$ then $r=\frac{k-j}{j-i}$; (2) if $ik = j^2$ then $r=\frac{j}{i}$.
We have $r =\frac{a+jd}{a+id} =\frac{a+kd}{a+jd}$. Cross-multiplying, $(a+id)(a+kd) =(a+jd)^2$ so $a^2+(i+k)ad+ikd^2 =a^2+2ajd+j^2d^2$ or $d(d(j^2-ik)+a(2j-(i+k))) = 0$.
Therefore either $d = 0$ or $d(j^2-ik)+a(2j-(i+k))=0$.
As a check, with $i=1, j=2, k=5$ in the OP, this is $-d-2a=0$, which matches.
So we must have $d(j^2-ik)+a(2j-(i+k))=0$.
There are two cases, depending if $ik \ne j^2$ or $ik = j^2$.
If $ik \ne j^2$, $d =\frac{a(2j-(i+k))}{ik-j^2}$. Then,
$\begin{array}\\ r &=\dfrac{a+jd}{a+id}\\ &=\dfrac{a+j\frac{a(2j-(i+k))}{ik-j^2}}{a+i\frac{a(2j-(i+k))}{ik-j^2}}\\ &=\dfrac{a(ik-j^2)+j(a(2j-(i+k)))}{a(ik-j^2)+i(a(2j-(i+k)))}\\ &=\dfrac{ik-j^2+2j^2-j(i+k)}{ik-j^2+2ij-i(i+k)}\\ &=\dfrac{j^2-ij-jk+ik}{-j^2+2ij-i^2}\\ &=\dfrac{j(j-i)-k(j-i)}{-(i-j)^2}\\ &=\dfrac{(k-j)(j-i)}{(j-i)^2}\\ &=\dfrac{k-j}{j-i}\\ \end{array}$
In the OP's case, we get $\frac{5-2}{2-1} =\frac{3}{1} = 3$, which, again, matches.
If $ik = j^2$, then $a(2j-(i+k))=0$.
If $a=0$, then $r = \frac{j}{i} = \frac{k}{j}$ and any $d$ works.
If $a \ne 0$, then $2j = i+k$. Then $(i+k)^2 = 4j^2 =4ik$ so $i^2+2ik+k^2 =4ik$ or $(i-k)^2 = 0$. This contradicts our original assumption of $i < j < k$, so this can not happen.
Therefore we can always get the ratio, without knowing the initial or increment values.
let 2nd, 3rd and 6th term of AP are a+d, a+2d and a+5d. a+d = A, a+2d = Ar and a+5d =ar^2 a+2d = Ar or (a+d) +d = Ar A + d = Ar A=d/(r-1)-----------1) a+5d =Ar^2 a+2d + 3d = Ar^2 Ar + 3d = A*r^2 A = 3d/r(r-1) ----------2) by eq 1) and 2) d/(r-1) = 3d/r(r-1) 1 = 3/r r =3 |
# Video: Finding the Area of a Triangle on the Cartesian Coordinate Using Determinants
Find the area of the triangle below using determinants.
03:02
### Video Transcript
Find the area of the triangle below using determinants.
Here, we’ve labelled the vertices. Before we find the area using determinants, there is a different way to solve this problem. The area of a triangle is equal to one-half times the base of the triangle times the height of the triangle. So if we would let this be our base, which is four, and then the height is exactly perpendicular has to go straight down or straight up, if it was upwards, from the base, so the height of this triangle would be nine. So we have one-half times four times nine. And one-half of four is two. So two times nine means our area should be 18.
Now, this is not how we were supposed to solve it we’re supposed to solve it using determinants, which we will do. But it’s always a good idea if you know another way to solve a problem; maybe just do it just to double check. To find the area of a triangle using determinants, area is equal to one-half times the determinate of this three-by-three matrix.
So what are 𝘢, 𝘣, 𝘤, 𝘥, 𝘦, and 𝘧? Well, the vertices of the triangle are 𝑥, which is the point 𝘢, 𝘣, 𝑦, which is the point 𝘤, 𝘥, and 𝑧, which is the point 𝘦, 𝘧. So we can go ahead and just choose whichever points we want to be 𝑥, 𝑦, and 𝑧. So 𝑥 is zero, five, 𝑦 is four, five, and 𝑧 is three, negative four. So we can replace 𝘢 and 𝘣 with zero, five; 𝘤, 𝘥 with four, five; and 𝘦, 𝘧 with three, negative four.
So now how do we evaluate the determinant of this three-by-three matrix? Let’s first write down the one-half, so we don’t forget to multiply by at the end. So we take zero times the determinant of the numbers that are not in the row or column with the zero. Then we subtract five times the determinant of the numbers that are not in the row or column of the five. And then we add one times the determinant of the numbers that are not in the row or column of the one.
So how do we find the determinant of a matrix? We take 𝘢 times 𝘥 minus 𝘣 times 𝘤; it’s almost like cross multiplying and subtracting them. So bring down the one-half. Now before we evaluate that first determinant in the pink, it’s zero times that determinant and zero times anything will be zero. So there’s no point of wasting our time evaluating that because we’ll get zero for that anyway. So let’s begin with negative five times that determinant.
So we’ve negative five and then we start to evaluate the determinant. Four times one minus one times three plus one times four times negative four minus five times three. After multiplying the numbers on the innermost parentheses, which are brackets, now we need to subtract these numbers. Next, we need to multiply. So we’ve negative five minus 31. And now, we need to subtractive those. So we have one-half times negative 36, which is negative 18. However, this is an area, and an area is a measurement which needs to be positive. So the area of this triangle will be 18 units squared. |
# Inner Product, Orthogonality and Length of Vectors
The inner product , the condition of orthogonality and the length of vectors are presented through examples including their detailed solutions.
## Definition of the Inner Product of two Vectors
Let vectors x and y be two column vectors (or an n 1 matrix) defined by
The inner product of x and y is a scalar quantity written as x · y defined by
## Properties of the Inner Product
If x, y and z are vectors in Rn and k1 and k2 are scalars, then
1. x · y = y · x
2. (x + y)z = x · z + y · z
3. (k1 x) · (k2 y) = k1 k2 (x · y)
## Orthogonal Vectors
Vectors x and y are called orthogonal if
x · y = 0
## Definition of the Length (or Norm) of a Vector and Unit Vector
The length (or norm ) of vector $$\textbf x = \begin{bmatrix} x_1 \\ x_2 \\ . \\ . \\ . \\ x_n \end{bmatrix}$$ written as $$|| \textbf x ||$$ is given by
$|| \textbf x || = \sqrt {x_1^2 + x_2^2 + .... + x_n^2} = \sqrt {\textbf x \cdot \textbf x}$
From the above definition, we can easily conclude that
$$|| \textbf x || \ge 0$$ and $$|| \textbf x ||^2 = \textbf x \cdot \textbf x$$
A unit vector is a vector whose length (or norm) is equal to 1.
## Pythagorean Theorem
Vectors $$\textbf x$$ and $$\textbf y$$ are orthogonal if and only if
$||x+y||^2 = ||x||^2 + ||y||^2$
## Distance Between two Vectors
The distance between vectors $$\textbf x$$ and $$\textbf y$$ is defined as
$dist(\textbf x,\textbf y) = || \textbf x - \textbf y ||$
## Examples with Solutions
Example 1
Given the vectors $$\textbf x = \begin{bmatrix} -2 \\ 3 \\ 0 \\ -1 \end{bmatrix}$$ , $$\textbf y = \begin{bmatrix} 3 \\ -1 \\ 4 \\ 0 \end{bmatrix}$$ , $$\textbf z = \begin{bmatrix} 2 \\ -1 \\ 4 \\ -7 \end{bmatrix}$$, find
a) $$\textbf x \cdot \textbf y$$ and $$\textbf y \cdot \textbf x$$
b) $$\textbf x \cdot \textbf y + \textbf x \cdot \textbf z$$ and $$\textbf x \cdot (\textbf y + \textbf z)$$
c) $$(3 \textbf x ) \cdot (-2\textbf z)$$
Solution to Example 1
Use the definition given above
a)
$$\textbf x \cdot \textbf y = (-2)(3) + 3(-1) + 0(4) + (-1)0 = - 9$$
Using property 1 of the inner product above
$$\textbf y \cdot \textbf x = \textbf x \cdot \textbf y = - 9$$
b)
$$\textbf x \cdot \textbf y + \textbf x \cdot \textbf z = - 9 + 23 = 14$$
According to property 2 of the inner product
$$\textbf x \cdot (\textbf y + \textbf z) = \textbf x \cdot \textbf y + \textbf x \cdot \textbf z = 14$$
c)
According to property 3 of the inner product
$$(3 \textbf x ) \cdot (-2\textbf z) = (3)(-2) \textbf x \cdot \textbf y = -6 (-9) = 54$$
Example 2
a) Show that vectors $$\textbf x = \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix}$$ and $$\textbf y = \begin{bmatrix} 3 \\ 2 \\ 4 \end{bmatrix}$$ are orthogonal.
b) Find the constant $$a$$ and $$b$$ so that the vector $$\textbf z = \begin{bmatrix} a \\ b \\ 4 \end{bmatrix}$$ is orthogonal to both vectors $$\textbf x$$ and $$\textbf y$$
Solution to Example 2
Calculate the inner product of vectors $$\textbf x$$ and $$\textbf y$$
a)
$$\textbf x \cdot \textbf y = (-2)(3) + 3(2) + 0(4) = 0$$
Since the inner product of vectors $$\textbf x$$ and $$\textbf y$$ is equal to zero, the two vectors are orthogonal.
b)
We first calculate the following inner product
$$\textbf x \cdot \textbf z = -2(a) + 3(b) + 0(4) = -2a + 3b$$
$$\textbf y \cdot \textbf z = 3(a) + 2(b) + 4(4) = 3a + 2b + 16$$
For vector $$\textbf z$$ to be orthogonal to both $$\textbf x$$ and $$\textbf y$$, both inner product calculated above must be equal to zero. Hence the system of equations to solve
$$-2a + 3b =0 \\ 3a + 2b + 16 = 0$$
Solve the above system to to obtain
$$a = -\dfrac{48}{13} , b = - \dfrac{32}{13}$$
Example 3
Let $$\textbf x = \begin{bmatrix} \sqrt 2 \\ 0 \\ 1 \end{bmatrix}$$ and $$\textbf y = \begin{bmatrix} 0 \\ \sqrt 5 \\ 0 \end{bmatrix}$$.
