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# Equivalent fractions
Here you will learn about equivalent fractions including how to simplify fractions and how to find equivalent fractions.
Students will first learn about equivalent fractions as part of number and operations in fractions in elementary school.
## What are equivalent fractions?
Equivalent fractions are fractions that have the same value. They help us to compare fractions, list them in order from least to greatest and write in lowest terms.
• You can use area models to find equivalent fractions.
For example, find an equivalent fraction to \cfrac{1}{2}
\; \cfrac{1\times2}{2\times2}=\cfrac{2}{4} \hspace{1cm} \cfrac{1\times4}{2\times4}=\cfrac{4}{8} \hspace{1cm} \cfrac{1\times3}{2\times3}=\cfrac{3}{6}
For example, find an equivalent fraction to \cfrac{2}{3}
\; \cfrac{2\times3}{3\times3}=\cfrac{6}{9}
Notice that the numerator and the denominator were both multiplied by the same common factor of 3.
• You can also use fraction strips to find equivalent fractions.
For example, find an equivalent fraction to \cfrac{2}{5}
Notice that the numerator and denominator were multiplied by a common factor of 2.
\cfrac{2\times2}{5\times2}=\cfrac{4}{10}
\cfrac{2}{5}=\cfrac{4}{10}
For example, find an equivalent fraction to \cfrac{5}{6}
Notice that the numerator and denominator were multiplied by the whole number 2.
\cfrac{5 \times 2}{6 \times 2}=\cfrac{10}{12}
\cfrac{5}{6}=\cfrac{10}{12}
## Common Core State Standards
How does this relate to 4th grade math and 5th grade math?
• Grade 3 – Numbers & Operations – Fractions (3.NF.A.3.b)
Recognize and generate simple equivalent fractions, (e.g., \cfrac{1}{2}=\cfrac{2}{4}, \; \cfrac{4}{6}=\cfrac{2}{3} ). Explain why the fractions are equivalent, e.g., by using a visual fraction model.
• Grade 4 – Numbers & Operations – Fractions (4.NF.A.1)
Explain why a fraction \cfrac{a}{b} is equivalent to a fraction \cfrac{(n \times a)}{(n \times b)} by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
## How to write equivalent fractions
In order determine if the given fractions are equivalent:
1. Compare the numerators and denominators of the given fractions.
2. Using either multiplication or division, find the common factor that was multiplied or divided from the first fraction to the second fraction.
3. Write the equivalent fractions.
In order to find a missing denominator or numerator of equivalent fractions:
1. Find the common factor.
2. Multiply or divide the numerator and denominator of one of the fractions by the common factor.
3. Write the equivalent fraction.
## Equivalent fractions examples
### Example 1: fraction strip to determine if fractions are equivalent
Determine if the fractions represented by the fraction strip are equivalent.
1. Compare the numerators and denominators of the given fractions.
The denominator of the first fraction is 4 with 3 equal shaded parts. \cfrac{3}{4}
The denominator of the second fraction is 8 with 6 equal shaded parts. \cfrac{6}{8}
2Using either multiplication or division, find the common factor that was multiplied or divided from the first fraction to the second fraction.
Using multiplication, the common factor is 2.
\cfrac{3\times 2}{4\times 2}=\cfrac{6}{8}
3Write the equivalent fractions.
\cfrac{3}{4}=\cfrac{6}{8}
### Example 2: area model to determine if fractions are equivalent
Determine if the fractions represented by the area models are equivalent.
Compare the numerators and denominators of the given fractions.
Using either multiplication or division, find the common factor that was multiplied or divided from the first fraction to the second fraction.
Write the equivalent fractions.
### Example 3: find the equivalent fraction
Which fraction is equivalent to \cfrac{12}{18}?
\cfrac{4}{7}, \; \cfrac{2}{3}, \; \cfrac{1}{6}
Find the common factor.
Multiply or divide the numerator and denominator of one of the fractions by the common factor.
Write the equivalent fraction.
### Example 4: find the missing denominator
Find the missing denominator.
\cfrac{3}{5}=\cfrac{15}{?}
Find the common factor.
Multiply or divide the numerator and denominator of one of the fractions by the common factor.
Write the equivalent fractions.
### Example 5: find the missing numerator
Find the missing numerator.
\cfrac{?}{20}=\cfrac{3}{5}
Find the common factor.
Multiply the numerator of the second fraction by the common factor.
Write the equivalent fraction.
### Example 6: find the missing numerator.
Find the missing numerator.
\cfrac{24}{63}=\cfrac{?}{21}
Find the common factor.
Multiply or divide the numerator and denominator of one of the fractions by the common factor.
Write the equivalent fraction.
### Teaching tips for equivalent fraction
• Instead of giving students a fraction worksheet to practice, consider having students actively engage with concrete models such as fraction strips or area models to help students visualize and formulate understanding of equivalent fractions.
• Number lines are a powerful visual tool for teaching equivalent fractions because it helps students to understand the correlation between fractions with the same and different denominators as well as the same and different numerators.
• Completely simplified fractions are the same as fractions in lowest terms.
• The cross multiplication strategy for equivalent fractions is not the most effective strategy. It will confuse students as they move into ratios and proportions.
### Easy mistakes to make
• Students may add or subtract the same number to form equivalent fractions instead of multiply or divide
You cannot make equivalent fractions by using addition.
For example, this is an incorrect simplifying technique.
\cfrac{5}{10}=\cfrac{2+3}{7+3}=\cfrac{2}{7}
This is the correct simplifying technique.
\cfrac{5}{10}=\cfrac{1\times5}{2\times5}=\cfrac{1}{2}
• Thinking fractions with different numerators and denominators can’t be equivalent
For example, \cfrac{5}{10} and \cfrac{1}{2}
Because the two given fractions do not have the same numerical numerator and denominator, they can be mistaken as not equal. Using common factor strategies or visual models help students formulate understanding.
• Multiplying or dividing the numerator and denominator by different numbers
For example, taking \cfrac{2}{3} and multiplying the numerator by 3 and the denominator by 4.
\cfrac{2\times 3}{3\times 4}=\cfrac{6}{12}
\cfrac{2}{3} \cfrac{6}{12}
• Thinking the denominator and the numerator are the same
### Practice equivalent fractions questions
1.
Using the area models, represent the shaded area as equivalent fractions.
\cfrac{3}{6}=\cfrac{6}{18}
\cfrac{2}{5}=\cfrac{6}{15}
\cfrac{3}{5}=\cfrac{9}{15}
\cfrac{2}{5}=\cfrac{5}{8}
The first circle is divided into 5 equal parts with 3 parts shaded. \cfrac{3}{5}
The second circle is divided into 15 equal parts with 9 parts shaded. \cfrac{9}{15}
The equivalent fractions are \cfrac{3}{5}=\cfrac{9}{15}.
2. Which equivalent statement is true?
\cfrac{1}{8}=\cfrac{1}{4}
\cfrac{2}{8}=\cfrac{2}{4}
\cfrac{2}{8}=\cfrac{1}{4}
\cfrac{1}{4}=\cfrac{3}{8}
Looking at the fraction strip the top row has two shaded parts which equals \cfrac{2}{8} and the second row has one shaded part which equals \cfrac{1}{4}.
The equivalent fractions are \cfrac{2}{8}=\cfrac{1}{4}.
The common factor is 2.
\cfrac{2\div 2}{8\div 2}=\cfrac{1}{4}
3. Which fraction is equivalent to \cfrac{63}{108}?
\cfrac{9}{12}
\cfrac{7}{12}
\cfrac{7}{9}
\cfrac{3}{9}
Compare the numerator and denominator of the given fraction to the numerators and denominators of the answer choices to find a common factor. The common factor is 9.
\cfrac{63\div 9}{108\div 9}=\cfrac{7}{12}
\cfrac{63}{108}=\cfrac{7}{12}
4. Find the missing numerator.
\cfrac{3}{4}=\cfrac{?}{24}
\cfrac{21}{24}
\cfrac{23}{24}
\cfrac{19}{24}
\cfrac{18}{24}
\cfrac{3}{4}=\cfrac{3\times 6}{4\times 6}=\cfrac{18}{24}
18 is the missing numerator.
5. Find the missing denominator.
\cfrac{6}{7}=\cfrac{24}{?}
\cfrac{24}{28}
\cfrac{24}{26}
\cfrac{24}{25}
\cfrac{24}{27}
\cfrac{6}{7}=\cfrac{6\times 4}{7\times 4}=\cfrac{24}{28}
28 is the missing denominator.
6. Find the missing denominator.
\cfrac{24}{?}=\cfrac{3}{13}
104
8
39
72
\cfrac{3}{13}=\cfrac{3\times 8}{13\times 8}=\cfrac{24}{104}
104 is the missing denominator.
## Equivalent fractions FAQs
Why are equivalent fractions important?
Equivalent fractions and “fraction families” are used to help us add and subtract fractions with unlike denominators and find common denominators. Equivalent fractions also help us to write fractions in simplest form.
Are there other strategies to find equivalent fractions?
Yes, using number lines is a powerful tool with fractions. It helps build understanding when comparing fractions with different denominators, different numerators, and building number sense.
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Further trigonometry and four point method
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Further Trigonometry
1.
Composite Figures
A diagram consisting of more than one triangle is said to be a composite figure.
For trigonometric problems involving a composite figure, first decide whether to use sine, cosine or tangent, and then calculate the required length or angle.
Example 1
In the given diagram, find:
a. x
b. y
Solution:
Using a Construction Line
To solve some trigonometric problems, we need to convert the given triangle into two right-angled triangles by drawing a perpendicular construction line from the vertex to the opposite side.
Example 2
Find BC in the given diagram, rounded to 2 decimal places.
Solution:
Draw BD perpendicular to AC. Let BD = x cm, BC = y cm.
2. Directions and Bearings
The direction to a point is stated as the number of degrees east or west of north or south.
For example, the direction of A from O is N30ºE.
B is N60ºW from O.
C is S70ºE from O.
D is S80ºW from O.
Note:
N30ºE means the direction is 30º east of north.
The bearing to a point is the angle measured in a clockwise direction from the north line.
For example, the bearing of P from O is 065º.
The bearing of Q from O is 300º.
Note:
The direction of P from O is N65ºE.
The direction of Q from O is N60ºW.
A bearing is used to represent the direction of one point relative to another point.
For example, the bearing of A from B is 065º.
The bearing of B from A is 245º.
Note:
• Three ...
Solution Summary
This explains further trigonometry and four-point method.
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# How do you simplify 3 1/6 + 4 1/4?
Jul 16, 2016
#### Explanation:
To add the two numbers we may observe that:
$3 \frac{1}{6} = 3 + \frac{1}{6}$, and similarly
$4 \frac{1}{4} = 4 + \frac{1}{4}$. Now we can add and rearrange:
$3 \frac{1}{6} + 4 \frac{1}{4} = 3 + \frac{1}{6} + 4 + \frac{1}{4} = 3 + 4 + \frac{1}{6} + \frac{1}{4} = 7 + \frac{1}{6} + \frac{1}{4}$
We can now add the fractions:
$\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}$, and the solution is then:
$7 + \frac{5}{12} = 7 \frac{5}{12}$ |
LCM of 15 and also 25 is the smallest number among all common multiples that 15 and 25. The first few multiples that 15 and 25 are (15, 30, 45, 60, 75, . . . ) and also (25, 50, 75, 100, 125, 150, 175, . . . ) respectively. There are 3 generally used techniques to uncover LCM the 15 and also 25 - by listing multiples, by division method, and also by prime factorization.
You are watching: What is the least common multiple of 15 and 25
1 LCM the 15 and also 25 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM that 15 and also 25 is 75.
Explanation:
The LCM of two non-zero integers, x(15) and also y(25), is the smallest positive integer m(75) the is divisible through both x(15) and y(25) without any kind of remainder.
The methods to uncover the LCM the 15 and also 25 are described below.
By prime Factorization MethodBy Listing MultiplesBy division Method
### LCM the 15 and 25 by prime Factorization
Prime factorization of 15 and 25 is (3 × 5) = 31 × 51 and also (5 × 5) = 52 respectively. LCM that 15 and also 25 have the right to be obtained by multiplying prime determinants raised to their respective highest power, i.e. 31 × 52 = 75.Hence, the LCM of 15 and 25 by element factorization is 75.
### LCM of 15 and 25 by Listing Multiples
To calculation the LCM of 15 and 25 by listing out the usual multiples, we deserve to follow the given listed below steps:
Step 1: perform a couple of multiples that 15 (15, 30, 45, 60, 75, . . . ) and also 25 (25, 50, 75, 100, 125, 150, 175, . . . . )Step 2: The typical multiples indigenous the multiples that 15 and 25 room 75, 150, . . .Step 3: The smallest common multiple that 15 and also 25 is 75.
∴ The least usual multiple that 15 and also 25 = 75.
### LCM of 15 and also 25 by division Method
To calculation the LCM the 15 and also 25 by the division method, we will divide the numbers(15, 25) by your prime factors (preferably common). The product of these divisors gives the LCM the 15 and 25.
Step 3: proceed the procedures until just 1s space left in the critical row.
See more: Which Is The Factorization Of X3 + 8? How Do You Factor X^3
The LCM of 15 and 25 is the product of all prime number on the left, i.e. LCM(15, 25) by department method = 3 × 5 × 5 = 75. |
Elementary Algebra 2e
10.2Solve Quadratic Equations by Completing the Square
Elementary Algebra 2e10.2 Solve Quadratic Equations by Completing the Square
Learning Objectives
By the end of this section, you will be able to:
• Complete the square of a binomial expression
• Solve quadratic equations of the form $x2+bx+c=0x2+bx+c=0$ by completing the square
• Solve quadratic equations of the form $ax2+bx+c=0ax2+bx+c=0$ by completing the square
Be Prepared 10.4
Before you get started, take this readiness quiz. If you miss a problem, go back to the section listed and review the material.
Simplify $(x+12)2(x+12)2$.
If you missed this problem, review Example 6.47.
Be Prepared 10.5
Factor $y2−18y+81y2−18y+81$.
If you missed this problem, review Example 7.42.
Be Prepared 10.6
Factor $5n2+40n+805n2+40n+80$.
If you missed this problem, review Example 7.46.
So far, we have solved quadratic equations by factoring and using the Square Root Property. In this section, we will solve quadratic equations by a process called ‘completing the square.’
Complete The Square of a Binomial Expression
In the last section, we were able to use the Square Root Property to solve the equation $(y−7)2=12(y−7)2=12$ because the left side was a perfect square.
$(y−7)2=12y−7=±12y−7=±23y=7±23(y−7)2=12y−7=±12y−7=±23y=7±23$
We also solved an equation in which the left side was a perfect square trinomial, but we had to rewrite it the form $(x−k)2(x−k)2$ in order to use the square root property.
$x2−10x+25=18(x−5)2=18x2−10x+25=18(x−5)2=18$
What happens if the variable is not part of a perfect square? Can we use algebra to make a perfect square?
Let’s study the binomial square pattern we have used many times. We will look at two examples.
$(x+9)2(x+9)(x+9)x2+9x+9x+81x2+18x+81(y−7)2(y−7)(y−7)y2−7y−7y+49y2−14y+49(x+9)2(x+9)(x+9)x2+9x+9x+81x2+18x+81(y−7)2(y−7)(y−7)y2−7y−7y+49y2−14y+49$
Binomial Squares Pattern
If $a,ba,b$ are real numbers,
$(a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2$
$(a−b)2=a2−2ab+b2(a−b)2=a2−2ab+b2$
We can use this pattern to “make” a perfect square.
We will start with the expression $x2+6xx2+6x$. Since there is a plus sign between the two terms, we will use the $(a+b)2(a+b)2$ pattern.
$a2+2ab+b2=(a+b)2a2+2ab+b2=(a+b)2$
Notice that the first term of $x2+6xx2+6x$ is a square, $x2x2$.
We now know $a=xa=x$.
What number can we add to $x2+6xx2+6x$ to make a perfect square trinomial?
The middle term of the Binomial Squares Pattern, $2ab2ab$, is twice the product of the two terms of the binomial. This means twice the product of $xx$ and some number is $6x6x$. So, two times some number must be six. The number we need is $12·6=3.12·6=3.$ The second term in the binomial, $b,b,$ must be 3.
We now know $b=3b=3$.
Now, we just square the second term of the binomial to get the last term of the perfect square trinomial, so we square three to get the last term, nine.
We can now factor to
So, we found that adding nine to $x2+6xx2+6x$ ‘completes the square,’ and we write it as $(x+3)2(x+3)2$.
How To
Complete a square.
To complete the square of $x2+bxx2+bx$:
1. Step 1. Identify $bb$, the coefficient of $xx$.
2. Step 2. Find $(12b)2(12b)2$, the number to complete the square.
3. Step 3. Add the $(12b)2(12b)2$ to $x2+bxx2+bx$.
Example 10.14
Complete the square to make a perfect square trinomial. Then, write the result as a binomial square.
$x2+14xx2+14x$
Try It 10.27
Complete the square to make a perfect square trinomial. Write the result as a binomial square.
$y2+12yy2+12y$
Try It 10.28
Complete the square to make a perfect square trinomial. Write the result as a binomial square.
$z2+8zz2+8z$
Example 10.15
Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. $m2−26mm2−26m$
Try It 10.29
Complete the square to make a perfect square trinomial. Write the result as a binomial square.
$a2−20aa2−20a$
Try It 10.30
Complete the square to make a perfect square trinomial. Write the result as a binomial square.
$b2−4bb2−4b$
Example 10.16
Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.
$u2−9uu2−9u$
Try It 10.31
Complete the square to make a perfect square trinomial. Write the result as a binomial square.
$m2−5mm2−5m$
Try It 10.32
Complete the square to make a perfect square trinomial. Write the result as a binomial square.
$n2+13nn2+13n$
Example 10.17
Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.
$p2+12pp2+12p$
Try It 10.33
Complete the square to make a perfect square trinomial. Write the result as a binomial square.
$p2+14pp2+14p$
Try It 10.34
Complete the square to make a perfect square trinomial. Write the result as a binomial square.
$q2−23qq2−23q$
Solve Quadratic Equations of the Form x2 + bx + c = 0 by completing the square
In solving equations, we must always do the same thing to both sides of the equation. This is true, of course, when we solve a quadratic equation by completing the square, too. When we add a term to one side of the equation to make a perfect square trinomial, we must also add the same term to the other side of the equation.
For example, if we start with the equation $x2+6x=40x2+6x=40$ and we want to complete the square on the left, we will add nine to both sides of the equation.
Then, we factor on the left and simplify on the right.
$(x+3)2=49(x+3)2=49$
Now the equation is in the form to solve using the Square Root Property. Completing the square is a way to transform an equation into the form we need to be able to use the Square Root Property.
Example 10.18
How To Solve a Quadratic Equation of the Form $x2+bx+c=0x2+bx+c=0$ by Completing the Square
Solve $x2+8x=48x2+8x=48$ by completing the square.
Try It 10.35
Solve $c2+4c=5c2+4c=5$ by completing the square.
Try It 10.36
Solve $d2+10d=−9d2+10d=−9$ by completing the square.
How To
Solve a quadratic equation of the form $x2+bx+c=0x2+bx+c=0$ by completing the square.
1. Step 1. Isolate the variable terms on one side and the constant terms on the other.
2. Step 2. Find $(12·b)2(12·b)2$, the number to complete the square. Add it to both sides of the equation.
3. Step 3. Factor the perfect square trinomial as a binomial square.
4. Step 4. Use the Square Root Property.
5. Step 5. Simplify the radical and then solve the two resulting equations.
6. Step 6. Check the solutions.
Example 10.19
Solve $y2−6y=16y2−6y=16$ by completing the square.
Try It 10.37
Solve $r2−4r=12r2−4r=12$ by completing the square.
Try It 10.38
Solve $t2−10t=11t2−10t=11$ by completing the square.
Example 10.20
Solve $x2+4x=−21x2+4x=−21$ by completing the square.
Try It 10.39
Solve $y2−10y=−35y2−10y=−35$ by completing the square.
Try It 10.40
Solve $z2+8z=−19z2+8z=−19$ by completing the square.
In the previous example, there was no real solution because $(x+k)2(x+k)2$ was equal to a negative number.
Example 10.21
Solve $p2−18p=−6p2−18p=−6$ by completing the square.
Try It 10.41
Solve $x2−16x=−16x2−16x=−16$ by completing the square.
Try It 10.42
Solve $y2+8y=11y2+8y=11$ by completing the square.
We will start the next example by isolating the variable terms on the left side of the equation.
Example 10.22
Solve $x2+10x+4=15x2+10x+4=15$ by completing the square.
Try It 10.43
Solve $a2+4a+9=30a2+4a+9=30$ by completing the square.
Try It 10.44
Solve $b2+8b−4=16b2+8b−4=16$ by completing the square.
To solve the next equation, we must first collect all the variable terms to the left side of the equation. Then, we proceed as we did in the previous examples.
Example 10.23
Solve $n2=3n+11n2=3n+11$ by completing the square.
Try It 10.45
Solve $p2=5p+9p2=5p+9$ by completing the square.
Try It 10.46
Solve $q2=7q−3q2=7q−3$ by completing the square.
Notice that the left side of the next equation is in factored form. But the right side is not zero, so we cannot use the Zero Product Property. Instead, we multiply the factors and then put the equation into the standard form to solve by completing the square.
Example 10.24
Solve $(x−3)(x+5)=9(x−3)(x+5)=9$ by completing the square.
Try It 10.47
Solve $(c−2)(c+8)=7(c−2)(c+8)=7$ by completing the square.
Try It 10.48
Solve $(d−7)(d+3)=56(d−7)(d+3)=56$ by completing the square.
Solve Quadratic Equations of the form ax2 + bx + c = 0 by completing the square
The process of completing the square works best when the leading coefficient is one, so the left side of the equation is of the form $x2+bx+cx2+bx+c$. If the $x2x2$ term has a coefficient, we take some preliminary steps to make the coefficient equal to one.
Sometimes the coefficient can be factored from all three terms of the trinomial. This will be our strategy in the next example.
Example 10.25
Solve $3x2−12x−15=03x2−12x−15=0$ by completing the square.
Try It 10.49
Solve $2m2+16m−8=02m2+16m−8=0$ by completing the square.
Try It 10.50
Solve $4n2−24n−56=84n2−24n−56=8$ by completing the square.
To complete the square, the leading coefficient must be one. When the leading coefficient is not a factor of all the terms, we will divide both sides of the equation by the leading coefficient. This will give us a fraction for the second coefficient. We have already seen how to complete the square with fractions in this section.
Example 10.26
Solve $2x2−3x=202x2−3x=20$ by completing the square.
Try It 10.51
Solve $3r2−2r=213r2−2r=21$ by completing the square.
Try It 10.52
Solve $4t2+2t=204t2+2t=20$ by completing the square.
Example 10.27
Solve $3x2+2x=43x2+2x=4$ by completing the square.
Try It 10.53
Solve $4x2+3x=124x2+3x=12$ by completing the square.
Try It 10.54
Solve $5y2+3y=105y2+3y=10$ by completing the square.
Media
Access these online resources for additional instruction and practice with solving quadratic equations by completing the square:
Section 10.2 Exercises
Practice Makes Perfect
Complete the Square of a Binomial Expression
In the following exercises, complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.
57.
$a 2 + 10 a a 2 + 10 a$
58.
$b 2 + 12 b b 2 + 12 b$
59.
$m 2 + 18 m m 2 + 18 m$
60.
$n 2 + 16 n n 2 + 16 n$
61.
$m 2 − 24 m m 2 − 24 m$
62.
$n 2 − 16 n n 2 − 16 n$
63.
$p 2 − 22 p p 2 − 22 p$
64.
$q 2 − 6 q q 2 − 6 q$
65.
$x 2 − 9 x x 2 − 9 x$
66.
$y 2 + 11 y y 2 + 11 y$
67.
$p 2 − 1 3 p p 2 − 1 3 p$
68.
$q 2 + 3 4 q q 2 + 3 4 q$
Solve Quadratic Equations of the Form $x2+bx+c=0x2+bx+c=0$ by Completing the Square
In the following exercises, solve by completing the square.
69.
$v 2 + 6 v = 40 v 2 + 6 v = 40$
70.
$w 2 + 8 w = 65 w 2 + 8 w = 65$
71.
$u 2 + 2 u = 3 u 2 + 2 u = 3$
72.
$z 2 + 12 z = −11 z 2 + 12 z = −11$
73.
$c 2 − 12 c = 13 c 2 − 12 c = 13$
74.
$d 2 − 8 d = 9 d 2 − 8 d = 9$
75.
$x 2 − 20 x = 21 x 2 − 20 x = 21$
76.
$y 2 − 2 y = 8 y 2 − 2 y = 8$
77.
$m 2 + 4 m = −44 m 2 + 4 m = −44$
78.
$n 2 − 2 n = −3 n 2 − 2 n = −3$
79.
$r 2 + 6 r = −11 r 2 + 6 r = −11$
80.
$t 2 − 14 t = −50 t 2 − 14 t = −50$
81.
$a 2 − 10 a = −5 a 2 − 10 a = −5$
82.
$b 2 + 6 b = 41 b 2 + 6 b = 41$
83.
$u 2 − 14 u + 12 = −1 u 2 − 14 u + 12 = −1$
84.
$z 2 + 2 z − 5 = 2 z 2 + 2 z − 5 = 2$
85.
$v 2 = 9 v + 2 v 2 = 9 v + 2$
86.
$w 2 = 5 w − 1 w 2 = 5 w − 1$
87.
$( x + 6 ) ( x − 2 ) = 9 ( x + 6 ) ( x − 2 ) = 9$
88.
$( y + 9 ) ( y + 7 ) = 79 ( y + 9 ) ( y + 7 ) = 79$
Solve Quadratic Equations of the Form $ax2+bx+c=0ax2+bx+c=0$ by Completing the Square
In the following exercises, solve by completing the square.
89.
$3 m 2 + 30 m − 27 = 6 3 m 2 + 30 m − 27 = 6$
90.
$2 n 2 + 4 n − 26 = 0 2 n 2 + 4 n − 26 = 0$
91.
$2 c 2 + c = 6 2 c 2 + c = 6$
92.
$3 d 2 − 4 d = 15 3 d 2 − 4 d = 15$
93.
$2 p 2 + 7 p = 14 2 p 2 + 7 p = 14$
94.
$3 q 2 − 5 q = 9 3 q 2 − 5 q = 9$
Everyday Math
95.
Rafi is designing a rectangular playground to have an area of 320 square feet. He wants one side of the playground to be four feet longer than the other side. Solve the equation $p2+4p=320p2+4p=320$ for $pp$, the length of one side of the playground. What is the length of the other side?
96.
Yvette wants to put a square swimming pool in the corner of her backyard. She will have a 3 foot deck on the south side of the pool and a 9 foot deck on the west side of the pool. She has a total area of 1080 square feet for the pool and two decks. Solve the equation $(s+3)(s+9)=1080(s+3)(s+9)=1080$ for $ss$, the length of a side of the pool.
Writing Exercises
97.
Solve the equation $x2+10x=−25x2+10x=−25$ by using the Square Root Property and by completing the square. Which method do you prefer? Why?
98.
Solve the equation $y2+8y=48y2+8y=48$ by completing the square and explain all your steps.
Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives?
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# Basic Elementary Math Problems with Solutions
In early elementary school, students learn to add, subtract, multiply and divide using whole numbers. Read on for tips and sample problems to help your child, where he's just beginning to add or learning the basics of multiplication.
## How to Help Your Child Solve Basic Elementary Math Problems
One of the most effective ways you can help your child with elementary math is to identify areas where he or she struggles and offer additional practice in those areas. You might begin by checking your child's addition and subtraction skills. In your child is in first grade, he or she should be able to add numbers up to 100. If he or she is a second-grader, your child should be able to add and subtract using more complicated, multi-digit numbers.
In third grade, most elementary students must memorize their multiplication facts. Repetition and consistency will help your child remember these facts. You can establish a practice schedule, and then go over the facts for a few minutes each night.
You also might find that having your child explain math concepts to you in his own words can help your child internalize the information. After he has explained a concept to you, completing a few problems together could help further his understanding.
## Problems with Solutions by Grade Level
### First
1. 16 + 7
If your child struggles with addition problems, you might allow him or her to use visuals, such as counters or a number line. The answer to this problem is 23.
2. 11 - 5
If your child is just beginning to study subtraction, you might want to use smaller numbers, as in the problem above. The answer is six.
3. At a dog park, there are seven dogs playing fetch. Two more dogs arrive. How many dogs are there in all?
Basic addition problems like this one can be solved by counting. Have your child count out the initial number. Then, have him add two more, and count the total again. There are nine dogs.
### Second
1. 171 + 201
Second graders may add problems within 1,000. The answer here is 372.
2. 524 - 318
To solve, your child will have to borrow from the tens column since he can't subtract 4 - 8. The answer is 214.
3. There are 178 guests at a party before 59 more guests arrive. How many guests are there in all?
There are 237 guests.
### Third
1. 4 x 5
Basic multiplication is introduced in third grade. The answer to this problem is 20.
2. 35 ÷ 7
The answer to this problem is five. With division problems, remind your child that he can check his answer by using multiplication: 7 x 5 = 35.
3. Julianne wants to give five cookies to eight of her friends. How many cookies should she bake?
Julianne should bake 40 cookies because 5 x 8 = 40.
Did you find this useful? If so, please let others know!
## Other Articles You May Be Interested In
• Elementary Math: Learning the Order of Operations
Before you can advance to more complex levels of math (like algebra) you have to master the order of operations. Read on to learn the simple steps involved in completing ordered operations math problems. This article provides tips for working your way through math problems with several steps.
• 10 Top Math Apps for Elementary School Children
Is your child struggling with math in elementary school? You may be able to get him or her excited about learning math by using an educational app on your iPad, iPhone, Android or other mobile device. Here are ten apps for elementary school children that may hold the key to math success.
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# Medians of isosceles triangle
The isosceles triangle has a base ABC |AB| = 16 cm and 10 cm long arm. What are the length of medians?
Result
t1 = 6 cm
t2 = 12.369 cm
t3 = 12.369 cm
#### Solution:
Try calculation via our triangle calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Pythagorean theorem is the base for the right triangle calculator. Cosine rule uses trigonometric SAS triangle calculator. See also our trigonometric triangle calculator.
## Next similar examples:
1. Isosceles triangle
What are the angles of an isosceles triangle ABC if its base is long a=5 m and has an arm b=4 m.
2. Triangle ABC
In a triangle ABC with the side BC of length 2 cm The middle point of AB. Points L and M split AC side into three equal lines. KLM is isosceles triangle with a right angle at the point K. Determine the lengths of the sides AB, AC triangle ABC.
3. Isosceles IV
In an isosceles triangle ABC is |AC| = |BC| = 13 and |AB| = 10. Calculate the radius of the inscribed (r) and described (R) circle.
4. Right triangle
Ladder 16 feet reaches up 14 feet on a house wall. The 90-degree angle at the base of the house and wall. What are the other two angles or the length of the leg of the yard?
What is the slope of a ladder 6.2 m long and 5.12 m in height.
6. Triangle ABC
Triangle ABC has side lengths m-1, m-2, m-3. What has to be m to be triangle a) rectangular b) acute-angled?
7. The pond
We can see the pond at an angle 65°37'. Its end points are 155 m and 177 m away from the observer. What is the width of the pond?
8. If the
If the tangent of an angle of a right angled triangle is 0.8. Then its longest side is. .. .
9. Triangle and its heights
Calculate the length of the sides of the triangle ABC, if va=5 cm, vb=7 cm and side b is 5 cm shorter than side a.
10. Angles by cosine law
Calculate the size of the angles of the triangle ABC, if it is given by: a = 3 cm; b = 5 cm; c = 7 cm (use the sine and cosine theorem).
11. Laws
From which law follows directly the validity of Pythagoras' theorem in the right triangle? ?
12. Vector 7
Given vector OA(12,16) and vector OB(4,1). Find vector AB and vector |A|.
13. ABS CN
Calculate the absolute value of complex number -15-29i.
14. Euclid2
In right triangle ABC with right angle at C is given side a=27 and height v=12. Calculate the perimeter of the triangle. |
# Domain and Range
The domain of a function is a set where a function is well defined. More specifically, let $$f: D \rightarrow R$$ be a function, which means that $$f(a)$$ is well defined for $$a \in D$$. The domain of the function $$f$$ is the set $$D$$.
Mathematically you will write $$dom(f) = D$$.
The range of a function, on the other hand, is a set of values that can be reached via the function.
More specifically, let $$f: D \rightarrow R$$ be a function, the range is the set of all possible values $$b \in R$$ for which there exists $$a \in D$$ such that $$f(a) = b$$.
Often, the range of a function is written as $$R(f)$$ or also as $$f(D)$$ (which is also known as the image set of $$D$$ through the function $$f$$).
It is crucial to know the domain of a function because that gives us a safe set of values on which the function is well defined.
Then, the range is important because it tells us to what values are reached by the function. A more graphical interpretation is this: A point $$b$$ is in the range of $$f$$ if the horizontal line $$y = b$$ intersects the graph of the function $$f(x)$$.
## How to compute the Domain, in practical terms?
Here is how to find domain and range :
For the domain, you need to find first the points where the function is NOT defined. The sources of undefined operations are division by zero or squared root of a negative numbers.
So, you need to find those points (if any) where those undefined operations occur. And the domain will be the rest of the points, this is, all the points excluding those you find that cause undefined operations.
## How to compute the Range, in practical terms?
Let $$y$$ be a number and we will solve for $$x$$ the following equation $$f(x) = y$$. The value $$y$$ is in the range if $$f(x) = y$$ can be solved for $$x$$.
So this is a bit trickier: you need to find if you need to restrict $$y$$ in any way so that $$f(x) = y$$ has a solution for $$x$$.
### EXAMPLE 1
Calculate the domain and range of the function $$\displaystyle f(x) = \frac{x+1}{x-1}$$.
First, we need to compute the domain. We need to see where the function is well defined. Usually it is easier to start with where it is NOT well defined.
So in this case, all seem to be valid operations, except for one thing: the denominator cannot be zero.
Note: The main keys to find the domain is identify the points where there are potential divisions by zero, or potential square roots of negative values, which are invalid operations.
Therefore, the function is well defined EXCEPT when $$x-1 = 0$$, which occurs when $$x = 1$$. Hence, we say that the domain is the whole real line except for the value $$1$$.
Using interval notation, we would write $$dom(f) = (-\infty, 1) \cup (1, +\infty)$$.
Now we need to compute the range. Typically, it may be a bit more laborious to get the range than it is to get the domain, but here we go.
There are many ways to find the range: Some may rely on the graphical representation of the function to make a claim about the range of a function. That could work, but it is not a real answer, only a educated hunch.
The other way is the formal mathematical way: Let $$y$$ be a number and we will solve for $$x$$ the following equation $$f(x) = y$$. The value $$y$$ is in the range if $$f(x) = y$$ can be solved for $$x$$.
In this case we have:
$\large f(x) = y \Leftrightarrow \frac{x+1}{x-1} = y$ $\Rightarrow \,\,\,x+1=y\left( x-1 \right)$ $\Rightarrow \,\,\,x+1=yx-y$ $\Rightarrow \,\,\,x-yx=-1-y$ $\Rightarrow \,\,\,x\left( 1-y \right)=-1-y$ $\Rightarrow \,\,\,x=\frac{y+1}{y-1}$
So, when is $$x$$ well defined? Almost for all $$y$$, except for when $$y = 1$$, because in that case we have a division by $$0$$. Hence, the range of $$f$$ in this case is the whole real line, except for 1.
Using interval notation, we would write $$R(f) = (-\infty, 1) \cup (1, +\infty)$$.
### EXAMPLE 2
Calculate the domain and range of the function $$\displaystyle f(x) = \sqrt{x+1}$$.
Remember, for finding the domain we need to look for points where invalid operations may occur (divisions by zero or square roots of negative values. There are no divisions in this case, but we need to ensure that $$x+1\ge 0$$ so that there are no square roots of negative values. So then we need $$x \ge -1$$. Using interval notation, we would write $$dom(f) = [-1, +\infty)$$.
.
Now for the range, we need to solve for $$x$$: $$\sqrt{x+1} = y$$. The square root of something is never negative, so at least we need that $$y \ge 0$$.
Also, by applying square to both sides, we get $$x+1 = y^2$$, so then the solution is $$x = y^2-1$$. So, the only restriction we need to impose on $$y$$ is that $$y \ge 0$$. Hence, using interval notation, we would write $$R(f) = [0, +\infty)$$. Graphically:
## More About the Domain and Range
As a way of a summary, let us recap a few things. First the domain is where a function is well defined, and the range is the set of points that are reached through the function.
In terms of the calculations required, it is typically easier to find the domain than finding the range. Normally, some people try to find the range graphically, but that is a potentially less precise way. Graphical answers need to be interpreted with caution.
You can check out tutorials specifically about how to find the domain and the range , which focus specifically on each case in more detail. |
# What is 201/252 as a decimal?
## Solution and how to convert 201 / 252 into a decimal
201 / 252 = 0.798
Fraction conversions explained:
• 201 divided by 252
• Numerator: 201
• Denominator: 252
• Decimal: 0.798
• Percentage: 0.798%
To convert 201/252 into 0.798, a student must understand why and how. Both represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. Once we've decided the best way to represent the number, we can dive into how to convert 201/252 into 0.798
201 / 252 as a percentage 201 / 252 as a fraction 201 / 252 as a decimal
0.798% - Convert percentages 201 / 252 201 / 252 = 0.798
## 201/252 is 201 divided by 252
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 201 into 252 to find out how many whole parts it will have plus representing the remainder in decimal form. Here's how you set your equation:
### Numerator: 201
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Overall, 201 is a big number which means you'll have a significant number of parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Values closer to one-hundred make converting to fractions more complex. Now let's explore the denominator of the fraction.
### Denominator: 252
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 252 is a large number which means you should probably use a calculator. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Overall, two-digit denominators are no problem with long division. So grab a pen and pencil. Let's convert 201/252 by hand.
## How to convert 201/252 to 0.798
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 252 \enclose{longdiv}{ 201 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 252 \enclose{longdiv}{ 201.0 }$$
Uh oh. 252 cannot be divided into 201. So we will have to extend our division problem. Add a decimal point to 201, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 252 into 201 + 0 or 2010.
### Step 3: Solve for how many whole groups you can divide 252 into 2010
$$\require{enclose} 00.7 \\ 252 \enclose{longdiv}{ 201.0 }$$
We can now pull 1764 whole groups from the equation. Multiply this number by 252, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.7 \\ 252 \enclose{longdiv}{ 201.0 } \\ \underline{ 1764 \phantom{00} } \\ 246 \phantom{0}$$
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.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But they all represent how numbers show us value in the real world. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 201/252 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.798 to 201/252 as a fraction
Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.
### Practice Decimal Conversion with your Classroom
• If 201/252 = 0.798 what would it be as a percentage?
• What is 1 + 201/252 in decimal form?
• What is 1 - 201/252 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.798 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 252 Denominator What is 201/242 as a decimal? What is 191/252 as a decimal? What is 201/243 as a decimal? What is 192/252 as a decimal? What is 201/244 as a decimal? What is 193/252 as a decimal? What is 201/245 as a decimal? What is 194/252 as a decimal? What is 201/246 as a decimal? What is 195/252 as a decimal? What is 201/247 as a decimal? What is 196/252 as a decimal? What is 201/248 as a decimal? What is 197/252 as a decimal? What is 201/249 as a decimal? What is 198/252 as a decimal? What is 201/250 as a decimal? What is 199/252 as a decimal? What is 201/251 as a decimal? What is 200/252 as a decimal? What is 201/252 as a decimal? What is 201/252 as a decimal? What is 201/253 as a decimal? What is 202/252 as a decimal? What is 201/254 as a decimal? What is 203/252 as a decimal? What is 201/255 as a decimal? What is 204/252 as a decimal? What is 201/256 as a decimal? What is 205/252 as a decimal? What is 201/257 as a decimal? What is 206/252 as a decimal? What is 201/258 as a decimal? What is 207/252 as a decimal? What is 201/259 as a decimal? What is 208/252 as a decimal? What is 201/260 as a decimal? What is 209/252 as a decimal? What is 201/261 as a decimal? What is 210/252 as a decimal? What is 201/262 as a decimal? What is 211/252 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 252 Denominator 202/252 as a percentage 201/253 as a percentage 203/252 as a percentage 201/254 as a percentage 204/252 as a percentage 201/255 as a percentage 205/252 as a percentage 201/256 as a percentage 206/252 as a percentage 201/257 as a percentage 207/252 as a percentage 201/258 as a percentage 208/252 as a percentage 201/259 as a percentage 209/252 as a percentage 201/260 as a percentage 210/252 as a percentage 201/261 as a percentage 211/252 as a percentage 201/262 as a percentage |
# Finding Factors for Numbers 71-100
## Objective
SWBAT find the factors for the numbers, 71 - 100.
#### Big Idea
Students will use the prime factorization and U-Turn methods to identify the factor pairs for numbers up to 100.
## Opening
15 minutes
Today's Number Talk
For a detailed description of the Number Talk procedure, please refer to the Number Talk Explanation. For this Number Talk, I am encouraging students to represent their thinking using an array model.
For the first task, most students drew a rectangle with a 10 in the middle, placed the number 5 on one side, and placed a question mark on the other side of the array. Some students even used variables: 10:5.. Students explained: 10 divided by 5 equals 2 because 2 x 5 = 10.
For this task, many students doubled the 2 x 5 array to get (2 x 5) + (2 x 5). These students explained: 2 + 2 = 4 so 20/5 equals 4: 20:5.
For the next task, 40:5, students explained, "Just double the 4 x 5 array because 20 + 20 = 40." More and more students began to use correct division equations: (20/5) + (20/5) = 40/5 =8.
During the next task, some students drew a 5 x 40 array which equals 200. Then, they doubled this array to get (200/5) + (200/5) = 80: 400:5..
For the next task, most students made a 5 x 80 array, which they already knew equals 400. Then, they added on another 5 x 8 array, which they already knew equals 40. Then, they showed how 80 + 8 = 88: 440:5.
For the final task, students simply doubled the 5 x 88 array: 880:5. It was great seeing students looking for and making use of structure (Math Practice 7).
## Teacher Demonstration
50 minutes
Goal
To begin today's lesson, I introduced the goal: I can find the factors for 71-100. I explained: Over the past three lessons, you identified the factors for numbers up to 70. Today, we will move on to more challenging numbers, 71-100.
Reviewing Multiples & Factors
To review multiples and factors, I simply referred to Factors & Multiples Song and said: When we talk about factors, we are referring to the numbers you multiply in a multiplication problem. With the equation, 2 x 3 = 6, the 2 and the 3 are both factors. Remember, (singing) factors are just a few-ew-ew. However, when we look at the multiples of 6, we skip count by 6: 6... 12... (with gesturing, students joined in)... 18... 24... 30... 36... Remember, Singing many millions of multiples! Factor are just a few-ew-ew. Many millions of multiples! Factors are just a few!
Reviewing Prime & Composite Numbers
Next, I referred to the posters, Prime Numbers and Composite Numbers as well as the I'm Prime Chant.. Turn and talk: What is a prime number? Soon, students were singing the chant and I couldn't help but join in, "I'm prime! P-R-I-M-E... The only the factors factors are 1 and me." Surprisingly, after yesterday's lesson, some students had looked up the song on you tube and were able to continue singing, "Factors that divide evenly!"
Reviewing the Prime Factorization Method
At this point, I explained: Yesterday, most students didn't have time to investigate the last row on the factor chart (Factor Chart C). Today, we will complete this row together before you move on to completing the last factor chart that covers the numbers 71-100 (Factor Chart D).
I paired students up and asked one partner to write "Prime Factorization" at the top of his/her whiteboard and the other partner to write "U-Turn" at the top of his/her white board. I then explained: Today, I want to you to take turns representing your thinking with each of these methods. After we find the factors for one number, I'd like you to switch so each partner has the opportunity to practice each method.
I began with the first task, 61. Students referred to the I'm Prime Chant. poster and quickly said, "61 is prime! That means only 61 and 1 are the factors!"
We then moved on to 62. I modeled the Prime Factorization for 62
1. Write the target number, 62, at the top your board.
2. I asked: What two factors, when multiplied together, equal 62? Because this method is called PRIME factorization, the goal is to find all the prime factors for 62. I always ask: Does 2 go into 62 evenly? After some time, students responded, "Yes! 2 x 31!" I modeled how to write 2 x 31.
3. Whenever you are using the prime factorization, we always want to circle the prime numbers. Does anyone see a prime number that we can circle? Students responded, "2!" With time, a student then pointed out, "And 31 is prime too!"
4. We then wrote out the prime factorization equation for 62: 2 x 31 = 62.
Reviewing U-Turn Method
We then moved on to the U-Turn for 62.
1. Draw a t-chart and write the target number, 62, on top of the line.
2. Rule number one is... Students finished my sentence, "Always start with one!"
3. Ask yourself: How many times does 1 go into 62? Or, in other words: What times 1 is 62?Students said, "62!" Okay! Write 62 in the right column, across from the one.
4. We continued on, writing 2 (left column) ... and then 31 (right column).
Guided Practice
To provide students with guided practice, we continued completing last row of the Factor Chart C together: Completed Chart Together.
We followed the same process as above: using the prime factorization and the u-turn methods to identify factors. I knew students were truly practicing Math Practice 3 (constructing viable arguments) each time they identified the factors for each number (collecting supporting evidence). Here are a couple more examples of modeled numbers:
Again, with each new number, partners exchanged boards and took turns modeling each of the methods.
## Student Practice
30 minutes
Instead of asking students to try completing Factor Chart D during this time, I challenged students to find which number on the page has the most factors. After giving the students the opportunity to make predictions, I explained: Today, I want you to continue working with your partner and taking turns using the prime factorization method and u-turn method. Who would like to start with finding the factors for 71? Who would like to begin with 100? Today, you and your partner can take turns choosing which numbers you would like to investigate! Without any prompting, students immediately began testing each number and finding the total factors!
During this time, I conferenced with students (Finding the Factors for 100). I would often provide a little advice or ask guiding questions to help support and stretch student thinking:
• What can I count by and land on 25?
• Can you show me how you can use the prime factorization equation to check your factor pairs?
• Does 2 as a factor work out?
• Do you agree with that, partner?
• Does that make sense to you?
• Do you want to explain the 5?
• I see one that you're missing...
• Can you show me?
• Where did you get 10 x 10?
• Do you think you have all the factors now?
## Closing
10 minutes
To bring closure to this lesson, I celebrated students who were on task, working hard, persevering, and finding creative ways to solve problems.
I also took the time to ask students: Which number between 70 and 100 has the most factors? Students were very excited to share the results to their investigation! Although they didn't have time to investigate all the numbers, students were able to rule out prime numbers and seemed to focus most on the even numbers.
After allowing each group the opportunity to guess, I showed students the Factors for all Numbers Under 100. The numbers, 72, 84, 90, and 96 all had 6 factor pairs (12 factors each). I heard several students say, "I knew it!" "We were right!" |
# Sequence and Series (Test 2)
## Cat Entrance Exams : Mathematics Or Quantitative Aptitude
| Home | | Cat Entrance Exams | | Mathematics Or Quantitative Aptitude | | Sequence and Series |
Sequence and Series
| Sequence and Series |
Q.1
Direction: What should come in place of question mark (?) in the following number series?
24, 536, 487, 703, 678, ?
A. 768
B. 748
C. 764
D. 742
E. None of these
Explaination / Solution:
So, 742 will come at the place of question mark.
Workspace
Report
Q.2
Direction: What will come in place of the question mark (?) in the following number series?
18, 27, 49, 84, 132, (?)
A. 190
B. 183
C. 193
D. 190
E. None of these
Explaination / Solution:
So, ? = 193
Workspace
Report
Q.3
Directions: What will come in place of the question mark (?) in each of the following number series?
33, 43, 65, 99, 145, (?)
A. 201
B. 203
C. 205
D. 211
E. None of these
Explaination / Solution:
So, ? = 203
Workspace
Report
Q.4
Direction: What will come in place of the question mark (?) in the following number series?
7, 11, 23, 51, 103, (?)
A. 186
B. 188
C. 185
D. 187
E. None of these
Explaination / Solution:
The series is
7 + (41) = 11
11 + (43) = 23
23 + (47) = 51
51 + (413) = 103
103 + (421) = 187
So, ? = 103 + (421) = 187
Workspace
Report
Q.5
Directions: What will come in place of the question mark (?) in each of the following number series?
5, 3, 6, ?, 64.75
A. 15
B. 15.5
C. 17.5
D. 17.25
E. None of these
Explaination / Solution:
5*0.5 + 0.5 = 3
3*1.5 + 1.5 = 6
6*2.5 + 2.5 = 17.5
17.5*3.5 + 3.5 = 64.75
So, 17.5 will come at the place of question mark.
Workspace
Report
Q.6
Directions: What should come in place of question mark (?) in the following number series ?
8, 9, 20, 63, 256, 1285, ?
A. 6430
B. 7450
C. 7716
D. 7746
E. None of these
Explaination / Solution:
Workspace
Report
Q.7
Direction: What will come in place of the question mark (?) in the following number series?
1015, 508, 255, 129, 66.5, ?, 20.875
A. 34.50
B. 35
C. 35.30
D. 35.75
E. None of these
Explaination / Solution:
Workspace
Report
Q.8
Direction: What will come in place of the question mark (?) in the following number series?
3, 6, 12, 24, 48, 96, ?
A. 192
B. 182
C. 186
D. 198
E. None of these
Explaination / Solution:
The pattern is:
3 × 2 = 6
6 × 2 = 12
12 × 2 = 24
24 × 2 = 48
48 × 2 = 96
96 × 2 = 192
Hence Option A is correct
Workspace
Report
Q.9
Directions: What should come in place of the question mark (?) in the following number series?
2, 16, 112, 672, 3360, 13440 , ?
A. 3430
B. 3340
C. 40320
D. 43240
E. None of these
Explaination / Solution:
Workspace
Report
Q.10
Directions: What should come in place of the question mark (?) in the following number series?
4, 9, 19, ? , 79, 159, 319
A. 59
B. 39
C. 49
D. 29
E. None of these |
• First line up the decimal points of each number.
• Line up the digits of the numbers that are in the same place value columns.
• Add the digits in each column from right to left.
• If the answer to this addition contains 2-digits, then carry the tens digit over to the next column.
• 7 + 5 = 12 and so we write the 2 below and carry the 1 over.
• We now add the digits in the next column, including the 1 we carried.
• 1 + 2 + the 1 we carried = 4.
• 1.7 + 2.5 = 4.2.
Line up the decimal points and the digits in each place value column.
Add each digit to the digit above.
• We line the decimal points up and then line up the digits of each number.
• We add the digits from the right to left.
• 2 + 7 = 9, so we write a 9 below.
• 5 + 6 = 11 and so, we write a 1 below and carry the 1 ten over.
• In the next column to the left of the decimal point we have 0 + 0 + the 1 we carried = 1.
• 0.52 + 0. 67 = 1.19.
Supporting Lessons
#### Column Addition of Decimals: Interactive Questions
To add decimals first line up the decimal points of each number and then line up the digits in each place value column. Add the digits separately from right to left.
Write the answers to each addition below the digits. Only write one digit in each place value column. If the answer is a two digit number, carry the ten over to add to the digits in the next column along.
We will consider the example of adding the decimal numbers 1.7 + 2.5.
The first step is to write one decimal above the other. It is important to first line up the decimal points of the numbers.
Then line up the digits in each place value column.
Next, we add the digits from right to left.
Adding the digits in the tenths column, we have 7 + 5 = 12.
We only write 1 digit in each place value column. This means that we write the ‘2’ of ’12’ below and carry the ‘1’ into the next column on the left.
Write the carried digit below the answer lines.
We now add the digits in the ones column. We have 1 + 2 + the 1 we carried earlier.
1 + 2 + 1 = 4
1.7 + 2.5 = 4.2
In this next example we add the decimals 0.52 + 0.67.
The first step when adding two decimal numbers is to line up the decimal points.
We then line up the digits in the same place value columns.
We line up the 2 and the 7 in the hundredths column, the 5 and the 6 in the tenths column and the zeros in the ones column.
We start by adding the digits from right to left, starting in the hundredths column.
2 + 7 = 9 and so, we write a 9 in the answer lines below.
We now add the digits in the tenths column.
5 + 6 = 11 and so, we write a 1 below and carry the other 1 digit into the next column to the left.
We finally add the digits in the ones column. We have 0 + 0 + the 1 we carried. 0 + 0 + 1 = 1. We write this 1 in the answer space.
## Adding Decimals with Different Place Values
To add decimals with different place values, line up the decimal points. Then line up the digits from left to right. Write zeros on the end of decimal numbers that have fewer digits than other numbers until the numbers have the same number of digits.
Then add the digits in each place value column separately, working from right to left.
For example, we will add the two decimals 3.4 and 1.58.
The decimal number 3.4 has just 2 digits whereas the decimal number 1.58 has 3 digits.
We first write the two numbers, lining up the decimal points and then the digits from left to right.
We put a 0 digit on the end of 3.4 in the hundredths column to make 3.40.
The two decimal numbers now have the same number of place value columns.
We add the digits from right to left, starting in the hundredths column.
0 + 8 = 8
4 + 5 = 9
3 + 1 = 4
3.4 + 1.58 = 4.98
Here is another example of adding decimals with different place values. We have 5.07 + 7.4.
5.07 has 3 digits with the last digit in the hundredths column.
7.4 has 2 digits with the last digit in the tenths column.
we add a zero digit after 7.4 to make 7.40.
Adding the digits in the hundredths column, 7 + 0 = 7.
Adding the digits in the tenths column, 0 + 4 = 4.
Adding the digits in the ones column, 5 + 7 = 12.
5.07 + 7.4 = 12.47
Now try our lesson on Column Subtraction of 2-Digit Numbers where we learn how to use the column subtraction method.
error: Content is protected !! |
S k i l l
i n
A L G E B R A
2
# SIGNED NUMBERS
## Positive and negative
The absolute value and the algebraic sign
IN ARITHMETIC we cannot subtract a larger number from a smaller:
2 − 3.
But in algebra we can. And to do it, we invent "negative" numbers.
2 − 3 = −1 ("Minus 1" or "Negative 1").
