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# The eccentricity of an ellipse $9{x^2} + 16{y^2} = 144$ is (a) $\dfrac{{\sqrt 3 }}{5}$(b) $\dfrac{{\sqrt 5 }}{3}$(c) $\dfrac{{\sqrt 7 }}{4}$(d) $\dfrac{2}{5}$
Last updated date: 24th Jul 2024
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Hint: Here the eccentricity of an ellipse is a measure of how nearly circular is the ellipse. Eccentricity is found by the formula eccentricity = c/a where ‘c’ is the distance from the centre to the focus of the ellipse and ‘a’ is the distance from the centre to the vertex for the standard form of the ellipse.
Given ellipse is $9{x^2} + 16{y^2} = 144$
Rewriting the ellipse, we get
$\dfrac{{9{x^2}}}{{144}} + \dfrac{{16{y^2}}}{{144}} = 1 \\ \\ \dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1 \\$
For the ellipse of the form $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ the eccentricity is given by $e = \dfrac{c}{a}$, where $c = \sqrt {{a^2} - {b^2}}$.
Comparing both the equations we have $a = 4,{\text{ }}b = 3{\text{ }}$
So, $c = \sqrt {16 - 9} = \sqrt 7$
Therefore, $e = \dfrac{c}{a} = \dfrac{{\sqrt 7 }}{4}$
Thus, the answer is option (c) $\dfrac{{\sqrt 7 }}{4}$.
Note: The eccentricity of the ellipse is always greater than zero but less than one i.e. $0 < e < 1$. The standard form of the ellipse is $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1$ with centre $\left( {h,k} \right)$. In this problem we have centre $\left( {0,0} \right)$ so we have used the ellipse form $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.
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# Lesson 8 Law of Total Probability
## Motivating Example
You watch a magician place 4 ordinary quarters and 1 double-headed quarter into a box. If you select a coin from the box at random and toss it, what is the probability that it lands heads?
## Theory
In some situations, calculating the probability of an event is easy, once you condition on the right information. For example, in the example above, if we knew that the coin we chose was ordinary, then: $P(\text{heads} | \text{ordinary}) = \frac{1}{2}.$ On the other hand, if the coin we chose was double-headed, then $P(\text{heads} | \text{double-headed}) = 1.$
How do we combine these conditional probabilities to come up with $$P(\text{heads})$$, the overall probability that the coin lands heads?
Theorem 8.1 (Law of Total Probability) Let $$A_1, ..., A_n$$ be a partition of the possible outcomes. Then: $P(B) = \sum_{i=1}^n P(A_i) P(B | A_i).$
A partition is a collection of non-overlapping events that cover all the possible outcomes. For example, in the example above, $$A_1 = \{ \text{ordinary}\}$$ and $$A_2 = \{ \text{double-headed} \}$$ is a partition, since the coin that we selected has to be one of the two and cannot be both.
Applying the Law of Total Probability to this problem, we have \begin{align*} P(\text{heads}) &= P(\text{ordinary}) P(\text{heads} | \text{ordinary}) + P(\text{double-headed}) P(\text{heads} | \text{double-headed}) \\ &= 0.8 \cdot \frac{1}{2} + 0.2 \cdot 1 \\ &= 0.6 \end{align*}
So the overall probability $$P(B)$$ is just a weighted average of the conditional probabilities $$P(B | A_i)$$, where the “weights” are $$P(A_i)$$. (Note that these “weights” have to add up to 1, since $$A_1, ..., A_n$$ are a partition of all the possible outcomes, whose total probability is $$1.0$$.) This means that the overall probability $$P(B)$$ will always lie somewhere between the conditional probabilities $$P(B | A_i)$$, with more weight given to the more probable scenarios. For example, we could have predicted that $$P(\text{heads})$$ would lie somewhere between $$\frac{1}{2}$$ and $$1$$; the fact that it is much closer to the former is because choosing the ordinary coin was more probable.
## Examples
1. Here are some things we already know about a deck of cards:
• The top card in a shuffled deck of cards has a $$13/52$$ chance of being a diamond.
• If the top card is a diamond, then the second card has a $$12/51$$ chance of being a diamond.
• If the top card is not a diamond, then the second card has a $$13/51$$ chance of being a diamond.
Now, suppose we “burn” (i.e., discard) the top card without looking at it. What is the probability that the second card is a diamond? Use the Law of Total Probability, conditioning on the top card. Does burning cards affect probabilities?
2. Anny is a fan of chess competitor Hikaru Nakamura, and tomorrow is the World Chess Championship. She is superstitious and believes that the weather influences how he will perform. Hikaru has a 60% chance of winning if it rains, a 25% chance if it is cloudy, and a 10% chance if it is sunny. Anny checks the weather the night before, and the forecast says that the chance of rain tomorrow is 40%; otherwise, it is equally likely to be cloudy as sunny. What is the probability that Hikaru wins the World Chess Championship?
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# 2.8: Expressions and the Distributive Property
Difficulty Level: Basic Created by: CK-12
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Practice Expressions and the Distributive Property
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Suppose a piece of gift-wrapping paper had a length of 5 feet and a width of x\begin{align*}x\end{align*} feet. When wrapping a present, you cut off a vertical strip of 1 foot so that the width of the piece of gift-wrapping paper decreased by 1 foot. What would be the area of the remaining paper in square feet? Could you write this expression in simplified form? In this Concept, you'll learn to use the Distributive property so that you can simplify expressions like the one representing this area.
### Guidance
At the end of the school year, an elementary school teacher makes a little gift bag for each of his students. Each bag contains one class photograph, two party favors, and five pieces of candy. The teacher will distribute the bags among his 28 students. How many of each item does the teacher need?
#### Example A
You could begin to solve this problem by deciding your variables.
Let p=photograph, f=favors,\begin{align*}p=photograph, \ f=favors,\end{align*} and c=candy.\begin{align*}c=candy.\end{align*}
Next you can write an expression to represent the situation: p+2f+5c.\begin{align*}p + 2f + 5c.\end{align*}
There are 28 students in class, so the teacher needs to repeat the bag 28 times. An easier way to write this is 28(p+2f+5c).\begin{align*}28 \cdot (p + 2f + 5c).\end{align*}
We can omit the multiplication symbol and write 28(p+2f+5c)\begin{align*}28(p + 2f + 5c)\end{align*}.
Therefore, the teacher needs 28p+28(2f)+28(5c)\begin{align*}28p + 28(2f) + 28(5c)\end{align*} or 28p+56f+140c.\begin{align*}28p + 56f + 140c.\end{align*}
The teacher needs 28 photographs, 56 favors, and 140 pieces of candy to complete the end-of-year gift bags.
When you multiply an algebraic expression by another expression, you apply the Distributive Property.
The Distributive Property: For any real numbers M, N,\begin{align*}M, \ N,\end{align*} and K\begin{align*}K\end{align*}:
M(N+K)=MN+MKM(NK)=MNMK\begin{align*}& M(N+K)= MN+MK\\ & M(N-K)= MN-MK\end{align*}
#### Example B
Determine the value of 11(2+6)\begin{align*}11(2 + 6)\end{align*} using both Order of Operations and the Distributive Property.
Solution: Using the Order of Operations: 11(2+6)=11(8)=88.\begin{align*}11(2 + 6) = 11(8)= 88.\end{align*}
Using the Distributive Property: 11(2+6)=11(2)+11(6)=22+66=88.\begin{align*}11(2 + 6) = 11(2) + 11(6)= 22 + 66 = 88.\end{align*}
Regardless of the method, the answer is the same.
#### Example C
Simplify 7(3x5).\begin{align*}7(3x - 5).\end{align*}
Solution 1: Think of this expression as seven groups of (3x5)\begin{align*}(3x -5)\end{align*}. You could write this expression seven times and add all the like terms.
(3x5)+(3x5)+(3x5)+(3x5)+(3x5)+(3x5)+(3x5)=21x35\begin{align*}(3x-5)+(3x-5)+(3x-5)+(3x-5)+(3x-5)+(3x-5)+(3x-5)=21x-35\end{align*}
Solution 2: Apply the Distributive Property.
7(3x5)=7(3x)+7(5)=21x35\begin{align*}7(3x-5)= 7(3x)+7(-5)= 21x-35\end{align*}
### Guided Practice
Simplify 27(3y211).\begin{align*}\frac{2}{7} (3y^2 - 11).\end{align*}
Solution: Apply the Distributive Property.
27(3y2+11)=27(3y2)+27(11)=27(3y2)1+27(11)1=2×3y27×1+2×(11)7×1=6y27+227=6y27227\begin{align*}&\frac{2}{7} (3y^2 + -11)= \frac{2}{7} (3y^2) + \frac{2}{7}(-11)=\\ &\frac{2}{7} \frac{(3y^2)}{1} + \frac{2}{7}\frac{(-11)}{1}=\frac{2 \times 3y^2}{7 \times 1} + \frac{2\times (-11)}{7\times 1}=\\ &\frac{6y^2}{7}+\frac{-22}{7}=\frac{6y^2}{7}-\frac{22}{7}\end{align*}
### Practice
Use the Distributive Property to simplify the following expressions.
1. (x+4)2(x+5)\begin{align*}(x + 4) - 2(x + 5)\end{align*}
2. 12(4z+6)\begin{align*}\frac{1}{2}(4z + 6)\end{align*}
3. (4+5)(5+2)\begin{align*}(4 + 5)-(5 + 2)\end{align*}
4. (x+2+7)\begin{align*}(x + 2 + 7)\end{align*}
5. 0.25(6q+32)\begin{align*}0.25 (6q + 32)\end{align*}
6. y(x+7)\begin{align*}y(x + 7)\end{align*}
7. 4.2(h11)\begin{align*}-4.2(h - 11)\end{align*}
8. 13x(3y+z)\begin{align*}13x(3y + z)\end{align*}
9. 12(xy)4\begin{align*}\frac{1}{2}(x - y) - 4\end{align*}
10. 0.6(0.2x+0.7)\begin{align*}0.6(0.2x + 0.7)\end{align*}
11. (2j)(6)\begin{align*}(2 - j)(-6)\end{align*}
12. 4(m+7)6(4m)\begin{align*}4(m + 7) -6(4 - m)\end{align*}
13. 5(y11)+2y\begin{align*}-5(y - 11) + 2y\end{align*}
Mixed Review
1. Translate the following into an inequality: Jacob wants to go to Chicago for his class trip. He needs at least $244 for the bus, hotel stay, and spending money. He already has$104. How much more does he need to pay for his trip?
2. Underline the math verb(s) in this sentence: 6 times a number is 4 less than 16.
3. Draw a picture to represent 334\begin{align*}3 \frac{3}{4}\end{align*}.
4. Determine the change in y\begin{align*}y\end{align*} in the equation y=16x4\begin{align*}y = \frac{1}{6} x-4\end{align*} between x=3\begin{align*}x=3\end{align*} and x=9\begin{align*}x=9\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English Spanish
Distributive Property
For any real numbers $M, \ N,$ and $K$: $&M(N+K)= MN+MK\\ &M(N-K)= MN-MK$
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# In a marriage ceremony of her daughter Poonam, Ashok has to make arrangements for the accommodation of 150 persons. For this purpose, he plans to build a conical tent in such a way that each person have $4\;{\text{sq}}{\text{.}}\,{\text{meters}}$ of the space on ground and $20\;{\text{cubic}}\;{\text{meters}}$ of air to breath. What would be the height of the conical tent? A. $20\;{\text{m}}$ B. $15\;{\text{m}}$ C. $12\;{\text{m}}$ D. $30\;{\text{m}}$
Last updated date: 21st Jun 2024
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Hint: To solve the given problem, first of all we will have to find the area of the tent for the 150 people as the area for each person is given. Then we will have to find the volume for 150 people as for the person it is given. Then by using the formula of volume of cone we can find the height of the conical tent.
Given:
The total number of people in the ceremony is $150$ .
The space required on the ground for each person is $4\;{\text{sq}}{\text{.}}\;{\text{meters}}$ .
Volume of air required by one person is $20\;{\text{cubic}}\;{\text{meters}}$ .
The following is the schematic diagram of the conical tent.
For the conical tent, the area of the base of the tent will be,
$\Rightarrow A = 150 \times 4\;{\text{sq}}{\text{.}}\;{\text{meters}}\\ \Rightarrow A = 600\;{\text{sq}}{\text{.}}\,\;{\text{meters}}$
The formula for the area of the base of the cone is $\pi {r^2}$ .
Therefore, $\pi {r^2} = 600\;{\text{sq}}{\text{.}}\;{\text{meters}}$.
So, the above expression of area can be written as:
${r^2} = \dfrac{{600\;{\text{sq}}{\text{.}}\;{\text{meters}}}}{\pi }\\ {r^2} = \dfrac{{2100}}{{11}}\;{\text{sq}}{\text{.}}\;{\text{meters}}$
Volume of air required by 150 people will be,
$v = 150 \times 20\;{\text{cubic meters}} = 3000\;{\text{cubic meters}}$
The formula for the volume of cone is $v = \dfrac{1}{3}\pi {r^2}h$ .
Substitute $3000\;{\text{cubicmeters}}$ for $v$ and $\dfrac{{2100}}{{11}}\;{\text{sq}}{\text{.}}\;{\text{meters}}$ for ${r^2}$ in the above expressions.
$3000\;{\text{cubic meters}} = \dfrac{1}{3}\pi \times \dfrac{{2100}}{{11}}\;{\text{sq}}{\text{.}}\;{\text{meters}} \times h$
Rearrange the above expression.
$\Rightarrow h=\dfrac{{3000\;{\text{cubic meters}}}}{{\dfrac{1}{3}\pi \times \dfrac{{2100}}{{11}}\;{\text{sq}}{\text{.}}\;{\text{meters}}}}\\ h=\dfrac{{3000}}{{\dfrac{1}{3} \times \dfrac{{22}}{7} \times \dfrac{{2100}}{{11}}}}{\text{meters}}\\ h = 15\;{\text{meters}}$
Hence, the height of the conical tent will be $15\;{\text{m}}$ .
So, the correct answer is “Option B”.
Note: In the question, the data is given for each person, so make sure to calculate the data for the 150 people as the total volume occupied will be the volume occupied by 150 people. The calculations should be done in the standard units only.
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Converting Fractions into Decimals and Percentages
Hello and welcome to this video about converting fractions to decimals and percents! In this video, we will explore:
Before we get started, let’s review a couple of key concepts that we will use to help the math make sense.
Consider the fraction $$\frac{10}{10}$$. One way to think about fractions is to think of them as division.
$$\frac{10}{10}=1$$
In other words, ten-tenths is one whole.
$$\frac{10}{10}$$ can be written equivalently as $$\frac{100}{100}$$. The fraction bar can be said as “per,” so this expression can be said as “100 per 100.” The word “percent” literally means “per 100,” so “100 per 100” means 100 percent.
Therefore, when the same number is divided by itself, the result as a decimal is 1 and as a percent is 100%.
But what happens when our fraction is less than 1? Let’s take a look.
Consider the fraction $$\frac{1}{4}$$. Visually, this is what’s happening:
We can see in the diagram that $$\frac{1}{4}$$ of the whole is $$\frac{25}{100}$$. $$\frac{25}{100}$$ means “25 per 100,” so it equals 25%.
Now let’s figure out how to convert this into a decimal.
We’re going to take our fraction $$\frac{1}{4}$$, which is the same as saying 1 divided by 4.
Dividing this way doesn’t look like it will work. But using our knowledge of place value, we can make it work:
First, rewrite the 1 as 1.0. Instead of 1, the dividend is now ten tenths.
Second, 4 ones will go into ten-tenths two-tenths times.
Third, $$4\times 0.2=0.8$$.
Fourth, $$1.0 -0.8 = 0.2$$.
Fifth, we’re gonna rewrite the original dividend as 100-hundredths and bring the new 0 down.
Sixth, 4 ones goes into 20-hundredths 5-hundredths times.
Seventh, we’re gonna multiply $$4\times 0.05=0.20$$. Then we’ll subtract to get 0.
This means $$\frac{1}{4}=0.25$$.
How to Convert Fraction to Percent
Let’s see this work with a non-unit fraction, like $$\frac{3}{16}$$.
Here’s what the division looks like. The sequence of adding a decimal and dividing repeats as often as necessary until either the remainder is 0 or the decimal begins to repeat.
So, $$\frac{3}{16}=0.1875$$ and $$0.1875=\frac{1,875}{10,000}=\frac{18.75}{100}$$. Therefore, 0.1875 is the same as 18.75% because it’s 18.75 per, 100.
How about a repeating decimal? Everything is the same and the process can be stopped when it is clear that the decimal repeats. Here’s a quick example: $$\frac{1}{3}$$.
The process of subtracting 9 units from 10 units repeats, causing the decimal to repeat.
Lastly, consider a fraction that is greater than 1, such as $$\frac{5}{2}$$.
Visually, here’s what we have:
We can see that the shaded quantity is $$\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}$$. Each $$\frac{1}{2}=50$$, so we can say equivalently $$\frac{250}{100}$$, which, as a decimal, is 2.5 (meaning $$2\frac{1}{2}$$ wholes, which the diagram shows). We can also equivalently say 250 per 100, or 250%.
And the same works for the division as well:
Thanks for watching and happy studying!
How do you change a fraction into a percent?
A
Converting a fraction to a percent can be accomplished in $$2$$ quick steps.
Step 1: Divide the numerator by the denominator.
Step 2: Multiply this decimal by $$100$$.
For example, $$\frac{2}{5}$$ shows a numerator of $$2$$, and a denominator of $$5$$.
Step 1:
$$5\phantom{.}$$ $$\phantom{.}2$$
is the same as
$$5\phantom{.}$$ $$\phantom{.}2$$ . $$0$$ $$0$$
, which equals $$0.4$$ when using the process of long division.
Step 2: $$0.4\times100=40$$. This means that the fraction $$\frac{2}{5}$$ represents $$40\%$$.
How do I turn fractions into decimals?
A
Fractions can be converted into decimals by using long division. For example, the fraction $$\frac{4}{25}$$ is expressing “$$4$$ divided by $$25$$”. This can be set up as 25⟌4. Place a decimal and zeros after the $$4$$ so that it becomes $$4.00$$. Now we have 25⟌4.00. $$25$$ does not go into $$4$$, but it does go into $$40$$, one time. From here, the standard rules of long division apply. The result will be $$0.16$$. The important thing to remember is that the fraction bar represents division, so fractions can be calculated as decimals by dividing the numerator by the denominator.
What is the meaning of a fraction bar?
A
A fraction represents “part” out of “whole”. The fraction bar separating the “part” and the “whole” represents division. For example, the fraction $$\frac{5}{9}$$ represents $$5$$ parts out of $$9$$ parts, or “$$5\text{ divided by }9$$”. Another way to think about this, is that the fraction bar represents “per”. For example, $$\frac{40}{100}$$ represents “$$40\text{ per }100$$”.
What does percent mean?
A
Percent means “$$\text{per }100$$”. For example, $$96\%$$ represents $$96$$ per $$100$$. This also represents the fraction $$\frac{96}{100}$$. All percents can be expressed as a fraction because percents are always out of $$100$$. Percents will sometimes be smaller than $$1$$, and larger than $$100$$. For example, the percent $$125\%$$ represents $$\frac{125}{100}$$, and the percent $$0.5\%$$ represents $$\frac{0.5}{100}$$.
Why do you divide a fraction to turn it into a decimal?
A
Fractions represent a “part” out of a “whole”. The fraction bar separating the “part” and the “whole” represents division. This means that all fractions can be converted into decimals by dividing the numerator by the denominator.
For example, the fraction $$\frac{4}{5}$$ represents $$4$$ out of $$5$$, or $$4$$ divided by $$5$$. This fraction can be converted into a decimal by dividing $$4$$ by $$5$$. 5⟌4 can be solved if a decimal point and zeros are added to the right of the $$4$$. 5⟌4 becomes 5⟌4.00. $$5$$ goes into $$40$$ $$8$$ times, and the rest of the process follows the standard rules of long division. The result is $$0.8$$.
Practice Questions
Question #1:
What is $$\frac{42}{100}$$ as a percent?
42%
4.2%
0.42%
420%
The correct answer is 42%. Percent means “per 100,” so to turn a fraction into a percent, convert the denominator to 100 and the numerator is your percentage.
Question #2:
What is $$\frac{16}{100}$$ as a decimal?
16.00
0.016
0.16
1.6
The correct answer is 0.16. Since 16 is over 100, write the decimal so that 6 is in the hundredths place, which means 1 will be in the tenths place.
Question #3:
Which of the following is not a proper way to rewrite $$\frac{96}{1,000}$$?
0.096
96%
9.6%
0.09600
The correct answer is 96%. $$\frac{96}{1,000}$$ as a decimal is 0.096 and adding zeroes to the end of a decimal number doesn’t change its value, so 0.09600 is equivalent to 0.096. $$\frac{96}{1,000}$$ as a percentage is 9.6%. 96% as a fraction is $$\frac{96}{100}$$.
Question #4:
What is $$\frac{7}{20}$$ as a percent?
7%
20%
28%
35%
The correct answer is 35%. Percent means “per 100,” so to turn a fraction into a percent, convert the denominator to 100 and the numerator is your percentage. To convert $$\frac{7}{20}$$ to have a denominator of 100, multiply both the numerator and the denominator by 5, which gives you $$\frac{35}{100}$$. 35 is the numerator, so $$\frac{7}{20}=35%$$.
Question #5:
What is $$\frac{23}{125}$$ as a decimal?
0.23
1.25
1.86
0.184
The correct answer is 0.184. To convert a fraction to a decimal, first convert the fraction so it has a denominator of a multiple of 10. This can be done by multiplying $$\frac{23}{125}$$ by $$\frac{8}{8}$$. $$\frac{23}{125}\times\frac{8}{8}=\frac{184}{1,000}$$, and since 184 is over 1,000, place the 4 in the thousandths place of the decimal, which puts 8 in the hundredths place, and 1 in the tenths place, 0.184.
Question #6:
Write the fraction $$\frac{4}{5}$$ as a decimal.
0.8
0.6
0.5
0.4
The correct answer is 0.8. To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 4 by 5. Long division can be used for this:
Question #7:
Write the fraction $$\frac{5}{8}$$ as a decimal.
5.8
0.58
0.625
0.562
The correct answer is 0.625. To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 5 by 8. Long division can be used for this:
Question #8:
Write the fraction $$\frac{1}{5}$$ as a decimal.
0.5
0.2
0.1
2.0
The correct answer is 0.2. To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 1 by 5. Long division can be used for this:
Question #9:
Write the fraction $$\frac{5}{6}$$ as a decimal.
0.3222
0.8333…
0.8314
0.2333…
The correct answer is 0.8333. To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 5 by 6. Long division can be used for this:
Question #10:
Write the fraction $$\frac{3}{11}$$ as a decimal.
0.3111…
0.3131…
0.3535…
0.2727…
The correct answer is 0.2727…To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 3 by 11. Long division can be used for this:
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# Problem Solving on the Coordinate Plane
Videos and solutions to help Grade 6 students learn how solve problems related to the distance between points that lie on the same horizontal or vertical line and the coordinate plane.
New York State Common Core Math Grade 6, Module 3, Lesson 19
Related Topics:
Lesson Plans and Worksheets for Grade 6
Lesson Plans and Worksheets for all Grades
### New York State Common Core Math Module 3, Grade 6, Lessons 19
Lesson 19 Student Outcomes
Students solve problems related to the distance between points that lie on the same horizontal or vertical line.
Students use the coordinate plane to graph points, line segments and geometric shapes in the various quadrants and then use the absolute value to find the related distances.
Opening Exercise
In the coordinate plane, find the distance between the points using absolute value.
Exercises
1. Locate and label (4, 5) and 4, –3 ) Draw the line segment between the endpoints given on the coordinate plane. How long is the line segment that you drew? Explain.
2. Draw a horizontal line segment starting at (4, -3) that has a length of units. What are the possible coordinates of the other endpoint of the line segment? (There is more than one answer.)
Which point do you choose to be the other endpoint of the horizontal line segment? Explain how and why you chose that point. Locate and label the point on the coordinate grid.
3. The two line segments that you have just drawn could be seen as two sides of a rectangle. Given this, the endpoints of the two line segments would be three of the vertices of this rectangle.
a. Find the coordinates of the fourth vertex of the rectangle. Explain how you find the coordinates of the fourth vertex using absolute value.
b. How does the fourth vertex that you found relate to each of the consecutive vertices in either direction? Explain.
c. Draw the remaining sides of the rectangle.
4. Using the vertices that you have found and the lengths of the line segments between them, find the perimeter of the rectangle.
5. Find the area of the rectangle.
6. Draw a diagonal line segment through the rectangle with opposite vertices for endpoints. What geometric figures are formed by this line segment? What are the areas of each of these figures? Explain.
7. Construct a rectangle on the coordinate plane that satisfies each of the criteria listed below. Identify the coordinate of each of its vertices.
• Each of the vertices lies in a different quadrant.
• Its sides are either vertical or horizontal.
• The perimeter of the rectangle is 28 units.
Using absolute value, show how the lengths of the sides of your rectangle provide a perimeter of 28 units.
Lesson Summary
The length of a line segment on the coordinate plane can be determined by finding the distance between its endpoints.
You can find the perimeter and area of figures such as rectangles and right triangles by finding the lengths of the line segments that make up their sides, and then using the appropriate formula.
Lesson 19 Classwork Lesson 19 Problem Set
1. One endpoint of a line segment is (-3, -6). The length of the line segment is 7 units. Find four points that could serve as the other endpoint of the given line segment.
2. Two of the vertices of a rectangle are (1, -6) and (-8, -6). If the rectangle has a perimeter of 26 units, what are the coordinates of its other two vertices?
3. A rectangle has a perimeter of 28 units, an area of 48 square units, and sides that are either horizontal or vertical. If one vertex is the point (-5,-7) and the origin is in the interior of the rectangle, find the vertex of the rectangle that is opposite (-5,-7).
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.
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# Class 9 Maths MCQ – Application of Heron’s Formula in finding Areas of Quadrilaterals
This set of Class 9 Maths Chapter 12 Multiple Choice Questions & Answers (MCQs) focuses on “Application of Heron’s Formula in finding Areas of Quadrilaterals”.
1. An umbrella is made by stitching 8 triangular pieces of cloth of two different colours, each piece measures 60cm, 60cm and 20cm. How much cloth of each colour is required for the umbrella?
a) $$50\sqrt{35} cm^2$$
b) $$25\sqrt{65} cm^2$$
c) $$50\sqrt{45} cm^2$$
d) $$25\sqrt{55} cm^2$$
Explanation: s = $$\frac{a+b+c}{2}=\frac{60+60+20}{2} = \frac{140}{2}$$ = 70
According to heron’s formula, area of the triangle = $$\sqrt{s*(s-a)*(s-b)*(s-c)}$$
= $$\sqrt{70*(70-60)*(70-60)*(70-20)}$$
= $$\sqrt{70*10*10*50}$$
= 100$$\sqrt{35} cm^2$$
This area is the combined area of both colours.
Hence, area of each colour = $$\frac{100\sqrt{35}}{2}$$
= $$50\sqrt{35} cm^2$$.
2. A triangle and a parallelogram has same base and same area as shown in the diagram below. Dimensions of triangle are 28cm, 26cm and 30cm with 28cm being the base. What is the height of the parallelogram?
a) 15cm
b) 10cm
c) 12cm
d) 18cm
Explanation: a = 28, b = 26 and c = 30 cm.
s = $$\frac{a+b+c}{2}=\frac{28+26+30}{2} = \frac{84}{2}$$ = 42
According to heron’s formula, area of the triangle = $$\sqrt{s*(s-a)*(s-b)*(s-c)}$$
= $$\sqrt{42*(42-28)*(42-26)*(42-30)}$$
= $$\sqrt{42*14*16*12}$$
= 336 cm2
It is given that area of the triangle and the parallelogram is same.
Now, we know that area of a parallelogram = base * perpendicular (height of a parallelogram)
Therefore, 28 * height = 336
height = $$\frac{336}{28}$$
height = 12cm.
Sanfoundry Global Education & Learning Series – Mathematics – Class 9.
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Functions and Relations
# Functions and Relations
## Sets
Consider two sets, which we will call S and T.
There are several ways in which the two sets S and T may be related:
• The sets may be identical, that is, every element of one set is also an element of the other set.
• The sets may be disjoint, that is, no element belongs to both sets. There is no overlap between the sets.
• Set S may be a proper subset of set T. That is, every element of S is also an element of T, but T has some elements that are not in S.
• Set T may be a proper subset of set S.
• The sets may overlap (some elements are in both S and T), without either being a proper subset of the other (each set contains some elements that are not in the other set).
## Relations
However, this page isn't really about sets. This page is really about relations and functions. For clarity, all our examples use disjoint sets (see Figure 1). However, the same definitions apply regardless of the relationship between the two sets. The only difference is that, if we didn't use disjoint sets, the examples would be harder to figure out.
Suppose we have two sets, S and T. A relation on S and T is a set of ordered pairs, where the first thing in each pair is an element of S, and the second thing is an element of T.
For example, suppose S is the set {A, B, C, D, E}, while T is the set {W, X, Y, Z}. Then one relation on S and T is {(A,Y), (C,W), (C,Z), (D,Y)}. There are four ordered pairs in the relation. We can draw this as four arrows going from S to T (see Figure 2). One arrow goes from A to Y; another goes from C to W; another goes from C to Z; and another goes from D to Y.
The purpose of this page is just to define terms. Giving names to things is not important unless you can later use those names to talk about the things. These terms we define here are used throughout mathematics, and are pretty important; but we don't go into any of that here. We just define the terms.
The most important of the terms we will define is function. You have probably seen this word defined in algebra or calculus, and you may think this is another meaning for the same word. It's important to realize that there is only one meaning in mathematics for the word "function", and this is it. Moreover, the definition given here is the "best" definition because, since the definition is given in terms of sets, it is the most general and most applicable definition. Any other definition is just a special case.
Anyway, to continue.
The domain of a relation on S and T is the set of elements of S that appear as the first element in an ordered pair of the relation. In the relation {(A,Y), (C,W), (C,Z), (D,Y)}, the domain is {A, C, D}. If you look at Figure 2, these are the elements of S that have arrows coming out of them.
The range of a relation on S and T is the set of elements of T that appear as the second element in an ordered pair of the relation. In the relation {(A,Y), (C,W), (C,Z), (D,Y)}, the range is {W, Y, Z}. If you look at Figure 2, these are the elements of T that have arrows pointing to them.
For some reason, the word codomain has become popular as a synonym for "range." I think it's an ugly word. If anyone has an explanation for why this word has become popular, I would very much like to hear it.
In Figure 2, not every element of T has an arrow pointing to it. That is, there are some elements of T (in particular, the element X) that do not occur as a second element of an ordered pair. We say that the relation is into T.
Suppose we have a relation in which every element occurs (at least once) as a second element of an ordered pair. For example, the relation {(A,Y), (B, X), (C,W), (C,Z), (D,Y)} (See Figure 3) is just like the previous relation, except that it also contains the ordered pair (B, X). The relation is onto: every element of set T has an arrow (at least one) pointing to it.
While the word "onto" has a precise definition (the range of the relation is the set T itself), the word "into" is not usually so well defined. "Into" could be used to mean "not onto" (there is at least one element of T that does not have an arrow pointing to it), or it could mean "not necessarily onto", that is, there might be some element of T that does not have an arrow pointing to it. Different authors might choose to define "into" in different ways, or not define it precisely, or just not define it at all.
Suppose we put in every possible arrow from S to T. That is, from each element of S we draw arrows to each element of T (see Figure 4). This "largest possible relation" is called the Cartesian product of S and T. Every other relation on S and T is a subset of the Cartesian product.
(How is it possible for a relation to be a subset of another relation? Remember that a relation is just a set of ordered pairs. Or, in our pictures, a relation is a set of arrows.)
Next we will say that a relation is one to one, or 1-1, if no element of S occurs more than once as the first element of an ordered pair, and no element of T occurs more than once as a second element of an ordered pair. In other words, no element of S or T has more than a single arrow attached to it. (See Figure 5).
(This definition holds even when S and T are not disjoint, but the picture is a little more confusing. An element that is in both S and T could have a single arrow attached to it, as before, but at both ends!)
## Functions
Now we get to what is perhaps the most important term. Suppose every element of S occurs exactly once as the first element of an ordered pair. In our pictures (see Figure 6), every element of S has exactly one arrow coming from it. This kind of relation is called a function.
Another word for function is mapping. A function is said to map an element in its domain to an element in its range.
Here are some important facts about a function from S to T:
1. Every element in S in the domain of the function; that is, every element of S is mapped to some element in the range. (If some element in S has no mapping (arrow), then the relation is sometimes called a partial function, but it is not a function!)
2. No element in the domain maps to more than one element in the range.
3. The mapping is not necessarily onto; some elements of T may not be in the range.
4. The mapping is not necessarily 1-1; some elements of T may have more than one element of S mapped to them.
5. S and T need not be disjoint.
To tell whether a relation on S and T is a function, you can ignore T altogether. Just look at S. If every element of S has one and only one element coming out of it, then the relation is a function.
## Kinds of functions
There are three kinds of functions that are important enough to have special names: surjection, injection, and bijection.
Since the domain of a function from S to T is the entire set S (by the definition of function), and since the range of a surjective function from S to T is the entire set T, this means that every element of both sets is in the relation (has an arrow connected to it).
Also note that if you have a surjection from S to T, S may have more elements than T, or it may have the same number of elements, but it cannot have fewer elements than T. This is because every element of S has exactly one arrow emanating from it, but each element of T has at least one arrow pointing to it, and could have many.
An injection or one-to-one function from S to T is a function that is one to one. That is, every element in both S and T has at least one arrow attached to it. By the definition of function, of course, every element of S has exactly one arrow emanating from it, no more, no less. A one-to-one function also requires that every element of T can have no more than one arrow pointing to it; but there could be elements of T that do not have any arrows pointing to them.
Figure 7 shows an injection. However, we had to modify the sets a little bit to get this example. A little thought will show that, with an injection from S to T, T must have at least as many elements as S. To get our example, we removed some elements from S. (We could equally as well have added elements to T.)
Finally, a function that is both 1-1 and onto is called a bijection. Such a function maps each and every element of S to exactly one element of T, with no elements left over. shows a bijection. Again, we had to modify the sets a little because, with a bijection, sets S and T have to have exactly the same number of elements.
A bijection is particularly interesting because this kind of function has an inverse. If you took a bijection from S to T and reversed the direction of all the arrows, you would have a bijection from T to S. This new function would be the inverse of the original function.
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By now, surely everyone knows that poker is a game of skill as well as a game of chance. And it’s hard to consider yourself a skilled poker player if you don’t understand how probability relates to the game and how you play it.
This post is meant to serve as an introduction to probability in general and to probability in poker. If you’re already an intermediate or advanced poker player, you’re probably familiar with most of these concepts. I remember how complicated I thought these concepts were when I started playing, so I tried to write the post I wish I’d been able to read.
Here’s how to understand probability in poker:
1- The 1st Step Is to Understand Probability in General
Probability is a specific branch of math, like algebra or geometry. It concerns itself with the study of how likely events are to happen. Probability is also the word used to describe how likely something is to happen or not happen. You can express an event’s probability in multiple ways, but understand that every event has a probability of between 0 and 1. 0 represents something that will never happen. The closer your probability gets to 0, the less likely it is to happen. 1 represents something that will always happen. The closer your probability gets to 1, the more likely it is to happen.
Most people understand probabilities at an almost intuitive level when they’re presented as percentages. This is how the weather is presented. When we’re told that there’s a 50% chance of rain, we know that it’s just as likely to be raining as not. If we say you have a 20% chance of winning a game, we know that means you’ll win 1 out of 5 times.
You could, though, represent a probability as a fraction. In gambling, that’s how it starts most of the time. The simplest probability formula is this: An event’s probability is the number of ways an event can happen by the total possible number of outcomes.
Here’s an example from poker:
You have a 52-card deck. You have 13 cards in each of 4 suits. If you draw a card at random and want to know the probability that it’s going to be a heart, the probability is 1/4. There are 4 suits, and there are an equal number of cards in each of those suits.
There are 4 cards of each rank, too, so you can also calculate the probability of drawing a card of a specific rank—an ace, for example. That probability is 1/13. Once you know the probability as a fraction, it’s easy to translate it to a percentage. You convert it to a decimal, and that’s done by dividing the numerator by the denominator. Then you move the decimal 2 places to the right.
For 1/13, we’re looking at 1 divided by 13, which is 0.076923077. Moving the decimal 2 places to the right gives me the following percentage: 7.6923077%. Most people round percentages off to the nearest whole number or to 1 or 2 digits after the decimal. Now we’re looking at 8% or 7.7%. An even more useful way of expressing a probability in poker is by expressing it in odds format. This is simply a comparison of the number of ways something can’t happen with the number of ways it can happen.
If the probability of something is 1/13, the odds are 12 to 1. There are 12 ways it can’t happen, and only 1 way it can happen. This becomes useful because you can compare the payout odds on a bet with the odds of winning that bet to estimate whether it’s a good or bad bet.
2- Probabilities for Multiple Events
When you’re calculating probabilities for multiple events, you either add those probabilities together or you multiply them by each other. The difference is based on the words “OR” and “AND.” If you want to know the probability that A “or” B will happen, you add the 2 probabilities together.
Here’s an example:
You want to know the probability that you’ll get an ace OR a king from a 52-card deck. That’s 1/13 + 1/13, or 2/13. If you want to know the probability that A “and” B will happen, you multiply the 2 probabilities together.
Here’s an example:
You want to calculate the probability that the 1st card you get will be an ace, and that the 2nd card you get will also be an ace. That’s 1/13 X 3/51, or 3/661. Most people would reduce that and say it’s roughly 1/221. You’ll notice that the 1st probability used 13 as the denominator and the 2nd used 51 as the denominator. That’s because once you’ve dealt a card out of the deck, there are only 51 cards left.
Most poker probability calculations are going to involved multiple cards which have left the deck. The only cards you subtract from the total are the ones you know the rank and suit of. Just because another player has those cards in the hole doesn’t mean you subtract them from the number in the deck for purposes of calculating the probability.
Here’s an example:
You’re playing Texas holdem with 3 other players. You have 2 hole cards, both of which are of the same suit. The probability of getting another card of the same suit is 11/50. There are 13 cards in each suit, but 2 of them are in your hand. There are 50 cards that are unknown. Those other hearts might be in your opponents’ hands. That’s why they’re including in the 50 cards.
Let’s look at another example:
You’re playing Texas holdem, and you have 2 hearts in the hole. On the flop, 2 more hearts show up, along with a 3rd card of another rank. Now you have 4 hearts, so you just need 1 more to hit your flush. What’s the probability that the next card will make your hand? 13 – 4 = 9. That’s how many hearts are left in the deck. 52 – 5 = 47. That’s how many cards are left in the deck. The probability of getting a heart on the next card is 9/47. Most gamblers would simplify that to about 1/5.
3- How Pot Odds Relate to the Odds in Poker
Pot odds are a measure of how much you’ll get paid if you win a bet in poker. It’s the odds offered by the pot. Here’s a simple example:
If there’s \$10 in the pot, and another player bets \$10, you have to call \$10 to play. The pot odds in this situation are the \$10 in the pot plus the \$10 bet you’re calling, or \$20. If you win this bet, you’re getting pot odds of 2 to 1 on your money. If you estimate that your odds of winning this pot are 2 to 1 or less, then this is a profitable (or at least break-even bet). But if the odds of winning are higher than that, this is a bet with a negative expectation.
One of the key skills in poker is estimating your odds of winning and comparing it to the payout odds for the bet. If you can do this well, you can profit from poker consistently. It’s just a matter of putting money into the pot when you have a positive expectation and folding when you have a negative expectation.
4- Outs in Poker and How They Relate to Probabilities
One of the ways to estimate your odds of winning a hand is by calculating your “outs.” An out, in poker, is a card that will improve your hand and make it the winner. The tricky part is accounting for what cards your opponents might be holding.
Here’s an example:
You’re playing heads up with another player. You have a pair of jacks in the hole, but there’s a queen on the flop. You’re pretty sure your opponent either has a queen in the hole or a pair of aces or kings. You’re dominated. But if you catch another jack on the turn or the river, you’ve got him beat. Since you already have 2 of the jacks, there are only 2 jacks left in the deck. This means you have 2 outs.
A quick and easy way of approximating the odds of getting your outs is by multiplying them by 2% for both the turn and for the river. In this case, you’re looking at 2 X 2 X 2, or 6%. That’s about the same as 16 to 1. If there’s \$40 in the pot, and it costs you \$10 to call a bet, you’re only getting 4 to 1 on your money. In the long run, you’re going to lose money in this situation.
On the other hand, if there’s \$400 in the pot, and it costs you \$20 to call a bet, you’re getting 20 to 1 on your money. You’ll profit in the long run from this bet. In some cases, you’ll also deal with implied odds. This takes into account future bets into the pot.
In other cases, you’ll also deal with partial outs. These are outs that might make you a winner of the hand some of the time. Or sometimes you’ll have an out that will improve your hand, but it might also improve your opponent’s hand enough so that he wins.
5- Probability and Bluffing
Another way probability can be useful in poker is when deciding whether to bluff. It’s especially useful when determining whether a bluff against multiple players is worthwhile. It’s also illustrative of the fact that you don’t need to win with a tactic most of the time to be profitable. Let’s say you’re heads-up with another player, and there’s \$20 in the pot already. You don’t have anything, but you suspect maybe he doesn’t either. You think there’s a 60% chance he’ll fold if you bet \$20 here.
If you’re right in your estimate, then you’ll win \$20 60% of the time, which is worth \$12 in expected value. You’ll also lose \$20 (or at least have to continue to play with it), 40% of the time. That’s worth \$8 in negative expected value. Since \$12 is more than -\$8, this is a profitable bluff.
If you reversed it, though, and estimated that the other player would only fold to a bluff 40% of the time, you’re making a negative expectation play. Bluffing multiple opponents makes it even harder to succeed. If you’re in a pot with 3 other players, and you want to raise, hoping they’ll fold, your probabilities need to be multiplied. Let’s say there’s a 50% chance that each of them will fold.
You want to calculate the probability that player 1 AND player 2 AND player 3 fold. To calculate that, you multiply: (50% X 50% X 50% = 12.5% ). You need roughly 8 to 1 pot odds to make that a break-even bluff. There’d have to be a lot of dead money in that pot.
If you semi-bluff, which is when you bet with a hand that might improve, you get to add the probability of making your hand to those odds. If you have a 12.5% chance of winning when the other players fold, AND a 12.5% chance of hitting a hand that will win, you have a 25% chance of winning the pot. You only need 4 to 1 from the pot to make this semi-bluff worthwhile. Of course, if you’re lousy at estimating these probabilities, you’ll probably not profit at all. That’s one of the things that makes poker such a great game.
6- Probability in Video Poker
I hesitate to even mention video poker in this post, as it’s not really a poker game. But it does use the poker hand rankings and the probabilities of a 52-card deck. If you understand probability in poker well enough, it can inform your video poker strategy very well.
Video poker is like a slot machine game with payouts based on combinations that come from poker hand rankings instead of just arbitrary symbols. When you look at probabilities related to video poker, you’re most interested in the game’s payback percentage. This is a function of the probability of getting each hand that pays out multiplied by the probability of winding up with that hand.
For example:
In a Jacks or Better game, you have a probability of about 21.5% of winning 1 unit on a pair of jacks or higher. That adds 21.5% to the payback percentage for the game. You also have a roughly 12.9% of getting 2 pair. That pays off at 2 for 1, so that adds (12.9% X 2) or 25.8% to the payback percentage for the game. As the hands get harder to hit, the payoffs for those hands go up. When you add them all together, you have the payback percentage for the game.
Most video poker games have better payback percentages than slot machines, but only if you know how to make close to the right decisions. Understanding the odds of catching various cards versus keeping various other cards, and taking into account the payouts for those hands, is crucial to getting the best possible payback percentage.
Conclusion
This post was meant to serve as an introduction to probability in poker. You can and should delve deeper into the subject. Any good book on poker will cover pot odds versus payoff odds. It will also cover calculating outs.
The great thing about poker, of course, is that it’s a game of incomplete information, so success is tied to learning how to account for the potential range of hands your opponent might have, too.
Related Articles
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# What is the distance between (–10, –2, 2) and (4, –4, –2)?
Apr 8, 2017
See the entire solution process below:
#### Explanation:
The formula for calculating the distance between two points is:
$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2} + {\left(\textcolor{red}{{z}_{2}} - \textcolor{b l u e}{{z}_{1}}\right)}^{2}}$
Substituting the values from the points in the problem gives:
$d = \sqrt{{\left(\textcolor{red}{4} - \textcolor{b l u e}{- 10}\right)}^{2} + {\left(\textcolor{red}{- 4} - \textcolor{b l u e}{- 2}\right)}^{2} + {\left(\textcolor{red}{- 2} - \textcolor{b l u e}{2}\right)}^{2}}$
$d = \sqrt{{\left(\textcolor{red}{4} + \textcolor{b l u e}{10}\right)}^{2} + {\left(\textcolor{red}{- 4} + \textcolor{b l u e}{2}\right)}^{2} + {\left(\textcolor{red}{- 2} - \textcolor{b l u e}{2}\right)}^{2}}$
$d = \sqrt{{14}^{2} + {\left(- 2\right)}^{2} + {\left(- 4\right)}^{2}}$
$d = \sqrt{196 + 4 + 16}$
$d = \sqrt{216} = 14.697$ rounded to the nearest thousandth.
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Saturday, December 10, 2022
1.1. Revision
1.1. Revision
A number means something you can count, just like counting the number of coins one has or the number of students in a class. These are natural numbers which are normally used for counting and sometimes called counting numbers
1.1.1. Natural Numbers
Natural numbers start from 1 to infinity ∞. Natural numbers are numbers you will use when counting things. For instance if you have to count the number of Oranges, you will start from 1.
Natural Numbers (N) 1,2,3,4,5,6,7,8,9,10,11,…………
1.1.2. Whole Numbers
Whole numbers start from 0 to infinity ∞. Whole numbers are Natural Numbers together with a zero.
Whole Number ( N): 0,1,2,3,4,5,6,7,8,….
1.1.3. Integers
Integers are Whole Numbers plus negatives.
Integers (Z): …-4,-3,-2,-1,0,1,2,3,4,…
1.1.4. Rational Numbers
Rational Numbers are all numbers of the form a/b, where both a and b are integers. b cannot be zero. Rational numbers include fractions.
Fractions can be numbers which are smaller than 1, i.e 1/2 ; 1/4 and 4/7. These are called proper fractions. Fractions can also be numbers bigger than 1 and they are called improper fractions), i.e. 4/2 ; 5/3 ; 7/2.
1.1.5. Irrational Numbers
Irrational Numbers cannot be expressed as ratio of integers. As decimals they never repeat or terminate.
1.1.6. The Real Numbers
The real numbers is the set of numbers containing all of the rational numbers and all of the irrational numbers.
1.1.7. Non Real Numbers ( Imaginary Numbers)
Not all numbers are real numbers. For instance all square roots of the negative numbers are imaginary numbers or non real numbers. For example √-2 , √-4, √-1 and √-22.
1.2. Rational Exponents and Surds
The laws of exponents can also be extended to include the rational numbers. A rational
number is any number that can be written as a fraction with an integer in the numerator
and in the denominator. We also have the following definitions for working with
rational exponents.
Videos (English)
Exercise 1-2: Laws of Exponents
Exercise 1-3: Rational Exponents and Surds
Exercise 1-4: Simplification of Surds
Exercise 1-5: Rationalising the Denominator
Exercise 1-6: Solving Surd Equations
Exercise 1-8: End of the Chapter Exercises
Videos (Venda)
Exercise 1-2: Laws of Exponents
Exercise 1-3: Rational Exponents and Surds
Exercise 1-4: Simplification of Surds
Exercise 1-5: Rationalising the Denominator
Exercise 1-6: Solving Surd Equations
Videos (Sepedi)
Exercise 1-2: Laws of Exponents
Exercise 1-3: Rational Exponents and Surds
Exercise 1-4: Simplification of Surds
Exercise 1-5: Rationalising the Denominator
Exercise 1-6: Solving Surd Equations
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## Examples and Practice
There are some simple tricks for doing percents problems using mental math. In particular, this lesson will show you how to do mental math for...
• 50%
• 25%
• 10%
• 20%
• 1%
## 50%
Since $50\mbox{%}=\frac{1}{2}$, then the shortcut for finding 50% of a number is to divide by 2.
Example: What is 50% of 850?
Solution: 850 ÷ 2 = 425. So, 50% of 850 is 425.
## 25%
Since $25\mbox{%}=\frac{1}{4}$, then the shortcut for finding 25% of a number is to divide by 4.
Example: What is 25% of 68?
Solution: 68 ÷ 4 = 17. So, 25% of 68 is 17.
## 10%
Since $10\mbox{%}=\frac{1}{10}$, then the shortcut for finding 10% of a number is to divide by 10. NOTE: A trick for dividing by 10 is to move the decimal one place to the left.
Example: What is 10% of 192?
Solution: 192 ÷ 10 = 19.2. So, 10% of 192 is 19.2.
## 20%
Since 20% is twice as much as 10%, the shortcut for finding 20% is to find 10% and then double it.
Example: What is 20% of 41?
Solution: 10% of 41 is 4.1, so 20% is 8.2.
## 1%
Since $1\mbox{%}=\frac{1}{100}$, then the shortcut for finding 1% of a number is to divide by 100. NOTE: A trick for dividing by 100 is to move the decimal two places to the left.
Example: What is 1% of 720?
Solution: 720 ÷ 100 = 7.2, so 1% of 720 is 7.2.
Look at these shopping examples.
50% 25% 10% 20% 1% Divide by 2 Divide by 4 Divide by 10 Or Move decimal to the left ONE space Divide by 5 Or Find 10% and then DOUBLE it Divide by 100 Or Move decimal to the left TWO spaces $80 80 ÷ 2 =$40 80 ÷ 4 = $20 80 ÷ 10 =$8 80 ÷ 10 = 8 8 x 2 = $16 80 ÷ 100 =$0.80 $250 250 ÷ 2 =$125 250 ÷ 4 = $62.50 250 ÷ 10 =$25 250 ÷ 10 = 25 25 x 2 = $50 250 ÷ 100 =$2.50 $4 4 ÷ 2 =$2 4 ÷ 4 = $1 4 ÷ 10 =$0.40 4 ÷ 10 = 0.40 0.40 x 2 = $0.80 4 ÷ 100 =$0.04
# Self-Check
Question 1
What is 25% of 32?