Find $$|| \textbf x ||$$, $$|| \textbf y ||$$ and $$|| \textbf x + \textbf y||$$ and compare $$|| \textbf x ||^2 + || \textbf y ||^2$$ and $$|| \textbf x + \textbf y||^2$$
Solution to Example 3
Use formula for the definition of the length of a vector
$$|| \textbf x || = \sqrt { (\sqrt 2)^2 + 0^2 + 1^2 } = \sqrt 3$$
$$|| \textbf y || = \sqrt { 0^2 + (\sqrt 5)^2 + 0^2 } = \sqrt 5$$
$$|| \textbf x + \textbf y|| = \sqrt { (\sqrt 2)^2 + (\sqrt 5)^2 + 1} = \sqrt 8$$
$$|| \textbf x ||^2 + || \textbf y ||^2 = 3 + 5 = 8$$
$$|| \textbf x + \textbf y||^2 = 8$$
We notice that $$|| \textbf x + \textbf y||^2 = || \textbf x ||^2 + || \textbf y ||^2$$ and that is becaues vectors $$\textbf x$$ and $$\textbf y$$ are orthogonal (the inner product of two vectors is equal to 0) which verify the Pythagorean theorem above. |
# Walk-a-thon 2
Alignments to Content Standards: 7.RP.A.2
Julianna participated in a walk-a-thon to raise money for cancer research. She recorded the total distance she walked at several different points in time, but a few of the entries got smudged and can no longer be read. The times and distances that can still be read are listed in the table below.
1. Assume Julianna walked at a constant speed. Complete the table and plot Julianna’s progress in the coordinate plane.
2. What was Julianna’s walking rate in miles per hour? How long did it take Julianna to walk one mile? Where do you see this information on the graph?
3. Write an equation for the distance $d$, in miles, that Julianna walked in $n$ hours.
4. Next year Julianna is planning to walk for seven hours. If she walks at the same speed next year, how many miles will she walk?
Time in hrs Miles walked
1 Â
2 6.4
 8
5 Â
## IM Commentary
The purpose of this task is for students to translate information about a context involving constant speed into information presented in a table and to find the time it takes to travel a unit distance as well as the distance traveled per unit time. Students then have to translate the information to equations and graphs and then use these mathematical tools to make predictions about the future.
If we only used integer values, the first few parts of this task could come from a 6th grade task. In fact, a 6th grade version of this task is 6.RP Walk-a-thon.
In 7th grade, students work with rational numbers, they are also expected to identify the unit rate as a coordinate of a point on the graph $(1,r)$ and to write an equation that relates the quantities that are in a proportional relationship. The task directly addresses parts (b), (c), and (d) of 7.RP.A.2, and it indirectly addresses part (a) because it requires students to work in the opposite direction. Part (a) requires more complex reasoning than simply producing a table for two quantities in a proportional relationship because it requires them to use information in the table to generate other specific ordered pairs rather than writing any correct ordered pairs they want.
The teacher may wish to explicitly ask students to answer part (d) using all three representations as tools and ask them to comment on the usefulness and limitations of each representation. In this case, the task could be used to launch a whole-class discussion about the connections between tables, equations, and graphs.
The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, the commentary will spotlight one practice connection in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.
Mathematical Practice 5, “Use appropriate tools strategically,†emphasizes a student’s ability to consider a tool’s usefulness and constraints as well as knowing how to use it appropriately. During this task, students engage in MP.5 by considering the connections among tables, graphs, and equations (tools) to find the time it takes to travel a unit distance in addition to the distance traveled per unit time. They extend their understanding by using these same tools to predict how many miles Julianna will walk next year if she maintains the same speed. Although this particular task directs students to use specific tools, it does not prevent them from trying additional tools such as a double number line. This lends itself nicely into a class discussion regarding the pros and cons of using each of these tools.
## Solution
1. We see that the only complete time-distance pair indicates that Julianna walked 6.4 miles in 2 hours. If she walked at a constant speed, we can conclude that Julianna walked 3.2 miles in 1 hour. Using this speed, we can find the remaining values in the table. The easier entry to complete is finding out how far Julianna walked in 5 hours: $5 \text{ hours} \cdot 3.2 \text{ miles/hour} = 16 \text{ miles}$. To find out how long it took to walk 8 hours we can either notice that 8 is half of 16, so it took half the time to walk half the distance or we can divide 8 miles by 3.2 miles per hour to find that it took 2.5 hours (but this calculation is not obvious).
We get the following table.
Time in hrs Miles walked
1 3.2
2 6.4
2.5 8
5 16
Let's plot her progress in the coordinate plane by having her time, in hours, be represented on the horizontal axis, and the number of miles she walked be represented on the vertical axis. Using the rows in the table above as our coordinate points, we get the following graph.
2. We found in part (a) that Julianna walks 3.2 miles in 1 hour.
We find the part of the graph that holds this information, i.e. the point with coordinates (1,3.2).
We can also use the graph to estimate out how long it takes Julianna to walk one mile. Since Julianna walks at a constant rate, we can connect the points with a line to visualize how far she has walked at any time. This allows us to see that it took a little bit less than $\frac13$ of an hour to walk 1 mile.
To find the exact value we can use that it takes Julianna 1 hour to walk 3.2 miles or $\frac{1 \text{ hour}}{3.2 \text{ miles}}$. We would like to know how much time it took to walk 1 mile or $\frac{\text{?? hours}}{1 \text{ mile}}$. Starting with $\frac{1 \text{ hour}}{3.2 \text{ mile}}$ we have $$\frac{1 \text{ hour}\cdot\frac{1}{3.2}}{3.2 \text{ mile}\frac{1}{3.2}} = \frac{0.3125\text{ hour}}{1 \text{ mile}}$$
We can also use a double number line to find the answer.
If we divide the 3.2 mile line segment into 16 pieces of 0.2 mile length we can see that we have 5 of the 0.2 mile pieces to get to 1 mile. We also divide the 1 hour line segment into 16 equal pieces, each of which has length 1/16 mile. We see that 5 of the 1/16 mile pieces will get us to line up with the 1 mile mark. Therefore, in 1 hour Julianna walks $5/16=0.3125$ mile.
3. Julianna walks 3.2 miles every hour, so if we let $n$ be the number of hours walked, we have $d=3.2n$.
4. We have three different representations that we can use to find out how far Julianna would walk in 7 hours. We found in parts (a) and (b) that Julianna’s rate of travel was 3.2 miles per hour. If this stayed constant, we can find how many miles she would walk in 7 hours by extending our table.
Time in hrs Miles walked
1 3.2
2 6.4
2.5 8
5 16
6 19.2
7 22.4
Thus, we see that in 7 hours, Julianna will walk 22.4 miles at that rate.
We can also use the equation $d=3.2n$ with $n=7$ to find that $d=3.2\cdot7=22.4$ miles.
Finally, we can extend the graph to 7 hours on the horizontal axis to estimate that the line passes through the point (7,22.4), although reading off the exact distance is not very easy. |
# How do you solve x² - 2x = 15 by completing the square?
Jun 10, 2018
$x = 5$ or $x = - 3$
#### Explanation:
Although there is a better method, completing the square method is as such:
rearrange the formula to equate 0:
${x}^{2} - 2 x - 15 = 0$
${\left(x - 1\right)}^{2} - 1 - 15 = 0$
Rearrange to get:
${\left(x - 1\right)}^{2} = 16$
Square root on both sides to get:
$\left(x - 1\right) = \pm 4$
Therefore adding 1 to both sides gives us either: $x = 5$ or $x = - 3$
Jun 10, 2018
$x = 5$ and $x = - 3$
#### Explanation:
${x}^{2} - 2 x - 15 = 0$ is in the format of $a {x}^{2} + b x + c = 0$.
As $c$ has already been moved to the left, let's add ${\left(\frac{b}{2}\right)}^{2}$ to both sides and factor:
${x}^{2} - 2 x = 15$
${x}^{2} - 2 x + {\left(\frac{b}{2}\right)}^{2} = 15 + {\left(\frac{b}{2}\right)}^{2}$
Your value of $b$ is the coefficient before the $x$ in $b x$ (from $a {x}^{2} + b x + c = 0$):
${x}^{2} - 2 x + {\left(- \frac{2}{2}\right)}^{2} = 15 + {\left(- \frac{2}{2}\right)}^{2}$
${x}^{2} - 2 x + {\left(- 1\right)}^{2} = 15 + {\left(- 1\right)}^{2}$
${x}^{2} - 2 x + 1 = 15 + 1$
$\left(x - 1\right) \left(x - 1\right) = 16$
${\left(x - 1\right)}^{2} = 16$
Now solve for $x$:
$x - 1 = \pm \sqrt{16}$
$x - 1 = \pm 4$
$x = 4 + 1$ and $x = - 4 + 1$
Therefore, $x = 5$ and $x = - 3$
Jun 10, 2018
$x = - 3 , x = 5$
#### Explanation:
${x}^{2} - 2 x - 15 = 0$
$\Rightarrow {\left(x - 1\right)}^{2} - 1 - 15$
$\Rightarrow {\left(x - 1\right)}^{2} - 16 = 0$
$\Rightarrow {\left(x - 1\right)}^{2} = 16$
$\Rightarrow x - 1 = \pm \sqrt{16}$
$\Rightarrow x - 1 = \pm 4$
$\Rightarrow x = 1 \pm 4$
$\Rightarrow x = 1 - 4$ or $x = 1 + 4$
$\Rightarrow x = - 3 , x = 5$ |
# Evaluating and Graphing Polynomials
## Evaluating Polynomial Functions: A Comprehensive Guide
Polynomial functions are a fundamental concept in mathematics, and understanding how to evaluate them is crucial for solving equations and recognizing different types of graphs. In this article, we will delve into the method of solving polynomials by graphing and identifying the various types of graphs based on their degree.
### What Are Polynomials?
Before diving into evaluation techniques, let's define what a polynomial is. A polynomial is a mathematical expression containing a variable raised to positive whole-number exponents, with each term being multiplied by a coefficient.
### The Standard Form of Polynomial Functions
Polynomial functions follow a standard form, as shown below:
P(x) = axn + bxn-1 + ... + cx + d
It's important to note that polynomial functions are typically written in descending order, with the highest exponent first. Additionally, the exponents in polynomial functions must be positive whole numbers. Negative exponents, such as x-2, do not fit the criteria for polynomial functions.
### Evaluating Polynomial Functions: The Two Methods
When evaluating polynomial functions, there are two primary approaches: direct substitution and synthetic substitution.