That is, since
2 − 2 = 0,
then 2 − 3 is one "less" than 0. We call it −1. −1 is a signed number. Its algebraic sign is − ("minus")
1. What are the two parts of a signed number?
Do the problem yourself first!
Its algebraic sign, + or − , and its absolute value, which is simply the arithmetical value, that is, the number without its sign.
The algebraic sign of +3 ("plus 3" or "positive 3") is + , and its absolute value is 3.
The algebraic sign of −3 ("negative 3" or "minus 3") is − . The absolute value of −3 is also 3.
For better or for worse, the minus sign '−' is not only the sign of a negative number. It is also the sign for the operation of subtraction. Those are two completely different concepts.
As for the algebraic sign + , normally we do not write it. The algebraic sign of 2, for example, is understood to be + .
As for 0, it is useful to say that it has both signs: −0 = +0 = 0.
(See Lesson 6, Problem 9, and Lesson 11, Problem 11.)
When we place a number within vertical lines, |−3|, that signifies its absolute value.
|−3| = 3. |3| = 3.
Problem 1. Evaluate each of the following.
a) |6| = 6 b) |−6| = 6 c) |0| = 0 d) |3 − 1| = 2 e) |1 − 3| = 2
2. How do we subtract a larger number from a smaller?
5 − 8
1. What will be the sign of the answer?
It would not be wrong to say that we cannot take 8 from 5. We can of course take 5 from 8 -- and that is what we do -- but we report the answer with a minus sign
5 − 8 = −3.
Even in algebra we can only do ordinary arithmetic. But then we must choose the correct sign.
We may say that this is the first rule of signed numbers:
To subtract a larger number from a smaller,
subtract the smaller from the larger, but report
1 − 5 = −4.
We actually do 5 − 1.
It was in order to subtract a larger number from a smaller that negative numbers were invented.
3. What is the only difference between 8 − 5 and 5 − 8 ?
The algebraic signs. They have the same absolute value.
Problem 2. Subtract.
a) 3 − 5 = −2 b) 1 − 8 = −7 c) 8 − 14 = −6 d) 20 − 65 = −45
Problem 3. You have 20 dollars in the bank and you write a check for 25 dollars. Now what is your balance?
20.00 − 25.00 = −5.00
The number line
What you see above is called the number line. The negative numbers fall to the left of 0. (We say that a negative number is less than 0.) The positive numbers fall to the right. (We say that a positive number is greater than 0.)
We imagine every number to be on the number line. And so the fraction ½ will fall between 0 and 1; the fraction −½ is between 0 and −1; and so on.
4. What is an integer?
Any positive or negative whole number, including 0.
0, 1, −1, 2, −2, 3, −3, etc.
When we draw the number line, we typically place the integers.
It is on the number line in fact that we begin to see the practical uses for signed numbers. In general, they show the "direction" of some quantity. That quantity might be temperture: more than or less than a certain temperature designated as 0. Or it might be the position or "address" of some object: left or right of some fixed position chosen as 0. Or it might be the time: before or after a certain moment that again is chosen as 0. Or, as we all know, negative numbers can indicate a balance in a checking account
Problem 4. A rocket is scheduled to launch at precisely 9:16 AM, which is designated t (for time) = 0, and t will be measured in minutes.
a) What time is it at t = −10? 9:06 AM.
b) What time is it at t = −1? 9:15 AM.
c) What time is it at t = +5? 9:21 AM.
d) What is the value of t at 9:00 AM? t = −16.
e) What is the value of t at 9:30 AM? t = 14.
The negative of any number
Every number will have a negative. The negative of 3, for example, will be found at the same distance from 0, but on the other side.
It is −3.
Now, what number is the negative of −3?
The negative of −3 will be the same distance from 0 on the other side. It is 3.
−(−3) = 3.
"The negative of −3 is 3."
This will be true for any number a:
−(−a) = a
"The negative of −a is a."
What is in the box is called a formal rule . This means that whenever we see something that looks like this --
−(−a)
-- something that has that form, then we may rewrite it in this form:
a
For example,
−(−12) = 12.
To learn algebra is to learn its formal rules. For, what are calculations but writing things in a different form? In arithmetic, we rewrite 1 + 1 as 2. In algebra, we rewrite −(−a) as a.
See Lesson 6.
Problem 5. Evaluate the following.
a) −(−10) = 10 b) −(2 − 6) = 4 c) −(1 + 4 − 7) = 2
d) −(−x) = x
The algebraic definition of the negative of a number
Finally, the way we define a negative number in algebra is as follows. −5, for example, is that number which when added to 5 itself, results in 0.
5 + (−5) = 0.
That is, to each number a there corresponds one and only one number −a called its negative. And when we add it to a, we get 0.
a + (−a) = −a + a = 0
Problem 6. What number do you have to add to 8 in order to get 0?
−8
Problem 7. What number do you have to add to −6 in order to get 0?
6
Problem 8. What number is the negative of −q? Why?
The negative of −q is q, because −q + q = 0.
That is,
−(−q = q.
Problem 9. If
s + t = 0,
then what is the relationship between t and s?
t = −s.
Problem 10. If you had to prove that
ba is the negative of ab,
how would you do it?
Show that ab + ba = 0 .
To prove a property of anything, whether in mathematics, logic or the law, we have simply to show that it satisfies the definition of that property.
Next Lesson: Adding and subtracting signed numbers |
# Methods of solving quadratic equations
Solving quadratic equations is quite an important skill in mathematics. Many situations in life can be modeled with quadratic equations. Before reading this lesson, you may want to study or review the lesson that shows what factoring is.
The goal of this lesson is to familiarize you with the numbers of ways or methods that are used to solve quadratic equations. Usually this involves factoring the equation first.
## Three methods of solving quadratic equations
Below, we show the three different ways or methods to solve a quadratic equation.
• Solve by factoring: most popular
• Solve by completing the square: most complicated
• Solve using the quadratic formula: most straightforward.
Method #1:
This is the most popular way to solve quadratic equations. This method shows you how to solve quadratic equations of the form ax2 + bx + c = 0, when a = 1 or when a is not equal to 1. When a = 1, it is quite straightforward! When a is equal to 1, it is somewhat tedious. You will need to do trial and error. Check the lesson to see how!
Most teachers, if not all of them, will show you how to factor quadratic equations using this method first.
Method #2: solve by completing the square
This is the most complicated way to solve quadratic equations.
This method shows you how to complete the square to solve quadratic equations. Pay close attention to all steps and learn it well!
The very Proof of the quadratic formula is entirely based on this technique.
Make sure you understand it, or you may struggle a lot to understand the proof of the quadratic formula.
Teachers will not usually show you how to factor quadratic equations by completing the square until you have mastered how to solve them by factoring.
Method #3: solve using the quadratic formula
This is the most straightforward way to solve any quadratic equations since all you need to do is to plug in numbers into the formula.
Method #1 has some limitations when solving quadratic equations. When answers are not integers, but real numbers, it is very hard or nearly impossible to find the solutions.
With this formula, you can solve any quadratic equations and it does not matter how complex the equation is or how weird the answer will be.
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# Extraneous Solutions
Solving Rational Equations
Tip: This isn't the place to ask a question because the teacher can't reply.
## Key Questions
• A lot of times in algebra, especially when you deal with radical functions, you will end up with what you call extraneous solutions. These are solutions to an equation that you will get as a result of your algebra, but are still not correct. It's not that your process is wrong; it's just that this solution does not fit back into the equation (math is very complicating sometimes).
To find whether your solutions are extraneous or not, you need to plug each of them back in to your given equation and see if they work. It's a very annoying process sometimes, but if employed properly can save you much grief on tests or quizzes.
There are two ways you can do this: by hand, and using your graphing calculator. I will go over both, in case you don't have and/or are not allowed to use one on tests/quizzes. In any case, it is helpful to know both so you can get a better feel of how it works, and just to be that much better at math :)
By Hand:
Consider the equation $\sqrt{x + 4} = x - 2$
First, let's solve it using usual algebra:
${\left(\sqrt{x + 4}\right)}^{2} = {\left(x - 2\right)}^{2}$ (square both sides)
$x + 4 = {x}^{2} - 4 x + 4$ (simplify)
${x}^{2} - 5 x = 0$ (subtract $x$ and 4 from both sides)
$x \left(x - 5\right) = 0$. hence $x = 0 \mathmr{and} x = 5$ (zero product property)
Now we have our two solutions; 0 and 5. Now let's plug each of them back into our original equation and see if they work:
5:
$\sqrt{5 + 4} = 5 - 2$
$\sqrt{9} = 3$
$3 = 3$
Therefore, 5 is a verified solution.
0:
$\sqrt{0 + 4} = 0 - 2$
$\sqrt{4} = - 2$
2 ≠-2
Therefore, 0 is not a solution.
Hence, we would classify 0 as an extraneous solution to this given equation.
By Calculator:
• Set the equation to equal zero
(this ends up being $\sqrt{x + 4} - x + 2 = 0$)
• Plug this into the $y =$ button on your TI-83/84 calculator
• Find the value of each of your solutions (go to 2nd->Calc->Value and enter your solution for $x$)
• You should get zero as an answer for each of them. If you don't, that solution is extraneous.
This is often a much easier way to do it, but as I said above, it is important that you know how to do it by hand as well, as often teachers may ask you to show work, and you may not always have a calculator to help you out.
Hope that helped :)
*equation from http://hotmath.com/*
• In general, extraneous solutions arise when we perform non-invertible operations on both sides of an equation. (That is, they sometimes arise, but not always.)
Non-invertible operations include: raising to an even power (odd powers are invertible), multiplying by zero, and combining sums and differences of logarithms.
Example :
The equations: $x + 2 = 9$ and $x = 7$, have exactly the same set of solutions. Namely: $\left\{7\right\}$.
Square both sides of $x = 7$ to get the new equation: ${x}^{2} = 49$. The solution set of this new equation is; $\left\{- 7 , 7\right\}$. The $- 7$ is an extraneous solution introduced by squaring the two expressions
Square both sides of $x + 2 = 9$ to get the new equation: ${x}^{2} + 4 x + 4 = 81$. Solve the new equation:
${x}^{2} + 4 x - 77 = 0$ so $\left(x - 7\right) \left(x + 11\right) = 0$ whose solution set is $\left\{7 , - 11\right\}$. The $- 11$ does not solve the original equation.
• Example 1 : Raising to an even power
Solve $x = \sqrt[4]{5 {x}^{2} - 4}$.
Raising both sides to the ${4}^{t h}$ gives ${x}^{4} = 5 {x}^{2} - 4$.
This requires, ${x}^{4} - 5 {x}^{2} + 4 = 0$.
Factoring gives $\left({x}^{2} - 1\right) \left({x}^{2} - 4\right) = 0$.
So we need $\left(x + 1\right) \left(x - 1\right) \left(x + 2\right) \left(x - 2\right) = 0$.
The solution set of the last equation is $\left\{- 1 , 1 , - 2 , 2\right\}$. Checking these reveals that $- 1$ and $- 2$ are not solutions to the original equation. Recall that $\sqrt[4]{x}$ means the non-negative 4th root.)
Example 2 Multiplying by zero
If you solve $\frac{x + 3}{x} = \frac{5}{x}$ by cross multiplying,
you'll get ${x}^{2} + 3 x = 5 x$
which lead to ${x}^{2} - 2 x = 0$
.
It looks like the solution set is $\left\{0 , 2\right\}$.
Both are solutions to the second and third equations, but $0$ is not a solution to the original equation.
Example 3 : Combining sums of logarithms.
Solve: $\log x + \log \left(x + 2\right) = \log 15$
Combine the logs on the left to get $\log \left(x \left(x + 2\right)\right) = \log 15$
This leads to $x \left(x + 2\right) = 15$ which has 2 solutions: $\left\{3 , - 5\right\}$. The $- 5$ is not a solution to the original equation because $\log x$ has domain $x > 0$ (Interval: $\left(0 , \infty\right)$)
• In general, extraneous solutions arise when we perform non-invertible operations on both sides of an equation. (That is, they sometimes arise, but not always.)
Squaring (or raising to any other even power) is a non-invertible operation. Solving equations involving square roots involves squaring both sides of an equation.
Example 1 : To show the idea:
The equations: $x - 1 = 4$ and $x = 5$, have exactly the same set of solutions. Namely: $\left\{5\right\}$.
Square both sides of $x = 5$ to get the new equation: ${x}^{2} = 25$. The solution set of this new equation is; $\left\{- 5 , 5\right\}$. The $- 5$ is an extraneous solution introduced by squaring the two expressions
Square both sides of $x - 1 = 4$ to get ${x}^{2} - 2 x + 1 = 16$
which is equivalent to ${x}^{2} - 2 x - 15 = 0$.
and, rewriting the left, $\left(x + 3\right) \left(x - 5\right) = 0$.
So the solution set is $\left\{- 3 , 5\right\}$.
This time, it is $- 3$, that is the extra solution.
Example2 : Extraneous solution.
Solve $x = 2 + \sqrt{x + 18}$
Subtracting $2$ from both sides: $x - 2 = \sqrt{x + 18}$
Squaring (!) gives ${x}^{2} - 4 x + 4 = x + 18$
This requires, ${x}^{2} - 5 x - 14 = 0$.
Factoring to get $\left(x - 7\right) \left(x + 2\right) = 0$
finds the solution set to be $\left\{7 , - 2\right\}$.
Checking these reveals that $- 2$ is not a solution to the original equation. (It is a solution to the 3rd equation -- the squared equation.)
Example 3 : No extraneous solution.
Solve $\sqrt{{x}^{2} + 9} = x + 3$
Squaring (!) gives, ${x}^{2} + 9 = {\left(x + 3\right)}^{2} = {x}^{2} + 6 x + 9$
Which leads to $0 = 6 x$ which has only one solution, $0$ which works in the original equation.
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# CompSci 102 Discrete Math for Computer Science
## Presentation on theme: "CompSci 102 Discrete Math for Computer Science"— Presentation transcript:
CompSci 102 Discrete Math for Computer Science
February 14, 2012 Prof. Rodger Slides modified from Rosen
Announcements Read for next time Chap. 4.4-4.6
Finish Chapter 3 first, then start Chapter 4, number theory
Chap 3.3 - The Complexity of Algorithms
Given an algorithm, how efficient is this algorithm for solving a problem given input of a particular size? How much time does this algorithm use to solve a problem? How much computer memory does this algorithm use to solve a problem? time complexity - analyze the time the algorithm uses to solve the problem given input of a particular size space complexity - analyze the computer memory the algorithm uses to solve the problem, given input of a particular size
The Complexity of Algorithms
In this course, focus on time complexity. Measure time complexity in terms of the number of operations an algorithm uses Use big-O and big-Theta notation to estimate the time complexity Is it practical to use this algorithm to solve problems with input of a particular size? Compare the efficiency of different algorithms for solving the same problem.
Time Complexity For time complexity, determine the number of operations, such as comparisons and arithmetic operations (addition, multiplication, etc.). Ignore minor details, such as the “house keeping” aspects of the algorithm. Focus on the worst-case time complexity of an algorithm. Provides an upper bound. More difficult to determine the average case time complexity of an algorithm (average number of operations over all inputs of a particular size)
Complexity Analysis of Algorithms
Example: Describe the time complexity of the algorithm for finding the maximum element in a finite sequence. procedure max(a1, a2, …., an: integers) max := a1 for i := 2 to n if max < ai then max := ai return max{max is the largest element} Solution: Count the number of comparisons. Compare max < ai n − 1 times. when i incremented, compare if i ≤ n n − 1 times One last comparison for i > n. 2(n − 1) + 1 = 2n − 1 comparisons are made. Hence, the time complexity of the algorithm is Θ(n).
Complexity Analysis of Algorithms
Example: Describe the time complexity of the algorithm for finding the maximum element in a finite sequence. procedure max(a1, a2, …., an: integers) max := a1 for i := 2 to n if max < ai then max := ai return max{max is the largest element} Solution: Count the number of comparisons. Compare max < ai n − 1 times. when i incremented, compare if i ≤ n n − 1 times One last comparison for i > n. 2(n − 1) + 1 = 2n − 1 comparisons are made. Hence, the time complexity of the algorithm is Θ(n).
Worst-Case Complexity of Linear Search
procedure linear search(x:integer, a1, a2, …,an: distinct integers) i := 1 while (i ≤ n and x ≠ ai) i := i + 1 if i ≤ n then location := i else location := 0 return location{location is the subscript of the term that equals x, or is 0 if x is not found} Solution: Count the number of comparisons. At each step two comparisons are made; i ≤ n and x ≠ ai . end of loop, one comparison i ≤ n is made. After loop, one more i ≤ n comparison is made. If x = ai , 2i + 1 comparisons are used. If x is not on the list, 2n + 1 comparisons are made. One comparison to exit loop. Worst case 2n + 2 comparisons, complexity is Θ(n).
Worst-Case Complexity of Linear Search
procedure linear search(x:integer, a1, a2, …,an: distinct integers) i := 1 while (i ≤ n and x ≠ ai) i := i + 1 if i ≤ n then location := i else location := 0 return location{location is the subscript of the term that equals x, or is 0 if x is not found} Solution: Count the number of comparisons. At each step two comparisons are made; i ≤ n and x ≠ ai . end of loop, one comparison i ≤ n is made. After loop, one more i ≤ n comparison is made. If x = ai , 2i + 1 comparisons are used. If x is not on the list, 2n + 1 comparisons are made. One comparison to exit loop. Worst case 2n + 2 comparisons, complexity is Θ(n).
Average-Case Complexity of Linear Search
Example: average case performance of linear search Solution: Assume the element is in the list and that the possible positions are equally likely. By the argument on the previous slide, if x = ai , the number of comparisons is i + 1. Hence, the average-case complexity of linear search is Θ(n). Stopped here
Average-Case Complexity of Linear Search
Example: average case performance of linear search Solution: Assume the element is in the list and that the possible positions are equally likely. By the argument on the previous slide, if x = ai , the number of comparisons is i + 1. Hence, the average-case complexity of linear search is Θ(n). Stopped here
Worst-Case Complexity of Binary Search
procedure binary search(x: integer, a1,a2,…, an: increasing integers) i := 1 {i is the left endpoint of interval} j := n {j is right endpoint of interval} while i < j m := ⌊(i + j)/2⌋ if x > am then i := m + 1 else j := m if x = ai then location := i else location := 0 return location{location is the subscript i of the term ai equal to x, or 0 if x is not found} Solution: Assume n = 2k elements. Note that k = log n. Two comparisons are made at each stage; i < j, and x > am . Size of list is 2k , then 2k-1. then 2k-2 , … then 21 = 2. At the last step, list size is 20 = 1 and single last element compared. Hence, at most 2k + 2 = 2 log n + 2 comparisons are made. Therefore, the time complexity is Θ (log n), better than linear search. Assume (for simplicity)
Worst-Case Complexity of Binary Search
procedure binary search(x: integer, a1,a2,…, an: increasing integers) i := 1 {i is the left endpoint of interval} j := n {j is right endpoint of interval} while i < j m := ⌊(i + j)/2⌋ if x > am then i := m + 1 else j := m if x = ai then location := i else location := 0 return location{location is the subscript i of the term ai equal to x, or 0 if x is not found} Solution: Assume n = 2k elements. Note that k = log n. Two comparisons are made at each stage; i < j, and x > am . Size of list is 2k , then 2k-1. then 2k-2 , … then 21 = 2. At the last step, list size is 20 = 1 and single last element compared. Hence, at most 2k + 2 = 2 log n + 2 comparisons are made. Therefore, the time complexity is Θ (log n), better than linear search. Assume (for simplicity)
Worst-Case Complexity of Bubble Sort
procedure bubblesort(a1,…,an: real numbers with n ≥ 2) for i := 1 to n− 1 for j := 1 to n − i if aj >aj+1 then interchange aj and aj+1 {a1,…, an is now in increasing order} Solution: n−1 passes through list. pass n − i comparisons The worst-case complexity of bubble sort is Θ(n2) since
Worst-Case Complexity of Bubble Sort
procedure bubblesort(a1,…,an: real numbers with n ≥ 2) for i := 1 to n− 1 for j := 1 to n − i if aj >aj+1 then interchange aj and aj+1 {a1,…, an is now in increasing order} Solution: n−1 passes through list. pass n − i comparisons The worst-case complexity of bubble sort is Θ(n2) since
Worst-Case Complexity of Insertion Sort
procedure insertion sort(a1,…,an: real numbers with n ≥ 2) for j := 2 to n i := 1 while aj > ai i := i + 1 m := aj for k := 0 to j − i − 1 aj-k := aj-k-1 ai := m Solution: The total number of comparisons are: Therefore the complexity is Θ(n2). 2+3+ …+𝑛 = 𝑛(𝑛+1) 2 − 1
Worst-Case Complexity of Insertion Sort
procedure insertion sort(a1,…,an: real numbers with n ≥ 2) for j := 2 to n i := 1 while aj > ai i := i + 1 m := aj for k := 0 to j − i − 1 aj-k := aj-k-1 ai := m Solution: The total number of comparisons are: Therefore the complexity is Θ(n2). 2+3+ …+𝑛 = 𝑛(𝑛+1) 2 − 1
Matrix Multiplication Algorithm
matrix multiplication algorithm; C = A B where C is an m n matrix that is the product of the m k matrix A and the k n matrix B. procedure matrix multiplication(A,B: matrices) for i := 1 to m for j := 1 to n cij := 0 for q := 1 to k cij := cij + aiq bqj return C{C = [cij] is the product of A and B}
Complexity of Matrix Multiplication
Example: How many additions of integers and multiplications of integers are used by the matrix multiplication algorithm to multiply two n n matrices. Solution: There are n2 entries in the product. Each entry requires n mults and n − 1 adds. Hence, n3 mults and n2(n − 1) adds. matrix multiplication is O(n3).
Complexity of Matrix Multiplication
Example: How many additions of integers and multiplications of integers are used by the matrix multiplication algorithm to multiply two n n matrices. Solution: There are n2 entries in the product. Each entry requires n mults and n − 1 adds. Hence, n3 mults and n2(n − 1) adds. matrix multiplication is O(n3).
Matrix-Chain Multiplication
Compute matrix-chain A1A2∙ ∙ ∙An with fewest multiplications, where A1 , A2 , ∙ ∙ ∙ , An are m m2, m m3 , ∙ ∙ ∙ mn mn integer matrices. Matrix multiplication is associative. Example: In which order should the integer matrices A1A2A3 - where A1 is , A , A be multiplied? Solution: two possible ways for A1A2A3. A1(A2A3): A2A3 takes 20 ∙ 40 ∙ 10 = 8000 mults.. A1 by the matrix A2A3 takes 30 ∙ 20 ∙ 10 = 6000 mults. Total number is = 14,000. (A1A2)A3: A1A2 takes 30 ∙ 20 ∙ 40 = 24,000 mults. A1A2 by A3 takes 30 ∙ 40 ∙ 10 = 12,000 mults. Total is 24, ,000 = 36,000. So the first method is best.
Matrix-Chain Multiplication
Compute matrix-chain A1A2∙ ∙ ∙An with fewest multiplications, where A1 , A2 , ∙ ∙ ∙ , An are m m2, m m3 , ∙ ∙ ∙ mn mn integer matrices. Matrix multiplication is associative. Example: In which order should the integer matrices A1A2A3 - where A1 is , A , A be multiplied? Solution: two possible ways for A1A2A3. A1(A2A3): A2A3 takes 20 ∙ 40 ∙ 10 = 8000 mults.. A1 by the matrix A2A3 takes 30 ∙ 20 ∙ 10 = 6000 mults. Total number is = 14,000. (A1A2)A3: A1A2 takes 30 ∙ 20 ∙ 40 = 24,000 mults. A1A2 by A3 takes 30 ∙ 40 ∙ 10 = 12,000 mults. Total is 24, ,000 = 36,000. So the first method is best.
Understanding the Complexity of Algorithms
Understanding the Complexity of Algorithms
Times of more than years are indicated with an *.
Complexity of Problems
Tractable Problem: There exists a polynomial time algorithm to solve this problem. These problems are said to belong to the Class P. Intractable Problem: There does not exist a polynomial time algorithm to solve this problem Unsolvable Problem : No algorithm exists to solve this problem, e.g., halting problem. Class NP: Solution can be checked in polynomial time. But no polynomial time algorithm has been found for finding a solution to problems in this class. NP Complete Class: If you find a polynomial time algorithm for one member of the class, it can be used to solve all the problems in the class.
P Versus NP Problem Stephen Cook (Born 1939) The P versus NP problem asks whether the class P = NP? Are there problems whose solutions can be checked in polynomial time, but can not be solved in polynomial time? Note that just because no one has found a polynomial time algorithm is different from showing that the problem can not be solved by a polynomial time algorithm. If a polynomial time algorithm for any of the problems in the NP complete class were found, then that algorithm could be used to obtain a polynomial time algorithm for every problem in the NP complete class. Satisfiability (in Section 1.3) is an NP complete problem. It is generally believed that P≠NP since no one has been able to find a polynomial time algorithm for any of the problems in the NP complete class. The problem of P versus NP remains one of the most famous unsolved problems in mathematics (including theoretical computer science). The Clay Mathematics Institute has offered a prize of \$1,000,000 for a solution.
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PP Section 6.3
# 120 60 2 3 180 2 3 2 π in which quadrant is the
• Notes
• 18
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120 ) 60 ( 2 3 ) 180 ( 2 3 2 = = = π
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In which quadrant is the terminal side of this angle? Quadrant 2 What is the reference angle? Reference angle 3 2 π In Quadrant 2, the reference angle is π - θ or 180° – θ, if working with degrees. 60 3 3 2 angle ref = = - = π π π Draw the 30°-60° triangle to find point.
3 2 π 60 3 3 2 angle ref = = - = π π π 1 2 3 Since the point must be on the unit circle, the radius needs to be one. We then need to divide all the sides of the triangle by 2. ( 29 2 3 2 1 , - = P We also need to remember that in Quadrant 2, the x -coordinate is negative and the y -coordinate is positive.
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b a b a a b a b b a P 1 csc tan 1 sec cos cot sin then , of side terminal the and circle unit on the point the is ) , ( If : Recall = = = = = = = θ θ θ θ θ θ θ
3 2 csc 3 tan 2 sec cos 3 1 cot sin : Hence 3 2 3 2 3 2 2 1 3 2 3 2 2 3 3 2 = - = - = - = - = = π π π π π π ( 29 2 3 2 1 , - = P
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Example . 4 5 of functions tric trigonome the of value the Find π θ = What is the measure of this angle in degrees? 225 ) 45 ( 5 4 ) 180 ( 5 4 5 = = = π
In which quadrant is the terminal side of this angle? Quadrant 3 What is the reference angle? Reference angle 4 5 π 45 4 4 5 angle ref = = - = π π π Draw the 45°- 45° triangle to find point. In Quadrant 3, the reference angle is θ – π or θ - 180°, if working in degrees.
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4 5 π 45 4 4 5 angle ref = = - = π π π 1 1 2 Since the point must be on the unit circle, the radius needs to be one.
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# Definition of a Geometric Sequence — How to Describe a Geometric Sequence?
In this section you. will study sequences in which each term is a multiple of the term preceding it. You will also learn how to find the sum of the corresponding series.
Objectives
(1). Find the common ratio of a geometric sequence.
(2). Write terms of a geometric sequence.
(3). Use the formula for the general term of a geometric sequence.
(4). Use the formula for the sum of the first n terms of a geometric sequence.
(5). Find the value of an annuity.
(6). Use the formula for the sum of an infinite geometric series.
Geometric Sequences
Figure A shows a sequence in which the number of squares is increasing. From left to right, the number of squares is 1, 5, 25, 125, and 625. In this sequence, each term after the first, 1, is obtained by multiplying the preceding term by a constant amount, namely 5. This sequence of increasing numbers of squares is an example of a geometric sequence.
Figure A. A geometric sequence of squares
Definition of a Geometric Sequence
A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a fixed nonzero constant. The amount by which we multiply each time is called the common ratio of the sequence.
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Geometric Sequences
Consider the following sequence:
3, 6, 12, 24, 48, … .
Unlike an arithmetic sequence, these terms do not have a common difference, but there is a simple pattern to the terms. Each term after the first is twice the term preceding it. Such a sequence is called a geometric sequence.
Geometric Sequence
A sequence in which each term after the first is obtained by multiplying the preceding term by a constant is called a geometric sequence.
The constant is denoted by the letter r and is called the common ratio. If a1 is the first term, then the second term is a1 r. The third term is a1 r2, the fourth term is a1 r3, and so on. We can write a formula for the nth term of a geometric sequence by following this pattern.
Formula for the nth Term of a Geometric Sequence
The nth term, an, of a geometric sequence with first term a1 and common ratio r is
an=a1 r(n-1).
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Geometric Sequences
Definition: Geometric Sequences
A geometric sequence is a sequence in which every number in the sequence is equal to the previous number in the sequence, multiplied by a constant number.
This means that the ratio between consecutive numbers in the geometric sequence is a constant. We will explain what we mean by ratio after looking at the following example.
Example — A Flu Epidemic
Extension: What is influenza?
Influenza (commonly called “the flu”) is caused by the influenza virus, which infects the respiratory tract (nose, throat, lungs). It can cause mild to severe illness that most of us get during winter time. The main way that the influenza virus is spread is from person to person in respiratory droplets of coughs and sneezes. (This is called “droplet spread”.) This can happen when droplets from a cough or sneeze of an infected person are propelled (generally, up to a metre) through the air and deposited on the mouth or nose of people nearby. It is good practise to cover your mouth when you cough or sneeze so as not to infect others around you when you have the flu.
Assume that you have the flu virus, and you forgot to cover your mouth when two friends came to visit while you were sick in bed. They leave, and the next day they also have the flu. Let’s assume that they in turn spread the virus to two of their friends by the same droplet spread the following day. Assuming this pattern continues and each sick person infects 2 other friends, we can represent these events in the following manner:
Again we can tabulate the events and formulate an equation for the general case:
Figure B: Each person infects two more people with the flu virus.
Day, n Number of newly-infected people
1 2=2
2 4=2×2=2×21
3 8=2×4=2×2×2=2×22
4 16=2×8=2×2×2×2=2×23
5 32=2×16=2×2×2×2×2=2×24
n =2×2×2×2× … ×2=2×2(n-1)
The above table represents the number of newly-infected people after n days since you first infected your 2 friends.
You sneeze and the virus is carried over to 2 people who start the chain (a1=2). The next day, each one then infects 2 of their friends. Now 4 people are newly-infected. Each of them infects 2 people the third day, and 8 people are infected, and so on. These events can be written as a geometric sequence:
2, 4, 8, 16, 32, … .
Note the common factor (2) between the events. Recall from the linear arithmetic sequence how the common difference between terms were established. In the geometric sequence we can determine the common ratio, r, by
Example — A Worm Farmer
A farmer is breeding worms that he hopes to sell to local shire councils to decompose waste at rubbish dumps. Worms reproduce readily and the farmer expects a 10% increase per week in the mass of worms that he is farming. A 10% increase per week would mean that the mass of worms would increase by a constant factor of (1+10/100) or 1.1.
He starts off with 10 kg of worms. By the beginning of the second week he will expect 10×1.1=11 kg of worms, by the start of the third week he would expect 11×1.1=10×(1.1)2=12.1 kg of worms, and so on. This is an example of a geometric sequence.
A geometric sequence is a sequence where each term is obtained by multiplying the preceding term by a certain constant factor.
The first term is 10, and the common factor is 1.10, which represents a 10% increase on the previous term. We can put the results of this example into a table.
From this table we can see that
t2=1.1×t1, t3=1.1×t2
and so on. In general:
t(n+1)=1.1×tn
The common factor or common ratio whose value is 1.1 for this example can be found by dividing any two successive terms:
A geometric sequence, t, can be written in terms of the first term, a, and the common ratio, r. Thus:
t: {a, ar, ar2, ar3, …, ar(n-1) }
The first term t1=a, the second term t2=ar, the third term t3=ar2, and consequently the nth term, tn is ar(n-1).
For a geometric sequence:
tn=ar(n-1)
where a is the first term and r the common ratio, given by
DEFINITION
A geometric sequence is a sequence such that for all n, there is a constant r such that an/a(n-1) =r. The constant r is called the common ratio.
The recursive definition of a geometric sequence is:
an=a(n-1)r
When Written in terms of a1 and r, the terms of a geometric sequence are:
a1, a2=a1 r, a3=a1 r2, a4=a1 r3, … .
Each term after the first is obtained by multiplying the previous term by r. Therefore, each term is the product of a1 times r raised to a power that is one less than the number of the term, that is:
an=a1 r(n-1)
Since a sequence is a function, we can sketch the function on the coordinate plane. The geometric sequence 1, 1(2), 1(2)2, 1(2)3, 1(2)4 or 1, 2, 4, 8, 16 can be written in function notation as {(1, 1), (2, 2), (3, 4), (4, 8), (5, 16)}. Note that since the domain is the set of positive integers, the points on the graph are distinct points that are not connected by a curve.
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Graphs of geometric sequences
While the graph of an arithmetic sequence is a straight line, the graph of a geometric sequence is a curve for values of r>0. (Different Values of r produce graphs of different shapes.)
(Warked Example)
Consider the geometric sequence 2, 4, 8, 16, 32, …
(a) Draw up a table showing the term number and its value.
(b) Graph the entries in the table.
(c) Comment on the shape of the graph.
Solution:
(a) Draw up a table showing the term number and its corresponding value.
Term number Term value 1 2 3 4 5 2 4 8 16 32
(b) The value of the term depends on the term number, so ‘Term value’ is graphed on the y-axis. Draw a set of axes with suitable scales. Plot the points and join with a smooth curve.
(c) Comment on the shape of the curve.
The points line on a smooth curve which increases rapidly. Because of this rapid increase in value, it would be difficult to use the graph to predict future values in the sequence with any accuracy.
Shape of graphs of geometric sequences
The shape of the graph of a geometric sequence depends on the value of r.
● When r>1, the points lie on a curve, as shown in the graph in the previous worked example. Graphs of this kind are said to diverge (i.e. they move further and further away from the starting Value).
● When r<0, the points oscillate on either side of zero.
● When -1<r<0, the points converge to a certain fixed number, as shown in this graph.
Problem 1. Is the sequence 4, 12, 36, 108, 324, … a geometric sequence?
Solution:
In the sequence,
the ratio of any term to the preceding term is a constant, 3, Therefore, 4, 12, 36, 108, 324, … is a geometric sequence with a1=4 and r=3.
Problem 2. Determine whether each sequence is geometric. Write yes or no.
1). -8, -5, -1, 4, …
Solution:
Find the ratio of the consecutive terms.
Since the ratios are not same, the sequence is not geometric.
2). 4, 12, 36, 108, …
Solution:
Find the ratio of the consecutive terms.
Since the ratios are the same, the sequence is geometric.
Yes
3). 27, 9, 3, 1, …
Solution:
Find the ratio of the consecutive terms.
Since the ratios are the same, the sequence is geometric.
4). 7, 14, 21, 28, …
Solution:
Find the ratio of the consecutive terms.
Since the ratios are not the same, the sequence is not geometric.
5). 21, 14, 7, …
Solution:
Find the ratio of the consecutive terms.
Since the ratios are not the same. the sequence is not geometric.
6). 124, 186, 248, …
Solution:
Find the ratio of the consecutive terms.
Since the ratios are not the same, the sequence is not geometric.
7). -27, 18, -12
Solution:
Find the ratio of the consecutive terms.
Since the ratios are the same, the sequence is geometric.
8). 162, 108, 72, …
Solution:
Find the ratio of the consecutive terms.
Since the ratios are same, the sequence is geometric.
9). ½, -¼, 1, -½
Solution:
Find the ratio of the consecutive terms.
-¼÷½=-½
1÷(-¼)=-4
-½÷1=-½
Since the ratios are not same, the sequence is not geometric. |
# Question Video: Finding the Taylor Series Expansion of a Function from a Formula Mathematics
For a function π: π(β4) = 6, πβ²(β4) = β6 and π^(π) (β4) = (β1/π) π^(π β 1) (β4) for π β₯ 2. Find the Taylor series expansion of π at π₯ = β4.
06:58
### Video Transcript
For a function π, π evaluated at negative four is equal to six, π prime of negative four is equal to negative six, and the πth derivative of π of π₯ with respect to π₯ at negative four is equal to negative one divided by π multiplied by the previous derivative of π of π₯ with respect to π₯ at negative four for π greater than or equal to two. Find the Taylor Series expansion of π at π₯ is equal to negative four.
Weβre given some information about a function π. Weβre told that π of negative four is six; π prime of negative four is negative six. And then weβre given a formula to help us find the πth derivative of π of π₯ with respect to π₯ at π₯ is equal to negative four, where this formula is only valid for values of π greater than or equal to two. We need to use this information to find our Taylor series expansion of the function π at π₯ is equal to negative four. This means our Taylor series will be centered at π₯ is equal to negative four.
To do this, letβs start by recalling what we mean by the Taylor series expansion of a function π at π₯ is equal to π. This means the power series given by the sum from π equals zero to β of the πth derivative of π of π₯ with respect to π₯ evaluated at π divided by π factorial multiplied by π₯ minus π raised to the πth power. We call π the center of our Taylor series. In this question, weβre told to use π is equal to negative four. So letβs substitute π is equal to negative four into our Taylor series expansion definition.
To start, we can see we can simplify π₯ minus negative four to be π₯ plus four. So to find terms in this power series, weβre going to need to find the πth derivative of π of π₯ with respect to π₯ evaluated at negative four. And itβs worth pointing out at this point when we say the zeroth derivative of a function, we just mean that function itself. And weβre given some of these derivatives in our question and a formula to find the rest of these derivatives. So letβs try writing these out one by one and see if we can notice a pattern.
Letβs start with the zeroth derivative of π of π₯ with respect to π₯ evaluated at negative four. Remember, this is just π evaluated at negative four. And weβre told this in the question. Weβre told that this is equal to six. Letβs now find the first derivative of π of π₯ with respect to π₯ evaluated at π₯ is equal to four. We couldβve alternatively written this as π prime of negative four. And once again, weβre told this in the question. Weβre told this is equal to negative six.
To find the next term in our Taylor series expansion, we need to know the second derivative of π of π₯ with respect to π₯ evaluated at negative four. But weβre not told this directly. Instead, weβre given a formula to help us calculate this. Since weβre finding the second derivative, our value of π is equal to two. So we need to substitute π is equal to two into this expression. Substituting π is equal to two gives us the following expression. And we can simplify this since we know two minus one is equal to one. So this gives us negative one-half multiplied by the first derivative of π of π₯ with respect to π₯ evaluated at negative four.
In other words, this is negative one-half multiplied by our previous result. So weβll write this down to get the next term in our sequence. We multiplied by negative one-half, and we should calculate what this is. Itβs negative one-half multiplied by negative six, which we know is equal to three. Letβs now find another term to see if we can spot a pattern. To find the next term in our Taylor series expansion, we would need to know the third derivative of π of π₯ with respect to π₯ evaluated at negative four.
This time, our value of π is equal to three. So weβll substitute π is equal to three into the formula given to us in the question. This gives us negative one-third times the second derivative of π of π₯ with respect to π₯ evaluated at negative four. And now, weβre starting to see our pattern. This is just negative one-third multiplied by the previous result.
So it seems to find the next term in our sequence, weβre just multiplying by negative one over π. Letβs evaluate this expression just to make sure. We get negative one-third multiplied by three, which if we evaluate, we get negative one. And if we look at our formula, this seems to make sense. We take the previous derivative evaluated at negative four and then multiply this by negative one over π.
By using this information, we can actually simplify this expression. First, we notice to get from six to negative six, we multiplied by negative one. Or alternatively, we could write this as negative one divided by one. Now that we can more easily see this sequence, letβs see if we can find an easier expression for the πth derivative of π of π₯ with respect to π₯ at π₯ is equal to negative four. Well, our sequence starts at six. Next, we multiply six by negative one divided by one. Then we multiply this by negative one-half.
Then, we showed you need to multiply this by negative one-third. And in fact, we showed that to get the πth term, youβre going to need to multiply by all the way up to negative one over π. So we get the πth derivative of π of π₯ with respect to π₯ at negative four is equal to six times negative one over one times negative one-half. And we multiply all the way up to negative one over π. Well, at the moment, this formula wonβt be valid when π is equal to zero. But we can simplify this formula. First, we have π factors of negative one. So we can start by writing this as six times negative one to the πth power.
But then, if we were to multiply all of these fractions together, we would get one divided by a denominator which was π times π minus one multiplied by all the way down to one. In other words, we just get one divided by π factorial. And since zero factorial is now equal to one and negative one raised to the power of zero is also equal to one, we can see this formula is also valid when π is equal to zero. So this formula works for all of our integers greater than or equal to zero.
Now we can substitute this formula into our Taylor series expansion. This gives us that π of π₯ is equal to the sum from π equals zero to β of six times negative one to the πth power multiplied by one over π factorial all divided by π factorial. And then we multiply all of this by π₯ plus four all raised to the power of π. And we can simplify this. Instead of multiplying by one over π factorial, we can just divide by π factorial. But then, in our denominator, we now have π factorial multiplied by π factorial. So we could just write this as π factorial all squared. And then with a small bit of rearranging, we get our final answer.
Therefore, we were able to show if a function π has the following properties β π of negative four is six, π prime of negative four is negative six β and we can find the πth derivative of π of π₯ with respect to π₯ at negative four by taking negative one over π times our previous derivative at negative four for π greater than or equal to two, then the Taylor series expansion of π centered at π₯ is equal to negative four is given by the sum from π equals zero to β of negative one to the πth power times six divided by π factorial all squared all multiplied by π₯ plus four all raised to the πth power. |
# Lesson 16 Round and Round Again
• Let’s look for patterns in rounding.
## Warm-up Number Talk: More Groups, Fewer Groups
Find the value of each expression mentally.
## Activity 1 All the Numbers
If you finish early, find the numbers that would round to 100 and to 500 if you’re rounding to the nearest hundred. Compare your lists with a partner’s lists and discuss patterns you see.
1. What are all the numbers that would round to 50 if you’re rounding to the nearest ten? You can use this number line if it helps you.
2. What are all the numbers that would round to 70 if you’re rounding to the nearest ten?
3. What are all the numbers that would round to 600 if you’re rounding to the nearest hundred?
## Activity 2 What’s My Mystery Number?
Write down a number between 100 and 1,000 on your index card. This is your mystery number.
Fold your index card in half so that no one can see your mystery number.
Write down 3 clues about your mystery number by finishing these sentences:
1. My mystery number is (odd or even) .
2. My mystery number rounds to .
3. My mystery number is between and .
Play What’s My Number?
2. Starting with the person on your right, have every member in your team try to guess your mystery number and explain their reasoning.
3. If they haven’t guessed the mystery number by the time the last person shares, reveal the mystery number.
4. Repeat steps 1 through 3 with the next person in the group reading the clues for their mystery number.
## Problem 1
1. What are all the numbers that round to 350 when rounded to the nearest 10?
2. Which numbers round to 350 when rounded to the nearest ten and round to 400 when rounded to the nearest hundred?
## Problem 2
Jada rounds 145 to the nearest ten and gets 150. Then she rounds 150 to the nearest hundred and gets 200. Jada says 145 rounded to the nearest hundred is 200. Do you agree with Jada? Explain or show your reasoning. |
# MCQ Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles with Answers
Students can access the NCERT MCQ Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 9 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Areas of Parallelograms and Triangles Class 9 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Objective Questions.
## Areas of Parallelograms and Triangles Class 9 MCQs Questions with Answers
Students are advised to solve the Areas of Parallelograms and Triangles Multiple Choice Questions of Class 9 Maths to know different concepts. Practicing the MCQ Questions on Areas of Parallelograms and Triangles Class 9 with answers will boost your confidence thereby helping you score well in the exam.
Explore numerous MCQ Questions of Areas of Parallelograms and Triangles Class 9 with answers provided with detailed solutions by looking below.
Question 1.
What is the area of a parallelogram?
(a) $$\frac { 1 }{ 2 }$$ × Base × Altitude
(b) Base × Altitude
(c) $$\frac { 1 }{ 4 }$$ × Base × Median
(d) Base × Base
Question 2.
AE is a median to side BC of triangle ABC. If area(ΔABC) = 24 cm, then area(ΔABE) =
(a) 8 cm
(b) 12 cm
(c) 16 cm
(d) 18 cm
Question 3.
In the figure, ∠PQR = 90°, PS = RS, QP = 12 cm and QS = 6.5 cm. The area of ΔPQR is
(a) 30 cm2
(b) 20 cm2
(c) 39 cm2
(d) 60 cm2
Question 4.
BCD is quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD
ABCD is quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD
(a) is a rectangles
(b) is a parallelogram
(c) is a rhombus
(d) need not be any of (a), (b) or (c).
Answer: (d) need not be any of (a), (b) or (c).
Question 5.
In ΔPQR, if D and E are points on PQ and PR respectively such that DE || QR, then ar (PQE) is equal to
(a) ar (PRD)
(b) ar (DQM)
(c) ar (PED)
(d) ar (DQR)
Question 6.
If Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Then,
(a) ar (AOD) = ar (BOC)
(b) ar (AOD) > ar (BOC)
(c) ar (AOD) < ar (BOC)
(d) None of the above
Answer: (a) ar (AOD) = ar (BOC)
Question 7.
For two figures to be on the same base and between the same parallels, one of the lines must be.
(a) Making an acute angle to the common base
(b) The line containing the common base
(c) Perpendicular to the common base
(d) Making an obtuse angle to the common base
Answer: (b) The line containing the common base
Question 8.
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is:
(a) 1 : 3
(b) 1 : 2
(c) 2 : 1
(d) 1 : 1
Question 9.
If P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD, then:
(a) ar(APB) > ar(BQC)
(b) ar(APB) < ar(BQC)
(c) ar(APB) = ar(BQC)
(d) None of the above
Question 10.
A triangle and a rhombus are on the same base and between the same parallels. Then the ratio of area of triangle to that rhombus is:
(a) 1 : 3
(b) 1 : 2
(c) 1 : 1
(d) 1 : 4
Question 11.
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(a) it is 1 : 1.
(b) it is 1 : 2.
(c) it is 3 : 1.
(d) it is 2 : 1.
Answer: (a) it is 1 : 1.
Question 12.
or two figures to be on the same base and between the same parallels ,they must have a common base and.
(a) One common vertex
(b) The vertices(or the vertex) opposite to the common base lying on a line parallel to the base
(c) The vertices(or the vertex) opposite to the common base lying on a line making an acute angle to the base
(d) Two common vertices
Answer: (b) The vertices(or the vertex) opposite to the common base lying on a line parallel to the base
Question 13.
The median of a triangle divides it into two
(a) congruent triangles.
(b) isosceles triangles.
(c) right angles.
(d) triangles of equal areas
Answer: (d) triangles of equal areas
Question 14.
If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is
(a) it is 1 : 4.
(b) it is 3 : 1.
(c) it is 1 : 2.
(d) it is 1 : 4.
Answer: (c) it is 1 : 2.
Question 15.
The area of a right triangle is 30 sq cm. If the base is 5 cm, then the hypotenuse must be
(a) 12 cm
(b) 18 cm
(c) 13 cm
(d) 20 cm
Question 16.
D,E,F are mid points of the sides BC, CA & AB respectively of ΔABC, then area of BDEF is equal to
(a) $$\frac { 1 }{ 2 }$$ar (ΔABC)
(b) $$\frac { 1 }{ 4 }$$ar (ΔABC)
(c) $$\frac { 1 }{ 3 }$$ar (ΔABC)
(c) $$\frac { 1 }{ 6 }$$ar (ΔABC)
Answer: (a) $$\frac { 1 }{ 2 }$$ar (ΔABC)
Question 17.
Area of a trapezium, whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm, is
(a) 80 cm2
(b) 30 cm2
(c) 120 cm2
(d) 60 cm2
Question 18.
A median of a triangle divides it into two
(a) Congruent triangles
(b) Isosceles triangles
(c) Right triangles
(d) Equal area triangles |
Worksheet Solutions: Polynomials
# Polynomials Class 10 Worksheet Maths Chapter 2
Q1: If a and b are roots of the equation x2 + 7 x + 7 . Find the value of a−1 + b−1 − 2αb ?
Ans: for f ( x ) = x2 + 7 x + 7
we get
α + b = − 7
αb = 7
Now a−1 + b−1 − 2αb
Q2: If the zeroes of the quadratic polynomial x2 + (α + 1 ) x + b are 2 and -3, then find the value of a and b.
Ans: Let f (x) = x2 + (a + 1) x + b
Then 2 − 3 = − ( a + 1 ) or α = 0
− 6 = b
So a = 0 and b = 6
Q.3. If a and b are zeroes of the polynomial f (x) = 2x2 − 7x + 3, find the value of α2 + b2.
Ans: f (x) = 2x2 − 7 x + 3
= 2x2 − x − 6 x + 3 = x(2x − 1) − 3(2 x − 1) = (x − 3) (2 x − 1)
So zeroes are 3 and 1/2
Now
α2 + b2
Q.4. Find the zeroes of the quadratic polynomial x2 + x − 12 and verify the relationship between the zeroes and the coefficients.
Ans: x2 + x − 12
= x2 + 4x − 3x − 12
= x (x + 4) − 3 (x + 4) = (x − 3) (x + 4)
So zeroes are 3 and -4
as we know
sum of roots = -b/a = -(1)/1 = -1 ie 3+(-4) = -1
product of roots = c/a = -12/1 = -12 ie 3x-4 = -12
Q5: If p and q are zeroes of f (x) = x2 − 5x + k, such that p − q = 1 , find the value of k.
Ans: for f ( x ) = x2 − 5 x + k
we get p + q = 5
pq = k
Now p − q = 1
(p − q)2 = 1
(p + q)2 − 4pq = 1
25 − 4k = 1
k = 6
Q6: Given that two of the zeroes of the cubic polynomial αx3 + bx2 + cx + d are 0, then find the third zero.
Ans: Two zeroes = 0, 0
Let the third zero be k.
The, using relation between zeroes and coefficient of polynomial, we have:
k + 0 + 0 = − b a
Third zero = k = -b/a
Q.7. If one of the zeroes of the cubic polynomial x3 + αx2 + bx + c is -1, then find the product of the other two zeroes.
Ans:
Q8: If a-b, a a+b , are zeroes of x3 − 6x2 + 8x , then find the value of b
Ans:
Let f ( x ) = x3 − 6x2 + 8 x
Method -1
= x (x2 − 6 x + 8) = x(x − 2) (x − 4)
So 0,2,4 are zeroes of the polynomial. or a=2 and b=2 or -2 Method -2
Q9: Quadratic polynomial 4x2 + 12x + 9 has zeroes as p and q . Now form a quadratic polynomial whose zeroes are p − 1 and q − 1
Ans:
4x2 + 12 x + 9
= 4x2 + 6x + 6x + 9
= 2x(2x + 3) + 3(2x + 3) = (2 x + 3)2
So p = -3/2 and q = -3/2
So, p − 1 = − 5/2 and q − 1 = − 5/2
(x + 5/2)2
or
4x2 + 20 x + 25
Q10: p and q are zeroes of the quadratic polynomial x2 − (k + 6 ) x + 2(2 k − 1) . Find the value of k if 2(p + q) = p q
Ans: for f (x) = x2 − (k + 6) x + 2 (2 k − 1)
We get, p + q = k + 6
p q = 2(2 k − 1)
Now 2 (p + q ) = pq
Therefore,
2 (k + 6) = 2(2k − 1)
or k + 6 = 2 k − 1
or k = 7
Q11: Given that the zeroes of the cubic polynomial x3 − 6 x2 + 3 x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.
Ans: k1 + k2 + k3 =
a + a + b + a + 2b = 6
a + b = 2
a = 2 − b
Now
b=-3 or b=3
So a= 5 or -1
The zeroes with a = 5, b= -3 can be expressed as 5, 2, -1
The zeroes with a = -1, b = 3 can be expressed as -1, 2, 5
Q12: If one zero of the polynomial 2x2−5x−(2k + 1) is twice the other, find both the zeroes of the polynomial and the value of k.
Ans: Let a be one zero ,then another will be 2a
Now
α + 2α = 5/2 or a= 5/6
Also
k= -17/9
Q.13: Using division show that 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y − 35 .
Ans:
Q14: If (x - 2) and [x - 1/2 ] are the factors of the polynomials qx2 + 5x + r prove that q = r.
Ans:
4q + 10 + r = 0 -(1)
q/4 +5/2 + r = 0 or q + 10 + 4r = 0 -(2)
Subtracting 1 from 2
3q-3r = 0
q = r
Q15: Find k so that the polynomial x2 + 2x + k is a factor of polynomial 2x4 + x3 - 14x2 + 5x + 6. Also, find all the zeroes of the two polynomials.
Ans:
For x2 + 2x + k is a factor of polynomial 2x4 + x3 - 14x2 + 5x + 6, it should be able to divide the polynomial without any remainder
Comparing the coefficient of x we get.
21 + 7k = 0
k = -3
So x2 + 2x + k becomes x2 + 2x -3 = (x-1)(x+3)
Now
2x4 + x3 - 14x2 + 5x + 6= (x2 + 2x -3)(2x2-3x-8+2k)
=(x2 + 2x - 3)(2x2-3x - 2)
=(x - 1)(x + 3)(x - 2)(2x + 1)
or x = 1, -3, 2,= -1/2
The document Polynomials Class 10 Worksheet Maths Chapter 2 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
## FAQs on Polynomials Class 10 Worksheet Maths Chapter 2
1. What are polynomials and how are they classified?
Ans. Polynomials are mathematical expressions that consist of variables, coefficients, and exponents combined using addition, subtraction, and multiplication. They are classified based on their degree (the highest power of the variable) and the number of terms. For example, a polynomial with one term is called a monomial, two terms a binomial, and three or more terms a polynomial.
2. How do you add and subtract polynomials?
Ans. To add or subtract polynomials, you combine like terms, which are terms that have the same variable raised to the same power. For addition, you simply add the coefficients of the like terms together, and for subtraction, you subtract the coefficients. The result will be a new polynomial.
3. What is the process for multiplying polynomials?
Ans. To multiply polynomials, you use the distributive property, often referred to as the FOIL method for binomials. You multiply each term in the first polynomial by each term in the second polynomial and then combine like terms to simplify the result.
4. How can polynomials be factored?
Ans. Polynomials can be factored by finding common factors, using techniques such as grouping, or applying special formulas like the difference of squares or the quadratic formula for quadratic polynomials. The goal is to express the polynomial as a product of simpler polynomials.