Question 2
How much is 10% of 549?
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# Package 4 - Year 1 and 2 Mathematics - Maths explosion 7 feet follow up
This task offers your child the opportunity to measure length using uniform informal units.
Week 6 - Package 4 - Year 1 and 2 Mathematics - Maths explosion 7 feet follow up
## Things you need
### Ideal
• Cardboard
• Marker
• Activity sheet 1
• Pencil
## Why is this activity important?
• conduct mathematical investigations in their home
• measure lengths using uniform informal units (the cut out of their foot) by placing them end to end with no overlaps
• explore that when an item is longer, more units will be required to measure it.
## Before you start
Check that the video is working and the audio settings are correct for your child.
## What your child needs to know and do
This task encourages your child to conduct mathematical investigations at home where they will need to explore to find solutions. They need to place the foot cut out end to end to get an accurate measure.
## What to do next
View the MathXplosion 7 feet follow up video.
Help your child trace around their foot on cardboard or paper and have them cut it out. This will be the unit that they will use to measure around the home.
Help your child to measure their own height using their foot measure. Are they 7 feet (their own foot) tall?
• Can you find some things that are more than 7 of your feet tall? Write and draw your findings in Activity sheet 1.
• Can you find some things less than 7 of your feet tall? Write and draw your findings in Activity sheet 1.
### Activity too hard?
Make multiple feet so that they don’t have to keep the place of the unit and then move it up.
### Activity too easy?
Measure more things with their foot or and cut out a measure of your foot so that they can investigate what happens when the unit used to measure gets bigger.
• Why is it important to iterate (place them end to end with no gaps) the unit?
• How many of your feet do you think your bed will be? Why?
Measure the same item using different units from the feet of different members of your family.
## Activity sheet 1:
Trace around your foot and then cut it out:
Use your cut out to measure your height. Are you 7 feet tall?
Can you find some things that are more than 7 of your feet tall? Write and draw your findings.
Can you find some things that are less than 7 of your feet tall? Write and draw your findings.
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# Segmenting Data: Quartiles, Deciles, Percentiles
We often talk about the top 25% or top 10% or even top 1% of something. When we are segmenting data into percentages we commonly are talking about quartiles, deciles and percentiles. Quartiles divide the data into four parts; deciles divide the data into 10 parts; percentiles divide the data into 100 parts. Let’s take a look at how these different types of divisions are used.
## Quartiles
The quartiles of a data set divide the data into four equal parts, with one-fourth of the data values in each part. The second quartile position is the median of the data set, which divides the data set in half. To find the median position of the data set, divide the total number of data values (n) by 2. If there are an even number of data values, the median is the value that is the average of the value in the position and the + 1 position. (If there are an odd number of data values, the median is the value in the position.) For example, if the data set has 20 values, then the median is the average of the data values in the = 10th and + 1 = 10 + 1 = 11th position. For example, in the data set below, with 20 values, the median is the average of 9 and 11, which is 10.
The first quartile is the median of the first half of the data set and marks the point at which 25% of the data values are lower and 75% are higher. The third quartile is the median of the second half of the data set and marks the point at which 25% of the data values are higher and 75% lower. In the data set above, there are ten data values in each half, so the first quartile is the average of the values in the fifth and sixth positions (both of which are 5, so the first quartile is 5) and the third quartile is the average of the values in the fifteenth and sixteenth positions (17 and 20, so the third quartile is 18.5).
Quartiles are often used as a measure of spread of the data in what is called the interquartile range (IQR). The IQR is simply the difference between the third quartile and first quartile. Thus, in the sample data set given above, the IRQ is 18.5 – 5 = 13.5. While on its own the IQR is not a very useful measure, it can be useful when comparing the spread of two different data sets that measure the same phenomenon.
## Deciles and Percentiles
Deciles and percentiles are usually applied to large data sets. Deciles divide a data set into ten equal parts. One example of the use of deciles is in school awards or rankings. Students in the top 10% — or highest decile – may be given an honor cord or some other recognition. If there are 578 students in a graduating class, the top 10%, or 58 students, may be given the award. At the opposite end of the scale, students who score in the bottom 10% or 20% on a standardized test may be given extra assistance to help boost their scores.
Percentiles divide the data set into groupings of 1%. Standardized tests often report percentile scores. These scores help compare students’ performances to that of their peers (often across a state or country). The meaning of a percentile score is often misunderstood. A percentile score in this situation reflects the percentage of students who scored at or above that particular group of students. For example, students who receive a percentile ranking of 87 on a particular test received scores that were equal to or higher than 87% of students who took the test. For those who do not understand these scores, they often mistake them for the score the student received on the test.
Growth charts are another common example of an application of percentiles. To help doctors and parents determine if a child is developing normally, his or her measurements are compared to others in the same sex and age groups. The figure below shows a growth chart (from the My Growth Charts website) that gives the percentiles for height and weight for boys ages 0 to 5 years. A two-year-old boy who is 33 inches long, for example, is in the 12th percentile, meaning he is taller than or the same height as only 12% of all boys of his age. However, he weighs 31 pounds, putting him in the 89th percentile, making him heavier than or as heavy as 89% of his peers.
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Problem 1
# Find the point of intersection of each pair of straight lines. $$\begin{array}{l} y=3 x+4 \\ y=-2 x+14 \end{array}$$
Expert verified
The point of intersection of the given pair of straight lines is $$(2, 10)$$.
See the step by step solution
## Step 1: Set the y's equal to each other
Since both equations are in the form y = mx + b, we can set the y's equal to each other to find the value of x: $$3x + 4 = -2x + 14$$
## Step 2: Solve for x
Now we need to solve this equation to find the value of x: Add $$2x$$ to both sides of the equation: $$3x + 2x + 4 = -2x + 2x + 14$$ Simplifying, we get: $$5x + 4 = 14$$ Subtract 4 from both sides of the equation: $$5x + 4 - 4 = 14 - 4$$ Simplifying, we get: $$5x = 10$$ Now, divide by 5: $$\frac{5x}{5} = \frac{10}{5}$$ Simplifying, we get: $$x = 2$$
## Step 3: Substitute x into one of the equations to find y
Now that we have the value of x, we can substitute it into one of the equations to find the value of y. Let's use the first equation: $$y = 3x + 4$$ Substitute x = 2: $$y = 3(2) + 4$$
## Step 4: Solve for y
Now we solve for y: $$y = 6 + 4$$ $$y = 10$$
## Step 5: Write the solution as a coordinate point
Now that we have the values of x and y, we can write the solution as a coordinate point: The point of intersection of the given pair of straight lines is $$(2, 10)$$.
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TEXT
# Reading: Using the Definition of a Logarithm to Solve Logarithmic Equations
We have already seen that every logarithmic equation logb(x) = y is equivalent to the exponential equation by = x. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression. For example, consider the equation log 2(2) + log2(3x – 5) = 3. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x:
$\displaystyle{{log}_{{2}}{({2})}}+{{log}_{{2}}{({3}{x}-{5})}}={3}$ Apply the product rule of logarithms $\displaystyle{{log}_{{2}}{({2}{({3}{x}-{5})})}}={3}$ Distribute $\displaystyle{{log}_{{2}}{({6}{x}-{10})}}={3}$ Apply the definition of a logarithm $\displaystyle{2}^{{3}}={6}{x}-{10}$ Calculate 2 3 $\displaystyle{8}={6}{x}-{10}$ Add 10 to both sides $\displaystyle{18}={6}{x}$ Divide by 6 $\displaystyle{x}={3}$
## Using the Definition of a Logarithm to Solve Logarithmic Equations
For any algebraic expression S and real numbers b and c, where b > 0, b ≠ 1, log b(S) = c if and only if bc = S
### Example 1
Using algebra to solve a logarithmic equation
1. Solve $\displaystyle{2}{ln{{x}}}+{3}={7}$
1. Subtract 3: $\displaystyle{2}{ln{{x}}}={4}$
2. Divide by 2: $\displaystyle{ln{{x}}}={2}$
3. Rewrite in exponential form: $\displaystyle{x}={e}^{{2}}$
2. Solve $\displaystyle{6}+{ln{{x}}}={10}$
Using algebra before and after using the definition of the natural logarithm
1. Solve $\displaystyle{2}{ln{{({6}{x})}}}={7}$
1. Divide by 2: $\displaystyle{ln{{({6}{x})}}}=\frac{{7}}{{2}}$
2. Use the definition of ln: $\displaystyle{6}{x}={e}^{{{(\frac{{7}}{{2}})}}}$
3. Divide by 6: $\displaystyle{x}=\frac{{1}}{{6}}{e}^{{{(\frac{{7}}{{2}})}}}$
2. Solve $\displaystyle{2}{ln{{({x}+{1})}}}={10}$
3. Using a graph to understand the solution to a logarithmic equation, solve $\displaystyle{ln{{x}}}={3}$The graphs of y = lnxand y = 3 cross at the point (e3,3) which is approximately (20.0855, 3).
4. Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2x = 1000, solve to 2 decimal places. x ≈ 9.97
# Using the One-to-One Property of Logarithms to Solve Logarithmic Equations
As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0, S > 0, T > 0 and any positive real number b, where b ≠ 1, logbS = logbT if and only if S = T For example, If log2(x 1) = log2(8), then x 1 = 8. So, if x – 1 =8, then we can solve for x, and we get x = 9. To check, we can substitute x = 9 into the original equation: log2(9 – 1) = log2(8) = 3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown. For example, consider the equation $\displaystyle{log{{({3}{x}-{2})}}}-{log{{({2})}}}={log{{({x}+{4})}}}$. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for [latex-display]\displaystyle{log{{({3}{x}-{2})}}}-{log{{({2})}}}={log{{({x}+{4})}}}[/latex-display]
Apply the quotient rule of logarithms $\displaystyle{log{{(\frac{{{3}{x}-{2}}}{{2}})}}}={log{{({x}+{4})}}}$ Apply the one to one property of a logarithm. $\displaystyle\frac{{{3}{x}-{2}}}{{2}}={x}+{4}$ Multiply both sides of the equation by 2 $\displaystyle{3}{x}-{2}={3}{x}+{8}$ Subtract 2 x and add 2 $\displaystyle{x}={10}$
To check the result, substitute into [latex-display]\displaystyle{log{{({3}{({10})}-{2})}}}-{log{{({2})}}}={log{{({({10})}+{4})}}}[/latex-display] [latex-display]\displaystyle{log{{({28})}}}-{log{{({2})}}}={log{{({14})}}}[/latex-display] [latex-display]\displaystyle{log{{(\frac{{28}}{{2}})}}}={log{{({14})}}}[/latex-display] The solution checks.
## Using the One-to-One Property of Logarithms to Solve Logarithmic Equations
For any algebraic expressions S and T and any positive real number b, where b ≠ 1, logbS = logbT if and only if S = T Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution. Given an equation containing logarithms, solve it using the one-to-one property.
1. Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form logbS = logbT.
2. Use the one-to-one property to set the arguments equal.
3. Solve the resulting equation, S = T, for the unknown.
### Example 2
1. Using the one-to-one property of logarithms, solve $\displaystyle{ln{{({x}^{{2}})}}}={ln{{({2}{x}+{3})}}}$Use the one-to one property of the logarithm: $\displaystyle{x}^{{2}}={2}{x}+{3}$Get zero on one side before factoring: $\displaystyle{x}^{{2}}-{2}{x}-{3}={0}$Factor using FOIL: $\displaystyle{({x}-{3})}{({x}+{1})}={0}$ If a product is zero, one of the factors must be zero: $\displaystyle{x}-{3}={0}{\text{ or }}{x}+{1}={0}$ Solve for x: $\displaystyle{x}={3}{\text{ or }}{x}=-{1}$ Analysis There are two solutions: x = 3 or x = –1. The solution x = –1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.
2. Solve $\displaystyle{ln{{({x}^{{2}})}}}={ln{{1}}}$
OpenStax, Precalculus, "Exponential and Logarithmic Equations," licensed under a CC BY 3.0 license.
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# Volumes – Surface Area & Volumes | Class 9 Maths
• Difficulty Level : Hard
• Last Updated : 12 May, 2021
If an object is solid then space occupied by such an object is measured and termed as the Volume of the object. In short, the Volume of an object is the measure of the space it occupies, and the capacity of an object is the volume of substance its interior can accommodate. So, if we were to talk of the volume of a cube, we would be considering the measure of the space occupied by the cube.
The Unit of measurement of Volume is the cubic unit.
## Volume of Cube
A cube is a region of space formed by six identical square faces joined along their edges. Three edges join at each corner to form a vertex.
### Volume of Cube = edge x edge x edge = (a3) cubic units
Example 1: Find the volume of the cube, whose edge length is 3 cm?
Solution:
Volume of Cube = a3 (where a = edge or side length)
Volume = 3 x 3 x 3 = 27 cm
Example 2: Find the length of the side of the cube whose volume is 1331 cm3?
Solution:
Volume = side x side x side
=> 1331 = a3
Hence, side a = 11 cm.
## Volume of Cuboid
A cuboid is a box-shaped object. It has six flat faces and all angles are right angles. And all of its faces are rectangles.
### Where, l = length, w = width, h = height of Cuboid
Example 1: Find the volume of the cuboid whose length, width, and height are 10cm,11cm, and 13cm respectively?
Solution:
Volume of Cuboid = l x w x h = 10 x 11 x 13 = 1430 cm
Example 2: Calculate the length of the cuboid whose volume is given as 270-meter cube. And width and height are 6 and 9 meters respectively?
Solution:
Volume of cuboid = l x w x h
=> 270 = l x 6 x 9
=> l = 5 m
## Volume of Cylinder
Solids like measuring jars, circular pillars, circular pencils, circular pipes, road rollers, etc., are said to have a cylindrical shape.
### Where, r = base radius, h = height of cylinder
Example 1: Find the volume of a cylinder whose radius is 10cm and height is 15cm. Take π = 22/7?
Solution:
Volume of Cylinder = πr2h = 22/7(10 x 10 x 15)
= 4714.28 cm3
Example 2: Find the height of a cylinder whose radius is 7cm and volume is 1540cm3.Take π = 22/7?
Solution:
Volume of Cylinder = πr2h
=> 1540 = 22/7(7 x 7 x h)
=> h = 10cm
## Volume of Cone
Solids like ice-cream cones, conical tents, funnels, etc., are having the shape of a cone.
### Where, r = radius of cone and, h = height of the cone.
Example 1: Find the volume of a cone whose radius is 7cm and height is 12cm. (π = 3.14)?
Solution:
Volume of Cone = 1/3(πr2h)
= 1/3(3.14 x 7 x 7 x 12)
= 615.44 cm3
Example 2: Determine the height of the cone whose volume and radius 308cm3, 7cm respectively. (π = 22/7)?
Solution:
Volume of cone = 1/3(πr2h)
=> 308 = 1/3(22/7 x 7 x 7 x h)
=> h = 6cm
## Volume of Sphere
Objects like a football, a cricket ball, etc., are said to have the shape of a sphere.
### Where, r is the radius of the sphere
Example 1: Find the volume of a sphere whose radius is 14cm. (π = 22/7)?
Solution:
Volume of Sphere = 4/3(πr3)
= 4/3(22/7 x 14 x 14 x 14)
= 11498.66 cm3
Example 2: Determine the radius of the sphere whose volume is 38808 m3?
Solution :
The volume of Sphere = 4/3(πr3)
=> 38808 = 4/3(22/7 x r3)
=> r = 21m
## Volume of Hemisphere
A plane through the centre of a sphere cuts it into two equal parts. Each part is called a hemisphere.
### Where, r is the radius of the hemisphere
Example 1: Find the volume of a hemisphere whose radius is 14 cm. (π = 22/7)?
Solution:
Volume of hemisphere = 2/3(πr3)
= 2/3(22/7 x 14 x 14 x 14)
= 5749.34 cm3
Example 2: Find the volume of a hemisphere whose radius is 7 cm. (π = 22/7)?
Solution:
Volume of hemisphere = 2/3(πr3)
= 2/3(22/7 x 7 x 7 x 7)
= 718.66 cm3
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Associated Topics || Dr. Math Home || Search Dr. Math
### Find the Length of a Carpet
```
Date: 4/22/96 at 8:18:2
From: Stefan Isberg
Subject: Find the Length of a Carpet
You have a rectanguler room with sides of 3 and 5 meters.
You put a carpet on the diagonal (not exactly, of course!)
The carpet's corners must touch each wall in the room,
so the problem will not be as easy as it seems.
We have tried to illustrate how the carpet shall lie.
The carpet lies approximately on the diagonal.
5 meter
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
! \ !
! \ !
! \ ! 3 meter
!\ !
! \ !
! \ !
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
The carpet's width is 1 meter (The carpet is rectanguler) .
What is the exact length of this carpet?
Stefan and Anders, Math students at Umea University.
```
```
Date: 5/3/96 at 4:15:3
From: Dr. Alain
Subject: Re: Find the Length of a Carpet
Draw the room like this:
(-5,0) (0,0)
________________________________________
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(-5,-3) (0,-3)
Say the carpet hits the top line at (-a,0) and the right line at
(0,-b). Then it must hit the left wall at (-5, -3 + b), and hit the
bottom wall at (-5 + a, -3).
We know the width of the carpet is 1, so:
a^2 + b^2 = 1.
We also have that the carpet is rectangular, so the line from
(-5 + a, -3) to (0,-b) makes a right angle with the line from
(0,-b) to (-a,0).
This is equivalent to the product of the slope of these two lines
which is -1.
The slope of the first line is m1= (3-b)/(5-a).
The slope of the second line is m2= -b/a.
So: m2*m1=-b/a*(3-b)/(5-a) = -1.
Or b^2 - 3b = a^2 - 5a.
So we have
(1) a^2 + b^2 =1.
(2) b^2 - 3b = a^2 - 5a.
These two equations are enough to find a and b. These equations
bring up a fourth degree polynomial equation. It is possible to
find the exact solution to such an equation (when the solution
exists) but the formula for fourth degree polynomials is quite
horible. The best way to treat this polynomial equation is with
numerical methods such as the Newton-Raphson algorithm.
From the endpoints of the carpet we can find its length, L:
L^2 = (-5 + a)^2 + (3 - b)^2.
Now we have:
L^2 = 25 - 10a + a^2 + 9 - 6b + b^2.
Once you found a and b you can find L, which is the answer to your
question.
-Doctor Alain, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Triangles and Other Polygons
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## Finding Gifted ESP Participants An ESP test is conducted by randomly selecting one of five video clips and playing it in one building while a participant in another building tries to describe what is playing.
Finding Gifted ESP Participants An ESP test is conducted by randomly selecting one of five video clips and playing it in one building while a participant in another building tries to describe what is playing. Later, the participant is shown the five video clips and is asked to determine which one best matches the description he or she had given. By chance, the participant would get this correct with probability 1/5. Individual participants are each tested eight times, with five new video clips each time. They are identified as “gifted” if they guess correctly at least five times out of the eight tries. Suppose that people actually do have some ESP and can guess correctly with probability .30 (instead of the .20 expected by chance). What is the probability that a participant will be identified as “gifted”?
In Chapter 8, you will learn how to solve this kind of problem, but we can simulate the answer using a random number generator that produces the digits 0, 1, 2, … , 9 with equal likelihood. Many calculators and computers will simulate these digits. Here are the steps needed for one “repetition”:
• Each “guess” is simulated with a digit, equally likely to be 0 to 9.
• For each participant, we simulate eight “guesses” resulting in a string of eight digits.
• If a digit is 7, 8, or 9, we count that guess as “correct,” so P(correct) = 3/10 = .3, as required in the problem. If the digit is 0 to 6, the guess is “incorrect.” (There is nothing special about 7, 8, 9; we could have used any three digits.)
• If there are five or more “correct” guesses (digits 7, 8, 9), we count that as “gifted.”
The entire process is repeated many times, and the proportion of times that the process results in a “gifted” participant provides an estimate of the desired probability. Let’s use Minitab to simulate the experiment for 1000 participants, each making eight guesses.
### It was one of the most emotional legal meetings I have ever had with a client.
It was one of the most emotional legal meetings I have ever had with a client. Under the pressure of a major lawsuit, a husband and wife were desperate to….
### Assignment reusable straws – Marketing Management In the past couple of years, the global shift toward selling and buying more eco-friendly products seems to have accelerated.
Assignment reusable straws – Marketing Management In the past couple of years, the global shift toward selling and buying more eco-friendly products seems to have accelerated. In virtually any industry,….
### What is the primary difference between Backup Speed and Restore Speed?
What is the primary difference between Backup Speed and Restore Speed? Q20: Design and performance of a Call Center. In a call center there are M customer service representatives. Arriving calls are sent to….
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# Math Snap
## A wooden block ( كتل ختّب4) of mass $M=400$ gram and $m=100$ gram is the total mass just starts to move when the hanger mass 200 gram. The tension| الثند in the cord is:
#### STEP 1
Assumptions 1. The mass of the wooden block $M$ is 400 grams. 2. The additional mass $m$ on top of the block is 100 grams. 3. The total mass of the block system is $M + m = 400 \, \text{grams} + 100 \, \text{grams} = 500 \, \text{grams}$. 4. The hanging mass is 200 grams. 5. The system just starts to move, indicating that the tension in the cord equals the force of friction. 6. The acceleration due to gravity $g$ is $9.8 \, \text{m/s}^2$. 7. The frictional force is equal to the weight of the hanging mass.
#### STEP 2
Convert all masses to kilograms to use standard SI units. $M = 400 \, \text{grams} = 0.4 \, \text{kg}$ $m = 100 \, \text{grams} = 0.1 \, \text{kg}$ $\text{Total mass} = M + m = 0.4 \, \text{kg} + 0.1 \, \text{kg} = 0.5 \, \text{kg}$ $\text{Hanging mass} = 200 \, \text{grams} = 0.2 \, \text{kg}$
#### STEP 3
Calculate the weight of the hanging mass, which is the force due to gravity acting on it. $\text{Weight of hanging mass} = \text{Hanging mass} \times g = 0.2 \, \text{kg} \times 9.8 \, \text{m/s}^2$
#### STEP 4
Compute the weight of the hanging mass. $\text{Weight of hanging mass} = 0.2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1.96 \, \text{N}$
##### SOLUTION
Since the system just starts to move, the tension in the cord equals the weight of the hanging mass. $\text{Tension} = \text{Weight of hanging mass} = 1.96 \, \text{N}$ The tension in the cord is $1.96 \, \text{N}$.
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2009 AIME II Problems/Problem 1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Problem
Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.
Solution
After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to $130$ ounces each. Say $a$ is the fraction of the pink paint that is red paint and $x$ is the size of each stripe. Then equations can be written: $ax = 164 - 130 = 34$ and $(1-a)x = 188 - 130 = 58$. The second equation becomes $x - ax = 58$ and substituting the first equation into this one we get $x - 34 = 58$ so $x = 92$. The amount of each color left over at the end is thus $130 - 92 = 38$ and $38 * 3 = \boxed{114}$.
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Instructional Unit 2
Horizontal and Vertical Translations of a Parabola
by Sandy Cederbaum
When we connect the data points in Unit 1 by a smooth curve, we get a shape called a parabola. The figure below is the graph of the data from Unit 1.
Here are several other graphs of parabolas:
Write a short paragraph outlining the similarities and differences of the graphs above. Which of these graphs are functions?
At this time, we will only concern ourselves with those parabolas that are functions. The group of mathematical rules that create parabolas that are functions are called QUADRATIC FUNCTIONS. Much of the remaining units will deal with the manipulation of these rules so that we can find out certain information or behavior about a particular quadratic function. We will do this both graphically and algebraically.
It is often helpful to find the mathematical rule that defines a parabolic data set. For example, if we know that the rule for the parabola that corresponds to the data in Unit 1 is given by where h(t) is the height of our projectile in meters at time t (in seconds), we could determine the height of the projectile at any time. Or, we could find the time it will take for the projectile to reach a specified height. We will address each of these problems in subsequent units.
First, however, we will address the problem of finding a rule for a data set (like the data set in Unit 1) that appears parabolic in nature.
The most basic rule for a parabola is or using function notation,. We will call this rule the "parent function" for quadratics. Sketch the graph of the parent function by plotting points. Use an Excel Spreadsheet and Excel's Chart Wizard to create a connected scatterplot. Does your graph have a maximum or a minimum? The maximum or minimum point of a parabola is called the VERTEX. Write the coordinates of the vertex of the parent function. Does this function have any x or y intercepts? If so, write them down as well.
Graph each of the following functions using a graphing utility and write a brief description of how each compares to the parent function.
1.
2.
3.
4.
Now compare each of the following graphs to the parent function and write a brief description of the effect.
1.
2.
3.
4.
Describe the translations of the following graphs as compared to the parent function and sketch each graph on a piece of graph paper.
1 2
Give the coordinates of the vertices of each of the graphs above.
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# Poker probability (Texas hold 'em)
In poker, the probability of many events can be determined by direct calculation. This article discusses computing probabilities for many commonly occurring events in the game of Texas hold 'em and provides some probabilities and odds for specific situations. In most cases, the probabilities and odds are approximations due to rounding.
When calculating probabilities for a card game such as Texas Hold 'em, there are two basic approaches. The first approach is to determine the number of outcomes that satisfy the condition being evaluated and divide this by the total number of possible outcomes. For example, there are six outcomes (ignoring order) for being dealt a pair of aces in Hold 'em: A A, A A, A A, A A, A A and A A. There are 52 ways to pick the first card and 51 ways to pick the second card and two ways to order the two cards yielding $\tfrac{52 \times 51}{2} = 1326$ possible outcomes when being dealt two cards (also ignoring order). This gives a probability of being dealt two aces of $\begin{matrix} \frac{6}{1326} = \frac{1}{221} \end{matrix}$.
The second approach is to use conditional probabilities, or in more complex situations, a decision tree. There are 4 ways to be dealt an ace out of 52 choices for the first card resulting in a probability of $\begin{matrix} \frac{4}{52} = \frac{1}{13} \end{matrix}.$ There are 3 ways of getting dealt an ace out of 51 choices on the second card after being dealt an ace on the first card for a probability of $\begin{matrix} \frac{3}{51} = \frac{1}{17} \end{matrix}.$ This value is the conditional probability that second card dealt is an ace given the condition that the first card dealt is an ace. The joint probability of being dealt two aces is the product of the two probabilities: $\begin{matrix} \frac{1}{13} \times \frac{1}{17} = \frac{1}{221} \end{matrix}.$ This article uses both of these approaches.
The odds presented in this article use the notation x : 1 which translates to x to 1 odds against the event happening. The odds are calculated from the probability p of the event happening using the formula: odds = [(1 − p) ÷ p] : 1, or odds = [(1 ÷ p) − 1] : 1. Another way of expressing the odds x : 1 is to state that there is a 1 in x+1 chance of the event occurring or the probability of the event occurring is 1/(x+1). So for example, the odds of a roll of a fair six-sided die coming up three is 5 : 1 against because there are 5 chances for a number other than three and 1 chance for a three; alternatively, this could be described as a 1 in 6 chance or $\begin{matrix}\frac{1}{6}\end{matrix}$ probability of a three being rolled because the three is 1 of 6 equally-likely possible outcomes.
## Starting hands
### Single hand
The probability of being dealt various starting hands can be explicitly calculated. In Texas Hold 'em, a player is dealt two down (or hole or pocket) cards. The first card can be any one of 52 playing cards in the deck and the second card can be any one of the 51 remaining cards. This gives 52 × 51 ÷ 2 = 1,326 possible starting hand combinations. Since the order in which you receive the first two cards is not significant, the 2,652 permutations are divided by the 2 ways of ordering two cards. Alternatively, the number of possible starting hands is represented as the binomial coefficient
${52 \choose 2} = 1,326$
which is the number of possible combinations of choosing 2 cards from a deck of 52 playing cards.
The 1,326 starting hands can be reduced for purposes of determining the probability of starting hands for Hold 'em—since suits have no relative value in poker, many of these hands are identical in value before the flop. The only factors determining the strength of a starting hand are the ranks of the cards and whether the cards share the same suit. Of the 1,326 combinations, there are 169 distinct starting hands grouped into three shapes: 13 pocket pairs (paired hole cards), 13 × 12 ÷ 2 = 78 suited hands and 78 unsuited hands; 13 + 78 + 78 = 169. The relative probability of being dealt a hand of each given shape is different. The following shows the probabilities and odds of being dealt each type of starting hand.
Hand shapeNumber
of hands
Suit combinations
for each hand
CombinationsDealt specific handDealt any hand
ProbabilityOddsProbabilityOdds
Pocket pair13${4 \choose 2} = 6$13 × 6 = 78$\begin{matrix} \frac{6}{1326} \approx 0.00452 \end{matrix}$220 : 1$\begin{matrix} \frac{78}{1326}=\frac{3}{51} \approx 0.0588 \end{matrix}$16 : 1
Suited cards78${4 \choose 1} = 4$78 × 4 = 312$\begin{matrix} \frac{4}{1326} \approx 0.00302 \end{matrix}$331 : 1$\begin{matrix} \frac{312}{1326}=\frac{12}{51} \approx 0.2353 \end{matrix}$3.25 : 1
Unsuited cards non paired78${4 \choose 1}{3 \choose 1} = 12$78 × 12 = 936$\begin{matrix} \frac{12}{1326} \approx 0.00905 \end{matrix}$110 : 1$\begin{matrix} \frac{936}{1326}=\frac{36}{51} \approx 0.7059 \end{matrix}$0.417 : 1
Here are the probabilities and odds of being dealt various other types of starting hands.
HandProbabilityOdds
AKs (or any specific suited cards)0.00302331 : 1
AA (or any specific pair)0.00452220 : 1
AKs, KQs, QJs, or JTs (suited cards)0.012181.9 : 1
AK (or any specific non-pair incl. suited)0.012181.9 : 1
AA, KK, or QQ0.013672.7 : 1
AA, KK, QQ or JJ0.018154.3 : 1
Suited cards, jack or better0.018154.3 : 1
AA, KK, QQ, JJ, or TT0.022643.2 : 1
Suited cards, 10 or better0.030232.2 : 1
Suited connectors0.039224.5 : 1
Connected cards, 10 or better0.048319.7 : 1
Any 2 cards with rank at least queen0.049819.1 : 1
Any 2 cards with rank at least jack0.090510.1 : 1
Any 2 cards with rank at least 100.1435.98 : 1
Connected cards (cards of consecutive rank)0.1575.38 : 1
Any 2 cards with rank at least 90.2083.81 : 1
Not connected nor suited, at least one 2-90.5340.873 : 1
For any given starting hand, there are 50 × 49 ÷ 2 = 1,225 hands that an opponent can have before the flop. (After the flop, the number of possible hands an opponent can have is reduced by the three community cards revealed on the flop to 47 × 46 ÷ 2 = 1,081 hands.) Therefore, there are
${52 \choose 2}{50 \choose 2} \div 2 = 812,175$
possible head-to-head match ups in Hold 'em. (The total number of match ups is divided by the two ways that two hands can be distributed between two players to give the number of unique match ups.) However, since there are only 169 distinct starting hands, there are 169 × 1,225 = 207,025 distinct head-to-head match ups.[Note 1]
It is useful to know how two starting hands compete against each other heads up before the flop. In other words, we assume that neither hand will fold, and we will see a showdown. This situation occurs quite often in no limit and tournament play. Also, studying these odds helps to demonstrate the concept of hand domination, which is important in all community card games.
This problem is considerably more complicated than determining the frequency of dealt hands. To see why, note that given both hands, there are 48 remaining unseen cards. Out of these 48 cards, we can choose any 5 to make a board. Thus, there are
${48 \choose 5} = 1,712,304$
possible boards that may fall. In addition to determining the precise number of boards that give a win to each player, we also must take into account boards which split the pot, and split the number of these boards between the players.
The problem is trivial for computers to solve by brute force search; there are many software programs available that will compute the odds in seconds. A somewhat less trivial exercise is an exhaustive analysis of all of the head-to-head match ups in Texas Hold 'em, which requires evaluating each possible board for each distinct head-to-head match up, or 1,712,304 × 207,025 = 354,489,735,600 (≈354 billion) results.[Note 1]
When comparing two starting hands, the head-to-head probability describes the likelihood of one hand beating the other after all of the cards have come out. Head-to-head probabilities vary slightly for each particular distinct starting hand matchup, but the approximate average probabilities, as given by Dan Harrington in Harrington on Hold'em [p. 125], are summarized in the following table.
Favorite-to-underdog matchupProbabilityOdds for
Pair vs. 2 undercards0.831 : 4.9
Pair vs. lower pair0.821 : 4.5
Pair vs. 1 overcard, 1 undercard0.711 : 2.5
2 overcards vs. 2 undercards0.631 : 1.7
Pair vs. 2 overcards0.551 : 1.2
These odds are general approximations only derived from averaging all of the hand matchups in each category. The actual head-to-head probabilities for any two starting hands vary depending on a number of factors, including:
• Suited or unsuited starting hands;
• Shared suits between starting hands;
• Connectedness of non-pair starting hands;
• Proximity of card ranks between the starting hands (lowering straight potential);
• Proximity of card ranks toward A or 2 (lowering straight potential);
• Possibility of split pot.
For example, A A vs. K Q is 87.65% to win (0.49% to split), but A A vs. 7 6 is 76.81% to win (0.32% to split).
The mathematics for computing all of the possible matchups is simple. However, the computation is tedious to carry out by hand. A computer program can perform a brute force evaluation of the 1,712,304 possible boards for any given pair of starting hands in seconds.
### Starting hands against multiple opponents
When facing two opponents, for any given starting hand the number of possible combinations of hands the opponents can have is
${50 \choose 2}{48 \choose 2} = 1,381,800$
hands. For calculating probabilities we can ignore the distinction between the two opponents holding A J and 8 8 and the opponents holding 8 8 and A J. The number of ways that hands can be distributed between $n$ opponents is $n!$ (the factorial of n). So the number of unique hand combinations $H$ against two opponents is
$H = {50 \choose 2}{48 \choose 2} \div 2! = 690,900$
and against three opponents is
$H = {50 \choose 2}{48 \choose 2}{46 \choose 2} \div 3! = 238,360,500$
and against $n$ opponents is
$H = \prod_{k=1}^n {52 - 2k \choose 2} \div n!,$ or alternatively $H = {50 \choose 2n} \times (2n-1)!!,$
where $(2n-1)!!$ is the number of ways to distribute $2n$ cards between $n$ hands of two cards each.[Note 2] [!! is the double factorial operator: (2n-1)!! is not ((2n-1)!)!.] The following table shows the number of hand combinations for up to nine opponents.
OpponentsNumber of possible hand combinations
11,225
2690,900
3238,360,500
456,372,258,250
5≈9.7073 × 1012 (more than 9 trillion)
6≈1.2620 × 1015 (more than 1 quadrillion)
7≈1.2674 × 1017 (more than 126 quadrillion)
8≈9.9804 × 1018 (almost 10 quintillion)
9≈6.2211 × 1020 (more than 622 quintillion)
An exhaustive analysis of all of the match ups in Texas Hold 'em of a player against nine opponents requires evaluating each possible board for each distinct starting hand against each possible combination of hands held by nine opponents, which is
$169 \times {50 \choose 18} \times 17!! \times {32 \choose 5} \approx 2.117 \times 10^{28}$ (more than 21 octillion).
If you were able to evaluate one trillion (1012) combinations every second, it would take over 670 million years to evaluate all of the hand/board combinations. While it is possible to significantly reduce the total number of combinations by pruning combinations with identical properties, the total number of situations is still well beyond the number that can be evaluated by brute force. For this reason, most software programs compute probabilities and expected values for Hold 'em poker hands against multiple opponents by simulating the play of thousands or even millions of hands to determine statistical probabilities.
### Dominated hands
When evaluating a hand before the flop, it is useful to have some idea of how likely the hand is dominated. A dominated hand is a hand that is beaten by another hand (the dominant hand) and is unlikely to win against it. Often the dominated hand has only a single card rank that can improve the dominated hand to beat the dominant hand (not counting straights and flushes). For example, KJ is dominated by KQ—both hands share the king, and the queen kicker is beating the jack kicker. Barring a straight or flush, the KJ will need a jack on the board to improve against the KQ (and would still be losing if a queen appears on the board along with the jack). A pocket pair is dominated by a pocket pair of higher rank.
#### Pocket pairs
Barring a straight or flush, a pocket pair needs to make three of a kind to beat a higher pocket pair. See the section "After the flop" for the odds of a pocket pair improving to three of a kind.
To calculate the probability that another player has a higher pocket pair, first consider the case against a single opponent. The probability that a single opponent has a higher pair can be stated as the probability that the first card dealt to the opponent is a higher rank than the pocket pair and the second card is the same rank as the first. Where $r$ is the rank of the pocket pair (assigning values from 2–10 and J–A = 11–14), there are (14 − r) × 4 cards of higher rank. Subtracting the two cards for the pocket pair leaves 50 cards in the deck. After the first card is dealt to the player there are 49 cards left, 3 of which are the same rank as the first. So the probability $P$ of a single opponent being dealt a higher pocket pair is
\begin{align} P & = \frac{(14 - r) \times 4}{50} \times \frac{3}{49}\\ & = \frac{84 - 6r}{1225}.\\ \end{align}
The following approach extends this equation to calculate the probability that one or more other players has a higher pocket pair.
1. Multiply the base probability for a single player for a given rank of pocket pairs by the number of opponents in the hand;
2. Subtract the adjusted probability that more than one opponent has a higher pocket pair. (This is necessary because this probability effectively gets added to the calculation multiple times when multiplying the single player result.)
Where $n$ is the number of other players still in the hand and $P_{ma}$ is the adjusted probability that multiple opponents have higher pocket pairs, then the probability that at least one of them has a higher pocket pair is
$P = \left(\frac{84 - 6r}{1225}\right) \times n - P_{ma}= \frac{n(84-6r)-1,225P_{ma}}{1,225}.$
The calculation for $P_{ma}$ depends on the rank of the player's pocket pair, but can be generalized as
$P_{ma} = P_2 + 2P_3 + \cdots + (n-1)P_n,$
where $P_2$ is the probability that exactly two players have a higher pair, $P_3$ is the probability that exactly three players have a higher pair, etc. As a practical matter, even with pocket 2s against 9 opponents, $P_4 < 0.0015$ and $P_5 < 0.00009$, so just calculating $P_2$ and $P_3$ gives an adequately precise result.
The following table shows the probability that before the flop another player has a larger pocket pair when there are one to nine other players in the hand.
Probability of facing a
larger pair when holding
Against 1Against 2Against 3Against 4Against 5Against 6Against 7Against 8Against 9
KK0.00490.00980.01470.01960.02440.02930.03420.03910.0439
QQ0.00980.01950.02920.03880.04840.05790.06730.07660.0859
JJ0.01470.02920.04360.05770.07170.08560.09920.11270.1259
TT0.01960.03890.05780.07640.09460.11240.12990.14700.1637
990.02450.04840.07180.09460.11680.13840.15930.17950.1990
880.02940.05800.08570.11250.13840.16340.18730.21010.2318
770.03430.06740.09940.13010.15950.18740.21380.23870.2619
660.03920.07690.11300.14730.17990.21040.23890.26510.2890
550.04410.08620.12630.16420.19960.23240.26230.28920.3129
440.04900.09560.13950.18060.21860.25320.28410.31090.3334
330.05390.10480.15260.19670.23700.27290.30400.33000.3503
220.05880.11410.16540.21240.25460.29140.32220.34640.3633
The following table gives the probability that a hand is facing two or more larger pairs before the flop. From the previous equations, the probability $P_m$ is computed as
$P_m = P_2 + P_3 + \cdots + P_n.$
Probability of facing multiple
larger pairs when holding
Against 2Against 3Against 4Against 5Against 6Against 7Against 8Against 9
KK< 0.000010.000010.000030.000040.000070.000090.000120.00016
QQ0.000060.000180.000370.000610.000910.001280.001710.00220
JJ0.000170.000510.001020.001710.002570.003600.004820.00621
TT0.000330.000990.002000.003350.005040.007090.009500.01226
990.000540.001640.003300.005530.008360.011770.015800.02045
880.000810.002440.004930.008280.012530.017690.023780.03084
770.001120.003410.006890.011600.017580.024870.033510.04353
660.001490.004540.009180.015500.023530.033350.045030.05861
550.001910.005830.011820.019980.030400.043180.058400.07619
440.002390.007280.014800.025060.038210.054380.073710.09635
330.002910.008900.018120.030750.046980.066990.090990.11919
220.003490.010680.021800.037060.056730.081070.110340.14484
From a practical perspective, however, the odds of out drawing a single pocket pair or multiple pocket pairs are not much different. In both cases the large majority of winning hands require one of the remaining two cards needed to make three of a kind. The real difference against multiple overpairs becomes the increased probability that one of the overpairs will also make three of a kind.
#### Hands with one ace
When holding a single ace (referred to as Ax), it is useful to know how likely it is that another player has a better ace—an ace with a higher second card, since a weaker ace is dominated by a better ace. The probability that a single opponent has a better ace is the probability that he has either AA or Ax where x is a rank other than ace that is higher than the player's second card. When holding Ax, the probability that a chosen single player has AA is $\begin{matrix} \frac{3}{50} \times \frac{2}{49} \approx 0.00245 \end{matrix}$. In the case of a table with $n$ opponents, the probability of one of them holding AA is $(1-(1-0.00245)^n)$. If the player is holding Ax against 9 opponents, there is a probability of approximately 0.0218 that one opponent has AA.
Where $x$ is the rank 2–K of the second card (assigning values from 2–10 and J–K = 11–13) the probability that a single opponent has a better ace is calculated by the formula
\begin{align} P & = \left(\frac{3}{50} \times \frac{2}{49}\right) + \left(\frac{3}{50} \times \frac{(13 - x) \times 4}{49} \times 2\right)\\ & = \frac{3}{1225} + \frac{12 \times (13 - x)}{1225}\\ & = \frac{159 - 12x}{1225}.\\ \end{align}
The probability $\begin{matrix} \frac{3}{50} \times \frac{(13 - x) \times 4}{49} \end{matrix}$ of a player having Ay, where y is a rank such that x < y <= K, is multiplied by the two ways to order the cards A and y in the hand.
The following table shows the probability that before the flop another player has an ace with a larger kicker in the hand.
Probability of facing an ace
with larger kicker when holding
Against 1Against 2Against 3Against 4Against 5Against 6Against 7Against 8Against 9
AK0.002450.004890.007330.009760.012190.014600.017020.019420.02183
AQ0.012240.024340.036290.048090.059740.071260.082630.093860.10496
AJ0.022040.043600.064680.085290.105450.125170.144450.163310.18175
AT0.031840.062660.092500.121390.149370.176450.202670.228050.25263
A90.041630.081530.119770.156420.191540.225200.257450.288370.31799
A80.051430.100210.146490.190380.232020.271520.308980.344520.37823
A70.061220.118700.172660.223310.270860.315500.357410.396750.43369
A60.071020.137000.198290.255230.308120.357260.402910.445310.48471
A50.080820.155100.223380.286150.343840.396870.445610.490410.53160
A40.090610.173010.247950.316090.378060.434420.485670.532270.57465
A30.100410.190730.271990.345090.410850.470000.523220.571090.61416
A20.110200.208260.295520.373150.442230.503700.558400.607060.65037
## The flop
The value of a starting hand can change dramatically after the flop. Regardless of initial strength, any hand can flop the nuts—for example, if the flop comes with three 2s, any hand holding the fourth 2 has the nuts. Conversely, the flop can undermine the perceived strength of any hand—a player holding A A would not be happy to see 8 9 10 on the flop because of the straight and flush possibilities.
There are
${50 \choose 3} = 19,600$
possible flops for any given starting hand. By the turn the total number of combinations has increased to
${50 \choose 4} = 230,300$
and on the river there are
${50 \choose 5} = 2,118,760$
possible boards to go with the hand.
The following are some general probabilities about what can occur on the board. These assume a "random" starting hand for the player.
Board consisting ofMaking on flopMaking by turnMaking by river
Prob.OddsProb.OddsProb.Odds
Three or more of same suit (other suit can have two)0.0517718.3 : 10.175374.70 : 10.371071.69 : 1
Four or more of same suit0.0105693.7 : 10.0449021.3 : 1
Rainbow flop (all different suits)0.397651.51 : 10.105508.48 : 1
Three cards of consecutive rank (but not four consecutive)0.0347527.8 : 10.105448.48 : 10.199104.02 : 1
Four cards to a straight (but not five)0.0104095.1 : 10.0376325.6 : 1
Three or more cards of consecutive rank and same suit0.00217459 : 10.00675147 : 10.0130575.6 : 1
Three of a kind (but not a full house or four of a kind)0.00235424 : 10.00922107 : 10.0211346.3 : 1
A pair (but not two pair or three or four of a kind)0.169414.90 : 10.304252.29 : 10.422571.37 : 1
Two pair (but not a full house)0.0103795.4 : 10.0475420.0 : 1
One can see from the table above that more than 60% of the flops will have at least two of the same suit.
### Flopping overcards when holding a pocket pair
It is also useful to look at the chances different starting hands have of either improving on the flop, or of weakening on the flop. When holding a pocket pair, cards of higher rank than the pair weaken the hand because of the potential that such a card has paired a card in an opponent's hand. The hand gets worse the more such cards there are on the board and the more opponents that are in the hand because the probability that one of the overcards has paired a hole card increases. To calculate the probability of no overcard, take the total number of outcomes without an overcard divided by the total number of outcomes.
Where $x$ is the rank 3–K of the pocket pair (assigning values from 3–10 and J–K = 11–13), then the number of overcards is $\begin{matrix}(14 - x) \times 4\end{matrix}$ and the number of cards of rank $x$ or less is $\begin{matrix}50 - (14 - x) \times 4 = 4x - 6\end{matrix}$. The number of outcomes without an overcard is the number of combinations that can be formed with the remaining cards, so the probability $P$ of no overcard on the flop is
$P = {(4x - 6) \choose 3} \div {50 \choose 3},$
and on the turn and river are
$P = {(4x - 6) \choose 4} \div {50 \choose 4}$ and $P = {(4x - 6) \choose 5} \div {50 \choose 5},$ respectively.
The following table gives the probability that no overcards will come on the flop, turn and river, for each of the pocket pairs from 3 to K.
Holding pocket pairNo overcard on flopNo overcard by turnNo overcard by river
Prob.OddsProb.OddsProb.Odds
KK0.77450.29 : 10.70860.41 : 10.64700.55 : 1
QQ0.58570.71 : 10.48601.06 : 10.40151.49 : 1
JJ0.43041.32 : 10.32052.12 : 10.23693.22 : 1
TT0.30532.28 : 10.20143.97 : 10.13136.61 : 1
990.20713.83 : 10.11907.40 : 10.067313.87 : 1
880.13276.54 : 10.064914.40 : 10.031031.21 : 1
770.078611.73 : 10.031830.48 : 10.012479.46 : 1
660.041623.02 : 10.013374.26 : 10.0040246.29 : 1
550.018652.85 : 10.0043229.07 : 10.00091,057.32 : 1
440.0061162.33 : 10.00091,095.67 : 10.00018,406.78 : 1
330.0010979.00 : 10.000115,352.33 : 10.0000353,125.67 : 1
Notice that there is a better than 35% probability that an ace will come by the river if holding pocket kings, and with pocket queens, the odds are slightly in favor of an ace or a king coming by the turn, and a full 60% in favor of an overcard to the queen by the river. With pocket jacks, there's only a 43% chance that an overcard will not come on the flop and it is better than 3 : 1 that an overcard will come by the river.
Notice, though, that those probabilities would be lower if we consider that at least one opponent happens to hold one of those overcards.
## After the flop – outs
During play—that is, from the flop and onwards—drawing probabilities come down to a question of outs. All situations which have the same number of outs have the same probability of improving to a winning hand over any unimproved hand held by an opponent. For example, an inside straight draw (e.g. 3-4-6-7 missing the 5 for a straight), and a full house draw (e.g. 6-6-K-K drawing for one of the pairs to become three-of-a-kind) are equivalent. Each can be satisfied by four cards—four 5s in the first case, and the other two 6s and other two kings in the second.
The probabilities of drawing these outs are easily calculated. At the flop there remain 47 unseen cards, so the probability is (outs ÷ 47). At the turn there are 46 unseen cards so the probability is (outs ÷ 46). The cumulative probability of making a hand on either the turn or river can be determined as the complement of the odds of not making the hand on the turn and not on the river. The probability of not drawing an out is (47 − outs) ÷ 47 on the turn and (46 − outs) ÷ 46 on the river; taking the complement of these conditional probabilities gives the probability of drawing the out by the river which is calculated by the formula
$P = 1 - \left(\frac{47 - outs}{47} \times \frac{46 - outs}{46}\right) = \frac{93outs-outs^2}{2,162}.$
For reference, the probability and odds for some of the more common numbers of outs are given here.
Example drawing toOutsMake on turnMake on riverMake on turn or river
Prob.OddsProb.OddsProb.Odds
Inside straight flush; Four of a kind10.021346.0 : 10.021745.0 : 10.042622.5 : 1
Open-ended straight flush; Three of a kind20.042622.5 : 10.043522.0 : 10.084210.9 : 1
High pair30.063814.7 : 10.065214.3 : 10.12497.01 : 1
Inside straight; Full house40.085110.8 : 10.087010.5 : 10.16475.07 : 1
Three of a kind or two pair50.10648.40 : 10.10878.20 : 10.20353.91 : 1
Either pair60.12776.83 : 10.13046.67 : 10.24143.14 : 1
Full house or four of a kind;[A]
Inside straight or high pair
70.14895.71 : 10.15225.57 : 10.27842.59 : 1
Open-ended straight80.17024.88 : 10.17394.75 : 10.31452.18 : 1
Flush90.19154.22 : 10.19574.11 : 10.34971.86 : 1
Inside straight or pair100.21283.70 : 10.21743.60 : 10.38391.60 : 1
Open-ended straight or high pair110.23403.27 : 10.23913.18 : 10.41721.40 : 1
Inside straight or flush; Flush or high pair120.25532.92 : 10.26092.83 : 10.44961.22 : 1
130.27662.62 : 10.28262.54 : 10.48101.08 : 1
Open-ended straight or pair140.29792.36 : 10.30432.29 : 10.51160.955 : 1
Open-ended straight or flush; Flush or pair;
Inside straight, flush or high pair
150.31912.13 : 10.32612.07 : 10.54120.848 : 1
160.34041.94 : 10.34781.88 : 10.56980.755 : 1
170.36171.76 : 10.36961.71 : 10.59760.673 : 1
Inside straight or flush or pair;
Open-ended straight, flush or high pair
180.38301.61 : 10.39131.56 : 10.62440.601 : 1
190.40431.47 : 10.41301.42 : 10.65030.538 : 1
200.42551.35 : 10.43481.30 : 10.67530.481 : 1
Open-ended straight, flush or pair210.44681.24 : 10.45651.19 : 10.69940.430 : 1
1. ^ When drawing to a full house or four of a kind with a pocket pair that has hit a set (three of a kind) on the flop, there are 6 outs to get a full house by pairing the board and one out to make four of a kind. This means that if the turn does not pair the board or make four of a kind, there will be 3 additional outs on the river, for a total of 10, to pair the turn card and make a full house. This makes the probability of drawing to a full house or four of a kind on the turn or river 0.334 and the odds are 1.99 : 1. This makes drawing to a full house or four of a kind by the river about 8½ outs.