#### Direct Substitution
Direct substitution involves substituting a given value for the variable and solving for the result. Here's an example:
• Evaluate P(x) when x = 2:
• P(2) = 3(2)3 + 14(2)2 + 5(2) + 1
• P(2) = 24 + 56 + 10 + 1 = 91
• Thus, P(x) = 91 when x = 2.
#### Synthetic Substitution
Synthetic substitution is a more efficient method for evaluating large polynomial functions. Let's use the same example from above to illustrate this technique:
• Step 1. Write down the coefficients of each term, with the given value of x to the left of the leading coefficient:
• | 2 | 3 | 14 | 5 | 1
• Remember to add any missing terms with a coefficient of zero (0).
• Step 2. Multiply the leading coefficient by the given value of x and bring it down:
• | 2 | 3 | 14 | 5 | 1
• | | 6 | | |
• | 2 | 9 | 14 | 5 | 1
• Add the results and write the sum below the horizontal line:
• | 2 | 9 | 14 | 5 | 1
• | | 6 | | |
• | 2 | 9 | 20 | 5 | 1
• Continue this process until you reach the final value below the line, which is the solution for P(x) when x = 2:
• | 2 | 9 | 20 | 5 | 1
• | | 6 | | |
• | 2 | 9 | 62 | 5 | 1
• | | 12 | | |
• | 2 | 9 | 74 | 5 | 1
• Therefore, P(x) = 74 when x = 2.
### Graphing Polynomial Functions
Graphs are visual representations of polynomial functions, and there are various types based on the degree of the function, which is the highest exponent. Here are some examples:
• Linear graphs have a degree of 1 and appear as straight lines.
• Quadratic graphs have a degree of 2 and form parabolas.
• Cubic graphs have a degree of 3 and are more curved than quadratic graphs.
• Quartic graphs have a degree of 4 and exhibit even more curvature than cubic graphs.
• And so on, with the degree increasing as the graph becomes more curved and changes direction.
### Evaluating Polynomial Functions by Graphing
As the degree of a polynomial function increases, its graph becomes more complex. It's crucial to use key features to accurately represent these functions. These key features include finding the x-intercepts (or zeros) of the function and determining the number of direction changes in the graph.
## Understanding Zeros, Y-Intercepts, and End Behavior of Polynomial Functions
Polynomial functions are expressions that consist of multiple terms with a variable raised to positive whole-number exponents and each term may have a coefficient. These functions can be evaluated, graphed, and analyzed to determine important characteristics such as zeros, y-intercepts, and end behavior.
### Finding the Zeros of a Polynomial Function
The zeros of a function are the values of x that make the function equal to zero. In order to find the zeros, the function is set equal to zero and various methods like factoring, division of polynomials, completing the square, or using the quadratic formula can be used.
For example, let's consider the polynomial function P(x) = 2x3 - 11x2 + 20x - 12. Using the factoring method, we can rewrite this as P(x) = (x-2)(2x-3)(x+2). Thus, the zeros or x-intercepts of this function are 2, 3/2, and -2.
If a zero appears multiple times, the graph will touch the x-axis at that value and then change direction.
### Finding the Y-Intercept of a Polynomial Function
The y-intercept of a polynomial function can be found by substituting x = 0 into the original function. This will give us the y-coordinate where the curve crosses the y-axis.
### Determining the End Behavior of a Polynomial Function
The end behavior of a polynomial function refers to the behavior of the graph on either end. This can be determined using the leading coefficient test.
The leading coefficient of a polynomial is the term with the highest exponent. To determine the end behavior, we need to look at the exponent of this term and the sign of its coefficient.
Odd function (i.e. )
• Positive leading coefficient: The function will point downwards on the left and upwards on the right end of the curve.
• Negative leading coefficient: The function will point upwards on the left and downwards on the right end of the curve.
Even function (i.e. )
• Positive leading coefficient: The function will point upwards on both ends of the curve.
• Negative leading coefficient: The function will point downwards on both ends of the curve.
The end behavior of a polynomial graph can be summarized using the table below:
## Sketching a Polynomial Graph
To graph a polynomial, we can either use the leading coefficient test to determine the end behavior and draw a continuous and smooth curve, or we can make a table of values using direct substitution, plot the points on a coordinate plane, and connect them with a curve.
## Evaluating Polynomials
There are two methods for evaluating polynomials: direct substitution and synthetic substitution. The degree of a polynomial corresponds to the number of direction changes in its graph and the number of x-intercepts. The leading term can also help us determine the end behavior of the curve.
## What Are Polynomials in Math?
Polynomials are expressions with multiple terms containing a variable raised to positive whole-number exponents, and each term may have a coefficient. They can be evaluated, graphed, and analyzed to determine their characteristics.
## How to Evaluate a Polynomial?
Evaluating a polynomial means finding its solution for a given value of x. This can be done using direct or synthetic substitution.
## Examples of Evaluating Polynomials
Problem: Evaluate f(x) = 3x^2 + 4x - 5 at x = 2
Solution: f(x) = 3(2)^2 + 4(2) - 5 = 12 + 8 - 5 = 15
## What Does a Polynomial Graph Look Like?
The degree of a polynomial determines the shape of its graph. A polynomial with a degree of 2 or more will have a continuous and smooth line with possible maximum or minimum points in the middle and approaching positive or negative infinity on either end.
## Mastering Polynomial Functions: A Step-by-Step Guide to Graphing
Graphing polynomial functions may seem intimidating, but with these straightforward steps, you can confidently graph any polynomial function.
• Step 1: Find the Roots
The first step in graphing a polynomial function is to find its roots. This can be done through various methods such as factoring, using the quadratic formula, or completing the square.
• Step 2: Determine the y-Intercept
After finding the roots, the next step is to determine the y-intercept. This is where the graph intersects with the y-axis.
• Step 3: Conduct the Leading Coefficient Test
To determine the end behavior of the polynomial function, conduct the leading coefficient test. This involves considering the degree and leading coefficient of the function.
• Step 4: Sketch the Function
Using the information gathered in the previous steps, sketch the function on a graph. Begin by plotting the roots and y-intercept, then use the leading coefficient test to determine the end behavior of the graph.
Alternatively, you can also create a table of values by substituting various values of x and plotting the corresponding points on the graph. This method helps identify the middle section of the graph, and the curve can then be drawn smoothly, using the leading coefficient test to determine the end behavior.
It is essential to note that the polynomial function does not need to be factored for this approach to work.
Now equipped with these steps, you can confidently conquer polynomial functions. Happy graphing!
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# How to find the length and width of a rectangle when the sides are given in terms of x (making equations).
Sometimes you will be asked to work out the length and width of a rectangle when the sides are given in terms of x and all that you are given is the value of the perimeter. To solve these types of perimeter questions you will need to set up an equation and solve it.
Since you are given the perimeter you need to make an equation for the perimeter of the rectangle. To do this add up the four sides of the rectangle so that you have an expression that involves x and make this equal the perimeter that is stated in the question. All you need to do now is solve this equation and this will tell you the value of x. Finally, sub this value of x back into the expressions for the length and width to give the lengths of the sides that are needed. To summarise:
1. First add up all the lengths of the rectangle to give an expression in terms of x.
2. Next make this equal to the perimeter of the rectangle to give an equation in terms of x.
3. Solve this equation to find x.
4. Sub this value of x into the expressions for the length and width.
5. Check your values by finding the perimeter of the rectangle using your calculated values (this step is optional).
Example
A rectangle has side lengths of 5x + 3 and x – 1. If the perimeter of the rectangle is 64cm work out the actual length and width of the rectangle.
First add up all the lengths of the rectangle to give an expression in terms of x:
5x + 3 + x – 1 + 5x + 3 + x – 1
= 12x + 4
Next make this equal to the perimeter of the rectangle to give an equation in terms of x:
12x + 4 = 64
Next solve this equation to find x:
12x + 4 = 64
12x = 64 -4
12x = 60
x = 60/12
x = 5
Now sub x =5 into the expressions 5x + 3 and x – 1 to give the values of the length and width of the rectangle:
Length = 5x + 3 = 5 × 5 + 3 = 28 cm
Width = x – 1 = 5 -1 = 4 cm
You can also check that these values are correct by adding them up and seeing if they give the perimeter of the rectangle given in the question. If they don’t give the perimeter in the question then you have probably set up the wrong equation or solved the equation incorrectly.
28 + 4 + 28 + 4 = 64cm (which is the perimeter of the rectangle)
So the length of the rectangle is 28 cm and the width of the rectangle is 4 cm. |
When dealing with quadrilaterals, one crucial fact to understand is that in certain cases, a circle can be circumscribed around the shape. This article will delve into the conditions that need to be met for a circle to be circumscribed around a given quadrilateral, as well as the properties of such a circle. We will also discuss the relevance of this concept in various fields, including mathematics, engineering, and architecture.
## Conditions for a Circle to be Circumscribed around a Quadrilateral
In order for a circle to be circumscribed around a quadrilateral, it is necessary for the quadrilateral to be cyclic. What does it mean for a quadrilateral to be cyclic? Let’s break it down:
• A cyclic quadrilateral: A quadrilateral is considered cyclic if all its four vertices lie on the circumference of a circle. This implies that the opposite angles of the quadrilateral are supplementary, i.e., they add up to 180 degrees. In other words, if we draw the diagonals of the quadrilateral, they intersect at a point that lies on the circle.
• Concyclic points: The four vertices of the quadrilateral form what we call concyclic points, which means they all lie on the same circle.
It is important to note that not all quadrilaterals are cyclic; only specific types of quadrilaterals meet this condition. The most well-known cyclic quadrilaterals are the following:
• Rectangle: If a quadrilateral is a rectangle, then it is cyclic because the opposite angles are 90 degrees each, which add up to 180 degrees.
• Square: A square is cyclic for the same reason as a rectangle, as it has 90-degree angles.
• Isosceles Trapezoid: An isosceles trapezoid is cyclic because the base angles are congruent, and the non-parallel sides are extended to meet at a point on the circle.
• Kite: A kite is cyclic because the intersection of its diagonals lies on the circle, and the opposite angles are congruent.
## Properties of the Circumscribed Circle
Once a circle is circumscribed around a quadrilateral, it possesses unique properties that are essential to understanding its geometric significance. Some of these properties include:
• Center: The center of the circumscribed circle is the point of concurrency of the diagonals of the quadrilateral. This property is inherent to all cyclic quadrilaterals.
• Radius: The radius of the circumscribed circle is the distance from the center of the circle to any of the vertices of the quadrilateral. In a cyclic quadrilateral, the radius remains constant, provided the quadrilateral remains unchanged.