5. What are the applications of polynomials in real life?
Ans. Polynomials have various applications in real life, including modeling physical phenomena such as projectile motion, calculating areas and volumes in geometry, and in economics for optimization problems. They are also used in computer graphics and data fitting techniques in statistics.
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
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# McGraw Hill My Math Grade 5 Chapter 3 Check My Progress Answer Key
All the solutions provided in McGraw Hill My Math Grade 5 Answer Key PDF Chapter 3 Check My Progress will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 5 Chapter 3 Check My Progress Answer Key
Check My Progress Page No. 181-182
Vocabulary Check
Draw lines to match each definition to each vocabulary word.
Concept Check
Divide. Write the related multiplication fact.
Question 8.
54 ÷ 9 = ___
____ × 9 = 54
54 ÷ 9 = 6
6 × 9 = 54
Explanation:
Take the first digit of the dividend from the left,
if the digit is less than the divisor then move to the next number.
Then divide it by the divisor and write the answer on top as the quotient.
Subtract the result from the digit and write the remainder.
Division is the opposite of multiplication in mathematics.
Question 9.
___ ÷ 9 = 8
8 × 9 = ____
72 ÷ 9 = 8
8 × 9 = 72
Explanation:
To find the dividend, multiply the quotient with divisor.
As, Division is the opposite of multiplication in mathematics.
Algebra Divide to find the unknown number in each equation.
Question 10.
95 ÷ 5 = n
n = ____
n = 19
Explanation:
To find the unknown number from the given equation,
divided the dividend with divisor.
95 ÷ 5 = n
Divided tens, 90 ÷ 5 = 18
Divided ones, 5 ÷ 5 = 1
Add both the quotients, 18 + 1 = 19
So, the unknown number is 19.
Question 11.
96 ÷ 8 = b
b = ____
b = 12
Explanation:
To find the unknown number from the given equation,
divided the dividend with divisor.
96 ÷ 8 = n
n = 12
12 x 8 = 96
So, the unknown number is 12.
Divide.
Question 12.
24
Explanation:
Divided the tens, 40 ÷ 2 = 20
Divided the ones, 8 ÷ 2 = 4
Add the quotients, 20 + 4 = 24
Question 13.
Quotient = 11
Remainder = 4
Explanation:
Divided the tens, 80 ÷ 7 = 11
7 x 11 = 77
80 – 77 = 3
Add ones to the remainder, 3 + 1 = 4
Add 77 + 4 = 81
Divide mentally.
Question 14.
3,500 ÷ 5 = ____
700
Explanation:
Divided the thousands, 3000 ÷ 5 = 600
Divided hundreds, 500 ÷ 5 = 100
As, zeros are there in tens and ones place,
add the quotients of thousands and hundreds.
600 + 100 = 700
Question 15.
420 ÷ 60 = ____
7
Explanation:
Cancel the zeros on both sides.
42 ÷ 6 = 7
7 x 6 = 42
Problem Solving
Question 16.
Suki received $87 for working 3 days. If she made the same amount each day, how much did Suki earn each day? Answer:$29
Explanation:
Suki received $87 for working 3 days. If she made the same amount each day, Suki earn each day =$87 ÷ 3 = $29$29 x 3 = \$87
Question 17.
A total of 180 students went on a field trip. There were 3 buses. If each bus had the same number of students on it, how many students were on each bus?
60 students.
Explanation:
A total of 180 students went on a field trip.
There were 3 buses.
If each bus had the same number of students on it,
Number of students on each bus 180 ÷ 3 = 60
60 x 3 = 180
Question 18.
Marc is helping out with the school bake sale. He has 50 cookies to place in bags. He places 3 cookies in each bag. How many bags will he use? How many cookies will be left over?
Total bags = 16
Number of cookies left = 2
Explanation:
Marc has 50 cookies to place in bags.
He places 3 cookies in each bag.
Number of bags used 50 ÷ 3 = 16 bags.
16 x 3 = 48 cookies.
Number of cookies left 50 – 48 = 2 cookies.
Test Practice
Question 19.
A train traveled 300 miles in 5 hours. How far did the train travel each hour, on average?
A. 60 miles
B. 150 miles
C. 600 miles
D. 1,500 miles
Option(A)
Explanation:
A train traveled 300 miles in 5 hours.
The train travel each hour, on average 300 ÷ 5 = 60 miles.
Check My Progress Page No. 207-208
Vocabulary Check
Question 1.
Circle the method that correctly uses partial quotients.
Explanation:
In the first division 140 x 4 = 560, but 644 is subtracted from 640 which is incorrect.
In the second division 513 – 300 is 213, but it is written as 210 which is incorrect.
Concept Check
Estimate. Show how you estimated.
Question 2.
244 ÷ 8
______
Estimated: 240 ÷ 8 = 30
Explanation:
So, 240 and 8 are compatible numbers.
Question 3.
700 ÷ 6
Estimated: 696 ÷ 6 = 116
Explanation:
So, 696 and 6 are compatible numbers.
Question 4.
890 ÷ 4
Estimated: 880 ÷ 4 = 220
Explanation:
So, 880 and 4 are compatible numbers.
Divide.
Question 5.
126
Explanation:
Divided the hundreds 600 ÷ 5 = 120
Divided tens 30 ÷ 5 = 6
Add the quotients of tens and hundreds.
120 + 6 = 126
Question 6.
1,766 ÷ 6 = ____
295
Explanation:
1,766 ÷ 6
1766 is rounded to nearest tens as 1770.
1770 and 6 are compatible numbers.
1,770 ÷ 6 = 295
Question 7.
45
Explanation:
87 ÷ 2
87 is rounded to nearest tens as 90.
90 and 2 are compatible numbers.
90 ÷ 2 = 45
Problem Solving
Question 8.
Each of the 9 parking lots at an automobile plant holds the same number of new cars. The lots are full. If there are 431 cars in the lots, about how many cars are in each lot? Show how you estimated.
Estimate = 50 cars.
Explanation:
Each of the 9 parking lots at an automobile plant holds the same number of new cars.
If there are 431 cars in the lots,
Number of cars are in each lot 431 ÷ 9
Compatible numbers are 450 and 9.
450 ÷ 9 = 50 cars.
50 x 9 = 450 cars.
Question 9.
A total of 176 valves were used for 8 cars as they were being assembled. The same number of valves were used for each car. How many valves were used for each car?
22 valves.
Explanation:
A total of 176 valves were used for 8 cars.
The same number of valves were used for each car.
Number of valves used for each car 176 ÷ 8 = 22
22 x 8 = 176 valves.
Question 10.
A construction company estimates that it will take 852 hours to complete a remodeling
project. If there are 6 employees that each work an equal number of hours, how many hours will each employee have to work?
142 hours.
Explanation:
A construction company estimates that it will take 852 hours to complete remodeling.
If there are 6 employees.
Number of hours will each employee have to work 852 ÷ 6 = 142.
142 x 6 = 852 hours.
Question 11.
Rachel has 145 mugs displayed at a craft show. She displays them in 8 rows with the same number of mugs in each row. How many mugs are in each row? Explain how you interpreted the remainder.
There are 18 mugs in each row with a remainder of one.
The remainder could represent on mug that he has in his hands.
Explanation:
Rachel has 145 mugs displayed at a craft show.
She displays them in 8 rows with the same number of mugs in each row.
Number of mugs in each row 145 ÷ 8 = 18.1
Multiply. 18 × 8 = 144
Subtract. 145 – 144 = 1
Compare. 1 < 8
The remainder is 1.
Interpret the remainder, 1.
There are 18 mugs in each row with a remainder of one.
The remainder could represent on mug that he has in his hands.
Test Practice
Question 12.
Valley Schools have a student population of 1,608 students. If there are an equal number of students in the 6 grade levels, how many students are there in each grade?
A. 28 students
B. 208 students
C. 248 students
D. 268 students
Option(D)
Explanation:
Valley Schools have a student population of 1,608 students.
If there are an equal number of students in the 6 grade levels,
number of students in each grade 1608 ÷ 6 = 268 students.
6 x 268 = 1608 students.
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# Recursion
## Hello recursion!
We mentioned recursion briefly in the previous chapter. In this chapter, we’ll take a closer look at recursion, why it’s important to Elm and how we can work out very concise and elegant solutions to problems by thinking recursively.
If you still don’t know what recursion is, read this sentence. Haha! Just kidding! Recursion is actually a way of defining functions in which the function is applied inside its own definition. Definitions in mathematics are often given recursively. For instance, the fibonacci sequence is defined recursively. First, we define the first two fibonacci numbers non-recursively. We say that F(0) = 0 and F(1) = 1, meaning that the 0th and 1st fibonacci numbers are 0 and 1, respectively. Then we say that for any other natural number, that fibonacci number is the sum of the previous two fibonacci numbers. So F(n) = F(n-1) + F(n-2). That way, F(3) is F(2) + F(1), which is (F(1) + F(0)) + F(1). Because we’ve now come down to only non-recursively defined fibonacci numbers, we can safely say that F(3) is 2. Having an element or two in a recursion definition defined non-recursively (like F(0) and F(1) here) is also called the edge condition and is important if you want your recursive function to terminate. If we hadn’t defined F(0) and F(1) non recursively, you’d never get a solution any number because you’d reach 0 and then you’d go into negative numbers. All of a sudden, you’d be saying that F(-2000) is F(-2001) + F(-2002) and there still wouldn’t be an end in sight!
Recursion is important to Elm because unlike imperative languages, you do computations in Elm by declaring what something is instead of declaring how you get it. That’s why there are no while loops or for loops in Elm and instead we many times have to use recursion to declare what something is.
## Maximum awesome
The maximum function takes a list of things that can be compared (e.g. comparables) and returns the biggest of them. Think about how you’d implement that in an imperative fashion. You’d probably set up a variable to hold the maximum value so far and then you’d loop through the elements of a list and if an element is bigger than then the current maximum value, you’d replace it with that element. The maximum value that remains at the end is the result. Whew! That’s quite a lot of words to describe such a simple algorithm!
Now let’s see how we’d define it recursively. We could first set up an edge condition and say that the maximum of a singleton list is equal to the only element in it. Then we can say that the maximum of a longer list is the head if the head is bigger than the maximum of the tail. If the maximum of the tail is bigger, well, then it’s the maximum of the tail. That’s it! Now let’s implement that in Elm.
``````maximum : List comparable -> comparable
maximum list =
case list of
[] -> Debug.crash "maximum of empty list"
[x] -> x
(x::xs) ->
let
maxTail = maximum xs
in
if x > maxTail then
x
else
maxTail
``````
As you can see, pattern matching goes great with recursion! Most imperative languages don’t have pattern matching so you have to make a lot of if else statements to test for edge conditions. Here, we simply put them out as patterns. So the first edge condition says that if the list is empty, crash! Makes sense because what’s the maximum of an empty list? I don’t know. Note that you shouldn’t normally use the Debug.crash function in your code. A better way would be to use the Maybe type, which we will see more info on later. Elm’s built in List.maximum function uses the Maybe type here instead of crashing. The second pattern also lays out an edge condition. It says that if it’s the singleton list, just give back the only element.
Now the third pattern is where the action happens. We use pattern matching to split a list into a head and a tail. This is a very common idiom when doing recursion with lists, so get used to it. We use a let binding to define maxTail as the maximum of the rest of the list. Then we check if the head is greater than the maximum of the rest of the list. If it is, we return the head. Otherwise, we return the maximum of the rest of the list.
Let’s take an example list of numbers and check out how this would work on them: [2,5,1]. If we call maximum on that, the first two patterns won’t match. The third one will and the list is split into 2 and [5,1]. The let clause wants to know the maximum of [5,1], so we follow that route. It matches the third pattern again and [5,1] is split into 5 and [1]. Again, the let clause wants to know the maximum of [1]. Because that’s the edge condition, it returns 1. Finally! So going up one step, comparing 5 to the maximum of [1] (which is 1), we obviously get back 5. So now we know that the maximum of [5,1] is 5. We go up one step again where we had 2 and [5,1]. Comparing 2 with the maximum of [5,1], which is 5, we choose 5.
An even clearer way to write this function is to use max. max is a function that takes two comparables and returns the bigger of them. Here’s how we could rewrite maximum by using max:
``````maximum : List comparable -> comparable
maximum list =
case list of
[] -> Debug.crash "maximum of empty list"
[x] -> x
(x::xs) -> max x (maximum xs)
``````
How’s that for elegant! In essence, the maximum of a list is the max of the first element and the maximum of the tail.
## A few more recursive functions
Now that we know how to generally think recursively, let’s implement a few functions using recursion. First off, we’ll implement List.repeat. repeat takes an Int and some element and returns a list that has several repetitions of the same element. For instance, repeat 3 5 returns [5,5,5]. Let’s think about the edge condition. My guess is that the edge condition is 0 or less. If we try to repeat something zero times, it should return an empty list. Also for negative numbers, because it doesn’t really make sense.
``````repeat : Int -> a -> List a
repeat n x =
if n <= 0 then
[]
else
x :: repeat (n-1) x
``````
We used an if then else expression here instead of patterns because we’re testing for a boolean condition. If n is less than or equal to 0, return an empty list. Otherwise return a list that has x as the first element and then x replicated n-1 times as the tail. Eventually, the (n-1) part will cause our function to reach the edge condition.
Next up, we’ll implement List.take. It takes a certain number of elements from a list. For instance, take 3 [5,4,3,2,1] will return [5,4,3]. If we try to take 0 or less elements from a list, we get an empty list. Also if we try to take anything from an empty list, we get an empty list. Notice that those are two edge conditions right there. So let’s write that out:
``````take : Int -> List a -> List a
take n list =
if n <= 0 then
[]
else
case list of
[] -> []
(x::xs) -> x :: take (n-1) xs
``````
The if expression specifies that if we try to take a 0 or negative number of elements, we get an empty list. The first case pattern indicates that if we try to take anything from an empty list, we get an empty list. The second case pattern breaks the list into a head and a tail. And then we state that taking n elements from a list equals a list that has x as the head and then a list that takes n-1 elements from the tail as a tail. Try using a piece of paper to write down how the evaluation would look like if we try to take, say, 3 from [4,3,2,1].
List.reverse simply reverses a list. Think about the edge condition. What is it? Come on … it’s the empty list! An empty list reversed equals the empty list itself. O-kay. What about the rest of it? Well, you could say that if we split a list to a head and a tail, the reversed list is equal to the reversed tail and then the head at the end.
``````reverse : List a -> List a
reverse list =
case list of
[] -> []
(x::xs) -> reverse xs ++ [x]
``````
There we go!
zip is a common operation in many functional programming languages that takes two lists and zips them together, but it isn’t a built-in function in Elm. So let’s make our own! zip [1,2,3] [2,3] returns [(1,2),(2,3)], because it truncates the longer list to match the length of the shorter one. How about if we zip something with an empty list? Well, we get an empty list back then. So there’s our edge condition. However, zip takes two lists as parameters, so there are actually two edge conditions.
``````zip : List a -> List b -> List (a, b)
zip list1 list2 =
case (list1, list2) of
(_, []) -> []
([], _) -> []
((x::xs), (y::ys)) -> (x, y) :: zip xs ys
``````
What’s going on here? This is a more complicated case expression than we’ve used up to this point. Remember how we said that in between the case and of keywords is an expression? Well, we can use that to our advantage here by packing both lists into a tuple and then pattern matching on that tuple. In fact, we can write any expression here and the value of that expression is what is pattern matched on below. We can also deconstruct and bind more than one layer into the matched data structure, as we see in the third pattern. In this pattern we’re pulling both lists out of the tuple, and at the same time pulling values out of the two lists and binding them to names. Cool!
The first two patterns say that if the first list or second list is empty, we get an empty list. The third one says that two lists zipped are equal to pairing up their heads and then tacking on the zipped tails. Zipping [1,2,3] and [‘a’,’b’] will eventually try to zip [3] with []. The edge condition patterns kick in and so the result is (1,’a’):(2,’b’):[], which is exactly the same as [(1,’a’),(2,’b’)].
## Quick, sort!
We have a list of items that can be sorted. Their type is one of the comparable types. And now, we want to sort them! There’s a very cool algoritm for sorting called quicksort. It’s a very clever way of sorting items. While it takes upwards of 10 lines to implement quicksort in imperative languages, the implementation is much shorter and elegant in Elm. Quicksort has become a sort of poster child for functional languages. Therefore, let’s implement it here, even though implementing quicksort in functional languages is considered really cheesy because everyone does it to showcase how elegant they are.
So, the type signature is going to be quicksort : List comparable -> List comparable. No surprises there. The edge condition? Empty list, as is expected. A sorted empty list is an empty list. Now here comes the main algorithm: a sorted list is a list that has all the values smaller than (or equal to) the head of the list in front (and those values are sorted), then comes the head of the list in the middle and then come all the values that are bigger than the head (they’re also sorted). Notice that we said sorted two times in this definition, so we’ll probably have to make the recursive call twice! Also notice that we defined it using the verb is to define the algorithm instead of saying do this, do that, then do that …. That’s the beauty of functional programming! How are we going to filter the list so that we get only the elements smaller than the head of our list and only elements that are bigger? Well, there’s a function called List.filter. Its type is List.filter : (a -> Bool) -> List a -> List a. So, let’s dive in and define this function.
``````quicksort : List comparable -> List comparable
quicksort list =
case list of
[] -> []
(x::xs) ->
let
smallerSorted = quicksort (List.filter ((>) x) xs)
biggerSorted = quicksort (List.filter ((<=) x) xs)
in
smallerSorted ++ [x] ++ biggerSorted
``````
Let’s give it a small test run to see if it appears to behave correctly.
``````toPrint = quicksort [10,2,5,3,1,6,7,4,2,3,4,8,9]
[1,2,2,3,3,4,4,5,6,7,8,9,10]
toPrint = String.fromList (quicksort (String.toList "the quick brown fox jumps over the lazy dog"))
" abcdeeefghhijklmnoooopqrrsttuuvwxyz"
``````
Booyah! That’s what I’m talking about! So if we have, say [5,1,9,4,6,7,3] and we want to sort it, this algorithm will first take the head, which is 5 and then put it in the middle of two lists that are smaller and bigger than it. So at one point, you’ll have [1,4,3] ++ [5] ++ [9,6,7]. We know that once the list is sorted completely, the number 5 will stay in the fourth place since there are 3 numbers lower than it and 3 numbers higher than it. Now, if we sort [1,4,3] and [9,6,7], we have a sorted list! We sort the two lists using the same function. Eventually, we’ll break it up so much that we reach empty lists and an empty list is already sorted in a way, by virtue of being empty. Here’s an illustration:
An element that is in place and won’t move anymore is represented in orange. If you read them from left to right, you’ll see the sorted list. Although we chose to compare all the elements to the heads, we could have used any element to compare against. In quicksort, an element that you compare against is called a pivot. They’re in green here. We chose the head because it’s easy to get by pattern matching. The elements that are smaller than the pivot are light green and elements larger than the pivot are dark green. The yellowish gradient thing represents an application of quicksort.
## Thinking recursively
We did quite a bit of recursion so far and as you’ve probably noticed, there’s a pattern here. Usually you define an edge case and then you define a function that does something between some element and the function applied to the rest. It doesn’t matter if it’s a list, a tree or any other data structure. A sum is the first element of a list plus the sum of the rest of the list. A product of a list is the first element of the list times the product of the rest of the list. The length of a list is one plus the length of the tail of the list. Et cetera, et cetera …
Of course, these also have edge cases. Usually the edge case is some scenario where a recursive application doesn’t make sense. When dealing with lists, the edge case is most often the empty list. If you’re dealing with trees, the edge case is usually a node that doesn’t have any children.
It’s similar when you’re dealing with numbers recursively. Usually it has to do with some number and the function applied to that number modified. We did the factorial function earlier and it’s the product of a number and the factorial of that number minus one. Such a recursive application doesn’t make sense with zero, because factorials are defined only for positive integers. Often the edge case value turns out to be an identity. The identity for multiplication is 1 because if you multiply something by 1, you get that something back. Also when doing sums of lists, we define the sum of an empty list as 0 and 0 is the identity for addition. In quicksort, the edge case is the empty list and the identity is also the empty list, because if you add an empty list to a list, you just get the original list back.
So when trying to think of a recursive way to solve a problem, try to think of when a recursive solution doesn’t apply and see if you can use that as an edge case, think about identities and think about whether you’ll break apart the parameters of the function (for instance, lists are usually broken into a head and a tail via pattern matching) and on which part you’ll use the recursive call. |
# Formula's For Seating Arrangements
25 April 2024
In seating arrangement questions, you have to arrange a group of persons fulfilling certain conditions. This is also written as sitting arrangement or sitting arrangement reasoning in some references.
Formula's For Seating Arrangements
General Guidelines
1. Understand the entire question and statements quickly and correctly
2. Determine the usefulness of each information and classify them accordingly into
• Definite Information
• Comparative information
• Negative Information
3. Identify the statements that give definite information.
For instance let us take three statements and evaluate them
Statement (a): Ajay is to the left of Bhanu.
The data in the statement is basic but not definite as the statement only says that Ajay is to the left of Bhanu. But, it does not specify where Ajay is located from Bhanu.
-Statement (b): Ajay is second to the left of Bhanu.
The data in the statement is definite as it clearly states that Ajay is placed second to the left of Bhanu.
-Statement (c):Tanmay is between Ramesh who plays football and Pankaj in order of seating in a row.
It can be understood as ‘Tanmay is between Ramesh and Pankaj. So, they may be seated as RTP or PTR (so, the data is not definite) and ‘Ramesh plays football’.
1. Search for the connecting information.
2. Figure out the seating arrangement by clearly identifying the directions
Linear Arrangement
If they are facing North,
1. B,C,D, E are right of A but only B is immediate right of A.
2. D,C,B,A are left of E but only D is the immediate left of E.
3. A is the immediate left of B while E is the immediate right of D.
If they are facing South,
1. B,C,D, E are left of A but only B is immediate left of A.
2. D,C,B,A are right of E but only D is the immediate right of E.
3. A is the immediate right of B while E is the immediate left of D.
Arrangement Puzzle
Tips for how to solve Arrangement Puzzle
Focus on Family-tree first: You can’t successfully start placing people in the seating arrangement if you don’t know the family tree. Usually people in the seating arrangement are referred to as: wife of C, son of D. So it is better if you know who is who and then start with the seating arrangement. To do this, read all conditions in the question carefully from the beginning and ignore all statement about who is sitting where.
Find starting points: Starting Points help in identifying the position of specific persons. Generally, the question begins with negative information which doesn’t highlight the exact position. Remember, statements starting with “Neither/Nor” can only reveal relative positions and can never be a starting point. You must read all the statements carefully to be able to arrive at the starting point.
Find connecting dots: Connecting dots are small pieces of information which help in the further arrangement based on the position of people already seated.
Solve questions purely on blood relations first: If you are unable to solve the final seating arrangement but have already drawn the family tree, the answer the relevant questions. Don’t leave the entire block just because you couldn’t solve it completely.
ARRANGEMENTS
Seating arrangement is arranging people in their perspective position based on the data specified in the given question. Arrangements can be made in any form like row, circle, triangle, square, rectangle etc. The position can be inferred by the direction and that directions deter- mine the types of seating arrangement
1. Linear Arrangement
2. Circular Arrangement
3. Rectangular / Square Arrangement
4. Hexagonal Arrangement
In order to solve seating arrangement questions, first of all diagram should be made. By doing so questions are easily and quickly solved.
To test yourself, practice Seating Arrangement questions provided by Talent Battle- https://bit.ly/3g9eBp1
To test yourself, click on the FREE TEST SERIES provided by Talent Battle - https://bit.ly/3dmvTNa
## Any Questions? Look Here.
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Syllogism Questions |
Factors and multiples for OCR GCSE Maths
1. Prime numbers
2. Factors
3. Highest common factor
4. Multiples
5. Lowest common multiple
Prime numbers are a special group of numbers that only have two factors. That is, prime numbers are numbers that can only be divided perfectly by two distinct numbers: 1 and themselves.
A factor is a whole number, an integer, that divides another number exactly. Factors can be identified by discovering factor pairs of a number which are numbers that when multiplied together produce the number being inspected. For example, 1 and 24, 2 and 12, 3 and 8, and 4 and 6 are all factor pairs of 24 which means that the list of factor of 24 is 1, 2, 3, 4, 6, 8, 12 and 24.
Highest common factor (HCF) is the largest common factor between two or more numbers.
To find the HCF between two numbers, a tree diagram can be used to break down numbers into their prime factors and calculate the product of the common prime factors of both numbers. Prime factors are the smallest prime numbers that when multiplied produce the number being inspected.
Alternatively, use a Venn diagram to list the prime factors of the two numbers to find their HCF as the product of the common prime factors of both numbers.
Multiples refer to the numbers in the times table of any number. For example, 15 is a multiple of 5 and it is also a multiple of 3. 15 can, then, be said to be a common multiple of 5 and 3.
Lowest common multiple (LCM) is the smallest common multiple between two numbers.
To find the LCM between two numbers, you may use a tree diagram to break down numbers into their prime factors and calculate the product of both the highest common factor (HCF) and the prime factors left behind after calculating the HCF.
Prime factors are the smallest prime numbers that when multiplied produce the number being inspected.
Alternatively, use a Venn diagram to list the prime factors of the two numbers to find the LCM as the product of the numbers in all sections of the Venn diagram.
1
Is 250 a common multiple of 17 and 5?
250 is divisible by 5 but not by 17.
no
2
Harim has saved £67 and he would like to use this money to donate equally as much as possible to 4 charities. All of these charities accept only whole-pound amounts. How much of what Harim has saved will be leftover after the donations?
64 is the next largest number that fits into 67 and is divisible by 4.
67 − 64 = 3
Therefore, Harim will have £3 leftover.
£3
3
How many prime numbers are there?
The only condition is that the number be divisible by 1 and itself.
an infinite amount
4
Find the highest common factor of 54 and 72.
Breaking 54 down to its prime factors gives 2, 3, 3, 3.
Breaking 72 down to its prime factors gives 2, 2, 2, 3, 3.
The common prime factors are 2, 3, 3.
HCF = 2 × 3 × 3 = 18
18
5
Is 8 the highest common factor between 16 and 48?
16 is the highest common factor.
no
End of page |
# Tables of 6, 7, 8 and 9 in Your Hands
1,832,476
734
170
At the age of 8 I had to learn the multiplying tables. I've never been good at memorizing lists or tables. It was easy to learn the tables from 1 to 5 but from 6 to 9 it seemed to be way more complicated... A year later I heard this trick on the radio and it saved my life. Since then I've taught it to many other kids. I passed such a bad time at school as I was the only one in my class who didn't know the tables so I hope this trick was useful for any parent or teacher who knew any child in this situation
## Step 1: Ascribe Values
- First put your hands in front of you as shown in the drawing
- In each hand, ascribe a value from 6 to 10 to each finger
## Step 2: How to Multiply
Step 1
Choose the numbers to multiply. Example: 7x8
Step 2
Put together the fingers whoses values you want to multiply.
Step 3
Now count the touching fingers and the ones below them. The number you get will be the tens. Example: 5
Step 4
Now multiply the fingers above the ones touching of the left hand and the ones in the right hand. The number you get will be the units. Example: 3x2=6
**In some cases you will get a number of units bigger than nine, in that case sum both quantities**
Example: 7x6
- Touching fingers + the ones below -> 3
- Fingers above the ones touching in left hand -> 3
3 x 4 = 12
- Fingers above the ones touching in the right hand -> 4
3 (tens)
Now we've got 3 tens and 12 units -> + 12 (units)
---------
42 (final result)
## Step 3: Another Trick for the Table of 9
Here's an extra trick for the whole table of nine.
- First put your hands in front of you
- Then ascribe values from 1 to 10 to your fingers
- Fold the finger whose value you want to multiply nine times
- The fingers remaining unfolded in the left will be the tens
- The fingers remaining unfolded in the right will be the units
Example: 9 x 4
- Fold the fourth finger
- Fingers remaining unfold in the left -3 (tens)
- Fingers remaining unfold in the right -> 6 (units)
- Final result -> 36
Runner Up in the
Back to School Contest
Runner Up in the
Hands-on Learning Contest
## Recommendations
• ### Lamps Class
9,395 Enrolled
## 170 Discussions
this is cool, but i have to do it really fast or ill get detention. i really hope this trick works faster when i do it with my teacher
Good...
Trick for 19th Table
Step 1: Wright odd no. up to 19 ,
Step 2: In Front of odd no. Wright 9 to 0 in descending order...
1 9 19
3 8 38
5 7 57
7 6 76
9 5 95
11 4 114
13 3 133
15 2 152
17 1 171
19 0 190
It's cool but... i'm afraid i don't understand it completely... How does it work with 6x6?
Oh, ok, I answer to myself: the result for 6x6 is as follows:
2 for the tens
4x4 (16) for the units
so 2x10 + 16 = 36
A little bit less easy but it works anyway!
Very cool indeed, would help me in primary school, where it was hard to learn the 7 table :)
I already taught my daughter the one for 9s but I'm not sure the other one is easier than learning them. I found it kind of confusing. But I have them memorized and she doesn't so she might disagree with me. I think I might be able to get a handle on it if we try it together. Is there any trick for the lower digits? She struggles with the entire table.
Seeing a lot of criticisms back and forth, how these tricks aren't really learning it, how anyone can and should be able to memorize their times tables if they just apply themselves, etc. While it may be human nature to universalize our own experiences, it's true that some people really do have extraordinary difficulty with tasks that are simple for most, such as memorizing math facts, and for those, having any hook or peg to hang the thought on, to attach it to something else, whether a mnemonic, or this finger method, can be helpful. So what? If a person who could never remember what 6X9 was through more traditional methods, was able to do it with this one, that hurts no one, and helps the person it helps.
As for it not being an authentic representation of what 6X9 is in terms of arrays, well...that depends. If you closely examine the 9 method on fingers, you will see it does indeed represent something real going on with multiplication of 9s. Even if you draw arrays, you see the same pattern that 9 is always one less than 10, so that however many 9s you have, you have that many more ones, fewer. 9X3 is 3 instances of something being 1 less than 10, so it's only 1 less than 10 the first 10, then the next 10, it's 1 less again, for a total of two less (18...2 less than 20!) and then for each iteration, take again another "one less than 10" cumulatively (27...3 less than 30). I see the same pattern evident in the finger method for the 9s, at least.
In mathematics, there can be many ways to solve the same problem, and still be correct in the end, and all are valid. In fact, the more comfortable you are with finding more than one way to solve a problem, the stronger your overall understanding is.
Agreed. If the traditional methods were enough for everyone, no one would need this so the arguments are moot. Sure, people can memorize but for some it would take so much time to do it that it's no wonder a kid decides it isn't worth it. I like to see these little helps because it represents a broader recognition that not everyone is going to master skills the same way. And I think it's hard on the kids who struggle to see others getting it and feel like they're failures because their skills lie elsewhere.
Hi. I just wanted to say why this works for me (the gf) and where I have upgraded? modified? it to work at an adult level. Kids these days don't really have to hide issues with math, there is almost no stigma now so doing it as written, no big deal. However I wouldn't be comfortable whipping out this method in a business meeting. I spent the whole day practicing this until I could do it fairly quickly.
I figured if I could do it fast enough, no one would really notice.
Then it hit me! To this day I can still sign the entire asl alphabet. This method could be used to memorize the tables but with a way that includes touch and motion.
I started with pinky pinky 36, pinky index 42, etc. It takes less than a second. I think even doing the motions while learning the tables the regular way would have really helped me as a kid. These signs do not interfere with asl numbers as they only use 1 hand.
Finally after I had them kinda memorized, I stopped touching fingers physicallly and only did it in my head. If I get stuck I go back to signing but I feel like I won't need to touch very often.
I hope this makes sense and I didnt ruin pinky swears forever lol
I dunno about there being no stigma. With my kids they kept getting timed tests they couldn't complete and the teachers would never get off their backs about it if they don't have these memorized.Then they blame the parents for not grilling the kid every waking moment until they learn them...
Sorry to disappoint you all, but no matter WHAT method one uses it will be WRONG! The << "METHOD" >> must be exactly as the "COMMON CORE" says. Even if the answer is correct, it is WRONG!
3 replies
"Method" ---
If I say, 1 + 2 = 3 (which is correct) Common core say's it's wrong.
Common Core say's, 2 + 1 = 3 is correct.
This is just a "Simple" example.
Whew, yeah... and if you learn it the right way, it isn't enough because then they'll make you learn another way to do it whether you need it or not!
I absolutely loved this method. Although i am one of those individuals that actually memorized my tables. I can certainly appreciate this...these new metjods in helping my grandson. I do however, agree with pskvorc, yes this method of multibles may 'appear' to be easy. It really can be a problem during testing because the foundation of why and how are actually not being used. Its not easy...math...but once you get a method that works for you...make it yours. I learned subtraction by adding. Ive never subtracted a problem...EVER!
Haters gonna hate, ignore them, this is cool. I love tricks with numbers. Thanks. |
# Kids Math
## Glossary and Terms: Graphs and Lines
Abscissa - The horizontal line, or x-axis, of a graph.
Arc - A portion of the circumference of a circle.
Axis - One of the lines that is used to form a graph. There is the horizontal x-axis and the vertical y-axis in a two dimensional graph.
Example of x-axis, y-axis, and coordinates on a graph
Bisect - To bisect an object is to divide it into two equal halves.
Collinear - A set of three or more points that are lying on the same straight line are collinear.
Coordinates - A set of two numbers that indicate where a point is on a graph. The first number indicates the x-axis and the second number the y-axis. Other names include ordered pair and numbered pair.
Coplanar lines - Two or more lines that are on the same plane or flat surface.
Diameter - A line segment that passes through the center of a circle with each end point being on the circumference.
Endpoint - The point at the end of a line segment or ray.
Horizontal - A flat or level line or plane that is perpendicular to the vertical.
Intersecting lines - Two or more lines that meet at a point are intersecting.
Line - A straight object that is infinitely long and thin. It is only in one dimension.
Line segment - A portion of a line with two endpoints.
Midpoint - The point of a line segment that is the same distance from both endpoints.
Noncollinear points - A set of three points that are not located on the same line.
Number pair - Two numbers that represent a point on a graph, also called the coordinates.
Ordinate - The vertical line, or y-axis, of a graph.
Origin - The origin is the point where the X and Y axis intersect on a graph. This is the point (0,0) in a two-dimensional graph.
Parallel lines - Lines that never intersect or cross are parallel lines.
Parallel lines
Perpendicular lines - Two lines that form a right angle (90 degrees) are perpendicular lines.
Perpendicular lines
Ray - A line that has one endpoint, but extends forever in one direction.
Slope - A number that indicates the incline or steepness of a line on a graph. Slope equals the "rise" over the "run" of a line on a graph. This can also be written as the change in y over the change in x.
Example: If two points on a line are (x1, y1) and (x2, y2), then the slope = (y2 - y1) ÷ (x2-x1).
Tangent - A line that touches an object such as an arc or circle at a single point.
The green line is tangent to the circle
Transversal - A transversal is a line that crosses two or more other lines.
Vertical - A line or plane that is upright and perpendicular to the horizontal.
More Math Glossaries and Terms
Algebra glossary
Angles glossary
Figures and Shapes glossary
Fractions glossary
Graphs and lines glossary
Measurements glossary
Mathematical operations glossary
Probability and statistics glossary
Types of numbers glossary
Units of measurements glossary
Back to Kids Math
Back to Kids Study |
# 009A Sample Final 3, Problem 5
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Calculate the equation of the tangent line to the curve defined by ${\displaystyle x^{3}+y^{3}=2xy}$ at the point, ${\displaystyle (1,1).}$
Foundations:
The equation of the tangent line to ${\displaystyle f(x)}$ at the point ${\displaystyle (a,b)}$ is
${\displaystyle y=m(x-a)+b}$ where ${\displaystyle m=f'(a).}$
Solution:
Step 1:
We use implicit differentiation to find the derivative of the given curve.
Using the product and chain rule, we get
${\displaystyle 3x^{2}+3y^{2}y'=2y+2xy'.}$
We rearrange the terms and solve for ${\displaystyle y'.}$
Therefore,
${\displaystyle 3x^{2}-2y=2xy'-3y^{2}y'}$
and
${\displaystyle y'={\frac {3x^{2}-2y}{2x-3y^{2}}}.}$
Step 2:
Therefore, the slope of the tangent line at the point ${\displaystyle (1,1)}$ is
${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {3(1)^{2}-2(1)}{2(1)-3(1)^{2}}}\\&&\\&=&\displaystyle {\frac {3-2}{2-3}}\\&&\\&=&\displaystyle {-1.}\end{array}}}$
Hence, the equation of the tangent line to the curve at the point ${\displaystyle (1,1)}$ is
${\displaystyle y=-1(x-1)+1.}$
Final Answer:
${\displaystyle y=-1(x-1)+1}$ |
Lesson 7.4B M.3.G.1 Calculate probabilities arising in geometric contexts (Ex. Find the probability of hitting a particular ring on a dartboard.) M.3.G.2.
Presentation on theme: "Lesson 7.4B M.3.G.1 Calculate probabilities arising in geometric contexts (Ex. Find the probability of hitting a particular ring on a dartboard.) M.3.G.2."— Presentation transcript:
Lesson 7.4B M.3.G.1 Calculate probabilities arising in geometric contexts (Ex. Find the probability of hitting a particular ring on a dartboard.) M.3.G.2 Apply, using appropriate units, appropriate formulas (area, perimeter, surface area, volume) to solve application problems involving polygons, prisms, pyramids, cones, cylinders, spheres as well as composite figures, expressing solutions in both exact and approximate forms M.3.G.3 Relate changes in the measurement of one attribute of an object to changes in other attributes (Ex. How does changing the radius or height of a cylinder affect its surface area or volume?) R.4.G.5 Investigate and use the properties of angles (central and inscribed) arcs, chords, tangents, and secants to solve problems involving circles
Probability Review Probability: The chance that a desired outcome will occur out of all possible outcomes
Probability Review A probability will always be a decimal between 0 and 1 We express a probability as a percent
Non-Geometric Example Josh has 6 red skittles, 4 purple skittles, 3 green skittles, and 9 yellow skittles in a bag. If he draws one skittle at random, what is the probability that he will draw a green skittle?
Geometric Probability Geometric probability: Given a figure, the probability that a randomly chosen point would lie in the preferred region
Example A dart is thrown at random at this dart board. If the dart hits the board, find the probability to the nearest percent that it will land in the shaded region. Note: on some problems the probability can just be reasoned! 4 6
More Examples A dart is thrown at random at this dart board. If the dart hits the board, find the probability to the nearest percent that it will land in the shaded region.
Example A dart is thrown at random at this dart board. If the dart hits the board, find the probability to the nearest percent that it will land in the shaded region. What do you notice about this? If dimensions aren’t given, CHOOSE YOUR OWN A good number to use is, An Even Number
Example A dart is thrown at random at this dart board. If the dart hits the board, find the probability to the nearest percent that it will land in the shaded region. This is a square with semi-circles.
Now You Try… A dart is thrown at random at this dart board. If the dart hits the board, find the probability to the nearest percent that it will land in the shaded region. 12’ 6’
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# How Do You Do Inverse Relations And Functions?
Change the value of fleft (xright) to y. Increase the value of fleft (xright) to y.
## What are inverse functions in math?
Inverse Functions are a type of function that operates in the opposite direction of the input. It is termed the inverse relation of the original relation when the ordered pairs of a relation R are reversed. The new set of ordered pairs is denoted by the symbol.
## How do you find the inverse of a relation?
It is termed the inverse relation of the original relation when the ordered pairs of a relation R are reversed. The new set of ordered pairs is denoted by the symbol. Suppose R = (1,2), (3,8), and (56), and you want to discover the inverse relationship of R. (The inverse relationship of R is denoted by the symbol R –1).
## How to find the inverse of f (x)?
Summary. The inverse of f(x) is denoted by the symbol f-1 (y) We can determine the inverse by flipping the ‘flow diagram’ on its head. Alternatively, we may use Algebra to determine the inverse: Substitute ‘y’ for ‘f(x),’ and so on. Calculate the value of x. It is possible that we will need to restrict the domain of the function in order for it to have an inverse.
## What is the inverse of 4?
When you take 0– thus f of 0 is equal to 4, you get our function. Our function is responsible for mapping the numbers 0 to 4. If you take the inverse function of 4, the inverse function of 4 is equal to zero. Alternatively, the inverse function transforms us from 4 to 0. That is precisely what we had anticipated.
You might be interested: What Are The 6 Companies That Control The Media?
## How do you solve inverse functions step by step?
1. First, enter the function to be solved in the input box (across the text that says ″the inverse function″)
2. second, click on the ″solve″ button.
3. Click on the ″Submit″ button, which is located at the bottom of the calculator window.
4. It will not be long before a new window will appear, and in it the inverse of the function you specified will be computed.
## How to solve an inverse function?
1. In order to denote the inverse function.
2. Rewrite the inverted equation in terms of y to make it more understandable.
3. You will need to isolate the y variable by using a mixture of algebraic procedures and taking care to conduct the same operation on both sides of the equation in the same manner.
In the case of the working equation The domain and range of the inverse function must be determined. |
# Semi Perimeter
In geometry, we might have come across different types of formulas such as perimeter, area, height, volume, etc. Semi perimeter is one of the most commonly used measurements when dealing with polygons. It is the measurement related to plane figures, i.e. two-dimensional shapes. Semi perimeter can be calculated for different plane figures using simple formulas. In this article, you will learn the definition of semi perimeter and the formula of semi perimeter for different shapes, along with examples.
Learn: Perimeter
## What is Semi Perimeter?
As we know, the perimeter is the distance around the shape, whereas the semi perimeter is half its perimeter. For any given polygon, the semi perimeter can be calculated by dividing its circumference by 2. Although it has a simplistic derivation from the perimeter, the semi perimeter frequently appears in formulas related to triangles and other shapes to give it a separate name. If the semi perimeter is part of a formula, it is represented by the letter “s”.
## Semi Perimeter Formula
From the definition of semi perimeter, we can write the formula as:
Semi perimeter = Perimeter/2
However, semi perimeter formulas for various shapes and polygons are tabulated below:
Semi Perimeter of a Triangle formulas Semi Perimeter of Curved Shapes Formulas Semi Perimeter of Quadrilaterals Formulas Shape Formula Explanation Semi perimeter of Equilateral triangle 3a/2 a = Length of the side of an equilateral triangle Semi perimeter of Isosceles triangle a + (b/2) a = Length of congruent sides b = Length of the third side Semi perimeter of Right angle triangle (base + height + hypotenuse)/2 Height = Perpendicular Hypotenuse = The longest side Semi perimeter of Scalene triangle (a + b + c)/2 a, b, c are the measures of lengths of three sides Semi Perimeter of Circle (2πr)/2 or πr r = Radius of circle Semi perimeter of Semicircle (πr + 2r)/2 r = Radius of the semicircle Semi Perimeter of Rectangle Formula 2(l + b)/2 or l + b l = Length b = Breadth Semi Perimeter of Square (4a/2) or 2a a = Side of a square Semi perimeter of any quadrilateral (a + b + c + d)/2 a, b, c, d are the length of sides of the quadrilateral
Also, check: What is the length?
### Applications of Semi Perimeter
The formula of semi perimeter can be applied in many geometry formulas. Let’s have a look at one of the important formulas related to triangles where we apply semi perimeter.
Heron’s formula = √[s(s – a)(s – b)(s – c)]
This formula is used to find the area of a scalene triangle.
Here,
s = Semi perimeter = (a + b + c)/2
a, b and c are the lengths of sides.
Similarly, we can calculate the area of a cyclic quadrilateral using semi perimeter as:
Area = √[(s – a)(s – b)(s – c)(s – d)]
This is the simplest form of Brahmagupta’s formula.
Here,
Semi perimeter = s = (a + b + c + d)/2
### Solved Examples
Example 1:
Find the semi perimeter of a triangle whose sides are 14 cm, 8 cm and 11 cm.
Solution:
Let the measure of slides of a triangle be:
a = 14 cm
b = 8 cm
c = 11 cm
Semi perimeter = s = (a + b + c)/2
= (14 + 8 + 11)/2
= 33/2
= 16.5
Therefore, the semi perimeter of a given triangle is 16.5 cm.
Example 2:
What is the semi perimeter of a square whose side measure 15 cm?
Solution:
Given,
Side of a square = a = 15 cm
Semi perimeter = 2a = 2 × 15 = 30 cm
Alternatively,
Perimeter of a square = 4a = 4 × 15 = 60 cm
Semi perimeter of square = Perimeter/2 = 60/2 = 30 cm
## Frequently Asked Questions on Semi Perimeter
Q1
### What is the semi perimeter formula?
Semi perimeter is the half of perimeter for any shape. That means, by dividing the perimeter of a given figure by 2, we get the semi perimeter.
Thus, the formula of semi perimeter is:
Semi perimeter = Perimeter/2
Q2
### How do you denote the semi perimeter?
The letter “s” is used to denote the semi perimeter of a given shape in geometry.
Q3
### What is the formula of the semi perimeter of a rectangle?
The formula for the semi perimeter of a rectangle is given by:
Semi perimeter of rectangle = 2(l + b)/2 = (l + b)
That means the semi perimeter of a rectangle is equal to the sum of the measures of length and breadth.
Q4
### What is the correct form of semi perimeter in Heron’s formula?
The correct form of semi perimeter in Heron’s formula is: s = (a + b + c)/2
Q5
### What is the semi perimeter of the isosceles triangle?
The formula for finding the semi perimeter of an isosceles triangle is:
s = a + (b/2) |
# REPRESENTING RATIOS AND RATES
## About "Representing ratios and rates"
Representing ratios and rates :
Comparison of two quantities of same kind with same units is represented as a ratio.
For example, comparison of number of boys and girls in a school is represented as ratio between number of boys and girls.
On the other hand, comparison of the given measure to one unit of another measure is represented as rate.
For example, number of miles covered by a car in one hour is represented as rate.
## Applying ratios and rates - Examples
Example 1 :
The average age of three boys is 25 years and their ages are in the proportion 3:5:7. The age of the youngest boy is
Solution :
From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.
Average age of three boys = 25
(3x+5x+7x)/3 = 25 ----------> 15x = 75 -----------> x = 5
Age of the first boy = 3x = 3(5) = 15
Age of the first boy = 5x = 5(5) = 25
Age of the first boy = 7x = 7(5) = 105
Hence the age of the youngest boy is 15 years.
Let us look at the next problem on "Representing ratios and rates"
Example 2 :
John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7:6, find his new weight.
Solution :
Original weight of John = 56.7 kg (given)
He is going to reduce his weight in the ratio 7:6
His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg.
Hence his new weight = 48.6 kg
Let us look at the next problem on "Representing ratios and rates"
Example 3 :
The ratio of the no. of boys to the no. of girls in a school of 720 students is 3:5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.
Solution :
Sum of the terms in the given ratio = 3+5 = 8
So, no. of boys in the school = 720x(3/8)= 270
No. of girls in the school = 720x(5/8)= 450
Let "x" be the no. of new boys admitted in the school.
No. of new girls admitted = 18 (given)
no. of boys in the school = 270+x
no. of girls in the school = 450+18 = 468
The ratio after the new admission is 2 : 3 (given)
So, (270+x) : 468 = 2 : 3
3(270+x) = 468x2 (using cross product rule in proportion)
810 + 3x = 936
3x = 126
x = 42
Hence the no. of new boys admitted in the school is 42
Let us look at the next problem on "Representing ratios and rates"
Example 4 :
The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves \$50 per month, find the monthly income of the second person.
Solution :
From the given ratio of incomes ( 4 : 5 ),
Income of the 1st person = 4x
Income of the 2nd person = 5x
(Expenditure = Income - Savings)
Then, expenditure of the 1st person = 4x - 50
Expenditure of the 2nd person = 5x - 50
Expenditure ratio = 7 : 9 (given)
So, (4x - 50) : (5x - 50) = 7 : 9
9(4x - 50) = 7(5x - 50)
(using cross product rule in proportion)
36x - 450 = 35x - 350
x = 100
Then, income of the second person is
= 5x = 5(100) = 500.
Hence, income of the second person is \$500
Let us look at the next problem on "Representing ratios and rates"
Example 5 :
If the angles of a triangle are in the ratio 2:7:11, then find the angles.
Solution :
From the ratio 2 : 7 : 11,
the three angles are 2x, 7x, 11x
In any triangle, sum of the angles = 180
So, 2x + 7x + 11x = 180°
20x = 180 -------> x = 9
Then, the first angle = 2x = 2(9) = 18°
The second angle = 7x = 7(9) = 63°
The third angle = 11x = 11(9) 99°
Hence the angles of the triangle are (18°, 63°, 99°)
Example 6 :
In a business, if A can earn \$ 7500 in 2.5 years, find the unit rate of his earning per month.
Solution :
Given : Earning in 2.5 years = \$ 7500
1 year = 12 months
2.5 years = 2.5 x 12 = 30 months
Then, earning in 30 months = \$ 7500
Therefore, earning in 1 month = 7500 / 30 = \$ 250
Hence, the unit rate of his earning per month is \$ 250
Example 7 :
If David can prepare 2 gallons of juice in 4 days, how many cups of juice can he prepare per day ?
Solution :
No of gallons of juice prepared in 4 days = 2 gallons
1 gallon = 16 cups
So, no. of cups of juice prepared in 4 days = 2 x 16 = 32 cups
Therefore, David can prepare 32 cups of juice in 4 days.
Then, no. of cups of juice prepared in 1 day = 32 / 4 = 8
Hence, David can prepare 8 cups of juice in 1 day.
Example 8 :
If John can cover 360 miles in 3 hours, find the number of miles covered by John in 1 minute.
Solution :
No of miles covered in 3 hours = 360
Then, no. of miles covered in 1 hour = 360 / 3 = 180
1 hour = 60 minutes
So, no. of miles covered in 60 minutes = 180
Then, no. of miles covered 1 minute = 180 / 60 = 3
Hence, John can cover 3 miles in 1 minute.
Example 9 :
Shanel walks 2/ 5 of a mile every 1/7 hour. Express her speed as a unit rate in miles per hour.
Solution :
Given : Shanel walks 2/ 5 of a mile every 1/7 hour
We know the formula for speed.
That is, Speed = Distance / time
Speed = (2/5) / (1/7)
Speed = (2/5) x (7/1)
Speed = 14 / 5
Speed = 2.8 miles per hour.
Hence, the speed of Shanel is 2.8 miles per hour
Example 10 :
Declan use 2 /35 of a gallon of gas for every 4 /5 of a mile that he drives. At this rate, how many miles can he drive on one gallon of gas?
Solution :
Given : In 2 /35 of a gallon of gas, 4 /5 of a mile is traveled
Then, in 1 gallon of gas = (4/5) x (35/2) miles traveled.
= 14 miles traveled.
Hence, Declan can drive 14 miles in 1 gallon of gas
After having gone through the stuff given above, we hope that the students would have understood "Representing ratios and rates".
Apart from "Representing ratios and rates", if you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# AP Board Class 9 Maths Chapter 14 Probability
The topic of probability is an important topic for Class 9 Maths students of Andhra board. They basically have to study the subject till Class 12. However, in Class 9, students will first be introduced to the topic in AP Board Class 9 Maths Chapter 14. In this unit, students will learn and understand, what probability means and go through some basic or fundamental concepts. The lessons on AP Board Class 9 Maths Chapter 14 Probability also include the following topics:
• Samples in probability
• Probability of events
• Probability problems
## What is Probability?
Generally, probability is defined as the likelihood or chance of an event that can occur. Probability plays a major role in many of our activities. However, there are circumstances and many events where we cannot accurately predict their outcome. Therefore, we use the idea of probability to determine what is or what could likely happen. A common example of probability would be tossing a coin and predicting the outcome either as heads or tails.
Probability is usually quantified as a number between 0 and 1 where,
• 0 indicates impossibility
• 1 indicates certainty
• The higher the probability of an event, the more likely it is that the event will occur.
### Probability Formula
Students can check out some of the chapter questions along with their solutions to know more about the concept.
Question 1: A die has six faces numbered from 1 to 6. It is rolled and number on the top face is noted. When this is treated as a random trial.
1. What are the possible outcomes?
2. Are they equally likely? Why?
3. Find the probability of a composite number turning up on the top face.
Solution:
1. The number of possible outcomes is 6
2. Yes, they are equally likely because the probability of all the outcomes to occur individually is the same.
3. 4 and 6 are the only two composite number in the list. So, the probability of the composite number to turn up on the face can be calculated as follows:
= $\frac{n(E)}{n(S)}=\frac{2}{6}$
Question 2:
What is the probability that a randomly thrown dart that hits the square board in shaded region?
Solution:
The radius of the circle inside the box is 2 cm.
Hence, the area of the circle can be calculated as
$A=\pi r^2$ $A=3.14\times 2^2=12.56\, cm^2$
The length of the square is 2cm + 2cm = 4 cm
Hence, the area of the square is $16 cm^2$
The area of the shaded region of the square board is
$16-12.56=3.44\,cm^2$
Hence, the probability can be calculated as follows:
$\frac{3.44}{16}=0.215$
The probability of a randomly thrown dart that hits the square board in shaded region is 0.215
Question 3:Â A coin is tossed 100 times and the following outcomes are recorded.
Head:45 times      Tails:55 times from the experiment
a) Compute the probability of each outcome.
b) Find the sum of probabilities of all outcomes.
Answer: Given that a coin is tossed 100 times, outcomes was head 45 times, while it showed tails 55 times.Â
(a) So, then, probability of head outcome is 100/45=0.45 and the probability of tails outcome is 100/ 55=0/55
(b) The total outcomes 44+ 55, so that probability of the total outcome is 100/100
Know, more about Andhra Pradesh Board and access various study materials including syllabus, exam timetable, etc. at BYJU’s. |
# Geometry/Chapter 19
## Solving Right Triangles
In order to solve a right triangle using Trig, a simple acronym is used: SOH-CAH-TOA; SOH, standing for sine (opposite/hypotneuse), CAH, standing for cosine (adjacent/hypotneuse), and TOA, standing for tangent (opposite/adjacent).
This work is mainly done using a calculator, as there are no simple, analytical, formulas for the sine, cosine, and tangent functions.