If a player doesn't fold before the river, a hand with at least 14 outs after the flop has a better than 50% chance to catch one of its outs on either the turn or the river. With 20 or more outs, a hand is a better than 2 : 1 favorite to catch at least one out in the two remaining cards.
See the article on pot odds for examples of how these probabilities might be used in gameplay decisions.
### Estimating probability of drawing outs - The rule of four and two
Many poker players do not have the mathematical ability to calculate odds in the middle of a poker hand. One solution is to just memorize the odds of drawing outs at the river and turn since these odds are needed frequently for making decisions. Another solution some players use is an easily calculated approximation of the probability for drawing outs, commonly referred to as the "Rule of Four and Two". With two cards to come, the percent chance of hitting x outs is about (x × 4)%. This approximation gives roughly accurate probabilities up to about 12 outs after the flop, with an absolute average error of 0.9%, a maximum absolute error of 3%, a relative average error of 3.5% and a maximum relative error of 6.8%. With one card to come, the percent chance of hitting x is about (x × 2)%. This approximation has a constant relative error of an 8% underestimation, which produces a linearly increasing absolute error of about 1% for each 6 outs.
A slightly more complicated, but significantly more accurate approximation of drawing outs after the flop is to use (x × 4)% for up to 9 outs and (x × 3 + 9)% for 10 or more outs. This approximation has a maximum absolute error of less than 1% for 1 to 19 outs and maximum relative error of less than 5% for 2 to 23 outs. A more accurate approximation for the probability of drawing outs after the turn is (x × 2 + (x × 2) ÷ 10)%. This is easily done by first multiplying x by 2, then rounding the result to the nearest multiple of ten and adding the 10's digit to the first result. For example, 12 outs would be 12 × 2 = 24, 24 rounds to 20, so the approximation is 24 + 2 = 26%. This approximation has a maximum absolute error of less than 0.9% for 1 to 19 outs and a maximum relative error of 3.5% for more than 3 outs. The following shows the approximations and their absolute and relative errors for both methods of approximation.
OutsMake on turn or riverMake on river
Actual(x × 4)%(x × 3 + 9)%Actual(x × 2)%(x × 2 + (x × 2) ÷ 10)%
Est.Error % ErrorEst.Error % ErrorEst.Error % ErrorEst.Error % Error
14.2553%4%−0.26%6.00%4%−0.26%6.00%2.1739%2%−0.17%8.00%2%−0.17%8.00%
28.4181%8%−0.42%4.97%8%−0.42%4.97%4.3478%4%−0.35%8.00%4%−0.35%8.00%
312.4884%12%−0.49%3.91%12%−0.49%3.91%6.5217%6%−0.52%8.00%7%+0.48%7.33%
416.4662%16%−0.47%2.83%16%−0.47%2.83%8.6957%8%−0.70%8.00%9%+0.30%3.50%
520.3515%20%−0.35%1.73%20%−0.35%1.73%10.8696%10%−0.87%8.00%11%+0.13%1.20%
624.1443%24%−0.14%0.60%24%−0.14%0.60%13.0435%12%−1.04%8.00%13%−0.04%0.33%
727.8446%28%+0.16%0.56%28%+0.16%0.56%15.2174%14%−1.22%8.00%15%−0.22%1.43%
831.4524%32%+0.55%1.74%32%+0.55%1.74%17.3913%16%−1.39%8.00%18%+0.61%3.50%
934.9676%36%+1.03%2.95%36%+1.03%2.95%19.5652%18%−1.57%8.00%20%+0.43%2.22%
1038.3904%40%+1.61%4.19%39%+0.61%1.59%21.7391%20%−1.74%8.00%22%+0.26%1.20%
1141.7206%44%+2.28%5.46%42%+0.28%0.67%23.9130%22%−1.91%8.00%24%+0.09%0.36%
1244.9584%48%+3.04%6.77%45%+0.04%0.09%26.0870%24%−2.09%8.00%26%−0.09%0.33%
1348.1036%52%+3.90%8.10%48%−0.10%0.22%28.2609%26%−2.26%8.00%29%+0.74%2.62%
1451.1563%56%+4.84%9.47%51%−0.16%0.31%30.4348%28%−2.43%8.00%31%+0.57%1.86%
1554.1166%60%+5.88%10.87%54%−0.12%0.22%32.6087%30%−2.61%8.00%33%+0.39%1.20%
1656.9843%64%+7.02%12.31%57%+0.02%0.03%34.7826%32%−2.78%8.00%35%+0.22%0.62%
1759.7595%68%+8.24%13.79%60%+0.24%0.40%36.9565%34%−2.96%8.00%37%+0.04%0.12%
1862.4422%72%+9.56%15.31%63%+0.56%0.89%39.1304%36%−3.13%8.00%40%+0.87%2.22%
1965.0324%76%+10.97%16.86%66%+0.97%1.49%41.3043%38%−3.30%8.00%42%+0.70%1.68%
2067.5301%80%+12.47%18.47%69%+1.47%2.18%43.4783%40%−3.48%8.00%44%+0.52%1.20%
2169.9352%84%+14.06%20.11%72%+2.06%2.95%45.6522%42%−3.65%8.00%46%+0.35%0.76%
2272.2479%88%+15.75%21.80%75%+2.75%3.81%47.8261%44%−3.83%8.00%48%+0.17%0.36%
2374.4681%92%+17.53%23.54%78%+3.53%4.74%50.0000%46%−4.00%8.00%51%+1.00%2.00%
Either of these approximations is generally accurate enough to aid in most pot odds calculations.
### Runner-runner outs
Some outs for a hand require drawing an out on both the turn and the river—making two consecutive outs is called a runner-runner. Examples would be needing two cards to make a straight, flush, or three or four of a kind. Runner-runner outs can either draw from a common set of outs or from disjoint sets of outs. Two disjoint outs can either be conditional or independent events.
#### Common outs
Drawing to a flush is an example of drawing from a common set of outs. Both the turn and river need to be the same suit, so both outs are coming from a common set of outs—the set of remaining cards of the desired suit. After the flop, if $x$ is the number of common outs, the probability $P$ of drawing runner-runner outs in Texas hold 'em is
$P = \frac{x}{47} \times \frac{x-1}{46}= \frac{x^2-x}{2,162}.$
Since a flush would have 10 outs, the probability of a runner-runner flush draw is $\begin{matrix} \frac{10}{47} \times \frac{9}{46} = \frac{90}{2162} \approx \frac{1}{24} \approx 0.04163 \end{matrix}$. Other examples of runner-runner draws from a common set of outs are drawing to three or four of a kind. When counting outs, it is convenient to convert runner-runner outs to "normal" outs (see "After the flop"). A runner-runner flush draw is about the equivalent of one "normal" out.
The following table shows the probability and odds of making a runner-runner from a common set of outs and the equivalent normal outs.
Likely drawing toCommon outsProbabilityOddsEquivalent outs
Four of a kind (with pair)
Inside-only straight flush
20.000931,080 : 10.02
Three of a kind (with no pair)30.00278359 : 10.07
40.00556179 : 10.13
50.00925107 : 10.22
Two pair or three of a kind (with no pair)60.0138871.1 : 10.33
70.0194350.5 : 10.46
80.0259037.6 : 10.61
90.0333029.0 : 10.78
Flush100.0416323.0 : 10.98
#### Disjoint outs
Two outs are disjoint when there are no common cards between the set of cards needed for the first out and the set of cards needed for the second out. The outs are independent of each other if it does not matter which card comes first, and one card appearing does not affect the probability of the other card appearing except by changing the number of remaining cards; an example is drawing two cards to an inside straight. The outs are conditional on each other if the number of outs available for the second card depends on the first card; an example is drawing two cards to an outside straight.
After the flop, if $x$ is the number of independent outs for one card and $y$ is the number of outs for the second card, then the probability $P$ of making the runner-runner is
$P = \frac{x}{47} \times \frac{y}{46} \times 2 = \frac{xy}{1081}.$
For example, a player holding J Q after the flop 9 5 4 needs a 10 and either a K or 8 on the turn and river to make a straight. There are 4 10s and 8 Ks and 8s, so the probability is $\begin{matrix}\frac{4 \times 8}{1081} \approx 0.0296\end{matrix}$.
The probability of making a conditional runner-runner depends on the condition. For example, a player holding 9 10 after the flop 8 2 A can make a straight with J Q, 7 J or 6 7. The number of outs for the second card is conditional on the first card—a Q or 6 (8 cards) on the first card leaves only 4 outs (J or 7, respectively) for the second card, while a J or 7 (8 cards) for the first card leaves 8 outs ({Q, 7} or {J, 6}, respectively) for the second card. The probability $P$ of a runner-runner straight for this hand is calculated by the equation
$P = \left(\frac{8}{47} \times \frac{4}{46}\right) + \left(\frac{8}{47} \times \frac{8}{46}\right) = \frac{96}{2162} \approx 0.0444$
The following table shows the probability and odds of making a runner-runner from a disjoint set of outs for common situations and the equivalent normal outs.
Drawing toProbabilityOddsEquivalent outs
Outside straight0.0444021.5 : 11.04
Inside+outside straight0.0296032.8 : 10.70
Inside-only straight0.0148066.6 : 10.35
Outside straight flush0.00278359 : 10.07
Inside+outside straight flush0.00185540 : 10.04
The preceding table assumes the following definitions.
Outside straight and straight flush
Drawing to a sequence of three cards of consecutive rank from 3-4-5 to 10-J-Q where two cards can be added to either end of the sequence to make a straight or straight flush.
Inside+outside straight and straight flush
Drawing to a straight or straight flush where one required rank can be combined with one of two other ranks to make the hand. This includes sequences like 5-7-8 which requires a 6 plus either a 4 or 9 as well as the sequences J-Q-K, which requires a 10 plus either a 9 or A, and 2-3-4 which requires a 5 plus either an A or 6.
Inside-only straight and straight flush
Drawing to a straight or straight flush where there are only two ranks that make the hand. This includes hands such as 5-7-9 which requires a 6 and an 8 as well as A-2-3 which requires a 4 and a 5.
#### Compound outs
The strongest runner-runner probabilities lie with hands that are drawing to multiple hands with different runner-runner combinations. These include hands that can make a straight, flush or straight flush, as well as four of a kind or a full house. Calculating these probabilities requires adding the compound probabilities for the various outs, taking care to account for any shared hands. For example, if $P_s$ is the probability of a runner-runner straight, $P_f$ is the probability of a runner-runner flush, and $P_{sf}$ is the probability of a runner-runner straight flush, then the compound probability $P$ of getting one of these hands is
$P = P_s + P_f - P_{sf}.$
The probability of the straight flush is subtracted from the total because it is already included in both the probability of a straight and the probability of a flush, so it has been added twice and must therefore be subtracted from the compound outs of a straight or flush.
The following table gives the compound probability and odds of making a runner-runner for common situations and the equivalent normal outs.
Drawing toProbabilityOddsEquivalent outs
Flush, outside straight or straight flush0.0832611.0 : 11.98
Flush, inside+outside straight or straight flush0.0693813.4 : 11.65
Flush, inside-only straight or straight flush0.0555017.0 : 11.30
Some hands have even more runner-runner chances to improve. For example, holding the hand J Q after a flop of 10 J 7 there are several runner-runner hands to make at least a straight. The hand can get two cards from the common outs of {J, Q} (5 cards) to make a full house or four of a kind, can get a J (2 cards) plus either a 7 or 10 (6 cards) to make a full house from these independent disjoint outs, and is drawing to the compound outs of a flush, outside straight or straight flush. The hand can also make {7, 7} or {10, 10} (each drawing from 3 common outs) to make a full house, although this will make four of a kind for anyone holding the remaining 7 or 10 or a bigger full house for anyone holding an overpair. Working from the probabilities from the previous tables and equations, the probability $P$ of making one of these runner-runner hands is a compound probability
$P = 0.08326 + 0.00925 + \frac{2 \times 6}{1081} + (0.00278 \times 2) \approx 0.1092$
and odds of 8.16 : 1 for the equivalent of 2.59 normal outs. Almost all of these runner-runners give a winning hand against an opponent who had flopped a straight holding 8, 9,[Note 3] but only some give a winning hand against A 2 (this hand makes bigger flushes when a flush is hit) or against K Q (this hand makes bigger straights when a straight is hit with 8 9). When counting outs, it is necessary to adjust for which outs are likely to give a winning hand—this is where the skill in poker becomes more important than being able to calculate the probabilities.
1. ^ a b By removing reflection and applying aggressive search tree pruning, it is possible to reduce the number of unique head-to-head hand combinations from 207,025 to 47,008. Reflection eliminates redundant calculations by observing that given hands $h_1$ and $h_2$, if $w_1$ is the probability of $h_1$ beating $h_2$ in a showdown and $s$ is the probability of $h_1$ splitting the pot with $h_2$, then the probability $w_2$ of $h_2$ beating $h_1$ is $w_2 = 1 - (s + w_1)$, thus eliminating the need to evaluate $h_2$ against $h_1$. Pruning is possible, for example, by observing that Q♥ J has the same chance of winning against both 8 7 and 8 7 (but not the same probability as against 8 7 because sharing the heart affects the flush possibilities for each hand).
2. ^ See "Capital Pi notation for multiplication" for a description of the $\prod$ (capital π or pi) symbol.
|
7
Q:
# The average age of a couple and their son was 40 years, the son got married and a child was born just two years after their marriage. When child turned to 10 years, then the average age of the family becomes 38 years. What was the age of the daughter in law at the time of marriage ?
A) 12 years B) 10 years C) 14 years D) 13 years
Explanation:
Given the average age of couple and their son = 40
=> Sum of age (H +W +S) = 40×3
Let the age of daughter in law at the time of marriage = D years
Now after 10 years,
(H + S +W) + 3×12 + D +12 + 10 = 38×5
178 + D = 190
D = 190 -178 = 12 years
Q:
Sum of present ages of P and Q is 41. Age of P 2 year hence is equal to age of R, 1 year ago. Age of P, 4 year hence is equal to age of Q 1 year ago and ratio of present age of P and S is 3 : 4. Find the difference of age of R and S.
A) 2 B) 3 C) 4 D) 5
Explanation:
According to the given data,
P + Q = 41 ......(1)
R - 1 = P + 2
R = P + 3
and
P + 4 = Q - 1
=> Q = P + 5 ....(2)
From (1)&(2)
P = 18
Q = 18 + 5 = 23
R = 18 + 3 = 21
=> P/S = 3/4
=> S = 4/3 x 18 = 24
Required Difference = S - R = 24 - 21 = 3.
1 318
Q:
Present ages of Deepa and Hyma are in the ratio of 5:6 respectively. After four years, this ratio becomes 6: 7. What is the difference between their present ages?
A) 6 yrs B) 4 yrs C) 8 yrs D) 2 yrs
Explanation:
Let the present age of Deepa = 5p
Let the present age of Hyma = 6p
After four years their ratio = 6 : 7
=>
=> 35p + 28 = 36p + 24
=> p = 4
Therefore, the difference between their ages = 24 - 20 = 4 years.
6 451
Q:
The ratio between present ages of Renuka and Sony is 5 : 4 . 4 years ago, Sony's age was 24 years. What will be Renuka's age after 5 years ?
A) 25 B) 30 C) 35 D) 40
Explanation:
Present age of Sony = 24 + 4 = 28 years
After 5 years Renuka's age = 7×5 + 5 = 40 years
8 427
Q:
Hari Ram’s present age is three times his son’s present age and two fifth of his father’s present age. The average of the present age of all of them is 46 years. What is the difference between Hari Ram’s son’s present age and Hari Ram’s father’s present age?
A) 44 yrs B) 56 yrs C) 67 yrs D) 78 yrs
Explanation:
Let Hari Ram's present age = x
Then, his son's age = x/3
Father's age = 5x/2
Now the required difference =
6 415
Q:
Sweety is 54 years old and her mother is 80. How many years ago was Sweety's mother three times her age?
A) 20 B) 41 C) 26 D) 33
Explanation:
Given P = 54 and M = 80
According to the question,
Let ‘A’ years ago, M = 3P
=> M - A/P - A = 3
=> 80 - A = 3(54 - A)
=> 80 - A = 162 - 3A
=> 2A = 82
=> A = 41
Therefore, A = 41 years ago Sweety’s mother was 3 times of Sweety’s age.
Hence, at the age S = 13 and M = 39.
6 529
Q:
7 years ago, the ratio of the ages of Anirudh to that of Bhavana, was 7:9. Chandhu is 12 years older than Anirudh and 12 years younger than Bhavana. What is Chandhu's present age?
A) 91 years B) 115 years C) 103 years D) Can't be determined
Explanation:
Let present Anirudh's age be 'A' and Bhavana's age be 'B'
Given seven years, their ratio is 7:9
=> A-7 : B-7 = 7 : 9
=> 9A - 7B = 14 .......(1)
And given,
Chandhu is 12 years older than A nad 12 years younger than B
=> C = A + 12....(2)
=> B = C + 12
=> B = A + 12 + 12 .....(From (2))
=> B = A + 24 ....(3)
Put (3) in (1)
9A - 7(A+24) = 14
9A - 7A - 168 = 14
2A = 182
=> A = 91 years
=> C = A + 12 = 91 +12 = 103 years.
7 464
Q:
The average age of A and B, 2 years ago was 26. If the age of A, 5 years hence is 40 yrs, and B is 5 years younger to C, then find the difference between the age of A and C?
A) 11 yrs B) 9 yrs C) 7 yrs D) 13 yrs
Explanation:
Let the present ages of A and B be 'x' and 'y' respectively
From the given data,
[(x-2) + (y-2)]/2 = 26
=> x+y = 56
But given the age of A, 5 years hence is 40 yrs => present age of A = 40 - 5 = 35 yrs
=> x = 35 => y = 56 - 35 = 21
=> Age of B = 21 yrs
Given B is 5 years younger to C,
=> Age of C = 21 + 5 = 26 yrs
=> Required Difference between ages of A and C = 35 - 26 = 9 yrs.
5 578
Q:
The average age of husband, wife and their child 3 years ago was 24 years and that of wife and child 5 years ago was 25 years. The present age of the husband is:
A) 21 yrs B) 27 yrs C) 29 yrs D) 31 yrs
|
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## ASVAB Arithmetic Reasoning Sample Questions
With 15 questions to solve in 55 minutes, (CAT-ASVAB) and 30 questions in 36 minutes (P&P ASVAB) – there’s a significant difference – which can be explained by the increasing difficulty in the computerized version. Let’s look at a few different examples.
Q1: It took Sarah six minutes to drive to the post office, eight minutes to find parking and five minutes to choose stationery; she then waited on line for seventeen minutes. How many minutes passed from the time Sarah left her house until she completed her errand at the post office?
39
33
36
31
A1: The correct answer is 36.
If we add up all of the time that it took Sarah from the time she left her house until she completed her errand, we will find that 6+8+5+17 = 36.
Q2: There are six printers at "Today's news" newspaper, all printing at the same constant rate. When working together, the six printers can complete the printing of tomorrow's newspaper within 12 hours.
If the owner wishes to reduce the printing time to 8 hours, how many other printers will he need to purchase, each with the same constant rate?
One
Three
Six
Five
Explanation
The approach to this type of question is to work in two steps:
(1) Finding the individual rate of work per printer
(2) Finding the additional number of printers needed
Step 1- The work rate of 6 printers together is one newspaper/ 12 hours => 1/12. Since each printer works at the same rate, we can say that each printer contributes 1/6th of the total work rate, meaning: (1/12)/6 = 1/72, or one printer/ 72 hours.
Step 2- We will insert what we have learned in Step 1 and the required time (hours) into the basic formula:
Work = rate*time.
1 newspaper = number of printers*(1/72) *8 hours
=> 1 = number of printers*1/9
=> Number of printers = 9
The number of printers includes the original six printers + the additional printers required. Thus, 9 - 6 = 3 additional printers.
More Arithmetic Reasoning Questions
Q3: What was its original price if a shirt costs \$20 after a 20% discount?
\$24
\$25
\$28
\$30
A3: The correct answer is B.
Let's assume the original price is X. 80% of X (because there's a 20% discount) is \$20. So, 0.8X = \$20. Dividing both sides by 0.8 gives X = \$25.
Q4: The scalding summer sun dried up 983.45 gallons of water from the 4078 gallons that were in the tank. How many gallons of water remain?
3104.55
3094.55
3095.45
5051.45
A4: Straightforward solution: Use long subtraction to arrive to 4078-983.45=3094.55. Ah, and you might as well pack your stuff, because test time is probably over..
Pick a trick: With this type of question, one must find ways to dismiss answer choices. Firstly, you can narrow the possible answer choices to B and C, since subtraction of 0.45 from a whole number will result in 0.55. Secondly, determine whether the correct answer is higher or lower than 3100, since this is the rounded up number close to both possibilities. Since it is lower than 3100 – mark B and move on.
Arithmetic Reasoning questions assess basic calculations and simple formulas while reading through passages. The most important key to success in this section is the ability to quickly build a formula with all the relevant data or use the answers to assess which applies to the data in the question. Practice different answering methods while seeing hundreds of different questions in time restrictions in our All Inclusive ASVAB PrepPack.
Let’s continue to the first of two verbal sections of the ASVAB and AFQT – Word Knowledge.
## ASVAB Mathematics Knowledge Sample Questions
With 15 questions to solve in 31 minutes (CAT-ASVAB) and 25 questions in 24 minutes (P&P ASVAB), you will have a minute or two for each question. That means solid control of mathematical formulas and quick calculations are keys to success. Let’s look at a few examples.
Q1: For any value of x, (x - 4)(x + 4) = ?
x2 + 2x - 16
x2 - 2x + 16
x2 + 16
x2 - 16
±4
A1: This is solved using the connection a2 - b2 = (a + b)(a - b).
Hence (x - 4)(x + 4) = x2 - 16.
If you do not remember the formula by heart this can also be solved by opening the brackets:
(x - 4)(x + 4) = x2 + 4x - 4x – 16 = x2 - 16
Q2: 7.4 × 102 decimetres equal:
74 centimeters
740 centimeters
7,400 centimeters
74,000 centimeters
740,000 centimeters
A2: The correct answer is 7,400 centimeters.
A decimetre is a tenth of a meter or ten centimeters.
7.4 × 102 = 740 (decimetres).
740 × 10 = 7,400 (centimetres).
More Mathematical Knowledge Questions
Q3: If y=2x+3, what is the value of y when x = 5?
10
13
15
25
A3: Plugging in x = 5, we get y=2(5)+3=10+3=13.
Q4: The table is 90 cm long. What is the length of the table in metres?
0.09
9
900
0.9
A4: 100 centimetres (cm) are 1 metre. Therefore, we should divide the length of the table in centimetres by 100 for the length in metres.
90 ÷ 100 = 0.9
As can be seen, Mathematical Knowledge assesses basic arithmetic operations, mathematical formulas, working with units of measurement, and more. With 16 questions in 20 minutes (CAT-ASVAB) and 25 questions in 24 minutes (P&P ASVAB) – you have about a minute for each question. That means solid control of mathematical formulas and quick calculations are keys to success. Mathematical questions are best improved by repetitive practice – and you can find many practice tests, simulations, and guides in our All Inclusive ASVAB PrepPack.
Next, we’ll cover another mathematical section –Arithmetic Reasoning. Approaching word problems that require calculations to succeed poses new challenges, different from those described in MK. Let’s continue!
## ASVAB Word Knowledge Sample Questions
This section of the test assesses your vocabulary and grammar skills. With 15 questions in 9 minutes (CAT-ASVAB) and 35 questions in 11 minutes (P&P ASVAB), the questions are clearly on a very narrow time frame.
Q1: HUNGRY is the opposite of:
Tasty
Full
Happy
Ugly
A1: The correct answer is Full.
Hungry is a word used to describe someone who has a desire for food, while full is a word used to describe someone whose appetite is satisfied.
Q2: The new program was sanctioned by the board of education.
You can replace the bold word with:
Explained
Authorized
Divided
Offered
A2: The correct answer is Authorized.
The word sanctioned has two meanings: (1) An action that has been given official approval (in this case, by the educational board), (2) The imposition of a sanction or a penalty.
Option B (authorized) adheres to the first meaning.
More Word Knowledge Questions
Q3: Which of the following is an antonym for "benevolent"?
Malevolent
Kind
Generous
Altruistic
A3: The correct answer is A.
"Benevolent" means showing kindness or goodwill. "Malevolent" means having or showing a wish to harm others, making it the antonym.
Q4: Vend most nearly means:
Vouch
Gamble
Peddle
Instrument
A4: The correct answer is C.
Vend means to sell (you are probably familiar with vending machines which sell food or drinks). Peddle refers to people who sell by going from door to door, offering their goods. Thus, Peddle is close in meaning to vend.
The Word Knowledge section contains three types of questions: Synonyms, Antonyms, and Context. Reading, attempting to use different words in different contexts – and of course, specific, and broad practice is the key to success. Find hundreds of Work Knowledge questions with detailed explanations in our All Inclusive ASVAB PrepPack.
Let’s move on to the final section to be included in the AFQT score – another verbal section, Paragraph Comprehension. While WK assesses straightforward knowledge, PC assess your ability to understand, summarize and apply reason for short paragraphs of text.
## ASVAB Paragraph Comprehension Sample Questions
With 10 questions in 27 minutes (CAT-ASVAB) and 15 questions in 13 minutes (P&P ASVAB, the questions may vary – as it is more possible that the questions in the P&P version may contain more than a single question per text – requiring less time to read and understand a new text.
The brain is our most complex organ, in terms of connections and microscopic structure. It is heterogeneous, with many areas and networks differing from one another in function. And, what is more, the brain is a ‘hidden entity’, embedded in an envelope made of bones, the skull. Brain imaging really came to age in medicine 40 years ago, thanks to computers. The technologies of structural anatomy like computerized tomography and magnetic resonance imaging have brought about a revolution in neurology by showing the lesion and its topography.
Q1: What is not true about the human brain?
Microscopic structure
Heterogeneous
Hidden entity
The largest organ
A1: The correct answer is D.
Note that this question asks what is NOT found in the passage. Reviewing the features of the brain mentioned:
Heterogeneous (answer B), microscopic structure (A), and hidden entity (C) are all mentioned. Only the fact that it is our largest organ (D) isn't mentioned; so, this is the correct answer.
Pay television is now under threat, especially in America. Prices have been driven so high at a time of economic malaise that many people cannot afford it. Disruptive, deep-pocketed firms like Amazon and Netflix lurk, whispering promises of internet-delivered films and television shows for little or no money. It is unclear whether the lure of such alternatives or poverty is causing people to cancel their subscriptions. But the proportion of Americans who pay for TV is falling.
Q2: How many possible explanations are offered for pay-TV's fall?
None
One
Two
Three
A2: The correct answer is C.
The passage offers two explanations for the decline of pay TV: the rise of internet-delivered films and television firms such as Netflix and Amazon and the high rate of subscriptions.
Q3: "The library is a place of learning. Beyond books, it offers resources like computers and classes to the community."
Which statement is true based on the paragraph?
The library only has books
The library offers computers to the community
The library dislikes technology
The library only serves students
A3: The correct answer is B.
The paragraph mentions that the library offers resources like computers to the community.
Greece – the first EU country of entry for many thousands of asylum-seekers does not have – and has not had for several years – asylum-determination procedures or adequate reception conditions in line with international law. It has also not ensured that such asylum-seekers are treated with respect for their dignity on arrival. Many of those seeking international protection have been detained in horrendous conditions, then released and left to live destitute on the streets. Some have been removed from Greece despite the fact that such removal placed them at real risk of further grave human rights violations.
Q4: What did Greece's inadequate policies prevent from people on their arrival?
Basic international law rights
A place to sleep
Basic food
Employment
A4: The correct answer is A.
Many of those seeking international protection have been detained in horrendous conditions, then released and left to live destitute on the streets. Some have been removed from Greece despite the fact that such removal placed them at real risk of further grave human rights violations.
Paragraph Comprehension questions assess your ability to quickly read, understand and apply reason/summarization/responses to different questions regarding short text. While reading books and articles is a great way to improve your abilities – it is possible to shorten your answering time and improve your scores using different methods, such as reading the question first and then looking for the answer in the text. Find hundreds of Paragraph Comprehension texts and questions with detailed explanations in our All Inclusive ASVAB PrepPack.
You can further practice your skills with our Free Civil Service Exam.
## The Line (Composite) Scores
The line (Composite) Scores: a combination of scores from various individual ASVAB subtests. These composite scores are used to determine qualification for specific military jobs, known as Military Occupational Specialties (MOS) in the U.S. Army or ratings in the Navy, for example.
Each branch of the U.S. military has its own set of line scores calculated from specific combinations of ASVAB subtests. These composite scores provide more detailed information about a candidate's aptitudes in particular areas beyond the general measure provided by the AFQT score.
Subjects Line Scores AS+AR+MC Combat – CO AR+MK+VE Clerical- CL AR+ GS+ EI+ MK Electronics- EL AS+ GS+ EI+ MK General Maintenance- GM AR+ MK+ MC Field Artillery- FA AS+ EI+ MC Mechanical Maintenance- MM AR+ VE General Technical- GT AS+ MC+ VE Operators and Food- OF GS+ MC+MK+ VE Skilled Technical- ST AR+ AS+ MC+ VE Surveillance and Communications- SC
Reminder:
AS - Auto & Shop Information
GS -General Science
AR - Arithmetic Reasoning
MC - Mechanical Comprehension
VE - Verbal Expression(combination of the grades of PC+WK)
EI - Electronics Information
MK - Mathematics Knowledge
PC - Paragraph Comprehension
WK - Word Knowledge
The higher your score in the composite scores – the better the chances to qualify for more prestigious and demanded positions. Therefore, we recommend focusing your practice on specific sections which are part of your preferred line score.
Understand your ASVAB score with our ASVAB Scores guide.
We will now continue to the additional five sections which are included in the composite score. Make sure to check in the above table which sections are relevant to your preferred position and begin your practice. We will start with General Science, which assesses your basic knowledge in various scientific fields.
## ASVAB General Science Sample Questions
General Science offers a broad overview of both the physical and biological sciences. With 15 questions in 12 minutes (CAT-ASVAB) and 25 questions in 11 minutes (P&P ASVAB) both versions provide 30 seconds or less for each question. Let’s dive in with a few example questions.
Q1: What does an amperemeter measure?
The voltage
The current
The power
The resistance
A1: The correct answer is B.
An amperemeter measures amperes (A). Since the ampere measures current, the correct answer is B.
Q2: A ball thrown horizontally slows due to:
Mass
Gravity
Inertia
Friction
Mass: Mass is a measure of the amount of matter in an object or substance, typically expressed in units like kilograms or grams.
Gravity: Gravity is the natural force that attracts objects toward the center of the Earth or toward any other massive body.
Inertia: Inertia is an object's inherent resistance to changes in its state of motion or rest.
Friction: Friction is the force that opposes the motion of two surfaces sliding or trying to slide across each other. Therefore, the correct answer is D.
More General Science Questions
Q3: Which of the following is NOT a mammal?
Whale
Dolphin
Bat
Shark
A3: The correct answer is D.
While whales, dolphins, and bats are all mammals, sharks are fish, and therefore, not mammals.
Q4: Which of the following elements is essential for the formation of haemoglobin in the blood?
Iron
Calcium
Sodium
Potassium
A4: The correct answer is A.
Iron is a key component of haemoglobin, a protein in red blood cells responsible for transporting oxygen throughout the body.
General Science questions vary and require vast, yet basic, knowledge in the physical, chemical, biological, and other common scientific fields. You can find many General Science questions with explanations in our All-Inclusive ASVAB PrepPack.
The next section, Mechanical Comprehension, includes mostly Physical subjects in basic mechanics. Unlike General Science – this section may include a large portion of calculative questions regarding common equations in mechanics. Let's continue.
## ASVAB Mechanical Comprehension Sample Questions
Mechanical Comprehension delves into principles associated with simple machines, structural support, properties of materials, and mechanical motion. With 15 questions in 22 minutes (CAT-ASVAB) and 25 questions in 19 minutes (P&P ASVAB), you must have a great knowledge of the basic formulas and concepts to answer knowledge questions quickly and have enough time to calculate.
Q1: A hockey puck is sliding on an infinite plane of ice. The friction between the puck and the ice is zero, as well as the friction between the puck and the air. What will affect the time required for the hockey puck to come to a full stop?
The puck's weight
Gravity
The puck's initial velocity
Nothing, it will never stop
A1: The correct answer is D.
With no friction, the puck is sliding with no forces applied to it apart from its weight. Since the direction of the weight is downwards, it will not affect the velocity which is perpendicular to it (the ice is a horizontal plane). With no other forces applied to the puck, it will never stop due to its inertia (its resistance to a change in its motion). According to inertia, the puck will slide forever without a decelerating force.
Observe the following diagram:
Q2: A perfectly smooth tube with a constant rate of flow of water is given. A segment of the tube with two pressure gauges attached is shown in the diagram below.
Which pressure gauge will indicate a higher value?
1
2
Both will indicate the same
A2: The correct answer is B.
Following Bernoulli's principle, in a steady flow of a fluid (with negligible compression due to pressure and negligible friction due to viscous forces - like water), we have the equation:
where P = pressure, ρgh = potential energy per volume, ½ρv2 = kinetic energy per volume.
Since the area of the tube's cross-section is unchanged throughout, and the rate of water flow is constant, we may conclude that the velocity of the water remains constant (Rate of flow = Velocity*Area). Thus, the kinetic energy remains constant throughout the tube.
The pressure gauges are situated at an identical height; therefore, their potential energy is equal. According to Bernoulli’s principle, the pressure at point 1 must be equal to the pressure at point 2.
More Mechanical Comprehension Questions
Q3: A pulley system is used to:
Increase friction
Increase electrical conductivity
Amplify sound
Change the direction or magnitude of a force
A3: The correct answer is D.
Pulleys are mechanical devices that change the direction or magnitude of a force applied to a load.
Observe the following diagram:
Q4: When the following crate is slipping down the ramp, which of the following arrows describes the direction of the crate's friction?
A
B
C
D
A4: The force of friction is always in the opposite direction to the object's movement. Thus, when the crate is sliding downhill (A), the direction of the friction will be uphill (D).
Let's review the other arrows:
(A) represent a component of the crate's weight. This component acts as the motivating force that causes the crate to slide downhill.
(C) represents a component of the crate's weight, in the direction of gravity. Due to this force the crate adheres to the ramp's surface.
(B) represents a force called the normal force. This force is perpendicular to the surface and prevents the crate from penetrating into the surface. It is identical in size and opposite in direction to C.
Mechanical Comprehension questions may include either knowledge of basic rules and concepts in mechanics and physics or the ability to utilize them for calculations, like the second example. You can find a comprehensive Mechanics guide, questions, and detailed explanations in our All-Inclusive ASVAB PrepPack.
Next up is Electronics Information. Very similar to Mechanical Comprehension – it encompasses the electricity portion of physics in the ASVAB test. Concepts, knowledge, tools, and calculations must be performed in this section.
You can further practice your skills with our Mechanical Aptitude Practice Test.
## ASVAB Electronics Information Sample Questions
Electronics Information focuses on electronic systems and devices' fundamental principles and concepts. With 15 questions in 10 minutes (CAT-ASVAB) and 20 questions in 9 minutes (P&P ASVAB), your time frame is very narrow, and you must come to the test with broad knowledge of electricity concepts, as well as practice many calculation questions prior, and be able to perform them as quickly as possible.
Q1: In which direction does the electric current flow in an electric circuit?
From the negative to the positive terminal
From the positive to the negative terminal
In the opposite direction to the positive-charge flow
In the same direction as the positive-charge flow
A1: The correct answer is B.
Conventional current is designed to flow in the same direction as positive charges. In metals, where the charge carriers (electrons) are negative, conventional current flows in the opposite direction of the electrons. In conductors, where the charge carriers are positive, conventional current flows in the same direction as charge carriers.
Options C and D present the exact opposite of that reasoning.
In a common electrical wired circuit, negatively charged electrons flow from the negative terminal to the positive terminal. According to the above definition, the electric current flows in the oppo
Observe the following diagram:
Q2: A force of 10 N(Newton) compresses two identical springs in parallel for 8 cm. What will be the total distance that four identical springs in series are compressed?
32cm
16cm
20cm
64cm
A2: The correct answer is 64 cm.
An external force exerted on a spring causes it to compress or extend.
The distance of compression or extension is directly and linearly proportional to the force exerted on the spring, i.e., doubling the force causes the spring to stretch or compress twice as much.
When identical springs are connected in parallel, the applied force is divided equally between them. Consider the image below:
Each of the two springs connected in parallel is subjected to a force of 5 N that compresses each spring by 8 cm (1). On the other hand, had only one spring been subjected to a 10N force, it would compress by 16 cm (2).
When identical springs are connected in series, each spring's entire force is applied separately. Consider the image below:
Each of the four springs in the series is subjected to a force of 10 N, and therefore resembles the case in image (2), compressing 16 cm. Thus, the total compression is the combined compression of all four springs: 4 X 16 cm = 64 cm.
More Electronics Information Questions
Q3: A power supply outputs 15V at 3A. How much power does it deliver?
5W
18W
45W
50W
A3: The correct answer is C.
Power P=V × I = 15V × 3A = 45W.
Observe the following diagram:
Q4: What is the total capacitance in the circuit?
1.875 μF
2 μF
8 μF
15 μF
A4: For capacitors in parallel,
Ct=C1+C2.
So, Ct=3+5=8 μF.
Electronic Information questions test your knowledge of electricity and your ability to perform calculations. You can find many Electrical Information questions with explanations in our All-Inclusive ASVAB PrepPack.
## How to Pass the ASVAB With a High Score
After months of thorough research and sifting through feedback from 100’s customers, our experts developed a 3-step formula to ace the ASVAB test. It includes a proven study plan to help you pass every subtest of the ASVAB, even if you’ve been out of school or college for several years.
Step 1: Take our Full ASVAB Introduction Simulation to identify your strengths and weaknesses. Once you finish the test, you get an instant score report highlighting your weaker areas. You can then form a personalized prep plan based on your score and preferred lines.
Step 2: Start prepping with focused practice drillsfor each ASVAB section you need help with. These includestep-by-step explanations that demonstrate how to solve every question. Use our guides and detailed explanations to increase your knowledge further.
Step 3: Finish your preparation with full ASVAB and AFQT Practice Simulations and see your amazing progress. If topics still need improvement, you’ll continue practicing until your final score is excellent.
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The next section comprises two subjects combined – Auto and Shop (in the CAT ASVAB they may be in two different sections – each is as short as half of a full section). The section tests knowledge of automobiles and workshops, which is on the one hand very specific, yet requires wide knowledge in those fields. Let's continue.
## ASVAB Auto and Shop Information Sample Questions
These sections, often combined as Auto & Shop Information in the paper-and-pencil version, assess knowledge and understanding of automobiles, tools, and shop terminology and practices. With 11 questions in 7 minutes for each individual section (CAT-ASVAB) and 25 questions in 11 minutes for the unified section (P&P ASVAB), you must be able to decide on each question quickly.
Q1: What is the primary function of a car's alternator?
Starting the engine
Charging the battery
Filtering the oil
Cooling the engine
A1: The correct answer is B.
The alternator recharges the battery while the vehicle is running and powers other electrical systems.
Q2: A "Phillips" and a "flathead" refer to types of what tool?
You can replace the bold word with:
Wrenches
Pliers
Screwdrivers
Hammers
A2: The correct answer is C.
These are types of screwdriver heads. Phillips has a cross shape, while flathead is a straight line.
More Auto and Shop Questions
Q3: A car's “check engine” light comes on. What tool can diagnose the reason?
OBD-II scanner
Torque wrench
Jumper cable
Tire Pressure gague
A3: The correct answer is A.
An OBD-II scanner can be plugged into a vehicle's OBD-II port to read diagnostic codes, explaining the "check engine" light.
Q4: Which of the following pairs correctly describes a tool and its primary usage?
Hammer: Driver nails; Pliers: Grabs and holds objects
Screwdriver: cuts through metal; hacksaw: measure length
A4: The correct answer is A.
Hammers are primarily used to drive nails, and pliers are used to grab, hold, or manipulate objects.
Auto and Shop questions require broad knowledge in both fields – which on the one hand, can be considered basic but, on the other, requires a lot of studying and practice. While there are few calculations, questions can be tricky and require you not to dwell too much on each question and be able to perform calculative guesses if required. You can find many Auto and Shop questions with explanations in our All-Inclusive ASVAB PrepPack.
The final section remaining on the ASVAB is Assembling Objects. The section is more like the first four sections – in the term that it does not test specific knowledge but abilities and skills. Visualizing and working with images and figures will require thorough yet quick work. Let's continue.
## Applying For An Officer Position in The Air Force?
You will need to take the AFOQT Exam. We have got you covered with everything you will need to ace the test and get your dream job! Please check out our page on the test and learn more about it o our AFOQT Test PrepPack or study with sample questions with our Free AFOQT Practice.
## ASVAB Assembling Objects Sample Questions
Assembling Objects is designed to test spatial awareness and visualization skills. With 15 questions in 18 minutes (CAT-ASVAB) and 25 questions in 15 minutes (P&P ASVAB), time is not sparse to answer.
Observe the following objects:
Q1: Which of the following figures includes all the objects assembled?
A1: The correct answer is B.
Observe the following diagram:
Q2: Which of the figures shown includes all the objects in the top rectangle assembled? Note that the labels (x,y,z) must be paired appropriately, x to x, and so on.
A
B
C
D
E
A2: The correct answer is E.
More Assembling Objects Questions
Observe the following diagram:
Q4: Which of the following figures includes all the objects assembled? Note that the labeled (A,B,C) must be paired appropriately.
A3: The correct answer is B (figure 2).
Look at the ends marked A. If the ends marked A were put together, how would they look? Of the five pictures, only pictures 1 and 2 have the ends marked A put together. Now look at the ends marked C. Which of the pictures 1 and 2 show that the two places marked C are put together? Of the two, only picture 2 has the places marked C put together.
Therefore, picture 2 is the correct answer.
Observe the following diagram:
Q4: The letters near the sides of each shape point out where exactly should the different shapes be joined together. Which of the following options shows the joint shape?
A
B
C
D
A4: The correct answer is B.
Assembling Objects questions are unique – and with 2D and 3D assembling requirements – you must be able to visualize the object and quickly decide on the correct answer. Prior practice is required to be able to visualize images on screen and assemble them to respond quickly and accurately. You can find hundreds of Assembling Objects questions with explanations in our All-Inclusive ASVAB PrepPack.
## AFQT Scoring:
The scoring of the AFQT calculates four sections – Mathematical Knowledge (MK), Arithmetic Reasoning (AR), Word Knowledge (WK), and Paragraph Comprehension (PC). The minimal score required for each branch is as follows:
• Air Force – 31
• Army – 31
• Marine Corps – 32
• Navy – 35
• Coast Guard - 40
The aforementioned required scores are for High School Seniors / High School Diploma Recipients. For GED Holder, the minimal score for all the branches is 50.
Not what you are looking for? Please contact us, and we will do our best to ensure you get the most accurate preparation for your upcoming assessment.
Let’s continue to the first of two verbal sections of the ASVAB and AFQT – Word Knowledge.
## Types of ASVAB Tests
1. CAT-ASVAB (Computer Adaptive Test): This version of the test adjusts the difficulty of questions based on the test taker's responses. It allows for a tailored test experience and provides immediate scores upon completion.
2. P&P-ASVAB (Paper and Pencil): A traditional version of the test administered with paper and pencil. The test-taker answers all the questions in a fixed format.
3. ASVAB Career Exploration Program (CEP): an extension of the ASVAB test, aimed not just at those considering military service but also at high school and post-secondary students who are exploring career options.
Also checkout our ASVAB Test Guide.
### Get to Know the ASVAB Inside Out
The ASVAB, or Armed Services Vocational Aptitude Battery, consists of several subtests designed to evaluate specific skills and abilities. Here's a brief breakdown of those sections and how scoring works:
Sections of the ASVAB:
1. General Science (GS): Assesses physical and biological sciences knowledge.
2. Arithmetic Reasoning (AR): Tests ability to solve arithmetic word problems.
3. Word Knowledge (WK): Measures understanding of word meanings.
4. Paragraph Comprehension (PC): Evaluates the ability to understand written material.
5. Mathematics Knowledge (MK): Assesses high school math principles knowledge.
6. Electronics Information (EI): Measures knowledge of electronic systems and devices.
7. Auto and Shop Information (AS): Tests knowledge of automobile mechanics and shop practices.
8. Mechanical Comprehension (MC): Evaluates understanding of basic mechanical principles.
9. Assembling Objects (AO): Assesses spatial visualization and tool recognition.
The test can be divided into the AFQT (Armed Forces Qualifications Score), comprised of the MK, AR, WK, and PC sections, and must be successfully passed to qualify for service in the different branches. The other sections and the four AFQT sections comprise the composite scores, which determine qualification for specific military occupational specialties (MOS).
You can also study for the un-timed at-home version of the ASVAB test with our PiCAT PrepPack, or learn more on our PiCAT Free Practice page.
## ASVAB PrepPacks
Below, you'll find popular ASVAB PrepPacks. Choose the one that suits you the most and begin your path to a successful career in the service.
### 🪖Army ASVAB Practice
The Army is famous for its special teams like the Army Rangers, Green Berets (special forces), SOAR, ISA, AWG, and others. The National Guard is also a part of the Army. Find out more about what the Army needs and the roles you can take on. Begin your preparation with JobTestPrep's Army ASVAB practice pack to get the best score possible.
### 🛫Air Force ASVAB Practice
To enlist in the Air Force, you need a minimum of 31 on the ASVAB AFQT exam, or 50 if you possess a GED.
Yet, aiming for premier Air Force positions (AFSC) demands more effort. This can be challenging, particularly for those who've been away from academics or testing environments for a while. However, proper Air Force ASVAB preparation can turn this aim into reality and help improve your overall Air Force ASVAB Score.
### 🚢Navy ASVAB Practice
If you want to join the Navy, you need to score at least 35 on the AFQT part of your ASVAB test. This is the basic requirement for everyone.
To qualify for the top jobs in the Navy (NEC) and achieve your full potential, you need a high score. Considering the ASVAB is a challenging 3-hour test, scoring high isn't easy so Navy ASVAB preparation is vital and can help increase your overall score.
### 💣Marine Corps ASVAB Practice
The Marine Corps is well-known for its excellent units, including Marine Recon, USMC FAST, USMC Anglico (special forces), USMC MARSOC, and others. To join, you need determination, drive, good physical condition, and especially a high score on the Marine Corps ASVAB Test. Start your Marine Corps preparation by pressing the button below.
We rely on customer feedback to ensure our PrepPacks stay accurate and suited to match test-taker needs. Do you have questions regarding which PrepPack is best for you? Can't find the PrepPack you're looking for? Let us help! Reach out at info@jobtestprep.com.
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Q 1-In an examination, 80% of the students passed in English, 85% in Mathematics and 75 % in both English and Mathematics. If 40 students failed in both the subjects , find the total number of students.A - 350B - 375C - 400D - 450
```Answer - C
Explanation
"Let the total number of students be x
Let A and B represent the sets of students
who passed in English and Mathematics respectively.
Then, number of students passed in one or both the subjects
= n (A∪ B) = n(A) + n(B) - n(A∩ B)
= 80% of x + 85% of x - 75% of x
=(80/100x+85/100x−75/100x) = 90/100x
= 9/10x
Students who failed in both the subjects = (x − 9x / 10)= x / 10
So, x/10= 40 of x = 400.
Hence total number of students = 400."
```
Q 2-In an examination 35% of total students failed in Hindi , 45% failed English and 20% in both .Find the percentage of those who passed in both the subjects .A - 40%B - 42%C - 43%D - 44%
```Answer - A
Explanation
"Then A and B be the sets of students
who failed in Hindi and English respectively
Then , n(A) = 35, n(B) = 45,
n(A∩B) = 20
so, n(A∪B) = n(A) + n(B) - n(A∩B)
= (35 + 45 - 20) = 60
∴ Percentage failed in Hindi or English or both = 60%
Hence ,percentage passed = (100 - 60)% = 40%."
```
Q 3-Due to a reduction of 6(1/4)% in the price of sugar , a man is able to buy 1 kg more for Rs. 120. Find the original and reduced rate of sugar .A - 6.50 kgB - 7 kgC - 7.25kgD - 7,50 kg
```Answer - D
Explanation
"Le original rate be Rs. x per kg.
Reduced rate = Rs.[(100−[25/4])×[1/100x]]
=Rs. 15x/16 per kg.
∴120/[15x/16]−120/x=1
⇔128/x−(120/x)= 1
⇔ x=8
So , original rate = Rs. 8 per kg.
Reduced rate = Rs.(15/16×8) per kg. = Rs. 7.50 per kg.
"
```
Q 4-How many kg of pure salt must be added to 30kg of 2% solution of salt and water to increase it to a 10% solution ?A - 1(2/3)B - 2(1/2)C - 2(1/3)D - 2(2/3)
```Answer - D
Explanation
"Amount of salt in 30 g solution = (2/100×30) kg = 0.6 kg
Let x kg of pure salt be added .
Then , 0.6+x/30+x= 10/100
⇔ 60+100x= 300+10x
⇔ 90x=240
⇔x=8/3=2(2/3)
"
```
Q 5-If A's salary is 20% less than B's salary, by how much percent is B's salary more than A'sA - 25%B - 26%C - 27%D - 28%
```Answer - A
Explanation
"Required percentage = [20/(100−20)×100]%
= 25%.