• Opposite Angles: The opposite angles of a cyclic quadrilateral are supplementary, meaning they add up to 180 degrees. This property is significant in proving the cyclicity of a quadrilateral.
• Sum of Opposite Angles: The sum of the opposite angles of a cyclic quadrilateral is 360 degrees. This theorem is known as the “Cyclic Quadrilateral Theorem.”
Understanding the properties of the circumscribed circle provides valuable insights into the relationships between the sides, angles, and diagonals of a quadrilateral. Additionally, these properties are foundational in the field of geometry and have practical applications in various disciplines.
## Relevance and Applications
The concept of a circle being circumscribed around a quadrilateral has far-reaching implications and applications in different fields. Some notable applications include:
• Mathematics: In mathematics, the study of cyclic quadrilaterals and their properties is fundamental in geometry and trigonometry, particularly in proving geometric theorems and solving complex problems related to circles and quadrilaterals.
• Engineering: Cyclic quadrilaterals and circumscribed circles are relevant in engineering disciplines, such as structural and civil engineering. Engineers use these concepts to analyze the stability and forces acting on various structures, including bridges, towers, and trusses.
• Architecture: Architects often utilize the principles of circumscribed circles and cyclic quadrilaterals in designing aesthetically pleasing and structurally sound buildings. These concepts help architects understand the relationships between different components of a structure to achieve balance and stability.
• Surveying and Mapping: Surveyors and cartographers rely on the geometric properties of circumscribed circles to accurately measure and map land areas, as well as in defining boundaries and angles in surveying projects.
• Computer-Aided Design (CAD): In CAD software, understanding the concept of circumscribed circles is essential for creating precise and efficient geometric designs, as well as for simulating real-world scenarios in virtual environments.
The practical implications of circumscribed circles and cyclic quadrilaterals underscore their importance in various professional fields, making them not only theoretical concepts but also valuable tools for problem-solving and decision-making.
## Conclusion
In conclusion, the concept of a circle being circumscribed around a quadrilateral is a fundamental aspect of geometry with wide-ranging applications in mathematics, engineering, architecture, and other disciplines. Understanding the conditions for a quadrilateral to be cyclic, as well as the properties of the circumscribed circle, provides valuable insights into the relationships and behaviors of geometric shapes. Moreover, the practical relevance of these concepts underscores their significance in solving real-world problems and advancing technological innovations. As such, the study and application of circumscribed circles and cyclic quadrilaterals continue to be essential in the fields of science, technology, and design.
Android62 is an online media platform that provides the latest news and information about technology and applications. |
# Make x The Subject
Here we will learn about making x the subject of an equation or a formula.
There are also rearranging equations worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What does it mean to make x the subject?
Making x the subject of a formula or equation means rearranging the equation or formula so that we have a single x variable equal to the rest of it.
For example,
Make x the subject.
Step-by-step guide: Rearranging equations
## How to make x the subject
In order to make x the subject:
1. Isolate the variable by:
– Removing any fractions by multiplying by the denominator(s).
– Dividing by the coefficient of the variable.
– Adding or subtracting terms near to the variable.
– Taking a root or power of both sides of the equation.
2. Rearrange the equation so each term containing x is on the same side of the =.
3. Factorisation may be needed if there are multiple terms containing x.
E.g. factorise 2x + 3xy to x(2+3y)
*not always required*
4. Perform an operation to ensure only a single x variable is left as the subject.
## Make x the subject examples
### Example 1: multiple step but with single variable
Make x the subject.
Step 1:
Divide each side of the equation by 3
Step 2:
Subtract a from each side of the equation
### Example 2: questions involving x2
Step 1:
Subtract t from both sides of the equation
Step 2:
Square root each side
Remember the square root can be + or \;−
### Example 3: questions involving √x
Step 1:
Add 9a to both sides of the equation
Step 2:
The inverse operation of ‘square root’ is to ‘square’ each side
Step 3:
The inverse operation of multiply is divide, so divide both sides by 5
### Example 4: factorisation of the variable is required
Step 1:
Multiply each side of the equation by the denominator
Step 2:
Expand the bracket on the left hand side of the equation and rearrange the equation. This will help us to get all terms with x onto one side of the equation
Step 3:
Factorise the left side of the equation so we have a single variable x.
Step 4: Divide by (a - 5b)
This will leave x as the subject of the equation
### Example 5: factorisation of the variable is required
Step 1:
Multiply each side of the equation by the denominator of the other side.
Step 2:
Expand the bracket on the LHS and RHS of the equation and rearrange. This will help to get all terms with x onto one side of the equation
Step 3:
Factorise the left side of the equation so that we are left with only one of the variable x.
Step 4:
Divide both sides by (b + 8) to leave x as the subject
### Common misconceptions
• Perform the same operation to both sides
When we perform an operation to the left hand side of the equation we have to perform the same operation to the right hand side.
E.g
$\frac{x}{2}=3 y+5$
To isolate the variable x we need to multiply both sides by 2.
$x=6y+5$
This is wrong because we have only multiplied the 3y by 2.
The correct answer should be:
$x=2(3y+5)=6y+10$
This is correct because we have multiplied everything by 2 using brackets.
• Inverse operations
E.g.
To isolate the variable x from 5x we need to perform the inverse operation.
5x = 5 × x, so the inverse operation is divide by 5.
E.g.
$\frac{x}{5}=5 \div x$
so the inverse operation is × 5.
E.g.
To isolate the variable x from x+5 we need to perform the inverse operation:
The inverse operation of +5 is −5.
E.g.
To isolate the variable x from x−5 we need to perform the inverse operation:
The inverse operation of −5 is +5.
• Factorising
To make x the subject there can only be one x visible at the end. We need to get all x’s on one side of the equal sign and then factorise.
E.g.
$x = ax+y$
\begin{aligned} x-a x &=y \\ x(1-a) &=y \\ x &=\frac{y}{1-a} \end{aligned}
• Square rooting a term
When we square rooting a number/variable as an inverse operation the answer can be positive or negative.
E.g.
\begin{align} & {{x}^{2}}=4 \\ & x=\pm \sqrt{4}=\pm 2 \\ \end{align}
$\sqrt{x} \text { should be written as } \pm \sqrt{x}$
### Practice make x the subject questions
1.Make x the subject of the formula.
y = 6(x+8)
x = y – \frac{3}{4}
x = \frac{y}{6} + 8
x = \frac{y}{6} – 8
x = y – 8
y = 6(x + 8)
Divide both sides by 6
\frac{y}{6} = x + 8
Then subtract 8 from both sides
x = \frac{y}{6} – 8
2.Make x the subject of the formula.
3p={x}^2-4b
x=\pm\sqrt{3b+4p}
x=\pm\sqrt{4b-3p}
x=4b+3p
x=\pm\sqrt{4b+3p}
3p={x}^2-4b
Add 4b to both sides
3p+4b=x^{2}
Square root both sides
x=\pm\sqrt{4b+3p}
3. Make x the subject of the formula.
6g=\sqrt{7x-8}
x=\frac{6{g}^2+8}{7}
x=\frac{36{g}^2-8}{7}
x=\frac{36{g}^2+8}{7}
x=\frac{6g+8}{7}
6g=\sqrt{7x-8}
Square both sides
(6g)^{2}=(\sqrt{7x-8})^{2}
36g^{2}=7x-8
Add 8 to both sides
36g^{2}+8=7x
Divide both sides by 7
x=\frac{36{g}^2+8}{7}
4.Make x the subject of the formula.
y=\frac{4x-f}{5x}
x=\frac{f}{5y-4}
x=\frac{-f}{5y+4}
x=\frac{f}{5y+4}
x=\frac{-f}{5y-4}
y=\frac{4x-f}{5x}
Multiply both sides by 5x
5xy=4x-f
Subtract 4x from both sides
5xy-4x=-f
Factorise the left hand side
x(5y-4)=-f
Divide both sides by the quantity in the bracket
x=\frac{-f}{5y-4}
5. Make x the subject of the formula.
\frac{y}{3}=\frac{6-2x}{x+3}
x=\frac{18+3y}{y+6}
x=\frac{18-3y}{y-6}
x=\frac{18-3y}{y+6}
x=\frac{18+3y}{y-6}
\frac{y}{3}=\frac{6-2x}{x+3}
Multiply each side of the equation by the denominator of the other side.
xy+3y=18-6x
To both sides, add 6x and subtract 3y
xy+6x=18-3y
Factorise the left hand side
x(y+6)=18-3y
Divide by the quantity in the bracket
x=\frac{18-3y}{y+6}
### Make x the subject GCSE Questions
1. Make x the subject of the formula
y=5x-7
(2 marks)
y+7=5x
(1)
\frac{y+7}{5} = x
(1)
2. Make x the subject of the formula
z^{2}=x^{2}-5 a y
(2 marks)
z^{2}+5 a y=x^{2}
(1)
\pm\sqrt{z^{2}+5 a y}=x
(1)
3. Make x the subject of the formula
y=\frac{3(t+5 x)}{x}
(4 marks)
y x=3 t+15 x
(1)
y x-15 x=3 t
(1)
x(y-15)=3 t
(1)
(1)
## Learning checklist
You have now learned how to:
• Understand and use standard mathematical formulae
• Rearrange formulae to change the subject
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme. |
Summary of trigonometric identities
You have seen quite a few trigonometric identities in the past few pages. It is convenient to have a summary of them for reference. These identities mostly refer to one angle denoted θ, but there are some that involve two angles, and for those, the two angles are denoted α and β.
The more important identities.
You don’t have to know all the identities off the top of your head. But these you should.
Defining relations for tangent, cotangent, secant, and cosecant in terms of sine and cosine.
The Pythagorean formula for sines and cosines. This is probably the most important trig identity.
Identities expressing trig functions in terms of their complements. There's not much to these. Each of the six trig functions is equal to its co-function evaluated at the complementary angle.
Periodicity of trig functions. Sine, cosine, secant, and cosecant have period 2π while tangent and cotangent have period π.
Identities for negative angles. Sine, tangent, cotangent, and cosecant are odd functions while cosine and secant are even functions.
Ptolemy’s identities, the sum and difference formulas for sine and cosine.
Double angle formulas for sine and cosine. Note that there are three forms for the double angle formula for cosine. You only need to know one, but be able to derive the other two from the Pythagorean formula.
The less important identities.
You should know that there are these identities, but they are not as important as those mentioned above. They can all be derived from those above, but sometimes it takes a bit of work to do so.