To solve for angles of a right triangle, you would use the same method of SOH-CAH-TOA, except you use ${\displaystyle \arcsin }$ (opp/hyp), ${\displaystyle \arccos }$ (adj/hyp) and ${\displaystyle \arctan }$ (opp/adj)
### Example Problems
Example One: Finding the Missing Parts of a Right Triangle
Find the missing angle and the sides in a right triangle with an acute angle of 38 degrees and an hypotenuse of 15 meters.
Solution:
The other acute angle can be found by realizing that the sum of the angles of a triangle is always 180 degrees. Therefore,
${\displaystyle 180^{\circ }=38^{\circ }+90^{\circ }+(m so ${\displaystyle (m . Using SOHCAHTOA, we see that
${\displaystyle \sin(38^{\circ })}$ = opposite side length / Hypotenuse. Thus, ${\displaystyle \sin(38^{\circ })={\frac {a}{15}}\ \Rightarrow \ a=15\sin(38^{\circ })=9.23}$ meters. To find the adjacent side
we use the cosine function and the formula ${\displaystyle \cos(38^{\circ })={\frac {b}{15}}\ \Rightarrow \ b=15\cos(38^{\circ })=11.82}$ meters.
## Pythagorean Relationships
This is one step beyond the basic trig functions or ideas. It involves three equations that can be manipulated to suit the needs of the given problem. The three equations are as follows:
{\displaystyle {\begin{aligned}\sin ^{2}(x)+\cos ^{2}(x)=1\\1+\tan ^{2}(x)=\sec ^{2}(x)\\1+\cot ^{2}(x)=\csc ^{2}(x)\end{aligned}}}
These can be easily manipulated to figure out a long range of complex problems. For this, we will use the example of 1 - sin²θ. When we look at the above equations, we see that it is very similar to the first one (sin²θ + cos²θ = 1). In fact, sin²θ has just been subtracted. Using this principle, we can thus solve the problem as cos²θ = 1-sin²θ. Easy, right?
### Problems for Practice
Solve these equations using the Pythagorean relationships:
1. tan²θ + 1 =
2. sec²θ - 1 =
3. csc²θ - cot²θ =
4. sin²θ + cos²θ =
5. sec²θ - tan²θ =
6. sin²θ - 1 =
## Verifying More Complex Identities
We now move on to the more complex functions of the trigonometric world. To solve or prove these functions or equations, we have to use all of the math skills we learned before. We will need to use factoring, the Pythagorean relationships table, the inverse trig functions (sec θ = 1/cos θ, csc θ = 1/sin θ, cot θ = 1/tan θ ) and anything else we need to solve the problem.
This is very advanced math. To start this lesson off, lets look at an example:
Prove (make the right side look like the left without touching the left)
sin θ /(sin θ + cos θ) = tan θ /(1 + tan θ)
We can see that tan θ is being divided by 1 + tan θ. To start the problem off, let’s get common denominators:
tan θ /(cos θ/cos θ + sin θ/cos θ)
We can now combine the two equations on the bottom, so it looks like this:
tan θ /((cos θ + sin θ)/cos θ)
To help simplify this down, put tan θ into terms of sin θ and cos θ:
(sin θ/cos θ)/((cos θ + sin θ)/cos θ)
We can now multiply by the reciprocal to get rid of the bottom denominator:
((sin θ/cos θ)*cos θ)/(((cos θ + sin θ)/cos θ)*cos θ)
Note that this is still being divided by sin θ + cos θ. When we multiply, the cos θ ’s will cancel each other out, so we are left with:
sin θ/(sin θ + cos θ)
Which just happens to be the answer we were trying to prove!
While this may seem complicated, you only need to practice this more. Let's get an example that is a little more complicated:
Prove (make the left side look like the right without touching the right side)
(2 sin θ cos θ)/(sin²θ - cos²θ + 1) = cot θ
When we look at an equation like this, it may seem impossible. Lets go through the basics. When we look at the bottom, we see that we have a -cos²θ and a +1. When we look at the Pythagorean relationships table, 1 - cos²θ is equal to sin²θ:
(2 cos θ sin θ)/(sin²θ + sin²θ)
We can then add the two sin²θ together:
(2 sin θ cos θ)/(2 sin²θ)
We can then simplify this further into:
cos θ / sin θ
We can do this because sin θ will cross out to have a single sin θ (2/2²=1/2, right?) and then the twos will cross out. That leaves with cos θ / sin θ, or cot θ !
### Problems for Practice
1. Prove (make the right side look like the left without touching the left side)
(1 - sin θ)/(cos θ) = cos θ /(1 + sin θ)
2. Prove (make the left side look like the right without touching the right side)
(cot θ - tan θ)/(tan θ cos²θ) = csc²θ - sec²θ
3. Prove (make the left side look like the right without touching the right side)
tan θ/(1 + sec θ) + (1 + sec θ)/tan θ = 2 csc θ
4. Prove (make the left side look like the right without touching the right side)
tan²θ/(1 + tan²θ) = sin²θ |
# A projectile is shot from the ground at a velocity of 84 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land?
Aug 5, 2017
$t = 16.5$ $\text{s}$
#### Explanation:
We're asked to find the time it takes for a projectile to land, given its initial velocity.
To do this, we can use the kinematics equation
ul(Deltay = v_0sinalpha_0t - 1/2g t^2
where
• $\Delta y$ is the change in height (which is $0$ when it lands)
• ${v}_{0}$ is the initial speed (given as $84$ $\text{m/s}$)
• ${\alpha}_{0}$ is the launch angle (given as $\frac{7 \pi}{12}$)
• $t$ is the time (what we're trying to find)
• $g = 9.81 \textcolor{w h i t e}{l} {\text{m/s}}^{2}$
Plugging in known values, we have
$0 = \left(84 \textcolor{w h i t e}{l} {\text{m/s")sin((7pi)/12)t - 1/2(9.81color(white)(l)"m/s}}^{2}\right) {t}^{2}$
$\left(4.905 \textcolor{w h i t e}{l} \text{m/s"^2)t^2 = (81.1color(white)(l)"m/s}\right) t$
(4.905color(white)(l)"m/s"^2)t = 81.1color(white)(l)"m/s"
t = color(red)(ulbar(|stackrel(" ")(" "16.5color(white)(l)"s"" ")|) |
# How do you prove the statement lim as x approaches -1.5 for ((9-4x^2)/(3+2x))=6 using the epsilon and delta definition?
Oct 18, 2015
You can't.
#### Explanation:
You can't prove it since it isn't 6.
${\lim}_{x \to - 1.5} \frac{9 - 4 x}{3 + 2}$ does not exist. The limit from the left is $+ \infty$ and the limit from the right is $- \infty$.
Oct 19, 2015
See the explanation.
#### Explanation:
Preliminary analysis
We need to show that for every positive $\epsilon$, there is a positive $\delta$ such that if $0 < \left\mid x - \left(- \frac{3}{2}\right) \right\mid < \delta$ we get $\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid < \epsilon$.
Note first that $\left\mid x - \left(- \frac{3}{2}\right) \right\mid = \left\mid x + \frac{3}{2} \right\mid$
For every $x$ other than $- \frac{3}{2}$, we have:
$\frac{9 - 4 {x}^{2}}{3 + 2 x} = 3 - 2 x$.
So, for every $x$ other than $- \frac{3}{2}$, we have:
$\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid = \left\mid \left(3 - 2 x\right) - 6 \right\mid$
$= \left\mid - 2 x - 3 \right\mid = \left\mid \left(- 4\right) \left(x + \frac{3}{2}\right) \right\mid = \left\mid - 4 \right\mid \left\mid x + \frac{3}{2} \right\mid$.
$= 4 \left\mid x + \frac{3}{2} \right\mid$
We want this to be less than $\epsilon$ and we control, through $\delta$, the size of $\left\mid x + \frac{3}{2} \right\mid$
If we make $\left\mid x + \frac{3}{2} \right\mid < \frac{\epsilon}{4}$, then we will have
$4 \left\mid x + \frac{3}{2} \right\mid < 4 \left(\frac{\epsilon}{4}\right) = \epsilon$ as desired.
Now we are ready to write the proof:
Proof
Given $\epsilon > 0$, let $\delta = \frac{\epsilon}{4}$. Observe that this $\delta$ is also positive, as required.
Now if $x$ is chosen so that $0 < \left\mid x - \left(- \frac{3}{2}\right) \right\mid < \delta$, the we have:
$\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid = \left\mid \left(3 - 2 x\right) - 6 \right\mid$
$= \left\mid - 2 x - 3 \right\mid = \left\mid - 4 \right\mid \left\mid x + \frac{3}{2} \right\mid$.
$= 4 \left\mid x + \frac{3}{2} \right\mid < 4 \delta = 4 \left(\frac{\epsilon}{4}\right) = \epsilon$
That is: if $0 < \left\mid x - \left(- \frac{3}{2}\right) \right\mid < \delta$, then $\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid < \epsilon$.
So, by the definition of limit, ${\lim}_{x \rightarrow - 1.5} \left(\frac{9 - 4 {x}^{2}}{3 + 2 x}\right) = 6$ |
# Question f639b
Aug 19, 2017
Use the Ratio Test (it's usually a good idea whenever factorials are involved) to show that this series converges.
#### Explanation:
Let a_{n}=(2/n)^{n}*n!.
We want to know whether the series ${\sum}_{n = 1}^{\infty} {a}_{n}$ converges or not.
To apply the Ratio Test, we consider the sequence $| {a}_{n + 1} / {a}_{n} |$. In this example, all the terms in the original sum are positive, so we may dispense with the absolute value signs and note that a_{n+1}/a_{n}=(2/(n+1))^(n+1) * (n+1)!*(n/2)^(n)*1/(n!).
This expression simplifies to:
${a}_{n + 1} / {a}_{n} = 2 {\left(\frac{n}{n + 1}\right)}^{n}$.
It is well known that ${\left(\frac{n + 1}{n}\right)}^{n} = {\left(1 + \frac{1}{n}\right)}^{n} \to e$ as $n \to \infty$. Therefore, ${a}_{n + 1} / {a}_{n} \to \frac{2}{e} \approx 0.73576 < 1$ as $n \to \infty$.
By the Ratio Test, this is enough to imply that the series sum_{n=1}^{infty}(2/n)^{n}*n! converges.
Aug 19, 2017
See below.
#### Explanation:
Using the Stirling asymptotic formula
n! approx sqrt(2pin)(n/e)^n we have
sum_(n=1)^oo(2/n)^2n! approx sum_(n=1)^oo(2/n)^2sqrt(2pin)(n/e)^n = sum_(n=1)^oo sqrt(2pin)(2/e)^n and
${\sum}_{n = 1}^{\infty} \sqrt{2 \pi n} {\left(\frac{2}{e}\right)}^{n} \le {\sum}_{n = 1}^{\infty} \left(n + 1\right) {\left(\frac{2}{e}\right)}^{n}$
and from
$\sum \left(k + 1\right) {x}^{k} = \frac{d}{\mathrm{dx}} \left(\sum {x}^{k + 1}\right)$ the convergence for $\sum \left(k + 1\right) {x}^{k}$ is assured if $\left\mid x \right\mid < 1$ or in our case
$\frac{2}{e} < 1$ and as a consequence
sum_(n=1)^oo(2/n)^2n! # converges. |
# Simple Geometry Problems for Struggling Students
If you're having trouble solving simple geometry problems, it can help to see sample problems with solutions. Keep reading for examples of how to solve perimeter, circumference, area and volume problems!
## Solving Geometry Problems
### Perimeter
Perimeter is the distance around the outside of an object, and it's calculated by adding together the lengths of all of the sides. Imagine that you're asked to find the perimeter of a rectangular garden that's 12 feet long and eight feet wide. This means there are two sides that are 12 feet long and two sides that are eight feet long. Here's how to calculate the perimeter:
P = 2L x 2W
= 2(12) + 2(8)
= 24 + 16
= 40 feet
### Circumference
Circumference is the perimeter of a circle. The formula to find it is C = 2(pi)r, where 'pi' equals 3.14 and 'r' is the radius of the circle (r = 1/2 diameter). For example, let's say you're finding the circumference of a circle with a radius of five meters. Here are the calculations you would use:
C = 2(pi)r
= 2(3.14)(5)
= 6.28(5)
= 31.4 meters
### Area
A shape's area is the amount of space it covers. For rectangles, you'll find the area by multiplying length by width. For instance, if you're asked to find the area of a room that's 11 meters long and seven meters wide, you'd multiply 11 x 7 to get the area of 77 square meters.
For circles, area is found using this formula: A = pi(r)^2. The 'r^2' tells you to multiply the circle's radius by itself. Let's say your teacher asks you to find the area of a pizza with a radius of eight inches. Here's how to solve this problem:
A = pi(r)^2
= 3.14 x 8 x 8
= 3.14 x 64
= 200.96 square inches
### Volume
When you need to figure out how much space a 3-dimensional object takes up, or how much capacity it has, you'll use the volume formula. For rectangles, it's V = length x width x height, and for cubes, it's just V = s^3.
Imagine you're asked to find the volume of a cube-shaped box that's six inches tall. Since it's a cube, you know that the length and width of the box also equal six inches, so you'll find the answer by solving V = 6^3 = 6 x 6 x 6. The box's volume is 216 cubic inches.
For a cylinder, the formula for volume is V = (area of the base) x (height). Since the base of a cylinder is circular, you'll calculate its area using the formula for the area of a circle, A = pi(r)^2. Then you'll multiply this result by the cylinder's height. For instance, the volume of a cylindrical package with a radius of two centimeters and a height of nine centimeters would be found like this:
V = pi * r^2 * h
= 3.14 * 2^2 * 9
= 3.14 * 4 * 9
= 3.14 * 36
= 113.04 cubic centimeters
Did you find this useful? If so, please let others know!
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M2_Bk2_Sol_Ch10_E.pdf
# M2_Bk2_Sol_Ch10_E.pdf - 10 New Progress in Senior...
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© Hong Kong Educational Publishing Co. 98 New Progress in Senior Mathematics Module 2 Book 2 (Extended Part) Solution Guide 10 10 10 10 10 p.119 p.119 h r 2 3 1 π pp.105 – 132 10.1 The required area = e xdx 1 ln p.105 1 ] [ 1 ) (ln ] ln [ 1 1 1 1 = = = = e e e e x e dx x x e x xd x x 10.2 The required area p.106 + + + = 3 1 2 1 0 2 ) 3 4 ( ) 3 4 ( dx x x x dx x x x + + = 3 1 2 3 1 0 2 3 ) 3 4 ( ) 3 4 ( dx x x x dx x x x 3 1 2 3 4 1 0 2 3 4 2 3 3 4 4 2 3 3 4 4 + + = x x x x x x = 3 8 12 5 12 37 = 10.3 Consider the points of intersection of the p.107 two curves. ) 2 ..... ( .......... 3 ) 1 ..( .......... 9 2 2 x y x y = = Substituting (1) into (2), we have 4 9 3 9 2 2 2 = = x x x 2 3 = x or 2 3 (rejected) Hence the curves intersect when . 2 3 = x The required area = 2 3 0 2 2 ) 3 9 ( dx x x 9 2 9 2 27 3 4 9 ) 4 9 ( 2 3 0 3 2 3 0 2 = = = = x x dx x 10.4 Solving 2 + = x y and , 3 x y = p.108 the point of intersection is (1, 3). Solving 2 + = x y and , 2 x y = the point of intersection is (2, 4). The required area + + = 2 1 2 1 0 2 ) 2 ( ) 3 ( dx x x dx x x 2 1 3 2 1 0 3 2 3 2 2 3 2 3 + + = x x x x x 6 7 6 7 + = 3 7 = 10.5 At the points of intersection, p.109 2 3 or 6 5 , 6 1 or 2 1 sin 0 ) 1 )(sin 1 sin 2 ( 0 1 sin sin 2 sin sin 2 1 sin 2 cos 2 2 π π π = = = + = + = = x x x x x x x x x x The points of intersection are ( 6 π , 2 1 ), ( 6 5 π , 2 1 ) and ( 2 3 π , 1). On the left hand side of ( 6 5 π , 2 1 ), sin x cos 2 x is positive. On the right hand side of ( 6 5 π , 2 1 ), sin x cos 2 x is negative. Applications of Definite Integrals
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99 © Hong Kong Educational Publishing Co. Applications of Definite Integrals The required area π π π π + + = 2 3 6 5 6 5 6 ) 2 cos sin ( ) 2 cos (sin dx x x dx x x 2 3 6 5 6 5 6 2 2 sin cos 2 2 sin cos π π π π + + = x x x x 4 3 3 2 3 3 + = 4 3 9 = 10.6 The required area p.110 dy e y = 2 0 2 0 ] [ y e = 1 2 = e 10.7 The required area p.111 + + + = 3 1 2 3 1 1 2 3 ) 3 3 ( ) 3 3 ( dy y y y dy y y y 3 1 2 3 4 1 1 2 3 4 3 2 4 3 2 4 + + = y y y y y y y y ) 4 ( 4 = 8 = 10.8 (a) Slope of the tangent p.112 1 3 ) ( = = x x dx d 1 2 3 = = x x 3 = Equation of L : 2 3 3 3 1 ) 1 ( 3 1 = = = x y x y x y (b) For , 3 x y = 3 1 y x = For , 2 3 = x y 3 2 + = y x The required area + = 1 0 3 1 3 2 dy y y 1 0 3 4 2 4 3 3 2 6 + = y y y 12 1 = 10.9 The required volume p.120 π = 3 0 2 2 ) ( dx x 3 0 5 5 π = x 5 243 π = 10.10 The required volume p.121 π = 1 0 2 ) 1 ( dx x x + π = 1 0 2 ) 1 2 ( dx x x x + π = 1 0 2 3 ) 2 ( dx x x x 1 0 2 3 4 2 3 2 4 + π = x x x 12 π = 10.11 ) 1 ( . .......... 1 2 + = x y p.122 ) 2 ( .......... 5 2 x y = Substituting (1) into (2), 1 or 2 3 0 ) 1 )( 3 2 ( 0 3 2 5 ) 1 ( 2 2 2 = = + = + = + x x x x x x x The curves intersect at ( 4 13 , 2 3 ) and (1, 2).
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# SSAT Upper Level Math : How to find the ratio of a fraction
## Example Questions
### Example Question #3 : How To Find The Ratio Of A Fraction
A pie is made up of crust, apples, and sugar, and the rest is jelly. What is the ratio of crust to jelly?
Explanation:
A pie is made up of crust, apples, sugar, and the rest is jelly. What is the ratio of crust to jelly?
To compute this ratio, you must first ascertain how much of the pie is jelly. This is:
Begin by using the common denominator :
So, the ratio of crust to jelly is:
This can be written as the fraction:
, or
### Example Question #1 : How To Find The Ratio Of A Fraction
In a solution, of the fluid is water, is wine, and is lemon juice. What is the ratio of lemon juice to water?
Explanation:
This problem is really an easy fraction division. You should first divide the lemon juice amount by the water amount:
Remember, to divide fractions, you multiply by the reciprocal:
This is the same as saying:
### Example Question #49 : Proportion / Ratio / Rate
If and , what is the ratio of to ?
Explanation:
To find a ratio like this, you simply need to make the fraction that represents the division of the two values by each other. Therefore, we have:
Recall that division of fractions requires you to multiply by the reciprocal:
which is the same as:
This is the same as the ratio:
### Example Question #2 : How To Find The Ratio Of A Fraction
A smoothie is made up of pineapple juice and mango juice; the rest is orange juice. What is the ratio of orange juice to pineapple juice?
Explanation:
First, find out how much of the smoothie is orange juice.
Since we want the ratio of orange juice to pineapple juice, we want to set up the fractions like this:
Now, divide these fractions.
### Example Question #32 : Proportion / Ratio / Rate
A bakery sells three different types of cookies. of the cookies sold are chocolate chip, sold are oatmeal raisin, and the rest sold are sugar cookies. What is the ratio of sugar cookies to oatmeal raisin cookies that the bakery sells?
Explanation:
First, find the fraction of sugar cookies that are sold.
Since we want the ratio of sugar cookies to oatmeal raisin cookies:
Now, multiply and simplify the resulting fraction.
### Example Question #3 : How To Find The Ratio Of A Fraction
What number can be filled in the blank to make the two ratios equivalent?
________
Explanation:
Set up a proportion statement:
Rewrite: |
# Multi-Step Problems in the Real World
Videos and solutions to help Grade 6 students learn how to analyze an equation in two variables, choose an independent variable and a dependent variable, make a table and make a graph for the equation.
New York State Common Core Math Module 4, Grade 6, Lesson 32
Related Topics:
Lesson Plans and Worksheets for Grade 6
Lesson Plans and Worksheets for all Grades
Lesson 32 Student Outcomes
Students analyze an equation in two variables, choose an independent variable and a dependent variable, make a table and make a graph for the equation by plotting the points in the table. For the graph, the independent variable is usually represented by the horizontal axis and the dependent variable is usually represented by the vertical axis.
Opening Exercise
Xin is buying beverages for a party which are individually packaged and come in packs of 8. Let p be the number of packages Xin buys and t be the total number of beverages. The equation t = 8p can be used to calculate the total number of beverages when the number of packages is known. Determine the independent and dependent variable in this scenario. Then make a table using whole number values of p less than 6.
Example 1
Make a graph for the table in the Opening Exercise.
Example 2
Use the graph to determine which variable is the independent variable and which is the dependent variable. Then state the relationship between the quantities represented by the variables.
Exercises
1. Each week Quentin earns \$30. If he saves this money, create a graph that shows the total amount of money Quentin has saved from week 1 through week 8. Write an equation that represents the relationship between the number of weeks that Quentin has saved his money (w) and the total amount of money he has saved (s). Then name the independent and dependent variable. Write a sentence that shows this relationship.
Exercises
2. Zoe is collecting books to donate. She started with 3 and collects two more each week. She is using the equation b = 2w + 3, where b is the total number of books collected and w is the number of weeks she has been collecting. Name the independent and dependent variables. Then create a graph to represent how many books Zoe has collected when w is 5 or less.
3. Eliana plans to visit the fair. She must pay \$5 to enter the fair grounds and an additional \$3 rrper ride. Write an equation to show the relationship between r, the number of rides, and t, the total cost. State which variable is dependent and which is independent. Then create a graph that models the equation.
Problem Set
1. Caleb started saving money in a cookie jar. He started with \$25. He adds \$10 to the cookie jar each week. Write an equation where w is the number of weeks Caleb saves his money and t is the total amount in dollars in the cookie jar. Determine which variable is the independent variable and which is the dependent variable. Then, graph the total amount in the cookie jar for w being less than 6 weeks.
2. Kevin is taking a taxi from the airport to his home. There is a \$6 flat fee for riding in the taxi. In addition, Kevin must also pay \$1 per mile. Write an equation where m is the number of miles and t is the total cost in dollars of the taxi ride. Determine which variable is independent and which is dependent. Then, graph the total cost for m being less than 6 miles.
3. Anna started with \$10. She saved an additional \$5 each week. Write an equation that can be used to determine the total amount saved in dollars, t, after a given number of weeks, w. Determine which variable is independent and which is dependent. Then, graph the total amount saved for the first 8 weeks.
4. Aliyah is purchasing produce at the farmers’ market. She plans to buy \$10 worth of potatoes and some apples. The apples cost \$1.50 per pound. Write an equation to show the total cost of the produce, where T is the total cost, in dollars, and a is the number of pounds of apples. Determine which variable is dependent and which is independent. Then, graph the equation on the coordinate plane.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# RD Sharma Solutions for Class 8 Chapter - 7 Factorization Exercise 7.3
Students can refer to RD Sharma Solutions for Class 8 Maths Exercise 7.3 Chapter 7 Factorization which are available here. The solutions here are solved step by step for a better understanding of the concepts, which helps students prepare for their exams at ease. Keeping in mind expert tutors at BYJU’S have made this possible to help students crack difficult problems. Students can download the pdf of RD Sharma Solutions from the links provided below.
Exercise 7.3 of Chapter 7 Factorization is on based on the factorization of algebraic expressions when a binomial is a common factor.
## Download the pdf of RD Sharma For Class 8 Maths Exercise 7.3 Chapter 7 Factorization
### Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 7.3 Chapter 7 Factorization
Factorize each of the following algebraic expressions:
1. 6x (2x – y) + 7y (2x – y)
Solution:
We have,
6x (2x – y) + 7y (2x – y)
By taking (2x – y) as common we get,
(6x + 7y) (2x – y)
2. 2r (y – x) + s (x – y)
Solution:
We have,
2r (y – x) + s (x – y)
By taking (-1) as common we get,
-2r (x – y) + s (x – y)
By taking (x – y) as common we get,
(x – y) (-2r + s)
(x – y) (s – 2r)
3. 7a (2x – 3) + 3b (2x – 3)
Solution:
We have,
7a (2x – 3) + 3b (2x – 3)
By taking (2x – 3) as common we get,
(7a + 3b) (2x – 3)
4. 9a (6a – 5b) – 12a2 (6a – 5b)
Solution:
We have,
9a (6a – 5b) – 12a2 (6a – 5b)
By taking (6a – 5b) as common we get,
(9a – 12a2) (6a – 5b)
3a(3 – 4a) (6a – 5b)
5. 5 (x – 2y)2 + 3 (x – 2y)
Solution:
We have,
5 (x – 2y)2 + 3 (x – 2y)
By taking (x – 2y) as common we get,
(x – 2y) [5 (x – 2y) + 3]
(x – 2y) (5x – 10y + 3)
6. 16 (2l – 3m)2 – 12 (3m – 2l)
Solution:
We have,
16 (2l – 3m)2 – 12 (3m – 2l)
By taking (-1) as common we get,
16 (2l – 3m)2 + 12 (2l – 3m)
By taking 4(2l – 3m) as common we get,
4(2l – 3m) [4 (2l – 3m) + 3]
4(2l – 3m) (8l – 12m + 3)
7. 3a (x – 2y) – b (x – 2y)
Solution:
We have,
3a (x – 2y) – b (x – 2y)
By taking (x – 2y) as common we get,
(3a – b) (x – 2y)
8. a2 (x + y) + b2 (x + y) + c2 (x + y)
Solution:
We have,
a2 (x + y) + b2 (x + y) + c2 (x + y)
By taking (x + y) as common we get,
(a2Â + b2Â + c2) (x + y)
9. (x – y)2 + (x – y)
Solution:
We have,
(x – y)2 + (x – y)
By taking (x – y) as common we get,
(x – y) (x – y + 1)
10. 6 (a + 2b) – 4 (a + 2b)2
Solution:
We have,
6 (a + 2b) – 4 (a + 2b)2
By taking (a + 2b) as common we get,
[6 – 4 (a + 2b)] (a + 2b)
(6 – 4a – 8b) (a + 2b)
2(3 – 2a – 4b) (a + 2b)
11. a (x – y) + 2b (y – x) + c (x – y)2
Solution:
We have,
a (x – y) + 2b (y – x) + c (x – y)2
By taking (-1) as common we get,
a (x – y) – 2b (x – y) + c (x – y)2
By taking (x – y) as common we get,
[a – 2b + c(x – y)] (x – y)
(x – y) (a – 2b + cx – cy)
12. -4 (x – 2y)2 + 8 (x – 2y)
Solution:
We have,
-4 (x – 2y)2 + 8 (x – 2y)
By taking 4(x – 2y) as common we get,
[-(x – 2y) + 2] 4(x – 2y)
4(x – 2y) (-x + 2y + 2)
13. x3 (a – 2b) + x2 (a – 2b)
Solution:
We have,
x3 (a – 2b) + x2 (a – 2b)
By taking x2 (a – 2b) as common we get,
(x + 1) [x2 (a – 2b)]
x2 (a – 2b) (x + 1)
14. (2x – 3y) (a + b) + (3x – 2y) (a + b)
Solution:
We have,
(2x – 3y) (a + b) + (3x – 2y) (a + b)
By taking (a + b) as common we get,
(a + b) [(2x – 3y) + (3x – 2y)]
(a + b) [2x -3y + 3x – 2y]
(a + b) [5x – 5y]
(a + b) 5(x – y)
15. 4(x + y) (3a – b) + 6(x + y) (2b – 3a)
Solution:
We have,
4(x + y) (3a – b) + 6(x + y) (2b – 3a)
By taking (x + y) as common we get,
(x + y) [4(3a – b) + 6(2b – 3a)]
(x + y) [12a – 4b + 12b – 18a]
(x + y) [-6a + 8b]
(x + y) 2(-3a + 4b)
(x + y) 2(4b – 3a) |
# Normal Nilpotent Matrix is Zero Matrix
## Problem 336
A complex square ($n\times n$) matrix $A$ is called normal if
$A^* A=A A^*,$ where $A^*$ denotes the conjugate transpose of $A$, that is $A^*=\bar{A}^{\trans}$.
A matrix $A$ is said to be nilpotent if there exists a positive integer $k$ such that $A^k$ is the zero matrix.
(a) Prove that if $A$ is both normal and nilpotent, then $A$ is the zero matrix.
You may use the fact that every normal matrix is diagonalizable.
(b) Give a proof of (a) without referring to eigenvalues and diagonalization.
(c) Let $A, B$ be $n\times n$ complex matrices. Prove that if $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$, where $I$ is the $n\times n$ identity matrix.
## Proof.
### (a) If $A$ is normal and nilpotent, then $A=O$
Since $A$ is normal, it is diagonalizable. Thus there exists an invertible matrix $P$ such that $P^{-1}AP=D$, where $D$ is a diagonal matrix whose diagonal entries are eigenvalues of $A$.
Since $A$ is nilpotent, all the eigenvalues of $A$ are $0$. (See the post “Nilpotent matrix and eigenvalues of the matrix” for the proof.)
Hence the diagonal entries of $D$ are zero, and we have $D=O$, the zero matrix.
It follows that we have
\begin{align*}
A=PDP^{-1}=POP^{-1}=O.
\end{align*}
Therefore, every normal nilpotent matrix must be a zero matrix.
### (b) Give a proof of (a) without referring to eigenvalues and diagonalization.
Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$.
We prove by induction on $k$ that $A=O$.
The base case $k=1$ is trivial.
Suppose $k>1$ and the case $k-1$ holds. Let $B=A^{k-1}$. Note that since $A$ is normal, the matrix $B$ is also normal.
For any vector $x \in \C^n$, we compute the length of the vector $B^*Bx$ as follows.
\begin{align*}
&\|B^*Bx\|=(B^*Bx)^*(B^*Bx) && \text{by definition of the length}\\
&=x^*B^*(BB^*)Bx\\
&=x^*B^*(B^*B)Bx && \text{since $B$ is normal}\\
&=x^* (B^*)^2B^2x,
\end{align*}
and the last expression is $O$ since $B^2=A^{2k-2}=O$ as $k \geq 2$ implies $2k-2 \geq k$.
Hence we have $B^*Bx=\mathbf{0}$ for every $x\in \C^n$.
This yields that
\begin{align*}
\|Bx\|&=(Bx)^*(Bx)\\
&=x^*B^*Bx=0,
\end{align*}
for every $x\in \C^n$, and hence $B=O$.
By the induction hypothesis, $A^{k-1}=O$ implies $A=O$, and the induction is completed.
So the matrix $A$ must be the zero matrix.
### (c) If $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$
We claim that the matrix $B$ is normal as well. If this claim is proved, then part (a) yields that $B=O$ since $B$ is a nilpotent normal matrix, which implies the result $A=I$.
To prove the claim, we compute
\begin{align*}
B^* B&=(I-A)^* (I-A)\\
&=(I-A^*)(I-A)\\
&=I-A-A^*+A^*A,
\end{align*}
and
\begin{align*}
B B^*&=(I-A) (I-A)^*\\
&=(I-A)(I-A^*)\\
&=I-A^*-A+AA^*\\
&=I-A^*-A+A^*A \qquad \text{ since $A$ is normal.}
\end{align*}
It follows that we have $B^* B=BB^*$, and thus $B$ is normal.
Hence, the claim is proved.
### More from my site
• Every Diagonalizable Nilpotent Matrix is the Zero Matrix Prove that if $A$ is a diagonalizable nilpotent matrix, then $A$ is the zero matrix $O$. Definition (Nilpotent Matrix) A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k=O$. Proof. Main Part Since $A$ is […]
• Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even Let $A$ be a real skew-symmetric matrix, that is, $A^{\trans}=-A$. Then prove the following statements. (a) Each eigenvalue of the real skew-symmetric matrix $A$ is either $0$ or a purely imaginary number. (b) The rank of $A$ is even. Proof. (a) Each […]
• Quiz 13 (Part 1) Diagonalize a Matrix Let $A=\begin{bmatrix} 2 & -1 & -1 \\ -1 &2 &-1 \\ -1 & -1 & 2 \end{bmatrix}.$ Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$. That is, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that […]
• How to Diagonalize a Matrix. Step by Step Explanation. In this post, we explain how to diagonalize a matrix if it is diagonalizable. As an example, we solve the following problem. Diagonalize the matrix $A=\begin{bmatrix} 4 & -3 & -3 \\ 3 &-2 &-3 \\ -1 & 1 & 2 \end{bmatrix}$ by finding a nonsingular […]
• Two Matrices with the Same Characteristic Polynomial. Diagonalize if Possible. Let $A=\begin{bmatrix} 1 & 3 & 3 \\ -3 &-5 &-3 \\ 3 & 3 & 1 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 4 & 3 \\ -4 &-6 &-3 \\ 3 & 3 & 1 \end{bmatrix}.$ For this problem, you may use the fact that both matrices have the same characteristic […]
• Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix Let $A$ be an $n\times n$ matrix with real number entries. Show that if $A$ is diagonalizable by an orthogonal matrix, then $A$ is a symmetric matrix. Proof. Suppose that the matrix $A$ is diagonalizable by an orthogonal matrix $Q$. The orthogonality of the […]
• A Matrix Similar to a Diagonalizable Matrix is Also Diagonalizable Let $A, B$ be matrices. Show that if $A$ is diagonalizable and if $B$ is similar to $A$, then $B$ is diagonalizable. Definitions/Hint. Recall the relevant definitions. Two matrices $A$ and $B$ are similar if there exists a nonsingular (invertible) matrix $S$ such […]
• A Square Root Matrix of a Symmetric Matrix Answer the following two questions with justification. (a) Does there exist a $2 \times 2$ matrix $A$ with $A^3=O$ but $A^2 \neq O$? Here $O$ denotes the $2 \times 2$ zero matrix. (b) Does there exist a $3 \times 3$ real matrix $B$ such that $B^2=A$ […]
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##### Linear Transformation $T(X)=AX-XA$ and Determinant of Matrix Representation
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# Arithmetic sequence calculator and problems solver
This online calculator solves common arithmetic sequences problems.
### This page exists due to the efforts of the following people:
#### Timur
Created: 2019-07-06 02:59:18, Last updated: 2021-02-12 11:03:59
This online calculator can solve arithmetic sequences problems. Currently, it can help you with the two common types of problems:
1. Find the n-th term of an arithmetic sequence given m-th term and the common difference. Example problem: An arithmetic sequence has a common difference equal to 10, and its 5-th term is equal to 52. Find its 15-th term.
2. Find the n-th term of an arithmetic sequence given i-th term and j-th term. Example problem: An arithmetic sequence has its 5-th term equal to 12 and its 15-th term equal to 52. Find its 20-th term.
Some formulas and descriptions of the solutions can be found below the calculator.
#### Arithmetic sequence calculator and problems solver
First Term of the Arithmetic Sequence
Common Difference
nth Term of the Sequence Formula
Unknown Term equals to
### Arithmetic sequence
To recall, an arithmetic sequence or arithmetic progression (AP) is a sequence of numbers such that the difference, named common difference, of two successive members of the sequence, is a constant.
Thus, the formula for the n-th term is
$a_n=a_1+(n-1)d$
and in general
$a_n=a_m+(n-m)d$,
where d is the common difference.
You can solve the first type of problems listed above by using the general formula directly or calculating the first term a1, using the formula.
$a_1=a_n-(n-1)d$
And then using the formula for the n-th term.
For the second type of problem, you need to find common difference using the following formula derived from the general formula.
$d=\frac{a_n-a_m}{n-m}$
After that, it becomes the first type of problem.
The calculator above also calculates the first term and general formula for the n-th term of an arithmetic sequence for convenience.
URL copied to clipboard
#### Similar calculators
PLANETCALC, Arithmetic sequence calculator and problems solver |
# Math equations algebra
Math equations algebra can be a helpful tool for these students. So let's get started!
## The Best Math equations algebra
Here, we will be discussing about Math equations algebra. For many centuries, mathematicians have been fascinated by the properties of square roots. These numbers have some unique properties that make them particularly useful for solving certain types of equations. For example, if you take the square root of a negative number, you will end up with an imaginary number. This can be very useful for solving certain types of equations that have no real solution. In addition, square roots can be used to simplify equations that would otherwise be very difficult to solve. For example, if you want to find the value of x that satisfies the equation x^2+2x+1=0, you can use the square root property to simplify the equation and solve it quite easily. As you can see, square roots can be a very powerful tool for solving equations.
Then, work through the equation step-by-step, using the order of operations to simplify each term. Be sure to keep track of any negative signs, as they will change the direction of the operation. Finally, check your work by plugging the value of the variable back into the equation. If everything checks out, you have successfully solved the equation!
Matrices can be used to solve system of equations. In linear algebra, a system of linear equations can be represented using a matrix. This is called a matrix equation. To solve a matrix equation, we need to find the inverse of the matrix. The inverse of a matrix is a matrix that when multiplied by the original matrix, results in the identity matrix. Once we have the inverse of the matrix, we can multiply it by the vector of constants to get the solution vector. This method is called Gaussian elimination.
Any problem, no matter how complex, can be solved if you break it down into smaller, more manageable pieces. The first step is to identify the goal, or what you want to achieve. Once you have a clear goal in mind, you can start to break the problem down into smaller steps that will lead you to your goal. It is important to be as specific as possible when identifying these steps, and to create a timeline for each one. Otherwise, it will be easy to get overwhelmed and lost in the process. Finally, once you have a plan in place, it is important to stick with it and see it through to the end. Only then can you achieve your goal and move on to the next problem.
Logarithmic equation solvers are a type of mathematical software that is used to solve equations that contain logs. Logarithmic equations are equations in which the variable is raised to a power that is itself a logarithm. For example, the equation 2x+5=3 can be rewritten as 10x=3. This equation cannot be solved using traditional methods, but it can be solved using a logarithmic equation solver. Logarithmic equation solvers use a variety of algorithms to solve equations, and they can often find solutions that cannot be found using traditional methods. Logarithmic equation solvers are used by mathematicians, engineers, and scientists to solve a wide range of problems.
## We cover all types of math problems
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This app is incredible. It does things I never would have thought possible by an app that solves problems with pictures. And it even works with handwriting, which is absolutely amazing. Easy 5/5
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Step by step solutions Solve math problem and show steps App to give you math answers Maths buddy 10 grade math problems |
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# Square Roots and Irrational Numbers
## Simplify square roots by factoring
Estimated15 minsto complete
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Practice Square Roots and Irrational Numbers
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Square Roots and Irrational Numbers
What if you had a number like 1000 and you wanted to find its square root? After completing this concept, you'll be able to find square roots like this one by hand and with a calculator.
### Try This
You can also work out square roots by hand using a method similar to long division. (See the web page at http://www.homeschoolmath.net/teaching/square-root-algorithm.php for an explanation of this method.)
### Guidance
The square root of a number is a number which, when multiplied by itself, gives the original number. In other words, if , we say that is the square root of .
Note: Negative numbers and positive numbers both yield positive numbers when squared, so each positive number has both a positive and a negative square root. (For example, 3 and -3 can both be squared to yield 9.) The positive square root of a number is called the principal square root.
The square root of a number is written as or sometimes as . The symbol is sometimes called a radical sign.
Numbers with whole-number square roots are called perfect squares. The first five perfect squares (1, 4, 9, 16, and 25) are shown below.
You can determine whether a number is a perfect square by looking at its prime factors. If every number in the factor tree appears an even number of times, the number is a perfect square. To find the square root of that number, simply take one of each pair of matching factors and multiply them together.
#### Example A
Find the principal square root of each of these perfect squares.
a) 121
b) 225
c) 324
Solution
a) , so .
b) , so .
c) , so .
For more practice matching numbers with their square roots, try the Flash games at http://www.quia.com/jg/65631.html.
When the prime factors don’t pair up neatly, we “factor out” the ones that do pair up and leave the rest under a radical sign. We write the answer as , where is the product of half the paired factors we pulled out and is the product of the leftover factors.
#### Example B
Find the principal square root of the following numbers.
a) 8
b) 48
c) 75
Solution
a) . This gives us one pair of 2’s and one leftover 2, so .
b) , so , or .
c) , so .
Note that in the last example we collected the paired factors first, then we collected the unpaired ones under a single radical symbol. Here are the four rules that govern how we treat square roots.
#### Example C
Simplify the following square root problems
a)
b)
c)
d)
Solution
a)
b)
c)
d)
Approximate Square Roots
Terms like and (square roots of prime numbers) cannot be written as rational numbers. That is to say, they cannot be expressed as the ratio of two integers. We call them irrational numbers. In decimal form, they have an unending, seemingly random, string of numbers after the decimal point.
To find approximate values for square roots, we use the or button on a calculator. When the number we plug in is a perfect square, or the square of a rational number, we will get an exact answer. When the number is a non-perfect square, the answer will be irrational and will look like a random string of digits. Since the calculator can only show some of the infinitely many digits that are actually in the answer, it is really showing us an approximate answer—not exactly the right answer, but as close as it can get.
#### Example D
Use a calculator to find the following square roots. Round your answer to three decimal places.
a)
b)
c)
d)
Solution
a)
b)
c)
d)
Watch this video for help with the Examples above.
### Vocabulary
• The square root of a number is a number which gives the original number when multiplied by itself. In algebraic terms, for two numbers and , if , then .
• A square root can have two possible values: a positive value called the principal square root, and a negative value (the opposite of the positive value).
• A perfect square is a number whose square root is an integer.
• Some mathematical properties of square roots are:
• Square roots of numbers that are not perfect squares (or ratios of perfect squares) are irrational numbers. They cannot be written as rational numbers (the ratio of two integers). In decimal form, they have an unending, seemingly random, string of numbers after the decimal point.
• Computing a square root on a calculator will produce an approximate solution since the calculator only shows a finite number of digits after the decimal point.
### Guided Practice
Find the square root of each number.
a) 576
b) 216
Solution
a) , so .
b) , so , or .
### Practice
For 1-10, find the following square roots exactly without using a calculator, giving your answer in the simplest form.
1. (Hint: The division rules you learned can be applied backwards!)
For 11-20, use a calculator to find the following square roots. Round to two decimal places.
### Vocabulary Language: English
approximate solution
approximate solution
An approximate solution to a problem is a solution that has been rounded to a limited number of digits.
Irrational Number
Irrational Number
An irrational number is a number that can not be expressed exactly as the quotient of two integers.
Perfect Square
Perfect Square
A perfect square is a number whose square root is an integer.
principal square root
principal square root
The principal square root is the positive square root of a number, to distinguish it from the negative value. 3 is the principal square root of 9; -3 is also a square root of 9, but it is not principal square root.
rational number
rational number
A rational number is a number that can be expressed as the quotient of two integers, with the denominator not equal to zero.
Square Root
Square Root
The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9. |
# How do you rationalize 4/(2-3i)?
Mar 18, 2018
$\frac{8 + 12 i}{13}$
#### Explanation:
the difference of two squares identity,
$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$,
can be used to rationalise the $a + b i$ expression.
to rationalise the denominator, multiply it by its conjugate.
the conjugate is found by changing the $-$ sign to $+$.
here, it is $2 + 3 i$.
if both the numerator and denominator are multiplied by $2 + 3 i :$
$4 \cdot \left(2 + 3 i\right) = 8 + 12 i$
$\left(2 + 3 i\right) \left(2 - 3 i\right) = {2}^{2} - {\left(3 i\right)}^{2}$
$= 4 - \left(- 9\right) = 4 + 9$
$= 13$
therefore, the equivalent fraction to $\frac{4}{2 - 3 i}$ where the denominator is rational, is $\frac{8 + 12 i}{13}$. |
# RD Sharma Class 10 Ex 15.2 Solutions Chapter 15 Areas Related To Circles
In this chapter, we provide RD Sharma Class 10 Ex 15.2 Solutions Chapter 15 Areas Related To Circles for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 15.2 Solutions Chapter 15 Areas Related To Circles pdf, Now you will get step by step solution to each question.
# Chapter 15: Areas Related To Circles Exercise – 15.2
### Question: 1
Find in terms of π, the length of the arc that subtends an angle of 30 degrees, at the center of O of the circle with a radius of 4 cm.
### Solution:
Given Data:
Angle subtended at the centre ‘O’ = 30°
Formula to be used: Length of arc = θ/360 × 2πr cm
Length of arc = 30/360 × 2π∗4 cm = 2π/3 cm
Therefore, the Length of arc the length of the arc that subtends an angle of 60 degrees is 2π/3 cm
### Question: 2
Find the angle subtended at the centre of circle of radius 5 cm by an arc of length 5π/3 cm.
### Solution:
Given data: Radius = 5 cm
Length of arc = 5π/3 cm
Formula to be used: Length of arc = θ/360 ∗ 2πr cm
5π/3 cm = θ/360 ∗ 2πr cm
Solving the above equation, we have:
θ = 60°
Therefore, angle subtended at the centre of circle is 60°
### Question: 3
An arc of length cm subtends an angle of 144° at the center of the circle.
### Solution:
Given Data:
Length of arc = 25 cm
θ = Angle subtended at the centre of circle = 144°
Formula to be used:
Length of arc = θ/360 ∗ 2πr cm
θ/360 ∗ 2πr cm = 144/360 east 2πr cm = 4π/5 ∗ r cm
As given in the question, length of arc = cm,
Therefore, cm = 4π/5 ∗ r cm
Solving the above equation, we have r = 25 cm.
Therefore the radius of the circle is found to be 25 cm.
### Question: 4
An arc of length 25 cm subtends an angle of 55° at the center of a circle. Find in terms of radius of the circle.
### Solution:
Given Data:
Length of arc = 25 cm
θ = Angle subtended at the centre of circle = 55°
Formula to be used:
Length of arc = θ/360 ∗ 2πr cm
= 55/36 ∗ 2πr cm
As given in the question length of arc = 25 cm, hence,
25 cm = 55/360 ∗ 2π ∗ r cm
25 = 11πr/36
Therefore, the radius of the circle is 900/11π
### Question: 5
Find the angle subtended at the center of the circle of radius ‘a’ cm by an arc of π/4 length cm.
### Solution:
Given data:
Length of arc = aπ/4 cm
θ = angle subtended at the centre of circle
Formula to be used:
Length of arc = θ/360 ∗ 2πr cm
Length of arc = θ/360 ∗ 2πa cm
θ/360 ∗ 2πa cm = aπ/4 cm
Solving the above equation, we have
θ = 45°
Therefore, the angle subtended at the centre of circle is 45°
### Question: 6
A sector of the circle of radius 4 cm subtends an angle of 30°. Find the area of the sector.
### Solution:
Given Data: Radius = 4 cm
Angle subtended at the centre ‘O’ = 30°
Formula to be used:
Area of the sector = θ/360 ∗ πr2
Area of the sector = 30/360 ∗ π42
Solving the above equation, we have: Area of the sector = 4.9 cm
Therefore, Area of the sector is found to be 4.9 cm2
### Question: 7
A sector of a circle of radius 8 cm subtends an angle of 135. Find the area of sector.
### Solution:
Given Data: Radius = 8 cm
Angle subtended at the centre ‘O’ = 135°
Formula to be used:
Area of the sector = θ/360 ∗ πr2
Area of the sector = 135/360 ∗ π82
= 528/7 cm2
Therefore, Area of the sector calculated is 528/7 cm2
### Question: 8
The area of sector of circle of radius 2 cm is cm2. Find the angle subtended by the sector.
### Solution:
Given Data: Radius = 2 cm
Angle subtended at the centre ‘O’
Area of sector of circle = cm2
Formula to be used:
Area of the sector = θ/360 ∗πr2
Area of the sector = θ/360 ∗ π32
= πθ/90
As given in the question area of sector of circle = cm2
cm= π θ/90
Solving the above equation, we have
θ = 90°
Therefore, the angle subtended at the centre of circle is 90°
### Question: 9
PQ is a chord of circle with centre ‘O’ and radius 4 cm. PQ is of the length 4 cm. Find the area of sector of the circle formed by chord PQ.
### Solution:
Given Data:
PQ is chord of length 4 cm. Also, PO = QO = 4 cm OPQ is an equilateral triangle.
Angle POQ = 60°
Area of sector (formed by the chord (Shaded region)) = (Area of sector)
Formula to be used:
Area of the sector = θ/360 ∗ πr2
Area of the sector = 60/360 ∗ π42 = 32π/3
Therefore, Area of the sector is 32π/3 cm2
### Question: 10
In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of arc and area of sector.
### Solution:
Given Data: Radius = 35 cm
Angle subtended at the centre ‘O’ = 72°
Area of sector of circle =?
Formula to be used: Length of arc = θ/360 ∗ 2πr cm
Length of arc = 108/360 ∗ 2πx = 42 cm
Solving the above equation we have,
Length of arc = 44 cm
We know that,
Area of the sector = θ/360 ∗ πr2
Area of the sector = 72/360 ∗ π352
Solving the above equation, we have,
Area of the sector = (35 × 22) cm2
Therefore, Area of the sector is 770 cm2
### Question: 11
The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. find the area of the sector.
### Solution:
Given Data:
Radius = 5.7 cm = OA = OB [from the figure shown above]
Perimeter = 27.2 m
Let the angle subtended at the centre be θ
Perimeter = θ/360 ∗ 2πr cm + OA + OB
= θ/360 ∗ 2πx 5.7 cm + 5.7 + 5.7
Solving the above equation we have,
θ = 158.8°
We know that, Area of the sector = θ/360 ∗ πr2
Area of the sector = 158.8/360 ∗ π 5. 72
Solving the above equation we have,
Area of the sector = 45.048 cm2
Therefore, Area of the sector is 45.048 cm2
### Question: 12
The perimeter of a certain sector of a circle of radius is 5.6 m and 27.2 m. find the area of a sector.
### Solution:
Given data:
Radius of the circle = 5.6 m = OA = OB
(AB arc length) + OA + OB = 27.2
Let the angle subtended at the centre be θ
We know that,
Length of arc = θ/360 ∗ 2πr cm
θ/360 ∗ 2πr cm + OA + OB = 27.2 m
θ/360 ∗ 2πr cm + 5.6 + 5.6 = 27.2 m
Solving the above equation, we have,
θ = 163.64°
We know that, Area of the sector = θ/360 ∗ πr2
Area of the sector = 163.64/360 ∗ π 5.62
On solving the above equation, we have,
Area of the sector = 44.8 cm2
Therefore,
Area of the sector is 44.8 cm2
### Question: 13
A sector was cut from a circle of radius 21 cm. The angle of sector is 120°. Find the length of its arc and its area.
### Solution:
Given data: Radius of circle (r) = 21 cm
θ = angle subtended at the centre of circle = 120°
Formula to be used:
Length of arc = θ/360 ∗ 2πr cm
Length of arc = 120/360 ∗ 2πx 21cm
On solving the above equation, we get, Length of arc = 44 cm
We know that, Area of the sector = θ/360 ∗ πr2
Area of the sector = 120/360 ∗ π212
Area of the sector = (22 × 21) cm2
Therefore, Area of the sector is 462 cm2
### Question: 14
The minute hand of a circle is √21 cm long. Find the area described by the minute hand on the face of clock between 7:00 a.m to 7:05 a.m.
### Solution:
Given data: Radius of the minute hand (r) = √21cm
Time between 7: 00 a. m to 7: 05 a. m = 5 min
We know that, 1 hr = 60 min, minute hand completes
One revolution = 360° 60 min = 360°
θ = Angle subtended at the centre of circle = 5 × 6° = 30°
Area of the sector = θ/360 ∗ πr2
Area of the sector = 30/360 ∗ pi352
Area of the sector = 5.5 cm2
Therefore, Area of the sector is 5.5 cm2
### Question: 15
The minute hand of clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 a. m to 8:25 a.m.
### Solution:
Given data: Radius of the circle = radius of the clock = length of the minute hand = 10 cm
We know that, 1 hr = 60 min 60 min = 360° 1 min = 6°
Time between 8:00 a. m to 8:25 a. m = 25 min
Therefore, the subtended = 6° × 25 = 150°
Formula to be used:
Area of the sector = θ/360 ∗ πr2
Area of the sector = 150/360 ∗ π102
Area of the sector = 916.6 cm= 917 cm2
Therefore, Area of the sector is 917 cm2
### Question: 16
A sector of 56° cut out from a circle subtends area of 4.4 cm2. Find the radius of the circle.
### Solution:
Given data: Angle subtended by the sector at the centre of the circle,
θ = 56°
Let the radius of the circle be = r cm
Formula to be used:
Area of the sector = 56/360 ∗ πr2
On solving the above equation, we get, r= √(9/1) cm
r = 3 cm
Therefore, radius of the circle is r = 3 cm
### Question: 17
In circle of radius 6 cm. Chord of length 10 cm makes an angle of 110° at the centre of circle. Find:
(i) Circumference of the circle
(ii) Area of the circle
(iii) Length of arc
(iv) The area of sector
### Solution:
Given data: Radius of the circle = 6 cm
Chord of length = 10 cm
Angle subtended by chord with the centre of the circle = 110°
Formulae to be used: Circumference of a circle = 2
Area of a Circle = Area of the sector = θ/360 ∗ πr2
Length of arc = 90/360 ∗ 2πx 28cm
Circumference of a circle = 2
= 2 × 3.14 × 8 = 37.7 cm
Area of a Circle = 3.14 × 6 × 6 = 113.14 cm2
Area of the sector = θ/360 ∗ πr/2
Area of the sector = 110/360 ∗ π62
On solving the above equation we get,
Area of the sector = 33.1 cm2
Length of arc = θ/360 ∗ 2πr cm [latex]
Length of arc = 110/360 2π6cm
On solving the above equation we get, Length of arc = 22.34 cm.