"
```
Q 6-If A earns 33(1/3)% more than B,how much percent does B earn less then A?A - 24%B - 25%C - 26%D - 27%
```Answer - B
Explanation
"Required percentage = ( {[ 100 / 3] / [100 + [100/3]} x 100 )%
=([100 / 400] x 100) %
= 25%."
```
Q 7-During one year , the population of a town increased by 5% and during the next year , the population decreased by 5% . If the total population is 9975 at the end of the second year, then what was the population size in the beginning of the first year ?A - 10000B - 10010C - 10100D - 11000
```Answer - A
Explanation
"population in the beginning of the first year
= 9975 / [(1+{5/100})(1−{5/100})]
= (9975× [20/21] × [20/19] )
= 10000
"
```
Q 8-In the new budget , the price of kerosene oil rose by 25% . By how much percent must a persn reduce his consumption so that his expenditure on it does not increase ?A - 20%B - 21%C - 22%D - 23%
```Answer - A
Explanation
"Reduction in consumption = [R/(100+R)×100]%
= ([25/125]×100)
= 20%.
"
```
Q 9-If the numerator of a fraction be increased by 15% and its denominator be diminished by 8% , the value of the fraction is 15/16. Find the original fraction.A - 2/3B - 3/4C - 3/5D - 2/5
```Answer - B
Explanation
"Let the original fraction be. x/y
Then
115%of x/92%of y= 15/16
⇒ 115x/92y= 15/16 ⇒ x/y
= (15/16×92/115)
= 3/4.
"
```
Q 10-When the price of a product was decreased by 10% , the number sld increased by 30% . What was the effect on the total revenue ?A - 17%B - 18%C - 19%D - 20%
```Answer - A
Explanation
"Let the price of the product be Rs. 100
and let original sale be 100 pieces.
Then , total Revenue = Rs. 100 x 100 = 10000
New revenue = Rs. 90 x 130 = Rs. 11700
∴ Increase in revenue = ([1700/10000]×100))%
= 17%
"
```
Q 11-The salary of a person was reduced by 10% . By what percent should his reduced salary be raised so as to bring it at per with his original salary ?A - 11(1/3)%B - 11(2/5)%C - 11(1/8)%D - 11(1/9)%
```Answer - D
Explanation
"Let the original salary be Rs. 100. New salary = Rs. 90.
Increase on 90 = 10. Increase on 100 = ([10/90]×100)%
= 11(1/9)%.
"
```
Q 12-Paulson spends 75% of his income . His income is increased by 20% and he increased his expenditure by 10%. Find the percentage increase in his savings.A - 45%B - 47%C - 49%D - 50%
```Answer - D
Explanation
"Let original income = Rs. 100 .
Then , expenditure = Rs. 75 and saving = 25
New income = Rs 120, New expenditure = Rs.([110/100]×75)
= Rs. 165/2
New saving = Rs. (120−[165/2] )
= Rs. 75/2
Increase in savings = Rs. ([75/2−25]
= Rs. 25/2
∴ Increase % = ([25/2×1/25]×100)%
= 50%
"
```
Q 13-Raman's salary was decreased by 50% and subsequently increased by 50% Bow much percent does he lose ?A - 25%B - 26%C - 27%D - 28%
```Answer - A
Explanation
"Let original salary = Rs. 100
New final salary = 150% of (50% of 100)
= Rs. ([150/100[×[50/100]×100)
= Rs.75
∴Decrease = 25%.
"
```
Q 14-A salesman's commission is 5 on all sales upto Rs. 10,000 and 4% on all sales exceeding this . He remits Rs 31,100 to his parent company after deducting his commission. Find the total sales.A - 31500B - 32000C - 32500D - 32650
```Answer - A
Explanation
"Let his total sales be Rs. x . Now, (total sales ) - (Commission)
= Rs. 31,100
∴ x - [5% of 10000 + 4% of (x- 10000)] = 31100
⇔ x−[(5/100)×10000+4/100(x−10000)]
= 31100
⇔ x−500−(x−10000/25)
= 31100
⇔ x−(x/25)=31200
⇔ 24x/25=31200
⇔ x=(31200×25/24) = 32500.
∴Total sales = Rs. 32,500.
"
```
Q 15-10% of the inhabitants of a village having died of cholera, a panic set in, during which 25% of the remaining inhabitants left the village . The population is them reduced to 4050. Find the number of original inhabitants.A - 6000B - 6100C - 6200D - 6300
```Answer - A
Explanation
"Let the total number of original inhabitants be x
Then , (100 - 25)% of (100 - 10)% of x = 4050
⇔([75/100]×[90/100]*x)= 4050
⇔ (27/40)x =4050
⇔ x= (4050×40/27) = 6000.
∴Number of original inhabitants = 6000.
"
```
|
Need the Area of a Triangle? The Pope Can Help! – Explorations for Students, Part 1
Author(s):
Betty Mayfield (Hood College)
Explorations for Students 1
1. The arithmetical rule for finding area, as we have seen, comes from the description for a triangular number, usually written $1 + 2 + \ldots + n = \frac{n(n + 1)}{2},$ where $n$ is a positive integer. But could the rule also be used for triangles with non-integer sides?
• Use the arithmetical rule $A = \frac{b(b + 1)}{2}$ to find the (approximate) area of an equilateral triangle of side 2. (We will omit units for the sake of simplicity.)
• Now use it for an equilateral triangle of side 3.
• What result does it give for an equilateral triangle of side $2\frac{1}{2}$? Given your previous answers, does it seem reasonable?
• What result does it give for an equilateral triangle of side $\sqrt{5}$? Again, does it seem to work?
1. Exact and approximate answers
• Sketch an equilateral triangle of side 7. Draw in its altitude. What is its exact value? How do you know?
• What is the true area of an equilateral triangle of side 7?
• You have access to two tools which Gerbert did not have: the decimal representation of a real number, and a hand-held calculator. Find a decimal approximation for the area of an equilateral triangle of side 7. Round to 3 decimal places.
• What result did Gerbert get, using his $\frac{6}{7}$ rule? How far off was he?
• What about for the triangle with side 30? Find a decimal representation for the correct area and compare it to Gerbert’s result. How far off was he this time?
It appears that things get worse, the bigger the triangle. But there is another way to look at error: the relative error is defined as $\frac{error}{true value}$.
• For the triangle of side 7, the error was _____ and the exact area (to 3 decimal places) was _____, so the relative error was the ratio of those two numbers, or __________.
• Find the relative error for the triangle of side 30. What do you notice?
• In general, we might say that Gerbert’s special $\frac{6}{7}$ rule for finding the area of an equilateral triangle has an error of about _____%. Not bad!
1. Why $\bf{\frac{6}{7}}$?
All equilateral triangles are similar, so if we can find the altitude of an equilateral triangle of side 1, we can find the altitude of any such triangle, simply by multiplying the result by the side length $b$.
• Use the Pythagorean Theorem to find the (exact) measure of the altitude of an equiangular triangle of side 1.
• Use a calculator to find decimal approximations to $\frac{6}{7}$ and $\frac{\sqrt{3}}{2}$, say to five decimal places. How close was Gerbert’s approximation?
• Consider all the simple fractions between 0 and 1 of the form $\frac{n}{n + 1}$: $\frac{1}{2}, \frac{2}{3}, \ldots, \frac{9}{10}.$
Which of them is closest to $\frac{\sqrt{3}}{2}$? Do you think Gerbert knew that?
Betty Mayfield (Hood College), "Need the Area of a Triangle? The Pope Can Help! – Explorations for Students, Part 1," Convergence (November 2022)
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# Four pipes can fill a tank in 70 minutes. How long will it take to fill the tank if 7 pipes are used?
Last updated date: 21st Jul 2024
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Hint: We must first assume a variable (say $x$) that depicts the rate at which fluid flows from one pipe. We can then use the information that 4 pipes take 70 minutes to fill the tank, to find the volume of the tank. We will again have to assume a variable (say $t$) that is the time taken to fill the tank by 7 pipes. Hence, by equating the volume in two cases, we can find the value of variable $t$.
Complete step-by-step solution:
Let us assume that a pipe can fill $x$ litres in 1 minute, that is, $x$ litres per minute.
So, we can say that 4 pipes can fill $4x$ litres in 1 minute, that is, $4x$ litres per minute.
We are given that 4 pipes take 70 minutes to fill the tank.
So, we can easily write that in 70 minutes, 4 pipes will fill = $70\times 4x$ litres.
Thus, we can say that the total volume of the tank = $280x$ litres.
Now, since we know that 1 pipe fills $x$ litres per minute, so we can very well say that 7 pipes will fill the tank at the rate of $7x$ litres per minute.
Again, let us assume that the time taken to fill the tank by 7 pipes is $t$ minutes.
So, we can say that in $t$ minutes, 7 pipes will fill = $7xt$ litres.
But we had assumed that in $t$ minutes, the tank is completely filled. So, $7xt$ litres must be equal to the volume of the tank, which is $280x$ litres.
Thus, we can write
$7xt=280x$
Cancelling the terms from both sides, we get
$t=40$ minutes.
Hence, we can say that 7 pipes can fill the tank in 40 minutes.
Note: We can also solve this problem without having to assume any extra variable, using the following approach. We know that 4 pipes can fill the tank in 70 minutes. So, we can say that 1 pipe will tank in $70\times 4=280$ minutes. And thus, 7 pipes can fill the tank in $\dfrac{280}{7}=40$ minutes.
|
# FORM FOUR MATHEMTICS STUDY NOTES TOPIC 2: AREA AND PERIMETER
## AREA AND PERIMETER
Area of any Triangle
The Formula for the Area of any Triangle
Derive the formula for the area of any triangle
Area of triangle is given by½bh, whereby b is the base of the triangle and h is the height of the given triangle. Consider the illustrations below:
From the figure above, we see where the base and height are located.
Applying the Formula to find the Area of any Triangle
Apply the formula to find the area of any triangle
Example 1
The base of a triangle is 12cm long. If the corresponding height is 7cm, find the area of the triangle.
Solution
Consider the figure below:
The area of a triangle is given by½bh.
Area =½×12cm×7cm
Area = 42cm2
Therefore area of a triangle is 42cm2
Example 2
The lengths of two sides of a triangle are 6cm and 8cm. Find the area of a triangle if the included is
Solution:
Consider the triangle below, name it triangle ABC.
The area of a triangle above is given by½b×h
So, Area = ½× 8cm× 6cm × sin 45°
=24cm2× sin45°
= 16.97cm2
Therefore the area of ABC = 16.97cm2
### 2 thoughts on “FORM FOUR MATHEMTICS STUDY NOTES TOPIC 2: AREA AND PERIMETER”
1. A simple and clear notes
|
# Vector Multiplication
The multiplication of a vector by a vector produces some interesting results, known as the vector inner product and as the vector outer product.
Prerequisite: This material assumes familiarity with matrix multiplication.
## Vector Inner Product
Assume that a and b are vectors, each with the same number of elements. Then, the inner product of a and b is s.
a'b = b'a = s
where
a and b are column vectors, each having n elements,
a' is the transpose of a, which makes a' a row vector,
b' is the transpose of b, which makes b' a row vector, and
s is a scalar; that is, s is a real number - not a matrix.
Note this interesting result. The product of two matrices is usually another matrix. However, the inner product of two vectors is different. It results in a real number - not a matrix. This is illustrated below.
a =
1 2 3
b =
4 5 6
Then,
a'b = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32
Thus, the inner product of a'b is equal to 32.
Note: The inner product is also known as the dot product or as the scalar product.
## Vector Outer Product
Assume that a and b are vectors. Then, the outer product of a and b is C.
ab'= C
where
a is a column vector, having m elements,
b is a column vector, having n elements,
b' is the transpose of b, which makes b' a row vector, and
C is a rectangular m x n matrix
Unlike the inner product, the outer product of two vectors produces a rectangular matrix, not a scalar. This is illustrated below.
a =
v w
b =
x y z
Then,
C = ab' =
v * x v * y v * z w * x w * y w * z
Notice that the elements of Matrix C consist of the product of elements from Vector A crossed with elements from Vector B. Thus, Matrix C winds up being a matrix of cross products from the two vectors.
## Test Your Understanding of This Lesson
Problems
Consider the matrices shown below - a, b, and c
a =
0 1
b =
2 3
c =
4 5 6
Using a, b, and c, answer the questions below.
1. Find a'b, the inner product of a and b.
2. Find bc', the outer product of b and c.
3. True or false: bc' = cb'
Solutions
1. The term a'b is an inner product, which is equal to 3. The solution appears below.
a'b =
0 1
*
2 3
= 0*2 + 1*3 = 3
2. The term bc' is an outer product, which results in the 2 x 3 matrix D.
bc' =
2 3
*
4 5 6
=
2*4 2*5 2*6 3*4 3*5 3*6
=
8 10 12 12 15 18
= D
3. The statement bc' = cb' is false.
Note that b is a 2 x 1 vector and c is a 3 x 1 vector. Therefore, bc' is a 2 x 3 matrix, and cb' is a 3 x 2 matrix. Because bc' and cb' have different dimensions, they cannot be equal.
|
### Permutation And Combination
Permutations and Combination, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called permutation when the order of selection is a factor, a combination when order is not a factor.
### Factorial:
The product of ‘n’ natural number 1, 2, 3.—-n is denoted by n! and read as n factorial
0! is defined to equal 1.
n! = 1 x 2 x 3 x …….x (n – 1) x n
1! = 1
0! = 1
n! = n. (n-1)! = n (n-1)(n-2)….etc
### Permutation
The concepts and differences between permutations and combinations can be illustrated by an examination of all the different ways in which a pair of objects can be selected from five different objects—
such as the letters A, B, C, D, and E. If both the letters selected and the order of selection are considered, then the following 20 outcomes are possible:
Here, each of these 20 different possible selections is called a permutation. They are called the permutations of five objects taken two at a time, and the number of such permutations possible is denoted by the symbol:
5P2, read – 5 permute 2.
Permutations are the different ways in which a collection of items can be arranged.
If there are n objects available from which to select, and permutations (P) are to be formed using k of the objects at a time, the number of different permutations possible is denoted by the symbol nPk or P(n ,r) .
nPk = n!(n − k)!
where n! is “n factorial
### Properties of Permutation
1). nPk = n(n – 1)(n – 2)……1 = n!
2). nP0 = 1
3). nP1 = n
4). nPn-1 = n!
5). nPrnPr-1 = n – r + 1
For Example:
12!= 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
The permutations of 2, 3, 4, 5 are:
• 5432, 5423, 5324, 5342, 5234, 5243, 4532, 4523, 4325, 4352, 4253, 4235, 3542, 3524, 3425, 3452, 3254, 3245, 2543, 2534, 2435, 2453, 2354, 2345.
### Some Examples:
1). A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?
Solution: The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 9 x 12 = 108 ways.
2). How many three-letter words are formed using the letters of the word TIME?
Solution: The number of letters in the given word is four.
The number of three-letter words that can be formed using these four letters is ⁴P₃ = 4x 3 x 2 = 24.
3). Using all the letters of the word “THURSDAY”, how many different words can be formed?
Solution: Total number of letters = 8
Using these letters the number of 8 letters words formed is ⁸P₈ = 8!.
### Combination
For combinations, k objects are selected from a set of n objects to produce subsets without ordering. The previous permutation example:
The corresponding combination, the AB and BA subsets are no longer different selections;
So by eliminating such cases there remain only 10 different possible subsets—AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE.
The number of such subsets is denoted by:
nCk or C(n ,r) , read n choose k. For combinations, since k objects have k! arrangements, there are k! identical permutations for each choice of k objects;
hence dividing the permutation formula by k! : the following combination formula:
nCk = n!k! (n − k)! , (0 ≤ k ≤ n)
### Properties of Combination
1). nCn = 1
2). nC0 = 1
3). nC1 = n
4). nCr = nC(n-r)
5). nCr = nPrr!
6). nCr + nCr-1 = n+1Cr
7). nC0 + nC1 + nC2 + —-+ nCn = 2n
8). nC0 + nC2 + nC4+—- = nC1 + nC3 + nC5+—- = 2n-1
9). If nCr = nCp then either
r = p or r + p = n
### Some Examples:
1).In how many ways can a coach choose three swimmers from among five swimmers?
Solution:
There are 5 swimmers to be taken 3 at a time.
C(5,3) = P(5,3)/3! =5 x 4 x 3/3 x 2 x1= 10
The coach can choose the swimmers in 10 ways.
2). Six friends want to play enough games of chess to be sure everyone plays everyone else. How many games will they have to play?
Solution:
There are 6 players to be taken 2 at a time.
C(6,2)=P(6,2)/2!=6 x 5/2 x 1=15.
They will need to play 15 games.
3). Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Solution:
Number of ways of selecting 3 consonants from 7 = 7C3
Number of ways of selecting 2 vowels from 4 = 4C2
Number of ways of selecting 3 consonants from 7 and 2 vowels from 4
= 7C3 × 4C2
=(7×6×5/3×2×1)×(4×3/2×1)=210
we can have 210 groups where each group contains total 5 letters :
3 consonants and 2 vowels.
Number of ways of arranging 5 letters among themselves
=5!=5×4×3×2×1=120
Hence, the number of ways
=210×120=25200
4). In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Solution: In a group of 6 boys and 4 girls, four children are to be selected such that at least one boy should be there.
We can select 4 boys …(1)
Number of ways to this = 6C4
We can select 3 boys and 1 girl …(2)
Number of ways to this = 6C3 × 4C1
We can select 2 boys and 2 girls …(3)
Number of ways to this = 6C2 × 4C2
We can select 1 boy and 3 girls …(4)
Number of ways to this = 6C1 × 4C3
Total number of ways:
= 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3
= 6C2 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C1 [∵ nCr = nC(n-r)]
=(6×5/2×1)+(6×5×4/3×2×1)×4+(6×5/2×1)×(4×3/2×1)+6×(4)
=15+80+90+24=209
|
# Algebra 1 : How to find the whole from the part with percentage
## Example Questions
### Example Question #1 : Whole And Part
70% of a quantity is 35. What is the quantity?
Explanation:
We can write this as an equation:
### Example Question #2 : Whole And Part
80% of ____ = 72?
Explanation:
Divide by fractions:
### Example Question #3 : Whole And Part
Malcolm spent 15% of his money on a bicycle that costs $300. How many dollars does Malcolm have left? Possible Answers: Correct answer: Explanation: If$300 was 15% of Malcolm's money, then we can figure out how much money Malcolm had by creating this equation:
In this case, Malcolm had $2,000. Since he spent$300 of it on a bicycle, he has only $1,700 left. ### Example Question #1 : Whole And Part If 45 is 15% of , find the value of . Possible Answers: Correct answer: Explanation: The key to this problem is identifying that "15% of " is the same as . With this information, we can write out the simple equation Dividing both sides by gives us ### Example Question #1 : How To Find The Whole From The Part With Percentage % of what number is 900? Possible Answers: Correct answer: Explanation: % of a number, or, equivalently, 1.5% of a number, is the same as 0.015 multiplied by that number. If we call that number , then ### Example Question #1 : Whole And Part Dana spent 24% of her savings on a laptop that costs$900. How much savings does she have left?
Explanation:
We know that prior to her purchase, $900 was 24% of Dana's Savings. Therefore, Dana's total savings prior to her purchase can be modeled as , where is Dana's total savings. Solving for would give you , which indicates that Dana's savings was$3750 prior to her purchase. After her purchase, she will have \$2850 left.
### Example Question #4 : Whole And Part
There are 36 blue marbles in a bag. If blue marbles made up 24% of the marbles in the bag, what is the total number of marbles are in the bag?
Explanation:
If is the total number of marbles in the bag, then , since 24% of marbles in the bag are blue and there are 36 marbles. Solving for this equation will give you , which means there must be a total of 150 marbles in the bag.
### Example Question #8 : How To Find The Whole From The Part With Percentage
12% of the students at a certain high school have perfect attendance. If 27 students have perfect attendance, how many total students does the school contain?
Explanation:
From the given information here, we know that 12% of the total number equals 27. Mathematically, if we use to represent the total that we are looking for, we can write this as
The next step is dividing both sides by .
so there are 225 students in the school.
### Example Question #1 : Whole And Part
is of what?
Explanation:
To figure out the value, translate the question into an equation, knowing that "is" means equals, and "of" means multiply:
To solve, turn the percentage into a decimal:
now divide both sides by 0.48
### Example Question #1 : How To Find The Whole From The Part With Percentage
is of what number?
|
# 7.1 Linear momentum and force
Page 1 / 4
• Define linear momentum.
• Explain the relationship between momentum and force.
• State Newton’s second law of motion in terms of momentum.
• Calculate momentum given mass and velocity.
## Linear momentum
The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as
$\mathbf{p}=m\mathbf{\text{v}}.$
Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its velocity, the greater its momentum. Momentum $\mathbf{p}$ is a vector having the same direction as the velocity $\mathbf{\text{v}}$ . The SI unit for momentum is $\text{kg}·\text{m/s}$ .
## Linear momentum
Linear momentum is defined as the product of a system’s mass multiplied by its velocity:
$\mathbf{p}=m\mathbf{\text{v}}.$
## Calculating momentum: a football player and a football
(a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s.
Strategy
No information is given regarding direction, and so we can calculate only the magnitude of the momentum, $p$ . (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes
$p=\text{mv}$
when only magnitudes are considered.
Solution for (a)
To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation.
${p}_{\text{player}}=\left(\text{110 kg}\right)\left(8\text{.}\text{00 m/s}\right)=\text{880 kg}·\text{m/s}$
Solution for (b)
To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation.
${p}_{\text{ball}}=\left(\text{0.410 kg}\right)\left(\text{25.0 m/s}\right)=\text{10.3 kg}·\text{m/s}$
The ratio of the player’s momentum to that of the ball is
$\frac{{p}_{\text{player}}}{{p}_{\text{ball}}}=\frac{\text{880}}{\text{10}\text{.}3}=\text{85}\text{.}9.$
Discussion
Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player’s motion is only slightly affected if he catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections.
## Momentum and newton’s second law
The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. Using symbols, this law is
${\mathbf{F}}_{\text{net}}=\frac{\Delta \mathbf{p}}{\Delta t},$
where ${\mathbf{F}}_{\text{net}}$ is the net external force, $\Delta \mathbf{p}$ is the change in momentum, and $\Delta t$ is the change in time.
## Newton’s second law of motion in terms of momentum
The net external force equals the change in momentum of a system divided by the time over which it changes.
${\mathbf{F}}_{\text{net}}=\frac{\Delta \mathbf{p}}{\Delta t}$
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Berger describes sociologists as concerned with
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How to Multiply and Divide Fractions? (+FREE Worksheet!)
Multiplying fractions is easier than adding and subtracting them. It only has two simple steps and that's it! To learn it join us.
Multiplying fractions may be easier for you than multiplying integers because the numbers you work with in fractions are usually smaller. The other good news is that dividing fractions is as simple as multiplying them. Use the following easy steps to multiply or divide fractions.
Step by step guide to multiply and divide fractions
Multiplying fractions:
• Step 1: Multiply the numerators of the fractions by each other to get the numerator of the new fraction.
• Step 2: Multiply the denominators of the fractions by each other to get the denominator of the new fraction.
• Tip: Sometimes when multiplying fractions, you have the opportunity to simplify fractions. For example, when the numerator and denominator of a fraction are both even numbers, this fraction can be simplified.When multiplying fractions, you can usually simplify the fraction by removing equal factors in the numerator and denominator of the fraction. Crossing out equal factors will make the numbers smaller for you, making it easier to work with those numbers. You also get rid of simplifying the fraction at the end of the multiplication.
How to simplify fractions during multiplication operations:
• Step 1: If the numerator of one fraction and the denominator of another fraction are the same number, change both of these numbers to $$1$$ and crossing out the corresponding numbers.
• Step 2: When the numerator of one fraction and the denominator of another fraction are both divisible by the same number, factor that number from both (both the numerator and the denominator). In other words, divide the numerator and the denominator by that common factor.
Dividing fractions:
You know that inverse division is multiplication. So to divide fractions:
• Step 1: Write the first fraction in the same way as the original
• Step 2: The division sign is converted to multiplication sign
• Step 3: The second fraction is reversed and fliped the numerator and denominator
• Step 4: Now you can slove like multiplication of fractions
Multiplying and Dividing Fractions – Example 1:
Multiply fractions. $$\frac{2}{5} \times \frac{3}{4}=$$
Solution:
Multiply the top numbers and multiply the bottom numbers.
$$\frac{2}{5} \times \frac{3}{4}= \frac{2 \ \times \ 3}{5 \ \times \ 4}=\frac{6}{20}$$ , simplify: $$\frac{6}{20} = \frac{6 \ \div \ 2}{20 \ \div \ 2}=\frac{3}{10}$$
Multiplying and Dividing Fractions – Example 2:
Divide fractions. $$\frac{1}{2} \div \frac{3}{5}=$$
Solution:
Keep first fraction, change division sign to multiplication, and flip the numerator and denominator of the second fraction. Then: $$\frac{1}{2} \times \frac{5}{3} = \frac{1 \ \times \ 5}{2 \ \times \ 3}=\frac{5}{6}$$
Multiplying and Dividing Fractions – Example 3:
Multiply fractions. $$\frac{5}{6} \times \frac{3}{4}=$$
Solution:
Multiply the top numbers and multiply the bottom numbers.
$$\frac{5}{6}×\frac{3}{4}=\frac{5×3}{6×4}=\frac{15}{24}$$, simplify: $$\frac{15}{24}=\frac{15÷3}{24÷3}=\frac{5}{8}$$
Multiplying and Dividing Fractions – Example 4:
Divide fractions. $$\frac{1}{4} \div \frac{2}{3}=$$
Solution:
Keep the first fraction, change the division sign to multiplication, and flip the numerator and denominator of the second fraction. Then: $$\frac{1}{4}×\frac{3}{2}=\frac{1×3}{4×2}=\frac{3}{8 }$$
Exercises for Multiplying and Dividing Fractions
Calculate. Simplify if necessary.
1. $$\color{blue}{\frac{1}{5} \times \frac{2}{3}}$$
2. $$\color{blue}{\frac{3}{4} \times \frac{2}{3}}$$
3. $$\color{blue}{\frac{2}{5} \times \frac{3}{7}}$$
4. $$\color{blue}{\frac{2}{9} \div \frac{1}{4}}$$
5. $$\color{blue}{\frac{1}{2} \div \frac{1}{3}}$$
6. $$\color{blue}{\frac{6}{11} \div \frac{3}{4}}$$
1. $$\color{blue}{\frac{2}{15}}$$
2. $$\color{blue}{\frac{1}{2}}$$
3. $$\color{blue}{\frac{6}{35}}$$
4. $$\color{blue}{\frac{8}{9}}$$
5. $$\color{blue}{\frac{3}{2}}$$
6. $$\color{blue}{\frac{8}{11}}$$
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# Basic Analysis I & II: Introduction to Real Analysis, Volumes I & II
## Section0.3Basic set theory
Note: 1–3 lectures (some material can be skipped, covered lightly, or left as reading)
Before we start talking about analysis, we need to fix some language. Modern 1 analysis uses the language of sets, and therefore that is where we start. We talk about sets in a rather informal way, using the so-called “naïve set theory.” Do not worry, that is what the majority of mathematicians use, and it is hard to get into trouble. The reader has hopefully seen the very basics of set theory and proof writing before, and this section should be a quick refresher.
### Subsection0.3.1Sets
#### Definition0.3.1.
A set is a collection of objects called elements or members. A set with no objects is called the empty set and is denoted by $$\emptyset$$ (or sometimes by $$\{ \}$$).
Think of a set as a club with a certain membership. For example, the students who play chess are members of the chess club. The same student can be a member of many different clubs. However, do not take the analogy too far. A set is only defined by the members that form the set; two sets that have the same members are the same set.
Most of the time we will consider sets of numbers. For example, the set
\begin{equation*} S \coloneqq \{ 0, 1, 2 \} \end{equation*}
is the set containing the three elements 0, 1, and 2. By “$$\coloneqq$$”, we mean we are defining what $$S$$ is, rather than just showing equality. We write
\begin{equation*} 1 \in S \end{equation*}
to denote that the number 1 belongs to the set $$S\text{.}$$ That is, 1 is a member of $$S\text{.}$$ At times we want to say that two elements are in a set $$S\text{,}$$ so we write “$$1,2 \in S$$” as a shorthand for “$$1 \in S$$ and $$2 \in S\text{.}$$
Similarly, we write
\begin{equation*} 7 \notin S \end{equation*}
to denote that the number 7 is not in $$S\text{.}$$ That is, 7 is not a member of $$S\text{.}$$
The elements of all sets under consideration come from some set we call the universe. For simplicity, we often consider the universe to be the set that contains only the elements we are interested in. The universe is generally understood from context and is not explicitly mentioned. In this course, our universe will most often be the set of real numbers.
While the elements of a set are often numbers, other objects, such as other sets, can be elements of a set. A set may also contain some of the same elements as another set. For example,
\begin{equation*} T \coloneqq \{ 0, 2 \} \end{equation*}
contains the numbers 0 and 2. In this case all elements of $$T$$ also belong to $$S\text{.}$$ We write $$T \subset S\text{.}$$ See Figure 1 for a diagram.
#### Definition0.3.2.
1. A set $$A$$ is a subset of a set $$B$$ if $$x \in A$$ implies $$x \in B\text{,}$$ and we write $$A \subset B\text{.}$$ That is, all members of $$A$$ are also members of $$B\text{.}$$ At times we write $$B \supset A$$ to mean the same thing.
2. Two sets $$A$$ and $$B$$ are equal if $$A \subset B$$ and $$B \subset A\text{.}$$ We write $$A = B\text{.}$$ That is, $$A$$ and $$B$$ contain exactly the same elements. If it is not true that $$A$$ and $$B$$ are equal, then we write $$A \not= B\text{.}$$
3. A set $$A$$ is a proper subset of $$B$$ if $$A \subset B$$ and $$A \not= B\text{.}$$ We write $$A \subsetneq B\text{.}$$
For the example $$S$$ and $$T$$ defined above, $$T \subset S\text{,}$$ but $$T \not= S\text{.}$$ So $$T$$ is a proper subset of $$S\text{.}$$ If $$A = B\text{,}$$ then $$A$$ and $$B$$ are simply two names for the same exact set.
To define sets, one often uses the set building notation,
\begin{equation*} \bigl\{ x \in A : P(x) \bigr\} . \end{equation*}
This notation refers to a subset of the set $$A$$ containing all elements of $$A$$ that satisfy the property $$P(x)\text{.}$$ Using $$S = \{ 0, 1, 2 \}$$ as above, $$\{ x \in S : x \not= 2 \}$$ is the set $$\{ 0, 1 \}\text{.}$$ The notation is sometimes abbreviated as $$\bigl\{ x : P(x) \bigr\}\text{,}$$ that is, $$A$$ is not mentioned when understood from context. Furthermore, $$x \in A$$ is sometimes replaced with a formula to make the notation easier to read.
#### Example0.3.3.
The following are sets including the standard notations.
1. The set of natural numbers, $$\N \coloneqq \{ 1, 2, 3, \ldots \}\text{.}$$
2. The set of integers, $$\Z \coloneqq \{ 0, -1, 1, -2, 2, \ldots \}\text{.}$$
3. The set of rational numbers, $$\Q \coloneqq \bigl\{ \frac{m}{n} : m, n \in \Z \text{ and } n \not= 0 \bigr\}\text{.}$$
4. The set of even natural numbers, $$\{ 2m : m \in \N \}\text{.}$$
5. The set of real numbers, $$\R\text{.}$$
Note that $$\N \subset \Z \subset \Q \subset \R\text{.}$$
We create new sets out of old ones by applying some natural operations.
#### Definition0.3.4.
1. A union of two sets $$A$$ and $$B$$ is defined as
\begin{equation*} A \cup B \coloneqq \{ x : x \in A \text{ or } x \in B \} . \end{equation*}
2. An intersection of two sets $$A$$ and $$B$$ is defined as
\begin{equation*} A \cap B \coloneqq \{ x : x \in A \text{ and } x \in B \} . \end{equation*}
3. A complement of $$B$$ relative to $$A$$ (or set-theoretic difference of $$A$$ and $$B$$) is defined as
\begin{equation*} A \setminus B \coloneqq \{ x : x \in A \text{ and } x \notin B \} . \end{equation*}
4. We say complement of $$B$$ and write $$B^c$$ instead of $$A \setminus B$$ if the set $$A$$ is either the entire universe or if it is the obvious set containing $$B\text{,}$$ and is understood from context.
5. We say sets $$A$$ and $$B$$ are disjoint if $$A \cap B = \emptyset\text{.}$$
The notation $$B^c$$ may be a little vague at this point. If the set $$B$$ is a subset of the real numbers $$\R\text{,}$$ then $$B^c$$ means $$\R \setminus B\text{.}$$ If $$B$$ is naturally a subset of the natural numbers, then $$B^c$$ is $$\N \setminus B\text{.}$$ If ambiguity can arise, we use the set difference notation $$A \setminus B\text{.}$$
We illustrate the operations on the Venn diagrams in Figure 2. Let us now establish one of most basic theorems about sets and logic.
#### Proof.
The first statement is proved by the second statement if we assume the set $$A$$ is our “universe.”
Let us prove $$A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)\text{.}$$ Remember the definition of equality of sets. First, we must show that if $$x \in A \setminus (B \cup C)\text{,}$$ then $$x \in (A \setminus B) \cap (A \setminus C)\text{.}$$ Second, we must also show that if $$x \in (A \setminus B) \cap (A \setminus C)\text{,}$$ then $$x \in A \setminus (B \cup C)\text{.}$$ So let us assume $$x \in A \setminus (B \cup C)\text{.}$$ Then $$x$$ is in $$A\text{,}$$ but not in $$B$$ nor $$C\text{.}$$ Hence $$x$$ is in $$A$$ and not in $$B\text{,}$$ that is, $$x \in A \setminus B\text{.}$$ Similarly $$x \in A \setminus C\text{.}$$ Thus $$x \in (A \setminus B) \cap (A \setminus C)\text{.}$$ On the other hand suppose $$x \in (A \setminus B) \cap (A \setminus C)\text{.}$$ In particular, $$x \in (A \setminus B)\text{,}$$ so $$x \in A$$ and $$x \notin B\text{.}$$ Also as $$x \in (A \setminus C)\text{,}$$ then $$x \notin C\text{.}$$ Hence $$x \in A \setminus (B \cup C)\text{.}$$
The proof of the other equality is left as an exercise.
The result above we called a Theorem, while most results we call a Proposition, and a few we call a Lemma (a result leading to another result) or Corollary (a quick consequence of the preceding result). Do not read too much into the naming. Some of it is traditional, some of it is stylistic choice. It is not necessarily true that a Theorem is always “more important” than a Proposition or a Lemma.
We will also need to intersect or union several sets at once. If there are only finitely many, then we simply apply the union or intersection operation several times. However, suppose we have an infinite collection of sets (a set of sets) $$\{ A_1, A_2, A_3, \ldots \}\text{.}$$ We define
\begin{equation*} \begin{aligned} & \bigcup_{n=1}^\infty A_n \coloneqq \{ x : x \in A_n \text{ for some } n \in \N \} , \\ & \bigcap_{n=1}^\infty A_n \coloneqq \{ x : x \in A_n \text{ for all } n \in \N \} . \end{aligned} \end{equation*}
We can also have sets indexed by two natural numbers. For example, we can have the set of sets $$\{ A_{1,1}, A_{1,2}, A_{2,1}, A_{1,3}, A_{2,2}, A_{3,1}, \ldots \}\text{.}$$ Then we write
\begin{equation*} \bigcup_{n=1}^\infty \bigcup_{m=1}^\infty A_{n,m} = \bigcup_{n=1}^\infty \left( \bigcup_{m=1}^\infty A_{n,m} \right) . \end{equation*}
And similarly with intersections.
It is not hard to see that we can take the unions in any order. However, switching the order of unions and intersections is not generally permitted without proof. For instance,
\begin{equation*} \bigcup_{n=1}^\infty \bigcap_{m=1}^\infty \{ k \in \N : mk < n \} = \bigcup_{n=1}^\infty \emptyset = \emptyset . \end{equation*}
However,
\begin{equation*} \bigcap_{m=1}^\infty \bigcup_{n=1}^\infty \{ k \in \N : mk < n \} = \bigcap_{m=1}^\infty \N = \N. \end{equation*}
Sometimes, the index set is not the natural numbers. In such a case we require a more general notation. Suppose $$I$$ is some set and for each $$\lambda \in I\text{,}$$ there is a set $$A_\lambda\text{.}$$ Then we define
\begin{equation*} \bigcup_{\lambda \in I} A_\lambda \coloneqq \{ x : x \in A_\lambda \text{ for some } \lambda \in I \} , \qquad \bigcap_{\lambda \in I} A_\lambda \coloneqq \{ x : x \in A_\lambda \text{ for all } \lambda \in I \} . \end{equation*}
### Subsection0.3.2Induction
When a statement includes an arbitrary natural number, a common method of proof is the principle of induction. We start with the set of natural numbers $$\N = \{ 1,2,3,\ldots \}\text{,}$$ and we give them their natural ordering, that is, $$1 < 2 < 3 < 4 < \cdots\text{.}$$ By $$S \subset \N$$ having a least element, we mean that there exists an $$x \in S\text{,}$$ such that for every $$y \in S\text{,}$$ we have $$x \leq y\text{.}$$
The natural numbers $$\N$$ ordered in the natural way possess the so-called well ordering property. We take this property as an axiom; we simply assume it is true.
The principle of induction is the following theorem, which is in a sense 2 equivalent to the well ordering property of the natural numbers.
#### Proof.
Let $$S$$ be the set of natural numbers $$n$$ for which $$P(n)$$ is not true. Suppose for contradiction that $$S$$ is nonempty. Then $$S$$ has a least element by the well ordering property. Call $$m \in S$$ the least element of $$S\text{.}$$ We know $$1 \notin S$$ by hypothesis. So $$m > 1\text{,}$$ and $$m-1$$ is a natural number as well. Since $$m$$ is the least element of $$S\text{,}$$ we know that $$P(m-1)$$ is true. But the induction step says that $$P(m-1+1) = P(m)$$ is true, contradicting the statement that $$m \in S\text{.}$$ Therefore, $$S$$ is empty and $$P(n)$$ is true for all $$n \in \N\text{.}$$
Sometimes it is convenient to start at a different number than 1, all that changes is the labeling. The assumption that $$P(n)$$ is true in “if $$P(n)$$ is true, then $$P(n+1)$$ is true” is usually called the induction hypothesis.
#### Example0.3.7.
Let us prove that for all $$n \in \N\text{,}$$
\begin{equation*} 2^{n-1} \leq n! \qquad \text{(recall } n! = 1 \cdot 2 \cdot 3 \cdots n\text{)}. \end{equation*}
We let $$P(n)$$ be the statement that $$2^{n-1} \leq n!$$ is true. Plug in $$n=1$$ to see that $$P(1)$$ is true.
Suppose $$P(n)$$ is true. That is, suppose $$2^{n-1} \leq n!$$ holds. Multiply both sides by 2 to obtain
\begin{equation*} 2^n \leq 2(n!) . \end{equation*}
As $$2 \leq (n+1)$$ when $$n \in \N\text{,}$$ we have $$2(n!) \leq (n+1)(n!) = (n+1)!\text{.}$$ That is,
\begin{equation*} 2^n \leq 2(n!) \leq (n+1)!, \end{equation*}
and hence $$P(n+1)$$ is true. By the principle of induction, $$P(n)$$ is true for all $$n \in \N\text{.}$$ In other words, $$2^{n-1} \leq n!$$ is true for all $$n \in \N\text{.}$$
#### Example0.3.8.
We claim that for all $$c \not= 1\text{,}$$
\begin{equation*} 1 + c + c^2 + \cdots + c^n = \frac{1-c^{n+1}}{1-c} . \end{equation*}
Proof: It is easy to check that the equation holds with $$n=1\text{.}$$ Suppose it is true for $$n\text{.}$$ Then
\begin{equation*} \begin{split} 1 + c + c^2 + \cdots + c^n + c^{n+1} & = ( 1 + c + c^2 + \cdots + c^n ) + c^{n+1} \\ & = \frac{1-c^{n+1}}{1-c} + c^{n+1} \\ & = \frac{1-c^{n+1} + (1-c)c^{n+1}}{1-c} \\ & = \frac{1-c^{n+2}}{1-c} . \end{split} \end{equation*}
Sometimes, it is easier to use in the inductive step that $$P(k)$$ is true for all $$k = 1,2,\ldots,n\text{,}$$ not just for $$k=n\text{.}$$ This principle is called strong induction and is equivalent to the normal induction above. The proof of that equivalence is left as an exercise.
### Subsection0.3.3Functions
Informally, a set-theoretic function $$f$$ taking a set $$A$$ to a set $$B$$ is a mapping that to each $$x \in A$$ assigns a unique $$y \in B\text{.}$$ We write $$f \colon A \to B\text{.}$$ An example function $$f \colon S \to T$$ taking $$S \coloneqq \{ 0, 1, 2 \}$$ to $$T \coloneqq \{ 0, 2 \}$$ can be defined by assigning $$f(0) \coloneqq 2\text{,}$$ $$f(1) \coloneqq 2\text{,}$$ and $$f(2) \coloneqq 0\text{.}$$ That is, a function $$f \colon A \to B$$ is a black box, into which we stick an element of $$A$$ and the function spits out an element of $$B\text{.}$$ Sometimes $$f$$ is called a mapping or a map, and we say $$f$$ maps $$A$$ to $$B$$.
Often, functions are defined by some sort of formula; however, you should really think of a function as just a very big table of values. The subtle issue here is that a single function can have several formulas, all giving the same function. Also, for many functions, there is no formula that expresses its values.
To define a function rigorously, first let us define the Cartesian product.
#### Definition0.3.10.
Let $$A$$ and $$B$$ be sets. The Cartesian product is the set of tuples defined as
\begin{equation*} A \times B \coloneqq \bigl\{ (x,y) : x \in A, y \in B \bigr\} . \end{equation*}
For instance, $$\{ a,b \} \times \{ c , d\} = \bigl\{ (a,c), (a,d), (b,c), (b,d) \bigr\}\text{.}$$ A more complicated example is the set $$[0,1] \times [0,1]\text{:}$$ a subset of the plane bounded by a square with vertices $$(0,0)\text{,}$$ $$(0,1)\text{,}$$ $$(1,0)\text{,}$$ and $$(1,1)\text{.}$$ When $$A$$ and $$B$$ are the same set we sometimes use a superscript 2 to denote such a product. For example, $$[0,1]^2 = [0,1] \times [0,1]$$ or $$\R^2 = \R \times \R$$ (the Cartesian plane).
#### Definition0.3.11.
A function $$f \colon A \to B$$ is a subset $$f$$ of $$A \times B$$ such that for each $$x \in A\text{,}$$ there exists a unique $$y \in B$$ for which $$(x,y) \in f\text{.}$$ We write $$f(x) = y\text{.}$$ Sometimes the set $$f$$ is called the graph of the function rather than the function itself.
The set $$A$$ is called the domain of $$f$$ (and sometimes confusingly denoted $$D(f)$$). The set
\begin{equation*} R(f) \coloneqq \{ y \in B : \text{there exists an } x \in A \text{ such that } f(x)=y \} \end{equation*}
is called the range of $$f\text{.}$$ The set $$B$$ is called the codomain of $$f\text{.}$$
It is possible that the range $$R(f)$$ is a proper subset of the codomain $$B\text{,}$$ while the domain of $$f$$ is always equal to $$A\text{.}$$ We generally assume that the domain of $$f$$ is nonempty.
#### Example0.3.12.
From calculus, you are most familiar with functions taking real numbers to real numbers. However, you saw some other types of functions as well. The derivative is a function mapping the set of differentiable functions to the set of all functions. Another example is the Laplace transform, which also takes functions to functions. Yet another example is the function that takes a continuous function $$g$$ defined on the interval $$[0,1]$$ and returns the number $$\int_0^1 g(x) \,dx\text{.}$$
#### Definition0.3.13.
Consider a function $$f \colon A \to B\text{.}$$ Define the image (or direct image) of a subset $$C \subset A$$ as
\begin{equation*} f(C) \coloneqq \bigl\{ f(x) \in B : x \in C \bigr\} . \end{equation*}
Define the inverse image of a subset $$D \subset B$$ as
\begin{equation*} f^{-1}(D) \coloneqq \bigl\{ x \in A : f(x) \in D \bigr\} . \end{equation*}
In particular, $$R(f) = f(A)\text{,}$$ the range is the direct image of the domain $$A\text{.}$$
#### Example0.3.14.
Define the function $$f \colon \R \to \R$$ by $$f(x) \coloneqq \sin(\pi x)\text{.}$$ Then $$f\bigl([0,\nicefrac{1}{2}]\bigr) = [0,1]\text{,}$$ $$f^{-1}\bigl(\{0\}\bigr) = \Z\text{,}$$ etc.
Read the last line of the proposition as $$f^{-1}( B \setminus C) = A \setminus f^{-1} (C)\text{.}$$
#### Proof.
We start with the union. If $$x \in f^{-1}( C \cup D)\text{,}$$ then $$x$$ is taken to $$C$$ or $$D\text{,}$$ that is, $$f(x) \in C$$ or $$f(x) \in D\text{.}$$ Thus $$f^{-1}( C \cup D) \subset f^{-1} (C) \cup f^{-1} (D)\text{.}$$ Conversely if $$x \in f^{-1}(C)\text{,}$$ then $$x \in f^{-1}(C \cup D)\text{.}$$ Similarly for $$x \in f^{-1}(D)\text{.}$$ Hence $$f^{-1}( C \cup D) \supset f^{-1} (C) \cup f^{-1} (D)\text{,}$$ and we have equality.
The rest of the proof is left as an exercise.
For direct images, the best we can do is the following weaker result.
The proof is left as an exercise.
#### Definition0.3.17.
Let $$f \colon A \to B$$ be a function. The function $$f$$ is said to be injective or one-to-one if $$f(x_1) = f(x_2)$$ implies $$x_1 = x_2\text{.}$$ In other words, $$f$$ is injective if for all $$y \in B\text{,}$$ the set $$f^{-1}(\{y\})$$ is empty or consists of a single element. We call such an $$f$$ an injection.
If $$f(A) = B\text{,}$$ then we say $$f$$ is surjective or onto. In other words, $$f$$ is surjective if the range and the codomain of $$f$$ are equal. We call such an $$f$$ a surjection.
If $$f$$ is both surjective and injective, then we say $$f$$ is bijective or that $$f$$ is a bijection.
When $$f \colon A \to B$$ is a bijection, then the inverse image of a single element, $$f^{-1}(\{y\})\text{,}$$ is always a unique element of $$A\text{.}$$ We then consider $$f^{-1}$$ as a function $$f^{-1} \colon B \to A$$ and we write simply $$f^{-1}(y)\text{.}$$ In this case, we call $$f^{-1}$$ the inverse function of $$f\text{.}$$ For instance, for the bijection $$f \colon \R \to \R$$ defined by $$f(x) \coloneqq x^3\text{,}$$ we have $$f^{-1}(x) = \sqrt[3]{x}\text{.}$$
#### Definition0.3.18.
Consider $$f \colon A \to B$$ and $$g \colon B \to C\text{.}$$ The composition of the functions $$f$$ and $$g$$ is the function $$g \circ f \colon A \to C$$ defined as
\begin{equation*} (g \circ f)(x) \coloneqq g\bigl(f(x)\bigr) . \end{equation*}
For example, if $$f \colon \R \to \R$$ is $$f(x)\coloneqq x^3$$ and $$g \colon \R \to \R$$ is $$g(y) = \sin(y)\text{,}$$ then $$(g \circ f)(x) = \sin(x^3)\text{.}$$ It is left to the reader as an easy exericise to show that composition of one-to-one maps is one-to-one and composition of onto maps is onto. Therefore, composition of bijections is a bijection.
### Subsection0.3.4Relations and equivalence classes
We often compare two objects in some way. We say $$1 < 2$$ for natural numbers, or $$\nicefrac{1}{2} = \nicefrac{2}{4}$$ for rational numbers, or $$\{ a,c \} \subset \{ a,b,c \}$$ for sets. The $$<$$’, $$=$$’, and $$\subset$$’ are examples of relations.
#### Definition0.3.19.
Given a set $$A\text{,}$$ a binary relation on $$A$$ is a subset $$\sR \subset A \times A\text{,}$$ which are those pairs where the relation is said to hold. Instead of $$(a,b) \in \sR\text{,}$$ we write $$a \, \sR \, b\text{.}$$
#### Example0.3.20.
Take $$A \coloneqq \{ 1,2,3 \}\text{.}$$
Consider the relation $$<$$’. The corresponding set of pairs is $$\bigl\{ (1,2), (1,3), (2,3) \bigr\}\text{.}$$ So $$1 < 2$$ holds as $$(1,2)$$ is in the corresponding set of pairs, but $$3 < 1$$ does not hold as $$(3,1)$$ is not in the set.
Similarly, the relation $$=$$’ is defined by the set of pairs $$\bigl\{ (1,1), (2,2), (3,3) \big\}\text{.}$$
Any subset of $$A \times A$$ is a relation. Let us define the relation $$\dagger$$ via $$\bigl\{ (1,2), (2,1), (2,3), (3,1) \bigr\}\text{,}$$ then $$1 \dagger 2$$ and $$3 \dagger 1$$ are true, but $$1 \dagger 3$$ is not.
#### Definition0.3.21.
Let $$\sR$$ be a relation on a set $$A\text{.}$$ Then $$\sR$$ is said to be
1. Reflexive if $$a\, \sR \, a$$ for all $$a \in A\text{.}$$
2. Symmetric if $$a\, \sR \, b$$ implies $$b \, \sR \, a\text{.}$$
3. Transitive if $$a\, \sR \, b$$ and $$b \, \sR \, c$$ implies $$a\, \sR \, c\text{.}$$
If $$\sR$$ is reflexive, symmetric, and transitive, then it is said to be an equivalence relation.
#### Example0.3.22.
Let $$A \coloneqq \{ 1,2,3 \}\text{.}$$ The relation $$<$$’ is transitive, but neither reflexive nor symmetric. The relation $$\leq$$’ defined by $$\bigl\{ (1,1), (1,2), (1,3), (2,2), (2,3), (3,3) \bigr\}$$ is reflexive and transitive, but not symmetric. Finally, a relation $$\star$$’ defined by $$\bigl\{ (1,1), (1,2), (2,1), (2,2), (3,3) \bigr\}$$ is an equivalence relation.