The Pythagorean formula for tangents and secants. There’s also one for cotangents and cosecants, but as cotangents and cosecants are rarely needed, it’s unnecessary.
Identities expressing trig functions in terms of their supplements.
Sum, difference, and double angle formulas for tangent.
The half angle formulas. The ones for sine and cosine take the positive or negative square root depending on the quadrant of the angle θ/2. For example, if θ/2 is an acute angle, then the positive root would be used.
Truly obscure identities.
These are just here for perversity. No, not really. They have some applications, but they’re usually narrow applications, and they could just as well be forgotten until needed.
Product-sum identities. This group of identities allow you to change a sum or difference of sines or cosines into a product of sines and cosines.
Product identities. Aside: weirdly enough, these product identities were used before logarithms were invented in order to perform multiplication. Here’s how you could use the second one. If you want to multiply x times y, use a table to look up the angle α whose cosine is x and the angle β whose cosine is y. Look up the cosines of the sum α + β. and the difference α – β. Average those two cosines. You get the product xy! Three table look-ups, and computing a sum, a difference, and an average rather than one multiplication. Tycho Brahe (1546–1601), among others, used this algorithm known as prosthaphaeresis.
Triple angle formulas. You can easily reconstruct these from the addition and double angle formulas.
More half-angle formulas. These describe the basic trig functions in terms of the tangent of half the angle. These are used in calculus for a particular kind of substitution in integrals sometimes called the Weierstrass t-substitution. |
# 1983 AHSME Problems/Problem 4
## Problem 4
In the adjoining plane figure, sides $AF$ and $CD$ are parallel, as are sides $AB$ and $EF$, and sides $BC$ and $ED$. Each side has length $1$. Also, $\angle FAB = \angle BCD = 60^\circ$. The area of the figure is
$\textbf{(A)} \ \frac{\sqrt 3}{2} \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3}\qquad \textbf{(E)}\ 2$
## Solution
$[asy] pair A, B, C, D, E, F; A = (0, 1.732); B = (0.5, 0.866); C = (0,0); D = (1, 0); E = (1.5, 0.866); F = (1, 1.732); draw(A--B--C--D--E--F--A); label("A", A, NW); label("B", B, 3W); label("C", C, SW); label("D", D, SE); label("E", E, E); label("F", F, NE); draw(B--D, dashed+linewidth(0.5)); draw(B--E, dashed+linewidth(0.5)); draw(B--F, dashed+linewidth(0.5)); [/asy]$
By rotating the diagram and drawing the dotted lines, we see that the figure can be divided into four equilateral triangles, each of side length $1$. The area of one such equilateral triangle is $\frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}$, which gives a total of $4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}$, or $\boxed{D}$.
1983 AHSME (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions |
## Friday, August 30, 2002
### Experience Mathematics # 11 -- Counting
How can you tell if there are enough chairs in your classroom so that every student has a chair? One way is to count the chairs and the number of students in your class. A simpler way is to ask each student to sit down. If all the students are able to sit down, and no chair is left over, the number of students and chairs is equal.
This fundamental idea is at the heart of mathematics, because it deals with counting. Just like there is a weight that represents a kilogram, there is a unit in mathematics that represents a number $n$. Let $I(n)$ be the set containing the first $n$ natural numbers. The set $I(n)$ is the “unit” that represents the number $n$. For example, the set $A$ with elements $a, b, c, \dots, z$, has $26$ elements because this set can be put in one to one correspondence with $I(26)$.
But what about infinite sets? The set $N$ of natural numbers is a unit for infinite sets. We say a set $S$ is countable if it can be put in one to one correspondence with $N$. It is remarkable that the following subsets of $N$ are countable. Can you find the one to one correspondence between $N$ and these sets?
1. The set of all even numbers.
2. The set with elements $1, 4, 7, 10, \dots$
3. The set of all rational numbers.
Can you show that the set of prime numbers is countable?
When it comes to infinite sets, a part can be equal to the whole.
## Wednesday, August 21, 2002
### Experience Mathematics #10 -- Sets
A fundamental object in mathematics is a set. You can think of a set as a collection of objects. We are familiar with the set $N$ of natural numbers. The empty set is a set with no elements.
We say that a set $B$ is a subset of a set $A$ if each element of $B$ is also in $A$. Two sets are equal if they are subsets of each other. The empty set is a subset of every set. For example the set of even numbers $\left\{ 2, 4, 6, \dots\right\}$ is a subset of $N$.
Q1. How many subsets does the empty set have? (Hint: There is atleast one subset.)
The members of the set are called its elements.
Let $I(n)$ denote the set containing the first $n$ natural numbers. The set $I(n)$ has n elements.
Q2. List all the subsets of the set $I(1)$. How many subsets does $I(1)$ have?
Q3. List all the subsets of the set $I(2)$. How many subsets does $I(2)$ have?
Q4. List all the subsets of the set $I(3)$. How many subsets does $I(3)$ have?
Suppose you delete $3$ from each subset of $I(3)$. What do you get? Use this idea to do Q5.
Q5. List all the subsets of the set $I(4)$. How many subsets does $I(4)$ have?
Q6. How many subsets does $I(n)$ have? Guess the answer from your experiments above.
If you cannot find a pattern, you must have made a mistake in listing the sets earlier.
## Friday, August 16, 2002
### Experience Mathematics #9 -- Prime numbers
The prime numbers are numbers that only have $1$ and themselves as factors. The first few prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, \dots$. One of the most important ideas in the theory of numbers is that a number can be written in a unique way as a product of prime powers. For example,
$4=2^2$, $15$ is $3$ times $5$, and $20$ is $2\times 5$.
The uniqueness of the prime factorization is used to show that the square root of $2$ is not a rational number. A rational number is a number in the form $p/q$, where both $p$ and $q$ are integers, and $q$ is not equal to zero. By doing the following questions, you can prove that the square root of two is an irrational number. The proof is by contradiction. We assume that the square root of $2$ can be written as a fraction, and then show our assumption implies a false statement.
Q1. Suppose the square root of $2$ equals $p/q$ for some integers $p$ and $q$. Show that $2q^2=p^2$.
Q2. Show that the power of $2$ in the prime factorization of the number $2q^2$ is an odd number.
Q3. Show that the power of $2$ in the prime factorization of the number $p^2$ is an even number.
The above two statements are contradictory. Thus our assumption, that the square root of $2$ is a rational number, must be false.
Can you generalize this proof to prove that the square root of $3$ and $5$ are also irrational numbers?
## Friday, August 09, 2002
### Experience Mathematics #8 -- Finding patterns
Mathematics is all about finding patterns. Begin learning to spot patterns in numbers today, and maybe one day you can solve a big mathematical problem—like Maninder Agarwal, Neeraj Kayal, and Nitin Saxena, computer scientists at IIT, Kanpur. These mathematical stars have just announced a new algorithm, by which they can tell whether a number is a prime number. An implementation of this algorithm will mean that everyone will have to re-look at computer programs used to keep messages transferred on the Internet confidential.
In today’s activity, find the next three terms in the following sequences.
Q1. $1, 3, 5, 7,\dots$
Q2. $2, 4, 6, 8, \dots$
Q3. $1, 4, 9, 16, \dots$
Q4. $2, 3, 5, 7, 11, 13, 17, \dots$
Q5. $1, 2, 4, 8, 16, \dots$
Q6. $1, 4, 1, 4, 2, \dots$
Q7. $3, 6, 9, 12, 15, 18, \dots$
Q8. $0, 3, 8, 15, 24, \dots$
Q9. $0, 1, 1, 2, 3, 5, 8, \dots$
Q10. $1, 3, 6, 10, 15, 21, \dots$
Q11. $2, 8, 20, 40, 70, \dots$
Q12. $1, 5, 14, 20, 55, \dots$
## Friday, August 02, 2002
### Experience Mathematics # 7 -- A continued fraction
We all know that the square root of $2$ is an irrational number. That is to say, it cannot be written as a fraction $p/q$. Here $p$ and $q$ are integers, and $q$ is not zero. But it can be written in the form of a continued fraction.
Here is how you can discover the continued fraction representation of the square root of $2$. You will need a calculator to do the calculations. Carefully understand the following calculations.
Note that the $.41421\dots$ starts repeating, and we get a fraction that looks like
You can chop off the fraction at any point and get a fraction that is approximately equal to the square root of $2$.
As for this week’s activity, do a similar calculation (using a calculator) for the square root of $3$ and the square root of $5$ and find the continued fraction representation of these irrational numbers. |
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# How Accurate Is Parallax?
Home » Experiments » How Accurate Is Parallax?
Contents
## Introduction
In this science experiment, we will try to find out how accurate is parallax? To get the answer to this question, we will be determining the accuracy of the parallax equation in measuring the distance between the star and the earth. We will take stars as dependent variables and distance as an independent variable for performing this experiment.
We hypothesized that as the distance increases, accuracy decreases.
## Aim
To determine the accuracy of parallax in measuring the distance between the stars and the earth.
## Theory
1. The apparent difference in the position of an object on viewing from two different lines of sight is called parallax.
2. The method that uses two apparent points to measure the distance from the observer is called the parallax equation.
The distance measured to the star = 1/ parallax angle.
d = 1/p.
1. Candles
2. Balls
3. A Pole
4. Notebook
## Procedure
Step 1. For representing stars, use candles. And for representing the earth and the sun, use balls.
Step 2. Place four candles (four stars) each at a distance of 0.9398, 2.921, 4.9276, and 5.5372 from the ball (the earth).
Step 3. To show the relative stars, use a pole.
Step 4. Place the earth in the position of January. And obtain the degree of angle for each of the four stars.
Step 5. Place the earth in the position of July. And obtain the degree of angle for each of the four stars.
Step 6. Use the baseline as 1.8288 meters.
Step 7. Calculate the value by using the parallax equation.
Step 8. Record your observations.
## Observation
1. We calculated all the values in inches but later converted them to meters.
2. We placed our first star 0.9398 meters away from the earth. But by measuring with a parallax equation, it came out to be 0.87122 meters.
3. The star that we placed at a distance of 2.921 came out to be 3.302 meters on calculating with the parallax equation.
4. The star that we placed at a distance of 4.9276 meters from the earth came out to be 4.28752 meters with the parallax equation.
5. And, the last star that we placed at a distance of 5.5372 meters from the earth came out to be 3.556 meters with the parallax equation.
## Result
1. Our hypothesis, which suggested that on increasing the distance, the accuracy of the parallax equation decreases, was right.