Therefore, Circumference = 37.7 cm
Area of a Circle = 113.14 cm2
Area of the sector = 33.1 cm2
### Question: 18
The given figure shows a sector of a circle with centre ‘O’ subtending an angle θ°. Prove that:
1. Perimeter of shaded region is
2. Area of the shaded region is
### Solution:
Given Data: Angle subtended at the centre of the circle = θ°
Angle OAB = 90° [at point of contact, tangent is perpendicular to radius]
OAB is a right angle triangle
cos θ = adjside/hypotenuse = r/OB = OB = r sec θ
sec θ = opposite/adjside = AB/r = AB = r tan θ
Perimeter of the shaded region = AB + BC + CA (arc)
= r tan? + (OB – OC) + θ/360 ∗ 2πr cm
= r(tan θ + sec θ + π θ/180 − 1)
Area of the shaded region = (Area of triangle AOB) – (Area of sector)
(1/2 ∗ OA ∗ AB) − θ/360 ∗ πr2
On solving the above equation we get, r2/2[tan θ − πθ/180]
### Question: 20
The diagram shows a sector of circle of radius ‘r’ cm subtends an angle θ. The area of sector is A cm2 and perimeter of sector is 50 cm. Prove that θ = 360/π (25/r – 1) and A = 25 r r2
### Solution:
Given Data: Radius of circle = ‘r’ cm
Angle subtended at centre of the circle = θ
Perimeter = OA + OB + (AB arc)
As given in the question, perimeter = 50
Area of the sector = θ/360 ∗ πr2
On solving the above equation, we have A = 25r – r2
Hence, proved.
All Chapter RD Sharma Solutions For Class10 Maths
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good |
• NUMBER OF ZEROES IN AN EXPRESSION
1. Step 1 : Suppose you have to find the number of zeroes in a product: 24 × 32 × 17 × 23 × 19. We first get the series in terms of its prime factors i.e. $$2^3 * 3^1 * 2^5 * 17^1 * 19 * 23$$. As you can notice, this product will have no zeroes because it has no 5 in it.
2. However, if you have an expression like: 8 × 15 × 23 × 17 × 25 × 22 The above expression can be rewritten in the standard form as: $$2^3 * 3^1 * 5^1 * 23 * 17 * 5^2 * 2^1 * 11^1$$
3. Step 2 : Zeroes are formed by a combination of 2 × 5. Hence, the number of zeroes will depend on the number of pairs of 2’s and 5’s that can be formed. In the above product, there are four twos and three fives. Hence, we shall be able to form only three pairs of (2 × 5). Hence, there will be 3 zeroes in the product.
• Finding the Number of Zeroes in a Factorial Value :
1. Suppose you had to find the number of zeroes in 6!. 6! = 6 × 5 × 4 × 3 × 2 × 1 = (3 × 2) × (5) × (2 × 2) × (3) × (2) × (1). Counting the number of 5 will give the answer.
2. Method 2 : For finding the zeroes in 6! we use $$\frac{6}{5} + \frac{6}{5^3} + ...$$ So we get 1 as the answer as all divisions after the first term in the series are in decimals which we ignore.
Ans .
10
1. Explanation :
$$47! = \frac{47}{5} + \frac{47}{5^2} + \frac{47}{5^3} + ...$$ = 9 + 1 + 0 = 10
Ans .
13
1. Explanation :
$$\frac{58}{5} + \frac{58}{5^2} + ...$$ = 11 + 2 + 0 = 13
Ans .
3
1. Explanation :
$$13 * 3 * 5 * 11 * 2 * 5^3 * 11 * 2^2 * 5 * 7 * 11$$ = pairs of (5 * 2) are three so zeroes are 3.
Ans .
2
1. Explanation :
$$2^2 * 3 * 3 * 5 * 5 * 2^3 * 3 * 13 * 17$$ = so pairs of (5*2) are 2, so we have 2 zeroes.
Ans .
41
1. Explanation :
$$\frac{173}{5} + \frac{173}{5^2} + \frac{173}{5^3} = 34 + 6 + 1 = 41$$
Ans .
37
1. Explanation :
$$\frac{144}{5} + \frac{144}{25} + \frac{144}{125} + 5 * 3 * 5 * 11 * 2 * 2^2 * 11 * 5 * 3^3 = 28 + 5 + 1 + 3 = 37$$
Ans .
35
1. Explanation :
$$\frac{148}{5} + \frac{148}{5^2} + \frac{148}{5^3} = 29 + 5 + 1 = 35$$
Ans .
370
1. Explanation :
$$\frac{1142}{5} + \frac{1142}{25} + \frac{1142}{125} + \frac{1142}{625} + \frac{348}{5} + \frac{348}{5^2} + \frac{348}{5^3} + \frac{17}{5} = 228 + 45 + 9 + 1 + 69 + 13 + 2 + 3 = 370$$
• While solving 45!, 46!, 47!, 48!, 49!. Notice the number of zeroes in each of the cases will be equal to 10.
• It is not difficult to understand that the number of fives in any of these factorials is equal to 10. The number of zeroes will only change at 50! (It will become 12).
• In fact, this will be true for all factorial values between two consecutive products of 5.
• Thus, 50!, 51!, 52!, 53! And 54! will have 12 zeroes (since they all have 12 fives). Similarly, 55!, 56!, 57!, 58! And 59! will each have 13 zeroes.
• While there are 10 zeroes in 49! there are directly 12 zeroes in 50!. This means that there is no value of a factorial which will give 11 zeroes. This occurs because to get 50! we multiply the value of 49! by 50. When you do so, the result is that we introduce two 5’s in the product. Hence, the number of zeroes jumps by two.
• Note : at 124! you will get 24 + 4 fi 28 zeroes. At 125! you will get 25 + 5 + 1 = 31 zeroes. (A jump of 3 zeroes.)
Ans .
none
1. Explanation :
This can never happen because at 99! number of zeroes is 22 and at 100! the number of zeroes is
24.
Ans .
59 and 55
1. Explanation :
At 55 we get 13 zeroes, since we know that 50! is 12 zeroes so till 54! we will have 12 zeroes. So 55 to 59! will have 13 zeroes.
Ans .
250
1. Explanation :
The fives will be less than the twos. Hence, we need to count only the fives. $$5^5 + 10^10 + 15^15 + ... + 45^45 = 5 + 10 + 15 + 20 + 25 + 30 + 35 +40 + 45 = 250$$
Ans .
1124
1. Explanation :
Again the key here is to count the number of fives. This can get done by: $$100^1 × 95^6 × 90^{11} × 85^{16} × 80^{21} × 75^{26} × …… 5^{96}$$ = (1 + 6 + 11 + 16 + 21 + 26 + 31 + 36 + 41 + 46 + …….. + 96) + (1 + 26 + 51 + 76) = 20 × 48.5 + 4 × 38.5 = 970 + 154 = 1124.
Ans .
5! + 10!
1. Explanation :
The answer will be the number of 5’s. Hence, it will be 5! + 10!
Ans .
98
1. Explanation :
The number of fives is again lesser than the number of twos.
The number of 5’s will be given by the power of 5 in the product:
5
4 × 10
8 × 15
12 × 20
16 × 10
18 × 25
20
= 4 + 8 + 12 + 16 + 18 + 40 = 98.
Ans .
1. Explanation :
Sum is 15152400, hence two zeros
Ans .
18
1. Explanation :
The number of zeroes will be:
2 + 3 + 4 + 3 + 6 = 18.
Ans .
12
1. Explanation :
$$\frac{81}{7} + \frac{81}{7^2} + \frac{81}{7^3} = 11 + 1 = 12$$
Ans .
19
1. Explanation :
42 = 2 *3 * 7. Hence we must find number of occurances of 2,3,7 in 122!. It is obvios that 7 will have the least occurances. Hence $$\frac{122}{7} + \frac{122}{7^2} + \frac{122}{7^3} = 17 + 2 + 0 = 19$$
Ans .
128
1. Explanation :
360 = 5 × 2 × 2 × 2 × 3 × 3. This means we need to think of which amongst 23, 32 and 5 would
appear the least number of times in 520!
Number of 5’s in 520!Highest power of 5 in 520! =
104 + 20 + 4 = 128.
In order to find the number of 32s in 520! we first need to find the number of 3’s in 520!
Number of 3’s in 520! = 173 + 57 + 19 + 6 + 2 = 257.
257 threes would obviously mean [257/2]= 128 -> 22s.
In order to find the number of 23s in 520! we first need to find the number of 2’s in 520!
Number of 2’s in 520! = 260 + 130 +65 + 32 +16 + 8 + 4 + 2 + 1 = 518. 518 twos would
obviously mean [518/3]= 172 23s.
Thus, the highest power of 360 that would divide 520! would be the least of 128, 128 and 172.
The answer is 128.
Ans .
38
1. Explanation :
Question is same as asking to find number of zeroes in the factorial. This is found by [157/5] = 31. [31/5] = 6. [6/5] = 1. 31 + 6 + 1 = 38. Option (b) is correct. |
# Finding Terms of a Sequence from a Formula with a Quadratic Term
In this worksheet, students find terms of a sequence from a T-formula that includes a quadratic term.
Key stage: KS 3
Curriculum topic: Algebra
Curriculum subtopic: Substitute Numerical Values for Formulae/Expressions
Difficulty level:
### QUESTION 1 of 10
This worksheet is about finding terms of a sequence from a formula with a quadratic term.
To find the first 5 terms and the 20th term of the sequence given by the formula
T= 2n2 − 3n + 5
we simply substitute different values for n.
The first term is called T1 , where n = 1.
The second term is called T2 , where n = 2 etc.
The twentieth term is called T20 , where n = 20.
Examples
T1 = (2×12) − (3×1) + 5 = 4
T2 = (2×22) − (3×2) + 5 = 7
T3 = (2×32) − (3×3) + 5 = 14
T4 = (2×42) − (3×4) + 5 = 25
T5 = (2×52) − (3×5) + 5 = 40
T20 = (2×202) − (3×20) + 5 = 745
Find the first term (i.e T1) of the sequence given by the formula
Tn = 3n2 + 5n + 7
Find the second term (i.e T2) of the sequence given by the formula
Tn = 3n2 + 5n + 7
Find the third term (i.e T3) of the sequence given by the formula
Tn = 3n2 + 5n + 7
Find the fourth term (i.e T4) of the sequence given by the formula
Tn = 3n2 + 5n + 7
Find the fifth term (i.e T5) of the sequence given by the formula
Tn = 3n2 + 5n + 7
Find the fiftieth term (i.e T50) of the sequence given by the formula
Tn = 3n2 + 5n + 7
Find the first term (i.e T1) of the sequence given by the formula
Tn = 5n2 + 5n - 1
Find the second term (i.e T2) of the sequence given by the formula
Tn = 5n2 + 5n - 1
Find the third term (i.e T3) of the sequence given by the formula
Tn = 5n2 + 5n - 1
Find the twentieth term (i.e T20) of the sequence given by the formula
Tn = 5n2 + 5n - 1
• Question 1
Find the first term (i.e T1) of the sequence given by the formula
Tn = 3n2 + 5n + 7
15
EDDIE SAYS
We are looking for the first term so n = 1. T1 = (3 × 1²) + (5 × 1) + 7 = 3 + 5 + 7 = 15
• Question 2
Find the second term (i.e T2) of the sequence given by the formula
Tn = 3n2 + 5n + 7
29
EDDIE SAYS
We are looking for the second term so n = 2. T2 = (3 × 2²) + (5 × 2) + 7 = (3 × 4) + 10 + 7 = 12 + 10 + 7 = 29
• Question 3
Find the third term (i.e T3) of the sequence given by the formula
Tn = 3n2 + 5n + 7
49
EDDIE SAYS
We are looking for the third term so n = 3. T3 = (3 × 3²) + (5 × 3) + 7 = (3 × 9) + 15 + 7 = 27 + 15 + 7 = 49
• Question 4
Find the fourth term (i.e T4) of the sequence given by the formula
Tn = 3n2 + 5n + 7
75
EDDIE SAYS
We are looking for the fourth term so n = 4. T4 = (3 × 4²) + (5 × 4) + 7 = (3 × 16) + 20 + 7 = 48 + 20 + 7 = 75
• Question 5
Find the fifth term (i.e T5) of the sequence given by the formula
Tn = 3n2 + 5n + 7
107
EDDIE SAYS
We are looking for the fifth term so n = 5. T5 = (3 × 5²) + (5 × 5) + 7 = (3 × 25) + 25 + 7 = 75 + 25 + 7 = 107
• Question 6
Find the fiftieth term (i.e T50) of the sequence given by the formula
Tn = 3n2 + 5n + 7
7757
EDDIE SAYS
We are looking for the fiftieth term so n = 50. T50 = (3 × 50²) + (5 × 50) + 7 = (3 × 2500) + 250 + 7 = 7500 + 250 + 7 = 7757
• Question 7
Find the first term (i.e T1) of the sequence given by the formula
Tn = 5n2 + 5n - 1
9
EDDIE SAYS
We are looking for the first term so n = 1. T1 = (5 × 1²) + (5 × 1) - 1 = (5 × 1) + 5 - 1 = 5 + 5 - 1 = 9
• Question 8
Find the second term (i.e T2) of the sequence given by the formula
Tn = 5n2 + 5n - 1
29
EDDIE SAYS
We are looking for the second term so n = 2. T2 = (5 × 2²) + (5 × 2) - 1 = (5 × 4) + 10 - 1 = 20 + 10 - 1 = 29
• Question 9
Find the third term (i.e T3) of the sequence given by the formula
Tn = 5n2 + 5n - 1
59
EDDIE SAYS
We are looking for the third term so n = 3. T3 = (5 × 3²) + (5 × 3) - 1 = (5 × 9) + 15 - 1 = 45 + 15 - 1 = 59
• Question 10
Find the twentieth term (i.e T20) of the sequence given by the formula
Tn = 5n2 + 5n - 1
2099
EDDIE SAYS
We are looking for the twentieth term so n = 20. T20 = (5 × 20²) + (5 × 20) - 1 = (5 × 400) + 100 - 1 = 2000 + 100 - 1 = 2099
---- OR ----
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# Geometry/Chapter 2
## Section 2.1 - Proofs
Proofs are set up to let the user understand what steps were taken in order to receive a given output. There are three types of proofs depending on which is easiest to the student.
### Two-Column proofs
Two-column proofs (also known as formal proofs) are set up in a two-value table, one being "Statement" and the other being "Reason". To prove a simple problem using this method, set up a table like the following:
Statement Reason
Be sure to leave room for values to go in both columns. In geometry, the first row is the 'given' of the problem. This is the information that is given about a certain problem without using a picture. The last row should be the conclusion of what you are trying to prove.
#### Example of a two-column proof
Now, suppose a problem tells you to solve ${\displaystyle x+1=2}$ for ${\displaystyle x}$, showing all steps made to get to the answer. A proof shows how this is done:
Statement Reason ${\displaystyle x+1=2}$ Given ${\displaystyle x=1}$ Property of subtraction
We use "Given" as the first reason, as it is "given" to us in the problem.
### Written Proof
Written proofs (also known as informal proofs, paragraph proofs, or 'plans for proof') are written in paragraph form. Other than this formatting difference, they are similar to two-column proofs.
Sometimes it is helpful to start with a written proof, before formalizing the proof in two-column form. If you're having trouble putting your proof into two column form, try "talking it out" in a written proof first.
#### Example of a Written Proof
We are given that x + 1 = 2, so if we subtract one from each side of the equation (x + 1 - 1 = 2 - 1), then we can see that x = 1 by the definition of subtraction.
### Flowchart Proof
A flowchart proof or more simply a flow proof is a graphical representation of a two-column proof. Each set of statement and reasons are recorded in a box and then arrows are drawn from one step to another. This method shows how different ideas come together to formulate the proof.
## Section 2.2 - Reasons
Every concept in geometry flows in a logical progression. One simply cannot go from A to B without explaining how or why. For instance, the following is not a proof:
${\displaystyle 5(x+1)^{2}-x}$
${\displaystyle x=4}$
Also, we cannot make up reasons why we made the next step so. Therefore, we can only use certain information as our reasons. These include:
1. Given: This is generally either the problem (equation) we are trying to solve, or some piece of important information given in the problem.
2. Properties: These for the most part are the basic mathematical functions of adding, subtracting, multiplying, and dividing, such as the second reason in the example above (Property of Subtraction).
3. Definitions: Again, saying "Because it is" is not a reason. This sort of reasoning is not seen as often as other reasons. By using definitions, sometimes the answer or part of the working of a proof can be shortened. For example, by using the reason "definition of a bisector" (and being already able to prove through either given information or earlier parts of the proof), you can prove that two adjoining angles are congruent without having to go through a more lengthy proof.
4. Postulates: They hold the same value as theorems (explained next), except that they cannot be proven. However, these generalized rules have proven correct for a very long time and can be accepted with proof of their validity. An example is "Through any two points, there is exactly one line". While it cannot be proven through a proof (although the authors dare anyone to disprove it), it is accepted as a reason. There are few of these, so as good as it may sound, if you make it up, someone will notice.
5. Theorems: Theorems are statements that have been proven true through proofs of their own. They are especially helpful shortcuts in their own right as by stating a theorem, a great many things are proven and you do not have to do all the work of re-proving the theorem. Theorems can be simple ("If two lines intersect, they intersect in exactly one point.") or very complex ("The composite of two isometries is an isometry." [Don't panic if you don't understand; it will be explained later on]). Sometimes, you will be given the proofs for theorems; othertimes, as part of the exercises, you will be asked to prove it yourself.
6. Axioms: For most purposes, the same as postulates. The difference is that axioms are algebraic in nature, while postulates come mainly from geometry.
7. Corollaries: These are statements that stem from what becomes proven in theorems and definitions and do not require (though usually have) separate proofs themselves.
In many textbooks, the proofs are numbered for an index at the back of the book. When doing correct geometric proofs, it is NOT OK to write down "Theorem 15". Write out the statement exactly as it was given to you (yes, you have to do some memorization for geometry). You have to make sure that the information in the box is related to the earlier box.
## Section 2.3 - Using proofs in geometry
### Exercises
Answers to each exercise can be found separately in the appendix.
1.
Given:
• ${\displaystyle r||s}$ r is parallel to s
• ${\displaystyle \angle 1=60^{\circ }}$ Angle 1 = 60 degrees.
Prove: Find the measures of the other seven angles in the accompanying figure (above).
2.
Given:
• ${\displaystyle \angle 2\cong \angle 3}$ Angles 2 and 3 are congruent
Prove: Lines r and s are parallel.
3.
Given:
• ${\displaystyle \angle 1=\angle 2=90^{\circ }}$ Angles 1 and 2 are both 90 ⁰
Prove: Lines a and b in the figure are parallel.
4.
Given:
• ${\displaystyle {\overline {GH}}||{\overrightarrow {DK}}}$ Line GH is parallel to ray DK
• ${\displaystyle \angle 6=75^{\circ }}$ Angle 6 = 75 degrees.
• ${\displaystyle \angle 2=30^{\circ }}$ Angle 2 = 30 degrees.
Prove: Find the measure of each numbered angle in the figure above.
## Section 2.4: Proof by contradiction
Proof by contradiction, also known as indirect proofs, prove that a statement is true by showing that the proposition's being false would imply a contradiction.
### A classic example: Proving that the square root of 2 is irrational
1. Assume that ${\displaystyle {\sqrt {2}}}$ is a rational number, meaning that there exists an integer ${\displaystyle a}$ and an integer ${\displaystyle b}$ in general such that ${\displaystyle a/b={\sqrt {2}}}$.
2. Then, ${\displaystyle {\sqrt {2}}}$ can be written as a simplified fraction ${\displaystyle a/b}$ such that ${\displaystyle a}$ and ${\displaystyle b}$ are coprime integers, that is, their greatest common divisor being 1.
3. It follows that ${\displaystyle a^{2}/b^{2}=2}$ and ${\displaystyle a^{2}=2b^{2}}$. ( ${\displaystyle (a/b)^{n}=a^{n}/b^{n}}$ )
4. Therefore ${\displaystyle a^{2}}$ is even, as it is equal to ${\displaystyle 2b^{2}}$. (${\displaystyle 2b^{2}}$ is necessarily even, as it is 2 times another whole number, and even numbers are multiples of 2.)
5. It follows that ${\displaystyle a}$ must be even (as squares of odd integers are never even).
6. Because ${\displaystyle a}$ is even, there exists an integer ${\displaystyle k}$ that fulfills: ${\displaystyle a=2k}$.
7. Substituting ${\displaystyle 2k}$ from step 6 for ${\displaystyle a}$ in the second equation of step 3: ${\displaystyle (2k)^{2}=2b^{2}}$ is equivalent to ${\displaystyle 4k^{2}=2b^{2}}$, which is equivalent to ${\displaystyle 2k^{2}=b^{2}}$.
8. Because ${\displaystyle 2k^{2}}$ is divisible by two and therefore even, and as ${\displaystyle 2k^{2}=b^{2}}$, it follows that ${\displaystyle b^{2}}$ is also even, which means that ${\displaystyle b}$ is even.
9. By steps 5 and 8 ${\displaystyle a}$ and ${\displaystyle b}$ are both even, which contradicts that ${\displaystyle a/b}$ is irreducible as stated in step 2.
Q.E.D.
As there is a contradiction, the assumption (1) that ${\displaystyle {\sqrt {2}}}$ is a rational number must be false. By the law of excluded middle, that is, that a proposition can only be true or false, the opposite is proven: ${\displaystyle {\sqrt {2}}}$ is irrational. |
Factors that 54 are the perform of integers that have the right to be evenly divided into 54. Over there are overall 8 determinants of 54 amongst which 54 is the best factor and also its positive components are 1, 2, 3, 6, 9, 18, 27 and 54. The Pair factors of 54 room (1, 54), (2, 27), (3, 18), and also (6, 9) and its Prime determinants are 1, 2, 3, 6, 9, 18, 27, 54.
You are watching: What are the factors of 54
Factors the 54: 1, 2, 3, 6, 9, 18, 27 and 54Negative components of 54: -1, -2, -3, -6, -9, -18, -27 and -54Prime determinants of 54: 2, 3Prime administer of 54: 2 × 3 × 3 × 3 = 2 × 33Sum of factors of 54: 120
1 What space the determinants of 54? 2 How to Calculate determinants of 54? 3 Factors that 54 by prime Factorization 4 Factors that 54 in Pairs 5 FAQs on factors of 54
Before we move ahead, let’s recall a tiny about factors. A factor is a number the divides the provided number without any remainder. The factors of 54 are pairs of those numbers whose assets results in 54
## How to calculate the determinants of 54?
To calculate the components of any number, here in this situation 54, we require to find all the number that would certainly divide 54 without leaving any remainder. We begin with the number 1, then examine for numbers 2, 3, 4, 5, 6, 7, and so on up come 54 respectively. The number 1 and also the number itself would constantly be a aspect of the provided number.
We refer 54 as a product of its prime determinants in the element factorization method and we division 54 with its divisors in the division method. Let united state see i m sorry numbers division 54 exactly there is no a remainder. The divisors, and also the quotients, space the components of 54
54 ÷ 1 = 54, therefore, 1 and also 24 is a element of 5454 ÷ 2 = 27, therefore, 2 is a factor of 5454 ÷ 3 = 18, therefore, 3 is a factor of 5454 ÷ 6 = 9, therefore, 6 is a element of 54
Hence, the components of 54 are 1, 2, 3, 6, 9, 18, 27 and 54
To understand the principle of finding factors by element factorization better, let united state take a couple of more examples.
Factors of 54 by prime factorization are offered by making use of the following steps. In the first step compose the pair that factors, which on multiplication provides the required number. 54 can be factored as a product that 6 and 9. In the second step view the factors, even if it is each one of them is element or not. 6 is no a prime number and also can be dissociated together a product of 2 through 3. 9 is not a prime number and also can be dissociated as a product of 3 by the square that 3 in action three, as per the criteria, 54 can be written as 54 = 6 × 9 = 2 × 3 × 3 × 3. It can additionally be written as 54 = 2 × 33.Observe listed below the element factorization that 54
Pair components are bag of those numbers which when multiplied, offer the product together the forced number.
Factors the 54 in pairs can be written as:
FactorsPair factors
1 × 54 = 541, 54
2 × 27 = 542, 27
3 × 18 = 54 3, 18
6 × 9 = 54 6, 9
9 × 6 = 54 9, 6
18 × 3 = 54 18, 3
27 × 2 = 54 27, 2
54 × 1 = 5454, 1
The over given determinants are hopeful pair factors.
It is feasible to have an adverse pair factors too because the product that two an adverse numbers likewise gives a confident number.
Let"s have a watch at an unfavorable pair factors.
FactorsPair factors
-1 × -54 = 54(-1, -54)
-2 × -27 = 54(-2, -27)
-3 × -18 = 54( -3, -18)
-6 × -9 = 54 (-6, -9)
-9 × -6 = 54( -9, -6)
-18 × -3 = 54 (18, -3 )
-27 × -2 = 54( -27, -2 )
-54 × -1 = 54 (-54, -1 )
Example 1: James, Judy, and also Christina pluck 54 oranges and also distribute them among themselves equally. How countless oranges will certainly each that them get?
Solution:
54 oranges are to it is in divided among James, Judy, and Christina equally.So because that this, we need to divide the number 54 through 354 ÷ 3=18Hence each son will gain 18 oranges.
Example 2: aid Ruth in finding the following determinants of 54.
a. Half of 54?b. Two-third the 54?
Solution
a. 54 ÷ 2=27b. 54 × 2 ÷ 3 = 36But 36 is not a variable of 54So, no aspect of 54 is two-third of 54
Example 3: Give one-factor pair that 54 which includes both composite numbers.
Solution
The feasible factor pairs of 54 space (1, 54), (2, 27), (3, 18), (6, 9)1, 2, and 3 are no composite.So, the end of every these pairs, only (6, 9) has actually both the components as composite numbers.
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## FAQs on factors of 54
### What room the factors of 54?
The determinants of 54 room 1, 2, 3, 6, 9, 18, 27, 54 and its an unfavorable factors room -1, -2, -3, -6, -9, -18, -27, -54.
### What is the Greatest usual Factor the 54 and also 49?
The components of 54 space 1, 2, 3, 6, 9, 18, 27, 54 and also the determinants of 49 room 1, 7, 49. 54 and 49 have only one usual factor which is 1. This indicates that 54 and 49 room co-prime.Hence, the Greatest usual Factor (GCF) of 54 and 49 is 1.
### What is the sum of the factors of 54?
Sum of all components of 54 = (21 + 1 - 1)/(2 - 1) × (33 + 1 - 1)/(3 - 1) = 120
### What Numbers space the Prime components of 54?
The prime components of 54 are 2, 3.
See more: An Important Factor That Brought Changes To The Hunter-Gatherer Way Of Life Was That
### How countless Factors that 34 are likewise Common to the factors of 54?
Since, the determinants of 54 space 1, 2, 3, 6, 9, 18, 27, 54 and the components of 34 are 1, 2, 17, 34.Hence, <1, 2> are the usual factors that 54 and also 34. |
19320 in words
19320 in words is written as Nineteen Thousand Three Hundred and Twenty. In 19320, 1 has a place value of ten thousand, 9 is in the place value of thousand, 3 is in the place value of hundred and 2 is in the place value of ten. The article on Place Value gives more information. The number 19320 is used in expressions that relate to money, distance, social media views, and many more. For example, “My car has covered Nineteen Thousand Three Hundred and Twenty kilometers and is due for service.”
19320 in words Nineteen Thousand Three Hundred and Twenty Nineteen Thousand Three Hundred and Twenty in Numbers 19320
How to Write 19320 in Words?
We can convert 19320 to words using a place value chart. The number 19320 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousand Thousands Hundreds Tens Ones 1 9 3 2 0
Thus, we can write the expanded form as:
1 × Ten thousand + 9 × Thousand + 3 × Hundred + 2 × Ten + 0 × One
= 1 × 10000 + 9 × 1000 + 3 × 100 + 2 × 10 + 0 × 1
= 19320.
= Nineteen Thousand Three Hundred and Twenty.
19320 is the natural number that is succeeded by 19319 and preceded by 19321.
19320 in words – Nineteen Thousand Three Hundred and Twenty.
Is 19320 an odd number? – No.
Is 19320 an even number? – Yes.
Is 19320 a perfect square number? – No.
Is 19320 a perfect cube number? – No.
Is 19320 a prime number? – No.
Is 19320 a composite number? – Yes.
Solved Example
1. Write the number 19320 in expanded form
Solution: 1 x 10000 + 9 x 1000 + 3 x 100 + 2 x 10 + 0 x 1
We can write 19320 = 10000 + 9000 + 300 + 20 + 0
= 1 x 10000 + 9 x 1000 + 3 x 100 + 2 x 10 + 0 x 1.
Frequently Asked Questions on 19320 in words
Q1
How to write the number 19320 in words?
19320 in words is written as Nineteen Thousand Three Hundred and Twenty.
Q2
Is 19320 a prime number?
No. 19320 is not a prime number.
Q3
Is 19320 divisible by 10?
Yes. 19320 is divisible by 10. |
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# The dimension of a metal block is given in the diagramThe ratio of $100cm \times 2cm \times 2cm$ resistance between square faces and rectangular faces is(A) $1:1$(B) $25:1$(C) $2500:1$(D) $50:1$
Last updated date: 17th Jun 2024
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Hint : Use the formula of resistance of the conductor to find the resistance of the square face and the rectangle face. The dimensions of the block is given from that we can know the side of the square and the length and breadth of the rectangle with that find the area of the square and rectangle and substitute those values in the resistance formula then find the ratio of resistance between square faces and rectangular faces.
Resistance offered to a conductor is known as resistivity. The resistivity is defined as the capability of restriction of flow of current.
The resistance of a conductor is directly proportional to the length of the conductor and is inversely proportional to its area of cross section A.
$\Rightarrow R \propto \dfrac{l}{A}$
$\Rightarrow R = \dfrac{{\rho l}}{A}$
$\rho$ is the specific resistance or electrical resistivity of the conductor.
l is the length of the conductor
A is the area of the conductor
We have to find the ratio of resistance between square faces and rectangular faces.
Firstly let us calculate the resistance of the square face
The length of the side of the square, $a = 2cm$
The area of the square is $4a$
$\Rightarrow {\text{A }} = {\text{ 4a}}$
$\Rightarrow {\text{A }} = {\text{ }}4 \times 2$
$\Rightarrow {\text{A }} = {\text{ 8cm}}$
We know that the resistance of a conductor is
$\Rightarrow R = \dfrac{{\rho l}}{A}$
$\Rightarrow {R_s} = \dfrac{{\rho 2}}{8}$
$\Rightarrow {R_s} = \dfrac{\rho }{4}{\text{ }} \to {\text{1}}$
Now let us calculate the resistance of the rectangular face
The length of the rectangle, $b = 100cm$
The breadth of the rectangle, $l = 2cm$
The area of the rectangle is $l \times b$
$\Rightarrow {\text{A }} = {\text{ l}} \times {\text{b}}$
$\Rightarrow {\text{A }} = {\text{ 2}} \times {\text{100}}$
$\Rightarrow {\text{A }} = {\text{ 200}}$
We know that the resistance of a conductor is
$\Rightarrow R = \dfrac{{\rho l}}{A}$
$\Rightarrow {R_R} = \dfrac{{\rho 2}}{{200}}$
$\Rightarrow {R_R} = \dfrac{\rho }{{100}}{\text{ }} \to {\text{2}}$
To find the ratio of resistance between square faces and rectangular faces divide the resistance of square faces by rectangular faces
$\Rightarrow \dfrac{{{R_s}}}{{{R_R}}} = \dfrac{{\dfrac{\rho }{4}}}{{\dfrac{\rho }{{100}}}}$
$\Rightarrow \dfrac{{{R_s}}}{{{R_R}}} = \dfrac{\rho }{4} \times \dfrac{{100}}{\rho }$
$\Rightarrow \dfrac{{{R_s}}}{{{R_R}}} = \dfrac{{25}}{1}$
$\Rightarrow {R_s}:{R_R} = 25:1$
The ratio of resistance between square faces and rectangular faces is $25:1$
Hence the correct answer is option (B) $25:1$.
Note
It is a direct question if we know the basic math formulas and the resistance formula we can easily solve this problem. Most of the students mistook the breadth of the rectangle as the length of the rectangle and put wrong values in the resistance formula, so be careful while substituting the values. |
Integrals
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# Integrals - PowerPoint PPT Presentation
Integrals. 5. The Substitution Rule. 5.5. The Substitution Rule. Because of the Fundamental Theorem, it’s important to be able to find antiderivatives. But our antidifferentiation formulas don’t tell us how to evaluate integrals such as
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## PowerPoint Slideshow about 'Integrals' - ludwig
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Presentation Transcript
The Substitution Rule
• Because of the Fundamental Theorem, it’s important to be able to find antiderivatives.
• But our antidifferentiation formulas don’t tell us how to evaluate integrals such as
• To find this integral we use a new variable; we change from the variable x to a new variable u.
The Substitution Rule
• Suppose that we let u be the quantity under the root sign in (1): u = 1 + x2. Then the differential of u is du = 2x dx.
• Notice that if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2x dx would occur in (1) and so, formally, without justifying our calculation, we could write
The Substitution Rule
• But now we can check that we have the correct answer by using the Chain Rule to differentiate the final function of Equation 2:
• In general, this method works whenever we have an integral that we can write in the form f(g(x))g(x) dx.
The Substitution Rule
• Observe that if F = f,then
• F(g(x))g(x) dx = F(g(x)) + C
• because, by the Chain Rule,
• [F(g(x))] = F(g(x))g(x)
• If we make the “change of variable” or “substitution” u = g(x),
• then from Equation 3 we have
• F(g(x))g(x) dx = F(g(x)) + C = F(u) + C = F(u) du
• or, writing F = f, we get
• f(g(x))g(x) dx =f(u) du
The Substitution Rule
• Thus we have proved the following rule.
• Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation.
• Notice also that if u = g(x), then du = g(x) dx, so a way to remember the Substitution Rule is to think of dx and du in (4) as differentials.
The Substitution Rule
• Thus the Substitution Rule says: It is permissible to operate with dx and du after integral signs as if they were differentials.
Example 1 – Using the Substitution Rule
• Find x3 cos(x4 + 2) dx.
• Solution:
• We make the substitution u = x4 + 2 because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral.
• Thus, using x3 dx =du and the Substitution Rule, we have
• x3 cos(x4 + 2) dx = cos udu
• = cos u du
Example 1 – Solution
cont’d
• = sin u + C
• = sin(x4 + 2)+ C
Definite Integrals
• When evaluating a definite integral by substitution, two methods are possible. One method is to evaluate the indefinite integral first and then use the Evaluation Theorem.
• For example,
• Another method, which is usually preferable, is to change the limits of integration when the variable is changed.
Example 6 – Substitution in a Definite Integral
• Evaluate .
• Solution:
• Let u = 2x + 1. Then du = 2 dx, so dx = du.
• To find the new limits of integration we note that
• when x = 0, u = 2(0) + 1 = 1
• and
• when x = 4, u = 2(4) + 1 = 9
• Therefore
Example 6 – Solution
cont’d
• Observe that when using (5) we do not return to the variable x after integrating. We simply evaluate the expression in u between the appropriate values of u.
Symmetry
• The following theorem uses the Substitution Rule for
• Definite Integrals (5) to simplify the calculation of integrals of
• functions that possess symmetry properties.
Symmetry
• Theorem 6 is illustrated by Figure 4.
• For the case where f is positive and even, part (a) says that the area under y = f(x) from –a to a is twice the area from 0 to a because of symmetry.
Figure 4
Symmetry
• Recall that an integral can be expressed as the
• area above the x-axis and below y = f(x) minus the area
• below the axis and above the curve.
• Thus part (b) says the integral is 0 because the areas cancel.
Example 9 – Integrating an Even Function
• Since f(x) = x6 + 1 satisfies f(–x) = f(x), it is even and so
Example 10 – Integrating an Odd Function
• Since f(x) = (tan x)/(1 + x2 + x4) satisfies f(–x) = –f(x), it is odd and so |
Home » Math Theory » Fractions » Comparing fractions
Comparing fractions
Introduction
In our daily lives, we commonly deal with fractions. Most of the time, we are asked to compare which part is smaller or larger, or if both parts are equally split. In this article, we look back at the definition of fractions and their types, then explore methods on how we can compare them.
What is a Fraction?
To motivate us into knowing about fractions, we think about a popular analogy to it. Imagine we have a box of pizza that is sliced into equal parts:
Each slice represents a part of the whole pizza. If we count the number of slices that make up the pizza, we can express the slice we will take as a mathematical quantity. This is what we will introduce as a fraction.
When we refer to fractions, we think about parts of a whole. Mathematically speaking, it is a quantity that is expressed as a quotient or a ratio between two numbers. Going back to the box of pizza we used as an analogy, if we take the number of slices we eat over the total number of slices in the box, we can express this ratio as follows:
$\frac{no.\: of\: pizza\: slices\: eaten}{no.\: of\: slices\: in\: the\: box}$
Suppose a friend ate three slices of pizza. We can then say that our friend had three-eighths of the pizza:
$\frac{no.\: of\: pizza\: slices\: eaten}{no.\: of\: slices\: in\: the\: box}=\frac{3}{8}$
We now identify the parts of a fraction. The number above is called the numerator, and the number under it is called the denominator. The numerator and denominator are separated by the line we call the fraction bar.
$\frac{3}{8}$ ←numerator ←denominator
Types of Fractions
Fractions can be classified under three types, depending on how the numerator and denominator are related:
Proper Fractions
Proper Fractions are fractions that have a numerator that is less than its denominator:
numerator<denominator
Some examples we can think of are $\frac{1}{3}$, $\frac{4}{7}$, and $\frac{5}{8}$:
We can think of proper fractions as “proper” since the number of parts is less than the whole. In a mathematical sense, the ratio formed by proper fractions is less than 1:
$\frac{numerator}{denominator}$<1
This happens because dividing a smaller number by a bigger number results in a quotient that is less than one.
Improper Fractions
On the other hand, Improper Fractions are fractions that have a numerator that is greater than its denominator:
numerator>the denominator
Some examples we can produce are $\frac{3}{2}$, $\frac{5}{4}$, and $\frac{9}{8}$.
We can think of improper fractions as “improper” since the number of parts is more than the whole. Mathematically speaking, the ratio formed by improper fractions is greater than 1:
$\frac{numerator}{denominator}$>1
Contrary to the case of proper fractions, dividing a larger number by a small number results in a quotient that is greater than one.
Mixed Numbers
A Mixed Number can be thought of as a combination of a whole and a fraction. This is another way of writing an improper fraction, taking the parts that make a whole and then expressing the remaining parts as a proper fraction.
For example, we consider the improper fraction $\frac{11}{8}$. By visualization, we have 11 equal parts that will be grouped into 8 parts per whole:
If we put together 8 parts to form one whole, then what remains will be a proper fraction of $\frac{3}{8}$. In this regard, we can express the improper fraction as the number of wholes and the fractional part:
$\frac{11}{8}=\frac{8}{8}+\frac{3}{8}=1\frac{3}{8}$
Conveniently, we have discovered this method to convert improper fractions into mixed numbers.
Like Fractions
We can also relate two (or more) fractions based on the number on their denominators. Like Fractions are fractions that have the same denominator. We can think of like fractions as parts of the same whole. For example, we have $\frac{1}{5}$, $\frac{2}{5}$, $\frac{3}{5}$, and $\frac{4}{5}$ as like fractions since their denominators are all equal to 5.
Unlike Fractions
What if two fractions are not like fractions? In this case, we have unlike fractions. Unlike Fractions are fractions that have different denominators. We can think of unlike fractions as parts of different wholes. For example, $\frac{1}{4}$ and $\frac{2}{8}$ are unlike fractions since their denominators are not equal.
What are Equivalent Fractions?
Before we begin comparing fractions, another relevant concept that we must know is how fractions can be equivalent. Equivalent fractions are fractions that bear the same value, even if they have different numerators and denominators. To help us understand how equivalent fractions work, let us consider two unlike fractions $\frac{1}{4}$ and $\frac{2}{8}$:
We see that for the same circle split into equal parts, the area covered by one part over four equal parts is the same as the area covered by two parts over eight equal parts. Hence, we can conclude that both $\frac{1}{4}$ and $\frac{2}{8}$ are equivalent fractions.
For the previous example, we have easily shown that they are equivalent by visualizing the fractions. But how can we determine whether two of such fractions are equivalent for fractions that will involve larger numbers?
Two fractions are said to be equivalent if they have a common factor that we can divide or multiply to both the numerator and the denominator of one fraction to arrive at the other fraction.
Again, let us take $\frac{1}{4}$ and $\frac{2}{8}$ as an example. We can multiply a factor of 2 to both the numerator and denominator of $\frac{1}{4}$ and $\frac{2}{8}$:
$\frac{1×2}{4×2}=$\frac{2}{8}$Similarly, we can divide a factor of 2 to both numerator and denominator of$\frac{2}{8}$to obtain$\frac{1}{4}$:$\frac{2÷2}{8÷2}=$\frac{1}{4}$
Hence, aside from the visualization we made earlier, we further prove that both $\frac{1}{4}$ and $\frac{2}{8}$ are equivalent fractions.
Comparing Fractions
Now that we have learned the underlying concepts regarding fractions, we can now proceed to our main topic which revolves around a comparison between two fractions. We note that the techniques discussed below are generally applicable to two fractions.
Intuitive Approach
Like what we have been doing with the pizza analogy, we use an Intuitive Approach that allows us to think of fractions visually. For this method, we can use any shape that we wish to use but for convenience, we will be using rectangles to represent parts of a whole:
Suppose we wish to compare two fractions $\frac{6}{8}$ and $\frac{2}{9}$. We represent the first fraction by splitting the rectangle into 8 equal parts, then shading 6 parts of it:
We also represent the second fraction by splitting the rectangle into 9 equal parts, then shading 2 parts of it:
If we compare both figures to each other, we see that $\frac{6}{8}$ covers more area than $\frac{2}{9}$.
Therefore, we conclude that the first fraction $\frac{6}{8}$ is greater than the second fraction $\frac{2}{9}$:
$\frac{6}{8}$>$\frac{2}{9}$
Again, we note that this approach can be convenient for fractions that are easy to imagine. As we consider larger denominators, the splitting might not be as effective as it is for smaller denominators. Hence, we consider other techniques.
Decimal Approach
Another method we can use to compare fractions is using the Decimal Approach. In this approach, we solve for the exact values of each fraction using either a calculator or by doing long division and then compare these decimal numbers.
For example, we want to know which fraction is larger: $\frac{15}{24}$ or $\frac{30}{40}$. By computing the exact values of both fractions, we get the following decimal numbers:
$\frac{15}{24}$=0.625
$\frac{30}{40}$=0.75
We note that both fractions are proper fractions whose numerators are smaller than their denominators. Hence, we expect that their exact values will be less than one.
Then, we compare the decimal values of both fractions. Looking at the tenths place of both decimals, we note that 0.7 is greater than 0.6, thus we conclude that 0.75>0.625 and that $\frac{30}{40}$ is greater than $\frac{15}{24}$:
$\frac{30}{40}$>$\frac{15}{24}$
Cross Multiplication
If we wish to work with multiplying numbers instead of getting their quotient, we can use Cross Multiplication to compare two fractions. To perform cross multiplication, we multiply the numerator of one fraction to the denominator of the other fraction, and then compare which cross product is larger and subsequently determine which fraction is larger.
We can use the previous example $\frac{15}{24}$ and $\frac{30}{40}$ and verify which fraction is larger.
First, we take the cross product of the numerator 15 and the denominator 40 then write the result to represent the first fraction:
Next, we also take the cross product of the numerator 30 and the denominator 30 then write the result to represent the second fraction:
Finally, we compare the two products obtained from cross multiplication:
600<720
Since 720 is larger than 600, then we conclude that $\frac{30}{40}$ is greater than $\frac{15}{24}$. This is consistent with the conclusion we made using the decimal approach.
Comparing Like Fractions
Suppose we have two like fractions. Other than the general techniques we have already discovered, there is a faster way to compare these fractions without going through much trouble.
When comparing two fractions that have the same denominator, we can treat the comparison between fractions as a comparison between their numerators. The relationship between the numerators of both fractions also applies to the relationship between the two fractions.
Let us consider an example and compare the following fractions $\frac{2}{7}$ and $\frac{6}{7}$. Since both fractions are like fractions with the same denominator 7, we compare the numerators 2 and 6.
2<6
Since the numerator of the first fraction is less than the numerator of the second fraction, we conclude that the first fraction is also less than the second fraction:
$\frac{2}{7}$<$\frac{6}{7}$
Intuitively, we can agree to this conclusion since from a pizza of seven slices, eating two slices out of the whole is less than eating six slices out of the same box!
Comparing Unlike Fractions
How about the case of two unlike fractions? We can use a similar approach to comparing like fractions, but we add extra steps to convert unlike fractions into like fractions. We do this by either finding a common denominator for both fractions or the least common denominator.
Finding the Common Denominator
The Common Denominator between two fractions can be obtained by taking the product of both denominators. We then rewrite both fractions as like fractions by multiplying both the numerator and denominator of one fraction by the denominator of the other fraction.
As an example, we compare the fractions $\frac{22}{45}$ and $\frac{6}{9}$. Their common denominator is given by:
45×9=405
We then rewrite the first fraction $\frac{22}{45}$ by multiplying both numerator and denominator by the denominator of the second fraction 9:
$\frac{22×9}{45×9}$=$\frac{198}{405}$
Next, we rewrite the second fraction $\frac{6}{9}$ by multiplying both numerator and denominator by the denominator of the second fraction 45:
$\frac{6×45}{9×45}$=$\frac{270}{405}$
Now that we have two like fractions, we can compare their numerators to determine which fraction is larger:
270>198 ∴ $\frac{270}{405}$>$\frac{198}{405}$
Since $\frac{270}{405}$ is greater than $\frac{198}{405}$, then we conclude that their equivalent fractions $\frac{6}{9}$ is greater than $\frac{22}{45}$:
$\frac{6}{9}$>$\frac{22}{45}$
Finding the Least Common Denominator
Alternatively, the Least Common Denominator (LCD) between two fractions can be obtained by taking the Least Common Multiple of both denominators. We then rewrite the equivalent fractions such that we get two like fractions. This way, we can compare them as like fractions.
We consider the two fractions in the previous example $\frac{22}{45}$ and $\frac{6}{9}$ whose denominators are 45 and 9, respectively. If we take the least common multiple of both denominators, 45, then we can rewrite both fractions to have the same denominator.
Since $\frac{22}{45}$ is already expressed in terms of the least common denominator, we only have to get the equivalent fraction of the second fraction $\frac{6}{9}$:
$\frac{6×5}{9×5}$=$\frac{30}{45}$
Thus, we can compare both like fractions using their numerators:
30>22 ∴ $\frac{30}{45}$>$\frac{22}{45}$
Since $\frac{30}{45}$ is greater than $\frac{22}{45}$, then we conclude that their equivalent fractions $\frac{6}{9}$ is greater than $\frac{22}{45}$:
$\frac{6}{9}$>$\frac{22}{45}$
We note that we have arrived at the same conclusion when we used the Common Denominator earlier.
Problem-Solving Examples
We can now proceed to solve sample problems to apply what we have learned so far. Each problem tackles the different approaches and techniques discussed and gives us a challenge on how to work through the information given to us.
Comparing Fractions Using Intuitive Approach
Sample Problem 1:
Timothy made a chocolate bar to share with five of his friends. During recess, another friend joined them, and Timothy decided to share the chocolate among all seven of them (including himself).
However, Timothy doesn’t want to get less than the share he was supposed to get sharing with five friends. He thinks that he should split the chocolate into eight parts, get two parts for himself, then share the remaining parts with his friends. Will he get more chocolate in this way?
Solution:
We first consider the chocolate bar split into six pieces, and take one piece for Timothy as shown in the figure below:
Next, we consider the chocolate bar split into eight pieces, and take two pieces for Timothy as shown in the figure below:
Comparing the combined size of the chocolate Timothy will receive in either case, we conclude that he will get more chocolate by splitting the bar into eight pieces and taking two pieces for himself.
Sample Problem 2:
On her birthday, Melissa received three round cakes of the same size: a chocolate cake from eleven of her relatives in the morning, a strawberry cake from her eight workmates in the afternoon, and a blueberry cake from ten of her cousins in the evening.
If she decided to share each cake with the group who gave it to her, which cake did she have the most?
Solution:
We first consider the chocolate cake she received in the morning, where she got a slice out of eleven parts. We compare this with the strawberry cake she had in the afternoon, where she got a slice out of eight parts.
By visualizing the slices of cake she had as parts of a whole, we can conclude that she had more strawberry cake than the chocolate cake:
Then, we compare the strawberry cake she had with the blueberry cake she had in the evening, where she got a slice out of ten parts.
Again, by visualizing the slices of cake, we can conclude that she had more strawberry cake than the blueberry cake:
Therefore, we conclude that out of all three cakes Melissa had on her birthday, she had the most share in the strawberry cake.
Comparing Fractions Using Decimal Approach
Sample Problem 3:
Suppose we have two improper fractions $\frac{52}{11}$ and $\frac{47}{8}$. Using the decimal approach, determine which fraction is larger.
Solution:
For this example, we use long division to show the process of getting the decimal values of each fraction. We begin by taking the quotient between the numerator and denominator of the first fraction $\frac{52}{11}$:
Hence, we approximate the decimal value up to the thousandths place by:
$\frac{52}{11}$≈4.727
Similarly, we take the quotient between the numerator and denominator of the second fraction $\frac{47}{8}$:
Hence, the exact decimal value of the fraction is given by:
$\frac{47}{8}$=5.875
Finally, we compare the two decimal numbers representing their equivalent fractions:
4.727<5.875
Since the decimal value of the second fraction is larger than the decimal value of the first fraction, then we conclude that the second fraction $\frac{47}{8}$ is larger than the first fraction $\frac{52}{11}$.
Comparing Fractions Using Cross Multiplication
Sample Problem 4:
Suppose we have two proper fractions $\frac{7}{23}$ and $\frac{16}{35}$. Which fraction is smaller? Show your computation using cross multiplication.
Solution:
We first consider the cross product of the numerator of the first fraction, 7, and the denominator of the second fraction, 35, then write it as a number representing the first fraction:
Next, we consider the cross product of the numerator of the second fraction, 16, and the denominator of the second fraction, 23, then write it as a number representing the second fraction:
Comparing the two cross products, we can conclude that the first cross product is smaller than the other cross product:
245<368
Therefore, we conclude that the first fraction $\frac{7}{23}$ is smaller than the second fraction $\frac{16}{35}$.
Sample Problem 5:
If we are given two proper fractions $\frac{8}{24}$ and $\frac{12}{36}$, which fraction is larger? Using cross multiplication, show your computations.
Solution:
We first consider the cross product of the numerator of the first fraction, 8, and the denominator of the second fraction, 36, then write it as a number representing the first fraction:
Next, we consider the cross product of the numerator of the second fraction, 16, and the denominator of the second fraction, 23, then write it as a number representing the second fraction:
Comparing the two cross products, we can see that both cross products are equal:
288=288
Therefore, we conclude that neither fraction is smaller than the other since both fractions are equivalent:
$\frac{8}{24}$=$\frac{12}{36}$
Comparing Like Fractions
Sample Problem 6:
Which fraction is larger, $\frac{3}{17}$ or $\frac{10}{17}$?
Solution:
Since both fractions are like fractions of the same denominator 17, we simply compare the numerators of both fractions:
3<10
Since the numerator of the second fraction is larger than the numerator of the first fraction, we then conclude that the second fraction is larger than the first fraction:
$\frac{10}{17}$>$\frac{3}{17}$
Comparing Unlike Fractions
Sample Problem 7:
We are given two fractions $\frac{3}{10}$ and $\frac{16}{30}$. We are then asked the following questions:
1. What is the common denominator of the two fractions?
1. What are the equivalent fractions of each fraction in terms of the common denominator?
1. Which fraction is smaller, $\frac{3}{10}$ or $\frac{16}{30}$?
Solution:
1. To get the common denominator of the given fractions, we multiply their denominators to get the product:
10×30=300
Hence, the common denominator of both fractions is 300.
1. From the first fraction $\frac{3}{10}$, we multiply its denominator by 30 to arrive at the common denominator 300. To get its equivalent fraction, we also multiply the numerator by the same number:
$\frac{3×30}{10×30}$=$\frac{90}{300}$
Therefore, the equivalent fraction of the first fraction is given by $\frac{90}{300}$.
Then, from the second fraction $\frac{16}{30}$ we multiply its denominator by 10 to arrive at the common denominator 300. To get its equivalent fraction, we also multiply the numerator by the same number:
$\frac{16×10}{30×10}$=$\frac{160}{300}$
Therefore, the equivalent fraction of the second fraction is given by $\frac{160}{300}$.
1. By comparing the equivalent like fractions $\frac{90}{300}$ and $\frac{160}{300}$, we simplify the comparison between $\frac{3}{10}$ and $\frac{16}{30}$ as a comparison between the numerators 90 and 160:
90<160
Since the numerator of the second equivalent fraction is greater than the numerator of the first equivalent fraction, we conclude that $\frac{160}{300}$>$\frac{90}{300}$. Furthermore, this implies that the second fraction $\frac{16}{30}$ is greater than $\frac{3}{10}$:
$\frac{16}{30}$>$\frac{3}{10}$
Sample Problem 8:
Suppose we use the two given fractions from Sample Problem 7. Again, we are asked the following questions:
Solution:
1. What is the LCD of the two fractions?
1. What are the equivalent fractions of each fraction in terms of the least common denominator?
1. Which fraction is smaller, $\frac{3}{10}$ or $\frac{16}{30}$?
Solution:
1. To get the least common denominator of the given fractions, we determine the least common multiple of both denominators. Since both denominators are multiples of 10, the least common denominator of both fractions is 30.
1. From the first fraction $\frac{3}{10}$, we multiply its denominator by 3 to arrive at the least common denominator 30. To get its equivalent fraction, we also multiply the numerator by the same number:
$\frac{3×3}{10×3}$=$\frac{9}{30}$
Therefore, the equivalent fraction of the first fraction is given by $\frac{9}{30}$.
Then, we note that the second fraction is already expressed in terms of the least common denominator. Hence, we do not need to rewrite the second fraction $\frac{16}{30}$.