Equivalence relations are useful in that they divide a set into sets of “equivalent” elements.
#### Definition0.3.23.
Let $$A$$ be a set and $$\sR$$ an equivalence relation. An equivalence class of $$a \in A\text{,}$$ often denoted by $$[a]\text{,}$$ is the set $$\{ x \in A : a\, \sR \, x \}\text{.}$$
For example, given the relation $$\star$$’ above, there are two equivalence classes, $$[1] = [2] = \{ 1,2 \}$$ and $$[3] = \{ 3 \}\text{.}$$
Reflexivity guarantees that $$a \in [a]\text{.}$$ Symmetry guarantees that if $$b \in [a]\text{,}$$ then $$a \in [b]\text{.}$$ Finally, transitivity guarantees that if $$b \in [a]$$ and $$c \in [b]\text{,}$$ then $$c \in [a]\text{.}$$ In particular, we have the following proposition, whose proof is an exercise.
#### Example0.3.25.
The set of rational numbers can be defined as equivalence classes of a pair of an integer and a natural number, that is elements of $$\Z \times \N\text{.}$$ The relation is defined by $$(a,b) \sim (c,d)$$ whenever $$ad = bc\text{.}$$ It is left as an exercise to prove that $$\sim$$’ is an equivalence relation. Usually the equivalence class $$\bigl[(a,b)\bigr]$$ is written as $$\nicefrac{a}{b}\text{.}$$
### Subsection0.3.5Cardinality
A subtle issue in set theory and one generating a considerable amount of confusion among students is that of cardinality, or “size” of sets. The concept of cardinality is important in modern mathematics in general and in analysis in particular. In this section, we will see the first really unexpected theorem.
#### Definition0.3.26.
Let $$A$$ and $$B$$ be sets. We say $$A$$ and $$B$$ have the same cardinality when there exists a bijection $$f \colon A \to B\text{.}$$ We denote by $$\abs{A}$$ the equivalence class of all sets with the same cardinality as $$A$$ and we simply call $$\abs{A}$$ the cardinality of $$A\text{.}$$
For example, $$\{ 1,2,3 \}$$ has the same cardinality as $$\{ a,b,c \}$$ by defining a bijection $$f(1) \coloneqq a\text{,}$$ $$f(2) \coloneqq b\text{,}$$ $$f(3) \coloneqq c\text{.}$$ Clearly the bijection is not unique.
The existence of a bijection really is an equivalence relation. The identity, $$f(x) \coloneqq x\text{,}$$ is a bijection showing reflexivity. If $$f$$ is a bijection, then so is $$f^{-1}$$ showing symmetry. If $$f \colon A \to B$$ and $$g \colon B \to C$$ are bijections, then $$g \circ f$$ is a bijection of $$A$$ and $$C$$ showing transitivity. A set $$A$$ has the same cardinality as the empty set if and only if $$A$$ itself is the empty set: If $$B$$ is nonempty, then no function $$f \colon B \to \emptyset$$ can exist. In particular, there is no bijection of $$B$$ and $$\emptyset\text{.}$$
#### Definition0.3.27.
If $$A$$ has the same cardinality as $$\{ 1,2,3,\ldots,n \}$$ for some $$n \in \N\text{,}$$ we write $$\abs{A} \coloneqq n\text{.}$$ If $$A$$ is empty, we write $$\abs{A} \coloneqq 0\text{.}$$ In either case, we say that $$A$$ is finite. We say $$A$$ is infinite or “of infinite cardinality” if $$A$$ is not finite.
That the notation $$\abs{A} = n$$ is justified we leave as an exercise. That is, for each nonempty finite set $$A\text{,}$$ there exists a unique natural number $$n$$ such that there exists a bijection from $$A$$ to $$\{ 1,2,3,\ldots,n \}\text{.}$$
We can order sets by size.
#### Definition0.3.28.
We write
\begin{equation*} \abs{A} \leq \abs{B} \end{equation*}
if there exists an injection from $$A$$ to $$B\text{.}$$ We write $$\abs{A} = \abs{B}$$ if $$A$$ and $$B$$ have the same cardinality. We write $$\abs{A} < \abs{B}$$ if $$\abs{A} \leq \abs{B}\text{,}$$ but $$A$$ and $$B$$ do not have the same cardinality.
We state without proof that $$A$$ and $$B$$ have the same cardinality if and only if $$\abs{A} \leq \abs{B}$$ and $$\abs{B} \leq \abs{A}\text{.}$$ This is the so-called Cantor–Bernstein–Schröder theorem. Furthermore, if $$A$$ and $$B$$ are any two sets, we can always write $$\abs{A} \leq \abs{B}$$ or $$\abs{B} \leq \abs{A}\text{.}$$ The issues surrounding this last statement are very subtle. As we do not require either of these two statements, we omit proofs.
The truly interesting cases of cardinality are infinite sets. We will distinguish two types of infinite cardinality.
#### Definition0.3.29.
If $$\abs{A} = \abs{\N}\text{,}$$ then we say $$A$$ is countably infinite. If $$A$$ is finite or countably infinite, then we say $$A$$ is countable. If $$A$$ is not countable, then $$A$$ is said to be uncountable.
The cardinality of $$\N$$ is usually denoted as $$\aleph_0$$ (read as aleph-naught) 3 .
#### Example0.3.30.
The set of even natural numbers has the same cardinality as $$\N\text{.}$$ Proof: Let $$E \subset \N$$ be the set of even natural numbers. Given $$k \in E\text{,}$$ write $$k=2n$$ for some $$n \in \N\text{.}$$ Then $$f(n) \coloneqq 2n$$ defines a bijection $$f \colon \N \to E\text{.}$$
In fact, we mention without proof the following characterization of infinite sets: A set is infinite if and only if it is in one-to-one correspondence with a proper subset of itself.
#### Example0.3.31.
$$\N \times \N$$ is a countably infinite set. Proof: Arrange the elements of $$\N \times \N$$ as follows $$(1,1)\text{,}$$ $$(1,2)\text{,}$$ $$(2,1)\text{,}$$ $$(1,3)\text{,}$$ $$(2,2)\text{,}$$ $$(3,1)\text{,}$$ .... That is, always write down first all the elements whose two entries sum to $$k\text{,}$$ then write down all the elements whose entries sum to $$k+1$$ and so on. Define a bijection with $$\N$$ by letting 1 go to $$(1,1)\text{,}$$ 2 go to $$(1,2)\text{,}$$ and so on. See Figure 4.
#### Example0.3.32.
The set of rational numbers is countable. Proof: (informal) For positive rational numbers follow the same procedure as in the previous example, writing $$\nicefrac{1}{1}\text{,}$$ $$\nicefrac{1}{2}\text{,}$$ $$\nicefrac{2}{1}\text{,}$$ etc. However, leave out fractions (such as $$\nicefrac{2}{2}$$) that have already appeared. The list would continue: $$\nicefrac{1}{3}\text{,}$$ $$\nicefrac{3}{1}\text{,}$$ $$\nicefrac{1}{4}\text{,}$$ $$\nicefrac{2}{3}\text{,}$$ etc. For all rational numbers, include $$0$$ and the negative numbers: $$0\text{,}$$ $$\nicefrac{1}{1}\text{,}$$ $$\nicefrac{-1}{1}\text{,}$$ $$\nicefrac{1}{2}\text{,}$$ $$\nicefrac{-1}{2}\text{,}$$ etc.
For completeness, we mention the following statements from the exercises. If $$A \subset B$$ and $$B$$ is countable, then $$A$$ is countable. The contrapositive of the statement is that if $$A$$ is uncountable, then $$B$$ is uncountable. As a consequence, if $$\abs{A} < \abs{\N}\text{,}$$ then $$A$$ is finite. Similarly, if $$B$$ is finite and $$A \subset B\text{,}$$ then $$A$$ is finite.
We give the first truly striking result about cardinality. To do so we need a notation for the set of all subsets of a set.
#### Definition0.3.33.
The power set of a set $$A\text{,}$$ denoted by $$\sP(A)\text{,}$$ is the set of all subsets of $$A\text{.}$$
For example, if $$A \coloneqq \{ 1,2\}\text{,}$$ then $$\sP(A) = \bigl\{ \emptyset, \{ 1 \}, \{ 2 \}, \{ 1, 2 \} \bigr\}\text{.}$$ In particular, $$\abs{A} = 2$$ and $$\abs{\sP(A)} = 4 = 2^2\text{.}$$ In general, for a finite set $$A$$ of cardinality $$n\text{,}$$ the cardinality of $$\sP(A)$$ is $$2^n\text{.}$$ This fact is left as an exercise. Hence, for a finite set $$A\text{,}$$ the cardinality of $$\sP(A)$$ is strictly larger than the cardinality of $$A\text{.}$$ What is an unexpected and striking fact is that this statement is also true for infinite sets.
#### Proof.
An injection $$f \colon A \to \sP(A)$$ exists: For $$x \in A\text{,}$$ let $$f(x) \coloneqq \{ x \}\text{.}$$ Thus, $$\abs{A} \leq \abs{\sP(A)}\text{.}$$
To finish the proof, we must show that no function $$g \colon A \to \sP(A)$$ is a surjection. Suppose $$g \colon A \to \sP(A)$$ is a function. So for $$x \in A\text{,}$$ $$g(x)$$ is a subset of $$A\text{.}$$ Define the set
\begin{equation*} B \coloneqq \bigl\{ x \in A : x \notin g(x) \bigr\} . \end{equation*}
We claim that $$B$$ is not in the range of $$g$$ and hence $$g$$ is not a surjection. Suppose for contradiction that there exists an $$x_0$$ such that $$g(x_0) = B\text{.}$$ Either $$x_0 \in B$$ or $$x_0 \notin B\text{.}$$ If $$x_0 \in B\text{,}$$ then $$x_0 \notin g(x_0) = B\text{,}$$ which is a contradiction. If $$x_0 \notin B\text{,}$$ then $$x_0 \in g(x_0) = B\text{,}$$ which is again a contradiction. Thus such an $$x_0$$ does not exist. Therefore, $$B$$ is not in the range of $$g\text{,}$$ and $$g$$ is not a surjection. As $$g$$ was an arbitrary function, no surjection exists.
One particular consequence of this theorem is that there do exist uncountable sets, as $$\sP(\N)$$ must be uncountable. A related fact is that the set of real numbers (which we study in the next chapter) is uncountable. The existence of uncountable sets may seem unintuitive, and the theorem caused quite a controversy at the time it was announced. The theorem not only says that uncountable sets exist, but that there in fact exist progressively larger and larger infinite sets $$\N\text{,}$$ $$\sP(\N)\text{,}$$ $$\sP(\sP(\N))\text{,}$$ $$\sP(\sP(\sP(\N)))\text{,}$$ etc.
### Subsection0.3.6Exercises
#### Exercise0.3.1.
Show $$A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)\text{.}$$
#### Exercise0.3.2.
Prove that the principle of strong induction is equivalent to the standard induction.
#### Exercise0.3.4.
1. Find an example for which equality of sets in $$f( C \cap D) \subset f (C) \cap f (D)$$ fails. That is, find an $$f\text{,}$$ $$A\text{,}$$ $$B\text{,}$$ $$C\text{,}$$ and $$D$$ such that $$f( C \cap D)$$ is a proper subset of $$f(C) \cap f(D)\text{.}$$
#### Exercise0.3.5.
(Tricky) Prove that if $$A$$ is nonempty and finite, then there exists a unique $$n \in \N$$ such that there exists a bijection between $$A$$ and $$\{ 1, 2, 3, \ldots, n \}\text{.}$$ In other words, the notation $$\abs{A} \coloneqq n$$ is justified. Hint: Show that if $$n > m\text{,}$$ then there is no injection from $$\{ 1, 2, 3, \ldots, n \}$$ to $$\{ 1, 2, 3, \ldots, m \}\text{.}$$
#### Exercise0.3.6.
Prove:
1. $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\text{.}$$
2. $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}$$
#### Exercise0.3.7.
Let $$A \Delta B$$ denote the symmetric difference, that is, the set of all elements that belong to either $$A$$ or $$B\text{,}$$ but not to both $$A$$ and $$B\text{.}$$
1. Draw a Venn diagram for $$A \Delta B\text{.}$$
2. Show $$A \Delta B = (A \setminus B) \cup (B \setminus A)\text{.}$$
3. Show $$A \Delta B = (A \cup B) \setminus ( A \cap B)\text{.}$$
#### Exercise0.3.8.
For each $$n \in \N\text{,}$$ let $$A_n \coloneqq \{ (n+1)k : k \in \N \}\text{.}$$
1. Find $$A_1 \cap A_2\text{.}$$
2. Find $$\bigcup_{n=1}^\infty A_n\text{.}$$
3. Find $$\bigcap_{n=1}^\infty A_n\text{.}$$
#### Exercise0.3.9.
Determine $$\sP(S)$$ (the power set) for each of the following:
1. $$S = \emptyset\text{,}$$
2. $$S = \{1\}\text{,}$$
3. $$S = \{1,2\}\text{,}$$
4. $$S = \{1,2,3,4\}\text{.}$$
#### Exercise0.3.10.
Let $$f \colon A \to B$$ and $$g \colon B \to C$$ be functions.
1. Prove that if $$g \circ f$$ is injective, then $$f$$ is injective.
2. Prove that if $$g \circ f$$ is surjective, then $$g$$ is surjective.
3. Find an explicit example where $$g \circ f$$ is bijective, but neither $$f$$ nor $$g$$ is bijective.
#### Exercise0.3.11.
Prove by induction that $$n < 2^n$$ for all $$n \in \N\text{.}$$
#### Exercise0.3.12.
Show that for a finite set $$A$$ of cardinality $$n\text{,}$$ the cardinality of $$\sP(A)$$ is $$2^n\text{.}$$
#### Exercise0.3.13.
Prove $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}$$ for all $$n \in \N\text{.}$$
#### Exercise0.3.14.
Prove $$1^3 + 2^3 + \cdots + n^3 = {\left( \frac{n(n+1)}{2} \right)}^2$$ for all $$n \in \N\text{.}$$
#### Exercise0.3.15.
Prove that $$n^3 + 5n$$ is divisible by $$6$$ for all $$n \in \N\text{.}$$
#### Exercise0.3.16.
Find the smallest $$n \in \N$$ such that $$2{(n+5)}^2 < n^3$$ and call it $$n_0\text{.}$$ Show that $$2{(n+5)}^2 < n^3$$ for all $$n \geq n_0\text{.}$$
#### Exercise0.3.17.
Find all $$n \in \N$$ such that $$n^2 < 2^n\text{.}$$
#### Exercise0.3.19.
Give an example of a countably infinite collection of finite sets $$A_1, A_2, \ldots\text{,}$$ whose union is not a finite set.
#### Exercise0.3.20.
Give an example of a countably infinite collection of infinite sets $$A_1, A_2, \ldots\text{,}$$ with $$A_j \cap A_k$$ being infinite for all $$j$$ and $$k\text{,}$$ such that $$\bigcap_{j=1}^\infty A_j$$ is nonempty and finite.
#### Exercise0.3.21.
Suppose $$A \subset B$$ and $$B$$ is finite. Prove that $$A$$ is finite. That is, if $$A$$ is nonempty, construct a bijection of $$A$$ to $$\{ 1,2,\ldots,n \}\text{.}$$
#### Exercise0.3.22.
Prove Proposition 0.3.24. That is, prove that if $$\sR$$ is an equivalence relation on a set $$A\text{,}$$ then every $$a \in A$$ is in exactly one equivalence class. Then prove that $$a \, \sR \, b$$ if and only if $$[a] = [b]\text{.}$$
#### Exercise0.3.23.
Prove that the relation `$$\sim$$’ in Example 0.3.25 is an equivalence relation.
#### Exercise0.3.24.
1. Suppose $$A \subset B$$ and $$B$$ is countably infinite. By constructing a bijection, show that $$A$$ is countable (that is, $$A$$ is empty, finite, or countably infinite).
2. Use part a) to show that if $$\abs{A} < \abs{\N}\text{,}$$ then $$A$$ is finite.
#### Exercise0.3.25.
(Challenging) Suppose $$\abs{\N} \leq \abs{S}\text{,}$$ or in other words, $$S$$ contains a countably infinite subset. Show that there exists a countably infinite subset $$A \subset S$$ and a bijection between $$S \setminus A$$ and $$S\text{.}$$
#### Exercise0.3.26.
Prove the infinite versions of DeMorgan’s laws. Suppose $$A$$ is a set and $$B_\lambda$$ is a collection of sets for $$\lambda \in I\text{.}$$ Prove
\begin{equation*} A \setminus \biggl( \bigcup_{\lambda \in I} B_\lambda \biggr) = \bigcap_{\lambda \in I} ( A \setminus B_\lambda ) , \qquad A \setminus \biggl( \bigcap_{\lambda \in I} B_\lambda \biggr) = \bigcup_{\lambda \in I} ( A \setminus B_\lambda ) . \end{equation*}
#### Exercise0.3.27.
Suppose $$f \colon A \to B$$ is a function, and for $$\lambda \in I\text{,}$$ we have a collection of subsets $$C_\lambda \subset A$$ and $$D_\lambda \subset B\text{.}$$ Prove
\begin{equation*} f^{-1} \biggl( \bigcup_{\lambda \in I} D_\lambda \biggr) = \bigcup_{\lambda \in I} f^{-1}(D_\lambda) , \quad f^{-1} \biggl( \bigcap_{\lambda \in I} D_\lambda \biggr) = \bigcap_{\lambda \in I} f^{-1}(D_\lambda) , \end{equation*}
and
\begin{equation*} f \biggl( \bigcup_{\lambda \in I} C_\lambda \biggr) = \bigcup_{\lambda \in I} f(C_\lambda) , \quad f \biggl( \bigcap_{\lambda \in I} C_\lambda \biggr) \subset \bigcap_{\lambda \in I} f(C_\lambda) . \end{equation*}
The term “modern” refers to late 19th century up to the present.
To be completely rigorous, this equivalence is only true if we also assume as an axiom that $$n-1$$ exists for all natural numbers bigger than $$1\text{,}$$ which we do. In this book, we are assuming all the usual arithmetic holds.
For the fans of the TV show Futurama, there is a movie theater in one episode called an $$\aleph_0$$-plex.
Named after the German mathematician Georg Ferdinand Ludwig Philipp Cantor (1845–1918).
For a higher quality printout use the PDF versions: https://www.jirka.org/ra/realanal.pdf or https://www.jirka.org/ra/realanal2.pdf
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# A famous golfer tees off on a straight 390 yard par 4 and slices his drive to the right. The drive goes 280 yards from the tee. Using a 7-iron on his second shot, he hits the ball 180 yards and it lands inches from the hole. How many degrees (to the nearest degree) to the right of the line from the tee to the hole did he slice his drive?
## To find the number of degrees to the right of the line from the tee to the hole that the drive was sliced, we need to calculate the angle formed by the tee, the ball's landing spot, and the hole.
First, let's find the distance between the tee and the ball's landing spot. Since the golfer sliced his drive to the right, we need to subtract the distance the ball traveled from the total distance of the hole:
Total distance of the par 4 hole = 390 yards
Distance the ball traveled = 280 yards
Distance between the tee and the ball's landing spot = Total distance - Distance traveled
= 390 yards - 280 yards
= 110 yards
Next, we need to calculate the distance between the tee and the hole. Since it is a straight par 4, we can assume the golfer hit the ball directly towards the hole (ignoring the slice):
Distance between the tee and the hole = 390 yards
Now, we can use the trigonometric function "tan" to calculate the angle to the right of the line from the tee to the hole:
In this case, the opposite side is the distance between the tee and the ball's landing spot (110 yards), and the adjacent side is the distance between the tee and the hole (390 yards).
= tan^(-1)(110/390)
Calculating this using a calculator, we find that:
angle ≈ 16.8 degrees
Therefore, the golfer sliced his drive approximately 16.8 degrees to the right of the line from the tee to the hole.
## To find the number of degrees the golfer sliced his drive to the right of the line from the tee to the hole, we can use trigonometry.
First, let's calculate the distance from the tee to where the ball should have landed if it did not slice. We know the golfer hit a 280-yard drive and the hole is 390 yards away. So, the remaining distance the ball should have traveled is 390 - 280 = 110 yards.
Now, let's calculate the angle the golfer sliced the ball. We can use the sine of this angle to find the ratio of the distance sliced to the total distance that the ball should have traveled.
sin(angle) = distance sliced / total distance
In this case, the distance sliced is 280 yards (the distance the ball actually traveled) and the total distance should have been 110 yards (the straight line distance from the tee to where it should have landed).
sin(angle) = 280 / 110
Now, let's calculate the angle using the arcsin function on a calculator:
angle = arcsin(280 / 110)
The result is approximately 68.19 degrees.
Therefore, the golfer sliced his drive approximately 68 degrees to the right of the line from the tee to the hole.
## Instead of the triangle having sides 390, 280, and 180
let's use the similar triangle of 39, 28, 18 , the angles of course remain the same.
looks like we need the angle opposite the side 18,
using the cosine law:
18^2 = 39^2 + 28^2 - 2(39)(28) cosØ
2184cosØ = 1981
cosØ = 1981/2184 = ....
Ø =
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Goseeko blog
# What is Correlation? Why it is so important?
## Overview
Correlation is the measurement of the strength of a linear relationship between two variables say x and y.
In other words, we define it as if the change in one variable affects a change in other variables, then these two variables are said to be correlated.
For example:
1. The correlation between a person’s income and expenditures.
2. As the temperature goes up, the demand for ice cream also goes up.
## Types of correlation
Positive correlation- When both variables move in the same direction, or if the increase in one variable results in a corresponding increase in the other one is called positive correlation.
Negative correlation- When one variable increases and other decreases or vise-versa, then the variables said to be negatively correlated.
No correlation- When two variables are independent and do not affect each other then there will be no correlation between the two and said to be uncorrelated.
Note- (Perfect correlation)- When a variable changes constantly with the other variable, then these two variables are said to be perfectly correlated.
## Scatter plots or dot diagrams
A scatter or dot diagram is used to check the correlation between the two variables.
It is the simplest method to represent bivariate data.
When the dots in the diagram are very close to each other, then we can say that there is a fairly good correlation.
If the dots are scattered then we get a poor correlation.
## Karl Pearson’s coefficient of correlation
Karl Person’s coefficient of correlation is also called product-moment correlation coefficient.
It is denoted by ‘r’, and defined as-
Here and are the standard deviations of these series.
Alternate formula-
Note-
• Correlation coefficient always lies between -1 and +1.
• Correlation coefficient is independent of change of origin and scale.
• If the two variables are independent then correlation coefficient between them is zero.
## Solved example
Example: The data given below is about the marks obtained by a student and hours she studied.
Find the correlation coefficient between hours and marks obtained.
Solution:
Let hours = x and marks = y
Karl Person’s coefficient of correlation is given by-
The correlation coefficient between hours and marks obtained is-
r = 0.98
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Mbalisikerc
2022-07-17
Let ${X}_{n}$ be a geometric random variable with parameter $p=\lambda /n$. Compute $P\left({X}_{n}/n>x\right)$
$x>0$ and show that as n approaches infinity this probability converges to $P\left(Y>x\right)$, where Y is an exponential random variable with parameter $\lambda$. This shows that ${X}_{n}/n$ is approximately an exponential random variable.
Monastro3n
Step 1
For a geometric random variable X, we have that $P\left\{X>k\right\}={q}^{k}$ assuming that k is an integer. This can not be used directly for $P\left\{{X}_{n}>nx\right\}$ because x can be an irrational number. If this is the case, then nx is not an integer for any choice of n. However, since ${X}_{n}$ is a geometric random variable and $\left\{{X}_{n}>nx\right\}=\left\{{X}_{n}>i\right\}$ where i is a unique integer such that $i\le nx, then $P\left\{{X}_{n}>nx\right\}={q}^{i}$. For $P\left\{{X}_{n}/n>x\right\}$ we have:
$P\left\{{X}_{n}/n>x\right\}=P\left\{{X}_{n}>nx\right\}\approx {\left(1-\frac{\lambda }{n}\right)}^{nx}={\left(1-\frac{\lambda x}{nx}\right)}^{nx}\approx {\left(1-\frac{\lambda x}{i}\right)}^{i}$
Step 2
Since ${\left(1-\frac{\lambda x}{i}\right)}^{i}={e}^{-\lambda x}$ as $i\to \mathrm{\infty }$ and the probability of the exponential random variable, Y, is $P\left\{Y>x\right\}=1-F\left(x\right)={e}^{-\lambda x}$, this shows that $P\left\{{X}_{n}/n>x\right\}$ converges to $P\left\{Y>x\right\}$ as $i\to \mathrm{\infty }$. This reveals that ${X}_{n}/n$ is approximately an exponential random variable.
Do you have a similar question?
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# How do you solve 3x ^(2/3) + x^(1/3) - 2 = 0?
$x = \frac{8}{27}$ and $x = - 1$
#### Explanation:
From the given equation
$3 {x}^{\frac{2}{3}} + {x}^{\frac{1}{3}} - 2 = 0$
The solution goes like this
Let $w = {x}^{\frac{1}{3}}$
$3 {x}^{\frac{2}{3}} + {x}^{\frac{1}{3}} - 2 = 0$
we can write this equation this way
$3 {\left({x}^{\frac{1}{3}}\right)}^{2} + {\left({x}^{\frac{1}{3}}\right)}^{1} - 2 = 0$
also
$3 {\left(w\right)}^{2} + 1 \cdot {\left(w\right)}^{1} - 2 = 0$
also let $a = 3$ and $b = 1$ and $c = - 2$
$w = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$w = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 3 \cdot \left(- 2\right)}}{2 \cdot 3}$
$w = \frac{- 1 \pm \sqrt{1 + 24}}{6}$
$w = \frac{- 1 \pm \sqrt{25}}{6}$
$w = \frac{- 1 \pm 5}{6}$
we have two values for w:
${w}_{1} = \frac{- 1 + 5}{6} = \frac{4}{6} = \frac{2}{3}$ and ${w}_{2} = \frac{- 1 - 5}{6} = - \frac{6}{6} = - 1$
But $w = {x}^{\frac{1}{3}}$
and $x = {w}^{3}$
Using $w = \frac{2}{3}$
$x = {w}^{3} = {\left(\frac{2}{3}\right)}^{3} = \frac{8}{27}$
$x = \frac{8}{27}$
Using $w = - 1$
$x = {w}^{3} = {\left(- 1\right)}^{3} = - 1$
$x = - 1$
check at $x = \frac{8}{27}$
$3 {x}^{\frac{2}{3}} + {x}^{\frac{1}{3}} - 2 = 0$
$3 {\left(\frac{8}{27}\right)}^{\frac{2}{3}} + {\left(\frac{8}{27}\right)}^{\frac{1}{3}} - 2 = 0$
$3 \left(\frac{4}{9}\right) + \frac{2}{3} - 2 = 0$
$\frac{4}{3} + \frac{2}{3} - 2 = 0$
$\frac{6}{3} - 2 = 0$
$2 - 2 = 0$
$0 = 0$
$x = \frac{8}{27}$ is a root
check at $x = - 1$
$3 {x}^{\frac{2}{3}} + {x}^{\frac{1}{3}} - 2 = 0$
$3 {\left(- 1\right)}^{\frac{2}{3}} + {\left(- 1\right)}^{\frac{1}{3}} - 2 = 0$
$3 \left(1\right) - 1 - 2 = 0$
$3 - 3 = 0$
$0 = 0$
$x = - 1$ is a root
God bless....I hope the explanation is useful.
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# Question Video: Determining How Many Circles Pass through Three Points Mathematics
How many circles can pass through three points?
05:22
### Video Transcript
How many circles can pass through three points?
The mathematical definition of a circle is that it is a set of points in a plane that are a constant distance from a point in the center. We are going to consider how many circles can pass through three points. So let’s consider any three points, and we can define them as 𝐴, 𝐵, and 𝐶. A circle would pass through these three points if its center is equidistant from the three points. An example of a circle which wouldn’t go through points would have a center here. A circle with a center here wouldn’t work because we can see that it’s not the same distance from the center to each of the points 𝐴, 𝐵, and 𝐶.
And so we must ask, how do we find the center of a circle that would pass through these three points? How do we find the point which is equidistant from these three points? Let’s begin by drawing line segments between any two pairs of points. So here, we’ve got the line segment 𝐴𝐵 and the line segment 𝐵𝐶. The line segment 𝐴𝐶 would also work. We will now construct the perpendicular bisectors of each of these line segments. To do this accurately, we need one of these tools, which will be a compass or a pair of compasses, depending on where you live.
Let’s start with the perpendicular bisector of the line segment 𝐴𝐵. We take the sharp end of the compass and put it in point 𝐴, and then we stretch it open so that it’s longer than half of the length of the line segment 𝐴𝐵. Using the pencil, we draw an arc above the line segment and below it. We then repeat the process. This time, we put the pointy end into point 𝐵 and create another set of arcs, one below the line and one above the line.
We observe that each of the sets of arcs above and below the line segment has a point of intersection. Drawing the line between these two points of intersection creates the perpendicular bisector. Notice how this line is at 90 degrees to the line segment 𝐴𝐵, and we have created two congruent line segments. It’s usually very good to keep in our construction lines, but let’s remove them so that we can see exactly what’s happening.
The line that we have drawn here represents all the points that are equidistant from 𝐴 and 𝐵. We can now repeat the process by finding the perpendicular bisector of the line segment 𝐵𝐶. We can draw the first set of arcs from point 𝐵 and the second set from point 𝐶. And we have completed the second perpendicular bisector.
This line represents all the points which are equidistant from point 𝐵 and 𝐶. We don’t need to draw the line segment 𝐴𝐶 and find its perpendicular bisector. And that’s because the point where these two perpendicular bisectors intersect is the point which is equidistant from 𝐴, 𝐵, and 𝐶. This point can therefore be the center of a circle which passes through the three points. It would look something like this.
And so how many circles will pass through the three points? Well, we know that there’s one because we’ve drawn it. But will there be any more? We can alternatively ask, are there any other points which are also equidistant from 𝐴, 𝐵, and 𝐶? Remember that the first perpendicular bisector represents all the points equidistant from 𝐴 and 𝐵. The second bisector represents all the points which are equidistant from 𝐵 and 𝐶. The point where these two lines intersected is the point which is equidistant from all three points.
So could these two perpendicular bisectors intersect at a different point? How about here, above the three points, or maybe here, below the three points again? And of course, we should recognize that neither of these is possible. Two straight lines can only intersect at most one point. And therefore, there is only one circle which can pass through three points.
Before we finish by giving the answer, there is one really important thing to note. This only happens when the three points do not lie on a straight line. If we take the example of 𝐴, 𝐵, and 𝐶 on a straight line, then we can see that it’s not physically possible to draw a circle through all three points. We could draw a circle that goes through two of the points, but not all three. When the three points lie on a straight line, then zero circles could be drawn through them.
Excluding this case, however, we can give the answer of one circle. Finally, it’s worth mentioning that it’s not just that we might be able to draw a circle through three points that aren’t on a straight line. But there will always be a circle which we can draw through any three points.
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# BEST METHODS FOR SOLVING QUADRATIC INEQUALITIES.
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1 BEST METHODS FOR SOLVING QUADRATIC INEQUALITIES. I. GENERALITIES There are 3 common methods to solve quadratic inequalities. Therefore, students sometimes are confused to select the fastest and the best solving method. I generally explain below these 3 methods and then compare them through selected examples. Solving a quadratic inequality, in standard form f(x) = ax^2 + bx + c > 0 (or < 0), means finding all the values of x that make the inequality true. These values of x constitute the solution set of the inequality. The solution set of a quadratic inequality are expressed in the form of intervals. Examples of quadratic inequalities: x^2 8 x + 7 < 9 3x^2 + 4x 7 < 0 5x^2 12x > 17 (3x 5)(4x + 1) < 0 3x/(x -1) + 4x/(3 x) > 1 1/(x 2) + 2/(x 3) > 2 Examples of solution set expressed in terms of intervals: (-2, 3) (1.73, 3.42) (-infinity, 3] [-1/3, 13/5] [4, +infinity) II. STEPS IN SOLVING QUADRATIC INEQUALITIES In general, there are 4 common steps. Step 1. Transform the inequality into standard form: f(x) = ax^2 + bx + c < 0 (or > 0). Example. Solve: x(4x -3) > 7. In first step, transform the inequality into standard form: f(x) = 4x^2 3x 7 > 0. Example. Solve: (2x + 5)(x 1) < -3 In first step, transform the inequality into standard form: f(x) = 2x^2 + 3x 2 < 0. Step 2. Solve the quadratic equation f(x) = 0. You may use one of the 4 existing common methods (factoring ac method, completing the square, quadratic formula, graphing) or the new Diagonal Sum Method (Amazon e-book 2010). Before starting, find out if the equation has 2 real roots. How? Roughly estimate the Discriminant D s value. You don t need to calculate its exact value, unless you want to apply the quadratic formula. Just use mental math to see if D is positive (> 0) or negative (< 0). If D < 0, or D = 0, solve the inequality by referring to the first part of the Theorem explained in next paragraph. Page 1 of 7
2 If D > 0, solve the equation f(x) = 0 to get the 2 real roots x1 and x2. Find out if this equation can be factored. How? You may calculate D to see if it is a perfect square. Or, you may first use the new Diagonal Sum Method to solve the equation. It usually requires fewer than 3 trials. If it fails, meaning no diagonal sums is equal to b (or b), then the equation can t be factored, and therefore the quadratic formula must be used. Step 3. Solve the given quadratic inequality f(x) < 0 (or > 0), based on the 2 values x1 and x2, found in Step 2. You may choose one of the 3 common methods to solve quadratic inequalities described below. Step 4. Express the solution set of the quadratic inequality in terms of intervals. You must know how to correctly use the interval symbols. Examples: (2, 7): open interval between 2 and 7. The 2 end (critical) points are not included in the solution set. [-3, 5]: closed interval. The 2 end points -3 and 5 are included in the solution set. [2, +infinity): half-closed interval; only the end point 2 is included in the solution set (-infinity, -1]: half closed interval; only the end point is included in the solution set. III. COMMON METHODS TO SOLVE QUADRATIC INEQUALITIES There are 3 most common methods. Therefore, students are sometimes confused to select the best solving method. 1. The number line and test point method. Given a quadratic inequality in standard form f(x) = ax^2 + bx + c < 0 (or > 0), with a not equal zero. Suppose the Discriminant D > 0, and the given quadratic equation has 2 real roots x1 and x2. Plot them on a number line. They divide the number line into one segment (x1, x2) and 2 rays. The solution set of the quadratic equation should be either the segment, or the 2 rays. Always use the origin O as test point. Substitute x = 0 into the inequality. If it is true, then the origin O is located on the true segment (or the true ray). If one ray is a part of the solution set, then the other ray also belongs to the solution set, due to the symmetrical property of the parabola graph. Note 1. When D < 0, there are no real roots, this number line method can t be used. In this case you must solve the inequality by the algebraic method. When D = 0, there is a double root at x = -b/2a, this number line method can t be used. You must apply the algebraic method. Note 2. By this number-line method, you may use a double number line, or even a triple number line, to solve a system of two or three quadratic inequalities in one variable. See book titled New methods for solving quadratic equations and inequalities (Amazon e- book 2010). Page 2 of 7
3 Examples of solving by the test point method. Example. Solve: x^2 15x < 16. Solution. First step, write the inequality in standard form f(x) = x^2 15x 16 < 0. Second step, solve f(x) = 0. The 2 real roots are -1 and 16. Third step, solve the inequality f(x) < 0. Plot the 2 real roots -1 and 16 on a number-line. The origin O is located inside the segment (-1, 16). Use the origin O as test point. Substitute x = 0 into the inequality. We get -16 < 0. It is true, then the origin O is located on the true segment. Step 4, express the solution set in the form of open interval (-1, 16). The 2 endpoints -1 and 16 are not included in the solution set. Example. Solve: -3x^2 < -8x + 5 Solution. First step, write the inequality into the standard form: f(x) = -3x^2 + 8x 5 < 0. Second step, solve f(x) = 0. The 2 real roots are 1 and 5/3. Third step, plot these values on a number line. The origin O is located on the left ray. Use O as test point. Substitute x = 0 into the inequality. We get: -5 < 0. It is true, then the origin O is located on the true ray. By symmetry, the other ray also belongs to the solution set. Last step, express the solution set in the form of intervals: (-infinity, 1) and (5/3, +infinity). The 2 end points 1 and 5/3 are not included. Example. Solve: 9x^2 < 12x 1 Solution. First step: f(x) = 9x^2 12x + 1 < 0. Second step: Solve f(x) = 0. The new Diagonal Sum Method fails to solve it, this equation can t be factored. We must use the quadratic formula. The 2 real roots are x1 = (2 1.73)/3 = 0.09 and x2 = ( )/3 = Third step: Plot the numbers on a number line. The origin is located on the left ray. Substitute x = 0 into the inequality. We have: 1 < 0. It is not true. The origin O is not on the true ray. The solution set is the segment (0.09, 1.24). Step 4, solution set: open interval (0.09, 1.24); the end points not included. 2. The algebraic method. This solving method is popular in Europe. It is based on a theorem about the sign status of a trinomial f(x) = ax^2 + bx + c, with a not zero, and D = b^2 4ac Theorem on the sign status of a trinomial f(x). a. If D < 0, f(x) has the same sign as a regardless of the values of x. Example 1. The trinomial f(x) = 3x^2 x + 7 has D = b^2 4ac = 1 84 = -83 < 0. This trinomial f(x) is always positive, same sign as a = 3, regardless of the values of x. Page 3 of 7
4 Example 2. The trinomial f(x) = -5x^2 + 3x 8 has D = = < 0. This f(x) is always negative (< 0), same sign as a = -5, regardless of the values of x. b. When D = 0, f(x) has the same sign as a for any values of x different to ( b/2a). c. When D > 0, f(x) has the opposite sign of a between the 2 real roots x1 and x2, and f(x) has the same sign as a outside the interval (x1-x2). Example. The trinomial f(x) = x^2 8 x 9 has D = = 100 = 10^2 > 0. The equation f(x) = 0 has 2 real roots (-1) and (9). The trinomial f(x) is negative (< 0) within the interval (-1, 9). f(x) is positive (> 0) outside this interval. Example. The trinomial f(x) = -x^2 + 5x 4 = 0 has D = = 9 = 3^2 > 0. The equation f(x) = 0 has 2 real roots 1 and 4. The trinomial f(x) is positive, opposite to the sign of a = -1, within the interval (1, 4). The Theorem s proof a. When D < 0. We can write f(x) in the form f(x) = a(x^2 + bx/a + c/a). (1) Recall the equation developed to find the quadratic formula: (x^2 + bx/a + cx/a) = 0 x^2 + bx/a + b^2/4a^2 b^2/4a2 + c/a = 0 (x + b/2a)^2 (b^2 4ac)/4a^2 = 0 (Call D = b^2 4ac). (x + b/2a)^2 D/4a^2 = 0 Substitute this relation into the equation (1). f(x) = a [(x + b.2a)^2 D/4a^2] (2) a. When D < 0, the quantity inside the parenthesis is always positive. Therefore, f(x) has the same sign as a regardless of the values of x. b. When D = 0, the equation (2) reduces to: f(x) = a(x + b/2a) ^2. We see that f(x) has the same sign as a regardless of the values of x, since the quantity (x + b.2a)^2 is always positive. c. When D > 0. There are 2 real roots x1 and x2, with their sum x1 + x2 = -b/a, and their product x1.x2 = c/a. We can write f(x) in the form f(x) = a(x^2 + bx/a + c/a). The quantity in parenthesis is a quadratic equation that can be factored into 2 binomials in x with x1 and x2 as real roots. Page 4 of 7
5 f(x) = a(x x1)(x x2) (1) Now, setup a Sign Table (sign chart) to study the sign status of f(x) when x varies from infinity to +infinity and passes by the 2 values x1 and x2. Suppose x2 > x1. x -infinity x1 x2 +infinity x x x x (x x1)(x x2) We see from the relation (1) and the last line of the chart that f(x) has the opposite sign of a between the 2 real roots x1 and x2. Outside the interval (x1, x2), f(x) have the same sign as a. Graphic interpretation of the Theorem. We can easily understand this Theorem by looking at the parabola graph of a quadratic equation. If a > 0, the parabola is upward. Between the two x-intercepts (real roots) of the parabola, a part of the parabola is below the x-axis, meaning f(x) is negative (< 0) within this interval (x1, x2), opposite in sign to a (positive). If a < 0, the parabola is downward. Between the two x-intercepts (real roots), a part of the parabola is above the x-axis, meaning f(x) is positive within this interval (x1, x2), opposite in sign to a (negative). Examples about the Theorem. Example. The graph of the trinomial f(x) = x^2 7x 8 is an upward parabola whose x-intercepts are at -1 and 8. Within the interval (-1, 8) a part of the parabola is below the x-axis, meaning f(x) is negative (< 0) within this interval, opposite in sign to a (positive) Example. The graph of f(x) = - x^2 + 11x 10 is a downward parabola whose x- intercepts are at 1 and 10. Within the interval (1, 10), a part of the parabola is above the x-axis, meaning f(x) is positive (> 0) in this interval, opposite in sign to a (< 0). NOTE. Students study once the theorem s development, then they apply it to solve various types of quadratic inequalities. Page 5 of 7
6 3. The graphing method The concept of this method is simple. When the parabola graph of the trinomial f(x) is above the x-axis, the trinomial f(x) is positive, and vice-versa. You don t need to accurately graph the parabola. Based on the 2 real roots obtained by solving f(x) = 0, you may just draw a rough sketch of the parabola. Pay only attention to if the parabola is upward or downward, by considering the sign of a. After graphing, you get the solution set by looking at the graph of the parabola. Last, express the solution set in the form of intervals. By this graphing method, you may solve a system of two, or even three, quadratic inequalities, by graphing them on the same coordinate grid. IV. EXAMPLES OF SOLVING QUADRATIC INEQUALITIES Example. Solve: f(x) = 13x^2 7x + 8 > 0. Solution. The Discriminant D = < 0. There are no real roots. Use the algebraic method (Theorem) to solve it. Since a is positive, f(x) is always positive (> 0) regardless of the values of x. The inequality is always true. Example. Solve: f(x) = -5x^2 + 6x 10 > 0. Solution. D = < 0. There are no real roots. Use the Theorem to solve. The trinomial f(x) is always negative (< 0), same sign as a = -5, regardless of x. The equation is always not true. Example. Solve: f(x) = 3x^2 7x 10 > 0. Solution by the number line and test point method. The equation f(x) = 0 has 2 real roots -10/3 and 1. Plot these numbers on a number line. The origin O is located inside the segment (-10/3, 1). Substitute x = 0 into the inequality. We have: 10 > 0. It is not true, then O is not on the solution set. The solution set is the 2 rays. Express the solution set by intervals: (-infinity, -10/3) and (1, +infinity). Solution by the algebraic method. The equation f(x) has 2 real roots -10/3 and 1. Inside the segment (-10/3, 1) f(x) is negative (< 0), opposite in sign to a = 5. Therefore the 2 rays are the solution set. Answers: (-infinity, -10/3) and (1, +infinity). There is no need to draw the number line and doing the test point. Example. Solve: f(x) = -3x^2 + 18x 15 < 0. Solution by the number line. The equation f(x) = 0 has 2 real roots 1 and 5. Plot these numbers on a number line. The origin O is outside the segment (1, 5). Substitute x = 0 into the inequality. We get: -15 < 0. It is true then O is located on the true ray. The 2 rays are the solution set. Answers: (-infinity, 1) and (5, +infinity). Solution by the algebraic method. The 2 real roots are 1 and 5. Outside the interval (1, 5) f(x) is positive (> 0), opposite in sign to a = -3. The 2 rays are the solution set. Answers: (-infinity, 1) and (5, +infinity). Page 6 of 7
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# RBSE Solutions for Class 9 Maths Chapter 6 Rectilinear Figures Ex 6.1
RBSE Solutions for Class 9 Maths Chapter 6 Rectilinear Figures Ex 6.1 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Exercise 6.1.
Board RBSE Textbook SIERT, Rajasthan Class Class 9 Subject Maths Chapter Chapter 6 Chapter Name Rectilinear Figures Exercise Ex 6.1 Number of Questions Solved 11 Category RBSE Solutions
## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Ex 6.1
Question 1.
From the given figure, find the (RBSESolutions.com) three angles of the triangle ABC.
Solution.
Here ∠DCE = 53°
∴∠ACB = 53°
(Vertically opposite angle)
In ∆ABC
∠ABC + ∠ACB = ∠CAF
(By exterior angle property)
⇒ ∠ABC + 53°= 112°
⇒ ∠ABC = 112° – 53°
⇒ ∠ABC = 59°
Now in ∆ABC
∠BAC + ∠ABC + ∠BCA = 180°
(by angle sum property)
⇒ ∠BAC + 59° + 53° = 180°
⇒ ∠BAC = 68°
Hence, ∠BAC = 68°, ∠ABC = 59° and ∠ACB = 53°.
Question 2.
In figure, ∆ABC is an equilateral triangle. Find (RBSESolutions.com) the values of ∠x, ∠y and ∠z from the given figure.
Solution.
∵ ∆ABC is an equilateral triangle
∴AB = BC = CA
⇒ ∠ABC = ∠ACB = ∠BAC = 60°
60° = x + 22°
(exterior angle property)
x = 38°
Also
38° + 22° + ∠z = 180°
(by angle sum property of a ∆)
⇒ ∠z = 180° – 60° = 120°
Again by exterior angle property
⇒ ∠ACB = ∠y + 38°
⇒ 60° = ∠y + 38°
⇒ ∠y = 22°
Hence, ∠x = 38°, ∠y = 22° and ∠z = 120°
Question 3.
In the given figure, the sides AB and AC of ∆ABC are (RBSESolutions.com) produced to point E and D respectively. If the bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove
∠BOC = 90°- $$\frac { 1 }{ 2 }$$∠x.
Solution.
Ray BO is the (RBSESolutions.com) bisector of ∠CBE.
Therefore, ∠CBO = $$\frac { 1 }{ 2 }$$∠CBE
⇒ ∠CBO = $$\frac { 1 }{ 2 }$$(180° – y) = 90° – $$\frac { y }{ 2 }$$…(i)
Similarly, ray CO is the bisector of ∠BCD
Therefore, ∠BCO = $$\frac { 1 }{ 2 }$$∠BCD
= $$\frac { 1 }{ 2 }$$(180° – z)
= 90°- $$\frac { z }{ 2 }$$ …(ii)
Now in ∆BOC,
∠CBO + ∠BCO + ∠BOC = 180° …(iii)
(angle sum property of a triangle)
Using (i) and (ii) in (iii), we get
⇒ 90°- $$\frac { y }{ 2 }$$ + 90°- $$\frac { z }{ 2 }$$ + ∠BOC = 180°
⇒ ∠BOC = $$\frac { 1 }{ 2 }$$ – (y + z) …(iv)
But in ∆ABC,
x + y + z = 180°
(angle sum property of a triangle)
⇒ y + z = 180° – x
⇒ 2∠BOC = 180° – x
[using relation (iv)]
⇒ ∠BOC = 90° – $$\frac { 1 }{ 2 }$$ ∠x
Hence proved.
Question 4.
In figure, ∠P = 52°, ∠PQR = 64°. If QO and RO are the (RBSESolutions.com) bisectors of ∠PQR and ∠PRQ respectively of ∆PQR, find ∠x and
Solution.
Given: QO and RO are the (RBSESolutions.com) bisectors of ∠PQR and ∠PRQ respectively of ∆PQR and ∠P = 52°, ∠PQO = 64°.
To find: ∠x and ∠y
In ∆PQR,
∠P + ∠PQR + ∠PRQ = 180°
(angle sum property of a triangle)
⇒ 52° + 64° + ∠PRQ = 180°
⇒ ∠PRQ = 180° – 116°
⇒ ∠PRQ = 64°
⇒ ∠y = 32°
(as RD is bisector of ∠PRQ)
In ∆OQR,
∠OQR + ∠ORQ + ∠x = 180°
(reason as above)
⇒ 32° + 32° + ∠x= 180°
[QO and RO are bisector (RBSESolutions.com) of ∠PQR and ∠PRQ]
⇒ ∠x = 180° – 64°
⇒ ∠x = 116°
Hence, ∠x = 116°, ∠y = 32°.
Question 5.
In figure, AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Solution.
We are given that AB || DE
and ∠BAC = 35° and ∠CDE = 53°
AB || DE (given)
⇒ ∠BAE = ∠AED = 35°
(alternate angles)
Now in ∆CDE,
∠CDE + ∠E + ∠DCE = 180°
(angle sum property of a triangle)
⇒ 53° + 35° + ∠DCE = 180°
⇒ ∠DCE = 180° – 88° = 92°
⇒ ∠DCE = 92°
Hence, ∠DCE = 92°.
Question 6.
In the adjoining figure if lines PQ and RS intersect (RBSESolutions.com) at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Solution.
In ∆PRT,
∠RPT + ∠PRT + ∠RTP = 180°
(angle sum property of a triangle)
⇒ 95° + 40° + ∠RTP = 180°
⇒ 135° + ∠RTP = 180°
⇒ ∠RTP = 180° – 135°
= 45°
Now ∠RTP = ∠STQ = 45°
(vertically opposite angles)
In ∆STQ,
∠STQ + ∠TSQ + ∠SQT = 180°
(angle sum property of a triangle)
⇒ 45° + 75° + ∠SQT = 180°
⇒ ∠SQT = 180° – (45° + 75°)
= 180° – 120° = 60°
Hence, ∠SQT = 60°.
Question 7.
In figure, sides QP and RQ of ∆PQR are produced (RBSESolutions.com) to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Solution.
We are given that ∠SPR = 135° and ∠PQT =110°.
At point Q:
∠PQT + ∠PQR = 180°
(linear pair of angles)
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° – 110°
= 70° …(i)
At point P:
∠QPR + ∠SPR = 180°
(linear pair of angles)
⇒ ∠QPR + 135°= 180°
⇒ ∠QPR = 45° …(ii)
In ∆PQR,
∠PQR + ∠PRQ + ∠QPR = 180°
⇒ 70° + ∠PRQ + 45° = 180°
⇒ ∠PRQ = 180° – 115°
⇒ ∠PRQ = 65°
Hence, ∠PRQ = 65°.