2. Our first measurement was almost similar (of 0.6858 difference) with the parallax value.
3. The farthest star has the largest difference of 1.9812 between the parallax measurement and our measurement.
## Precautions
1. Put all the values correctly in the parallax equation.
2. Record your observation precisely.
## Conclusion
In this experiment, to find out how accurate is parallax. We determined the accuracy of the parallax equation in measuring the distance between the earth and the star.
## VIVA Questions With Answers
Q.1 What was the aim of your experiment?
ANS. We aimed to find out how accurate is parallax to check the accuracy of the parallax equation.
Q.2 What is the formula for the parallax equation?
ANS. The distance to a star, d = 1/p.
Where p is the parallax angle and measured in arc-second and
d in parsec.
Q.3 What was the result of your experiment?
ANS. We saw that on increasing the distance, the accuracy of the parallax equation decreases.
## Didgeridoo With Specific Base Frequency
In this science experiment, we will build a didgeridoo, a type of wind instrument that produces a...
## Determining the Type of Particle in an Air Sample by Using Laser Light Scattering
Introduction In this experiment(Determining the Type of Particle in an Air Sample by Using Laser...
## Cosmic Ray Shower Array Reconstruction
Introduction In this Cosmic Ray Shower Array Reconstruction experiment, we will analyze the event... |
8.17: Lines of Symmetry
Difficulty Level: At Grade Created by: CK-12
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Have you ever seen an art exhibit made of nuts?
Kasey saw an exhibit made up of walnuts. Here is a picture of a walnut. Looking at this picture, can you tell if this walnut has line symmetry or rotational symmetry?
Do you know the difference?
This Concept will teach you all about the different types of symmetry. You will know how to answer these questions by the end of the Concept.
Guidance
We have mentioned that reflected figures are symmetrical, and that the line that acts as a mirror is called the line of symmetry.
Symmetry means that when we divide a figure in half, the halves are congruent. In other words, a figure is symmetric if its outlines mirror each other.
Look at the figure below. Imagine you can fold it in half. When you fold it, do the outlines of each half match? They do, so this figure has symmetry.
When we “unfold” the figure, we have two congruent halves. The “fold” line is the line of symmetry. It divides the figure into halves that the two haves are mirror images of each other! Every part on one half “mirrors” or corresponds to a part on the other half.
Now let’s try another figure. Can we fold it perfectly in half?
We can try to fold this figure a bunch of different ways, but it does not have a line of symmetry.
Rotational symmetry is a different kind of symmetry. It means that when we rotate a figure, the figure appears to stay the same. The outlines do not change even as the figure turns. Look at the figure below.
We can tell the figure has been rotated because the dot moves clockwise. However, the outlines of the figure have not changed. This figure has rotational symmetry because every time we turn it, one of the arms of the star always faces up.
The figure below, on the other hand, does not have rotational symmetry. Can you see why?
No matter how we work with this figure it will look different each time it is rotated. Therefore, we know that it does not have rotational symmetry.
Does each of the following figures have line symmetry, rotational symmetry, both, or neither?
Example A
Cross
Solution: Line Symmetry
Example B
Arrow
Solution: Neither
Example C
Solution: Line Symmetry
Here is the original problem once again.
Kasey saw an exhibit made up of walnuts. Here is a picture of a walnut. Looking at this picture, can you tell if this walnut has line symmetry or rotational symmetry?
Do you know the difference?
Line symmetry involves dividing a figure in halves. You can split a figure horizontally, vertically or on the diagonals and one half is a mirror image of the other half.
This walnut does have one vertical line of symmetry. You can only divide it in half vertically and have it create a mirror image.
The walnut does not have rotational symmetry. If you turn the walnut, then the view and image of the walnut is different.
Vocabulary
Here are the vocabulary words in this Concept.
Rotation
a turn. The figure turns either clockwise or counterclockwise.
Line of Symmetry
the line that a figure reflects over. Also the line that divided a figure in half showing line symmetry.
Rotational Symmetry
the way a figure looks does not change no matter how you rotate it.
Symmetry
a figure that can be divided into two congruent halves is said to have symmetry.
Guided Practice
Here is one for you to try on your own.
Does this figure have line symmetry, rotational symmetry, both or neither.
This figure has line symmetry. Take a look.
When we can divide a figure or an object into two even matching halves, we say that the figure has line symmetry.
This figure can be divided in one way, vertically. If we tried to divide it horizontally, the two sides would not match.
Divided this way, the top half does not match the bottom half.
This figure does not rotate, but it does have line symmetry.
Video Review
Here is a video for review.
Practice
Directions: Answer each of the following questions true or false.
1. A reflection has rotational symmetry.
2. A square has line symmetry and rotational symmetry.
3. If a figure is a transformation than it always rotates clockwise.
4. A slide is also called a translation.
5. A flip has a line of symmetry because it is a reflection.
6. A rotation or turn always moves clockwise and never counterclockwise.
7. A star has rotational and line symmetry.
8. An regular octagon has rotational and line symmetry.
Directions: Tell whether the figures below have line symmetry, rotational symmetry, both, or neither.
Directions: Draw the second half of each figure.
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
TermDefinition
Bilateral Symmetry This occurs when a figure can only be divided into equal halves on one line. Such a figure has one line of symmetry.
Line Symmetry A figure has line symmetry or reflection symmetry when it can be divided into equal halves that match.
Lines of Symmetry Lines of symmetry are the lines that can be drawn to divide a figure into equal halves.
Reflection A reflection is a transformation that flips a figure on the coordinate plane across a given line without changing the shape or size of the figure.
Rotational Symmetry A figure has rotational symmetry if it can be rotated less than $360^\circ$ around its center point and look exactly the same as it did before the rotation.
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I. Calculating the squares of 2-digit numbers ending in 5:
examples: a) what is the square of 45?
b) what is 75 x 75?
the "shortcut" solution:
Step 1: take the "tens" place of the number we're calculating
Step 2: multiply that number with that number + 1
Step 3: append the "25" to the answer arrived from step 2
It's easier to illustrate:
a) what is the square of 45?
i) the tens place of "45" is the number "4"
ii) multiply "4" by "4+1" ==> 4 x 5 = "20"
iii) append "25" to "20" ==> 2025
the answer is 2025!
b) what is 75 x 75?
i) "7"
ii) 7 x 8 = 56
iii) 5625
c) what is the square of 55?
d) 95 x 95 = ?
II. Multiplying two numbers which are between 10 and 19:
examples: e) what is 12 x 14?
f) 18 x 13 = ?
the "shortcut" solution:
Step 1: take one of the numbers and add it to the "ones" place digit
of the other number (remember the answer: this number is
actually multiplied by 10, but you don't really need to
do that since multiplying by 10 is a trivial operation)
Step 2: multiply the "ones" place digits (remember the answer)
Step 3: "append" the result from step 2 to step 1 (see illustration below)
illustration:
e) what is 12 x 14?
i) we take "12" and add it to "4": 12 + 4 = 16
(we could have taken the other number "14" and added it to "2",
and we get the same answer of "16")
ii) we multiply "2" by "4": 2 x 4 = 8
iii) append the number "8" to "16": 168
16
(or you could think of it this way) + 8
=======
168
the answer is 168!
note: in (i), the answer of 16 actually implies 160
in (ii), it's probably safer to think of the answer as "08"
f) 18 x 13 = ?
i) 18 + 3 = 21 (or: 13 + 8 = 21)
ii) 8 x 3 = 24
iii) 21
+ 24
======
234
the answer is 234!
exercises (do these mentally!):
g) 16 x 12 = ?
h) 17 x 19 = ?
III. Multiplying two numbers whose tens place are the same number:
examples: i) what is 23 x 24?
j) 79 x 74 = ?
the "shortcut" solution: (well, it's not really a short "shortcut",
but this method should be easier for mental calculation)
Step 1: take one of the numbers and add it to the other number's
"ones"-digit (remember the answer)
Step 2: multiply the answer from step 1 by the "tens"-place of the
original numbers (remember the answer)
Step 3: multiply the "ones" place of each of the two numbers
Step 4: add (append) the answer from step 3 to the answer from step 2
(see illustration below)
i) what is 23 x 24?
i) "23" + "4" = "27" (or: 24 + 3 = 27)
ii) the "tens" place of the numbers "23" or "24" is the number "2",
multiply "27" (answer from step 1) by "2": 27 x 2 = 54
iii) multiply "3" x "4" (the "ones" digits of the two numbers):
3 x 4 = 12
iv) append/add the answers from step 2 and step 3:
54
+ 12
=======
552
the answer is 552!
j) 79 x 74 = ?
i) "79" + "4" = 83 (or: 74 + 9 = 83)
ii) 83 x "7" = 581 [I presume that you can do multiplications
of a 1-digit number by a 2-digit number
mentally with ease; if not, you SHOULD be
practicing mental multiplications of 1-digit
by 2-digit numbers before you proceed!!]
iii) "9" x "4" = 36
iv) 581
+ 36
=======
5846
the answer is 5846!
exercises (do these MENTALLY!):
k) 21 x 22 = ?
l) 46 x 45 = ?
m) 86 x 89 = ? |
# Rational Numbers Worksheet Grade 6 Word Problems
A Logical Numbers Worksheet can help your son or daughter be more acquainted with the ideas powering this ratio of integers. In this particular worksheet, individuals should be able to resolve 12 diverse difficulties relevant to realistic expressions. They will learn how to flourish several numbers, class them in pairs, and determine their products. They are going to also practice simplifying reasonable expression. As soon as they have enhanced these principles, this worksheet will be a valuable device for furthering their reports. Rational Numbers Worksheet Grade 6 Word Problems.
## Realistic Phone numbers are a ratio of integers
The two main kinds of phone numbers: irrational and rational. Realistic figures are described as complete numbers, in contrast to irrational figures tend not to perform repeatedly, and also have an limitless quantity of numbers. Irrational amounts are low-absolutely no, low-terminating decimals, and sq origins which are not excellent squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To define a rational variety, you need to understand such a logical variety is. An integer is a whole variety, along with a reasonable number is really a ratio of two integers. The percentage of two integers is the amount at the top divided up from the amount on the bottom. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be created in to a portion
A realistic quantity has a numerator and denominator which are not zero. This means that they may be indicated being a small fraction. In addition to their integer numerators and denominators, reasonable amounts can in addition have a unfavorable worth. The bad importance needs to be located left of and its particular total value is its length from no. To streamline this instance, we shall state that .0333333 is actually a fraction which can be created as being a 1/3.