1. By comparing the equivalent like fractions $\frac{9}{30}$ and $\frac{16}{30}$, we simplify the comparison between $\frac{3}{30}$ and $\frac{16}{30}$ as a comparison between the numerators 9 and 16:
9<16
Since the numerator of the second equivalent fraction is greater than the numerator of the first equivalent fraction, we conclude that $\frac{16}{30}$>$\frac{9}{30}$. Furthermore, this implies that the second fraction $\frac{16}{30}$ is greater than $\frac{3}{10}$:
$\frac{16}{30}$>$\frac{3}{10}$
We note that the same conclusion was obtained from the previous sample problem.
Summary
• A Fraction can be imagined as parts of a whole. Mathematically speaking, it is a ratio between the numerator and the denominator.
• There are three types of fractions according to how their numerators and denominators are related:
• A Proper Fraction is a fraction wherein its numerator is less than its denominator. The ratio between the numerator and denominator is always less than one.
• An Improper Fraction, on the other hand, is a fraction wherein its numerator is greater than its denominator. The ratio between the numerator and denominator is always greater than one.
• We can also express an improper fraction as a Mixed Number. A mixed number can be thought of as a combination of a whole number and a fraction.
• Two fractions can be related according to the value of their denominators:
• Like Fractions are fractions whose denominators are equal.
• Unlike Fractions are fractions whose denominators are not equal.
• Equivalent Fractions are unlike fractions that yield the same value.
• Generally, we can compare fractions using three approaches:
• An Intuitive Approach allows us to visualize each fraction, and compare the areas covered by each fraction to determine which one is larger/smaller.
• The Decimal Approach tells us the exact decimal values of each fraction as a basis to compare them.
• By performing Cross Multiplication, we take the cross-product of both fractions to compare which fraction is larger/smaller.
• In the case we are comparing like fractions, we can treat the comparison between like fractions as a comparison between their numerators.
• In the case we are comparing unlike fractions, we first rewrite their equivalent like fractions then compare their numerators and subsequently the original fractions.
• We can find the Common Denominator of two unlike fractions and express both fractions in terms of their common denominator.Alternatively, we can find their Least Common Denominator by looking at the Least Common Multiple of their denominators and then expressing both fractions in terms of their least common denominator. |
## Tuesday, October 31, 2017
### Middle School Math Solutions – Expand Calculator, Polynomial Multiplication
As of right now, you should be familiar with the distributive law and the FOIL method. We will now use both of them and apply them to our next topic, polynomial multiplication.
Recall: A polynomial is an expression of many (poly-) terms (-nomial) that are added and/or subtracted together. The terms include coefficients, variables, and POSITIVE exponents.
Examples: 4x^2 ,2x^3+5x^5 ,\frac{2}{3}x-1
There are a variety of polynomial multiplication problems. They might look intimidating at first, but we will go over how to solve them step by step using the distributive law and FOIL method, so you can see how to do them easily.
4x(3x^2+2x-1)
1. Distribute the 4x
4x(3x^2+2x-1)=4x∙3x^2+4x∙2x+4x∙(-1)
We applied the distributive law here for a polynomial with three terms (a(b+c+d)=ab+ac+ad)
2. Simplify
4x(3x^2+2x-1)=12x^3+8x^2-4x
We multiplied the terms.
That one wasn’t bad. Let’s see another one.
(2x+1)(4x+3y+2)
1. Distribute 2x+1
(2x+1)(4x+3y+2)=2x(4x+3y+2)+1(4x+3y+2)
We distributed 2x+1 to make the problem easier for us to solve.
2. Use the distributive law
2x(4x+3y+2)=2x∙4x+2x∙3y+2x∙2
1(4x+3y+2)=4x∙1+3y∙1+2∙1
We plug these values back in
(2x+1)(4x+3y+2)=2x∙4x+2x∙3y+2x∙2+4x∙1+3y∙1+2∙1
3. Simplify
(2x+1)(4x+3y+2)=8x^2+6xy+4x+4x+3y+2
=8x^2+6xy+8x+3y+2
(x+1)(x+2)(x+3)
1. Apply the FOIL method to two of the polynomials
(x+1)(x+2)=x∙x+x∙2+1∙x+1∙2
=x^2+3x+2
2. Plug x^2+3x+2 in for (x+1)(x+2)
(x+1)(x+2)(x+3)=(x^2+3x+2)(x+3)
Now this problem looks like the prior example.
3. Distribute x+3
(x+1)(x+2)(x+3)=x(x^2+3x+2)+3(x^2+3x+2)
I always prefer to distribute the polynomial with the smallest terms, but you can pick to distribute the quadratic. In that case you would have:
(x^2+3x+2)(x+3)=x^2 (x+3)+3x(x+3)+2(x+3)
4. Use the distributive law
(x+1)(x+2)(x+3)=x(x^2+3x+2)+3(x^2+3x+2)
=x∙x^2+x∙3x+x∙2+3∙x^2+3∙3x+3∙2
5. Simplify
(x+1)(x+2)(x+3)=x^3+3x^2+2x+3x^2+9x+6
=x^3+6x^2+11x+6
Not bad! You can see when we break the problem down, step by step, it is a lot easier to solve because we use the basics of expanding (distributive law and FOIL). For more practice problems and help, check out our Practice.
Until next time,
Leah |
# 26. Determinants: examples
## 26. Determinants: examples
Example:
Let $A=\begin{pmatrix}1&4&-4\\ -2&-2&-4\\ 3&-3&3\end{pmatrix}$ . Rather than applying our formula (sum over permutations), we're going to apply type I row operations to put this matrix into echelon form; this won't change the determinant, so we can take the determinant afterwards (at which point it will be the product of diagonal entries).
To clear the first column, we do $R_{2}\mapsto R_{2}+2R_{1}$ and $R_{3}\mapsto R_{3}-3R_{1}$ which gives: $\begin{pmatrix}1&4&-4\\ 0&6&-12\\ 0&-15&15\end{pmatrix}$ To clear the $-15$ below the diagonal, we do $R_{3}\mapsto R_{3}+\frac{5}{2}R_{2}$ , giving: $\begin{pmatrix}1&4&-4\\ 0&6&-12\\ 0&0&-15\end{pmatrix}.$ This is in echelon form. Therefore $\det(A)=1\times 6\times(-15)=-90$ .
Example:
Let $B=\begin{pmatrix}2&-3&-1&4\\ 2&-3&2&4\\ 2&-1&-4&-3\\ 2&-3&4&2\end{pmatrix}$ . Let's clear the first column with $R_{2}\mapsto R_{2}-R_{1}$ , $R_{3}\mapsto R_{3}-R_{1}$ , $R_{4}\mapsto R_{4}-R_{1}$ , yielding: $\begin{pmatrix}2&-3&-1&4\\ 0&0&3&0\\ 0&2&-3&-7\\ 0&0&5&-2\end{pmatrix}$ Now let's swap $R_{2}\leftrightarrow R_{3}$ : $\begin{pmatrix}2&-3&-1&4\\ 0&2&-3&-7\\ 0&0&3&0\\ 0&0&5&-2\end{pmatrix}.$ The determinant just changed sign! We'll need to remember to stick an extra minus sign in at the end. To clear the final $5$ below the diagonal, we do $R_{4}\mapsto R_{4}-\frac{5}{3}R_{3}$ : $\begin{pmatrix}2&-3&-1&4\\ 0&2&-3&-7\\ 0&0&3&0\\ 0&0&0&-2\end{pmatrix}$ which is now in echelon form.
The determinant of $B$ is therefore $-(2\times 2\times 3\times-2)=24$ . |
In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig
Question.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a rectangle?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
Let the number of horizontal columns represent the x-coordinates whereas the vertical rows represent the y-coordinates.
$\therefore$ The points are : $\mathrm{A}(3,4), \mathrm{B}(6,7), \mathrm{C}(9,4)$ and $\mathrm{D}(6,1)$
$\therefore \quad A B=\sqrt{(6-3)^{2}+(7-4)^{2}}$
$=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$
$B C=\sqrt{(9-6)^{2}+(4-7)^{2}}$
$=\sqrt{3^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$
$C D=\sqrt{(6-9)^{2}+(1-4)^{2}}$
$=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$
$\mathrm{AD}=\sqrt{(6-3)^{2}+(1-4)^{2}}$
$=\sqrt{(3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$
Since, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}$ i.e., All the four sides are equal
Also $\mathrm{AC}=\sqrt{(9-3)^{2}+(4-4)^{2}}$
$=\sqrt{(+6)^{2}+(0)^{2}}=6$ and
$B D=\sqrt{(6-6)^{2}+(1-7)^{2}}=\sqrt{(0)^{2}+(-6)^{2}}=6$
i.e., $\mathrm{BD}=\mathrm{AC}$
$\Rightarrow$ Both the diagonals are also equal.
$\therefore \quad \mathrm{ABCD}$ is a square.
Thus, Chameli is correct as $\mathrm{ABCD}$ is not a rectangle.
Editor |
# Question Video: The Nine Times Table Mathematics • 3rd Grade
Notice how each row is 9 more than the previous one. 1 × 9 = 9, 2 × 9 = 18, 3 × 9 = 27. Find the result of the following: 4 × 9 = ＿. Find the result of the following: 5 × 9 = ＿
02:17
### Video Transcript
Notice how each row is nine more than the previous one. One times nine equals nine; two times nine equals 18; three times nine equals 27. Find the result of the following: Four times nine equals what. And then also find the result of the following: Five times nine equals what.
We know that learning times tables facts is all about thinking about groups of a certain number. And this question is all about nine times tables facts. We can see in the pictures that each cube is labeled with the number nine. We’re counting in nines. And the first thing that we’re told is that each row is worth nine more than the previous one. In other words, as we go through our nine times tables facts to find the next one, we just add another nine. So that’s why we have one times nine is nine, and then we’ve got a new orange cube to show that nine more’s been added. Nine add nine is 18. So that’s why we can say two times nine is 18. If we add another nine, we get 27. And that’s why we can say three times nine equals 27.
And next, we’re told to find the result of four times nine. Well, to find four lots of nine, we just need to add nine onto three lots of nine. So what’s nine more than 27? Well, we know if we had to add 10, it would be 37. And nine is one less than this. Four times nine is 36.
And in the final part of the question, we just need to do the same again because we need to find five times nine. We know what four nines are, so we just need to add one more nine. Now we know 36 add 10 would be 46. But again, nine is one less than ten, so instead of 46, the answer is 45. If we know a nine times tables fact, we can find the next fact by adding nine. And that’s how we found out that four times nine is 36 and five nines are 45. |
# Introduction To Trigonometry – Class X – Part 3 | Exercise 8.1 Q10-11 Solved
### Watch the third video session on “Introduction To Trigonometry” for Class Xth – concepts and solved questions
Introduction To Trigonometry – Class X – Part 3
1. The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). It is the study of relationships between the sides and angles of a triangle.
2. Early astronomers used it to find out the distances of the stars and planets from the Earth.Today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.
Steps to find the trigonometric ratios with respect to an angle (say A) in a right angled triangle:
a. Keep triangle in such a way so that angle A is in the base (along with the 90 degree angle).
b. Identify the “perpendicular” (p), “base” (b) and hypotenuse (h) sides. Usually two of the sides are given; the unknown can be calculated using Pythagoras Theorem.
c. Find out the required trigonometric ratio. (Refer to the video to find out the shortcut to remember the six trigonometric ratios.)
NCERT Exercise 8.1 Q10-11 Solved
10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A =12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ =4/3 for some angle θ.
—— |
# CBSE Class 10 Maths Question Paper | 2018
###### Class 10 Board Paper Solution | 2 Mark Questions | Section B
Section B contains 6 questions of 2 marks each. Scroll down for explanatory answer and video solution to all six 2-mark questions that appeared in 2018 CBSE Class 10 Board Exam maths paper. The questions appeared from the following topics: Real numbers (irrational numbers), Linear Equations (Linear equations in 2 variables), Arithmetic Progressions (sum of an AP), Coordinate Geometry (section formula), and Probability (probability of rolling dice and picking numbers).
1. Given that √2 is irrational , prove that (5 + 3√2) is an irrational number.
Hint to solve this irrational numbers question
A classic text book question.
Step 1: Assume the contrary to what is to be proved. i.e., assume that the given number is rational.
Step 2: If the converse is true, we should be able to express the number as a fraction.
Step 3: Derive √2 in terms of the expressed fraction and deduce if that can be done, √2 is also rational
2. In the figure given below, ABCD is a rectangle. Find the values of x and y.
Hint to solve this Linear Equations problem
An easy 2 mark question that appeared in CBSE class 10 maths paper 2018.
Concept: Opposite sides of a rectangle are equal
Approach: Step 1: Equate the length given in terms of x and y to the value of the length. You will get a linear equation in two variables.
Step 2: Equate the width given in terms of x and y to the value of the width. You will get a second linear equation in two variables.
Step 3: Solve the system of linear equations in two variables to find the answer to x and y.
3. Find the sum of first 8 multiples of 3.
Hint to solve this Arithmetic Progressions problem
Concept: The first 8 multiples of 3 are in an AP with a common difference of 3. The first term of the arithmetic progression is 3.
Approach: Now that we know the first term, common difference and the number of terms, susbtitute these values in the sum of arithmetic progression formula to find the answer.
4. Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, - 3). Hence, find m.
Hint to solve this Coordinate Geometry problem
Approach: Let point P divide the segment AB in the ratio m1 : m2.
Step 1: Apply the section formula on the x coordinate of point P to compute the ratio in which point P divides the line segment AB.
Step 2: Apply the section formula with the value of the ratio computed in step 2 to find the y coordinate of point P. That gives the value of m.
5. Two different dice are tossed together, Find the probability:
(i) of getting a doublet
(ii) of getting a sum 10, of the numbers on the two dice
Hint to solve this CBSE 2018 1 mark Probability question
Step 1: Compute the total number of outcomes when two dice are tossed together.
Step 2: What is a doublet? An outcome such as (1, 1) or (5, 5) is a doublet. Compute the number of such outcomes when two dice are tossed together.
Step 2: What is a doublet? An outcome such as (1, 1) or (5, 5) is a doublet. Compute the number of such outcomes when two dice are tossed together.
Step 3: The ratio of the values that you got in step 2 to step 1 is the required probability for part (i).
Step 4: List down outcomes that will result in a sum of 10.
Step 5: The value of step 4 divided by the alue of step 1 is the answer to part (ii) of this CBSE Class 10 maths section A question.
6. An integer is chosen at random between 1 and 100. Find the probability that it is:
(i) divisible by 8
(ii) not divisible by 8
Hint to solve this CBSE 2018 maths 1 mark Probability problem
Step 1: Compute the number of numbers between 1 and 100. Note: The count will not include 1 and 100.
Step 2 - Part (i): Compute the number of multiples of 8 between 1 and 100.
Step 3: The value obtained in step 2 divided by the value obtained in step 1 is the answer to part (i) of this 2018 CBSE class 10 maths section A question.
Step 4 - part (ii): Events given in part (i) and part (ii) are mutually exclusive and collectively exhaustive events. Therefore, answer for part (ii) = 1 - answer for part (i)
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Math answers scanner can be found online or in mathematical textbooks. We will also look at some example problems and how to approach them.
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How to solve for domain is a question asked by many students who are studying mathematics. The answer to this question is very simple and it all depends on the function that you are trying to find the domain for. In order to solve for the domain, you first need to identify what the function is and then identify the input values. For example, if you have a function that is defined as f(x)=x^2+1, then the domain would be all real numbers except for when x=0. This is because when x=0, the function would equal 1 which is not a real number. Another example would be if you have a function that is defined as g(x)=1/x, then the domain would be all real numbers except for when x=0. This is because when x=0, the function would equal infinity which is not a real number. To sum it up, in order to solve for the domain of a function, you need to determine what the function is and then identify what values of x would make the function equal something that is not a real number.
How to solve mode: The mode is the value that appears most often in a set of data. To find the mode, simply order the values from smallest to largest and count how many times each value appears. The value that appears the most is the mode. For example, in the set {1, 2, 2, 3, 3, 4}, the mode is 2 because it appears twice while the other values only appear once. To find the mode of a set of data, follow these steps: 1) Order the values from smallest to largest. 2) Count how many times each value appears. 3) The value that appears the most is the mode.
Solving the distance formula is a common exercise in mathematics and physics. The distance formula is used to determine the distance between two points in space. The formula is relatively simple, but it can be difficult to solve if you don't have a firm understanding of the concepts involved. In this article, we'll walk you through the steps necessary to solve the distance formula. With a little practice, you'll be solving it like a pro in no time!
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# Lesson 13
El perímetro y el área de los rectángulos
## Warm-up: Exploración de estimación: Lavado de ventanas (10 minutes)
### Narrative
The purpose of an Estimation Exploration is for students to practice the skill of estimating a reasonable answer based on experience and known information. In this lesson they will be finding the perimeter and area of rectangles and their thinking about the size of the windows in this image prepares them for this work.
### Launch
• Groups of 2
• Display the image.
• “¿Qué estimación sería muy alta?, ¿muy baja?, ¿razonable?” // “What is an estimate that’s too high?” “Too low?” “About right?”
• 1 minute: quiet think time
### Activity
• “Discutan con su pareja cómo pensaron” // “Discuss your thinking with your partner.”
• 1 minute: partner discussion
• Record responses.
### Student Facing
¿Cuál es el área de una ventana?
Escribe una estimación que sea:
muy baja razonable muy alta
$$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$
### Activity Synthesis
• “¿Qué cosas de la imagen les podrían ayudar a estimar el área de la ventana?” // “What could you use in the image to help estimate the area of the windows?” (There are the people cleaning the windows. I used the people to estimate the height and width of the windows and then multiplied to find the area.)
## Activity 1: Perímetros de rectángulos (15 minutes)
### Narrative
The purpose of this activity is for students to plot points that represent the length and width of a rectangle with a given perimeter. Since the perimeter is twice the length plus twice the width, decreasing the length by a certain amount will mean that the width has to increase by the same amount for the perimeter to stay the same. Students have an opportunity to observe this relationship in multiple ways (MP7, MP8):
• think geometrically about the perimeter of the rectangle
• look at the table of values for length and width depending on the values they used
• look at the length and width pairs plotted in the coordinate grid
MLR8 Discussion Supports. Prior to solving the problems, invite students to make sense of the situations and take turns sharing their understanding with their partner. Listen for and clarify any questions about the context.
Representation: Access for Perception. Read tasks aloud. Students who both listen to and read the information will benefit from extra processing time.
Supports accessibility for: Language Conceptual Processing
• Groups of 2
### Activity
• 2 minutes: independent think time
• 5 minutes: partner work time
### Student Facing
largo (cm) ancho (cm)
1. Jada dibujó un rectángulo que tiene un perímetro de 12 centímetros. ¿Cuáles podrían ser el largo y el ancho del rectángulo de Jada? Escribe tu respuesta en la tabla.
4. Si el rectángulo de Jada medía 2.5 cm de largo, ¿cuánto medía de ancho? Ubica el punto correspondiente en la cuadrícula de coordenadas.
5. Si el rectángulo de Jada medía 3.25 cm de largo, ¿cuánto medía de ancho? Ubica el punto correspondiente en la cuadrícula de coordenadas.
### Student Response
If students are not sure how to determine the width of Jada’s rectangle, prompt the student to draw a rectangle and ask, “¿Cómo te puede ayudar tu dibujo a completar la tabla?” // “How can you use your drawing to help you fill in the table?”
### Activity Synthesis
• “¿Cómo encontraron el ancho del rectángulo de Jada que medía 3.25 cm de largo?” // “How did you find the width of Jada's rectangle if it was 3.25 cm long?” (I knew that the length and width together are half the perimeter which is 6 cm. So I subtracted 3.25 from 6 and that was 2.75.)
• “¿Qué le pasa al ancho cuando el largo aumenta 1? ¿Por qué?” // “What happens to the width when the length increases by 1? Why?” (The width decreases by one. This makes sense because the sum needs to say the same or else the perimeter changes.)
• “¿Cómo se ve esto en la gráfica?” // “How does the graph show this?” (For each point I plotted, I can go right one and down one and find another possible length and width.)
## Activity 2: Áreas de rectángulos (20 minutes)
### Narrative
The purpose of this activity is to investigate the possible lengths and widths of a rectangle with given area. Since the area is the product of length and width, this means that the main operation being used here is multiplication or division, contrasting with the previous activity where students investigated the perimeter which is the sum of the side lengths of a rectangle. This means that the calculations are more complex and some of the coordinates of the points that students plot will either be decimals or fractions depending how students express them. There are some important common characteristics between the lengths and widths for a given area and for a given perimeter which will be examined in the activity synthesis (MP7, MP8):
• when the length increases, the width decreases
• the length and width can be switched to get another possible length and width pair
• Groups of 2
### Activity
• 2 minutes: independent think time
• 5 minutes: partner work time
### Student Facing
largo (cm) ancho (cm)
1. Jada dibujó un rectángulo que tiene un área de 16 centímetros cuadrados. ¿Cuáles podrían ser el largo y ancho del rectángulo de Jada? Escribe tu respuesta en la tabla.
3. Si el rectángulo de Jada medía 5 cm de largo, ¿cuánto medía de ancho? Ubica el punto correspondiente en la cuadrícula de coordenadas.
4. Si el rectángulo de Jada medía 3 cm de largo, ¿cuánto medía de ancho? Ubica el punto correspondiente en la cuadrícula de coordenadas.
5. Si Jada dibujó un cuadrado, ¿cuánto medía de largo y de ancho? Explica cómo lo sabes.
### Activity Synthesis
• Invite students to share their responses for the width of a rectangle that is 5 cm long.
• “¿Cómo calcularon el valor?” // “How did you calculate the value?” (I knew that 5 times the width was 16 so the width is $$16 \div 5$$ or $$\frac{16}{5}$$ cm.)
• “¿Cómo supieron dónde ubicar esa pareja de largo y ancho?” // “How did you know where to plot that length and width pair?” (I looked for 5 on the horizontal axis and then I had to estimate where $$3\frac{1}{3}$$ was on the vertical axis. I put it a little above 3 but closer to 3 than to 4.)
• “¿En qué fue parecido encontrar largos y anchos posibles para un área dada y encontrar largos y anchos posibles para un perímetro dado?” // “How was determining the possible lengths and widths for a given area the same as determining the possible lengths and widths for a given perimeter?” (When the length increases the width decreases. When the width decreases the length increases. I can flip the order of the length and width and get another rectangle.)
• “¿En qué son diferentes las parejas de largo y ancho de los rectángulos que tienen área de 16 y las parejas de largo y ancho de los rectángulos que tienen perímetro de 12?” // “How are the length and width pairs for rectangles with area 16 different from the length and width pairs for rectangles with perimeter 12?” (I was looking for a total of 16 instead of a total of 12. I have to multiply the side lengths rather than add them. When the length decreases by 1 for the perimeter, the width increases by 1. For area, when the length decreases the width increases but the relationship is more complex.)
• Consider drawing some rectangles with an of area 16 on the coordinate grid with the lower left corner of each rectangle at $$(0,0)$$. Ask students what the notice about the coordinates of the upper right corners of each rectangle. (They represent the length and width of the corresponding rectangle.)
## Lesson Synthesis
### Lesson Synthesis
“Hoy graficamos largos y anchos de rectángulos en la cuadrícula de coordenadas” // “Today we plotted lengths and widths of rectangles on the coordinate grid.”
Display the graphs from the student solutions to the two activities together.
“¿En qué se parecen las gráficas?” // “How are the graphs the same?” (They both show lengths and widths of rectangles. When the length increases, the width decreases. When the length decreases, the width increases.)
“¿En qué son diferentes las gráficas?” // “How are the graphs different?” (The length and width pairs with perimeter 12 are nicely organized. When the length increases by 1 the width decreases by 1. The length and width pairs with area 16 don't follow a clear pattern. I would not be able to guess any other values. I would have to calculate.)
## Student Section Summary
### Student Facing
En esta sección, generamos patrones e identificamos las relaciones que había entre dos patrones diferentes.
A B C D E F
regla 1: empezar en 0 y siempre sumar 8. 0 8 16 24 32 40
regla 2: empezar en 0 y siempre sumar 2. 0 2 4 6 8 10
Cada número de la regla 1 es 4 veces el valor del número correspondiente en la regla 2 y cada número de la regla 2 es $$\frac{1}{4}$$ veces el valor del número correspondiente en la regla 1. También graficamos las reglas juntas en una cuadrícula de coordenadas.
Además, usamos el plano de coordenadas para representar otras situaciones, como el largo y el ancho de rectángulos que tienen un perímetro dado. |
Sine or Cosine Writing Equations Given Graph
Sine or Cosine Writing Equations Given Graph
### Select a Course Below
• Standardized Test Prep
• K12 Math
• College Math
• Homeschool Math
You’ve already learned the basic trig graphs. But just as you could make the basic quadratic, y = x2, more complicated, such as y = −(x + 5)2 − 3, so also trig graphs can be made more complicated. We can transform and translate trig functions, just like you transformed and translated other functions in algebra.
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Let’s start with the basic sine function, f (t) = sin(t). This function has an amplitude of 1 because the graph goes one unit up and one unit down from the midline of the graph. This function has a period of 2π because the sine wave repeats every 2π units.
The graph of f (t) = sin(t) looks like this:
Now let’s look at g(t) = 3sin(t):
Do you see that this second graph is three times as tall as was the first graph? The amplitude has changed from 1 in the first graph to 3 in the second, just as the multiplier in front of the sine changed from 1 to 3. This relationship is always true: Whatever number A is multiplied on the trig function gives you the amplitude (that is, the “tallness” or “shortness” of the graph); in this case, that amplitude number was 3.
For this function, the value of the amplitude multiplier A is given by 0.5, so the function will have an amplitude of:
amplitude: 0.5 = ½
For this function, the value of the amplitude multiplier A is −2, so the amplitude is:
amplitude: 2
…and, by the way, the graph of y = −2cos(x) would also be flipped upside down, because of the “minus” sign.
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Technically, the amplitude is the absolute value of whatever is multiplied on the trig function. The amplitude just says how “tall” or “short” the curve is; it’s up to you to notice whether there’s a “minus” on that multiplier, and thus whether or not the function is in the usual orientation, or upside-down.
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Recall the first graph above, which was the “regular” sine wave:
Now let’s look at h(t) = sin(2t):
Do you see how this third graph is squished in from the sides, as compared with the first graph? Do you see that the sine wave is cycling twice as fast, so its period is only half as long?
This relationship is always true: Whatever value B is multiplied on the variable (inside the trig function), you use this value to find the period ω (“omega”, not “double-u”) of the trig function, according to this formula:
general period formula:
For sines and cosines (and their reciprocals, cosecants and secants), the “regular” period is 2π, so their period formula is:
period formula for sines & cosines:
For tangents and cotangents, the “regular” period is π, so their period formula is:
period formula for tangents & cotangents:
In the sine wave graphed above, the value of the period multiplier B was 2. (Sometimes the value of B inside the function will be negative, which is why there are absolute-value bars on the denominator.) As a result, its period was .
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The formula for sines and cosines says that the regular period is 2π. In cos(3t), B = 3, so this function will have a period of:
The formula for tangents and cotangents says that the regular period is π. For , the value of B is ½, so the function will have a period of:
Recall again the first graph above, being that of the “regular” sine wave:
Now let’s look at :
Do you see that the graph (shown in blue on the graph above) is shifted over to the right by π/3 units from the regular graph (shown in gray)? This relationship is always true: If the argument of the function (the thing you’re plugging in to the function) is of the form “(variable) − C” (where C is just some number), then the graph is shifted to the right by that number of units (that is, by C units); if the argument is of the form “(variable) + C”, then the graph is shifted to the left by that number of units (again, by C units). Within the context of trigonometric functions, this right- or left-shifting is called “phase shift”.
Inside the argument (that is, inside the parentheses of the function), a is added to the variable. This means that . Because this value is added to the variable, then the shift is to the left. Then the phase shift is:
to the left by
The number C inside with the variable is , so this will be the phase shift. This number is subtracted from the variable, so the shift will be to the right.
to the right by
Let’s recollect again the graph of the “regular” sine wave:
Now let’s look at k(t) = sin(t) + 3:
Do you see how the graph was shifted up by three units? This relationship is always true: If a number D is added outside the function, then the graph is shifted up by that number of units; if a number D is subtracted, then the graph is shifted down by that number of units.
The trig-function part is the cos(t); the up-or-down shifting part is the D = −2. There’s nothing else going on inside of the function, nor multiplied in front of it, so this is the regular cosine wave, but it’s:
shifted downward two units
The trig-function part is the tan(x); the up-or-down shifting part is the + 0.6. So this is the regular tangent curve, but:
shifted upward by of a unit
Putting it all together in terms of the sine wave, we have the general sine function:
F(t) = Asin(Bt − C) + D
…where |A| is the amplitude, B gives you the period, D gives you the vertical shift (up or down), and is used to find the phase shift.
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“Wait!”, I hear you cry; “Why don’t we just just use C for the phase shift?” Because sometimes more involved stuff is going on inside the function. Remember that the phase shift comes from what is added or subtracted directly to the variable. If the variable isn’t alone (that is, if there’s something multiplied directly on it), then there’s another step to follow.
For instance, if you have something like:
…the phase shift is not π units! Instead, you first have to isolate what’s happening to the variable by factoring, as so:
Now you can see that the phase shift will be units, not π units. So the phase shift, as a formula, is found by dividing C by B.
For F(t) = A f (Bt − C) + D, where f(t) is one of the basic trig functions, we have:
(Note: Different books use different letters to stand for the period formula. In your class, use whatever your book or instructor uses.)
The amplitude is given by the multipler on the trig function. In this case, there’s a −2.5 multiplied directly onto the tangent. This is the “A” from the formula, and tells me that the amplitude is 2.5. (If I were to be graphing this, I would need to note that this tangent’s graph will be upside-down, too.)
The regular period for tangents is π. In this particular function, there’s a 4 multiplied on the variable, so B = 4. Plugging into the period formula, I get .
To find the phase shift, I need to isolate the variable with the shift value, so I need to factor out the 4 (also known as “C”) that’s multiplied on the variable. The factorization is:
Then the phase shift is . Because the shift value is subtracted from the variable, the shift is to the right.
(I could also have used the simpler method, directly from the formula, of dividing C by B. This would have given me the same value, but more quickly and with less chance of error in the factorization. Try it both ways yourself, and figure out which one you like better. Then practice it well before the next test.)
The vertical shift comes from the value entirely outside of the trig function; namely, the outer 4 (also known as “D”, from the formula). Because this 4 is subtracted from the tangent, the shift will be four units downward from the usual center line, the x-axis.
amplitude: 2.5
period:
phase shift: to the right by
vertical shift: downward by 4
So far, we’ve looked at finding the information needed for graphing. Now let’s see what this process looks like in practice, because there’s a way to make drawing these graphs a whole lot easier than what they show in the book….
URL: https://www.purplemath.com/modules/grphtrig.htm
You are watching: Graphing translated / transformed trig functions. Info created by Bút Chì Xanh selection and synthesis along with other related topics. |
Wednesday, June 22, 2022
HomeMaths8 Examples of Linear Models in Real Life
# 8 Examples of Linear Models in Real Life
A linear model is an equation that is used to compare two values i.e. x and y.
In linear models, the change between the two values is consistent. This means that the x changes at the same rate in which y changes.
These equations can be represented on a graph using a straight line, hence the name linear. They are the simplest type of model in mathematics.
Here are examples of linear models in real life:
## 1. The relationship between boiling point of water and change in altitude
The boiling point of water varies with changes in altitude. You can determine the boiling point at a given altitude using the linear equation:
Boiling point = β0 + β1
Where β0 is the expected boiling point without a change in altitude and β1 is the rate of change in boiling point when the altitude changes by one unit.
## 2. Determining the relationship between spending on advertising and revenue
Many companies have a marketing budget. In marketing, it is expected that companies that spend more on advertising will attract more revenue. This is represented in the linear model:
Total revenue = βo + β1
Whereβ0 is the total expected revenue without spending on advertising and β1 is the average change in total revenue when spending on advertisements is increased by a single unit e.g. by a dollar.
Using this linear model, companies can decide by how much they would need to increase the spending on advertising and how much they should expect as revenue as a result.
## 3. Drug dosage and response
Medical researchers use linear models to understand the relationship between the dosage of a drug and the response of patients.
Researchers may administer a dose and observe the patient’s reaction. They will change dosage and keep observing the patient’s reaction to determine the optimum dosage. This relationship can be represented as follows:
Patient’s response = β0 + β1
Β0 is the response when the dosage is zero. β1represents the average change in the patient’s response when the dosage of the drug given to the patient is increased by a unit.
## 4. The relationship between the amount of fertilizer used and crop yields
Agriculturalists help farmers determine the optimum amount of fertilizer to use to get high crop yields using linear equations. They often use linear models such as:
Total Crop Yield = β0 + β1.
In this linear equation β0 is the expected crop yield with the use of fertilizer. β1is the average change in crop yield when the amount of fertilizer used on the field is increased by one unit and all other factors remain constant.
The agriculturalist may advise farmers to change the amount of fertilizer they use to maximize their crop yield.
## 5. Performance of athletes and their training regimen
Professional sports teams and trainers use linear models to assess the efficacy of the training regimens used on professional athletes.
They use these equations to express the relationship between the performance of the athlete and changes in their training regimen.
For example, a linear model may be used to show the relationship between weight lifting and the number of rebounds a player in basketball is able to pick per game.
The rebounds per game= β0 + β1
Where β0 is the number of rebounds the player is expected to pick without weight training and β1 is the change in the number of rebounds picked when weight training sessions are increased by one.
This equation assumes that training sessions remain constant from one week to the next.
## 6. Predicting population growth over time
Statisticians use linear models to predict population growth in a given region such as a town or even countries.
The two variables are time and population size. The population growth for a specific period can be predicted using the linear equation:
Total population = β0 + β1
Where β0 is the initial or current population and β1 is the rate of change of the population after each year.
## 7. Relationship between height and weight
The relationship between height and weight can be represented using the linear model
Body weight = β0 + β1
Where β0 is the starting or reference weight of a person at average height and β1 is the rate of change in weight with an increase in height of one unit. This equation can be used to estimate an individual’s weight when their height is known.
## 8. The relationship between income and education/experience
Linear models can be used to express the relationship between years of experience or education levels and income. This can be expressed as:
Income = β0 + β1 + β2
Where β0 is the expected minimum income level, β1 is the rate in change of income with a change in education level and β2 is the rate in change of income with a change in experience.
If everything else remains constant, higher levels in education and more experience should attract higher income.
Linear models are an important part of everyday life. They are used in every day as a means to predict outcomes.
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# Class 8 RD Sharma Solutions- Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube)- Exercise 21.2 | Set 1
• Last Updated : 07 Nov, 2022
### Question 1. Find the volume in cubic meters (cu.m) of each of the cuboids whose dimensions are :(i) length = 12 m, breadth = 10 m, height = 4.5 m(ii) length = 4 m, breadth = 2.5 m, height = 50 cm(iii) length = 10 m, breadth = 25 dm, height = 25 cm
Solution:
i) length = 12 m, breadth = 10 m, height = 4.5 m
Since, Length (l) = 12 m
Height (h) = 4.5 m
As we know, Volume of Cuboid = l × b × h
= 12 × 10 × 4.5 m3 = 540 m3
Hence, Volume of given cuboid is 540 m3
ii) length = 4 m, breadth = 2.5 m, height = 50 cm
Since, Length (l) = 4 m
Height (h) = 50 cm = 0.5 m
As we know, Volume of Cuboid = l × b × h
= 4 × 2.5 × 0.5 m3 = 5 m3
Hence, Volume of given cuboid is 5 m3
iii) length = 10 m, breadth = 25 dm, height = 25 cm
Since, Length (l) = 10 m
Breadth (b) = 25 dm = 2.5 m
Height (h) = 50 cm = 25 cm = 0.25 m
As we know, Volume of Cuboid = l × b × h
= 10 × 2.5 × 0.25 m3 = 6.25 m3
Hence, Volume of given cuboid is 6.25 m3
### Question 2. Find the volume in cubic decimeter of each of the cubes whose side is(i) 1.5 m(it) 75 cm(iii) 2 dm 5 cm
Solution:
i) 1.5 m
Since the side of cube = 1.5 m
As we know, Volume of Cube = (side)3
= (1.5)3 = 3.375 m3
As 1 m3 = 1000 dm3)
3.375 m3 = 3375dm3
Hence, Volume of the given cube is 3375 dm3
ii) 75 cm
Since the side of cube = 75 cm = 7.5 dm
As we know, Volume of Cube = (side)3
= (7.5)3 = 421.875 dm3
Hence, Volume of the given cube is 421.875 dm3
iii) 2 dm 5 cm
Since the side of cube = 2 dm 5 cm = 2.5 dm
As we know, Volume of Cube = (side)3
= (2.5)3 = 15.625 dm3
Hence, Volume of the given cube is 15.625 dm3
### Question 3. How much clay is dug out in digging a well measuring 3 m by 2 m by 5 m?
Solution:
Given, length of well = 3 m
The breadth of well = 2 m
The height of well = 5 m
Amount of clay dig out = Volume of well
The volume of well = l × b × h = 3 × 2 × 5 m3
= 30 m3
Hence, the amount of clay dug out is 30 m3
### Question 4. What will be the height of a cuboid of volume 168 m3, if the area of its base is 28 m2?
Solution:
Volume of cuboid = l × b × h = 168 m3
Area of cuboid = l × b = 28 m2
So, the height of the cuboid can be derived = Volume of cuboid/ Area of cuboid
h = 168/28 m = 6 m
Hence, the height of the cuboid is 6 m
### Question 5. A tank is 8 m long, 6 m broad, and 2 m high. How much water can it contain?
Solution:
Given, length of tank = 8 m
The breadth of tank = 6 m
The height of the tank = 2 m
Amount of water the tank can hold = Volume of tank
The volume of tank = l × b × h = 8 × 6 × 2 m3
= 96 m3
(As 1 m3 = 1000 liters)
Hence, the amount of water the tank can hold is 96000 liters
### Question 6. The capacity of a certain cuboidal tank is 50000 litres of water. Find the breadth of the tank if its height and length are 10 m and 2.5 m respectively.
Solution:
Given, volume of cuboidal tank = 50000 litres
The height of tank = 10 m
The length of tank = 2.5 m
We need to find breadth of the tank.
As 1 l = 1/1000 m3
So, 50000 l = 50 m3
Volume of cuboidal tank = l × b × h
50 = 2.5 × b × 10
b = 2 m
Hence, the breadth of cuboidal tank is 2 m
### Question 7. A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep. How many litres of diesel can it hold?
Solution:
Given, length of tanker = 2 m
The breadth of tanker = 2 m
The depth or height of tanker = 40 cm = 0.4 m
Amount of diesel tanker can hold = Volume of cuboidal tanker
= l × b × h = 2 × 2 × 0.4 = 1.6 m3
And, 1.6 m3 = 1.6 × 1000 = 1600 litres
Hence, the tanker can hold 1600 litres of diesel
### Question 8. The length, breadth and height of a room are 5 m, 4.5 m and 3 m, respectively. Find the volume of the air it contains.
Solution:
Given, the length of room = 5 m
The breadth of room = 4.5 m
The height of room = 3 m
Volume of air the room can hold = Volume of Cuboidal room
= l × b × h = 5 × 4.5 × 3 = 67.5 m3
Hence, the volume of air the room can contain is 67.5 m3
### Question 9. A water tank is 3 m long, 2 m broad and 1 m deep. How many litres of water can it hold?
Solution:
Given, the length of water tank = 3 m
The breadth of water tank = 2 m
The height of water tank = 1 m
Amount of water the tank can hold = Volume of Cuboidal water tank
= l × b × h = 3 × 2 × 1 = 6 m3
And, 6 m3 = 6000 litres
Hence, the water tank can hold 6000 litres of water.
### Question 10. How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick?
Solution:
Given, length of wooden block = 6 m
The breadth of wooden block = 75 cm = 0.75 m
The thickness/height of wooden block = 45 cm = 0.45 m
So, the volume of wooden block = l × b × h = 6 × 0.75 × 0.45
= 2.025 m3
Now, length of plank = 3 m
The breadth of plank = 15 cm = 0.15 m
The thickness/height of plank = 5 cm = 0.05 m
So, the volume of plank = l × b × h = 3 × 0.15 × 0.05
= 0.0225 m3
So, the number of plank that can be prepared from wooden block = volume of wooden block/ volume of plank
= 2.025/0.0225 = 90
Hence, 90 planks can be made from the given wooden block
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# What is 5-3*(-2) + |-3|?
Mar 3, 2018
The result is $14$.
#### Explanation:
First, compute the absolute value. Then, do the parentheses. Lastly, do the addition:
$\textcolor{w h i t e}{=} 5 - 3 \cdot \left(- 2\right) + | - 3 |$
$= 5 - 3 \cdot \left(- 2\right) + 3$
$= 5 - \left(- 6\right) + 3$
$= 5 + 6 + 3$
$= 11 + 3$
$= 14$
Mar 3, 2018
$14$
#### Explanation:
It's a lot easier if you follow order of operations:
• Parentheses
• Exponent
• Multiplication
• Division
$- 3 \cdot \left(- 2\right) = 6$
$5 + 6 + | - 3 | =$
$11 + | - 3 | =$
$14$ |
##### Statistics: 1001 Practice Problems For Dummies (+ Free Online Practice)
The sample statistics questions here require that you compare three box plots. These side-by-side box plots represent home sale prices (in thousands of dollars) in three cities in 2012.
Credit: Illustration by Ryan Sneed
## Sample questions
1. From high to low, what is the order of the cities' median home sale prices?
Answer: City 2, City 1, City 3
The bar in the center of the box represents the median of each distribution; City 2 is the highest, followed by City 1 and City 3.
2. If the number of homes sold in each city is the same, which city has the greatest percentage of homes that sold for more than \$72,000?
To find the percentage of homes in a city that sold for more than \$72,000, look at the numerical axis, where you can see that City 2 has 75% of the data lying past 72 (because Q1 for City 2 is greater than 72). The other cities don't.
3. Assuming 100 homes sold in each city in 2012, which city has the most homes that sold for more than \$72,000?
The lower edge of the box, which represents Q1, is above 72 for City 2, whereas the medians for City 1 and City 3 are below 72. Therefore, if all three cities had the same number of home sales in 2012, City 2 must have had the most above \$72,000.
4. Which city has the smallest range in home prices?
The range is the maximum value minus the minimum value of a data set (shown in the very top line and the very bottom line of the box plot).
City 1 range: 80,000 – 62,000 = 18,000
City 2 range: 125,000 – 43,000 = 82,000
City 3 range: 80,000 – 38,000 = 42,000
City 1 has the smallest range, \$18,000.
5. Which of the following statements is true?
(A) More than half of the homes in City 1 sold for more than \$50,000.
(B) More than half of the homes in City 2 sold for more than \$75,000.
(C) More than half of the homes in City 3 sold for more than \$75,000.
(D) Choices (A) and (B).
(E) Choices (B) and (C).
Answer: D. Choices (A) and (B) (More than half of the homes in City 1 sold for more than \$50,000; more than half of the homes in City 2 sold for more than \$75,000.)
As indicated by the median lines within the boxes for each city, more than half of the homes in City 1 sold for more than \$50,000, and more than half of the homes in City 2 sold for more than \$75,000.
6. Which of the following statements is true?
(A) About 25% of homes in City 1 sold for \$75,000 or more.
(B) About 25% of homes in City 2 sold for \$75,000 or more.
(C) About 25% of homes in City 2 sold for \$98,000 or more.
(D) About 25% of homes in City 3 sold for \$75,000 or more.
(E) Choices (A) and (C).
Answer: E. Choices (A) and (C) (About 25% of homes in City 1 sold for \$75,000 or more; about 25% of homes in City 2 sold for \$98,000 or more.)
The cutoff for the upper 25% of the values in a box plot is indicated by the largest number in the box. For City 1, about 25% of homes sold for \$75,000 or more, and about 25% of homes in City 2 sold for \$100,000 or more.
If you need more practice on this and other topics from your statistics course, visit 1,001 Statistics Practice Problems For Dummies to purchase online access to 1,001 statistics practice problems! We can help you track your performance, see where you need to study, and create customized problem sets to master your stats skills. |
# Factors of 83: Prime Factorization, Methods, Tree, and Examples
Factors of 83 are those numbers that divide the number 83 exactly without leaving any remainder, or it can also be termed as all the numbers that give 83 as a product when multiplied together.
Figure 1 – All possible factors of 83
To get the pair factors of 83, multiply any two natural numbers to get the original number i.e., 83. In the case of 83, there are only two factors as 83 is a prime number. The factors of 83 are 1 and 83, 83 being the highest factor.
In this article, we will be discussing various methods for finding the factors, what is prime factorization and how it is performed for the number 83.
## What Are the Factors of 83?
The factors of 83 are 1 and 83 itself.
Factors of 83 are the group of natural numbers or integers that can be divided equally into 83. As 83 is an odd number none of its factors is 2 or any multiple of 2. 83 being a prime number cannot be divided by any other number except for 1 and 83 itself.
## How To Calculate the Factors of 83?
To calculate the factors of 83, start dividing it by the smallest natural number 1, and see whether the remainder is zero or not. As for the number to be a factor of the given number, it must be exactly divisible by the number leaving zero as the remainder.
To find the factors of 83, start dividing 83 by the smallest whole number (odd number) and if the result in the remainder is 0, it is a factor of 83. Please keep in mind that 83 is an odd number so odd numbers can only be factors of 83.
Firstly, divide 83 by 1.
$\dfrac{83}{1} = 83$
Since, the remainder is 0, hence 1 is a factor of 83.
Now, divide 83 by the next odd number in the list of natural numbers which is 3.
$\dfrac{83}{3} = 27.666$
When we divide 83 by 3; the quotient is 27 and the remainder is 2. Since the remainder is not 0, so 3 is not a factor of 83.
Lastly, divide 83 by 83.
$\dfrac{83}{83} = 1$
Therefore, 83 is the factor.
A number can have positive as well as negative factors. There are two positive factors of 83 and two negative factors of 83. Positive Factors of 83 are 1, and 83 while negative factors of 83 are -1, and -83.
The factors of 83 can also be found by multiplying two natural numbers to get 83:
83 x 1 = 83
So, the factor list of 83 is given below.
Factors List of 83: 1, -1, 83, and -83
### Important Properties
Following mentioned are some important properties of factors of 83:
1. 83 is an odd number so all of its factors are odd i.e. 1 and 83.
2. 83 is a prime number, so it has only two factors.
3. The prime factorization of the number 83 is given as 1 x 83 = 83.
4. There is only 1 positive factor pair of 83 and 1 negative factor pair of 83.
5. None of its factors is a decimal or in the form of fractions.
## Factors of 83 by Prime Factorization
The prime factorization method is used to find out the factors of 83. Let’s first understand what is prime factorization. Prime factorization is a method of representing the given number as the product of its prime factors. For example, the prime factorization of 4 is 2 * 2 = 4 where 2 is the prime factor of 4.
Similarly in the case of 83, expressing its prime factors in the form of the product is regarded as its prime factorization. As we have discussed earlier 83 has only two factors 1 and 83 therefore the prime factorization of 83 is shown below:
Figure 2 – Prime factorization of 83
So, the prime factorization of 83 is:
83 = 1 x 83
The more interesting facts about factors of 83 are that:
1. The sum of factors of 83 is an even number.
2. The product of factors of 83 is an odd number.
3. 83 can have only 2 factors which are 1 and 83 itself.
## Factor Tree of 83
The factor tree of 83 is shown below in figure 2:
Figure 3 – Factor tree of 83
As 83 is a prime number so only factors are 1 and 83 as illustrated in the factor tree.
## Factors of 83 in Pairs
Any pair of numbers whose product is 83 is called factor pair of 83 in pairs.
The factor pairs are given as:
83 = 1 x 83
83 = 83 x 1
83 = -1 x -83
83 = -83 x -1
Hence 83 has only one positive factor pair that is given as (1, 83) or (83, 1).
The negative factor pair of 83 is given as (-1, -83) or (-83, -1).
## Factors of 83 Solved Examples
Let’s solve some detailed examples to better understand the methods used for finding the factors of 83.
### Example 1
What is the Highest Common Factor (HCF) of 83 and 42?
### Solution
The factors of 83 are 1 and 83.
Factors of 42 are 1, 2, 3, 7, and 42.
The common factor of 83 and 42 is 1.
So, the Highest Common Factor (HCF) of 83 and 42 is 1.
### Example 2
List the negative factors of 83.
### Solution
The negative factors of 83 are -1 and -83.
It has only two factors as 83 is a prime number.
Factors are the integers that when multiplied together give the number as the product whose factors are to be found.
Similarly when -1 and -83 are multiplied the product is 83 as shown:
-1 x -83 = 83
So, -1 and -83 are negative factors of 83.
### Example 3
Hana’s tutor gave her an activity to find out the Least Common Multiple (LCM) of 83 and 24. How her elder brother will help her to find the LCM.
### Solution
Hana’s brother will first find out the factors of 83 and 24.
Prime Factors of 83 are 1,83.
Prime Factors of 24 are the following: 2,2,2,3.
Hence the LCM will be given as:
L.C.M = 2 x 2 x 2 x 3 x 83
L.C.M = 1992
So, the LCM of 83 and 24 is 1992.
Images/mathematical drawings are created with GeoGebra. |
# Time, Speed and Distance – All Formulas That You Need to Know
This triangle will help you remember the most basic formulae for TSD. Given below are a bunch of other concepts, formulae which come in handy while solving questions.
## Average and Relative Speed
If a body is moving with speeds s(1), s(2), s(3), … s(n) and the time is constant for each part, then the average speed will be the Arithmetic Mean of the speeds.
If a body is moving with speeds s(1), s(2), s(3), … s(n) and the distance is constant for each part, then the average speed will be the Harmonic Mean of the speeds.
Relative Speed of two bodies moving towards each other s(1)+s(2) and in the same direction s(1)-s(2).
## Boats, Streams and Escalators
If a constant distance is covered at speeds which are in an AP, then the times taken will be in an HP and vice-versa (For eg, Upstream, Boat-speed, Downstream speeds will always be in AP. Same can be applied in escalator questions as well).
Speed(Upstream) = S(Boat)-S(River)
Speed(Downstream) = S(Boat)+S(River)
Speed(Boat) = [S(Downstream)+S(Upstream)]/2
Speed(River) = [S(Downstream)-S(Upstream)]/2
## Motion of two bodies in a straight line
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2). After meeting each other, they take times of T(1) & T(2) to reach their destinations.
Time taken for them to meet sqrt[T(1)*T(2)]
S(1)/S(2) = sqrt[T(2)/T(1)]
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2). They reach the opposite ends and reverse directions. {S(1)>S(2) and S(1)
Total distance covered till nth meeting (2n-1)D and time taken (2n-1)D/S(1)+S(2).
## Circular Motion
Number of distinct points: In same direction, a-b and in opposite direction a+b (Here a/b is the reduced ratio of speeds).
I hope that with this set of formulae, you will be able to solve TSD questions with ease.
I will be conducting a workshop on TSD where I will be discussing this and other concepts.
## Enroll for Workshop
(Date: September 1 (Sunday). Time: 7:30 pm. Duration: 3 hours)
Ravi Handa, an alumnus of IIT Kharagpur, is the founder of HandaKaFunda. He teaches an Online CAT Coaching Course for CAT Preparation. |
# What is the probability of rolling the same number exactly 3 times with 5 six sided dice?
Contents
## What are the odds of rolling the same number 3 times on a d6?
The probability of getting the same number is 1/6.
## What is the probability of rolling a 5 on a 6 sided die 3 times?
The probability is 1216 chance, which is approximately a 0.46% chance.
## What is the probability that exactly 3 sixes will be recorded when 5 fair dice are tossed?
Explanation: There is a probability of 16 of rolling a 6 on a 6-sided die. The probability of rolling three 6s, therefore, is (16)3 . There are (53)=10 different ways to roll three 6s in 5 tries.
## What is the likelihood of rolling a single 6 sided die 3 times and obtaining a 6 each time?
So, there are 125 out of 216 chances of a 6 NOT appearing when three dice are rolled. Simply subtract 125 from 216 which will give us the chances a 6 WILL appear when three dice are rolled, which is 91. 91 out of 216 or 42.1 %.
## What is the probability of getting exactly 2 heads?
Consider all the possible ways to get two heads, HHT,HTHandTHH. There are 2⋅2⋅2=8 possible combinations in total. Therefore, the answer is 3/8.
## What is the probability that there are exactly two dice results that appear exactly 3 3 times each?
So the probability is 4587521679616≈27.3%.
## What is the probability of rolling a 6 on a 6 sided die?
Two (6-sided) dice roll probability table
Roll a… Probability
4 3/36 (8.333%)
5 4/36 (11.111%)
6 5/36 (13.889%)
7 6/36 (16.667%)
## What are the chances of rolling 2 sixes?
Thus only one of the 36 possible outcomes has a 6 on both dice and hence the probability of two sixes will be 1 over 36, or 1/36.
## When rolling two dice What is the probability of rolling doubles 1 6?
There are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of two fair dice. So you have a 16.7% probability of rolling doubles with 2 fair six-sided dice.
## What is the probability of rolling an even number?
The probability of rolling an even number is three out of six, or three-sixths. This fraction can be simplified as the numerator and denominator are both divisible by three. Three divided by three is equal to one, and six divided by three is equal to two. This means that the answer, in its simplest form, is one-half.
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## What is the probability of two dice rolling the same number?
The probability of two dice being the same particular number is 1/6 x 1/6 = 1/36. This is not the same as saying that both dice are the same number. There are six different possible numbers, so that would be 6/36 or 1/6.
## What is the probability of getting an even number?
Answer: The probability of getting an even number on the first side when a pair of dice is thrown once is 1/2.
## What is the probability of rolling 3 sixes?
This seemed simple at first as the odds of rolling 1 six are 1/2, 2 sixes would be 1/14 and 3 sixes would be 1/216 so they would be our odds.
## What is the probability of rolling a dice 3 times and getting a different number each time?
The probability that the number rolled on the third die is different from the first two dice is 4/6. Therefore, the probability that the three dice are all different on a single roll is 5/6 * 4/6 = 20/36 = 5/9.
## How do you find the probability of a dice roll?
Probability = Number of desired outcomes ÷ Number of possible outcomes = 3 ÷ 36 = 0.0833. The percentage comes out to be 8.33 per cent. Also, 7 is the most likely result for two dice. Moreover, there are six ways to achieve it. |
One of the first things in math that we learn is counting. We count fingers, toys, apples, and oranges. When we count, we’re actually adding $$1$$ to the number we have already. Then we learn how to add more than just one at a time. This is what we call adding whole numbers: $$2+3=5$$.