Question 8.
In figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find (RBSESolutions.com) the values of x and y.
Solution.
We are given that PQ ⊥ PS, PQ || SR
∠SQR = 28° and ∠QRT = 65°
SR and a transversal QR intersect them
∠PQR = ∠QRT = 65°
(alternate angles)
x + 28° = 65°
x = 65° – 28°
x = 37°
Now, in right angled triangle SPQ
x + y + ∠SPQ = 180°
(by angle sum property of a triangle)
x + y + 90° = 180°
⇒ 37° + y + 90°= 180°
⇒ 127° + y = 180°
⇒ y = 180 – 127°
⇒ y = 53°
Hence, x = 37° and y = 53°.
Question 9.
In the given figure, the side QR and APQR is produced to (RBSESolutions.com) a point S. If the bisector of ∠PQR and ∠PRS meet at a point T, then prove that ∠QTR = $$\frac { 1 }{ 2 }$$∠QPR.
Solution.
Side QR of ∆PQR is produced to S,
Exterior ∠PRS = ∠P + ∠Q …(i)
(Exterior angle is equal to sum of its opposite interior angles)
Dividing of both side (RBSESolutions.com) relation (i) by 2, we get
$$\frac { 1 }{ 2 }$$∠PRS = $$\frac { 1 }{ 2 }$$∠P + $$\frac { 1 }{ 2 }$$∠Q
⇒ x = $$\frac { 1 }{ 2 }$$∠P + y …(ii)
[∵ RT and QT are bisectors of ∠PRS and ∠PQS respectively]
Also, in ∆QRT,
Exterior ∠TRS = ∠T + ∠y
⇒ x = ∠T + y …(iii)
From (i) and (iii), we get
$$\frac { 1 }{ 2 }$$ ∠P + y = ∠T + y
$$\frac { 1 }{ 2 }$$∠P = ∠T
∠QTR = $$\frac { 1 }{ 2 }$$∠QPR
Hence proved.
Question 10.
In ∆ABC, ∠A = 90°, AL ⊥ BC, prove that ∠BAL = ∠ACB.
Solution.
Given: In ∆ABC, ∠A = 90° and AL ⊥ BC
To prove: ∠BAL = ∠ACB
Proof: Suppose ∠BAL = ∠1, ∠CAL = ∠2, ∠ABL = 3 and ∠ACL = ∠4
Now in ∆ABC
∠A + ∠B + ∠C = 180°
(angle sum property of a triangle)
⇒ ∠90° + ∠3 + ∠4 = 180°
(∵ ∠A = 90 given)
⇒ ∠3 + ∠4 = ∠90°
⇒ ∠4 = ∠90° – ∠3 …(i)
Now in ∠BAL
∠1 + ∠3 + ∠ALB = 180°
(Angle sum property of a triangle)
⇒ ∠1 + ∠3 + 90° = 180°
⇒ ∠1 + ∠3 = 90°
⇒ ∠1 = 90° – ∠3 …(ii)
From (i) and (ii), we get
∠1 = ∠4 => ∠BAL = ∠ACB
Hence proved.
Question 11.
The angles of a triangle are the ratio 2 : 3 : 4. Find all the (RBSESolutions.com) three angles of a triangle.
Solution.
Let angles of a triangle are 2x, 3x, 4x
2x + 3x + 4x = 180°
⇒ 9x = 180°
⇒ x = 20°
Angles are
2x = 2 x 20° = 40°,
3x = 3 x 20° = 60°,
4x = 4 x 20° = 80°.
We hope the given RBSE Solutions for Class 9 Maths Chapter 6 Rectilinear Figures Ex 6.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Exercise 6.1, drop a comment below and we will get back to you at the earliest.
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# Three resistors of equal resistance R are connected in series and then connected in parallel. What will be the ratio of equivalent resistance in series and parallel?
1. 1 : 9
2. 1 : 3
3. 3 : 1
4. 9 : 1
Option 4 : 9 : 1
## Combination of Resistors — Series and Parallel MCQ Question 1 Detailed Solution
CONCEPT:
• Resistance: The obstruction offered to the flow of current is known as the resistance. It is denoted by R.
• When two or more resistances are connected one after another such that the same current flows through them then it is called resistances in series.
• The equivalent resistance in series combination is will be
Rser = R1 + R2 + R3
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal then it is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
EXPLANATION:
Given that,
R1 = R2 = R3 = R
When resistor are connected in series, then the equivalent resistance is
Rser = R1 + R2 + R3 = R + R + R = 3R ---(1)
When the resistor is connected in parallel, then the equivalent resistance is
$$\frac{1}{{{R_{para}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_2}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}$$
∴ Rpara = R/3 ---(2)
Divide equation 1 and 2, we get
# The equivalent resistance of the given combination will be:
1. $$2 \ \Omega$$
2. $$4 \ \Omega$$
3. $$9 \ \Omega$$
4. $$11 \ \Omega$$
Option 2 : $$4 \ \Omega$$
## Combination of Resistors — Series and Parallel MCQ Question 2 Detailed Solution
CONCEPT:
• Resistance: The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.
There are mainly two ways of the combination of resistances:
• Resistances in series combination: When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
The net resistance/equivalent resistance (R) of resistances in series is given by:
Equivalent resistance, R = R1 + R2
• Resistances in parallel combination: When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
CALCULATION:
Here 3 Ω and 6 Ω are in parallel combination:
So 1/R' = 1/3 + 1/6 = (2 + 1)/6 = 3/6 = 1/2
R' = 2 Ω
Now R' and 2 Ω will be in series combination:
Equivalent resistance (R) = 2 + 2 = 4 Ω
Hence option 2 is correct.
# What is the minimum resistance which can be made using five resistors each of 1/5 Ω?
1. 1/5 Ω
2. 1/25 Ω
3. 1/10 Ω
4. 25 Ω
Option 2 : 1/25 Ω
## Combination of Resistors — Series and Parallel MCQ Question 3 Detailed Solution
CONCEPT:
• Resistance: The obstruction offered to the flow of current is known as the resistance. It is denoted by R.
• When two or more resistances are connected one after another such that the same current flows through them then it is called resistances in series.
• The equivalent resistance in series combination is will be
Rser = R+ R+ R3
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal then it is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
EXPLANATION:
Given - Resistance of 5 resistors = 1/5Ω
• When n such nR resistors are connected in parallel.
• Then, the effective resistance of the circuit would be
$$\Rightarrow \frac{1}{{{R_{net}}}} = \frac{1}{{n{R_1}}} + \frac{1}{{n{R_2}}} \pm - - - - - \; + n\;times = \frac{n}{R}$$
• Here, 5 resistors are connected in parallel, therefore the net resistance is given by
$$\Rightarrow {R_{net}} = \frac{R}{5} = \frac{{\frac{1}{5}}}{5} = \frac{1}{{25}}{\rm{Ω }}$$
# Two electrical resistances R and 2R are connected in parallel combination. This combination is connected in series with a battery of potential difference V. Find the ratio of heat dissipated in two resistances.
1. 2 : 1
2. 4 : 1
3. 1 : 4
4. 8 : 1
Option 1 : 2 : 1
## Combination of Resistors — Series and Parallel MCQ Question 4 Detailed Solution
The correct option is: 1
CONCEPT:
Resistances in parallel:
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
• Heating effect of electric current: When a current is flowing in a circuit having resistance there is heat dissipation due to the resistance. This is called the heating effect of electric current.
• The heat dissipated is given by:
$$\Rightarrow Heat (H)=I^2Rt=\frac{V^2t}{R}$$
Where I = the current flowing in the circuit, R = the resistance of the circuit, V is the potential difference, and t = the time taken
CALCULATION:
Given that: R1 = R Ω and R2 = 2R Ω
• Both are connected in parallel combination, so the potential difference (V) will be the same for both:
$$\Rightarrow Heat (H)=\frac{V^2t}{R}$$
Therefore,
$$\Rightarrow \frac{H_1}{H_2}=\frac{R_2}{R_1}=\frac{2R}{R}=\frac{2}{1}$$ (Since time t and Potential difference V are the same for both, hence those will cancel out in fraction)
# Which one of the following is the resistance that must be placed parallel with 12 Ω resistances to obtain a combined resistance of 4 Ω?
1. 2 Ω
2. 4 Ω
3. 6 Ω
4. 8 Ω
Option 3 : 6 Ω
## Combination of Resistors — Series and Parallel MCQ Question 5 Detailed Solution
CONCEPT:
• Parallel combination: Two or more resistors are said to be in parallel if one end of all the resistors is joined together and similarly the other ends joined together.
$$\frac{1}{{{{\rm{R}}_{{\rm{eq}}}}}} = \frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}} + \frac{1}{{{{\rm{R}}_3}}}$$
Where, R1, R2, and R3 are the resistance of the branches of a parallel circuit.
CALCULATION:
Given data- R1 = 12 Ω and Req = 4 Ω, R2 = ?
$$\frac{1}{{{{\rm{R}}_{{\rm{eq}}}}}} = \frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}$$
$$\frac{1}{{{\rm{\;}}4}} = \frac{1}{{12}} + \frac{1}{{{{\rm{R}}_2}}} \Rightarrow \frac{1}{{{{\rm{R}}_2}}} = \frac{1}{4} - \frac{1}{{12}} \Rightarrow \frac{1}{{{{\rm{R}}_2}}} = \frac{{3 - 1}}{{12}}$$
Then,
$$\frac{1}{{{{\rm{R}}_2}}} = \frac{2}{{12}}$$
Now,
R2 = 6Ω
# Three resistances of value 2Ω each are arranged in a triangle form. The total resistances between any two corners will be
1. 4 Ω
2. 2 Ω
3. 3/4 Ω
4. 4/3 Ω
Option 4 : 4/3 Ω
## Combination of Resistors — Series and Parallel MCQ Question 6 Detailed Solution
CONCEPT:
Resistance:
• The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.
Resistances in series:
• When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
• The net resistance/equivalent resistance (R) of resistances in series is given by:
• Equivalent resistance, R = R1 + R2
Resistances in parallel:
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
CALCULATION:
• It is given that, three 2 Ω resistors are connected to form a triangle. So, from the above figure, it is clear that resistors RAB and RAC are in series. Therefore, their equivalent resistance will be-
⇒ Rser = RAB + RBC
⇒ Rser = 2 Ω + 2 Ω = 4 Ω
• Now resistors Rser and RBC are in parallel with each other. Therefore, their equivalent resistance will be
$$\Rightarrow \frac{1}{{{R_{para}}}} = \frac{1}{{{R_BC}}} + \frac{1}{{{R_{ser}}}}$$
$$\Rightarrow\frac{1}{{{R_{para}}}} = \frac{1}{{2}} + \frac{1}{{4\;}} = \frac{3}{{4}}$$
$$\Rightarrow{R_{para}} = \frac{4}{{3}}{\rm{\Omega }}$$
# The resistance of a motor is 90 Ω, resistance of bulb is 60 Ω, and a fan of resistance 30 Ω are connected in parallel to a 240 V source. Find the total value (approx) of current flowing through all appliances?
1. 15 A
2. 10 A
3. 5 A
4. 12 A
Option 1 : 15 A
## Combination of Resistors — Series and Parallel MCQ Question 7 Detailed Solution
CONCEPT:
• Resistance: The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.
There are mainly two ways of the combination of resistances:
• Resistances in series combination: When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
The net resistance/equivalent resistance (R) of resistances in series is given by:
Equivalent resistance, R = R1 + R2
• Resistances in parallel combination: When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
• Ohm’s law: At constant temperature and other physical quantities, the potential difference across a current-carrying wire is directly proportional to the current flowing through it.
V= R I
Where V is the potential difference, R is resistance and I is current.
CALCULATION:
Given that:
Resistance of Motor (R1) = 90 ohm
Resistance of a bulb (R2) = 60 ohm
Resistance of a Fan (R3) = 30 ohm
Potential difference (V) = 220 volt
Total resistance of the three appliances in parallel:
1/R = 1/R1 + 1/R2 + 1/R3 = 1/90 + 1/60 + 1/30
⇒ 1/R = 11 / 180
So R = 16.36 Ohm
As we know that,
Ohm's Law, V = I x R
⇒ I = V/R
⇒ Electric current (I) = 240 / 16.36 = 14.66 A 15 A.
So option 1 is correct.
# Two resistors of R Ω and 20 Ω are connected in parallel to get an effective resistance of 15 Ω. Find R.
1. 40
2. 60
3. 50
4. 30
Option 2 : 60
## Combination of Resistors — Series and Parallel MCQ Question 8 Detailed Solution
CONCEPT:
• Resistances in parallel combination: When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
CALCULATION:
Given that,
R$$R\ Ω$$
R$$20\ Ω$$
Net resistance/effective resistance (R) = 15 Ω
Use the above-given formula for the parallel combination:
$${{1\over 15}={{1\over R}+{1\over 20}}}$$
$$20R= 300+15R$$
$$5R=300$$
$$R=60\ Ω.$$
# When two or more resistances are connected between the same two points, they are said to be connected in-
1. Across
2. Series
3. Parallel
4. Line
Option 3 : Parallel
## Combination of Resistors — Series and Parallel MCQ Question 9 Detailed Solution
CONCEPT:
• Parallel combination: When two or more resistances are connected between the same two points, they are said to be connected in parallel combination.
• A parallel circuit is a circuit in which the resistors are arranged with their heads connected together, and their tails connected together.
• The current in a parallel circuit breaks up, with some flowing along each parallel branch and re-combining when the branches meet again.
• The voltage across each resistor in parallel is the same.
The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
• Series combination: When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
The net resistance/equivalent resistance (R) of resistances in series is given by:
Equivalent resistance, R = R1 + R2
EXPLANATION:
• When two or more resistances are connected between the same two points, they are said to be connected in parallel combination. So option 3 is correct.
# An equilateral triangle is made of three wires of equal resistances 4 Ω. Find the equivalent resistance across any one side.
1. 4 Ω
2. 8 Ω
3. 4/3 Ω
4. 8/3 Ω
Option 4 : 8/3 Ω
## Combination of Resistors — Series and Parallel MCQ Question 10 Detailed Solution
CONCEPT:
Resistance:
• The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.
Resistances in series:
• When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
• The net resistance/equivalent resistance (R) of resistances in series is given by:
• Equivalent resistance, R = R1 + R2
Resistances in parallel:
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
CALCULATION:
• Here 4 Ω across AB and AC are in a series combination, the equivalent resistance is
⇒ S = R1 + R2 = 4 Ω + 4 Ω = 8 Ω
• Now 4 Ω and 8 Ω are connected in parallel combination, the equivalent resistance is
$$⇒ \frac{1}{R_{eq}} = \frac{1}{{{8}}} + \frac{1}{{{4}}}=\frac{3}{8}$$
⇒ Req = 8/3 Ω
# Keeping voltage constant, if more lamps are put into a series circuit, the overall current in the circuit:A. IncreasesB. DecreasesC. Remains the sameD. Becomes infinite
1. C
2. B
3. A
4. D
Option 2 : B
## Combination of Resistors — Series and Parallel MCQ Question 11 Detailed Solution
Option 2 is the correct answer.
CONCEPT:
• Resistances in series combination: When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
The net resistance/equivalent resistance (R) of resistances in series is given by:
Equivalent resistance, R = R1 + R2
• Ohm’s law: At constant temperature and other physical quantities, the potential difference across a current-carrying wire is directly proportional to the current flowing through it.
V= R I
Where V is the potential difference, R is resistance and I is current.
EXPLANATION:
• In a series circuit, the total resistance in the circuit is equal to the sum of all the resistances in the circuit.
• A lamp is nothing but resistance, hence with the addition of a lamp the total resistance of the circuit will increase.
• Now when the voltage is kept constant the current in the circuit will decrease with the increase of resistance in accordance with ohm’s Law (I = V/R) (with an increase in denominator value of fraction decreases).
• Thus on adding more lamps into a series circuit, the overall current in the circuit will decrease if the voltage is kept constant. So option 2 is correct.
• The opposite will happen in a circuit where lamps/resistors will be connected parallel.
*The wirings in houses are done in parallel.
# What is the equivalent resistance between the terminals A and B? (R = 3Ω )
1. 3 Ω
2. 8/3 Ω
3. 3/8 Ω
4. 8 Ω
Option 4 : 8 Ω
## Combination of Resistors — Series and Parallel MCQ Question 12 Detailed Solution
CONCEPT:
• Resistance is the measurement of the opposition of the flow of electric current through a conductor.
• It is represented by R.
• SI unit is the ohm ( Ω ).
There are two ways of a combination of resistance:
• Series Combination: When two or more than two resistance are connected in such a way that the same current is flowing through the resistance connected to each other then the combination of resistance are said to be in series combination.
• The equivalent resistance (R) is
RS = R1 + R2
• Parallel Combination: When two or more than two resistance are connected in such a way that the same voltage is applying on all the resistance connected each other then the combination of resistance is said to be in parallel combination
• The equivalent resistance (R) is
1/RP = 1/R1 + 1/R2
CALCULATION:
Given:
R = 3Ω
Redraw the given figure in the simple form we get
We know that the above two resistance are in series
Rs = R + R
Rs = 2R
Again redraw the figure we get
Clearly, the above two resistance are in parallel
1/RP = 1/R + 1/2R
1/RP = 3/2R
RP = 2R/3
Now all the resistance are in series solve by using a series combination
RE = R + R + 2R/3
RE = 8R/3
But the value of R is 3Ω
RE = 8Ω
So the equivalent resistance between terminal A and B is
Hence option 4 is correct.
• Ohms law state that voltage (V) is directly proportional to the electric current (I) flowing through a conductor
V ∝ I
# If a wire of resistance 2 R Ω, with uniform cross-section, is cut into two halves and the end points are joined together, the equivalent resistance obtained between these two points is:
1. R/2 Ω
2. R/4 Ω
3. 2R Ω
4. R Ω
Option 1 : R/2 Ω
## Combination of Resistors — Series and Parallel MCQ Question 13 Detailed Solution
Concept:
Resistance:
• The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.
There are mainly two ways of the combination of resistances:
1. Resistances in series:
• When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
• The net resistance/equivalent resistance (R) of resistances in series is given by:
• Equivalent resistance, R = R1 + R2
2. Resistances in parallel:
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{{{R_{eff}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
Calculation:
Given the resistance = 2R Ω
They are of uniform cross-sectional area and of the same material
Since they are cut into two equal parts each part has resistance = R Ω
So they are connected in parallel-
We know –
$$\frac{1}{{{R_{eff}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
$$\frac{1}{{{R_{eff}}}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$$
Reff = R/2
So the correct option is R/2.
# 2 Ω and 4 Ω resistors are joined in parallel across a 6 volt battery. The ratio of voltages across them will be
1. 1 : 1
2. 1 : 2
3. 2 : 1
4. 4 : 1
Option 1 : 1 : 1
## Combination of Resistors — Series and Parallel MCQ Question 14 Detailed Solution
CONCEPT:
Resistance:
• The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.
Resistances in parallel:
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$⇒ \frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
CALCULATION:
Given - R1 = 2 Ω, R= 4 Ω and V = 6 V
• Here 2 Ω and 4 Ω resistors are joined in parallel across a 6-volt battery and as we know that in parallel combination, the potential difference across the resistor is equal. Hence, the ratio of voltages across 2 Ω and 4 Ω is
⇒ V1 : V2 = 1 : 1
Resistances in series:
• When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
• The net resistance/equivalent resistance (R) of resistances in series is given by:
• Equivalent resistance, R = R1 + R2
# n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum resistance to the minimum resistance-
1. n
2. n2/2
3. n2
4. $$\dfrac{1}{n}$$
Option 3 : n2
## Combination of Resistors — Series and Parallel MCQ Question 15 Detailed Solution
Concept:
• Resistance is the ability of any electrical component to resist electric current across it.
• Equivalent Resistance can be defined as the resistance of the resistor which will replace all the resistors between two points and will draw the same current between these two points as it was flowing Earlier.
• Ohms Law: At constant temperature, a potential difference is the product of current and resistance.
Series and Parallel Connection:
Series Connection Parallel Connection Resistors are connected in such a way that the same current is passing across them. Resistors are connected in such a way that potential difference is the same Across them. Equivalent Resistance of n resistors connected in series is given as R = R1 + R2 + R3 .....Rn Equivalent Resistance of n resistors Connected in Parallel is given as $$\frac {1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} ....$$ 1R=1R1+1R2+1R3.....+1Rn" id="MathJax-Element-5-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}......\frac{1}{R_n}$$1R=1R1+1R2+1R3.....+1Rn1R=1R1+1R2+1R3.....+1Rn" id="MathJax-Element-15-Frame" role="presentation" style="position: relative;" tabindex="0">1R=1R1+1R2+1R3.....+1Rn$\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}.....+\frac{1}{{R}_{n}}$ Circuit Diagram: Circuit Diagram:
• For a combination of resistors, maximum equivalent resistance can be obtained when resistors are in series with each other and minimum equivalent resistance can be obtained if the resistors are in parallel with each other.
Calculation:
Let resistance of each resistor be r
For n resistors, the maximum resistance can be obtained when resistors are in series. So, for n resistors in series, the equivalent resistance is
Rs = r + r + r --- n times = nr
For n resistors, the minimum resistance can be obtained when resistors are in parallel. So, for n resistors in parallel, the equivalent resistance is
$$\frac {1}{R_{p}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} ....n \hspace{0.1cm}times$$
$$\implies \frac {1}{R_{p}} = \frac{n}{r}$$
$$\implies R_p = \frac{r}{n}$$
the ratio of the maximum resistance (Rs) to the minimum resistance (Rp)
$$\frac {R_{s}}{R_{p}} = \frac{nr}{r/n} = n^2$$
So, n2 is the correct option.
# Three resistors 80 Ω, 120 Ω and 240 Ω are connected in parallel. A 12 V battery is connect across combination of resistors. Find the current drawn from the battery.
1. 0.3 A
2. 0.09 A
3. 0.9 A
4. 3 A
Option 1 : 0.3 A
## Combination of Resistors — Series and Parallel MCQ Question 16 Detailed Solution
CONCEPT:
• Ohm’s law: At constant temperature and other physical quantities, the potential difference across a current-carrying wire is directly proportional to the current flowing through it.
V = RI
Where V is the potential difference, R is resistance and I is current.
• Resistances in parallel combination: When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
CALCULATION:
Given that:
Potential (V) = 12 V
R1 = 80 Ω, R2 = 120 Ω and R3 = 240 Ω
So equivalent resistance (R) =?
1/R = 1/R1 + 1/R2 + 1/R3 = 1/80 + 1/120 + 1/240 = 6/240 = 1/40
R = 40 Ω
According to Ohm’s law
⇒ I = V/R
Electric current drawn (I) = 12/40 = 0.3 A
So option 1 is correct.
# Two resistors of R Ω and 20 Ω are connected in parallel to get an effective resistance of 15 Ω. Find R.
1. 40
2. 60
3. 50
4. 30
5. None of the above/ More than one of the above.
Option 2 : 60
## Combination of Resistors — Series and Parallel MCQ Question 17 Detailed Solution
Key Points
CONCEPT:
• Resistances in parallel combination: When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
CALCULATION:
Given that,
R$$R\ Ω$$
R$$20\ Ω$$
Net resistance/effective resistance (R) = 15 Ω
Use the above-given formula for the parallel combination:
$${{1\over 15}={{1\over R}+{1\over 20}}}$$
$$20R= 300+15R$$
$$5R=300$$
∴ $$R=60\ Ω.$$
# There are n similar conductors each of resistance R. The resultant resistance comes out to be x when connected in parallel. If they are connected in series, the resistance comes out to be
1. x/n2
2. n2x
3. x/n
4. nx
Option 2 : n2x
## Combination of Resistors — Series and Parallel MCQ Question 18 Detailed Solution
CONCEPT:
Resistance:
The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.
There are mainly two ways of the combination of resistances:
1. Resistances in series:
• When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
• The net resistance/equivalent resistance (R) of resistances in series is given by:
• Equivalent resistance, R = R1 + R2
2. Resistances in parallel:
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
• $$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
Explanation:
Given - Resistance of n resistor = R and Resultant resistant = x
In parallel combination,
$$⇒ \frac{1}{R_{para}}=\frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \ldots \ldots ..$$
$$⇒ \frac{1}{x} = \frac{n}{R}$$
$$⇒ x = \frac{R}{n} ⇒ R = nx$$
In series combination,
⇒ Rseries = R + R + R ……n times
⇒ Rseries = n(R) = n(nx) = n2x
# Ten wires each of resistance 1 ohm are connected in parallel, then total resistance will be:
1. 10Ω
2. 1Ω
3. 0.1Ω
4. 0.0010Ω
Option 3 : 0.1Ω
## Combination of Resistors — Series and Parallel MCQ Question 19 Detailed Solution
CONCEPT:
Resistance:
• The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.
There are mainly two ways of the combination of resistances:
1. Resistances in series:
• When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
• The net resistance/equivalent resistance (R) of resistances in series is given by:
• Equivalent resistance, R = R1 + R2
2. Resistances in parallel:
• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:
• $$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
EXPLANATION:
Given - Resistance of n resistor = R and Resultant resistant = x
• In parallel combination,
$$⇒ \frac{1}{R_{para}}=\frac{1}{R_1} + \frac{1}{R_2}+------ + \frac{1}{R_{10}}$$
$$⇒ \frac{1}{R_{para}} = \frac{10}{1}$$
$$⇒ R_{para}=\frac{1}{10}=0.1\, \Omega$$
• In series combination,
⇒ Rseries = R + R + R ……n times
⇒ Rseries = n(R) = n(nx) = n2x
# What is the resultant resistance if two resistors of each 2 Ω resistance are connected in parallel?
1. 0.5 Ω
2. 1 Ω
3. 3 Ω
4. 2 Ω
Option 2 : 1 Ω
## Combination of Resistors — Series and Parallel MCQ Question 20 Detailed Solution
CONCEPT:
• Resistances in parallel combination: When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• A parallel circuit is a circuit in which the resistors are arranged with their heads connected together, and their tails connected together.
The net resistance/equivalent resistance(R) of resistances in parallel is given by:
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
CALCULATION:
Given that: R1 = R2 = 2 Ω
$$\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$
$$\frac{1}{R} = \frac{1}{{{2}}} + \frac{1}{{{2}}} = 2/2 = 1$$
So Resultant resistance (R) = 1 Ω
• The resultant resistance if two 2 Ω resistors are connected in parallel is 1 Ω. So option 2 is correct.
EXTRA POINTS:
• Resistances in series combination: When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.
The net resistance/equivalent resistance (R) of resistances in series is given by:
Equivalent resistance, R = R1 + R2
|
# Digital Roots of Sequences: Circular and Graphic Designs
The digital root of a number is the number obtained by adding all the digits, then adding the digits of that number, and then continuing until a single-digit number is reached.
Digital roots were known to the Roman bishop Hippolytos as early as the third century. It was employed by Twelfth-century Hindu mathematicians as a method of checking answers to multiplication, division, addition and subtraction.
For example, the digital root of 65,536 is 7, because 6 + 5 + 5 + 3 + 6 = 25 and 2 + 5 = 7.
The children will remember that two weeks ago we created fractals from Pascal’s triangles by coloring in the multiples of 3 and used the digital root test of adding the digits and if the sum was a multiple of 3, then the original number was also a multiple of 3.
This addition of the digits in a number is further simplified by first discarding or casting away any digits whose sum is 9. The remainder is set down, in each case as the digital root.
For example, the digital root of 972,632 is 2 because we cast out the nines or numbers who's sum is nine like the 7 + 2 and the 6 + 3. The remainder of 2 is the digital root.
When casting out nines, if there is a remainder of zero, the digital root is 9. The digital root of 927 is 9 because we cast out the 9 and the 7 + 2 leaving a remainder of zero.
FIND THE DIGITAL ROOT OF:
Number Calculation Digital Root
3 3 3
24 2 + 4 = 6 6
93 9 + 3 = 12 1 + 2 = 3 3
126 1 + 2 + 6 = 9 9
146 1 + 4 + 6 = 11 1 + 1 = 2 2
389,257 3 + 8 + 9 + 2 + 5 + 7 = 34 3 + 4= 7 7
96,871,565,493,528,698 9 + 6 + 8 + 7 + 1 + 5 + 6 + 5 + 4 + 9 + 3 + 5 + 2 + 8 + 6 + 9 + 8 = 101 1 + 0 + 1 = 2 2
Shortcut called "Casting Away Nines."
If we take the number 4,569,512,597,853, losing both 9s gives you 45,651,257,853. Then you can do the same with numbers that add to 9. So in the number we have now, we can also lose 4&5, 6&3, 1&8, 2&7 which leaves 555. Now we can find the digital root of 555 quite easily: 5+5+5=15, then 1+5=6. That's nice - no big additions to do to get the digital root of 4 569 512 597 853 to be 6!
This week I had the children list the digital roots of the multiples of the numbers from 1-12. They found amazing patterns of numbers from:
Multiples of 1: 1,2,3,4,5,6,7,8,9, repeat 1,2,3,4,5,6,7,8,9
Multiples of 2: 2,4,6,8,1,3,5,7,9, repeat 2,4,6,8,1,3,5,7,9
Multiples of 3: 3,6,9, repeat 3,6,9,
Multiples of 4: 4,8,3,7,2,6,1,5,9 repeat 4,8,3,7,2,6,1,5,9
Multiples of 5: 5,1,6,2,7,3,8,4,9 repeat 5,1,6,2,7,3,8,4,9
Multiples of 6: 6,3,9, repeat 6,3,9
Multiples of 7: 7,5,3,1,8,6,4,2,9 repeat 7,5,3,1,8,6,4,2,9
Multiples of 8: 8,7,6,5,4,3,2,1,9 repeat 8,7,6,5,4,3,2,1,9
Multiples of 9: 9, repeat 9
Multiples of 10: 1,2,3,4,5,6,7,8,9, repeat 1,2,3,4,5,6,7,8,9
Multiples of 11: 2,4,6,8,1,3,5,7,9, repeat 2,4,6,8,1,3,5,7,9
Multiples of 12: 3,6,9, repeat 3,6,9,
Square Numbers: 2,4,8,7,5,1 repeat 2,4,8,7,5,1
Powers of 3: 3,9, repeat 9
Powers of 4: 4,7,1, repeat 4,7,1
Powers of 5: 1,4,9,7,7,9,4,1,9, repeat 1,4,9,7,7,9,4,1,9,
Then I had the children graph the sequences on a circle graph with 9 points to create amazing designs. They noticed that the Multiples of 1, 8, and 10 were identical, so were Multiples of 2,7, and 11, so were Multiples of 3,6, and 9 and even Powers of 4, so were Multiples of 4 and 5, and Multiples of 9 and Powers of 3 were very close. Square Numbers had a pattern of a repeat of 6 digits and Powers of 5 was also a repeat of 9 digits but a little strange.
Talk about strange behavior, we then graphed the digital roots on a square grid but going in a direction of right and then taking a 90 degree turn to the right in a continuous clockwise fashion. These graphs made the most mundane sequence come alive.
AttachmentSize
Digital_Roots_of_Sequences_and_Designs.pdf3.11 MB
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Stat Trek
Teach yourself statistics
Teach yourself statistics
# Hypothesis Test for Regression Slope
This lesson describes how to conduct a hypothesis test to determine whether there is a significant linear relationship between an independent variable X and a dependent variable Y.
The test focuses on the slope of the regression line
Y = Β0 + Β1X
where Β0 is a constant, Β1 is the slope (also called the regression coefficient), X is the value of the independent variable, and Y is the value of the dependent variable.
If we find that the slope of the regression line is significantly different from zero, we will conclude that there is a significant relationship between the independent and dependent variables.
## Test Requirements
The approach described in this lesson is valid whenever the standard requirements for simple linear regression are met.
• The dependent variable Y has a linear relationship to the independent variable X.
• For each value of X, the probability distribution of Y has the same standard deviation σ.
• For any given value of X,
• The Y values are independent.
• The Y values are roughly normally distributed (i.e., symmetric and unimodal). A little skewness is ok if the sample size is large.
The test procedure consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.
## State the Hypotheses
If there is a significant linear relationship between the independent variable X and the dependent variable Y, the slope will not equal zero.
Ho: Β1 = 0
Ha: Β1 ≠ 0
The null hypothesis states that the slope is equal to zero, and the alternative hypothesis states that the slope is not equal to zero.
## Formulate an Analysis Plan
The analysis plan describes how to use sample data to accept or reject the null hypothesis. The plan should specify the following elements.
• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use a linear regression t-test (described in the next section) to determine whether the slope of the regression line differs significantly from zero.
## Analyze Sample Data
Using sample data, find the standard error of the slope, the slope of the regression line, the degrees of freedom, the test statistic, and the P-value associated with the test statistic. The approach described in this section is illustrated in the sample problem at the end of this lesson.
• Standard error. Many statistical software packages and some graphing calculators provide the standard error of the slope as a regression analysis output. The table below shows hypothetical output for the following regression equation: y = 76 + 35x . Predictor Coef SE Coef T P Constant 76 30 2.53 0.01 X 35 20 1.75 0.04
In the output above, the standard error of the slope (shaded in gray) is equal to 20. In this example, the standard error is referred to as "SE Coeff". However, other software packages might use a different label for the standard error. It might be "StDev", "SE", "Std Dev", or something else. If you need to calculate the standard error of the slope (SE) by hand, use the following formula:
SE = sb1 = sqrt [ Σ(yi - ŷi)2 / (n - 2) ] / sqrt [ Σ(xi - x)2 ]
where yi is the value of the dependent variable for observation i, ŷi is estimated value of the dependent variable for observation i, xi is the observed value of the independent variable for observation i, x is the mean of the independent variable, and n is the number of observations.
• Slope. Like the standard error, the slope of the regression line will be provided by most statistics software packages. In the hypothetical output above, the slope is equal to 35.
• Degrees of freedom. For simple linear regression (one independent and one dependent variable), the degrees of freedom (DF) is equal to:
DF = n - 2
where n is the number of observations in the sample.
• Test statistic. The test statistic is a t statistic (t) defined by the following equation.
t = b1 / SE
where b1 is the slope of the sample regression line, and SE is the standard error of the slope.
• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the test statistic. Use the degrees of freedom computed above.
## Interpret Results
If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.
Problem
The local utility company surveys 101 randomly selected customers. For each survey participant, the company collects the following: annual electric bill (in dollars) and home size (in square feet). Output from a regression analysis appears below.
Regression equation: Annual bill = 0.55 * Home size + 15 Predictor Coef SE Coef T P Constant 15 3 5.0 0.00 Home size 0.55 0.24 2.29 0.01
Is there a significant linear relationship between annual bill and home size? Use a 0.05 level of significance.
Solution
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Ho: The slope of the regression line is equal to zero.
Ha: The slope of the regression line is not equal to zero.
If the relationship between home size and electric bill is significant, the slope will not equal zero.
• Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a linear regression t-test to determine whether the slope of the regression line differs significantly from zero.
• Analyze sample data. To apply the linear regression t-test to sample data, we require the standard error of the slope, the slope of the regression line, the degrees of freedom, the t statistic, and the P-value of the test statistic.
We get the slope (b1) and the standard error (SE) from the regression output.
b1 = 0.55 SE = 0.24
We compute the degrees of freedom and the t statistic, using the following equations.
DF = n - 2 = 101 - 2 = 99
t = b1/SE = 0.55/0.24 = 2.29
where DF is the degrees of freedom, n is the number of observations in the sample, b1 is the slope of the regression line, and SE is the standard error of the slope.
Based on the t statistic and the degrees of freedom, we determine the P-value. The P-value is the probability that a t statistic having 99 degrees of freedom is more extreme than 2.29. Since this is a two-tailed test, "more extreme" means greater than 2.29 or less than -2.29. We use the t Distribution Calculator to find P(t > 2.29) is about 0.0121 and P(t < -2.29) is about 0.0121. Therefore, the P-value is 0.0121 + 0.0121 or 0.0242.
• Interpret results. Since the P-value (0.0242) is less than the significance level (0.05), we cannot accept the null hypothesis.
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# NCERT Solutions For Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.3) Exercise 2.3
An equation is a mathematical statement with an equal sign between two expressions with equal values. For example, 3x + 5 = 15. There are different types of equations, such as first-order, second-order, and third-order. Students could learn more about Mathematics Equations. An equation is a mathematical formula that contains two algebraic expressions on either side of an equal sign (=). It shows the equivalence relation between the formula written on the left and the formula written on the right. In any formula, L.H.S = R.H.S (left side = right side). Students can solve equations to find the values of unknown variables that represent unknown quantities. If a statement does not have an equal sign, it is not an equation. The difference between expressions and formulas is explained in a later section. For a thorough understanding of linear equations in one variable, students should consult the NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3. These solutions could help them understand the meaning of equations in Mathematics.
Equations have various parts, including coefficients, variables, operators, constants, terms, expressions, and equal signs. When writing an equation, students need the “=” symbol and the term on both sides. Both sides must be equal to each other. The equation does not have to have multiple terms on either side, using variables and operators. Students can also form equations without these. In contrast, equations involving variables are referred to as Algebraic Equations.
Solving a mathematical equation
The equation is like a pair of scales with equal weights on each side. This applies even if students add or subtract the same number on both sides of the equation. The rule applies even if both sides of the equation are multiplied or divided by the same number. To solve basic linear equations using one variable:
Step 1: Students should put all terms, including variables, on one side of the equation and all constants on the other side by doing arithmetic on both sides.
Step 2: Further, students should add, subtract, and combine all equal terms (terms containing the same variable with the same exponent).
Step 3: Students should finally simplify to get the answer. It would be enriching to look at yet another example of the basic equation: 3x – 20 = 7. To bring all the constants to the right side, students would need to add 20 to both sides. This means 3x – 20 + 20 = 7 + 20, which simplifies to 3x = 27. Now, students should divide both sides by 3. This gives x = 9, which is the desired solution of the equation.
Based on the order, Equations can be classified into three types. Below are three types of equations in mathematics.
• Linear Equation
• Cubic Equation
An equation of degree 1 is called a Linear Equation in mathematics. In such an equation, 1 is the highest exponent of the term. These can be further categorized as Linear Equations in One Variable, Linear Equations in Two Variables, Linear Equations in Three Variables and so on. The standard form of a Linear Equation with Two Variables x and y is ax + by – c = 0. b is the coefficient of x or y, and c is a constant.
## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.3) Exercise 2.3
Prior examination preparation is bound to be advantageous for students. Problems need to be solved in a simplified way so students can easily understand the solutions.
• A note about linear equations
A Linear Equation is defined as an algebraic equation containing Constants and Variables. It is an equation where the left expression has the same value as the right expression. A Linear Equation in One Variable is easily understood as an equation with only one variable. Example: px+q = 0, p and q are two integers, and x is a variable.
The equations studied in junior classes are Linear Equations in One Variable because the terms in such equations have only one variable. These equations are linear because the variables in the equations have a maximum power of 1. Any linear equation can have any rational number as its solution. This concept has been explained descriptively in NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 available on the website of Extramarks.
Students can download PDF versions of NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, which are compiled by experienced teachers according to the NCERT guidelines. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, consist of problems with comprehensive solutions that can help students revise their entire syllabus and achieve excellence in exams. Students can register on Extramarks to obtain NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, conveniently.
Extramarks is a platform that provides students with NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 and other learning materials. Students can register on the Extramarks learning portal and access the most reliable study materials created by mentors of Extramarks. Having access to NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, and solutions to other subjects makes learning academic disciplines like Science, Mathematics, and English easy.
### Access NCERT Solutions for Maths Chapter 2 – Linear Equations in One Variable
Choosing NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, is considered to be the best option for CBSE students when it comes to exam preparation. With the aid and assistance of the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, students can master the theme of Linear Equations in One Variable. Students can download these NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 at their convenience or study directly online via the Extramarks website or mobile application.
Extramarks’ subject matter experts have engineered the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, meticulously and in compliance with all CBSE guidelines. A Class 8 student who has a thorough understanding of all concepts implicit in the prescribed Mathematics textbook and a good grasp of all the problems from the exercises it contains will easily achieve the best possible score on the final exam. With the help of these NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, students can easily understand the various patterns of questions that may appear on exams pertaining to this chapter and also learn how to study this chapter.
Students can download the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, from the Extramarks website and get ready for their exam. If students have the Extramarks Learning App installed on their smartphone, they can also download NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, from the application. The best part about NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, is that they are accessible both online and offline.
### NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (Ex 2.3)
In addition to these NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, which pertain to Exercise 2.3, there are many exercises in this chapter with numerous problems. The solutions to all these exercises have been compiled by highly knowledgeable mentors at Extramarks. Therefore, they should all be of the highest quality and can be referenced by students while preparing for the exam. Understanding all the concepts in the textbook and solving NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 and the adjacent exercise problems is very important to getting the best score in the class.
### NCERT Solutions Class 8 Maths Chapter 2 All Exercises
The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 based on Linear Equations in One Variable, are available in PDF format for download. The NCERT solutions for the Linear Equations in One Variable chapter have been precisely designed by subject matter specialists of Extramarks. It is highly recommended that students ought to refer to the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 to prepare adequately for the exam. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 can also serve as a reference for students’ homework and assignments. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, contain systematic answers for all questions in Exercise 2.3, making them a great learning resource for Class 8 students. By referring to these NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, students will be able to perform well on their final exams and master the concepts. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, are developed based on the latest NCERT syllabus and cover all the key topics embedded in Exercise 2.3.
### 12 Questions with Solutions
The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 comprise a continuation of the explanation of concepts of algebraic expressions and equations that students learned in earlier classes. An equation with one variable will be presented to the student in this chapter. Some of the main topics or concepts covered in NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 are:
Algebraic equations are equations with variables. The word “equation” indicates that the value of the expression on one side of the equal sign is equal to the value of the expression on the other side. An equation may have, on both sides, linear expressions. Like numbers, variables can be transposed from one side of the equation to the other. Sometimes it is necessary to simplify the formulas that make up an equation before solving it in a general way. Some equations may not even be linear but can be made linear by multiplying both sides of the equation by a suitable formula. The usefulness of linear equations lies in their multiple applications. A variety of problems related to numbers, ages, sizes, combinations of bills, etc. are solved with linear equations.
### 10 Questions with Solutions
The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, focuses on Linear Equations in One Variable are based on solving equations with variables on both sides. The equation states that the values of the two expressions are equal. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, shows how to solve equations that have expressions with variables on both sides.
### NCERT Solutions for Class 8
The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 provides answers to all questions in Exercise 2.3 of Chapter 2 of the NCERT Class 8 Mathematics textbook. Many questions in the exam are derived from this exercise, and by solving NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, students can get a high score in the annual exam.
Q.1
$\begin{array}{l}\text{Solve the following equations and check your results.}\\ 1.\text{ }3\text{x}=2\text{x}+18\\ 2.\text{ }5\text{t}-3=3\text{t}-5\\ 3.\text{ }5\text{x}+9=5+3\text{x}\\ 4.\text{ }4\text{z}+3=6+2\text{z}\\ 5.\text{ }2\text{x}-1=14-\text{x}\\ 6.\text{ }8\text{x}+4=3\left(\text{x}-1\right)+7\\ 7.\text{ x}=\frac{4}{5}\left(\text{x}+10\right)\\ 8.\text{ }\frac{2\text{x}}{3}+1=\frac{7\text{x}}{15}+3\\ 9.\text{ }2\text{y}+\frac{5}{3}=\frac{26}{3}-\text{y}\\ 10.\text{ }3\text{m}=5\text{m}-\frac{8}{5}\end{array}$
Ans
$\begin{array}{l}1.3\mathrm{x}=2\mathrm{x}+18\\ \text{On transposing 2x to L.H.S,}\mathrm{we}\mathrm{get}\\ ⇒3\mathrm{x}-2\mathrm{x}=18\\ ⇒\mathrm{x}=18\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=3\mathrm{x}\\ =3×18\\ =54\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=2\mathrm{x}+18\\ =2×18+18\\ =36+18=54\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$
$\begin{array}{l}2.5\mathrm{t}-3=3\mathrm{t}-5\\ \text{On transposing 3t to L.H.S and}-\text{3 to R.H.S,}\text{we get}\\ ⇒5\mathrm{t}-3\mathrm{t}=3-5\\ ⇒2\mathrm{t}=-2\\ ⇒\mathrm{t}=-1\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=5\mathrm{t}-3\\ =5×\left(-1\right)-3\\ =-5-3\\ =-8\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=3\mathrm{t}-5\\ =3×\left(-1\right)-5\\ =-3-5\\ =-8\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}3.5\mathrm{x}+9=5+3\mathrm{x}\\ \text{On transposing 3x to L.H.S and}9\text{to R.H.S,}\text{we get}\\ ⇒5\mathrm{x}-3\mathrm{x}=5-9\\ ⇒2\mathrm{x}=-4\\ \text{On dividing both sides by 2,we get}\\ ⇒\frac{2\mathrm{x}}{2}=-\frac{4}{2}\\ ⇒\mathrm{x}=-2\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=5\mathrm{x}+9\\ =5×\left(-2\right)+9\\ =\left(-10\right)+9\\ =-1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=5+3\mathrm{x}\\ =5+3×\left(-2\right)\\ =5+\left(-6\right)\\ =-1\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}4.4\mathrm{z}+3=6+2\mathrm{z}\\ \text{On transposing 2z to L.H.S and}3\text{to R.H.S,}\text{we get}\\ ⇒4\mathrm{z}-2\mathrm{z}=6-3\\ ⇒2\mathrm{z}=3\\ \text{On dividing both sides by 2,}\text{we get}\\ ⇒\frac{2\mathrm{z}}{2}=\frac{3}{2}\\ ⇒\mathrm{z}=\frac{3}{2}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=4\mathrm{z}+3\\ =4×\left(\frac{3}{2}\right)+3\\ =6+3\\ =9\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=6+2\mathrm{z}\\ =6+2×\frac{3}{2}\\ =6+3\\ =9\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}5.\mathrm{}2\mathrm{x}-1=14-\mathrm{x}\\ \text{On transposing x to L.H.S and}1\text{to R.H}.\mathrm{S},\mathrm{we}\mathrm{get}\\ ⇒2\mathrm{x}+\mathrm{x}=14+1\\ ⇒3\mathrm{x}=15\\ \text{On dividing both sides by 3,we get}\\ ⇒\frac{3\mathrm{x}}{3}=\frac{15}{3}\\ ⇒\mathrm{x}=5\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=2\mathrm{x}-1\\ =2×5-1\\ =10-1\\ =9\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=14-\mathrm{x}\\ =14-5\\ =9\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}6.8\mathrm{x}+4=3\left(\mathrm{x}-1\right)+7\\ ⇒8\mathrm{x}+4=3\mathrm{x}-3+7\\ ⇒8\mathrm{x}+4=3\mathrm{x}+4\\ \text{On transposing 3x to L.H.S and}4\text{to R.H.S,we get}\\ ⇒8\mathrm{x}-3\mathrm{x}=4-4\\ ⇒5\mathrm{x}=0\\ ⇒\mathrm{x}=0\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=8\mathrm{x}+4\\ =8×0+4\\ =4\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=3\left(\mathrm{x}-1\right)+7\\ =3\left(0-1\right)+7\\ =-3+7\\ =4\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}7.\mathrm{}\mathrm{x}=\frac{4}{5}\left(\mathrm{x}+10\right)\\ \mathrm{Multiplying}\text{}\mathrm{both}\text{}\mathrm{sides}\text{}\mathrm{by}\text{}5,\mathrm{we}\text{}\mathrm{get}\\ ⇒5\mathrm{x}=4\left(\mathrm{x}+10\right)\\ ⇒5\mathrm{x}=4\mathrm{x}+40\\ \mathrm{Transposing}\text{}4\mathrm{x}\text{}\mathrm{to}\text{}\mathrm{L}.\mathrm{H}.\mathrm{S},\mathrm{we}\text{}\mathrm{get}\\ ⇒5\mathrm{x}-4\mathrm{x}=40\\ ⇒\mathrm{x}=40\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{x}\\ =40\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{4}{5}\left(\mathrm{x}+10\right)\\ =\frac{4}{5}\left(40+10\right)\\ =\frac{4}{5}\left(50\right)\\ =\frac{4}{5}×\left(50\right)\\ =4×10\right]\\ =40\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}8.\frac{2\mathrm{x}}{3}+1=\frac{7\mathrm{x}}{15}+3\\ \text{On transposing}\frac{7\mathrm{x}}{15}\text{to L.H.S and}1\text{to R.H.S,we get}\\ ⇒\frac{2\mathrm{x}}{3}-\frac{7\mathrm{x}}{15}=3-1\\ \text{Taking LCM of 3 and 15,we get}\\ ⇒\frac{10\mathrm{x}-7\mathrm{x}}{15}=2\\ ⇒\frac{3\mathrm{x}}{15}=2\\ ⇒3\mathrm{x}=30\\ ⇒\mathrm{x}=\frac{30}{3}=10\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{2\mathrm{x}}{3}+1\\ =\frac{2×10}{3}+1\\ =\frac{20}{3}+1\\ =\frac{20+3}{3}=\frac{23}{3}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{7\mathrm{x}}{15}+3\\ =\frac{7\mathrm{x}+45}{15}\\ =\frac{7×10+45}{15}\\ =\frac{115}{15}\\ =\frac{23}{3}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}9.2\mathrm{y}+\frac{5}{3}=\frac{26}{3}-\mathrm{y}\\ \text{On transposing}\mathrm{y}\text{to L.H.S and}\frac{5}{3}\text{to R.H.S,we get}\\ ⇒2\mathrm{y}+\mathrm{y}=\frac{26}{3}-\frac{5}{3}\\ ⇒3\mathrm{y}=\frac{21}{3}\\ ⇒\mathrm{y}=\frac{21}{9}\\ ⇒\mathrm{y}=\frac{7}{3}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}=2\mathrm{y}+\frac{5}{3}\\ =2×\frac{7}{3}+\frac{5}{3}\\ =\frac{14}{3}+\frac{5}{3}\\ =\frac{14+5}{3}=\frac{19}{3}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{26}{3}-\mathrm{y}\\ =\frac{26}{3}-\frac{7}{3}\\ =\frac{26-7}{3}\\ =\frac{19}{3}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}10.3\mathrm{m}=5\mathrm{m}-\frac{8}{5}\\ \text{On transposing}5\mathrm{m}\text{to L.H.S ,we get}\\ ⇒3\mathrm{m}-5\mathrm{m}=-\frac{8}{5}\\ ⇒-2\mathrm{m}=-\frac{8}{5}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}⇒\mathrm{m}=-\frac{8}{5×\left(-2\right)}\\ ⇒\mathrm{m}=\frac{4}{5}\end{array}$ $\begin{array}{l}\end{array}$
$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}=3\mathrm{m}\\ =3×\frac{4}{5}\\ =\frac{12}{5}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=5\mathrm{m}-\frac{8}{5}\\ =5×\frac{4}{5}-\frac{8}{5}\\ =4-\frac{8}{5}\\ =\frac{20-8}{5}\\ =\frac{12}{5}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\\ \end{array}$
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# If f(x) =tan^2x and g(x) = sqrt(5x-1 , what is f'(g(x)) ?