As well as negative integers, a reasonable quantity may also be made right into a small percentage. For example, /18,572 is actually a realistic number, whilst -1/ is not. Any small percentage comprised of integers is rational, provided that the denominator is not going to include a and will be written as an integer. Similarly, a decimal that ends in a point is yet another logical quantity.
## They can make perception
Even with their label, realistic figures don’t make significantly sensation. In math, they are single entities having a special duration on the number line. Which means that if we matter something, we can easily buy the shape by its ratio to its authentic volume. This keeps accurate even if you will find infinite reasonable numbers involving two certain numbers. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To get the duration of a pearl, as an example, we might matter its width. A single pearl weighs 10 kgs, which is a reasonable number. Moreover, a pound’s bodyweight means twenty kgs. Therefore, we will be able to split a pound by 15, without having be concerned about the size of an individual pearl.
## They could be depicted as being a decimal
If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal amount might be written as a several of two integers, so four times several is equivalent to seven. A similar problem necessitates the recurring fraction 2/1, and each side should be divided up by 99 to find the appropriate response. But how do you create the conversion process? Here are a few cases.
A logical quantity can also be printed in many forms, which include fractions as well as a decimal. A good way to symbolize a rational amount in the decimal is always to separate it into its fractional equal. There are actually 3 ways to separate a logical number, and each one of these methods produces its decimal comparable. One of these simple methods would be to separate it into its fractional counterpart, and that’s what’s referred to as a terminating decimal. |
$\newcommand{\dollar}{\} \DeclareMathOperator{\erf}{erf} \DeclareMathOperator{\arctanh}{arctanh} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
## Section3.2Properties of Logarithms
###### Motivating Questions
• What is a logarithmic function, and what properties do logarithmic functions have?
• How can we use logarithmic functions to solve exponential equations?
We defined an exponential function to be a function of the form
\begin{equation*} P=P(t)=a(b)^t. \end{equation*}
In this section we will discuss logarithmic functions which are inverses of exponential functions. In particular, the function
\begin{equation*} t=\log_{b}\left(\frac{P}{a}\right) \end{equation*}
is the inverse of the exponential function above. For $a=1\text{,}$ the following applet illustrates this inverse relationship:
While this inverse relationship is how we will think of logarithms in practice, let's give a formal definition:
###### The Logarithm
Let $b\neq 1$ be a positive number, then the function
\begin{equation*} f(t)=\log_b(t) \end{equation*}
is called a logarithm with base $b \text{.}$
Upon inputting a value $t$ the function $\log_b(t)$ will tell you the power of $b$ which will yield $t \text{.}$
Due to the relationship between logarithms and exponentials, we often say that the equations
\begin{equation*} x=\log_b(y)\ \ \ \ \text{and}\ \ \ \ b^x=y \end{equation*}
are equivalent.
###### Warning3.7
One of the more common bases of a logarithm is base $10 \text{.}$ Since it is used so often, we have developed a short hand notation for a logarithm of base $10\text{.}$ This short hand is shown below:
\begin{equation*} \log_{10}(y)=\log(y). \end{equation*}
In other words, we simply drop the subscript when referring to base $10 \text{.}$
### SubsectionProperties of Logarithms
Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Here are the laws we will need at present.
1. To multiply two powers with the same base, add the exponents and leave the base unchanged.
\begin{equation*} a^m \cdot a^n = a^{m+n} \end{equation*}
2. To divide two powers with the same base, subtract the exponents and leave the base unchanged.
\begin{equation*} \frac{a^m}{a^n} = a^{m-n} \end{equation*}
3. To raise a power to a power, keep the same base and multiply the exponents.
\begin{equation*} \left(a^m\right)^n = a^{mn} \end{equation*}
Each of these laws corresponds to one of three properties of logarithms.
###### Properties of Logarithms
If $x,y,b>0\text{,}$ and $b\neq 1\text{,}$ then
1. $\log_b(xy)=\log_b(x)+\log_b(y),$
2. $\log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),$
3. $\log_b\left(x^k\right)=k\cdot \log_b(x),$
4. $\log_b(b^y)=y,$
5. $b^{\log_b(x)}=x.$
We will examine the properties of logarithms closer in the Homework problems. For now, study the examples below, keeping in mind that a logarithm is is the inverse function of an exponential function.
1. Property (1): \begin{align*} \log_{2}{32}=\log_{2}{(4 \cdot 8)}=\log_2 4 +\log_2 8= 2+3=5 \\ \end{align*}
2. Property (2): \begin{align*} \log_{2}{\frac{16}{2}}=\log_2{16}-\log_2 2=4-1=3 \\ \end{align*}
3. Property (3): \begin{align*} \log_{2}{64}=\log_{2}{(4^3)}=3\log_2 4= 3 \cdot 2=6 \\ \end{align*}
### SubsectionUsing the Properties of Logarithms
Of course, these properties are useful not so much for computing logs but rather for simplifying expressions that contain variables. We will use them to solve exponential equations. But first, we will practice applying the properties. In Example3.8., we rewrite one log in terms of simpler logs.
###### Example3.8
Simplify $\log_{b}\sqrt{{xy}}\text{.}$
Solution
First, we write $\sqrt{xy}$ using a fractional exponent:
\begin{equation*} \log_{b}{\sqrt{xy}} = \log_{b}{\left((xy)^{1/2}\right)}. \end{equation*}
Then we apply Property (3) to rewrite the exponent as a coefficient:
\begin{equation*} \log_{b}{\left((xy)^{1/2}\right)} = \frac{1}{2}\log_{b}{(xy)}. \end{equation*}
Finally, by Property (1) we write the log of a product as a sum of logs:
\begin{equation*} \frac{1}{2}\left(\log_{b}{xy}\right) = \frac{1}{2}\left(\log_{b}{x} + \log_{b}{y}\right). \end{equation*}
Thus, $\log_{b}\sqrt{xy} = \frac{1}{2}\left(\log_{b}{x} + \log_{b}{y}\right)\text{.}$
###### Example3.9
Simplify $\log_{b}{xy^2}\text{.}$
$\log_b x + 2\log_b y$
###### Warning3.10
Be careful when using the properties of logarithms! Compare the statements below:
1. $\log_{b}{(2x)} = \log_{b}{2} + \log_{b}{x} \ \ \ \text{ by Property 1,}$
but
$\log_{b}{(2 + x)} \ne \log_{b}{2} + \log_{b}{x}.$
2. $\log_{b}{\left(\dfrac{x}{5}\right)}= \log_b x - \log_b 5 \ \ \ \text{ by Property 2,}$
but
$\log_{b}{\left(\dfrac{x}{5}\right)} \ne \dfrac{\log_b x}{\log_b 5}.$
We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.
###### Example3.11
Express $3(\log_b x - \log_b y)$ as a single logarithm with a coefficient of $1\text{.}$
Solution
We begin by applying Property (2) to combine the logs.
\begin{equation*} 3(\log_b x - \log_b y) = 3 \log_{b}\left(\frac{x}{y}\right) \end{equation*}
Then, using Property (3), we replace the coefficient $3$ by an exponent $3\text{.}$
\begin{equation*} 3 \log_{b}\left(\frac{x}{y}\right)=\log_{b}\left(\left(\frac{x}{y}\right)^3\right) \end{equation*}
###### Example3.12
Express $2\log_b x + 4\log_{b}{(x + 3)}$ as a single logarithm with a coefficient of $1\text{.}$
$\log_b \left(x^2(x+3)^4\right)$
### SubsectionThe Natural Exponential Function
There is another base for logarithms and exponential functions that is often used in applications. This base is an irrational number called $e\text{,}$ where
\begin{equation*} e \approx 2.71828182845. \end{equation*}
The number $e$ is essential for many advanced topics, and it is often called the natural base.
The base $e$ logarithm of a number $x\text{,}$ or $\log_ e x\text{,}$ is called the natural logarithm of $x$ and is denoted by $\ln x\text{.}$
###### The Natural Logarithm
The natural logarithm is the logarithm base $e\text{.}$
\begin{equation*} \ln x = \log_{e}{x}, ~~~~ x\gt 0 \end{equation*}
We use natural logarithms in the same way that we use logs to other bases. The properties of logarithms that we studied above also apply to logarithms base $e\text{.}$
### SubsectionSolving Exponential Equations
Suppose we want to solve the equation
\begin{equation*} 5^x = 7. \end{equation*}
We could rewrite the equation in logarithmic form to obtain the exact solution
\begin{equation*} x = \log_{5}{7}. \end{equation*}
However, sometimes we are stuck in a situation where we cannot evaluate $\log_{5}{7}\text{.}$ For example, some calculators do not have a log base $5$ button. So, if we want a decimal approximation for the solution, we begin by taking the base $10$ logarithm of both sides of the original equation, even though the base of the power is not $10\text{.}$ This gives us
\begin{equation*} \log_{10}{(5^x)} = \log_{10}{7}. \end{equation*}
Then we use Property (3) to rewrite the left side as
\begin{equation*} x \log_{10}{5} = \log_{10}{7}. \end{equation*}
Note how using Property (3) allows us to solve the equation: The variable, $x\text{,}$ is no longer in the exponent, and it is multiplied by a constant, $\log_{10}{5}\text{.}$ To finish the solution, we divide both sides by $\log_{10}{5}$ to get
\begin{equation*} x = \frac{\log_{10}{7}}{\log_{10}{5}}. \end{equation*}
On your calculator, enter the sequence
LOG( $7$ ) $\div$ LOG ( $5$ ) ENTER
to find that $x \approx 1.2091\text{.}$
###### Warning3.13
Do not confuse the expression $\dfrac{\log_{10}{7}}{\log_{10}{5}}$ with $\log_{10}{\left(\dfrac{7}{5}\right)}\text{;}$ they are not the same! Property (2) allows us to simplify $\log{\left(\dfrac{x}{y}\right)}\text{,}$ but not $\dfrac{\log x}{\log y}\text{.}$ We cannot rewrite $\dfrac{\log_{10}{7}}{\log_{10}{5}}\text{,}$ so we must evaluate it as $(\log 7)/(\log 5)\text{.}$ You can check on your calculator that
\begin{equation*} \dfrac{\log_{10}{7}}{\log_{10}{5}}\ne \log_{10}{\left(\dfrac{7}{5}\right)}= \log_{10}{1.4}\text{.} \end{equation*}
###### Example3.14
Solve $1640 = 80 \cdot 6^{0.03x}\text{.}$
Solution
First we divide both sides by $80$ to obtain
\begin{equation*} 20.5 = 6^{0.03x}. \end{equation*}
Next, we take the base $10$ logarithm of both sides of the equation and use Property (3) of logarithms to get
\begin{equation*} \log_{10}{20.5} = \log_{10}{(6^{0.03x})}= 0.03x \log_{10}{6}. \end{equation*}
On the right side of the equation, $x$ is multiplied by two constants, $0.03$ and $\log_{10}{6}\text{.}$ So, to solve for $x$ we must divide both sides of the equation by $0.03 \log_{10}{6}\text{.}$ We use a calculator to evaluate the answer:
\begin{equation*} x = \frac{\log_{10}{20.5}}{0.03 \log_{10}{6}}\approx 56.19. \end{equation*}
(On your calculator, remember to enclose the denominator, $0.03 \log_{10}{6}\text{,}$ in parentheses.)