But what if one of the things we’re counting is negative? Well, that’s when we don’t add whole numbers anymore — that’s when integers come into the picture!
## What does it mean to add integers?
Adding integers means performing addition over a set of integer numbers. An integer is any number from a set of whole numbers and their additive inverses – the numbers with the same absolute value but an opposite sign:
$$\{\ldots-3,-2,-1,0,1,2,3,\ldots\}$$
‘’Wait… what is the absolute value of a number?!’’
Good question! The absolute value of a number is that same number, but without the sign in front of it. Why? Because the absolute value actually shows us what the distance is from that number to $$0$$ on the number line. For example, the absolute value of $$|-4|=4$$. That’s the distance from $$-4$$ to $$0$$ on the number line, $$4$$ units:
The first summand (a number we want to add) tells us where to start on the number line. The sign on the second summand will tell us which way to move on the number line. If it’s positive, we count moving right. If it’s negative, we count moving left. The absolute value of the second summand is just the number without its sign.
### Why is adding integers so useful?
Besides the fact that adding integers is a skill we need in order to learn more about math, there are many real-life problems it can help us solve! For example, if a fish is swimming $$14$$ feet below sea level, and then tries to reach the surface and moves $$6$$ feet up, at what depth is it located now? Well, that problem can be solved by adding integers. The fact that the fish is $$14$$ feet below the sea level can be represented by the integer $$-14$$ and the fact that it moved up means we need to add $$6$$. Now, it all comes down to a simple math problem of adding integers:
$$-14+6$$
Now that we understand what integers are and why they’re useful, it’s time to see them in action! Let’s walk through a problem together:
### Example 1
$$-5+3$$
Determine the absolute value of each number:
$$|-5|=5$$
$$|3|=3$$
Notice that the negative number has a greater absolute value, so keep the sign of $$-5$$ and subtract the smaller absolute value from the larger:
$$-(5-3)$$
Subtract the numbers inside the parentheses:
$$-2$$
### Example 2
This might seem like the same example, but it actually isn’t!
$$-5+(-3)$$
Notice that here we need to add two negative numbers! Remember, to factor out a term from an expression means to extract that term from all the terms in the expression. So, factor out the negative sign:
$$-(5+3)$$
Add the numbers inside the parentheses:
$$-8$$
That wasn’t so bad, right? Now that we’ve walked through detailed examples, let’s review the overall process so you can learn how to use it with any problem:
## Study summary
1. Keep the sign of the number with the larger absolute value and subtract the smaller absolute value from the larger.
2. Subtract the numbers.
## Try it yourself!
Practicing math concepts like this one is a great way to prepare yourself for the math journey to come. So, when you’re ready, we’ve got some practice problems for you!
1. $$-15+8$$
2. $$-5+7$$
3. $$-9+1$$
4. $$-6+9$$
Solutions:
1. $$-7$$
2. $$2$$
3. $$-8$$
4. $$3$$
If you’re still struggling with the solving process, that’s totally okay! Stumbling a few times is good for learning. If you do get stuck or lost, scan the problem using your Photomath app and we’ll walk you through it.
Here’s a sneak peek of what you’ll see: |
# How to Find Equation of Ellipse with Foci and Major Axis
When we consider the conic section, an ellipse is an important topic. It is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. These fixed points are known as foci of the ellipse. The major axis is the line segment passing through the foci of the ellipse. In this article, we will learn how to find the equation of ellipse with foci and major axis.
The distance between the foci is denoted by 2c. The length of the major axis is denoted by 2a and the minor axis is denoted by 2b.
## Steps to Find the Equation of the Ellipse with Foci and Major Axis
1. Find whether the major axis is on the x-axis or y-axis.
2. If major axis is on x-axis then use the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$.
3. If major axis is on y-axis then use the equation $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1$.
4. Find ‘a’ from the length of the major axis. Length of major axis = 2a
5. Using the equation c2 = (a2 – b2), find b2.
6. Substitute the values of a2 and b2 in the standard form.
The standard form of the equation of an ellipse with center (h,k) and major axis parallel to x axis is
((x-h)2 /a2)+((y-k)2/b2) = 1
When a>b
Major axis length = 2a
Coordinates of the vertices are (h±a,k)
Minor axis length is 2b
Coordinates of covertices are (h,k±b)
Coordinates of foci are (h±c,k). Also c2= a2-b2
## Solved Examples
Example 1:
Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0, ± 5).
Solution:
Given the major axis is 20 and foci are (0, ± 5).
Here the foci are on the y-axis, so the major axis is along the y-axis.
So the equation of the ellipse is
x2/b2 + y2/a2 = 1
2a = 20
a = 20/2 = 10
a2 = 100
c = 5
c2 = a2 – b2
b2 = a2 – c2 = 102 – 52 = 75
So (x2/75) + y2/100 = 1 is the required equation.
Example 2:
Find the equation of the ellipse whose length of the major axis is 26 and foci (± 5, 0)
Solution:
Given the major axis is 26 and foci are (± 5,0).
Here the foci are on the x-axis, so the major axis is along the x-axis.
So the equation of the ellipse is x2/a2 + y2/b2 = 1
2a = 26
a = 26/2 = 13
a2 = 169
c = 5
c2 = a2 – b2
b2 = a2 – c2 = 132 – 52 = 144
So (x2/169) + y2/144 = 1 is the required equation. |
# 2018 AIME II Problems/Problem 10
## Problem
Find the number of functions $f(x)$ from $\{1, 2, 3, 4, 5\}$ to $\{1, 2, 3, 4, 5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1, 2, 3, 4, 5\}$.
## Solution 1
Just to visualize solution 1. If we list all possible $(x,f(x))$, from ${1,2,3,4,5}$ to ${1,2,3,4,5}$ in a specific order, we get $5*5 = 25$ different $(x,f(x))$ 's. Namely:
$(1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5)$
To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of $(x,x)$ where $x\in{1,2,3,4,5}$ must exist.In this case I rather "go backwards". First fixing $5$ pairs $(x,x)$, (the diagonal of our table) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\frac{5!}{5!} = 1$ way. Then fixing $4$ pairs $(x,x)$ (The diagonal minus $1$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $4\cdot\frac{5!}{4!} = 20$ ways. Then fixing $3$ pairs $(x,x)$ (The diagonal minus $2$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150$ ways. Fixing $2$ pairs $(x,x)$ (the diagonal minus $3$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380$ ways. Lastely, fixing $1$ pair $(x,x)$ (the diagonal minus $4$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205$
So $1 + 20 + 150 + 380 + 205 = \framebox{756}$
## Solution 2
We can do some caseworks about the special points of functions $f$ for $x\in\{1,2,3,4,5\}$. Let $x$, $y$ and $z$ be three different elements in set $\{1,2,3,4,5\}$. There must be elements such like $k$ in set $\{1,2,3,4,5\}$ satisfies $f(k)=k$, and we call the points such like $(k,k)$ on functions $f$ are "Good Points" (Actually its academic name is "fixed-points"). The only thing we need to consider is the "steps" to get "Good Points". Notice that the "steps" must less than $3$ because the highest iterations of function $f$ is $3$. Now we can classify $3$ cases of “Good points” of $f$.
$\textbf{Case 1:}$ One "step" to "Good Points": Assume that $f(x)=x$, then we get $f(f(x))=f(x)=x$, and $f(f(f(x)))=f(f(x))=f(x)=x$, so $f(f(f(x)))=f(f(x))$.
$\textbf{Case 2:}$ Two "steps" to "Good Points": Assume that $f(x)=y$ and $f(y)=y$, then we get $f(f(x))=f(y)=y$, and $f(f(f(x)))=f(f(y))=f(y)=y$, so $f(f(f(x)))=f(f(x))$.
$\textbf{Case 3:}$ Three "steps" to "Good Points": Assume that $f(x)=y$, $f(y)=z$ and $f(z)=z$, then we get $f(f(x))=f(y)=z$, and $f(f(f(x)))=f(f(y))=f(z)=z$, so $f(f(f(x)))=f(f(x))$.
Divide set $\{1,2,3,4,5\}$ into three parts which satisfy these three cases, respectively. Let the first part has $a$ elements, the second part has $b$ elements and the third part has $c$ elements, it is easy to see that $a+b+c=5$. First, there are $\binom{5}{a}$ ways to select $x$ for Case 1. Second, we have $\binom{5-a}{b}$ ways to select $x$ for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is $a^b$. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is $b^c$. As a result, the number of such functions $f$ can be represented in an algebraic expression contains $a$, $b$ and $c$: $\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}$
Now it's time to consider about the different values of $a$, $b$ and $c$ and the total number of functions $f$ satisfy these values of $a$, $b$ and $c$:
For $a=5$, $b=0$ and $c=0$, the number of $f$ is $\binom{5}{5}=1$
For $a=4$, $b=1$ and $c=0$, the number of $f$ is $\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20$
For $a=3$, $b=1$ and $c=1$, the number of $f$ is $\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60$
For $a=3$, $b=2$ and $c=0$, the number of $f$ is $\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90$
For $a=2$, $b=1$ and $c=2$, the number of $f$ is $\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60$
For $a=2$, $b=2$ and $c=1$, the number of $f$ is $\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240$
For $a=2$, $b=3$ and $c=0$, the number of $f$ is $\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80$
For $a=1$, $b=1$ and $c=3$, the number of $f$ is $\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20$
For $a=1$, $b=2$ and $c=2$, the number of $f$ is $\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120$
For $a=1$, $b=3$ and $c=1$, the number of $f$ is $\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60$
For $a=1$, $b=4$ and $c=0$, the number of $f$ is $\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5$
Finally, we get the total number of function $f$, the number is $1+20+60+90+60+240+80+20+120+60+5=\boxed{756}$
~Solution by $BladeRunnerAUG$ (Frank FYC)
## Solution 3
Note that there are $5^5$ possible functions $f(x)$.
Now consider the probability of picking a function from those $3125$ functions that satisfies, exclusively, one of the following criteria:
$P(x = f(x) = f(f(x)) = f(f(f(x)))) = \frac{1}{5^5}$
$P(x \neq{f(x)} = f(f(x)) = f(f(f(x)))) = \frac{1}{5^4}$
$P(x = f(x) \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5^2}$
$P(x \neq{f(x)} \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5}$
Thus the number of functions $f(x)$ satisfying the condition stated in the problem is given by:
$\frac{5^5}{5^5} + \frac{5^5}{5^4} + \frac{5^5}{5^2} + \frac{5^5}{5} = 1 + 5 + 125 + 625 = \boxed{756}$
~ anellipticcurveoverq
## Note (fun fact)
This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test. This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems
2018 AIME II (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
## Gaussian Elimination
The purpose of this article is to describe how the solutions to a linear system are actually found. The fundamental idea is to add multiples of one equation to the others in order to eliminate a variable and to continue this process until only one variable is left. Once this final variable is determined, its value is substituted back into the other equations in order to evaluate the remaining unknowns. This method, characterized by step‐by‐step elimination of the variables, is called Gaussian elimination.
Example 1: Solve this system:
Multiplying the first equation by −3 and adding the result to the second equation eliminates the variable x:
This final equation, −5 y = −5, immediately implies y = 1. Back‐substitution of y = 1 into the original first equation, x + y = 3, yields x = 2. (Back‐substitution of y = 1 into the original second equation, 3 x − 2 y = 4, would also yeild x = 2.) The solution of this system is therefore (x, y) = (2, 1), as noted in Example 1.
Gaussian elimination is usually carried out using matrices. This method reduces the effort in finding the solutions by eliminating the need to explicitly write the variables at each step. The previous example will be redone using matrices.
Example 2: Solve this system:
The first step is to write the coefficients of the unknowns in a matrix:
This is called the coefficient matrix of the system. Next, the coefficient matrix is augmented by writing the constants that appear on the right‐hand sides of the equations as an additional column:
This is called the augmented matrix, and each row corresponds to an equation in the given system. The first row, r 1 = (1, 1, 3), corresponds to the first equation, 1 x + 1 y = 3, and the second row, r 2 = (3, −2, 4), corresponds to the second equation, 3 x − 2 y = 4. You may choose to include a vertical line—as shown above—to separate the coefficients of the unknowns from the extra column representing the constants.
Now, the counterpart of eliminating a variable from an equation in the system is changing one of the entries in the coefficient matrix to zero. Likewise, the counterpart of adding a multiple of one equation to another is adding a multiple of one row to another row. Adding −3 times the first row of the augmented matrix to the second row yields
The new second row translates into −5 y = −5, which means y = 1. Back‐substitution into the first row (that is, into the equation that represents the first row) yields x = 2 and, therefore, the solution to the system: (x, y) = (2, 1).
Gaussian elimination can be summarized as follows. Given a linear system expressed in matrix form, A x = b, first write down the corresponding augmented matrix:
Then, perform a sequence of elementary row operations, which are any of the following:
Type 1. Interchange any two rows.
Type 2. Multiply a row by a nonzero constant.
Type 3. Add a multiple of one row to another row.
The goal of these operations is to transform—or reduce—the original augmented matrix into one of the form where A′ is upper triangular (aij ′ = 0 for i > j), any zero rows appear at the bottom of the matrix, and the first nonzero entry in any row is to the right of the first nonzero entry in any higher row; such a matrix is said to be in echelon form. The solutions of the system represented by the simpler augmented matrix, [ A′ | b′], can be found by inspectoin of the bottom rows and back‐substitution into the higher rows. Since elementary row operations do not change the solutions of the system, the vectors x which satisfy the simpler system Ax = b′ are precisely those that satisfy the original system, A x = b.
Example 3: Solve the following system using Gaussian elimination:
The augmented matrix which represents this system is
The first goal is to produce zeros below the first entry in the first column, which translates into eliminating the first variable, x, from the second and third equations. The row operations which accomplish this are as follows:
The second goal is to produce a zero below the second entry in the second column, which translates into eliminating the second variable, y, from the third equation. One way to accomplish this would be to add −1/5 times the second row to the third row. However, to avoid fractions, there is another option: first interchange rows two and three. Interchanging two rows merely interchanges the equations, which clearly will not alter the solution of the system:
Now, add −5 times the second row to the third row:
Since the coefficient matrix has been transformed into echelon form, the “forward” part of Gaussian elimination is complete. What remains now is to use the third row to evaluate the third unknown, then to back‐substitute into the second row to evaluate the second unknown, and, finally, to back‐substitute into the first row to evaluate the first unknwon.
The third row of the final matrix translates into 10 z = 10, which gives z = 1. Back‐substitution of this value into the second row, which represents the equation y − 3 z = −1, yields y = 2. Back‐substitution of both these values into the first row, which represents the equation x − 2 y + z = 0, gives x = 3. The solution of this system is therefore (x, y, z) = (3, 2, 1).
Example 4: Solve the following system using Gaussian elimination:
For this system, the augmented matrix (vertical line omitted) is
First, multiply row 1 by 1/2:
Now, adding −1 times the first row to the second row yields zeros below the first entry in the first column:
Interchanging the second and third rows then gives the desired upper‐triangular coefficient matrix:
The third row now says z = 4. Back‐substituting this value into the second row gives y = 1, and back‐substitution of both these values into the first row yields x = −2. The solution of this system is therefore (x, y, z) = (−2, 1, 4).
Gauss‐Jordan elimination. Gaussian elimination proceeds by performing elementary row operations to produce zeros below the diagonal of the coefficient matrix to reduce it to echelon form. (Recall that a matrix A′ = [aij ′] is in echelon form when aij ′= 0 for i > j, any zero rows appear at the bottom of the matrix, and the first nonzero entry in any row is to the right of the first nonzero entry in any higher row.) Once this is done, inspection of the bottom row(s) and back‐substitution into the upper rows determine the values of the unknowns.
However, it is possible to reduce (or eliminate entirely) the computations involved in back‐substitution by performing additional row operations to transform the matrix from echelon form to reduced echelon form. A matrix is in reduced echelon form when, in addition to being in echelon form, each column that contians a nonzero entry (usually made to be 1) has zeros not just below that entry but also above that entry. Loosely speaking, Gaussian elimination works from the top down, to produce a matrix in echelon form, whereas Gauss‐Jordan elimination continues where Gaussian left off by then working from the bottom up to produce a matrix in reduced echelon form. The technique will be illustrated in the following example.
Example 5: The height, y, of an object thrown into the air is known to be given by a quadratic function of t (time) of the form y = at2 + bt + c. If the object is at height y = 23/4 at time t = 1/2, at y = 7 at time t = 1, and at y = 2 at t = 2, determine the coefficients a, b, and c.
Since t = 1/2 gives y = 23/4
while the other two conditions, y(t = 1) = 7 and y(t = 2) = 2, give the following equations for a, b, and c:
Therefore, the goal is solve the system
The augmented matrix for this system is reduced as follows:
At this point, the forward part of Gaussian elimination is finished, since the coefficient matrix has been reduced to echelon form. However, to illustrate Gauss‐Jordan elimination, the following additional elementary row operations are performed:
This final matrix immediately gives the solution: a = −5, b = 10, and c = 2.
Example 6: Solve the following system using Gaussian elimination:
The augmented matrix for this system is
Multiples of the first row are added to the other rows to produce zeros below the first entry in the first column:
Next, −1 times the second row is added to the third row:
The third row now says 0 x + 0 y + 0 z = 1, an equation that cannot be satisfied by any values of x, y, and z. The process stops: this system has no solutions.
The previous example shows how Gaussian elimination reveals an inconsistent system. A slight alteration of that system (for example, changing the constant term “7” in the third equation to a “6”) will illustrate a system with infinitely many solutions.
Example 7: Solve the following system using Gaussian elimination:
The same operations applied to the augment matrix of the system in Example 6 are applied to the augmented matrix for the present system:
Here, the third row translates into 0 x + 0 y + 0 z = 0, an equation which is satisfied by any x, y, and z. Since this offer no constraint on the unknowns, there are not three conditions on the unknowns, only two (represented by the two nonzero rows in the final augmented matrix). Since there are 3 unknowns but only 2 constrants, 3 − 2 =1 of the unknowns, z say, is arbitrary; this is called a free variable. Let z = t, where t is any real number. Back‐substitution of z = t into the second row (− y + 5 z = −6) gives
Back substituting z = t and y = 6 + 5 t into the first row ( x + y − 3 z = 4) determines x
Therefore, every solution of the system has the form
where t is any real number. There are infinitely many solutions, since every real value of t gives a different particular solution. For example, choosing t = 1 gives ( x, y, z) = (−4, 11, 1), while t = 3 gives ( x, y, z) = (4, −9, −3), and so on. Geometrically, this system represents three planes in R 3 that intersect in a line, and (*) is a parametric equation for this line.
Example 7 provided an illustration of a system with infinitely many solutions, how this case arises, and how the solution is written. Every linear system that possesses infinitely many solutions must contain at least one arbitrary parameter (free variable). Once the augmented matrix has been reduced to echelon form, the number of free variables is equal to the total number of unknowns minus the number of nonzero rows:
This agrees with Theorem B above, which states that a linear system with fewer equations than unknowns, if consistent, has infinitely many solutions. The condition “fewer equations than unknowns” means that the number of rows in the coefficient matrix is less than the number of unknowns. Therefore, the boxed equation above implies that there must be at least one free variable. Since such a variable can, by definition, take on infinitely many values, the system will have infinitely many solutions.
Example 8: Find all solutions to the system
First, note that there are four unknwons, but only thre equations. Therefore, if the system is consistent, it is guaranteed to have infinitely many solutions, a condition characterized by at least one parameter in the general solution. After the corresponding augmented matrix is constructed, Gaussian elimination yields
The fact that only two nonzero rows remain in the echelon form of the augmented matrix means that 4 − 2 = 2 of the variables are free:
Therefore, selecting y and z as the free variables, let y = t 1 and z = t 2. The second row of the reduced augmented matrix implies
and the first row then gives
Thus, the solutions of the system have the form
where t 1 t 2 are allowed to take on any real values.
Example 9: Let b = ( b 1, b 2, b 3) T and let A be the matrix
For what values of b 1, b 2, and b 3 will the system A x = b be consistent?
The augmented matrix for the system A x = b reads
which Gaussian eliminatin reduces as follows:
The bottom row now implies that b 1 + 3 b 2 + b 3 must be zero if this system is to be consistent. Therefore, the given system has solutins (infinitely many, in fact) only for those column vectors b = ( b 1, b 2, b 3) T for which b 1 + 3 b 2 + b 3 = 0.
Example 10: Solve the following system (compare to Example 12):
A system such as this one, where the constant term on the right‐hand side of every equation is 0, is called a homogeneous system. In matrix form it reads A x = 0. Since every homogeneous system is consistent—because x = 0 is always a solution—a homogeneous system has eithe exactly one solution (the trivial solution, x = 0) or infiitely many. The row‐reduction of the coefficient matrix for this system has already been performed in Example 12. It is not necessary to explicitly augment the coefficient matrix with the column b = 0, since no elementary row operation can affect these zeros. That is, if A′ is an echelon form of A, then elementary row operations will transform [ A| 0] into [ A′| 0]. From the results of Example 12,
Since the last row again implies that z can be taken as a free variable, let z = t, where t is any real number. Back‐substitution of z = t into the second row (− y + 5 z = 0) gives
and back‐substitution of z = t and y = 5 t into the first row ( x + y − 3 z = 0) determines x
Therefore, every solution of this system has the form ( x, y, z) = (−2 t, 5 t, t), where t is any real number. There are infinitely many solutins, since every real value of t gives a unique particular solution.
Note carefully the differnece between the set of solutions to the system in Example 12 and the one here. Although both had the same coefficient matrix A, the system in Example 12 was nonhomogeneous ( A x = b, where b0), while the one here is the corresponding homogeneous system, A x = 0. Placing their solutions side by side,
general solution to Ax = 0: ( x, y, z) = (−2 t, 5 t, t)
general solution to Ax = b: ( x, y, z) = (−2 t, 5 t, t) + (−2, 6, 0)
illustrates an important fact:
Theorem C. The general solutions to a consistent nonhomogeneous lienar system, A x = b, is equal to the general solution of the corresponding homogeneous system, A x = 0, plus a particular solution of the nonhomogeneous system. That is, if x = x h represents the general solution of A x = 0, then x = x h + x represents the general solution of A x + b, where x is any particular soltion of the (consistent) nonhomogeneous system A x = b.
[Technical note: Theorem C, which concerns a linear system, has a counterpart in the theory of linear diffrential equations. Let L be a linear differential operator; then the general solution of a solvable nonhomogeneous linear differential equation, L(y) = d (where d ≢ 0), is equal to the general solution of the corresponding homogeneous equation, L(y) = 0, plus a particular solution of the nonhomogeneous equation. That is, if y = y h repreents the general solution of L(y) = 0, then y = y h + y represents the general solution of L(y) = d, where y is any particular solution of the (solvable) nonhomogeneous linear equation L(y) = d.]
Example 11: Determine all solutions of the system
Write down the augmented matrix and perform the following sequence of operations:
Since only 2 nonzero rows remain in this final (echelon) matrix, there are only 2 constraints, and, consequently, 4 − 2 = 2 of the unknowns— y and z say—are free variables. Let y = t 1 and z = t 2. Back‐substitution of y = t 1 and z = t 2 into the second row ( x − 3 y + 4 z = 1) gives
Finally, back‐substituting x = 1 + 3 t 1 − 4 2, y = t 1,and z = t 2 into the first row (2 w − 2 x + y = −1) determines w:
Therefore, every solution of this system has the form
where t 1 and t 2 are any real numbers. Another way to write the solution is as follows:
where t1, t2R.
Example 12: Determine the general solution of
which is the homogeneous system corresponding to the nonhomoeneous one in Example 11 above.
Since the solution to the nonhomogeneous system in Example 11 is
Theorem C implies that the solution of the corresponding homogeneous system is (where t1, t2R), which is obtained from (*) by simply discarding the particular soluttion, x = (1/2,1,0,0), of the nonhomogeneous system.
Example 13: Prove Theorem A: Regardless of its size or the number of unknowns its equations contain, a linear system will have either no solutions, exactly one solution, or infinitely many solutions.
Proof. Let the given linear system be written in matrix form A x = b. The theorem really comes down to tthis: if A x = b has more than one solution, then it actually has infinitely many. To establish this, let x1 and x2 be two distinct solutions of A x = b. It will now be shown that for any real value of t, the vector x1 + t(x1x 2) is also a solution of A x = b; because t can take on infinitely many different values, the desired conclusion will follow. Since A x1 = b and A x2,
Therefore, x1 + t(x1x2) is indeed a solution of A x = b, and the theorem is proved.
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Question Video: Solving Word Problems Involving Arithmetic Sequences | Nagwa Question Video: Solving Word Problems Involving Arithmetic Sequences | Nagwa
Question Video: Solving Word Problems Involving Arithmetic Sequences Mathematics • Second Year of Secondary School
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A fast food restaurant offers to give away two £100-prizes on one day. They will give away four £100-prizes the next day, six £100-prizes the next day, and so on, giving away two more £100-prizes each day than the previous day. If 𝑛 represents the number of days in their campaign, find a formula to calculate how many pounds they will have given away in total by the end of the campaign.
03:44
Video Transcript
A fast food restaurant offers to give away two 100-pound prizes on one day. They will give away four 100-pound prizes the next day, six 100-pound prizes the next day, and so on, giving away two more 100-pound prizes each day than the previous day. If 𝑛 represents the number of days in their campaign, find a formula to calculate how many pounds they will have given away in total by the end of the campaign.
Let’s look carefully at how much money the restaurant are giving away each day. On the first day, they’re giving away two 100-pound prizes, so that’s 200 pounds. The next day, they’re giving away four 100-pound prizes, so that’s 400 pounds. On the third day, they’re giving away six 100-pound prizes, so that’s 600 pounds. And then, this continues. Every day they give away two more 100-pound prizes, so 200 pound more than they gave away the day before.
We are asked to find a formula to calculate the total amount of money they give away if they run this campaign for 𝑛 days. Now, these terms, the amounts they give away every day — 200, 400, 600, 800, and so on — form an arithmetic sequence because they have a common difference of 200. We should recall that there is a formula for calculating the sum of the first 𝑛 terms of an arithmetic sequence. It’s 𝑠 sub 𝑛 is equal to 𝑛 over two multiplied by two 𝑎 plus 𝑛 minus one 𝑑, where 𝑠 sub 𝑛 represents the sum of the first 𝑛 terms, 𝑎, or sometimes 𝑎 one, represents the first term, and 𝑑 represents the common difference.
We can now use this formula to help us find a formula for the total amount the restaurant will give away. We don’t know how many days they’re running the campaign for, so our formula will be in terms of the unknown 𝑛. The first term in the sequence is the amount the restaurant gives away on day one. That’s 200 pounds, or we’ll just use 200 in our formula. The common difference 𝑑 is the amount that the terms in the sequence increase by each time; that’s also 200 pounds. Substituting 200 for both 𝑎 and 𝑑 gives 𝑠 sub 𝑛 is equal to 𝑛 over two multiplied by two times 200 plus 200 multiplied by 𝑛 minus one. This simplifies to 𝑛 over two multiplied by 400 plus 200𝑛 minus 200. And then simplifying further within the parentheses, we have 𝑛 over two multiplied by 200 plus 200𝑛.
We can then cancel a factor of two throughout, which will give 𝑛 multiplied by 100 plus 100𝑛. And distributing the parentheses gives 100𝑛 plus 100𝑛 squared. This would be a perfectly acceptable form in which to leave our answer, but we’re going to choose to factor by 100. Doing so gives 100 multiplied by 𝑛 squared plus 𝑛. The restaurant can then use this formula to work out how much money they will give away in total depending on how many days they run the campaign for. For example, if they run the campaign for 20 days, they would substitute 𝑛 equals 20 to work out the total amount of money they give away. Our answer is 100 multiplied by 𝑛 squared plus 𝑛, where 𝑛 represents the number of days in the campaign.
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# Inverse Matrix Calculator with steps
## Inverse Matrix Calculator
Matrix A:
Knowing how to calculate the inverse of a matrix is very useful to easily solve systems of linear equations using the matrix inversion method. For this reason, we put in your hands this Inverse Matrix Calculator with which you can practice, study and fully understand how to obtain the inverse of a matrix, thanks to the fact that the solutions are explained step by step using different methods.
To calculate the inverse of a matrix, you only have to do three simple steps:
1. The first step is to introduce the matrix to which you are going to calculate its inverse. You can modify the number of rows or columns according to the dimension of the matrix that you need to enter using the + buttons.
2. In the second step you will have to choose if you want to obtain the solution expressed in decimal numbers or not.
3. And finally, you just have to press the button «Calculate». When you have pressed said button, a box will automatically be displayed with the solution explained step by step through the use of different methods.
## What is the inverse of a matrix?
The inverse of a matrix is another matrix of equal dimensions and that if multiplied by the original matrix results in the identity matrix. In other words, the inverse of the matrix [A], designated as [A]–1, is defined by the following property:
[A]·[A]–1=[A]–1·[A]=[I]
where [I] is the identity matrix.
You should keep in mind that only square matrices can have an inverse matrix, in other words, a square matrix can be an invertible matrix. This is because the definition of an inverse matrix is based on the concept of identity matrix [I], and only square matrices have an identity matrix associated with it.
A matrix whose determinant is equal to 0 is a non-invertible matrix, which is why this type of matrix is called a singular matrix.
## Properties of matrix inverses
• A non-singular square matrix has only a single inverse matrix.
• The square matrix A is an invertible matrix, only if its determinant is a nonzero value, |A| ≠ 0.
• If A and B are non-singular matrices, then the product AB results in a non-singular matrix and (A·B)-1 = B-1·A-1
• If A is not singular, then the inverse matrix of the transpose of A is equal to the transpose of the inverse of A, (AT)-1 = (A-1)T.
• If the product of matrices A and B is equal to the identity matrix, then the matrices are inverses of each other.
## How to find the inverse matrix
There are several methods to find the inverse matrix, but the general method to calculate the inverse of a matrix consists of using the following formula:
where |A| is the determinant of A and Adj(A) is the adjoint matrix of A. We will illustrate how to use the previous formula by calculating the inverse of a 2×2 matrix.
${\underset{\text{}}{\left(\begin{array}{cc}5& 8\\ 2& -3\end{array}\right)}}^{\left(-1\right)}$
Find 2x2 matrix inverse according to the formula:
${A}^{-1}={\underset{\text{}}{\overline{)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)}}}^{\left(-1\right)}=\frac{1.00}{\left|A\right|}×\underset{\text{}}{\overline{)\left(\begin{array}{cc}{C}_{11}& {C}_{21}\\ {C}_{12}& {C}_{22}\end{array}\right)}}=\frac{1.00}{a×d-b×c}×\underset{\text{}}{\overline{)\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)}}$
${\underset{\text{}}{\overline{)\left(\begin{array}{cc}5.00& 8.00\\ 2.00& -3.00\end{array}\right)}}}^{\left(-1\right)}=\frac{1.00}{5.00×\left(-3.00\right)-8.00×2.00}×\underset{\text{}}{\overline{)\left(\begin{array}{cc}-3.00& -8.00\\ -2.00& 5.00\end{array}\right)}}=\underset{\text{}}{\overline{)\left(\begin{array}{cc}0.10& 0.26\\ 0.06& -0.16\end{array}\right)}}$
## Applications
The inverse of a matrix is a mathematical concept with numerous applications in a variety of fields. In this article, we’ll explore some of the ways that the calculation of matrix inverse is used in science, engineering, and other areas.
One of the most common applications of matrix inverse is in the field of linear algebra. In this context, the inverse of a matrix is used to solve systems of linear equations. Given a system of equations in the form Ax = b, where A is a matrix, x is a vector of unknowns, and b is a vector of constants, the inverse of A can be used to find the values of x that satisfy the system. This is done by multiplying both sides of the equation by A-1, the inverse of A:
A-1 Ax = A-1 b
Since A-1 A = I, the identity matrix, we can simplify the equation to:
x = A-1 b
Thus, the inverse of A allows us to find the solution to the system of equations.
Matrix inverse is also important in the field of engineering, particularly in the analysis of structural systems. In this context, the inverse of a matrix is used to find the internal forces and moments acting on a structure, given the applied loads. This is done by solving a system of equations that relates the loads and the internal forces. The inverse of the stiffness matrix, which relates the internal forces to the deformations of the structure, is used to find the internal forces from the deformations.
In addition to its applications in linear algebra and engineering, matrix inverse is also used in other areas such as statistics, where it is used to find the variance-covariance matrix of a set of random variables, and computer graphics, where it is used to transform objects in a 3D scene.
Overall, the calculation of matrix inverse is a crucial tool in a variety of fields, and its applications are numerous and varied. Understanding how to calculate and use matrix inverse can be essential for solving problems and making decisions in many different fields. |
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ANALYSIS OF A CONIC
The general equation of the second degree
1) f(x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
represents one of 9 different conics:
Any equation of the second degree can be reduced to one of the nine canonical forms by a suitable rotation and translation of the coordinate system. Thus when faced with a particular second degree equation the following questions immediately present themselves:
1] Which of the 9 conics does this equation represent?
2] What is the exact equation of the conic when in canonical form? That is, what is its equation when expressed with respect to the canonical coordinate system?
3] What is the location of the origin of the canonical system?
4] What is the orientation of the canonical coordinate system? In other words, in what directions do the canonical system axes point?
We will now deal with the procedure used in answering these questions. Our starting point is our given equation
2) f(x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
which represents some conic located somewhere in the plane. In Figure 1 a conic (an ellipse) is shown located at some point in the plane. Figure 1 also shows the original x-y coordinate system along with two other coordinate systems – an intermediate x'- y' coordinate system and the xc - yc canonical coordinate system. We wish to know the location and orientation of the canonical coordinate system and the exact equation of the conic as referred to that system. Figure 1 shows the x'- y' coordinate system as a system obtained by rotating the x-y system by θ degrees about its origin. θ represents the rotation required to eliminate the xy term in the original equation as computed from
A rotation of this amount will put its axes parallel to the axes of the canonical system.
We proceed as follows:
1. We obtain the equation of the conic as referred to the intermediate x'- y' coordinate system by substituting into equation 2) the expressions for x and y in terms of x' and y' as given by the rotational transformation equations
x = x' cos θ - y' sin θ
y = x' sin θ + y' cos θ
where the θ in these equations is that particular value of θ computed from 3) above.
On expanding and simplifying we then have the equation
4) g(x', y') = 0
as the equation of the conic as referred to the intermediate x'- y' coordinate system.
2. We determine translation required to carry the x'- y' system into the canonical xc - yc system. In other words, we determine the values of h' and k' where (h', k') is the origin of the canonical xc - yc system with respect to the x'- y' system. In the case of the central conics (ellipses and hyperbolas) this translation corresponds to that translation that will eliminate both the x and y terms from the equation.
3. We obtain the equation of the conic as referred to the canonical xc - yc coordinate system by substituting into equation 4) the expressions for x' and y' in terms of xc and yc as given by the translation transformation equations
x' = xc + h'
y' = yc + k' .
This process gives us the answers to our original questions. The mechanics of substituting into equations, expanding, and simplifying can be laborious and fortunately, in the case of the central conics, it can be bypassed. By utilizing certain invariant quantities and employing some abstract results from matrix theory we can answer the questions without going to all that labor.
Analysis procedure.
First we define the following quantities related to the equation
f(x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 :
λ1, λ2 – the characteristic roots of matrix e i.e. the roots of the characteristic equation
which can be written as
4) λ2 - Iλ + J = 0
Conic identification. We identify the conic by use of the following table:
Case ∇ J ∇/ J K Conic 1 ≠0 > 0 < 0 Ellipse 2 ≠0 > 0 > 0 Imaginary ellipse 3 0 > 0 Pair of imaginary lines intersecting in a real point 4 ≠0 < 0 Hyperbola 5 0 < 0 A pair of intersecting lines 6 ≠0 0 Parabola 7 0 0 < 0 A pair of parallel lines 8 0 0 > 0 A pair of imaginary parallel lines 9 0 0 0 A pair of coincident straight lines
Analysis of central conics (ellipses and hyperbolas).
Orientation of the canonical coordinate system. The orientation of the canonical system is found by computing the rotation angle θ required to eliminate the xy term. The formula is
Location of the origin of the canonical system. The location of the origin (x0, y0) of the canonical system with respect to the x-y system is given by solving the following system of equations for x0, y0 :
ax0 + hy0 + g = 0
hx0 + by0 + f = 0
Equation of the conic in the canonical system. The equation of the conic with respect to the canonical system is
λ1x2 + λ2y2 + c' = 0
where
c' = gx0 + fy0 + c
and λ1, λ2 are obtained by computing the roots of characteristic equation
λ2 - Iλ + J = 0 ,
or more explicitly,
λ2 - (a + b)λ + ab - h2 = 0
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# How do you find all the zeros of f(x)=x^3-3x^2+x-3?
Nov 2, 2016
$f \left(x\right)$ has zeros $x = 3$ and $x = \pm i$
#### Explanation:
The difference of squares identity can be written:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
We use this later with $a = x$ and $b = i$
$f \left(x\right) = {x}^{3} - 3 {x}^{2} + x - 3$
Note that the ratio of the first and second terms is the same as that of the third and fourth terms, so this cubic will factor by grouping:
${x}^{3} - 3 {x}^{2} + x - 3 = \left({x}^{3} - 3 {x}^{2}\right) + \left(x - 3\right)$
$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + x - 3} = {x}^{2} \left(x - 3\right) + 1 \left(x - 3\right)$
$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + x - 3} = \left({x}^{2} + 1\right) \left(x - 3\right)$
$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + x - 3} = \left({x}^{2} - {i}^{2}\right) \left(x - 3\right)$
$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + x - 3} = \left(x - i\right) \left(x + i\right) \left(x - 3\right)$
Hence zeros:
$x = \pm i$
$x = 3$ |
# 2.07 Estimation and multistep questions
Lesson
Do you remember how we can round numbers, to help us simplify some number problems?
Round $3322$3322 to the nearest thousand.
## Learn
When we solve addition and subtraction problems, estimating our answer gives us an idea of what to expect. In this video, we see how we can estimate the answer to addition or subtraction problems, using a number line to help us.
### Apply
##### Question 1
We want to estimate the answer to $258-26$25826.
1. Round each of the two numbers to the nearest $10$10:
$258-26$258−26 is rounded to $\editable{}-\editable{}$− $=$=$\editable{}$
2. Find the value of $258-26$25826.
## Learn
How can we solve a problem with more than one part? Have a look at this video to see some different ways to do this.
### Apply
##### Question 2
Find the value of $570+25+4$570+25+4.
Remember!
By adding in a different order, we may be able to use strategies such as Bridge to $10$10, to help us solve a problem. The number lines below show how we can solve $8+4+3$8+4+3 by changing the order, and still arrive at the same answer of $15$15.
### Outcomes
#### VCMNA153
Apply place value to partition, rearrange and regroup numbers to at least tens of thousands to assist calculations and solve problems
#### VCMNA163
Use equivalent number sentences involving addition and subtraction to find unknown quantities |
## What does the slope represent on a time graph?
The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s.
## What does the slope mean in a distance vs time graph?
The slope of a distance-time graph represents speed. The steeper the slope is, the faster the speed. Average speed can be calculated from a distance-time graph as the change in distance divided by the corresponding change in time.
## What does the slope of the line represent on a velocity vs time graph?
The principle is that the slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. If the acceleration is zero, then the slope is zero (i.e., a horizontal line). The slope of a velocity-time graph reveals information about an object’s acceleration.
You might be interested: Readers ask: What Does It Mean When Bottom Number Of Blood Pressure Is High?
## What is the slope of a speed time graph equal to?
A motion such as the one above further illustrates the important principle: the slope of the line on a velocity-time graph is equal to the acceleration of the object. This principle can be used for all velocity-time in order to determine the numerical value of the acceleration.
## What is the physical meaning of the slope?
Slope is the ‘steepness’ of the line, also commonly known as rise over run. We can calculate slope by dividing the change in the y-value between two points over the change in the x-value.
## How do you find the slope of a distance time graph?
Pick two points on the line and determine their coordinates. Determine the difference in y-coordinates of these two points (rise). Determine the difference in x-coordinates for these two points (run). Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).
## What does slope do to a line?
Slope describes the steepness of a line. The slope of any line remains constant along the line. The slope can also tell you information about the direction of the line on the coordinate plane. Slope can be calculated either by looking at the graph of a line or by using the coordinates of any two points on a line.
## What will be the slope of a distance time graph when the object is at rest?
In a distance-time graph, the slope of the graph is zero, the object is at rest or in motion & why?
## Why is the slope of a velocity time graph acceleration?
What does the slope represent on a velocity graph? The slope of a velocity graph represents the acceleration of the object. So, the value of the slope at a particular time represents the acceleration of the object at that instant.
You might be interested: Question: What Does The Tiny Number On The Back Of A Dollar Bill Mean?
## What is the slope of velocity time graph for retarded motion?
The slope of velocity-time graph denotes acceleration. For a retarded motion acceleration is negative.
## What graph is equal to velocity?
Normally, velocity is plotted on the y-axis (the vertical axis) and time is plotted on the x-axis (the horizontal axis). The area under the line on a velocity-time graph is equal to the displacement of the object.
## What is constant speed?
An object is travelling at a steady or constant speed when its instantaneous speed has the same value throughout its journey. For example, if a car is travelling at a constant speed the reading on the car’s speedometer does not change.
## How do you find the slope in a graph?
Find the slope from a graph
1. Locate two points on the line whose coordinates are integers.
2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
3. Count the rise and the run on the legs of the triangle.
4. Take the ratio of rise to run to find the slope. m=riserun. |
Question
1. # If Ab64Ab Is Divisible By 12, Then The Least Possible Value Of A + B Is?
Intuition tells us that the least possible value of a + B is 12. How can we know this for sure? The answer can be found by taking the quotient of each side of the equation. So, if Ab64Ab is divisible by 12, then the least possible value of a + B is 12.
## What is Ab64Ab?
There exists a number that is the sum of its two prime factors, but it is not divisible by any other number. This number is called AbAb (or sometimes simply Ab). What is the least possible value of a + b if AbAb is divisible by ?
The answer to this question can be found using the prime factorization of AbAb. The prime factorization of a number is the way that it can be broken down into smaller pieces, where each piece has only one prime factor. In the case of AbAb, the prime factorization is:
2 × 2 × 3 = 6
4 × 5 = 25
3 × 7 = 21
Note that because AbAb is divisible by 3, 5 and 7, the least possible value of a + b would be 18.
## The Method
There are a few ways to find the least possible value of a + b. The easiest way is to use the distributive property. This states that if a and b are both integers, then:
a + b = (a+b) ÷ 2
This equation can be solved for the least possible value of a+b by dividing both sides by 2. If ab is divisible by , then the least possible value of a+b is .
## Results
If AbAb is divisible by , then the least possible value of a + b is?
Consider the following equation:
a + b = c
We can simplify this equation by dividing both sides by . We get:
a + b /= c
This means that if a and b are both integers, then the least possible value of a + b is c.
## Conclusion
According to the theorem, if ab is divisible by 12 then the least possible value of a + b is 12. In other words, if you have two numbers and want to find out what their lowest possible sum would be, simply add them together and divide that number by 12.
2. If Ab64Ab is divisible by 12, then the least possible value of A and B is a common mathematical puzzle that has puzzled mathematicians and students alike. The puzzle requires you to figure out the least possible values for two variables, A and B, given that Ab64Ab is divisible by 12. Although it seems tricky at first glance, this puzzle can be solved in a matter of minutes using basic mathematics.
To solve this problem efficiently, one must first realize that 64 can be expressed as 8×8 or 4×16. Since 12 divides into both these numbers evenly, we know that both A and B must divide into either 8 or 16 with no remainders.
3. 🤔The age-old question of “if Ab64Ab is divisible by 12, then what is the least possible value of A + B?” has been vexing mathematicians for years.
But don’t worry, we’ve got the answer for you! After doing some number crunching and mathematical calculations, we can confidently say that the least possible value of A + B is 6.
Let’s break it down. If Ab64Ab is divisible by 12, then A must be divisible by 3 and B must be divisible by 4. This means that the least possible value of A is 3 and the least possible value of B is 4.
When we add these two numbers together, we get a total of 7. This is the least possible value of A + B when Ab64Ab is divisible by 12.
So there you have it – the least possible value of A + B when Ab64Ab is divisible by 12 is 6. 🤓
4. To determine the least possible value of A and B for the number Ab64Ab to be divisible by 12, we need to consider the divisibility rules for 12.
The rule for divisibility by 12 states that a number must be divisible by both 3 and 4 in order to be divisible by 12.
For a number to be divisible by 3, the sum of its digits must be divisible by 3. In this case, A + b + 6 + 4 + A + b = 2A + 2b +10 must be divisible by 3.
For a number to be divisible by 4, the last two digits of the number must form a multiple of 4. In this case, Ab must form a multiple of 4.
Considering these rules, we can find the least possible values for A and B that satisfy both conditions. |
# How do you solve the following system of equations by linear combination: 3x+2y=-7, 5x-4y=19?
Oct 4, 2017
Solution: $x = \frac{4}{11} , y = - \frac{46}{11}$
#### Explanation:
$3 x + 2 y = - 7 \left(1\right) , 5 x - 4 y = 19 \left(2\right)$ . Multiplying equation (1) by $2$
we get $6 x + 4 y = - 14 \left(3\right)$ Adding equation (2) and equation (3)
we get $11 x = 5 \mathmr{and} x = \frac{5}{11}$ . Putting $x = \frac{5}{11}$ in equation (1) we
get $3 \cdot \frac{5}{11} + 2 y = - 7 \mathmr{and} 2 y = - 7 - \frac{15}{11} \mathmr{and} 2 y = - \frac{92}{11}$
or $y = - \frac{92}{22} \mathmr{and} y = - \frac{46}{11} \mathmr{and}$
Solution: $x = \frac{4}{11} , y = - \frac{46}{11}$ [Ans] |
• # question_answer 9) A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (0 the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5 m long.
Let the height of the vertical pole be x m and the length of the shadow be y m. As the height of the vertical pole increases, the length of the shadow also increases in the same ratio. It is a case of direct proportion. We make use of the relation of the type $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}$. (i) Here ${{x}_{1}}=5\,m\,\,60\,cm\,=5.60\,m$ ${{y}_{1}}=3\,\,m\,20\,cm\,\,=3.20\,m$ ${{x}_{2}}=10\,m\,50\,cm\,=10.50\,m$ Therefore, $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}$ gives $\frac{5.6}{3.2}\,=\frac{10.5}{{{y}_{2}}}$ $\Rightarrow$ $5.6\,{{y}_{2}}=3.2\times 10.5$ $\Rightarrow$ ${{y}_{2}}=\frac{3.2\times 10.5}{5.6}$ $\Rightarrow$ ${{y}_{2}}=6$ Hence, the length of the shadow is 6 m. (ii) Here ${{x}_{1}}=5\,m\,\,60\,\,cm=560\,\,cm$ ${{y}_{1}}=\,3\text{ }m\text{ }20\text{ }cm=320\text{ }cm$ ${{y}_{3}}=5\text{ }m=500\text{ }cm$ Therefore, $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{3}}}{{{y}_{3}}}$ gives $\frac{560}{320}\,=\frac{{{x}_{3}}}{500}$ $\Rightarrow$ $320\,{{x}_{3}}=560\times 500$ $\Rightarrow$ ${{x}_{3}}=\frac{560\times 500}{320}$ $\Rightarrow$ ${{x}_{3}}=875$ Hence, the height of the pole is 8 m 75 cm. |
# How do you solve -31 + 8x= -5x +3(x+3)?
Feb 22, 2016
$x = 4$
#### Explanation:
$- 31 + 8 x = - 5 x + 3 \left(x + 3\right)$
Expand $3 \left(x + 3\right)$ to $3 x + 9$.
$- 31 + 8 x = - 5 x + 3 x + 9$
Add $31$ to both sides.
$8 x = - 5 x + 3 x + 9 + 31$
Simplify.
$8 x = - 5 x + 3 x + 40$
Combine $- 5 x + 3 x$ to $- 2 x$
$8 x = - 2 x + 40$
Add $2 x$ to both sides.
$2 x + 8 x = 40$
Simplify.
$10 x = 40$
Divide both sides by $10$.
$x = \frac{40}{10}$
Simplify.
$x = 4$ |
# Difference between revisions of "User:Vqbc/Testing"
## Statement
Given a triangle $\triangle ABC$ with sides of length $a, b, c$ opposite vertices are $A$, $B$, $C$, respectively. If cevian $AD$ is drawn so that $BD = m$, $DC = n$ and $AD = d$, we have that $b^2m + c^2n = amn + d^2a$. (This is also often written $man + dad = bmb + cnc$, a form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")
## Proof
Applying the Law of Cosines in triangle $\triangle ABD$ at angle $\angle ADB$ and in triangle $\triangle ACD$ at angle $\angle CDA$, we get the equations
• $n^{2} + d^{2} - 2\ce{nd}\cos{\angle CDA} = b^{2}$
• $m^{2} + d^{2} - 2\ce{md}\cos{\angle ADB} = c^{2}$
Because angles $\angle ADB$ and $\angle CDA$ are supplementary, $m\angle ADB = 180^\circ - m\angle CDA$. We can therefore solve both equations for the cosine term. Using the trigonometric identity $\cos{\theta} = -\cos{(180^\circ - \theta)}$ gives us
• $\frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}$
• $\frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}$
Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n$. However, $m+n = a$ so $$m^2n + n^2m = (m + n)mn = amn$$ and $$d^2m + d^2n = d^2(m + n) = d^2a.$$ This simplifies our equation to yield $c^2n + b^2m = amn + d^2a,$ or Stewart's theorem. |
# Ex.1.4 Q5 Integers Solution - NCERT Maths Class 7
Go back to 'Ex.1.4'
## Question
The temperature at $$12$$ noon was $$10^\circ \rm{c}$$ above zero. If it decreases at the rate of $$2^\circ \rm{c}$$ per hour until midnight, at what time would the temperature be $$8^\circ \rm{c}$$ below zero? What would be the temperature at mid night?
Video Solution
Integers
Ex 1.4 | Question 5
## Text Solution
Steps:
The temperature at $$12$$ noon $$=10^\circ \rm{c}$$ (given)
The temperature decreases $$2^\circ \rm{c} = 1$$ hour (given)
The temperature decreases $$1^\circ \rm{c} =$$ $$1 \over 2$$hour
The temperature decreases $$18^\circ \rm{c}=\frac{1}{2} \times 18$$
(From $$10^\circ \rm{c}$$ to $$8^\circ \rm{c}$$ below zero $$=9$$ hours)
Total time
\begin{align}&= 12\, \rm{noon}+9 \,\rm{hours}\\ &= 21 \,\rm{hours}\\&= 9\, \rm{pm}\end{align}
Thus, at $$9$$ pm temperature would be $$8^\circ \rm{c}$$ below zero.
ii) The temperature at $$12$$ noon $$=10^\circ \rm{c}$$
The temperature decreases $$=2^\circ \rm{c}$$ every hour
The temperature decreases in $$12$$ hours $$=\;–2^\circ \rm{c} ×12 =24 ^\circ \rm{c}$$
At midnight, the temperature will be $$=10^\circ \rm{c}+(–24)^\circ \rm{c} =–14^\circ \rm{c}$$
Therefore, the temperature at mid night will be $$14^\circ \rm{c}$$ below $$0.$$
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# 5.2 Solve systems of equations by substitution (Page 2/5)
Page 2 / 5
Solve the system by substitution. $\left\{\begin{array}{c}3x+y=5\hfill \\ 2x+4y=-10\hfill \end{array}$
## Solution
We need to solve one equation for one variable. Then we will substitute that expression into the other equation.
Solve for y . Substitute into the other equation. Replace the y with −3 x + 5. Solve the resulting equation for x . Substitute x = 3 into 3 x + y = 5 to find y . The ordered pair is (3, −4). Check the ordered pair in both equations: $\begin{array}{cccc}\begin{array}{ccc}\hfill 3x+y& =\hfill & 5\hfill \\ \hfill 3·3+\left(-4\right)& \stackrel{?}{=}\hfill & 5\hfill \\ \hfill 9-4& \stackrel{?}{=}\hfill & 5\hfill \\ \hfill 5& =\hfill & 5\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 2x+4y& =\hfill & -10\hfill \\ \hfill 2·3+4\left(-4\right)& =\hfill & -10\hfill \\ \hfill 6-16& \stackrel{?}{=}\hfill & -10\hfill \\ \hfill -10& =\hfill & -10\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$ The solution is (3, −4).
Solve the system by substitution. $\left\{\begin{array}{c}4x+y=2\hfill \\ 3x+2y=-1\hfill \end{array}$
$\left(1,-2\right)$
Solve the system by substitution. $\left\{\begin{array}{c}-x+y=4\hfill \\ 4x-y=2\hfill \end{array}$
$\left(2,6\right)$
In [link] it was easiest to solve for y in the first equation because it had a coefficient of 1. In [link] it will be easier to solve for x .
Solve the system by substitution. $\left\{\begin{array}{c}x-2y=-2\hfill \\ 3x+2y=34\hfill \end{array}$
## Solution
We will solve the first equation for $x$ and then substitute the expression into the second equation.
Solve for x . Substitute into the other equation. Replace the x with 2 y − 2. Solve the resulting equation for y . Substitute y = 5 into x − 2 y = −2 to find x . The ordered pair is (8, 5). Check the ordered pair in both equations: $\begin{array}{cccc}\begin{array}{ccc}\hfill x-2y& =\hfill & -2\hfill \\ \hfill 8-2·5& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill 8-10& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill -2& =\hfill & -2\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 3x+2y& =\hfill & 34\hfill \\ \hfill 3·8+2·5& \stackrel{?}{=}\hfill & 34\hfill \\ \hfill 24+10& \stackrel{?}{=}\hfill & 34\hfill \\ \hfill 34& =\hfill & 34\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$ The solution is (8, 5).
Solve the system by substitution. $\left\{\begin{array}{c}x-5y=13\hfill \\ 4x-3y=1\hfill \end{array}$
$\left(-2,-3\right)$
Solve the system by substitution. $\left\{\begin{array}{c}x-6y=-6\hfill \\ 2x-4y=4\hfill \end{array}$
$\left(6,2\right)$
When both equations are already solved for the same variable, it is easy to substitute!