Aug 2, 2017
$f ' \left(g \left(x\right)\right) = \frac{5 {\sec}^{2} \left(\sqrt{5 x - 1}\right) \tan \left(\sqrt{5 x - 1}\right)}{\sqrt{5 x - 1}}$
#### Explanation:
To find $f ' \left(g \left(x\right)\right)$, we can start by finding $f \left(g \left(x\right)\right)$ and then take the derivative. We substitute $\sqrt{5 x - 1}$ into $f \left(x\right)$, so we have:
$f \left(g \left(x\right)\right) = f \left(\sqrt{5 x - 1}\right) = {\tan}^{2} \left(\sqrt{5 x - 1}\right)$
Now we take the derivative using the chain rule.
$2 \tan \left(\sqrt{5 x - 1}\right) \cdot {\sec}^{2} \left(\sqrt{5 x - 1}\right) \cdot \frac{1}{2} {\left(5 x - 1\right)}^{- \frac{1}{2}} \cdot 5$
$\implies \frac{5 {\sec}^{2} \left(\sqrt{5 x - 1}\right) \tan \left(\sqrt{5 x - 1}\right)}{\sqrt{5 x - 1}}$
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# Trigonometric functions
We have informally used some of the trigonometric functions in examples so far. In this section we quickly review their definitions and some basic properties.
The trigonometric functions are used to describe relationships between triangles and circles as well as oscillatory motions. With such a wide range of utility it is no wonder that they pop up in many places and their origins date to Hipparcus and Ptolemy over 2000 years ago.
## The 6 basic trigonometric functions
We measure angles in radians, where $360$ degrees is $2\pi$ radians. By proportions, $180$ degrees is $\pi$ radian, $90$ degrees is $\pi/2$ radians, $60$ degrees is $\pi/3$ radians, etc. In general, $x$ degrees is $2\pi \cdot x / 360$ radians.
For a right triangle with angles $\theta$, $\pi/2 - \theta$, and $\pi/2$ we call the side opposite $\theta$ the "opposite" side, the shorter adjacent side the "adjacent" side and the longer adjacent side the hypotenuse.
With these, the basic definitions for the primary trigonometric functions are
~ \begin{align} \sin(\theta) &= \frac{\text{opposite}}{\text{hypotenuse}} &\quad(\text{the sine function})\\ \cos(\theta) &= \frac{\text{adjacent}}{\text{hypotenuse}} &\quad(\text{the cosine function})\\ \tan(\theta) &= \frac{\text{opposite}}{\text{adjacent}}. &\quad(\text{the tangent function}) \end{align} ~
Some algebra shows that $\tan(\theta) = \sin(\theta)/\cos(\theta)$. There are also 3 reciprocal functions, the cosecant, secant and cotangent.
These definitions in terms of sides only apply for $0 \leq \theta \leq \pi/2$. More generally, if we relate any angle taken in the counter clockwise direction for the $x$-axis with a point $(x,y)$ on the unit circle, then we can extend these definitions - the point $(x,y)$ is also $(\cos(\theta), \sin(\theta))$.
### The trigonometric functions in Julia
Julia has the 6 basic trigonometric functions defined through the functions sin, cos, tan, csc, sec, and cot.
Two right triangles - the one with equal, $\pi/4$, angles; and the one with angles $\pi/6$ and $\pi/3$ can have the ratio of their sides computed from basic geometry. In particular, this leads to the following values, which are usually committed to memory:
~ \begin{align} \sin(0) &= 0, \quad \sin(\pi/6) = \frac{1}{2}, \quad \sin(\pi/4) = \frac{\sqrt{2}}{2}, \quad\sin(\pi/3) = \frac{\sqrt{3}}{2},\text{ and } \sin(\pi/2) = 1\\ \cos(0) &= 1, \quad \cos(\pi/6) = \frac{\sqrt{3}}{2}, \quad \cos(\pi/4) = \frac{\sqrt{2}}{2}, \quad\cos(\pi/3) = \frac{1}{2},\text{ and } \cos(\pi/2) = 0. \end{align} ~
Using the circle definition allows these basic values to inform us of values throughout the unit circle.
These all follow from the definition involving the unit circle:
• If the angle $\theta$ corresponds to a point $(x,y)$ on the unit circle, then the angle $-\theta$ corresponds to $(x, -y)$. So $\sin(\theta) = - \sin(-\theta)$ (an odd function), but $\cos(\theta) = \cos(-\theta)$ (an even function).
• If the angle $\theta$ corresponds to a point $(x,y)$ on the unit circle, then rotating by $\pi$ moves the points to $(-x, -y)$. So $\cos(\theta) = x = - \cos(\theta + \pi)$, and $\sin(\theta) = y = -\sin(\theta + \pi)$.
• If the angle $\theta$ corresponds to a point $(x,y)$ on the unit circle, then rotating by $\pi/2$ moves the points to $(-y, x)$. So $\cos(\theta) = x = \sin(\theta + \pi/2)$.
The fact that $x^2 + y^2 = 1$ for the unit circle leads to the "Pythagorean identity" for trigonometric functions:
$~ \sin(\theta)^2 + \cos(\theta)^2 = 1. ~$
This basic fact can be manipulated many ways. For example, dividing through by $\cos(\theta)^2$ gives the related identity: $\tan(\theta)^2 + 1 = \sec(\theta)^2$.
Julia's functions can compute values for any angles, including these fundamental ones:
[cos(theta) for theta in [0, pi/6, pi/4, pi/3, pi/2]]
5-element Array{Float64,1}:
1.0
0.8660254037844387
0.7071067811865476
0.5000000000000001
6.123233995736766e-17
These are floating point approximations, as can be seen clearly in the last value. Symbolic math can be used if exactness matters:
using CalculusWithJulia # to load the SymPy package
using Plots
cos.([0, PI/6, PI/4, PI/3, PI/2])
$\left[ \begin{array}{r}1\\\frac{\sqrt{3}}{2}\\\frac{\sqrt{2}}{2}\\\frac{1}{2}\\0\end{array} \right]$
##### Example
Measuring the height of a tree may be a real-world task for some, but a typical task for trigonometry students. How might it be done? If a right triangle can be formed where the angle and adjacent side length are known, then the opposite side (the height of the tree) can be solved for with the tangent function. For example, if standing $100$ feet from the base of the tree the tip makes a 15 degree angle the height is given by:
theta = 15 * 180/pi
-466.6470644343919
Having some means to compute an angle and then a tangent of that angle handy is not a given, so the linked to article provides a few other methods taking advantage of similar triangles.
You can also measure distance with your thumb or fist. How, the fist takes up about $10$ degree of view when held straight out. So, pacing off backwards until the fist completely occludes the tree will give the distance of the adjacent side of a right triangle. If that distance is $30$ paces what is the height of the tree? Well, we need some facts. Suppose your pace is $3$ feet. Then the adjacent length is $90$ feet. The multiplier is the tangent of $10$ degrees, or:
tan(10 * pi/180)
0.17632698070846498
Which for sake of memory we will say is $1/6$ (a $5$ percent error). So that answer is roughly $15$ feet:
30 * 3 / 6
15.0
Similarly, you can use your thumb instead of your first. To use your first you can multiply by $1/6$ the adjacent side, to use your thumb about $1/30$ as this approximates the tangent of $2$ degrees:
1/30, tan(2*pi/180)
(0.03333333333333333, 0.03492076949174773)
This could be reversed. If you know the height of something a distance away that is covered by your thumb or fist, then you would multiply that height by the appropriate amount to find your distance.
### Basic properties
The sine function is defined for all real $\theta$ and has a range of $[-1,1]$. Clearly as $\theta$ winds around the $x$-axis, the position of the $y$ coordinate begins to repeat itself. We say the sine function is periodic with period $2\pi$. A graph will illustrate:
plot(sin, 0, 4pi)
The graph shows two periods. The wavy aspect of the graph is why this function is used to model periodic motions, such as the amount of sunlight in a day, or the alternating current powering a computer.
From this graph - or considering when the $y$ coordinate is $0$ - we see that the sine function has zeros at any integer multiple of $\pi$, or $k\pi$, $k$ in $\dots,-2,-1, 0, 1, 2, \dots$.
The cosine function is similar, in that it has the same domain and range, but is "out of phase" with the sine curve. A graph of both shows the two are related:
plot(sin, 0, 4pi)
plot!(cos, 0, 4pi)
The cosine function is just a shift of the sine function (or vice versa). We see that the zeros of the cosine function happen at points of the form $\pi/2 + k\pi$, $k$ in $\dots,-2,-1, 0, 1, 2, \dots$.
The tangent function does not have all $\theta$ for its domain, rather those points where division by $0$ occurs are excluded. These occur when the cosine is $0$, or again at $\pi/2 + k\pi$, $k$ in $\dots,-2,-1, 0, 1, 2, \dots$. The range of the tangent function will be all real $y$.
The tangent function is also periodic, but not with period $2\pi$, but rather just $\pi$. A graph will show this. Here we avoid the vertical asymptotes by keeping them out of the plot domain and layering several plots.
### Functions using degrees
Trigonometric function are functions of angles which have two common descriptions: in terms of degrees or radians. Degrees are common when right triangles are considered, radians much more common in general, as the relationship with arc-length holds in that $r\theta = l$, where $r$ is the radius of a circle and $l$ the length of the arc formed by angle $\theta$.
The two are related, as a circle of $2\pi$ radians and 360 degrees. So to convert from degrees into radians it takes multiplying by $2\pi/360$ and to convert from radians to degrees it takes multiplying by $360/(2\pi)$. The deg2rad and rad2deg functions are available for this task.
In Julia, the functions sind, cosd, tand, cscd, secd, and cotd are available to simplify the task of composing the two operations (that is sin(deg2rad(x)) is the same as sind(x)).
## The sum-and-difference formulas
Consider the point on the unit circle $(x,y) = (\cos(\theta), \sin(\theta))$. In terms of $(x,y)$ (or $\theta$) is there a way to represent the angle found by rotating an additional $\theta$, that is what is $(\cos(2\theta), \sin(2\theta))$?
More generally, suppose we have two angles $\alpha$ and $\beta$, can we represent the values of $(\cos(\alpha + \beta), \sin(\alpha + \beta))$ using the values just involving $\beta$ and $\alpha$ separately?
According to Wikipedia the following figure (from mathalino.com) has ideas that date to Ptolemy:
To read this, there are three triangles: the bigger (green with pink part) has hypotenuse $1$ (and adjacent and opposite sides that form the hypotenuses of the other two); the next biggest (yellow) hypotenuse $\cos(\beta)$, adjacent side (of angle $\alpha$) $\cos(\beta)\cdot \cos(\alpha)$, and opposite side $\cos(\beta)\cdot\sin(\alpha)$; and the smallest (pink) hypotenuse $\sin(\beta)$, adjacent side (of angle $\alpha$) $\sin(\beta)\cdot \cos(\alpha)$, and opposite side $\sin(\beta)\sin(\alpha)$.
This figure shows the following sum formula for sine and cosine:
~ \begin{align} \sin(\alpha + \beta) &= \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)\\ \cos(\alpha + \beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta). \end{align} ~
Using the fact that $\sin$ is an odd function and $\cos$ an even function, related formulas for the difference $\alpha - \beta$ can be derived.
Taking $\alpha = \beta$ we immediately get the "double-angle" formulas:
~ \begin{align} \sin(2\alpha) &= 2\sin(\alpha)\cos(\alpha)\\ \cos(2\alpha) &= \cos(\alpha)^2 - \sin(\alpha)^2. \end{align} ~
The latter looks like the Pythagorean identify, but has a minus sign. In fact, the Pythagorean identify is often used to rewrite this, for example $\cos(2\alpha) = 2\cos(\alpha)^2 - 1$ or $1 - 2\sin(\alpha)^2$.
Applying the above with $\alpha = \beta/2$, we get that $\cos(\beta) = 2\cos(\beta/2)^2 -1$, which rearranged yields the "half-angle" formula: $\cos(\beta/2)^2 = (1 + \cos(\beta))/2$.
##### Example
Consider the expressions $\cos((n+1)\theta)$ and $\cos((n-1)\theta)$. These can be re-expressed as:
~ \begin{align} \cos((n+1)\theta) &= \cos(n\theta + \theta) = \cos(n\theta) \cos(\theta) - \sin(n\theta)\sin(\theta), \text{ and}\\ \cos((n-1)\theta) &= \cos(n\theta - \theta) = \cos(n\theta) \cos(-\theta) - \sin(n\theta)\sin(-\theta). \end{align} ~
But $\cos(-\theta) = \cos(\theta)$, whereas $\sin(-\theta) = -\sin(\theta)$. Using this, we add the two formulas above to get:
$~ \cos((n+1)\theta) = 2\cos(n\theta) \cos(\theta) - \cos((n-1)\theta). ~$
That is the angle for a multiple of $n+1$ can be expressed in terms of the angle with a multiple of $n$ and $n-1$. This can be used recursively to find expressions for $\cos(n\theta)$ in terms of polynomials in $\cos(\theta)$.
## Inverse trigonometric functions
The trigonometric functions are all periodic. In particular they are not monotonic over their entire domain. This means there is no inverse function applicable. However, by restricting the domain to where the functions are monotonic, inverse functions can be defined:
• For $\sin(x)$, the restricted domain of $[-\pi/2, \pi/2]$ allows for the arcsine function to be defined. In Julia this is implemented with asin.
• For $\cos(x)$, the restricted domain of $[0,\pi]$ allows for the arccosine function to be defined. In Julia this is implemented with acos.
• For $\tan(x)$, the restricted domain of $(-\pi/2, \pi/2)$ allows for the arctangent function to be defined. In Julia this is implemented with atan.
For example, the arcsine function is defined for $-1 \leq x \leq 1$ and has a range of $-\pi/2$ to $\pi/2$:
plot(asin, -1, 1)
The arctangent has domain of all real $x$. It has shape given by:
plot(atan, -10, 10)
The horizontal asymptotes are $y=\pi/2$ and $y=-\pi/2$.
### Implications of a restricted domain
Notice that $\sin(\arcsin(x)) = x$ for any $x$ in $[-1,1]$, but, of course, not for all $x$, as the output of the sine function can't be arbitrarily large.
However, $\arcsin(\sin(x))$ is defined for all $x$, but only equals $x$ when $x$ is in $[-\pi/2, \pi/2]$. The output, or range, of the $\arcsin$ function is restricted to that interval.
This can be limiting at times. A common case is to find the angle in $[0, 2\pi)$ corresponding to a point $(x,y)$. In the simplest case (the first and fourth quadrants) this is just given by $\arctan(y/x)$. But with some work, the correct angle can be found for any pair $(x,y)$. As this is a common desire, the atan function with two arguments, atan(y,x), is available. This function returns a value in $(-\pi, \pi]$.
For example, this will not give back $\theta$ without more work to identify the quadrant:
theta = 3pi/4 # 2.35619...
x,y = (cos(theta), sin(theta)) # -0.7071..., 0.7071...
atan(y/x)
-0.7853981633974484
But,
atan(y, x)
2.356194490192345
##### Example
A (white) light shining through a prism will be deflected depending on the material of the prism and the angles involved (cf. the link for a figure). The relationship can be analyzed by tracing a ray through the figure and utilizing Snell's law. If the prism has index of refraction $n$ then the ray will deflect by an amount $\delta$ that depends on the angle, $\alpha$ of the prism and the initial angle ($\theta_0$) according to:
$~ \delta = \theta_0 - \alpha + \arcsin(n \sin(\alpha - \arcsin(\frac{1}{n}\sin(\theta_0)))). ~$
If $n=1.5$ (glass), $\alpha = \pi/3$ and $\theta_0=\pi/6$, find the deflection (in radians).
We have:
n, alpha, theta0 = 1.5, pi/3, pi/6
delta = theta0 - alpha + asin(n * sin(alpha - asin(sin(theta0)/n)))
0.8219769749498015
For small $\theta_0$ and $\alpha$ the deviation is approximated by $(n-1)\alpha$. Compare this approximation to the actual value when $\theta_0 = \pi/10$ and $\alpha=\pi/15$.
We have:
n, alpha, theta0 = 1.5, pi/15, pi/10
delta = theta0 - alpha + asin(n * sin(alpha - asin(sin(theta0)/n)))
delta, (n-1)*alpha
(0.10763338241545499, 0.10471975511965977)
The approximation error is about 2.7 percent.
##### Example
The AMS has an interesting column on rainbows the start of which uses some formulas from the previous example. Click through to see a ray of light passing through a spherical drop of water, as analyzed by Descartes. The deflection of the ray occurs when the incident light hits the drop of water, then there is an internal deflection of the light, and finally when the light leaves, there is another deflection. The total deflection (in radians) is $D = (i-r) + (\pi - 2r) + (i-r) = \pi - 2i - 4r$. However, the incident angle $i$ and the refracted angle $r$ are related by Snell's law: $\sin(i) = n \sin(r)$. The value $n$ is the index of refraction and is $4/3$ for water. (It was $3/2$ for glass in the previous example.) This gives
$~ D = \pi + 2i - 4 \arcsin(\frac{1}{n} \sin(i)). ~$
Graphing this for incident angles between $0$ and $\pi/2$ we have:
n = 4/3
D(i) = pi + 2i - 4 * asin(sin(i)/n)
plot(D, 0, pi/2)
Descartes was interested in the minimum value of this graph, as it relates to where the light concentrates. This is roughly at $1$ radian or about $57$ degrees:
rad2deg(1.0)
57.29577951308232
(Using calculus it can be seen to be $\arccos(((n^2-1)/3)^{1/2})$.)
##### Example: The Chebyshev Polynomials
Consider again this equation derived with the sum-and-difference formula:
$~ \cos((n+1)\theta) = 2\cos(n\theta) \cos(\theta) - \cos((n-1)\theta). ~$
Let $T_n(x) = \cos(n \arccos(x))$. Calling $\theta = \arccos(x)$ for $-1 \leq x \leq x$ we get a relation between these functions:
$~ T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x). ~$
We can simplify a few: For example, when $n=0$ we see immediately that $T_0(x) = 1$, the constant function. Whereas with $n=1$ we get $T_1(x) = \cos(\arccos(x)) = x$. Things get more interesting as we get bigger $n$, for example using the equation above we get $T_2(x) = 2xT_1(x) - T_0(x) = 2x\cdot x - 1 = 2x^2 - 1$. Continuing, we'd get $T_3(x) = 2 x T_2(x) - T_1(x) = 2x(2x^2 - 1) - x = 4x^3 -3x$.
A few things become clear from the above two representations:
• Starting from $T_0(x) = 1$ and $T_1(x)=x$ and using the recursive defintion of $T_{n+1}$ we get a family of polynomials where $T_n(x)$ is a degree $n$ polynomial. These are defined for all $x$, not just $-1 \leq x \leq 1$.
• Using the initial definition, we see that the zeros of $T_n(x)$ all occur within $[-1,1]$ and happen when $n\arccos(x) = k\pi + \pi/2$, or $x=\cos((2k+1)/n \cdot \pi/2)$ for $k=0, 1, \dots, n-1$.
Other properties of this polynomial family are not at all obvious. One is that amongst all polynomials of degree $n$ with roots in $[-1,1]$, $T_n(x)$ will be the smallest in magnitude (after we divide by the leading coefficient to make all polynomials considered to be monic). We can check this for one case. Take $n=4$, then we have: $T_4(x) = 8x^4 - 8x^2 + 1$. Compare this with $q(x) = (x+3/5)(x+1/5)(x-1/5)(x-3/5)$ (evenly spaced zeros):
T4(x) = (8x^4 - 8x^2 + 1) / 8
q(x) = (x+3/5)*(x+1/5)*(x-1/5)*(x-3/5)
plot(abs ∘ T4, -1,1)
plot!(abs ∘ q, -1,1)
## Hyperbolic trigonometric functions
Related to the trigonometric functions are the hyperbolic trigonometric functions. Instead of associating a point $(x,y)$ on the unit circle with an angle $\theta$, we associate a point $(x,y)$ on the unit hyperbola (with $x^2 - y^2 = 1$. We define the hyperbolic sin ($\sinh$) and hyperbolic cosine ($\cosh$) through $(\cosh(\theta), \sinh(\theta)) = (x,y)$.
These can be expressed using the exponential function as:
~ \begin{align} \sinh(x) &= \frac{e^x - e^{-x}}{2}\\ \cosh(x) &= \frac{e^x + e^{-x}}{2}. \end{align} ~
The hyperbolic tangent is then the ratio of $\sinh$ and $\cosh$. As well, three inverse hyperbolic functions can be defined.
## Questions
###### Question
What is bigger $\sin(1.23456)$ or $\cos(6.54321)$?
###### Question
Let $x=\pi/4$. What is bigger $\cos(x)$ or $x$?
###### Question
The cosine function is a simple tranformation of the sine function. Which one?
###### Question
Graph the secant function. The vertical asymptotes are at?
###### Question
A formula due to Bhaskara I dates to around 650AD and gives a rational function approximation to the sine function. In degrees, we have
$~ \sin(x^\circ) \approx \frac{4x(180-x)}{40500 - x(180-x)}, \quad 0 \leq x \leq 180. ~$
Plot both functions over $[0, 180]$. What is the maximum difference between the two to two decimal points? (You may need to plot the difference of the functions to read off an approximate answer.)
###### Question
Solve the following equation for a value of $x$ using acos:
$~ \cos(x/3) = 1/3. ~$
###### Question
For any postive integer $n$ the equation $\cos(x) - nx = 0$ has a solution in $[0, \pi/2]$. Graphically estimate the value when $n=10$.
###### Question
The sine function is an odd function.
• The hyperbolic sine is:
• The hyperbolic cosine is:
• The hyperbolic tangent is:
###### Question
The hyperbolic sine satisfies this formula:
$~ \sinh(\theta + \beta) = \sinh(\theta)\cosh(\beta) + \sinh(\beta)\cosh(\theta). ~$
Is this identical to the pattern for the regular sine function?
The hyperbolic cosine satisfies this formula:
$~ \cosh(\theta + \beta) = \cosh(\theta)\cosh(\beta) + \sinh(\beta)\sinh(\theta). ~$
Is this identical to the pattern for the regular sine function?
|
1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
2. Serial order wise
3. Ex.3.7 (Optional)
Transcript
Ex 3.7, 5 In a ABC, C = 3 B = 2 ( A + B). Find the three angles. Let A = x and B = y Given that C = 3 B C = 3y Also, C = 2 ( A + B) C = 2 (x + y) Applying Angle sum Property A + B + C = 180 For C = 3y Put A = x B = y and C = 3y Since ABC is a triangle, By angle sum property A + B + C = 180 x + y + 3y = 180 x + 4y = 180 For C = 2(x + y) Put A = x B = y and C = 2 (x + y) Since ABC is a triangle, By angle sum property A + B + C = 180 x + y + 2 (x + y) = 180 x + y + 2x + 2y = 180 3x + 3y = 180 3(x + y) = 180 (x + y) = 180/3 x + y = 60 Hence, the equations are x + 4y = 180 (1) x + y = 60 (2) From equation (1) x + 4y = 180 x = 180 4y Put x = 180 4y in equation (2) x + y = 60 180 4y + y = 60 180 60 = 4y y 120 = 3y 3y = 120 y = 120/3 y = 40 Put y = 40 in equation (1) x + 4y = 180 x + (4 40) = 180 x + 160 = 180 x = 20 Thus, A = x = 20 B = y = 40 C = 3y = 3 40 = 120
Ex.3.7 (Optional)
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Results of Equal Chords
Theorem 1: Equal chords are equidistant from the center.
Given: $OL = OM$
Construction: Join $OA$ and $OC$
Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.
Since $OL \perp AB \Rightarrow AL = \frac{1}{2} AB$
Similarly, $OM \perp CD \Rightarrow CM = \frac{1}{2} CD$
Given $AB = CD$
$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$
$\Rightarrow AL = CM$
Now consider $\triangle OMC$ and $\triangle OLA$
$AL = CM$ (proved above)
$OA = OC$ (radius)
$\angle OLA = \angle OMC = 90^o$
Therefore $\triangle OMC \cong \triangle OLA$ (By RHS criterion)
$\Rightarrow OL = OM$. Hence proved that equal chords are equidistant from the center.
$\\$
Theorem 2 (Converse of Theorem 1): Chords of a circle which are equidistant from the center are equal.
Given: Two chords $AB$ and $CD$ in circle $C(O, r)$
Also $OL = OM$ where $OL \perp AB$ and $OM \perp CD$
To Prove: $AB = CD$
Construction: Join $OA$ and $OC$
Proof: $OL \perp AB$ (given)
We know that the perpendicular from the center of a circle to a chord, bisects the chord.
$\Rightarrow AL = BL$
$\Rightarrow AL = \frac{1}{2} AB$
Similarly $OM \perp CD$
$\Rightarrow CM = DM$
$\Rightarrow CM = \frac{1}{2} CD$
Consider $\triangle OAL$ and $\triangle OCM$
$OA = OC$ (radius)
$\angle OLA = \angle OMC = 90^o$
$OL = OM$ (given)
Therefore $\triangle OAL \cong \triangle OCM$ (By S.A.S criterion)
$\Rightarrow AL = CM$
$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$
$\Rightarrow AB = CD$
Hence proved that chords of a circle which are equidistant from the center are equal.
$\\$
Theorem 3: Equal chords of concurrent circles are equidistant from the corresponding centers.
Given: Two concurrent circles $C(O,r)$ and $C(O',r)$ and $AB = CD$
Construction: Join $OA$ and $O'C$
To Prove: $OL = O'M$ where $OL \perp AB$ and $O'M \perp CD$
Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.
$AL = BL \Rightarrow AL = \frac{1}{2} AB$
Similarly, $CM = DM \Rightarrow CM = \frac{1}{2} CD$
Now $AB = CD$ (given)
$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$
$\Rightarrow AL = CM$
Consider $\triangle OAL$ and $\triangle O'CM$
$OA = O'C$ (congruent circles with radius $r$)
$\angle OLA = \angle O'MC = 90^o$
$AL = CM$ (proved above)
Therefore $\triangle OAL \cong \triangle OC'M$ (By RHS criterion)
Therefore $OL = O'M$ i.e. $AB$ and $CD$ are equidistant from $O$ and $O'$ respectively.
$\\$
Theorem 4 (Converse of Theorem 3): Chords of congruent circles which are equidistant from their corresponding centers are equal.
Given: Two circles $C(O,r)$ and $C(O',r)$. $AB$ and $CD$ are chords such that $OL = O'M$ . $OL \perp AB$ and $O'M \perp CD$
To Prove: $AB = CD$
Construction: Join $OA$ and $O'C$
Proof: Consider $\triangle OLA$ and $\triangle O'MC$
$OL = O'M$ (given)
$OA = O'C$ (equal radius)
$\angle OLA = \angle O'MC = 90^o$
Therefore $\triangle OLA \cong \triangle O'MC$ (By RHS criterion)
$\Rightarrow AL = CM$
$\Rightarrow 2 AL = 2 CM$
We know that the perpendicular from the center of a circle to a chord, bisects the chord.
$\Rightarrow AB = CD$
Hence proved that chords of congruent circles which are equidistant from their corresponding centers are equal.
$\\$
Theorem 5: Equal chords of a circle subtends equal angles at the center.
Given: $AB$ and $CD$ are two equal chords in circle $C(O,r)$
To Prove: $\angle AOB = \angle COD$
Proof: Consider $\triangle AOB$ and $\triangle COD$
$OA = OD$ (radius)
$OB = OC$ (radius)
$AB = CD$ (given)
Therefore $\triangle AOB \cong \triangle COD$
$\Rightarrow \angle AOB = \angle COD$.
Hence proved that equal chords of a circle subtend equal angles at the center.
$\\$
Theorem 6: (converse of Theorem 5) If angles subtended by two chords of a circle at the center are equal, then the chords are equal.
Given: $AB$ and $CD$ are chords of a circle $C(O,r)$ such that $\angle AOB = \angle COD$
To Prove: $AB = CD$
Proof: Consider $\triangle AOB$ and $\triangle COD$
$OA = OD$ (radius)
$OB = OC$ (radius)
$\angle AOB = \angle COD$ (given)
Therefore $\triangle AOB \cong \triangle COD$ (By S.A.S criterion)
Therefore $AB = CD$ (corresponding sides off congruent triangles are equal)
$\\$
Theorem 7: Equal chords of congruent triangles subtend equal angles at the center.
Given: Two concurrent circles $C(O, r)$ and $C(O', r)$. Chords $AB = CD$
To Prove: $\angle AOB = \angle CO'D$
Proof: Consider $\triangle AOB$ and $\triangle CO'D$
$OA = O'C$ (radius)
$OB = O'D$ (radius)
$AB = CD$ (given)
Therefore $\triangle AOB \cong \triangle CO'D$ (By S.S.S criterion)
Hence $\angle AOB = \angle CO'D$ (corresponding angles of congruent triangles are equal).
$\\$
Theorem 8 (converse of Theorem 7): If the angles subtended by the two chords of congruent circles at the corresponding centers of circles are equal, the chords are equal.
Given: Two congruent circles $C(O,r)$ and $C(O',r)$. AB and CD are chords such that $\angle AOB = \angle CO'D$
To Prove: $AB = CD$
Proof: Consider $\triangle AOB$ and $\triangle CO'D$
$OA = O'C$ (radius)
$OB = O'D$ (radius)
$\angle AOB = \angle CO'D$ (given)
Therefore $\triangle AOB \cong \triangle CO'D$ (By S.A.S criterion)
$\Rightarrow AB = CD$
$\\$
Theorem 9: Of any two chords of a circle, the chord which is larger is nearer to the circle.
Given: Two chords $AB$ and $CD$ of a circle with center $O$ such that $AB > CD$
Construction: Join $OA$ and $OC$
To Prove: $OL < OM$ or $OM > OL$
Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.
$OL \perp AB \Rightarrow AL = \frac{1}{2} AB$
And $OM \perp CD \Rightarrow CM = \frac{1}{2} CD$
In right triangles $\triangle OAL$ and $\triangle OCM$, we have
$OA^2 = OL^2 + AL^2$
And $OC^2 = OM^2 + CM^2$
Since $OA = OC$ (radius)
$OL^2 + AL^2 = OM^2 + CM^2$
Given $AB > CD$
$\Rightarrow \frac{1}{2} AB > \frac{1}{2} CD$
$\Rightarrow AL > CM$
$\Rightarrow AL^2 > CM^2$ (Adding $OL^2$ to both sides)
Therefore $OL^2 + AL^2 > OL^2 + CM^2$
$\Rightarrow OM^2 + CM^2 > OL^2 + CM^2$
$\Rightarrow OM^2 > OL^2$
$\Rightarrow OM > OL.$
Hence proved that of any two chords of a circle, the chord which is larger is nearer to the circle.
$\\$
Theorem 10 (converse of Theorem 9): Of any two chords of a circle, the chord which is nearer is larger.
Given: Two chords $AB$ and $CD$ of a circle $C(O, r)$ such that $OL < OM$, where $OL$ and $OM$ are perpendicular from $O$ on $AB$ and $CD$ respectively.
To Prove: $AB > CD$
Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.
Therefore $AL = \frac{1}{2} AB \ \ \ \& \ \ \ \ CM = \frac{1}{2} CD$
In right triangles $\triangle OAL$ and $\triangle OCM$ we have
$OA^2 = OL^2 + AL^2$
And $OC^2 = OM^2 + CM^2$
$\Rightarrow AL^2 = OA^2 - OL^2$ … … … … … i)
And $CM^2 = OC^2 - OM^2$ … … … … … ii)
Now $OL < OM$
$\Rightarrow OL^2 < OM^2$
$\Rightarrow = OL^2 > - OM^2$ (Adding $OA^2$ on both sides)
$OA^2 - OL^2 > OA^2 - OM^2$
$AL^2 > OA^2 - OM^2$
$AL^2 > OC^2 - OM^2$
Since $OA^2 = OC^2$ (radius of the circle)
$AL^2 > CM^2$
$\Rightarrow AL > CM$
$\Rightarrow 2AL > 2 CM$
$\Rightarrow AB > CD$. Hence proved that of any two chords of a circle, the chord which is nearer is larger.
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# RS Aggarwal Class 10 Solutions Chapter 14 - Height And Distance Ex 14A (14.1)
## RS Aggarwal Class 10 Chapter 14 - Height And Distance Ex 14A (14.1) Solutions Free PDF
Q.1: The angle of elevation of an airplane from a point on the ground is $60^{\circ}$. After flying for 15 seconds, the elevation changes to $30^{\circ}$. If the airplane is flying at a height of 1500 meters, find the speed of the airplane.
Sol:
Let O and B the two positions of the jet plane and let A be the point of observation.
Let AX be the horizontal ground.
Draw OC ⊥ AX and BD ⊥ AX.
Then, ∠CAO = 60°, ∠DAB = 30° and OC = BD = 1500√3 m
From right ∆ OCA, we have
$\frac{AC}{OC}$ = cot 60° = $\frac{1}{√3}$
$\frac{AD}{1500√3}$ = √3 =>AD = (1500√3 x √3) = 4500m
CD = (AD – AC) = (4500 – 1500)m =>3000m
OB = CD = 3000m
Thus, the airplane covers 3000 m in 15 seconds
= ($\frac{3000}{15}$ x $\frac{60 x 60}{1000}$) kmph
Hence the speed of the airplane is = 720kmph.
Q.2: From the top of a building AB, 60m high the angles of depression of the top and bottom of a vertical lamp post CD are observed to be $30^{\circ}\:and\:60^{\circ}$ respectively. Find
(i) The horizontal distance between AB and CD
(ii) The height of the lamp post
(iii) The difference between the heights of the building and the lamp post.
Sol:
Let AB be the building and CD be the light house.
AE is drawn the perpendicular to CD.
Now AB = 60 m
∠ADB = 60°, ∠CAE = 30°
Let BD = x m
AE = BD = x m
In right ∆ACE, let CE = h
$\frac{CE}{AE}$ = tan 30°
$\frac{h}{x}$ = $\frac{1}{√3}$
x = √3h . . . . . . . . . . (1)
In right ∆ABD,
$\frac{AB}{BD}$ = tan 60° => $\frac{60}{x}$ = √3
X = $\frac{60}{√3}$ => $\frac{60√3}{3}$ = 20√3
= 20 x 1.732 = 34.64m . . . . . . . . (2)
From (1) and (2),
20√3 = √3h
h = 20m
Hence,
(i) Difference of heights of light house and building = 20m.
(ii) The distance between light house and building = 34.64m.
Q.3: As observed from the top of a light house, 100m above sea level, the angle of depression of a ship, sailing directly towards it, changes from $30^{\circ}\:and\:60^{\circ}$. Determine the distance traveled by ship during the period of observation. [Use $\sqrt{3}=1.732$]
Sol:
Let AB be the light house and let C and D be the positions of the ship.
Let AD =x, CD = y
In ∆BDA,
$\frac{X}{100}$ = cot 60°
X = $\frac{100}{√3}$m
Similarly in ∆BCA, $\frac{x+y}{100}$ = cot 30°
=> (x+y) = 100√3m
y = (x+y) – x
= ( 100√3 – $\frac{100}{√3}$ )m
= ( $\frac{200}{√3}$ x $\frac{√3}{√3}$ )m = 115.46m
The distance travelled by the ship during the period of observation = 115.46 m
Q. 4: The angle of depression from the top of a tower of a point  A on the ground is $30^{\circ}$. On moving a distance of 20 metres from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is $60^{\circ}$. Find the height of the tower and its distance from the point A.
Sol:
Let CD be the tower. The angle of elevation from A and B are 60° and 30° respectively.
Let AD = x m and CD = h m
In right ∆ ACD,
$\frac{CD}{AD}$ = tan 60°
$\frac{h}{x}$ = √3
h = √3x   ……..(1)
In right angled ∆BCD,
$\frac{CD}{BD}$ = tan 30°
$\frac{h}{20 + x}$ = 1/√3
√3h = 20 + x  ………(2)
Eliminating x from (1) and (2),
√3h = 20 + $\frac{h}{√3}$
or 3h = 20√3 + h
or h = 10√3 => 17.32m
Height of the tower = 17.32 m
Q.5: From the top of a 7-metre-high building, the angle of elevation of the top of the cable tower is $60^{\circ}$ and the angle of depression of its foot is $30^{\circ}$. Determine the height of the tower. [Use $\sqrt{3}=1.732$]
Sol:
Let AB be the building 7 meters high. AE ⊥ CD, where CD is the cable tower
In ∆ AED,
∠EAD = 30° = Angle of depression
$\frac{AE}{ED}$ = cot 30°
=> $\frac{x}{7}$ = √3
x = 7√3 m
In  ∆ACE,
∠CAE = 60° => Angle of elevation of C
∠AEC = 90°
$\frac{CE}{AE}$ = tan 60°
$\frac{h}{x}$ = √3
h = √3x
h = √3 x 7√3 => 21m
Height of the tower = CD = CE + ED = (21 + 7) m = 28 m
Q.6: The angles of elevation of a tower from two points at distances of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Show that the height of the tower is 6 meters.
Sol:
Let AB be the tower and let C and D be the two positions of the observer. Then, AC = 9 meters,
Let ∆ ACB = θ
Then, ∠ADB = (90° – θ)
Let AB = h meters
From right ∆ CAB, we have
$\frac{AB}{AC}$ = tan θ
=> $\frac{h}{9}$ = tan θ
h = 9 tan θ
From right ∆ DAB, we have
$\frac{AB}{AD}$ = tan(90° – θ) => $\frac{h}{4}$ = cot θ
=> h = 4 cot θ
from (1) and (2), we get
$h^{2}$ = 36
=> h = 6
Hence, the height of tower is 6 meters.
Q.7: A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 10m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is $30^{\circ}$ and that of the top of the flagstaff is $45^{\circ}$. Find the height of the tower. [Use $\sqrt{3}=1.732$]
Sol:
Let P be the point of observation RQ is the building and BR is the flag staff of height h, ∠BPQ = 45°, ∠RPQ = 30°
$\frac{PQ}{QR}$ = cot 30° = √3
$\frac{PQ}{10}$ = √3
=> PQ = 10√3 m . . . . . . . . . (1)
From the right ∆PBQ, we have
$\frac{PQ}{QB}$ = $\frac{PQ}{10+h}$ = cot 45° = 1
PQ = 10 + h . . . . . . . . (2)
From (1) and (2), we have,
10 + h = 10√3
h = 10√3 – 10 = (10×1.73 – 10)m = (17.3 – 10) = 7.3m
Hence distance of building is and length of the flags staff is 7.3 m.
Q.8: From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^{\circ}\:and\:45^{\circ}$ respectively. If the bridge is at a height of 2.5m from the banks, find the width of the river. [Take $\sqrt{3}=1.732$]
Sol:
Let A and B be two points on the bank on opposite sides of the river. Let P be a point on the bridge at a height of 2.5 m
Thus, DP = 2.5 m
Then, ∠BAP = 30°, ∠ABP = 45° and PD = 2.5m
$\frac{DB}{PD}$ = cot 45° =$\frac{DB}{2.5}$ = 1
=> DB = 2.5m
$\frac{AD}{PD}$ = cot 30°
$\frac{AD}{2.5}$ = √3
Height of the river = AB
= (AD + DB) = 2.5(√3 + 1)m = 2.5( 1.732 + 1 )m => 6.83m
Q.9: Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as $30^{\circ}\:and\:45^{\circ}$ respectively. If the height of the tower is 50 metres, find the distance between the two men. [Take $\sqrt{3}=1.732$]
Sol:
Let AB be the tower. Let C and D be the positions of the two men.
Then, ∠ACB = 30°, ∠ADB = 45° and AB = 50 m
$\frac{AC}{AB}$ = cot 30° = √3
=> $\frac{AC}{50}$ = √3
=> AC = 50√3 m
$\frac{AD}{AB}$ = cot 45° = 1
=> $\frac{AD}{50}$ = 1
Distance between the two man = CD = (AC + AD)
= 50(√3 + 1) = 136.6m
Q.10: The horizontal distance between two towers is 60 meters. The angle of depression of the top of the first tower when seen from the top of the second tower is $30^{\circ}$. If the height of the second tower is 90 meters, find the height of the first tower. [Use$\sqrt{3}=1.732$]
Sol:
Let AB and CD be the first and second towers respectively.
Then, CD = 90 m and AC = 60 m.
Let DE be the horizontal line through D.
Draw BF ⊥ CD,
Then, BF = AC = 60 m
∠FBD = ∠EDB = 30°
Now, $\frac{FD}{BF}$ = tan 30° = $\frac{FD}{60}$ = $\frac{1}{√3}$
=> FD = ( 60 x $\frac{1}{√3}$ )m = 20√3 m
AB = FC = ( CD – FD ) = ( 90 – 20√3 )m = 55.36m
#### Practise This Question
A population will not exist in Hardy - Weinberg equilibrium if
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# Algebra II : Solving Expressions
## Example Questions
### Example Question #1 : Solving Expressions
Solve for x.
Explanation:
a. Simplify each side of the equation using the distributive property.
b. Add 6x to both sides of the equation to move all terms with "x" to the left side of the equation.
c. Add 5 to both sides of the equation to move all constants to the right side of the equation.
d. Divide both sides of the equation by 30 to isolate the variable. Simplify the resulting fraction
### Example Question #2 : Solving Expressions
If , simplify .
Explanation:
First, you substitute for
Next, use PEMDAS (Parentheses, Exponents, Multiplication, Dividion, Addition, and Subtraction) to preform the algebraic operations in the correct order. When we apply this rule to simplify we get the following:
### Example Question #3 : Solving Expressions
Solve for if .
Explanation:
First, substitute 2 for z:.
Then, simplify: .
Next, you must isolate y by moving all other numbers and variables to the other side of the equation: , which gives you .
And simplify: .
Here, we then take the square root of both sides: .
Simplfy: , because both and .
### Example Question #4 : Solving Expressions
Simplify given and .
Explanation:
First, substitute 1 for z, 2 for x and 3 for y: and simplify: .
Using PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction), we simplify the multiplication: .
Then add and subtract from left to right: .
### Example Question #5 : Solving Expressions
Emily buys a rose plant when it is inches tall. The tag indicates that it will grow inches every year. She also buys a tulip plant when it is inches tall. The tag indicates that it will grow inches a year. After how many years are the two plants the same height?
They will never be the same height at the same time.
Explanation:
We can express each plant's growth as a function of years in the following equations:
Rose height after x years =
Tulip height after x years =
Since we are looking for the year when the two plants are of equal height, we set these expressions for height equal to each other, and solve for x:
Combining like terms by subtracting 2x from both sides and subtracting 5 from both sides gives us:
The plants will reach the same height after 6 years of growth.
### Example Question #2011 : Algebra Ii
Evaluate the expression if , , and
Explanation:
After you plug in all of your given values the expression is as followed;
Since the numerator is zero, therefore, the entire fraction equals zero.
### Example Question #7 : Solving Expressions
Evaluate the expression if , , and
Explanation:
When you plug in your given values the expression should read as followed;
### Example Question #1 : Solving Expressions
Simplify the expression
-
Explanation:
### Example Question #9 : Solving Expressions
Evaluate the expression when and .
Explanation:
First, substitute for and for
Now, using the order of operations (Parentheses, Exponents, Multiplication, Division, Addittion, Subtraction), begin to simplify the expression:
Leaving you with,
### Example Question #1 : Solving Expressions
Evaluate the expression given .
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# Solving Linear Equations and 2 1 Inequalities Solving
• Slides: 32
Solving Linear Equations and 2 -1 Inequalities Solving Linear Equations and Inequalities Warm Up Lesson Presentation Holt Algebra 22
2 -1 Solving Linear Equations and Inequalities *Warm Up Simplify each expression. 2. –(w – 2) 1. 2 x + 5 – 3 x 3. 6(2 – 3 g) Graph on a number line. 4. t > – 2 – 4 – 3 – 2 – 1 0 1 2 3 4 5 5. Is 2 a solution of the inequality – 2 x < – 6? Explain. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Objectives Solve linear equations using a variety of methods. Solve linear inequalities. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Vocabulary equation solution set of an equation linear equation in one variable identify contradiction inequality Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities An equation is a mathematical statement that two expressions are equivalent. The solution set of an equation is the value or values of the variable that make the equation true. A linear equation in one variable can be written in the form ax = b, where a and b are constants and a ≠ 0. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Linear Equations in One variable 4 x = 8 3 x – = – 9 2 x – 5 = 0. 1 x +2 Nonlinear Equations + 1 = 32 + 1 = 41 3 – 2 x = – 5 Notice that the variable in a linear equation is not under a radical sign and is not raised to a power other than 1. The variable is also not an exponent and is not in a denominator. Solving a linear equation requires isolating the variable on one side of the equation by using the properties of equality. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation Do inverse operations in the reverse order of operations. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Example 1: Consumer Application The local phone company charges \$12. 95 a month for the first 200 of air time, plus \$0. 07 for each additional minute. If Nina’s bill for the month was \$14. 56, how many additional minutes did she use? Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Example 1 Continued Let m represent the number of additional minutes that Nina used. Model monthly plus charge 12. 95 Holt Algebra 2 + additional minute charge times 0. 07 * number of additional = total charge minutes m = 14. 56
2 -1 Solving Linear Equations and Inequalities Example 1 Continued Solve. 12. 95 + 0. 07 m = 14. 56 – 12. 95 0. 07 m = 0. 07 1. 61 0. 07 Subtract 12. 95 from both sides. Divide both sides by 0. 07. m = 23 Nina used 23 additional minutes. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Check It Out! Example 1 Stacked cups are to be placed in a pantry. One cup is 3. 25 in. high and each additional cup raises the stack 0. 25 in. How many cups fit between two shelves 14 in. apart? Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities * Check It Out! Example 1 Continued Let c represent the number of additional cups needed. You fill in the rest of the information. Model additional cup one cup plus height + Holt Algebra 2 times * number of total additional = height cups c =
2 -1 Solving Linear Equations and Inequalities Check It Out! Example 1 Continued Solve. 3. 25 + 0. 25 c = 14. 00 – 3. 25 0. 25 c = 10. 75 0. 25 Subtract 3. 25 from both sides. Divide both sides by 0. 25. c = 43 44 cups fit between the 14 in. shelves. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities *Example 2: Solving Equations with the Distributive Property Solve 4(m + 12) = – 36 Method 1 Holt Algebra 2 Divide both sides by 4 first, then solve.
2 -1 Solving Linear Equations and Inequalities * Example 2 Continued Solve 4(m + 12) = – 36 Method 2 Holt Algebra 2 Distribute the 4 first, then solve.
2 -1 Solving Linear Equations and Inequalities * Check your answer Example 2 Continued 4(m + 12) = – 36 4(____+ 12) 4(___) ___ Holt Algebra 2 – 36
2 -1 Solving Linear Equations and Inequalities * Check It Out! Example 2 a Solve 3(2 – 3 p) = 42. Method 1 Divide by 3 first. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities * Check It Out! Example 2 a Continued Solve 3(2 – 3 p) = 42. Method 2 Distribute the 3 first. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities * Check It Out! Example 2 b Solve – 3(5 – 4 r) = – 9. Choose your method; either divide by -3 or distribute -3. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities If there are variables on both sides of the equation: (1) simplify each side (2) collect all variable terms on one side and all constants terms on the other side (3) isolate the variables as you did in the previous problems. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Example 3: Solving Equations with Variables on Both Sides Solve 3 k– 14 k + 25 = 2 – 6 k – 12. Simplify each side by combining – 11 k + 25 = – 6 k – 10 like terms. +11 k Collect variables on the right side. Add. 25 = 5 k – 10 Collect constants on the left side. +10 + 10 35 = 5 k 5 5 7=k Holt Algebra 2 Isolate the variable.
2 -1 Solving Linear Equations and Inequalities * Check It Out! Example 3 Solve 3(w + 7) – 5 w = w + 12. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities You have solved equations that have a single solution. Equations may also have infinitely many solutions or no solution. An equation that is true for all values of the variable, such as x = x, is an identity. An equation that has no solutions, such as 3 = 5, is a contradiction because there are no values that make it true. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities * Example 4 A: Identifying Identities and Contractions Solve 3 v – 9 – 4 v = –(5 + v). The equation has no solution. The solution set is the empty set, which is represented by the symbol. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities *Example 4 B: Identifying Identities and Contractions Solve 2(x – 6) = – 5 x – 12 + 7 x. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities * Check It Out! Example 4 a Solve 5(x – 6) = 3 x – 18 + 2 x. The equation has no solution. The solution set is the empty set, which is represented by the symbol. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities * Check It Out! Example 4 b Solve 3(2 – 3 x) = – 7 x – 2(x – 3). Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities An inequality is a statement that compares two expressions by using the symbols <, >, ≤, ≥, or ≠. The graph of an inequality is the solution set, the set of all points on the number line that satisfy the inequality. The properties of equality are true for inequalities, with one important difference. If you multiply or divide both sides by a negative number, you must reverse the inequality symbol. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities These properties also apply to inequalities expressed with >, ≥, and ≤. Holt Algebra 2
2 -1 Solving Linear Equations and Inequalities Example 5: Solving Inequalities Solve and graph 8 a – 2 ≥ 13 a + 8 – 13 a – 5 a – 2 ≥ 8 +2 +2 – 5 a ≥ 10 – 5 a ≤ 10 – 5 a ≤ – 2 Holt Algebra 2 Subtract 13 a from both sides. Add 2 to both sides. Divide both sides by – 5 and reverse the inequality. – 10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1
2 -1 Solving Linear Equations and Inequalities * Check It Out! Example 5 Solve and graph x + 8 ≥ 4 x + 17. Holt Algebra 2
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### Section 10.9 : Absolute Convergence
2. Determine if the following series is absolutely convergent, conditionally convergent or divergent.
$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 3}}}}{{\sqrt n }}}$
Show All Steps Hide All Steps
Start Solution
Okay, let’s first see if the series converges or diverges if we put absolute value on the series terms.