###### Warning3.15
In Example3.14, do not try to simplify
\begin{equation*} 80 \cdot 6^{0.03x} \rightarrow 480^{0.03x}~~ \text{ Incorrect!} \end{equation*}
Remember that the order of operations tells us to compute the power $6^{0.03x}$ before multiplying by $80\text{.}$
###### Example3.16
Solve $5(1.2)^{2.5x} = 77\text{.}$
Hint
Divide both sides by 5.
Take the log of both sides.
Apply Property (3) to simplify the left side.
Solve for $x\text{.}$
$x=\dfrac{\log 15.4}{2.5\log 1.2}\approx 5.999$
• A logarithmic function is the inverse of an exponential function, and the function $f(t) = \log_b(t)$ will give the power of $b$ which yields $t\text{.}$
• There are two shorthand notations for logarithms: writing just $\log(x)$ means $\log_{10}(x)\text{,}$ and writing $\ln(x)$ means $\log_e(x)\text{.}$ |
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Numbers appear in different forms from whole numbers to fractions and percents. What is more, numbers are often connected and need to be analyzed in comparison to each other. This is why it is useful to know how to compare and order different forms of numbers. This topic will be covered and practiced in this lesson.
### Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Background to Help Understand Numbers
Background to Help Understand Multiples
Challenge
## Comparing Math Homework Progress
Tiffaniqua, LaShay, and Kevin are really good friends. They spend most of their free time together. On Wednesday evening, LaShay called Tiffaniqua and asked her how much of her math homework she completed.
Tiffaniqua said that she did about of the homework. LaShay replied that she had completed of the homework. Together they called Kevin, who told that he had finished of all the exercises. Who has completed the greatest part of the homework?
Discussion
## Can Any Two Numbers Be Compared?
Imagine that two numbers need to be compared. Consider this pair of numbers.
Which one is greater? Which is less? The first thing to check is whether the numbers have the same form. In this case, the first number is a percent, while the second is a decimal number.
They do not have the same form, so they cannot be directly compared. It is like comparing strawberries and dogs — they are just not comparable because they belong to totally different categories and have drastically different features.
It is the same way with numbers. Numbers can only be compared if they are written in the same format.
Therefore, to compare two numbers, always begin by making sure that they have the same form. If they do, go ahead and compare them! If they do not, first convert one or both numbers such that they are written in the same format.
Discussion
## Comparing Two Numbers in the Same Form
Consider two numbers in the same form. The good news is that they can totally be compared! But how? First, take a look at two decimal numbers.
The numbers look pretty similar. To determine which is greater, compare their digits one by one moving from left to right until a greater one is found. Remember, it is important to compare the corresponding digits — tenths versus tenths, hundredths versus hundredths, and so on.
Since the last digit of is greater, the first number is greater than the second one.
What about a pair of percents?
The same method can be used to determine which one is greater. Compare the numbers digit by digit, ignoring the sign. The second digit is greater than so the second percent is greater.
Pop Quiz
## Comparing Two Numbers in Decimal and Percent Forms
Consider a pair of numbers. The numbers might be a pair of decimals, a pair of percents, or one decimal number and one percent. Which number is greater? Compare the numbers and pick a correct sign.
Example
## Ordering Decimals and Percents From Least to Greatest
Tiffaniqua, LaShay, and Kevin, along with a couple of their friends, are mastering the skill of blind typing. After finishing their homework, they held a little competition to figure out who types the fastest at the moment.
Tiffaniqua said that her typing speed is characters per minute. LaShay said that her result is characters per minute. Kevin's speed is characters per minute.
a Order their speeds from least to greatest.
b Their friends told them that their results were given as percents or decimals that represent their speeds compared to the average speed of typing characters.
Order their results from least to greatest, too.
### Hint
a Compare the decimals digit by digit, or compare their locations on a number line.
b Start by rewriting the numbers so that they are in the same form.
### Solution
a Start by considering the typing speeds of Tiffaniqua, LaShay, and Kevin.
All these speeds are given as decimal numbers. One way to order these numbers from least to greatest is to compare them digit by digit. Another method is to plot them as points on a number line. On a number line, the farther to the right the number is, the greater it is.
Notice that point which represents Kevin's speed, is the least because it is farthest to the left. Next comes the point Tiffaniqua's speed. Finally, point is the farthest to the right, meaning that LaShay's speed is the greatest.
Therefore, Kevin is the slowest typist at the moment, Tiffaniqua is in the middle, and LaShay is the fastest. Not to worry — Kevin still has time to practice and beat Tiffaniqua and LaShay.
b Start by considering the given typing speeds. Some of them are percents and some are decimal numbers.
To be able to compare the numbers, they should be written in the same form — either all as percents or all as decimal numbers. Rewrite the decimals as percents by multiplying them by and adding the percent sign.
Now all the numbers are written as percents!
Compare the numbers by plotting them on a number line.
Finally, the percents can be ordered from least to greatest.
Discussion
## Comparing Fractions
Consider a pair of fractions. How can the fractions be compared when they have different numerators and denominators?
Both fractions represent a part of a whole, but it is difficult to say which one is greater straight away. The best way to compare these fractions would be to convert them to equivalent fractions that have the same denominator.
When two fractions have the same denominator, they are said to have a common denominator.
Concept
## Common Denominator
A common denominator is a denominator that is shared between two or more fractions. Consider a few examples.
Pair of Fractions Common Denominator
and
and
and
Fractions can always be rewritten to all have a common denominator. This process requires writing equivalent fractions by expanding or simplifying the fractions. As an example, take a look at a pair of fractions with different denominators.
These fractions are in their simplest form, which means that they can only be expanded. Write the multiples of their denominators, and to find the factor of expansion.
There are two potential common denominators in these lists. Expand the first fraction by and the second fraction by to make them have a common denominator of
Now the fractions share a common denominator.
Discussion
## The Most Convenient Common Denominator
There are a lot of possible common denominators that fractions can have. However, it is almost always easier to deal with smaller numbers. This is when the least common denominator comes in handy!
Concept
## Least Common Denominator
The least common denominator (LCD) of two fractions is the least common multiple (LCM) of the denominators of the fractions. In other words, the least common denominator is the smallest of all the common denominators. Some examples are provided in the table below.
Fractions Denominators Multiples of Denominators Common Denominators LCM of Denominators (LCD)
and and
and and
and and
The least common denominator is used when adding or subtracting fractions with different denominators.
To sum up, finding the least common denominator is the same as finding the least common multiple of the denominators. When fractions have a common denominator, which is most often the LCD, they are ready to be compared by comparing their numerators.
Fractions and
### Extra
Comparing Fractions With Common Numerators
There are cases when two fractions share the same numerator. In this case, the fractions can be compared based on their denominators. Consider a pair of fractions.
When fractions have the same denominator, the fraction with the greater numerator is greater. However, when fractions have the same numerator, it is the opposite — the fraction with the smaller denominator is greater.
Same Denominator Same Numerator
The greater numerator, the greater the fraction. The smaller denominator, the greater the fraction.
Since is less than the first fraction must be greater.
To understand why this is true, think of a whole represented by The denominators indicate how many pieces this whole is split into. Here, it is split into pieces for the first fraction and pieces for the second fraction.
Notice that the pieces of the first fraction are much larger than the pieces of the second. The numerator of each fraction indicates how many pieces get picked. Since the numerators are the same, pieces are selected from each whole. Which fraction has a larger selected section?
The total selected area is greater in the first fraction than in the second because each of the pieces is bigger. This is why the fraction with the smaller denominator is greater if the numerators are the same.
Pop Quiz
## Comparing Two Fractions
Consider the given pair of fractions. Which one is greater? Choose the correct sign to complete the expression. If necessary, rewrite the fractions such that they have a common denominator.
Example
## It All Started With a Ball
The next day, Kevin, LaShay, and Tiffaniqua excitedly discussed the details of a local story published in the morning newspaper.
Four fractions were mentioned in the article.
Order them from greatest to least.
### Hint
Rewrite the fractions so that they have a common denominator. Find it by determining the LCM of the denominators.
### Solution
Four fractions need to be ordered from greatest to least.
They have different numerators and denominators. Rewrite them such that they have a common denominator to be able to compare them. Start by finding the LCM of the denominators of the fractions to find a candidate for a common denominator.
First, list the multiples of all the numbers and try to find the least common one.
The LCM of the numbers is Next, divide by each of the numbers to find by which factor each fraction should be expanded.
Denominator Calculating the Quotient Factor
Now that the expansions have been found, they can be used to rewrite the fractions into equivalent fractions with a common denominator of Begin with the first fraction of
The rest of the fractions can be similarly expanded.
Fraction Equivalent Fraction
Finally, the fractions with the same denominator can be compared by looking at their numerators and ordering them from greatest to least.
Example
Tiffaniqua, LaShay, Kevin, and one of their classmates are working on a biology project comparing four different flowers. Each of them was assigned a specific flower. After some research time, they met up to discuss what they found out.
For their project, they need to compare various data points. However, they discovered that they had collected some of these values in different forms and they could not compare the values. Help the students by ordering the numbers from least to greatest.
### Hint
Rewrite the numbers so that they are in the same form. Then, plot the numbers on a number line to help place them in order.
### Solution
Consider the numbers found by the students during their research about different flowers. One number is a fraction, one is a decimal number, one is a mixed number, and one is a percent.
To compare the numbers, they need to be written in the same form. It does not matter which format is chosen, but all numbers will be rewritten as percents in this solution.
### Fraction
Start by rewriting the fraction as a percent. This can be done by multiplying the fraction by then dividing the new numerator by the denominator. Start with the multiplication.
Next, divide the numerator by the denominator and add a percent sign.
Therefore, corresponds to
### Decimal Number
Now consider the decimal number Rewrite it as a percent by multiplying it by and adding a percent sign. Remember that a number can be easily multiplied by by moving its decimal point two places to the right. |
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