Solve the system by substitution. $\left\{\begin{array}{c}y=-2x+5\hfill \\ y=\frac{1}{2}x\hfill \end{array}$
## Solution
Since both equations are solved for y , we can substitute one into the other.
Substitute $\frac{1}{2}x$ for y in the first equation. Replace the y with $\frac{1}{2}x.$ Solve the resulting equation. Start by clearing the fraction. Solve for x . Substitute x = 2 into y = $\frac{1}{2}x$ to find y . The ordered pair is (2,1). Check the ordered pair in both equations: $\begin{array}{cccc}\begin{array}{ccc}\hfill y& =\hfill & \frac{1}{2}x\hfill \\ \hfill 1& \stackrel{?}{=}\hfill & \frac{1}{2}·2\hfill \\ \hfill 1& =\hfill & 1\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill y& =\hfill & -2x+5\hfill \\ \hfill 1& \stackrel{?}{=}\hfill & -2·2+5\hfill \\ \hfill 1& =\hfill & -4+5\hfill \\ \hfill 1& =\hfill & 1\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$ The solution is (2,1).
Solve the system by substitution. $\left\{\begin{array}{c}y=3x-16\hfill \\ y=\frac{1}{3}x\hfill \end{array}$
$\left(6,2\right)$
Solve the system by substitution. $\left\{\begin{array}{c}y=\text{−}x+10\hfill \\ y=\frac{1}{4}x\hfill \end{array}$
$\left(8,2\right)$
Be very careful with the signs in the next example.
Solve the system by substitution. $\left\{\begin{array}{c}4x+2y=4\hfill \\ 6x-y=8\hfill \end{array}$
## Solution
We need to solve one equation for one variable. We will solve the first equation for y .
Solve the first equation for y . Substitute −2 x + 2 for y in the second equation. Replace the y with −2 x + 2. Solve the equation for x . Substitute $x=\frac{5}{4}$ into 4 x + 2 y = 4 to find y . The ordered pair is $\left(\frac{5}{4},-\frac{1}{2}\right).$ Check the ordered pair in both equations. $\begin{array}{cccc}\begin{array}{ccc}\hfill 4x+2y& =\hfill & 4\hfill \\ \hfill 4\left(\frac{5}{4}\right)+2\left(-\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 4\hfill \\ \hfill 5-1& \stackrel{?}{=}\hfill & 4\hfill \\ \hfill 4& =\hfill & 4\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{ccc}\hfill 6x-y& =\hfill & 8\hfill \\ \hfill 6\left(\frac{5}{4}\right)-\left(-\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill \frac{15}{4}-\left(-\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill \frac{16}{2}& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill 8& =\hfill & 8\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\hfill \end{array}$ The solution is $\left(\frac{5}{4},-\frac{1}{2}\right).$
Solve the system by substitution. $\left\{\begin{array}{c}x-4y=-4\hfill \\ -3x+4y=0\hfill \end{array}$
$\left(2,\frac{3}{2}\right)$
Solve the system by substitution. $\left\{\begin{array}{c}4x-y=0\hfill \\ 2x-3y=5\hfill \end{array}$
$\left(-\frac{1}{2},-2\right)$
In [link] , it will take a little more work to solve one equation for x or y .
Solve the system by substitution. $\left\{\begin{array}{c}4x-3y=6\hfill \\ 15y-20x=-30\hfill \end{array}$
## Solution
We need to solve one equation for one variable. We will solve the first equation for x .
Solve the first equation for x . Substitute $\frac{3}{4}y+\frac{3}{2}$ for x in the second equation. Replace the x with $\frac{3}{4}y+\frac{3}{2}.$ Solve for y .
Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.
Aziza is solving this equation-2(1+x)=4x+10
No. 3^32 -1 has exactly two divisors greater than 75 and less than 85 what is their product?
x^2+7x-19=0 has Two solutions A and B give your answer to 3 decimal places
3. When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes onthe elliptical trainer and 30 minutes circuit training she burned 473 calories. How manycalories does she burn for each minute on the elliptical trainer? How many calories doesshe burn for each minute of circuit training?
.473
Angelita
?
Angelita
John left his house in Irvine at 8:35 am to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50. At 3:30 pm, he left the meeting and drove home. He arrived home at 5:18.
p-2/3=5/6 how do I solve it with explanation pls
P=3/2
Vanarith
1/2p2-2/3p=5p/6
James
Cindy
4.5
Ruth
is y=7/5 a solution of 5y+3=10y-4
yes
James
Cindy
Lucinda has a pocketful of dimes and quarters with a value of $6.20. The number of dimes is 18 more than 3 times the number of quarters. How many dimes and how many quarters does Lucinda have? Rhonda Reply Find an equation for the line that passes through the point P ( 0 , − 4 ) and has a slope 8/9 . Gabriel Reply is that a negative 4 or positive 4? Felix y = mx + b Felix if negative -4, then -4=8/9(0) + b Felix -4=b Felix if positive 4, then 4=b Felix then plug in y=8/9x - 4 or y=8/9x+4 Felix Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. macadamia nuts cost$9 per pound and almonds cost $5.25 per pound. how many pounds of macadamia nuts and how many pounds of almonds should macario use for the mixture to cost$6.50 per pound to make?
Nga and Lauren bought a chest at a flea market for $50. They re-finished it and then added a 350 % mark - up Makaila Reply$1750
Cindy
the sum of two Numbers is 19 and their difference is 15
2, 17
Jose
interesting
saw
4,2
Cindy
Felecia left her home to visit her daughter, driving 45mph. Her husband waited for the dog sitter to arrive and left home 20 minutes, or 13 hour later. He drove 55mph to catch up to Felecia. How long before he reaches her?
hola saben como aser un valor de la expresión
NAILEA
integer greater than 2 and less than 12
2 < x < 12
Felix
I'm guessing you are doing inequalities...
Felix
Actually, translating words into algebraic expressions / equations...
Felix
hi
Darianna
hello
Mister
Eric here
Eric
6
Cindy |
# How do you write the standard form of a line given (1, 5), m = -3?
##### 1 Answer
Aug 18, 2017
$y = - 3 x + 8$
$3 x + y = 8 \text{ } \leftarrow$standard form
#### Explanation:
Use the point-slope formula: $y - {y}_{1} = m \left(x - {x}_{1}\right)$
We have $\left(1 , 5\right) = \left({x}_{1} , {y}_{1}\right) \mathmr{and} m = - 3$
$y - 5 = - 3 \left(x - 1\right)$
$y - 5 = - 3 x + 3$
$y = - 3 x + 3 + 5$
$y = - 3 x + 8 \text{ } \leftarrow y$-intercept/gradient form.
$3 x + y = 8 \text{ } \leftarrow$standard form
$3 x + y - 8 = 0 \text{ } \leftarrow$ general form
I hope that helps! |
Changing Percents to Decimals
CHANGING PERCENTS TO DECIMALS
Whenever computations need to be done with percents, the percents are first renamed as decimals.
Here are examples of changing percents to decimals.
Notice that the $\,\%\,$ symbol is replaced with a factor of $\,\frac{1}{100}\,$:
EXAMPLES:
$3.5\% = 3.5\cdot\frac{1}{100} = \frac{3.5}{100} = 0.035$
$25\% = 25\cdot \frac{1}{100} = \frac{25}{100} = 0.25$
$50\% = 50\cdot \frac{1}{100} = \frac{50}{100} = 0.5$
$100\% = 100\cdot \frac{1}{100} = \frac{100}{100} = 1$
$250\% = 250\cdot \frac{1}{100} = \frac{250}{100} = 2.5$
As these examples illustrate, the percent symbol instructs multiplication by $\,\frac{1}{100}\,$ (or, equivalently, division by $\,100\,$),
which is accomplished by moving the decimal point two places to the left.
Remember that if you don't see a decimal point, then it gets inserted just to the right of the ones place.
Now, you can go from percents to decimals in one easy step,
by moving the decimal point two places to the left:
EXAMPLES:
$3\% = 0.03$
$2.37\% = 0.0237$
$0.01\% = 0.0001$
$5032\% = 50.32$
Some of my students find this memory device helpful:
PuDdLe DiPpeR (Imagine a little kid, barefoot, dipping piggy-toes in puddles!)
Percent to Decimal, two places to the Left
Decimal to Percent, two places to the Right
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Introduction to Sets
Here, you will practice renaming percents as decimals.
For example, $\,54\%\,$ gets renamed as $\,0.54\,$.
Change to a decimal:
(an even number, please) |
# Linear Equations In Two Variables
Before knowing Linear equations in two variables, let us know what are linear equations in one variable.
Linear equations in one variable
Equations having degree ‘1’ is known as linear equations. Standard form of a linear equation in one variable is, ax+b=0, where a and b are real numbers and a≠0.
While solving a linear equation,
• Same number can be added to (or subtracted from) both sides of the equation.
• Both sides of the equation can be multiplied or divided by same non- zero number.
For example, $2x~ + ~4$ = $0$ is a linear equation in variable x.
The solution or root of the equation is,
$2x + 4$ = $0$,
⇒ $x$ = $-\frac{4}{2}$
⇒$x$ = $-2$
It is represented on the number line as follows,
Linear equations in two variables
Consider the situation,
In a football tournament, sum of the goals scored by two players ‘A’ and ‘B’ is 32. How is it represented in terms of an equation?
Let x and y be the number of goals scored by the players ‘A’ and ‘B’ respectively.
It is given that total number of goals scored by them is 32.
Therefore,
$x~ +~ y$ = $32$ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â — -(1)
The above equation is an example of linear equation in two variables, where x and y are the two variables. It is not compulsory that the two variables can only be written as x and y. It is customary to denote variables in terms of x and y in these equations.
Few more examples for linear equation in two variables are given below,
$5p ~+~ 4q$ = $12$, where p and q are the two variables.
$2.3u ~+~ 1.8v~ -~ 5$ = $0$, where u and v are the two variables.
$√2~t~ +~ 4v$ = $-5$, where t and u are the two variables.
Therefore, general form of a linear equation in two variables is
$ax~ +~ by ~+ ~c$ = $0$
where a, b and c are real numbers, and both a and b are not equal to zero.
An equation of the form $ax~+~b$ = $0$, where a and be are real numbers, and a≠0 can also be a linear equation in two variables because, it can be represented as,
$ax ~+~ 0~×~y ~+ ~b$ = $0$
For example;
$2x~ +~ 5$ = $0$ can be written as, $2x~ +~ 0~ ×~ y ~+ ~5$ = $0$.
Example: Write the equations given in the form of $ax~+~by~+~c$ = $0$ and find the values of a, b and c.
1. √3 x + y = 5
It can be written as, √3 x + y – 5 = 0
Therefore,a=√3, b=1 and c=-5
2 . 2y + 3 = 2
It can be written as, 0 × x + 2y + 1 = 0
Therefore,a=0, b=2 and c=1
3. 4y + 2x – 2 = 0
It can be written as, 2x + 4y – 2 = 0
Therefore,a=2, b=4 and c=-2
Â
Thus, Â linear equations in two variables are an important part of mathemics domain. Â To understand this topic, students are expected to practice with sample questions and answers like the ones provided here- NCERT solutions for linear equations in two variables.
#### Practise This Question
A linear equation in two variables |
# Real-World Area Problems
Related Topics:
Lesson Plans and Worksheets for Grade 7
Lesson Plans and Worksheets for all Grades
Videos, solutions and examples to help Grade 7 students learn how to determine the area of composite figures in real-life contextual situations using composition and decomposition of polygons and circular regions.
## New York State Common Core Math Grade 7, Module 6, Lesson 20
### Lesson 20 Student Outcomes
• Students determine the area of composite figures in real-life contextual situations using composition and decomposition of polygons and circular regions.
### Lesson 20 Summary
• The following are useful strategies when tackling area problems with real-world context:
• Decompose drawings into familiar polygons and circular regions, and identify all relevant
• Pay attention to the unit needed in a response to each question.
Lesson 20 Classwork
Opening Exercise
Find the area of each shape based on the provided measurements. Explain how you found each area.
Example 1
A landscape company wants to plant lawn seed. A lb. bag of lawn seed will cover up to sq. ft. of grass and costs plus the sales tax. A scale drawing of a rectangular yard is given. The length of the longest side is ft. The house, driveway, sidewalk, garden areas, and utility pad are shaded. The unshaded area has been prepared for planting grass.
How many lb. bags of lawn seed should be ordered, and what is the cost?
Exercise 1
A landscape contractor looks at a scale drawing of a yard and estimates that the area of the home and garage is the same as the area of a rectangle that is 100 ft. × 35 ft. The contractor comes up with 5,500 ft2. How close is this estimate?
Example 2
Ten dartboard targets are being painted as shown in the following figure. The radius of the smallest circle is in. and each successive, larger circle is in. more in radius than the circle before it. A “tester” can of red and of white paint is purchased to paint the target. Each oz. can of paint covers 16 ft2. Is there enough paint of each color to create all ten targets?
Lesson 20 Opening Exercise and Example 1 Lesson 20 Example 2
The square in this figure has a side length of inches. The radius of the quarter circle is inches.
b. What is the exact area of the shaded region?
c. What is the approximate area using π = 22/7 ? Lesson 20 Exit Ticket
Example:
A homeowner called in a painter to paint bedroom walls and ceiling. The bedroom is 18 ft. long, 12 ft. wide, and 8 ft. high. The room has two doors each 3 ft. by 7 ft. and three windows each 3 ft. by 5 ft. The doors and windows do not have to be painted. A gallon of paint can cover 300 ft2. A hired painter claims he will need 4 gal. Show that the estimate is too high.
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. |
# Whole Number Division
## Find quotients of multi-digit numbers.
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Practice Whole Number Division
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Standard MCC6.NS.2 - Whole Number Division
Did you figure out how many buckets of seafood Jonah will need?
Now that he knows how many pounds of seafood is needed, he will need to figure out how many buckets he needs to order. The seafood comes in 25 pound buckets. We know from the last Concept that Jonah will need to order 3,311 pounds of seafood. That will be enough to feed 43 seals for one week.
How many buckets should he order? Given that it comes in 25 pound buckets, will there be any seafood left over? This Concept will show you how to divide whole numbers. It is exactly what you will need to solve this problem.
### Guidance
You have learned how to add, subtract and multiply. The last operation that we will learn is division.
First, let’s talk about what the word “division” actually means. To divide means to split up into groups. Since multiplication means to add groups of things together, division is the opposite of multiplication.
\begin{align*}72 \div 9 = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
In this problem, 72 is the number being divided, it is the dividend. 9 is the number doing the dividing, it is the divisor. We can complete this problem by thinking of our multiplication facts and working backwards. Ask yourself "What number multiplied by 9 equals 72?" If you said "8", you're right! 9 x 8 = 72, so 72 can be split into 8 groups of 9.
The answer to a division problem is called the quotient.
Sometimes, a number won’t divide evenly. When this happens, we have a remainder.
\begin{align*}15 \div 2 =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
Hmmm. This is tricky, fifteen is not an even number. There will be a remainder here.
We can use an “\begin{align*}r\end{align*}” to show that there is a remainder. We can also divide larger numbers. We can use a division box to do this.
\begin{align*}8 \overline{)825 \;}\end{align*}
Here we have a one digit divisor, 8, and a three digit dividend, 825. We need to figure out how many 8’s there are in 825. To do this, we divide the divisor 8 into each digit of the dividend.
\begin{align*}& 8 \overline{)825 \;} \qquad How \ many \ 8's \ are \ there \ in \ 8?''\\ & \qquad \qquad \ \ The \ answer \ is \ 1.\end{align*}
We put the 1 on top of the division box above the 8.
\begin{align*}& \overset{\ 1}{8\overline{ ) 825}}\\ & \underline{-8} \Bigg \downarrow\\ & \quad 02\end{align*}
We multiply 1 by 8 and subtract our result from the dividend. Then we can bring down the next number in the dividend. Then, we need to look at the next digit in the dividend. “How many 8’s are there in 2?” The answer is 0.
We put a 0 into the answer next to the 1. \begin{align*}& \overset{\ 10}{8\overline{ ) 825}}\\ & \underline{-8} \;\; \Bigg \downarrow\\ & \quad \ 025\end{align*}
Because we couldn’t divide 8 into 2, now we can bring down the next number, 5, and use the two numbers together: 25
“How many 8’s are in 25?” The answer is 3 with a remainder of 1. We can add this into our answer.
\begin{align*}& \overset{\ 103r1}{8\overline{ ) 825 \;}}\\ & \ \underline{ -8 \ \ }\\ & \ \ \ 025\\ & \ \ \underline{-24}\\ & \qquad 1\end{align*} We can check our work by multiplying the answer by the divisor.
\begin{align*}& \qquad 103\\ & \ \underline {\times \quad \ \ 8 \ }\\ & \qquad 824 + r \ \text{of} \ 1 = 825\end{align*}
Let’s look at a problem with a two-digit divisor.
\begin{align*}& \overset{\ \hspace{2 mm} 2}{12\overline{ ) 2448}} && How \ many \ 12's \ are \ in \ 2? \ None.''\\ & \ \underline{-24} \Bigg \downarrow && How \ many \ 12's \ are \ in \ 24? \ Two. \ So \ fill \ that \ in.''\\ & \qquad \ 4 && \ Now \ bring \ down \ the \ "4".\\ \\ & \overset{\ \hspace{4 mm} 20}{12\overline{) 2448}} && How \ many \ 12's \ are \ in \ 4? \ None, \ so \ we \ add \ a \ zero \ to \ the \ answer.''\\ && &How \ many \ 12's \ are \ in \ 48?''\\ && &Four\\ && &There \ is \ not \ a \ remainder \ this \ time \ because \ 48 \ divides \ exactly \ by \ 12.\\ \\ &\overset{\ \hspace{6 mm} 204}{12\overline{ ) 2448}}\end{align*}
We check our work by multiplying: \begin{align*}204 \times 12\end{align*}.
\begin{align*}& \qquad \quad 204\\ & \ \underline {\times \qquad \ 12}\\ & \qquad \quad 408\\ & \ \underline {+ \quad \ 2040}\\ & \qquad \ 2448\end{align*}
We can apply these same steps to any division problem even if the divisor has two or three digits. We work through each value of the divisor with each value of the dividend. We can check our work by multiplying our answer by the divisor.
Now let's practice by dividing whole numbers
#### Example A
\begin{align*}4\overline{ ) 469 \;}\end{align*}
Solution: 117 r 1
#### Example B
\begin{align*}18\overline{ ) 3678 \;}\end{align*}
Solution: 204 r 6
#### Example C
\begin{align*}20\overline{ ) 5020 \;}\end{align*}
Solution: 251
Now back to Jonah and the buckets of seafood.
If the seafood comes in 25 lb. buckets, how many buckets will he need?
To complete this problem, we need to divide the number of pounds of seafood by the number of pounds in a bucket. Notice, that we divide pounds by pounds. The items we are dividing have to be the same.
Let’s set up the problem.
\begin{align*}& \overset{\ \ \ \hspace{2 mm} 132}{25\overline{) 3311 \;}}\\ & \ \ \underline{-25}\\ & \quad \ \ 81\\ & \quad \underline{-75 \ }\\ & \qquad \ 61\\ & \quad \ \ \underline{-50}\\ & \qquad \ \ 11\end{align*}
Uh oh, we have a remainder. This means that we are missing 11 pounds of fish. One seal will not have enough to eat if Jonah only orders 132 buckets. Therefore, Jonah needs to order 133 buckets. There will be extra fish, but all the seals will eat.
### Vocabulary
Here are the vocabulary words used in this Concept.
Dividend
the number being divided
Divisor
the number doing the dividing
Quotient
the answer to a division problem
Remainder
the value left over if the divisor does not divide evenly into the dividend
### Guided Practice
Here is a problem for you to solve on your own.
\begin{align*}25 \overline{)3075 \;}\end{align*}
Next, we divide twenty- five into 3075.
Division
### Video Review
Here are a few videos for review.
### Practice
Directions: Use what you have learned to solve each problem.
1. \begin{align*}12 \div 6 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
2. \begin{align*}13 \div 4 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
3. \begin{align*}132 \div 7 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
4. \begin{align*}124 \div 4 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
5. \begin{align*}130 \div 5 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
6. \begin{align*}216 \div 6 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
7. \begin{align*}1161 \div 43 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
8. \begin{align*}400 \div 16 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
9. \begin{align*}1827 \div 21 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
10. \begin{align*}1244 \div 40 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
11. \begin{align*}248 \div 18 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
12. \begin{align*}3264 \div 16 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
13. \begin{align*}4440 \div 20 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
14. \begin{align*}7380 \div 123 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
15. \begin{align*}102000 \div 200 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
16. \begin{align*}10976 \div 98 = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
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# Starting from rest a car moves with uniform acceleration and attains a velocity of $90{\text{km/h}}$ in $25\;{\text{s}}$. It then moves with uniform speed for $30\;{\text{s}}$ and is then brought to rest in $20\;{\text{s}}$ under uniform retardation. Find total distance travelled using velocity - time graph.
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Hint: In this question, the concept of the velocity time graph will be used, that is the area under the curve of the graph represents the total distance travelled. Plot the velocity-time graph according to the question given. Find the area of the graph which will give the result for total distance travelled using velocity - time graph.
Firstly, to solve this question, we have to convert the velocity given for the car in ${\text{m/s}}$format.
So,
$\Rightarrow 90{\text{km/h}} = 90 \times \dfrac{{1000}}{{360}}{\text{m/s}} = 25{\text{m/s}}$
Now, we plot the velocity-time graph (or the v-t graph) for the question, as shown below: -
The above diagram represents the velocity time graph (v-t graph) for the question.
As we have given in the question, velocity of the car is plotted in the $y$-axis and time is plotted in the $x$ -axis.
In case of $x$ -axis (that is for plotting time): The car moves with uniform acceleration and attains a velocity of $25{\text{m/s}}$ in$25{\text{ }}s$, so the first point is given as $25{\text{ }}s$. It then moves with uniform speed for$30s$, so the next point is given as $55\;{\text{s}}$ $(25 + 30 = 55s)$. Then the car is brought to rest in $20\;{\text{s}}$, so the last point is given as $75\;{\text{s}}$ $\left( {25 + 30 + 20 = 75\;{\text{s}}} \right)$.
Now, according to the formula for displacement, we have to calculate the Area under the velocity-time graph. (As Displacement= Area under velocity-time graph)
So we can clearly see in the figure that the figure forming in the velocity-time graph is a Trapezoid.
So,
$\Rightarrow {\text{Area of Trapezoid}} = \dfrac{1}{2} \times \left( {{\text{Sum of Parallel Sides}}} \right) \times \left( {{\text{Distance between them}}} \right)$
Here, Parallel sides (according to the v-t graph) are $AB$and $CD$. And the distance between them is $25{\text{m/s}}$.
$\Rightarrow {\text{Area of Trapezoid}} = \dfrac{1}{2} \times \left( {AB + CD} \right) \times 25$
Now, we substitute the given values as,
$\Rightarrow {\text{Area of Trapezoid}} = \dfrac{1}{2} \times (30 + 75) \times 25$
After simplification we get,
$\Rightarrow {\text{Area of Trapezoid}} = 1312.5\;{\text{m}}$
Hence, the total distance travelled using velocity-time graph is $1312.5\;{\text{m}}$.
Note: Mistakes occur while plotting the velocity-time graph. In $x$-axis, time plotted should be taken cautiously as the timestamps are summed up continuously and plotted. Conversion of units is done for the ease of calculation. |
Home | | Maths 9th std | Construction of the Incircle of a Triangle
# Construction of the Incircle of a Triangle
The incentre is (one of the triangle’s points of concurrency formed by) the intersection of the triangle’s three angle bisectors.
Construction of the Incircle of a Triangle
### Incentre
The incentre is (one of the triangle’s points of concurrency formed by) the intersection of the triangle’s three angle bisectors.
The incentre is the centre of the incircle ; It is usually denoted by I; it is the one point in the triangle whose distances to the sides are equal.
### Example 4.15
Construct the incentre of ΔABC with AB = 6 cm, B = 65° and AC = 7 cm Also draw the incircle and measure its radius.
Solution
Step 1 : Draw the ΔABC with AB = 6cm, B = 65° and AC = 7cm
Step 2 : Construct the angle bisectors of any two angles (A and B) and let them meet at I.
Then I is the incentre of ΔABC. Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 3: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally. |
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# What is the Difference Between Factors and Multiples?
Factors and multiples are fundamental concepts in mathematics, vital for understanding more complex equations and real-world applications. Although these terms are often introduced in elementary school, many adults may not remember the difference between the two. This comprehensive guide will explore factors and multiples, providing useful examples, intriguing applications, and fascinating facts.
## Factors Uncovered: Building Blocks of Numbers
Factors are the numbers that multiply together to obtain a specific product. To illustrate this concept, let’s examine the number 20. When working with factors, we consider 20 as our product—the result of a multiplication problem. Factors of 20 include:
1 x 20 = 20 2 x 10 = 20 4 x 5 = 20
These numbers are factor pairs, which means they can be multiplied together to produce the original number (20). The complete list of factors for 20 is 1, 2, 4, 5, 10, and 20.
Prime numbers, such as 2, 3, 5, and 7, are unique in that they have only two factors—themselves and 1. On the other hand, composite numbers, like 20, have more than two factors. Understanding factors can be useful in various real-life applications, such as dividing resources, organizing events, or solving puzzles.
## Multiples Unraveled: Infinite Possibilities
Multiples are the products of a number when multiplied by any other number. Unlike factors, there are an infinite number of multiples. Using 20 again as an example, its multiples include the products of multiplying 20 by any number:
1 x 20 = 20 2 x 20 = 40 3 x 20 = 60 4 x 20 = 80 …and so on.
Another way to think about multiples is “skip counting” by a specific number. The multiples of 20 are 20, 40, 60, 80, 100, 120, 140, 160, and so on.
Multiples have practical applications in various contexts, including financial calculations, scheduling, and understanding patterns in nature or technology.
## Fascinating Facts and Connections
1. The least common multiple (LCM) of two numbers is the smallest multiple they both share. LCMs are essential in solving problems related to syncing events, such as when two trains will meet at the same station.
2. The greatest common factor (GCF) of two numbers is the largest factor they both share. GCFs are crucial in reducing fractions to their simplest form and solving problems related to sharing resources or dividing objects evenly.
3. Factors can be used to determine whether a number is a perfect square, meaning it is the product of an integer multiplied by itself (e.g., 4, 9, 16). Perfect squares have an odd number of factors, while non-perfect squares have an even number of factors.
## Key Takeaways
Factors and multiples are essential mathematical concepts that have wide-ranging applications in everyday life. By understanding the difference between these two terms and their real-world significance, you can harness the power of mathematics to solve problems, organize events, and interpret patterns.
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# Continuity
Lesson Features »
Lesson Priority: Normal
Calculus $\longrightarrow$
Functions and Limits $\longrightarrow$
Objectives
• Understand continuity at a point or over an interval conceptually and visually based on graphs
• Define continuity of a function at a point using limits
• Define continuity of a function over an interval using limits
• Classify types of discontinuities, either based on graphs or based on function definitions
Lesson Description
We've talked in the past in Algebra Two and Pre-Calculus about a loose definition of what it means for a function to be continuous. Now we'll talk about a rigorous definition, using limits.
Practice Problems
Practice problems and worksheet coming soon!
## It's Good to Be Well-Connected
Continuity is a concept that could be studied and understood in Algebra, but is more acutely defined in here in Calculus, through the use of limits. So what exactly is Continuity all about? In short, it's a measure of whether a function is connected with itself. We'll see that this is easier to define visually / graphically than it is to define algebraically, though we will need to be experts on both for typical exam questions.Let's digest this concept first with a loose definition and a focus on the graphical approach, before moving to the more rigorous and tedious algebra based tasks.
## A Practical Definition
Functions are continuous along the parts of their graph that you can draw without picking up your pencil, and discontinuous at the place(s) where you had to pick up your pencil. This idea is the loose definition of continuity.Here is a graph of a function that has a point of discontinuity, followed by a graph that is continuous everywhere on the function's domain:Note that $f(x)$ has a gap when $x=3$, and how it would not be possible to draw this graph without picking up your pencil. This is not a rigorous definition of continuity, but it is true that $f(x)$ is not continuous at $x=3$. It is also true that $f(x)$ is continuous everywhere else that we can tell. This leads is to our first important point about continuity:Continuity is not a yes or no question - i.e. a function is not either continuous or discontinuous identically. A function is continuous on the intervals that are connected, and discontinuous at the points or intervals on the graph that we would have had to pick up our pencil when drawing.
## What's the Real Question?
Let's take a look at a few more graphs to better understand what continuity and discontinuity each look like.This graph has two discontinuities - at $x=-1$ there is a hole in the graph, and at $x=2$ there is a jump in the graph. The question we're often going to need to answer is - "where is the graph continuous"? Since most parts of most graphs you will see are continuous, this question is best answered by taking the perspective that the function is continuous everywhere except for the places we identify that are discontinuous. So the graph above looks to be continuous everywhere except for $x=-1$ and $x=2$. If we are asked to present our answer in interval notation, then we would say $x$ is continuous on the interval $(-\infty, -1) \,\, \bigcup \,\, (-1, 2) \,\, \bigcup \,\, [2, \infty)$.Here's another one:While much of this graph is smooth and connected, there are a few places that stand out as "break points" or as places where, when drawing the function, we had to lift our pencil up off the page. The function is discontinuous between $x=-1$ and $x=2$, not just at the specific locations of $x=-1$ and $x=2$, because the function is not defined at all on this interval. Therefore, when asked, we would say that this function is continuous on the interval $(-\infty, -1) \,\, \bigcup \,\, (2,5) \,\, \bigcup \,\, (5, \infty)$.
## The Four Flavors of Discontinuity
Hopefully with these few examples, it is becoming clear how to visually determine the places or intervals on a function that are continuous, and those that are discontinuous. Before we look at the limit definition of how to determine continuity, let's specify the four categories of discontinuities that we will encounter.Jump DiscontinuitiesThe first type of discontinuity is called a jump discontinuity. This is the classic "gap" situation where, at the point of discontinuity, each side of the function is approaching different values.Note that in terms of continuity at that point of separation, it doesn't matter which side has the solid dot where the function is actually defined, nor does it matter if neither is solid - a discontinuity will exist at this point regardless. This type of discontinuity is typically exclusive to piecewise defined functions.Removable DiscontinuitiesThe next type of discontinuity is called a removable discontinuity. Both of the following similar functions exhibit this type of discontinuity:The name comes from the idea that the function would be otherwise continuous in the vicinity of the point, were we to "plug the hole". In other words, each side of the function approaches the same value, but the function is not continuous because for whatever reason, the function value at that point is defined to be a different value, or not defined at that point at all (the first and second graphs above, respectively). The function is discontinuous either way, since the function has a puncture in it. Typically this happens naturally in rational functions that have a common factor in both the numerator and denominator (see the lesson on limits with zero in the denominator »).Infinite DiscontinuitiesUp third, we see one of the more common situations that occur in functions that we work with often.An infinite discontinuity occurs any time a vertical asymptote appears. This discontinuity category applies regardless of whether the graph approaches the same or different directions on each side of the asymptote. Both asymptotes in the following figure are infinite discontinuities.Typically, this type of behavior can appear naturally in rational functions that have a division by zero domain error. However, along with many of the stranger graphs in math, it can be a product of a piecewise defined function. In fact, it's possible to define a function piecewise such that one side of an asymptote has infinite behavior, while the other side has finite behavior. Here's an example of that usual case:Infinite Oscillation DiscontinuitiesLast but not least, we have one of the least frequent occurrences in functions we work with, as this issue is specific to trig functions. An infinitely oscillating discontinuity occurs when a periodic function oscillates with increasing frequency that becomes infinitely frequent at a specific instant. An example would be $f(x)=\sin(1/x)$ at $x=0$:Of all the discontinuity types, this one is easily the most unique, but it is also infrequent. Some teachers don't even cover it in fact, just to save time and energy.Note: while it is important and more palatable to digest the differences among these four classifications visually, we will also need to look at how to tell these four categories apart without graphs, using only the algebraic limit definition that we will turn to next. First we'll acquaint ourselves with the limit defintion approach, and then we will quickly revisit each of these four situations and see how to tell them apart with the limit method.
Pro Tip
Many teachers explicitly test your ability to classify discontinuities into one of these four categories. On a test is is possible to be asked to do this using the graph approach that we are using now, but it is even more likely that if you are asked to do this task, that you will need to use the limit definition approach that we are about to look at.
## The Proper Approach
While most students find it easier to describe continuity with a visual approach, we must ultimately define it symbolically, using limits.
Define: Function ContinuityLet $f(x)$ be a function with domain $D$. We say that $f(x)$ is continuous on every point in its domain such that$$\lim_{x \to a} f(x) = f(a)$$(1)for all $a \in D$, and only if $\lim_{x \to a} f(x)$ exists and is finite.
A word interpretation of this definition is to say that a function is continuous at all places where the function does what it says it will do, remembering the all-important idea that limits have to do solely with function behavior. If the limit at a point does not exist for any reason or is not finite, then the point fails this limit definition and therefore a discontinuity exists at this point.Let's take another pass through each of the four types and see how each is defined using the limit definition of continuity.Jump Discontinuities - Limit DefinitionLet's take another look at a function with a single jump discontinuity, and apply the limit definition.The limit as $x$ approaches $2$ from the left looks to be $4$, but the limit as $x$ approaches $2$ from the right looks to be $1$. Since the one sided limits do not agree, the limit of the function at $x=2$ does not exist, and therefore we know immediately that the function is discontinuous at $x=2$.Look at same function at any other point on the graph, and notice how limit does always equal the function value. This reinforces the idea that the easiest way to describe continuity is to say that the function is continuous on its domain except for the places that you can identify discontinuities, which in this case is $x=2$.
Define: Jump DiscontinuityIf, at any point $a$ on $f(x)$$\lim_{x \to a^-} f(x) = c_1$$$$\lim_{x \to a^+} f(x) = c_2$$where$c_1$and$c_2$are finite, and$c_1 \neq c_2$, then there exists a jump discontinuity in the function at$x = a$.Stated differently, when you examine a point of potential discontinuity, and you find that each of the one-sided limits are finite but do not agree, then you know specifically that you have a jump discontinuity. Removable Discontinuities - Limit DefinitionRecall that a removable discontinuity was the case when the function looked nearly continuous but has a puncture or hole, whether or not the function is defined at the point of interest:orIn terms of limits, the hallmark quality of a removable discontinuity is that each one side limit will agree, but the function value will not be equal to the limit. In the example above, we can see that$\lim_{x \to 3^{+}}$is$2$,$\lim_{x \to 3^{-}}$is$2$, but$f(3) \ne 2$. Define: Removable DiscontinuityIf, at any point on$f(x)$$\lim_{x \to a^-} f(x) = c_1$$$$\lim_{x \to a^+} f(x) = c_1$$$$f(a) \neq c_1$$where $c_1$ is finite, then there exists a removable discontinuity at in the function at $x=a$.In other words, any point where the two-sided limit exists but the function value evaluates to a different number is a place where the function will have a removable discontinuity.
Infinite Discontinuities - Limit DefinitionAs we've seen, the presence of any vertical asymptote creates an infinite discontinuity.If you find that either of the one-side limits is $+\infty$ or $-\infty$, then you know you're dealing with an infinite discontinuity (which is why they are named that way).
Define: Infinite DiscontinuityIf, at any point on $f(x)$$\lim_{x \to a^-} f(x) = \infty \,\, \mathrm{or} \,\, -\infty$$or$$\lim_{x \to a^+} f(x) = \infty \,\, \mathrm{or} \,\, -\infty$$then there exists an infinite discontinuity at in the function at$x=a$.In other words, any limit with an infinite result is an indicator that an infinite discontinuity exists. Oscillating Discontinuities - Limit DefinitionFinally, the oscillating discontinuity's hallmark is that the$\sin$or$\cos$function argument will approach infinity. For example, we looked at the function$y = \sin(1/x)$, which looks likeThe problem comes from the fact that as$x$approaches zero, the argument of the sine function approaches infinity. Since sine functions oscillate back and forth forever, no matter how far right you go, this behavior means that the function$\sin(1/x)$oscillates back and forth infinitely at$0$. The limit doesn't exist. Define: Oscillating DiscontinuityIf$f(x)$is a sinusoid, then an infinite oscillation will occur at any finite point$a$that makes the argument of the trig function approach infinity as$x$approaches$a$. Again, these infinite oscillations are distinguished from other types of discontinuities, and also uncommon. Remember! It is very possible that you are going to be asked to classify discontinuities without having a picture to guide you. Make sure you understand the limit definition, and can tell each of the four categories apart by their unique limit properties. Pro Tip If you can read the following table fluently and understand what it is saying and why, you're in good shape for quiz and exam questions. Type Description Jump Each of the one-sided limits are finite but disagree Removable The two one-sided limits agree but the function value does not agree Infinite One or both of the one-sided limits is either$+\infty$or$-\infty$Oscillating Typically examining some form of$\sin(\theta)$or$\cos(\theta)$at points where$\theta=1/0^+$or$\theta=1/0^-$## One-Sided Continuity Just like limits can be one-sided, so too can continuity. Knowing and understanding this nuance is less important than understanding everything we've mentioned up to this point, and many teachers will not even discuss one-sided continuity, so if you don't need to know about it, jump down to the next heading. If you do, read on!Jump discontinuities may be considered continuous on one side by a very similar definition to the general limit definition of continuity, but using one-sided limits: Definition:A function$f(x)$is continuous from the left if$$\lim_{x \to a^-} f(x) = f(a)$$and continuous from the right if$$\lim_{x \to a^+} f(x) = f(a)$$ Said a different way, jump discontinuities are one-sidedly continuous if one side has the filled in solid dot, and not one-sidedly continuous if the function is defined elsewhere.Note that this will only be possible for jump discontinuities. ## Well-Behaved Functions By now, you may have noticed that many of the continuity problems we are describing are specific to specialized functions, like piecewise, rational, or trigonometric functions. This is no coincidence. Many common functions, such as polynomials, have no instances of discontinuity at any point. In fact, several of the major function families have continuity behavior that can be summarized generally. Theorem 1:The following function families are continuous at every point in their domain: • Linear and Constant Functions • Polynomial Functions • Rational Functions • Exponential Functions • Logarithmic Functions • Trigonometric Functions But wait, didn't we just say that$f(x) = \sin(1/x)$had that strange infinite discontinuity at$x=0$? We sure did, but that's still in line with this theorem -$x=0$is not in the domain of$\sin(1/x)$, and so the theorem does not guarantee continuity for this function at$x=0$. It's noteworthy that the function promises continuity for every other real number$x$that could be input into$\sin(1/x)$, since$x=0$is the only number excluded from the domain.The takeaway is that, while it is possible to design funky and unusual variations of these typical function families, we can always guarantee continuity of these function types for any value of$x$that could be input into it to obtain a real answer (i.e. the domain). ## Mr. Math Makes It Mean The last consideration about continuity that we need to make is the case where some function$h(x)$is defined as some combination of two (or possibly more) other functions, say$f(x)$and$g(x)$. This could happen by adding, subtracting, multiplying, or dividing$f$and$g$, or it could be a result of composition, e.g.$h(x) = f(g(x))$. In any case, there is nothing too surprising happening here. Definition:Let$h(x)$be comprised of two functions,$f(x)$and$g(x)$, either by arithmetic or composition. It follows that$h(x)$will be continuous everywhere that$f(x)$and$g(x)$are both continuous. For function division, such as$h(x) = f(x) \, / \, g(x)$for example, we must also exclude$x$values that make$g(x)=0$, for$h(x) = f(x) \, / \, g(x)$would then be undefined. Similarly, we must exclude any$x$values that make$g(x)$equal to values that are not in the domain of$f(x)$for a composition of the form$f\big( g(x) \big)$, as to avoid causing domain issues with$f$. It is not a surprising fact that domain and continuity go hand-in-hand, but it's worth mentioning explicitly for clarity, and teachers will occasionally build quiz questions around this idea. ## Put It To The Test Identify Continuity - Graph BasedAmong the warm-up questions we should expect are questions asking us to identify which continuity type is which, based on the graph:Instructions for Examples 1-3: Determine the intervals on which the function is continuous, and determine the category of discontinuity that occurs anywhere that the function is discontinuous. Example 1 Show solution$\blacktriangleright$In this graph, not only must we identify the discontinuities at$x=-4$and$x=4$, but we must also be aware that the function doesn't appear to be defined at all between$x=2$and$x=4$. This is a good reminder that when we report the intervals of continuity, that we must not blindly expect the answer to be all real numbers excepting the discontinuity points. The answer is that the function is continuous on the following intervals:$$(-\infty, -4) \,\, \bigcup \,\, (-4, 2) \,\, \bigcup \,\, (4, \infty)$$Additionally, discontinuities exist at$x=-4$(of the jump variety) and$x=4$(of the infinite variety). Example 2 Show solution$\blacktriangleright$This function has two discontinuities - one at$-1$and one at$5$. The point labeled at$3$has no discontinuity - it was randomly labeled. Some teachers do that, so don't expect every labeled point to have a discontinuity issue. The discontinuity at$x=-1$is a jump type with right-side continuity, and the discontinuity at$x=3$is a removable discontinuity. Therefore we would answer the question by giving the union of the intervals of continuity:$$(-\infty, -1) \,\, \bigcup \,\, (-1, 5) \,\, \bigcup \,\, (5, \infty)$$Note that we do not notate the one-sided continuity unless we're asked about it, as including the point$-1$in the answer would suggest that the function is continuous at$-1$. Example 3 Show solution$\blacktriangleright$While you'd be correct to say this function isn't "smooth" (more to come on that soon, in the lesson that discusses differentiability »), it is not discontinuous at the corner point. In fact, this function looks to be continuous everywhere. Therefore, we will report that the function is continuous on the interval$$(-\infty, \infty)$$And because there are no discontinuities, we don't have to state types of discontinuities, or label anything further. Identify Continuity - Symbol BasedSimilarly yet more algebraically involved, we are often asked to classify based on function definition alone.Instructions for Examples 4-6: Determine the intervals on which the function is continuous, and determine the category of discontinuity that occurs anywhere that the function is discontinuous. Example 4$$f(x) = \frac{3x}{x^2-9}$$ Show solution$\blacktriangleright$Since rational functions are continuous everywhere in their domain (via Theorem 1 of this lesson), we only need to identify points that are not in the domain of$f(x)$when seeking points of potential discontinuity. In this case, the function is undefined when the denominator is zero, which happens at$x=-3$and$x=3$. Via our limit evaluation skills, you should conclude that$$\lim_{x \to -3^-} = -\infty$$$$\lim_{x \to -3^+} = \infty$$$$\lim_{x \to 3^-} = -\infty$$$$\lim_{x \to 3^+} = \infty$$Therefore we know that$f(x)$is continuous everywhere except for at$x=-3$and$x=3$, and that each of these discontinuities is of the infinite category. Example 5$$f(x) = \frac{2x^2 - 7x + 4}{3x^2 - 5x - 12}$$ Show solution$\blacktriangleright$Once again, faced with a rational function, we should seek to identify$x$values that are not in the domain of the function, since a rational function such as this one will be continuous at all other$x$values beside ones that are not in the domain (again, per Theorem 1 above). In the absence of logs, radicals, or anything funky, this rational function's domain will be all real numbers except for places that make the denominator zero.When is$3x^2 - 5x - 12 = 0$? Let's factor it and find out.$$3x^2 - 5x - 12 = (x-3)(3x+4) = 0$$$$\therefore x=3, \,\,\,\, x = \frac{-4}{3}$$Note that if we couldn't factor this denominator, that would be ok - we would just go straight to using the quadratic formula in that scenario.By Theorem 1, we conclude that there is a discontinuity in two places: at$x=3$and$x = -4/3$. Now we just need to determine what category of discontinuity each point falls into.If we factor both the numerator and the denominator, we will see that$$\frac{2x^2 - 7x + 4}{3x^2 - 5x - 12} = \frac{(x-3)(2x-1)}{(x-3)(3x+4)}$$$$= \frac{2x-1}{3x+4}$$From the knowledge we picked up when studying limits with zero in the denominator », this tells us that 1)$x=3$is a puncture, not an asymptote, and 2)$x=-4/3$is a vertical asymptote. Therefore, we can conclude that:$f(x)$has a removable discontinuity at$x=3$$f(x) has an infinite discontinuity at x = -4/3 Warning! While limit evaluation techniques dictate that we should seek to factor and cancel in a rational function to more quickly identify limit results, be careful not to jump to this process right away for questions regarding continuity. In Example 5, if you were to factor and cancel immediately, you would miss one of the places that is not in the domain of the function, because you removed the factor from the picture completely. Just because a denominator factor cancels with a numerator factor does not mean the associated root is in the domain of the original function - here x=3 is not in the domain of f(x), even though the (x-3) factors cancel, and it would be incorrect to ignore the discontinuity at that point. Example 6$$ f(x) = \left\{ \begin{array}{ll} -2x + 1 & : x \le -1\\ x^2 - 3 & : -1 \gt x \gt 3\\ \sqrt{x-3} & : x \ge 3 \end{array} \right. $$Show solution \blacktriangleright On their own, each of these three function definitions is continuous everywhere on their domain (again, via Theorem 1), so the only potential places we need to investigate are the breakpoints of the piecewise function.Let's start by examining the behavior at x=-1:$$\lim_{x \to -1^{-}} f(x) = \lim_{x \to -1^{-}} -2x + 1 = 3\lim_{x \to -1^{+}} f(x) = \lim_{x \to -1^{+}} x^2 - 3 = -2$$Since each one sided limit yields different but finite values, we know immediately that there is a jump discontinuity at x=-1. Now let's look at x = 3:$$\lim_{x \to 3^{-}} f(x) = \lim_{x \to 3^{-}} x^2 - 3 = 6\lim_{x \to 3^{+}} f(x) = \lim_{x \to 3^{+}} \sqrt{x-3} = 0$$Once again, we have a jump discontinuity. Therefore, we conclude that f(x) is continuous everywhere except at x=-1 and x=3, where in each case there exists a jump discontinuity.Combinations of FunctionsIn examples 7-9, use the given information and your knowledge of function combination to answer each question. Example 7The domain of f(x) is all real numbers, and it is continuous everywhere on its domain.The domain of g(x) is all real numbers except x=-2 and x=2, and is continuous everywhere on its domain.Find the intervals of continuity for the function (f + 2g)(x). Show solution \blacktriangleright Note that multiplying g(x) by 2 doesn't affect its domain - it does affect its range, but we will not need to consider range to answer this question.The given information about f and g implies that the combined function (f + 2g)(x) would have a domain of all real numbers except for x=-2 and x=2. And since both functions are continuous everywhere on their domain, it follows that the combined function is as well. Therefore, we can safely conclude that the combined function is continuous on the domain$$(-\infty, -2) \,\, \bigcup \,\, (-2, 2) \,\, \bigcup \,\, (2, \infty)$$Example 8The domain of f(x) is all real numbers except x=1, and f is continuous everywhere on its domain.The domain of g(x) is all real numbers except x=4, and g is continuous everywhere on its domain.Find the intervals of continuity for the function \bigg(\frac{f}{g}\bigg)(x). Show solution \blacktriangleright Similar to the last problem, the key is to follow the logic that governs the domain of a combination function, and let the continuity question follow the same logic. The domain of a quotient of functions is found by including only numbers that work in both functions simultaneously, which in this case would be all real numbers except x=1 and x=4. However, we also need to ensure that we do not create a division by zero error, and so we must also exclude any x that makes g(x) equal to zero, as this would cause f(x) / g(x) to be undefined. Since we do not have the functions written for us explicitly, the best way to answer the question is to say the following:The domain of \bigg(\frac{f}{g}\bigg)(x) is all real numbers except x=1, x=4, and any value of x such that g(x)=0. Because both functions are continuous at every point on their domain, we conclude that the function \bigg(\frac{f}{g}\bigg)(x) is continuous at all values of x except for x=1, x=4, and any value of x such that g(x)=0. Example 9The domain of g(x) is all real numbers except x=a, and f is continuous everywhere on its domain.The domain of h(x) is all real numbers except x=b, and g is continuous everywhere on its domain.a and b are real number constants.Find the intervals of continuity for the function g \big( h(x) \big). Show solution \blacktriangleright As with the last two problems, the domain is the key to making a statement about where the composite function is and is not continuous. The domain of g \big( h(x) \big) will be all real numbers except for x=a, x=b, and any x value that makes h(x)=a, since a cannot be input into the function g(x). Therefore, we will report our answer asThe function g \big( h(x) \big) is continuous at all values of x except for x=a, x=b, and any x value such that h(x)=a. Unknown Coefficient Problems Finally, we may be given piecewise functions with unknown coefficients. We can identify what coefficient values must be present to make the function continuous:Instructions for Examples 10-11: Determine the unknown coefficients a, b, etc such that the function is continuous everywhere on its domain.Example 10$$ f(x) = \left\{ \begin{array}{lr} \cos(x) & : x \le 0\\ \sin(x) + a & : x \gt 0 \end{array} \right. $$Show solution \blacktriangleright As we've seen with common functions that are covered by Theorem 1, we only need to worry about the piecewise function break point. Let's examine the limits from each side as x approaches 0.$$\lim_{x \to 0^-} \cos(x) = 1\lim_{x \to 0^+} \sin(x) + a = 0 + a = a$$All we need to do is make each one-sided limit equal to one another. There isn't too much to do in this particular example, so we have1 = aand therefore, the function will be continuous if a=1. Example 11$$ f(x) = \left\{ \begin{array}{lr} 7 & : x < -2\\ x^2 - 2ax + b & : -2 \leq x < 6\\ ax + b & : x \geq 6 \end{array} \right. $$Show solution \blacktriangleright As with the last example, we seek to find the constants that make each piece of the piecewise function equal to one another at the interval break points. Let's work left-to-right and see what we can discern: set the left-most function definition equal to the middle function definition.$$7 = x^2 - 2ax + b$$Furthermore, we want these two parts of the function definition to be equivalent specifically at x=-2, where the piecewise function "changes over" from the first definition to the second. So let's let x be -2.$$7 = 4 + 4a + b$$This equation is indeterminate, having 2 variables in it. Let's pause on it and do the same thing for the other cutoff point of the piecewise function - set the two pieces equal.$$x^2 - 2ax + b = ax + b$$At the place that we want these two definitions to be equal, x is 6. Let's plug in that value and see what remains.$$36 - 12a + \cancel{b} = 6a + \cancel{b}36 - 18a = 0\boxed{a = 2}$$Now that we know a=2, we can use that result in the first equation we had set up but had to put on pause:$$7 = 4 + 4a + b7 = 4 + 4(2) + b\boxed{b = -5}$$Therefore, b = -5, and the complete piecewise function that is continuous at all points is$$ f(x) = \left\{ \begin{array}{lr} 7 & : x < -2\\ x^2 - 4x - 5 & : -2 \leq x < 6\\ 2x -5 & : x \geq 6 \end{array} \right. If we wanted to double check, we can plug in this function into a graphing calculator to see that the various piecewise parts do indeed connect at the "break points".
You Should Know
Problems like these are not only intended to be worked without a graph, but may actually become incredibly confusing should you to try to graph them. Their true picture is not known at the start of the problem, since the graph would depend on the unknown constants that you are solving for.
Lesson Takeaways
• Understand what continuity is visually, conceptually, and symbolically
• Be able to classify discontinuity types based on graphs
• Be able to classify discontinuity types without graphs, by analyzing with limits
• Concisely describe the places or intervals on which a function is continuous
• If required, be able to identify one-sided continuity
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Lesson Metrics
At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).
Key Lesson Sections
Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.
Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).
Perils and Pitfalls - common mistakes to avoid.
Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!
Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!
Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!
Special Notes
Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.
Pro-Tip: Knowing these will make your life easier.
Remember! - Remember notes need to be in your head at the peril of losing points on tests.
You Should Know - Somewhat elective information that may give you a broader understanding.
Warning! - Something you should be careful about. |
# Complete the Equation: How to Finish Subtraction and Addition Sentences with Mixed Numbers
Mixed numbers, which combine whole numbers and fractions, are frequently encountered in various mathematical and real-world contexts.
When presented with incomplete addition or subtraction sentences involving mixed numbers, the challenge lies in determining the missing value. In this guide, we’ll explore how to finish such sentences, ensuring a clear understanding of the operations involved.
## Step-by-step Guide to Complete the Equation:
1. Understanding the Sentence Structure:
Incomplete sentences will typically have a missing value, represented by a blank or a variable (like $$x$$). Your task is to determine this missing value.
– Begin by adding/subtracting the fractional parts. If the fractions have different denominators, find a common denominator.
– Next, add/subtract the whole numbers.
– Combine the results to get the final mixed number.
3. Approaching Subtraction Sentences:
For subtraction:
– Start by subtracting the fractional parts. If the fraction from which you are subtracting is larger than the first fraction, borrow from the whole number.
– Then, subtract the whole numbers.
– Combine the results to get the final mixed number.
Complete the sentence: $$2 \frac{1}{4} + 3 \frac{3}{4} = \_\_\_\_$$
Solution:
Adding the fractions: $$\frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$$
Adding the whole numbers: $$2 + 3 + 1 (from the fractions) = 6$$
The completed sentence is: $$2 \frac{1}{4} + 3 \frac{3}{4} = 6$$
The Absolute Best Book for 5th Grade Students
### Example 2 (Subtraction):
Complete the sentence: $$5 \frac{3}{5} – 2 \frac{2}{5} = \_\_\_\_$$
Solution:
Subtracting the fractions: $$\frac{3}{5} – \frac{2}{5} = \frac{1}{5}$$
Subtracting the whole numbers: $$5 – 2 = 3$$
The completed sentence is: $$5 \frac{3}{5} – 2 \frac{2}{5} = 3 \frac{1}{5}$$
### Practice Questions:
1. Complete the sentence: $$4 \frac{2}{7} + 3 \frac{5}{7} = \_\_\_\_$$
2. Complete the sentence: $$7 \frac{3}{8} – 5 \frac{1}{8} = \_\_\_\_$$
3. Complete the sentence: $$6 \frac{1}{6} + 2 \frac{5}{6} = \_\_\_\_$$
A Perfect Book for Grade 5 Math Word Problems!
1. $$8$$
2. $$2 \frac{2}{8}$$ or $$2 \frac{1}{4}$$
3. $$9$$
The Best Math Books for Elementary Students
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