$\sum\limits_{n = 1}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^{n - 3}}}}{{\sqrt n }}} \right|} = \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt n }}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^{\frac{1}{2}}}}}}$
Now, by the by the $$p$$-series test we can see that this series will diverge.
Show Step 2
So, at this point we know that the series in the problem statement is not absolutely convergent so all we need to do is check to see if it’s conditionally convergent or divergent. To do this all we need to do is check the convergence of the series in the problem statement.
The series in the problem statement is an alternating series with,
${b_n} = \frac{1}{{\sqrt n }}$
Clearly the $${b_n}$$ are positive so we can use the Alternating Series Test on this series. It is hopefully clear that the $${b_n}$$ are a decreasing sequence and $$\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$$.
Therefore, by the Alternating Series Test the series from the problem statement is convergent.
Show Step 3
So, because the series with the absolute value diverges and the series from the problem statement converges we know that the series in the problem statement is conditionally convergent.
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# Outcomes and Probability (3)
In this worksheet, students determine the possible outcomes of two events by using a sample-space diagram.
Key stage: KS 3
Curriculum topic: Probability
Curriculum subtopic: Calculate Theoretical Probability
Difficulty level:
### QUESTION 1 of 10
When two events occur, there are several possible outcomes.
For example, when these two spinners are spun, there are 4 x 4 = 16 possible outcomes.
The two scores are added together and so we can create a Sample-Space diagram to show the possible totals.
1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
The possible totals are 2, 3, 4, 5, 6, 7 and 8.
The probability of getting a total of 7, P(7) = 2/16 = 1/8
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
How many different totals are possible?
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
Which total is most likely to occur?
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
What is the probability of getting a total score of 5?
(Write your answer with the / symbol, e.g. 2/3 and remember to reduce your fraction to its lowest terms)
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
True or False?
Scoring a total of 8 is more likely than scoring a total of 2.
True
False
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
True or False?
Scoring a total of 4 is more likely than scoring a total of 7.
True
False
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
How many different totals are possible?
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
In how many different ways can the total score of 10 be achieved?
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
In how many different ways can the total score of 5 be achieved?
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
Which total score occurs the most often?
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
What is the probability of scoring a total of 6?
(Write your answer with the / symbol, e.g. 2/3 and remember to reduce your fraction to its lowest terms)
• Question 1
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
How many different totals are possible?
7
• Question 2
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
Which total is most likely to occur?
5
• Question 3
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
What is the probability of getting a total score of 5?
(Write your answer with the / symbol, e.g. 2/3 and remember to reduce your fraction to its lowest terms)
1/4
• Question 4
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
True or False?
Scoring a total of 8 is more likely than scoring a total of 2.
False
• Question 5
These two spinners are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
True or False?
Scoring a total of 4 is more likely than scoring a total of 7.
True
• Question 6
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
How many different totals are possible?
11
• Question 7
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
In how many different ways can the total score of 10 be achieved?
3
EDDIE SAYS
6+4, 5+5, 4+6
• Question 8
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
In how many different ways can the total score of 5 be achieved?
4
EDDIE SAYS
4+1, 3+2, 2+3, 1+4
• Question 9
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
Which total score occurs the most often?
7
• Question 10
These two dice are spun and the two scores are added together.
Use a sample-space diagram to answer the question.
What is the probability of scoring a total of 6?
(Write your answer with the / symbol, e.g. 2/3 and remember to reduce your fraction to its lowest terms)
5/36
---- OR ----
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# Quick Answer: What Is The Difference Between Undefined And Zero Slope?
## What makes a line steeper?
The steepness, incline, or grade of a line is measured by the absolute value of the slope.
A slope with a greater absolute value indicates a steeper line.
The direction of a line is either increasing, decreasing, horizontal or vertical..
## What to do if the slope is undefined?
If the slope of a line is undefined, then the line is a vertical line, so it cannot be written in slope-intercept form, but it can be written in the form: x=a , where a is a constant. If the line has an undefined slope and passes through the point (2,3) , then the equation of the line is x=2 .
## What kind of slope is 0 6?
Explanation: An undefined slope means a vertical line. The equation of the vertical line passing through (0,6) is x=0 . The y axis has the equation x=0 too, so the line and the axis intercept at all points of the y axis.
## What is a positive slope?
A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y decreases also. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.
## Is 0 over undefined?
We can say, zero over zero equals x. … Just say that it equals “undefined.” In summary with all of this, we can say that zero over 1 equals zero. We can say that zero over zero equals “undefined.” And of course, last but not least, that we’re a lot of times faced with, is 1 divided by zero, which is still undefined.
## What is the slope of a diagonal line?
Lines that fall from left to right have a negative slope. So a diagonal line to the right ( / ) would have a positive slope. A diagonal line to the left ( \ ) would have a negative slope. A vertical line ( | )has an undefined slope and a horizontal line ( —- ) has a slope of 0.
## How do I know if a slope is undefined?
Note that when a line has a negative slope it goes down left to right. Note that when a line is horizontal the slope is 0. Note that when the line is vertical the slope is undefined.
## When can a slope of a line be equal to zero?
The slope of a line can be positive, negative, zero, or undefined. A horizontal line has slope zero since it does not rise vertically (i.e. y1 − y2 = 0), while a vertical line has undefined slope since it does not run horizontally (i.e. x1 − x2 = 0). because division by zero is an undefined operation.
## How do you know if a line is zero or undefined?
If one is walking up a hill, the slope is said to be positive; if one is walking down a hill, the slope is said to be negative; and if one is walking a flat line, the slope is said to be zero because there is no incline at all (i.e. no vertical change).
## What does it mean to have a slope of 0 and an undefined slope?
Remember, a horizontal (flat) line has a slope of 0. A Vertical line (one that forms a right angle or is perpendicular to a flat line) has an undefined slope.
## Is a slope of 0 4 undefined?
The “slope” of a vertical line. A vertical line has undefined slope because all points on the line have the same x-coordinate. As a result the formula used for slope has a denominator of 0, which makes the slope undefined..
## What happens when M is undefined?
1 Expert Answer Whenever the slope (m) is undefined it means that when you try to calculate the slope given any 2 points on the line you’ll end up with a zero in the denominator, which is undefined. That is, with an undefined slope you end up with a vertical line, which has the equation “x=some number”.
## How do I find the slope of the line?
To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .
## What is an undefined slope?
An undefined slope (or an infinitely large slope) is the slope of a vertical line! The x-coordinate never changes no matter what the y-coordinate is! There is no run! In this tutorial, learn about the meaning of undefined slope.
## What happens when the denominator is 0 in slope?
If the denominator of the fraction is 0, the slope is undefined. This occurs if the x value is the same for both points. The graph would be a vertical line and would indicate that the x value stays constant for every value of y. If the numerator of the fraction is 0, the slope is 0.
|
Associated Topics || Dr. Math Home || Search Dr. Math
### What is x?
```
Date: 01/07/2002 at 06:35:18
From: MARTIN TAYLOR
Subject: How do I work out 5x + 3 = 38
Please can you tell me how to work out 5x + 3 = 38 ?
```
```
Date: 01/07/2002 at 08:57:14
From: Doctor Ian
Subject: Re: How do I work out 5x + 3 = 38
Hi Martin,
Suppose that we're talking about bags of rocks, and x represents the
contents of one bag, which we'll show as () below. We put these on a
balance beam, and it looks like
r r r r r r r r
r r r r r r r r r r
r r r r r r r r r r
() () () () () r r r r r r r r r r r r r
\_____________________/ \_____________________/
That is, the left side contains 5 bags and 3 loose rocks, and the
right side contains 38 loose rocks. And since the problem tells us
that these are equal, the beam is balanced.
We can remove three rocks from each side of the balance beam without
changing the balance, right?
r r r r r
r r r r r r r r r r
r r r r r r r r r r
() () () () () r r r r r r r r r r
\_____________________/ \_____________________/
If the bags themselves weigh nothing, then each bag must be holding
1/5 of the number of rocks on the right side. So if we count the rocks
on the right, and divide by 5, we get the number of rocks in each bag.
() ----------> r r r r r r r
() ----------> r r r r r r r
() ----------> r r r r r r r
() ----------> r r r r r r r
() ----------> r r r r r r r
\_____________________/ \_____________________/
Using 'algebra', it looks like this:
5 * (? rocks per bag) + 3 = 38
5 * (? rocks per bag) + 3 - 3 = 38 - 3
5 * (? rocks per bag) = 35
5 * (? rocks per bag) 35
--------------------- = --
5 5
? rocks per bag = 7
As you can imagine, it gets a little tiring to write (or type)
? rocks per bag
over and over again. Since it doesn't matter too much what we call
this quantity - so long as we call it the same thing everywhere - we
can just pick something shorter, like 'rocks', or 'r', or (if we want
to be conventional) 'x'. Then it looks like this:
5x + 3 = 38
5x + 3 - 3 = 38 - 3
5x = 35
5x/5 = 35/5
x = 7
That looks a little more cryptic and mysterious, but once you get a
solid grasp on the idea of using letters as names for much longer
names of particular quantities, it becomes rather normal looking. And
that's mostly a matter of practice.
For an introduction to how to think about solving problems like this,
take a look at
Basic Tips on Solving for X
http://mathforum.org/dr.math/problems/megan.11.16.00.html
Does this help?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra
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# Graph
A set $V$ of vertices and a set $E$ of unordered and ordered pairs of vertices; denoted by $G(V,E)$. An unordered pair of vertices is said to be an edge, while an ordered pair is said to be an arc. A graph containing edges alone is said to be non-oriented or undirected; a graph containing arcs alone is said to be oriented or directed. A pair of vertices can be connected by two or more edges (arcs of the same direction) and such edges (arcs) are then said to be multiple. An arc (or edge) can begin and end at the same vertex, in which case it is known as a loop. (A "graph" is sometimes understood to be a graph without loops or multiple edges; in such a case a graph with multiple edges is said to be a multi-graph, whereas one containing both multiple edges and loops is said to be a pseudo-graph.)
Vertices connected by an edge or a loop are said to be adjacent. Edges with a common vertex are also called adjacent. An edge (arc) and any one of its two vertices are said to be incident. One says that an edge $\{u,v\}$ connects two vertices $u$ and $v$, while an arc $(u,v)$ begins at the vertex $u$ and ends at the vertex $v$. Each graph can be represented in Euclidean space by a set of points, corresponding to the vertices, which are connected by lines, corresponding to the edges (or the arcs) of the graph. In three-dimensional space any graph can be represented in such a way that the lines corresponding to edges (arcs) do not intersect at interior points.
There are various ways of specifying a graph. Let $u_1,\dots,u_n$ be the vertices of a graph $G(V,E)$ and let $e_1,\dots,e_m$ be its edges. The adjacency matrix corresponding to $G$ is the matrix $A=(a_{i,j})$ in which the element $a_{i,j}$ equals the number of edges (arcs) which join the vertices $u_i$ and $u_j$ (go from $u_i$ to $u_j$) and $a_{i,j}=0$ if the corresponding vertices are not adjacent. In the incidence matrix $B=(b_{i,j})$ of $G$ the element $b_{i,j}=1$ if the vertex $u_i$ is incident to the edge $e_j$, and $b_{i,j}=0$ if the vertex $u_i$ and the edge $e_j$ are not incident. A graph can be specified by lists, for example, by specifying pairs of vertices connected by edges (arcs) or by specifying the set of vertices adjacent to each vertex. Two graphs $G(V,E)$ and $H(W,I)$ are called isomorphic if there is a one-to-one correspondence between the sets of vertices $V,W$ and the sets of edges $E,I$ which preserves the incidence relationship (see also Graph isomorphism).
A subgraph $G'(V',E')$ of a graph $G(V,E)$ is defined as a graph with set of vertices $V'\subseteq V$ and set of edges (arcs) $E'\subseteq E$, each one of which is incident with vertices from $V'$ only. A subgraph $G'(V',E')$ is said to be generated or induced by the subset $V'\subseteq V$ if it is a graph with set of vertices $V'$ and ordered set of edges (arcs) $E'$, $E'$ consists of by all edges of $G$ which connect vertices of $V'$. A skeleton subgraph or spanning subgraph $G'(V,E')$ contains all vertices of $G$ and some subset of its edges (arcs) $E'\subseteq E$. A sequence of edges $(u_0,u_1),\dots,(u_{r-1},u_r)$ is called an edge progression or walk connecting the vertices $u_0$ and $u_r$. An edge progression is called a chain or trail if all its edges are different and a simple chain or path if all its vertices are different. A closed (simple) chain is also called a (simple) cycle. A graph is said to be connected if any pair of its vertices is connected by an edge progression. A maximal connected subgraph of a graph $G$ is said to be a connected component. A disconnected graph has at least two connected components (see also Graph, connectivity of a).
The length of an edge progression (chain, simple chain) is equal to the number of edges in the order in which they are traversed. The length of the shortest simple chain connecting two vertices $u_i$ and $u_j$ in a graph $G$ is said to be the distance $d(u_i,u_j)$ between $u_i$ and $u_j$. In a connected undirected graph the distance satisfies the axioms of a metric. The quantity $\min_{u_i} \max_{u_j} d(u_i,u_j)$ is called the diameter, while a vertex $u_0$ for which $\max_{u_j} d(u_i,u_j)$ assumes its minimum value is called a centre of $G$. A graph can contain more than one centre or no centre at all.
The degree of a vertex $u_i$ of a graph $G$, denoted by $d_i$, is the number of edges incident with that vertex. If a (loop-free) graph $G$ has $n$ vertices and $m$ edges, then $\sum_{i=1}^n d_i = 2m$. A vertex $u_i$ is said to be isolated if $d_i=0$ and terminal or pendant if $d_i=1$. A graph all vertices of which have the same degree (equal to $k$) is said to be regular of degree $k$. A complete graph has no loops and each pair of vertices is connected by exactly one edge. Let a graph $G(V,E)$ be free from loops or multiple edges; then the complementary graph to $G$ is the graph $\bar{G}(V,E)$ in which $\bar{V}=V$ and vertices are adjacent in $\bar{G}$ only if they are not adjacent in $G$. A graph which is complementary to a complete graph consists of isolated vertices and is known as empty. Many characteristics of a graph $G$ and of its complement $\bar{G}$ are related. In a directed graph $G$ one defines, for each vertex $u_i$, the output (or out) and the input (or in) (semi-) degree as the number of arcs issuing from and entering this vertex, respectively. A complete directed graph is known as a tournament. To each graph $G$ can be assigned a number of graphs which are derived from of $G$. Thus, the edge graph $L(G)$ of $G$ is the graph whose vertices correspond to the edges of $G$ and two vertices are adjacent in $L(G)$ if and only if the corresponding edges of $G$ are adjacent. In the total graph $T(G)$ of $G$ the vertices correspond to the elements of $G$, i.e. to vertices and edges, and two vertices in $T(G)$ are adjacent if and only if the corresponding elements in $G$ are adjacent or incident. Many properties of $G$ carry over to $L(G)$ and $T(G)$. Many generalizations of the concept of a "graph" are known, including that of a hypergraph and of a network graph.
With the aid of suitable operations it is possible to construct a graph from simpler graphs, to pass from a graph to simpler ones, to subdivide a graph into simpler ones, to pass from one graph to another in a given class of graphs, etc. The most common one-place operations include the removal of an edge (the vertices of the edge are preserved), the addition of an edge between two vertices of a graph, the removal of a vertex together with its incident edges (the graph obtained by removal of a vertex $v$ from a graph $G$ is often denoted by $G-v$), the addition of a vertex (which may be connected by edges with certain vertices of the graph), the contraction of an edge — identification of a pair of adjacent vertices, i.e. removal of a pair of adjacent vertices and addition of a new vertex which is adjacent to those vertices of the graph which were adjacent to at least one of the vertices which have been removed, and subdivision of an edge — removal of an edge and addition of a new vertex which is joined by an edge to each vertex of the edge which has been removed.
Two-place operations over a graph are employed in a number of problems in graph theory. Let $G_1(V_1,E_1)$ and $G_2(V_2,E_2)$ be graphs such that $V_1 \cap V_2 = \emptyset$ and $E_1 \cap E_2 = \emptyset$. The union of $G_1$ and $G_2$ is the graph $G = G_1 \cup G_2$ with set of vertices $V = V_1 \cup V_2$ and set of edges $E = E_1 \cup E_2$. The product of $G_1$ and $G_2$ is the graph $G = G_1 \times G_2$ whose set of vertices is the Cartesian product $V = V_1 \times V_2$, any two of the vertices $(u_1,u_2)$ and $(v_1,v_2)$ being adjacent if and only if either $u_1=v_1$ and $u_2$ is adjacent to $v_2$, or $u_2=v_2$ and $u_1$ is adjacent to $v_1$. For example, any graph is the union of its connected components; a graph known as the $n$-dimensional unit cube $Q_n$ can be recursively defined by the product operation
$Q_n = K_2 \times Q_{n-2}$
where $Q_1=K_2$ is the graph consisting of a pair of vertices connected by one edge. These operations can also be defined for intersecting graphs, in particular for subgraphs of a given graph. The addition modulo $2$ of two graphs $G_1$ and $G_2$ is defined as the graph $G$ with set of vertices $V = V_1 \cup V_2$ and set of edges $E = (E_1 \cup E_2) \setminus (E_1 \cap E_2)$. Other many-place operations on graphs are also employed.
For certain classes of graphs it is possible to find simple operations, a repeated application of which makes it possible to pass from any graph in the given class to any other graph in the same class. With the aid of the operation shown in Fig. A is possible to pass from any graph to any other graph within the class of graphs with the same set of degrees.
Figure: A
The operation shown in Fig. B makes it possible to pass from any triangulation to any other triangulation within the class of planar triangulations (cf. Graph, planar).
Figure: B
The description and study of certain classes of graphs also involves operations and sets of graphs making it possible to obtain any graph of a given class. Operations on graphs are also employed to construct graphs with given properties, to calculate numerical characteristics of graphs, etc. (cf. Graph, numerical characteristics of a).
The concept of a "graph" is employed in defining mathematical ideas such as a control system, in certain definitions of an algorithm, of a grammar, etc. The exposition of a number of mathematical theories becomes more easily understood if geometric representations of graphs are employed, e.g. the theory of Markov chains. The concept of a "graph" is widely employed in the formulation and description of various mathematical models in economics, biology, etc.
#### References
[1] C. Berge, "The theory of graphs and their applications" , Wiley (1962) (Translated from French) [2] O. Ore, "Theory of graphs" , Amer. Math. Soc. (1962) [3] A.A. Zykov, "The theory of finite graphs" , 1 , Novosibirsk (1969) (In Russian) [4] F. Harary, "Graph theory" , Addison-Wesley (1969) pp. Chapt. 9
There is as yet no universally accepted terminology in graph theory. In the English literature there are basically three schools of terminology: the French school typified by Berge's books [1] and [a1], the Canadian school typified by the books [a2] and [a3] and the American (especially Michigan) school typified by the books [4] and [a4]. Some of the terminology used in the present article differ from all these.
#### References
[a1] C. Berge, "Graphs and hypergraphs" , North-Holland (1973) (Translated from French) [a2] J.A. Bondy, U.S.R. Murthy, "Graph theory with applications" , Macmillan (1976) [a3] W.T. Tutte, "Graph theory" , Addison-Wesley (1984) [a4] M. Behzad, G. Chartrand, L.L. Foster, "Graphs and digraphs" , Prindle, Weber & Schmidt (1979) [a5] R.J. Wilson, "Introduction to graph theory" , Longman (1985)
How to Cite This Entry:
Graph. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Graph&oldid=38861
This article was adapted from an original article by V.P. Kozyrev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
|
Fractions and Mixed Numbers: Overview
Working with fractions can be difficult for students, because they may have a hard time relating their understanding of operations with whole numbers to operations with fractions. Students have learned that addition can be thought of as joining two or more groups, or sets, to form a new set. They have also learned to add units of like quantities, such as tens to tens, ones to ones, and so on. Addition of fractions should be an extension of these concepts. So, when students are trying to add and , they will learn that they need to add “like” quantities. To do this, they learn that they need to rewrite the fractions with common denominators or “parts equal in size” when adding or subtracting fractions. Note in the diagram below that = and = . So we can add the number of twelfths, + , and get .
Thus, + =
Understanding the need to use common denominators will help students remember to look for them when adding fractions or mixed numbers. After students have a visual concept for adding fractions, the next step is to have them practice adding fractions by using an algorithm. Remind students that in order to add fractions, they need to find a common denominator (a common multiple for the denominators). Some students will find it easier to add fractions if they use any common denominator. Other students may find it easier to use the least common denominator (by finding the least common multiple of the denominators). (See Grade 6, Factors and Fractions.) Let's look at adding 2 and 1. We can add the whole-number parts and then add the fractions. Both ways to do this are shown below. One shows using 24, the product of 6 and 4, as the common denominator. The other shows using the least common denominator, 12. Note that 12 is the LCM of 6 and 4.
The idea of “taking away” is one way of thinking about subtraction with whole numbers. When students subtracted whole numbers, they took away things that were alike. Students subtracted tens from tens and ones from ones. Similarly, when they want to subtract from , they need to find equivalent fractions with a common denominator, as shown below, and then subtract or take away quantities that are alike—in this case, fifteenths.
Once students have a good understanding of this visual model for the subtraction of fractions, they are ready to use an algorithm to subtract fractions. If the fractions to be subtracted do not already have a common denominator, then students should rewrite the given fractions as equivalent fractions with a common denominator as they did when adding fractions with unlike denominators.
When subtracting mixed numbers, it may be necessary to regroup in order to subtract the fractional parts. Relate this concept to the need to regroup when subtracting whole numbers, as when subtracting 28 from 72. Let's look at the following subtraction problem involving mixed numbers: 4 − 1. First, we need to rewrite the fractions as equivalent fractions with the same denominator. 4 = 4 and 1 = 1.
In this case, the LCD is the same as the common denominator found by multiplying the two denominators. Because you can't subtract from , you need to rename or regroup 4 as 3 so that we can easily subtract the two numbers.
Multiplication of fractions is probably the easiest of the fraction operations to perform. However, not many students have a good visual model for multiplying fractions. One way to think of multiplying x is to think of finding of . This will help students understand why the answer is less than . Let's look at a visual model of x .
Thus, the number of parts that are double shaded is 6, which is the product of the two numerators. The number of equal pieces is 12, which is the product of the two denominators. When having students multiply two fractions, it is a good idea to have them cancel any common factors before multiplying the two fractions, by writing each number in prime factorization form. This may be easier than having to simplify the fractions after multiplying. For example, when multiplying by we get ,
and we can cancel the common factors 2 x 2 in both the numerator and denominator and a common factor of 3 from both, leaving
A mistake that students often make when multiplying mixed numbers is to multiply the whole numbers and multiply the fractions and then add them. This results from a misuse of the Distributive Property. Remind students that when they multiply 47 x 23, they have to multiply the 3 x7 and then 3 x 40. Next, they multiply 20 x7 and 20 x 40 and they add all four products. Similarly, when multiplying 2x 3, students would have to multiply the 2 x3 and 2 x and then multiply x 3 and x . Fortunately, it is much simpler to change both numbers to improper fractions first and then multiply the fractions.
The algorithm for the division of fractions is relatively easy to learn. In order to help students understand it, relate it to the division of whole numbers. In 12 ÷ 4, we see how many of 12 items can go in 4 groups or sets. We can see from the diagram below that there are 3 items in each of 4 groups. Thus, 12 ÷ 4 = 3.
Similarly, to solve ÷ , we want to count how many sixths are in . In the diagram below, we see that there are 4 sixths in .
Providing students with a visual model to show the division of two fractions will help them better estimate the answer to a division problem and help them recognize the reasonableness of an answer. The procedure for dividing two fractions is relatively simple. You need only find the reciprocal of the divisor and multiply it by the dividend. Let's look at × .
Since is a little less than , the answer 1 seems reasonable.
To divide two mixed numbers, students should first change the mixed numbers to improper fractions and then follow the same procedure for dividing two fractions as described above, namely multiply the dividend by the reciprocal of the divisor. Let's look at 3 divided by 1.
|
Writing Equations
Contents
page 1 of 2
Page 1
Page 2
Horizontal Lines
Horizontal lines have a slope of 0 . Thus, in the slope-intercept equation y = mx + b , m = 0 . The equation becomes y = b , where b is the y -coordinate of the y -intercept.
Example 1: Write an equation for the following line:
Graph of a Line
Since y always takes the value -1 , an equation for the line is y = - 1 .
Example 2: Write an equation for the horizontal line that passes through (6, 2) .
Since the line is horizontal, y is constant--that is, y always takes the same value. Since y takes a value of 2 at the point (6, 2) , y always takes the value 2 . Thus, the equation is y = 2 .
Vertical Lines
Similarly, in the graph of a vertical line, x only takes one value. Thus, the equation for a vertical line is x = a , where a is the value that x takes.
Example 3: Write an equation for the following line:
Graph of a Line
Since x always takes the value 2 = , the equation for the line is x = .
Example 4: Write an equation for the vertical line that passes through (6, 2) .
Since the line is vertical, x is constant--that is, x always takes the same value. Since x takes a value of 6 at the point (6, 2) , x always takes the value 6 . Thus, the equation is x = 6 .
Page 1
Page 2
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# How to do synthetic division
This lesson will explain how to do synthetic division to quickly divide a polynomial by another.
A couple of synthetic division examples are shown below.
Divide x2 + 11x + 30 by x + 5
The quotient is x + 6
Explanation of the steps we took while using synthetic division to divide x2 + 11x + 30 by x + 5.
1.
Reverse the sign of the constant term in the divisor. For example, the constant term in the divisor is 5. Change it to -5. Remove the coefficients from the dividend and rewrite the division as shown above in blue.
2.
Bring down the first coefficient or 1. The 1 will begin the quotient.
3.
Multiply the first coefficient by the new divisor and add the answer to the next coefficient or 11. We get 6. write down 6 in the same position as 11 and -5.
4.
Keep multiplying and adding until the remainder is found. For example, multiply 6 by -5. We get -30. Add -30 to 30 to get 0 and 0 is the remainder. In fact, the last number on the right that you find after you add is the remainder whether it is 0 or not.
## How to do synthetic division when the degree of the polynomial is 3
Divide x3 + 5x2 -2x - 24 by x - 2
Step 1: Reverse the sign of the constant term in the divisor. In other words, change -2 in x - 2 into 2 and bring it down.
_____________________
x - 2 | x ³ + 5x² - 2x - 24
2 | 1 5 -2 -24
Step 2: Bring down the first coefficient or 1. The 1 will begin the quotient.
2 | 1 5 -2 -24
________________________
1
Step 3: Multiply 1 by 2 and add the result to 5. The result is 7. Bring 7 down.
2 | 1 5 -2 -24
2
________________________
1 7
Step 4: Multiply 7 by 2 and add the result to -2. Keep multiplying and adding until the remainder is found.
2 | 1 5 -2 -24
2 14
__________________________
1 7 12
2 | 1 5 -2 -24
2 14 24
__________________________
1 7 12 0
The quotient is x² + 7x + 12
Synthetic division is also called the method of detached coefficients. This is because the first step in synthetic division is to remove the coefficients from the polynomial that is being divided.
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The Central Slice Theorem: An Example
The central slice theorem is key to understanding tomography. In Intermediate Physics for Medicine and Biology, Russ Hobbie and I ask the reader to prove the central slice theorem in a homework problem. Proofs are useful for their generality, but I often understand a theorem better by working an example. In this post, I present a new homework problem that guides you through every step needed to verify the central slice theorem. This example contains a lot of math, but once you get past the calculation details you will find it provides much insight.
The central slice theorem states that taking a one-dimensional Fourier transform of a projection is equivalent to taking the two-dimensional Fourier transform and evaluating it along one direction in frequency space. Our “object” will be a mathematical function (representing, say, the x-ray attenuation coefficient as a function of position). Here is a summary of the process, cast as a homework problem.
Section 12.4
Problem 21½. Verify the central slice theorem for the object
(a) Calculate the projection of the object using Eq. 12.29,
Then take a one-dimensional Fourier transform of the projection using Eq. 11.59,
(b) Calculate the two-dimensional Fourier transform of the object using Eq. 12.11a,
Then transform (kx,ky) to (θ,k) by converting from Cartesian to polar coordinates in frequency space.
(c) Compare your answers to parts (a) and (b). Are they the same?
I’ll outline the solution to this problem, and leave it to the reader to fill in the missing steps.
Fig. 12.12 from IPMB, showing how to do a projection.
The Projection
Figure 12.12 shows that the projection is an integral of the object along various lines in the direction θ, as a function of displacement perpendicular to each line, x’. The integral becomes
Note that you must replace x and y by the rotated coordinates x’ and y’
You can verify that x2 + y2= x’2 + y’2.
After some algebra, you’re left with integrals involving eby’2 (Gaussian integrals) such as those analyzed in Appendix K of IPMB. The three you’ll need are
The resulting projection is
Think of the projection as a function of x’, with the angle θ being a parameter.
The One-Dimensional Fourier Transform
The next step is to evaluate the one-dimensional Fourier transform of the projection
The variable k is the spatial frequency. This integral isn’t as difficult as it appears. The trick is to complete the square of the exponent
Then make a variable substitution u = x’ + ik2b. Finally, use those Gaussian integrals again. You get
This is our big result: the one-dimensional Fourier transform of the projection. Our next goal is to show that it’s equal to the two-dimensional Fourier transform of the object evaluated in the direction θ.
Two-Dimensional Fourier Transform
To calculate the two-dimensional Fourier transform, we must evaluate the double integral
The variables kx and ky are again spatial frequencies, and they make up a two-dimensional domain we call frequency space.
You can separate this double integral into the product of an integral over x and an integral over y. Solving these requires—guess what—a lot of algebra, completing the square, and Gaussian integrals. But the process is straightforward, and you get
Select One Direction in Frequency Space
If we want to focus on one direction in frequency space, we must convert to polar coordinates: kx = k cosθ and ky = k sinθ. The result is
This is exactly the result we found before! In other words, we can take the one-dimensional Fourier transform of the projection, or the two-dimensional Fourier transform of the object evaluated in the direction θ in frequency space, and we get the same result. The central slice theorem works.
I admit, the steps I left out involve a lot of calculations, and not everyone enjoys math (why not?!). But in the end you verify the central slice theorem for a specific example. I hope this helps clarify the process, and provides insight into what the central slice theorem is telling us.
Source: http://hobbieroth.blogspot.com/2021/08/the-central-slice-theorem-example.html
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# Add the following:(a) $(-7)-8-(-25)$(b) $(-13)+32-8-1$(c) $(-7)+(-8)+(-90)$(d) $50-(-40)-(-2)$
To do:
We have to add the given expressions.
Solution:
We know that,
$[(-)\times(-)=(+)\times(+)=(+)]$
$[(-)\times(+)=(+)\times(-)=(-)]$
(a) $(-7)-8-(-25)=-7-8+25$
$=25-(7+8)$
$=25-15$
$=10$
(b) $(-13)+32-8-1=32-(13+8+1)$
$=32-22$
$=10$
(c) $(-7)+(-8)+(-90)=-(7+8+90)$
$=-105$
(d) $50-(-40)-(-2)=50+40+2$
$=92$
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Updated on: 10-Oct-2022
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# GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.3
Gujarat Board GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.3 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 10 Maths Chapter 14 Statistics Ex 14.3
Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Solution:
Median:
n = 68
⇒ $$\frac {n}{2}$$ = 34
⇒ Median class 125 – 145.
Mean:
Mean $$\bar{x}$$ = l + $$\frac{f_{i}-f_{0}}{2 f_{i}-f_{0}-f_{2}}$$ x h = 135 + $$\frac {7}{68}$$ x 20 = 135 + 2.06 = 137.06 units
Mode: Since the maximum frequency is 20.
Hence, the modal class is 125 – 145.
Mode = l + $$\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]$$ x h = 125 + $$\frac {7}{68}$$ x 20 = 125 + $$\frac{7}{40-27}$$ x 20
= 125 + $$\frac{7 x 20}{13}$$ = 125 + 10.77 = 135.77 units
Question 2.
If the median of the distribution given below is 28.5, find the values of x and y. ∑fi = 60.
Solution:
Here, n = 60
⇒ x + y + 45 = 60
⇒ x + y = 15 ……(1)
Since median = 28.5
Hence, the median class is 20-30.
28.5 = 20 + $$\left[\frac{30-(5+x)}{20}\right]$$ x 10
8.5 = $$\frac{25-x}{2}$$
⇒ 17 = 25 – x
⇒ x = 8
From (1) y = 15 – 8 = 7
∴ x = 8 and y = 7
Question 3.
A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if polices are only given to persons having age 18 years onwards but less than 60 years.
Solution:
n = 100
$$\frac{n}{2}$$ = $$\frac{100}{2}$$ = 50
Median class is 35 – 40.
= 35 + $$\left[ \frac { 50-45 }{ 33 } \right]$$ x 5 = 35 + $$\frac{25}{33}$$ = 35 + 0.76 = 35. 76 years
Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Find the median length of the leaves.
Solution:
Firstly, data needs to be converted into continuous classes.
Now, n = 40
So, $$\frac{n}{2}$$ = $$\frac{40}{2}$$ = 20
144.5 Р153.5 is the media̱ class.
= 144.5 + $$\left[ \frac { 20-17 }{ 12 } \right]$$ x 9 = 144.5 + 2.25 = 146.75 mm
Hence, the median length of the leaves is 146.75 mm.
Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps..
Find the median lifetime of a lamp.
Solution:
Now, n = 400
So, $$\frac{n}{2}$$ = $$\frac{400}{2}$$ = 200
= 3000 +1 1×500 = 3000 + 406.98 = 3406.98 hours
Hence, the median lifetime of a lamp is 3406.98 hours.
Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:
Median:
Now, n = 100
So, $$\frac{n}{2}$$ = $$\frac{100}{2}$$ = 50
7 – 10 is the median class.
= 7 + $\frac{50 – 36}{40}$ x 3 = 7 + $$\frac{21}{20}$$ = 7 + 1.05 = 8.05
Hence, the median number of letters in the surnames is 8.05.
Mean: Take a = 8.5, h = 3
$$\frac{21}{20}$$ = 8.5 + (-$$\frac{6}{100}$$) x 3 = 8.5 – 0.18 = 8.32
Hence, the mean number of letters in the surnames is 8.32.
Mode: Since the maximum number of surnames have a number of letters in the interval 7 – 10, the modal class is 7 – 10.
Therefore, l = 7, h = 3, f1 = 40, f0 = 30, f2 = 16
∴ Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ x h = 7 + $$\left(\frac{40-30}{2 \times 40-30-16}\right)$$ x 3 = 7 + $$\frac{30}{34}$$ = 7 + 0.88 = 7.88
Hence, the modal size of the surnames is 7.88.
Question 7.
The distribution below gives the weights of 30 students of a class, find the median weight of the students.
Solution:
Now, n = 30
So, $$\frac{n}{2}$$ = $$\frac{30}{2}$$ = 15
= 55 + $$\frac{15 – 13}{6}$$ x 5 = 55 + $$\frac{10}{6}$$ = 55 + $$\frac{5}{3}$$ = 55 + 1.67 = 56.67
Hence, the median weight of the students is 56.67 kg.
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A quadratic equation is an equation written in the form of:
$a{x}^{2}+bx+c=0$
where a, b and c are real numbers with a not equal to 0. A quadratic equation is also called a second-degree polynomial equation in x.
Solving quadratic equations may produce three possible outcomes: no solution, one solution or two solutions. These techniques are also a part of calculus, in which the solutions are the roots of a parabola crossing the x-axis.
There are a number of simple quadratic equations that are straightforward to solve, such as:
${ x }^{ 2 }=8\\ \\ x=\pm \sqrt { 8 }$
The sign $\pm$ is used because both the positive and negative square roots of 8 produce 8 when squared. Similarly:
${ (a-2 })^{ 2 }=7\\ \\ a-2=\pm \sqrt { 7 } \\ \\ a=\pm \sqrt { 7 } +2$
This section will discuss two different methods to solve quadratic equations, while another method is discussed in the Quadratic Formula section.
Solving by Factorisation
The first method require trinomial factorisation.
Example:
1. Solve ${ x }^{ 2 }+2x-3=0$
${ x }^{ 2 }+2x-3=0\\ \\ (x+3)(x-1)=0\\ \\ x+3=0\qquad or\qquad x-1=0\\ \\ x=-3\qquad or\qquad x=1$
As seen above, the solutions could only be obtained if either $x+3=0$ or $x-1=0$. Therefore there are two possible solutions for this quadratic equations.
Remember: To be certain that a solution is correct, substitute the solution into the original equation and see if it is satisfied.
2. Solve ${ x }^{ 2 }+6x+9=0$
${ x }^{ 2 }+6x+9=0\\ \\ (x+3)(x+3)=0\\ \\ { (x+3) }^{ 2 }=0\\ \\ x+3=\sqrt { 0 } =0\\ \\ x=-3$
Since both solutions are equal, then we only have one solution $x=-3$
3. Solve $3{ a }^{ 2 }-14a=-8$
$3{ a }^{ 2 }-14a=-8\\ \\ 3{ a }^{ 2 }-14a+8=0\\ \\ (3a-2)(a-4)=0\\ \\ 3a-2=0\qquad or\qquad a-4=0\\ \\ a=\frac { 2 }{ 3 } \qquad or\qquad a=4$
Solving by Completing the Square
Not all trinomials will factorise perfectly, so other methods such as completing the square needs to be used to solve quadratic equations. Starting from the basics, completing the square is basically re-writing a quadratic equation in another form:
$a{x}^{2}+bx+c=0\qquad to \qquad a{ (x+d) }^{ 2 }+e=0$
where $d=\frac { b }{ 2a } \qquad and\qquad e=c-\frac { { b }^{ 2 } }{ 4a }$
Using a simple example, the equation ${x}^{2}+bx$ can be rearranged into a square, and by completing the square with ${\frac { b }{ 2 }}^{2}$.
In geometry, it looks like:
${x}^{2}+bx \qquad \qquad \qquad + \qquad\qquad \qquad {\frac { b }{ 2 }}^{2} \qquad \qquad \qquad = \qquad \qquad \qquad{ \left( x+\frac { b }{ 2 } \right) }^{ 2 }$
(courtesy of mathsisfun.com)
Examples:
Complete the square on ${x}^{2}+6x$.
${ x }^{ 2 }+6x\qquad <-\quad \frac { 6 }{ 2 } =3\\ \\ { x }^{ 2 }+6x+{ 3 }^{ 2 }\quad =\quad { (x+3) }^{ 2 }\\ \\ { x }^{ 2 }+6x+9\quad =\quad { (x+3) }^{ 2 }$
For example, complete the square on ${ x }^{ 2 }+6x+7$.
${ x }^{ 2 }+6x+7\qquad <-\quad \frac { 6 }{ 2 } =3\\ \\ =\quad ({ x }^{ 2 }+6x+{ 3 }^{ 2 })+(7-{ 3 }^{ 2 })\\ \\ =\quad { (x+3) }^{ 2 }-2$
Now, to solve a quadratic equation in the form of $a{x}^{2}+bx+c=0$ by completing the square, there are steps to follow:
1. Divide all terms by a (the coefficient of ${x}^{2}$
2. Move the independent term $\frac{c}{a}$ to the right side.
3. Complete the square on the left side, and balance this by adding $\frac{b}{2a}$ to the right hand.
4. Take the square root on both sides.
5. Subtract the number that remains on the left side to finally obtain x.
Examples:
Solve by completing the square:
1. ${ x }^{ 2 }-6x+3=0$
${ x }^{ 2 }-6x=-3\\ \\ { x }^{ 2 }-6x+9=-3+9\\ \\ { (x-3) }^{ 2 }=6\\ \\ x-3=\pm \sqrt { 6 } \\ \\ x=\pm \sqrt { 6 } +3$
2. $5{ x }^{ 2 }-4x-2=0$
${ x }^{ 2 }-\frac { 4 }{ 5 } x-\frac { 2 }{ 5 } =0\\ \\ { x }^{ 2 }-\frac { 4 }{ 5 } x=\frac { 2 }{ 5 } \qquad <-\quad \cfrac { \frac { 4 }{ 5 } }{ 2 } =\frac { 4 }{ 10 } \\ \\ { x }^{ 2 }-\frac { 4 }{ 5 } x+{ \left( \frac { 4 }{ 10 } \right) }^{ 2 }=\frac { 2 }{ 5 } +{ \left( \frac { 4 }{ 10 } \right) }^{ 2 }\\ \\ { \left( x-\frac { 4 }{ 10 } \right) }^{ 2 }=\frac { 14 }{ 25 } \\ \\ x-\frac { 4 }{ 10 } =\pm \sqrt { \frac { 14 }{ 25 } } \\ \\ x=\pm \sqrt { \frac { 14 }{ 25 } } +\frac { 4 }{ 10 }$
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# How does the Law of Sines work?
My teacher gave me the formula for the Law of Sines and I know how to solve questions like this, but I don't see how the theorem below can actually work.
Can someone please explain it to me?
• Are you asking for a proof of the Law of Sines? – Ted Shifrin May 28 '14 at 0:16
• I don't understand how this law works. – IHeartBunnies May 28 '14 at 0:20
• Do you understand how the double equal signs work? – Brad May 28 '14 at 0:24
• What do you mean? – IHeartBunnies May 28 '14 at 0:40
Let's look at the right triangle below. I use a right triangle so that we can illustrate the law of sines by applying the familiar SOH-CAH-TOA.
As an example, let's look at angles $A$ and $C$.
• We see that $\displaystyle\sin(A)=\frac{a}{b}=\frac{\text{opposite side of A}}{\text{hypotenuse}}$. This gives $\displaystyle \frac{1}{b}=\frac{\sin(A)}{a}$
• We see that $\displaystyle\sin(C)=\frac{c}{b}=\frac{\text{opposite side of C}}{\text{hypotenuse}}$. This gives $\displaystyle \frac{1}{b}=\frac{\sin(C)}{c}$
Since we have two expressions for $\displaystyle\frac{1}{b}$, we can set the expressions equal to each other and obtain $$\frac{\sin(A)}{a}=\frac{\sin(C)}{c}$$ This is the kind of relation that is fundamental in the Law of Sines.
It turns out that this relation holds for any two angles that you pick (so not restricted to angles $A$ or $C$). It also turns out that the Law of Sines holds for not just right angles, but also acute and obtuse angles as well.
However, the approach for deriving the Law of Sines for acute and obtuse are different; I only showed the approach for right angles. But please ask further if you'd like to see more explanation of how this Law of Sines works for acute/obtuse angles.
Consider the hight dropped from the vertex $B$, giving two right triangles. Then $\sin(A)=\frac{h}{c}$ and $\sin(C)=\frac{h}{a}$ or $c\sin(A)=a\sin(C)$ this gives one of the three equalities, the others being the same.
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Chapter 1: Fundamental Concepts
# 1.4 Dot Product
A dot product produces a single number to describe the product of two vectors. If you haven’t taken linear algebra yet, this may be a new concept. This is a form of multiplication that is used to calculate work, unit vectors, and to find the angle between two vectors.
$\vec A\cdot \vec B=|\vec A||\vec B|\cos\theta$
A vector can be multiplied by another vector but may not be divided by another vector. There are two kinds of products of vectors used broadly in physics and engineering. One kind of multiplication is a scalar multiplication of two vectors. Taking a scalar product of two vectors results in a number (a scalar), as its name indicates. Scalar products are used to define work and energy relations. For example, the work that a force (a vector) performs on an object while causing its displacement (a vector) is defined as a scalar product of the force vector with the displacement vector. A quite different kind of multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, as its name suggests. Vector products are used to define other derived vector quantities. For example, in describing rotations, a vector quantity called torque is defined as a vector product of an applied force (a vector) and its distance from pivot to force (a vector). It is important to distinguish between these two kinds of vector multiplications because the scalar product is a scalar quantity and a vector product is a vector quantity.
Scalar multiplication of two vectors yields a scalar product.
Dot Product
The scalar product $\vec A\cdot \vec B$ of two vectors $\vec A \text{ and } \vec B$ is a number defined by the equation:
$\vec A\cdot \vec B=|\vec A||\vec B| \cos \phi$
where ϕ is the angle between the vectors. The scalar product is also called the dot product because of the dot notation that indicates it.
When the vectors are given in their vector component forms:
$$\vec A=A_x\underline{\hat{i}}+A_y\underline{\hat{j}}+A_z\underline{\hat{k}}\text{ and }\vec B=B_x\underline{\hat{i}}+B_y\underline{\hat{j}}+B_z\underline{\hat{k}}$$
we can compute their scalar product as follows:
$$\vec A\cdot\vec B=(A_x\underline{\hat{i}}+A_y\underline{\hat{j}}+A_z\underline{\hat{k}})\cdot(B_x\underline{\hat{i}}+B_y\underline{\hat{j}}+B_z\underline{\hat{k}})\\=A_xB_x\underline{\hat{i}}\cdot\underline{\hat{i}}+A_xB_y\underline{\hat{i}}\cdot\underline{\hat{j}}+A_xB_z\underline{\hat{i}}\cdot\underline{\hat{k}}\\+A_yB_x\underline{\hat{j}}\cdot\underline{\hat{i}}+A_yB_y\underline{\hat{j}}\cdot\underline{\hat{j}}+A_yB_z\underline{\hat{j}}\cdot\underline{\hat{k}}\\+A_zB_x\underline{\hat{k}}\cdot\underline{\hat{i}}+A_zB_y\underline{\hat{k}}\cdot\underline{\hat{j}}+A_zB_z\underline{\hat{k}}\cdot\underline{\hat{k}}$$
Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselves give one, there are only three nonzero terms in this expression. Thus, the scalar product simplifies to:
$\vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z$
We can use the equation below to find the angle between two vectors. When we divide $\vec A\cdot\vec B=|\vec A||\vec B| \cos\phi$ by $|\vec A || \vec B|$ , we obtain the equation for cos(ϕ), into which we substitute the equation from above:
$$\cos\phi=\frac{\vec A\cdot\vec B}{|\vec A||\vec B| }=\frac{A_xB_x+A_yB_y+A_zB_z}{|\vec A||\vec B| }$$
Angle between vectors $\vec A \text{ and }\vec B$ is obtained by taking the inverse cosine of the expression above.
Source: University Physics Volume 1, OpenStax CNX, https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/2-4-products-of-vectors (many examples at this page).
But what IS it?
The dot product is the component of vector A along B ( |A| cos Θ ) times the magnitude (size of B). OR, it’s the component of B on A times the magnitude of A. Visually this can be seen in the figure.[1]
There is a nice mathematical proof on page 169 of Calculus-Based Physics.
One neat thing about the dot product is that AB = BA
An example of a dot product is in a solar panel. To maximize efficiency, the rays coming from the sun should be perpendicular to the panels, that is, straight on. You could use the dot product between a vector of the sun’s rays (yellow in the image below) and the unit vector perpendicular to the surface (green in the image) to calculate what portion of a ray that comes in at an angle produces energy.
Key Takeaways
Basically: Dot product is a method to find a number that is a product of two vectors.
Application: Two ropes attached to a sign are being pulled in different directions. To find the angle between them, use the dot product of the two vectors.
Looking ahead: We will use the dot product in Section 2.3 on particle equilibrium equations (and more in dynamics next semester).
1. Source: https://en.wikipedia.org/wiki/Dot_product#/media/File:Dot_Product.svg
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# A, B and C Can Reap a Field in 15 3 4 Days; B, C and D in 14 Days; C, D and a in 18 Days; D, a and B in 21 Days. in What Time Can A, B, C and D Together Reap It? - Mathematics
Course
ConceptTime and Work
#### Question
AB and C can reap a field in $15\frac{3}{4}$ days; BC and D in 14 days; CD and A in 18 days; DA and B in 21 days. In what time can ABC and D together reap it?
#### Solution
$\text{ Time taken by } \left( A + B + C \right) \text{ to do the work } = 15\frac{3}{4} \text{ days } = \frac{63}{4} \text{ days }$
$\text{ Time taken by } \left( B + C + D \right) \text{ to do the work = 14 days}$
$\text{ Time taken by } \left( C + D + A \right) \text{ to do the work = 18 days }$
$\text{ Time taken by } \left( D + A + B \right) \text{ to do the work = 21 days }$
$\text{ Now, }$
$\text{ Work done by } \left( A + B + C \right) = \frac{4}{63}$
$\text{ Work done by } \left( B + C + D \right) = \frac{1}{14}$
$\text{ Work done by } \left( C + D + A \right) = \frac{1}{18}$
$\text{ Work done by } \left( D + A + B \right) = \frac{1}{21}$
$\therefore \text{ Work done by working together } = \left( A + B + C \right) + \left( B + C + D \right) + \left( C + A + D \right) + \left( D + A + B \right)$
$= \frac{4}{63} + \frac{1}{14} + \frac{1}{18} + \frac{1}{21}$
$= \frac{4}{63} + \left( \frac{9 + 7 + 6}{126} \right) = \frac{4}{63} + \frac{22}{126}$
$= \frac{4}{63} + \frac{11}{63} = \frac{15}{63}$
$\therefore \text{ Work done by working together } = 3\left( A + B + C + D \right) = \frac{15}{63}$
$\therefore \text{ Work done by } \left( A + B + C + D \right) = \frac{15}{63 \times 3} = \frac{5}{63}$
$\text{ Thus, together they can do the work in } \frac{63}{5} \text{ days or } 12\frac{3}{5} \text{ days } .$
Is there an error in this question or solution?
#### APPEARS IN
Solution A, B and C Can Reap a Field in 15 3 4 Days; B, C and D in 14 Days; C, D and a in 18 Days; D, a and B in 21 Days. in What Time Can A, B, C and D Together Reap It? Concept: Time and Work.
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# How to write a function rule
In this lesson, you will learn to write a function rule using information given in a table.
A function rule such as cost = p + 0.08p is an equation that describes a functional relationship.
If p is the price you pay for an item and 0.08 is the sales tax, the function rule above is the cost of the item.
If you are given a table, usually you have to carefully examine the table to see what the function rule is.
Sometimes, it is easy to get the rule. Other times, it is not so easy. It may require some keen observation or using the concept of slope.
We will now show how to do this with some examples:
Example #1:
Suppose you are washing your clothes at a laundromat.The table below shows the relationship between number of loads and cost.
Write a function rule showing the relationship between number of loads and cost.
Number of loads 1 2 3 4 5 Cost 3 6 9 12 15
First, show the relationship:
Cost is 3 × number of loads. For instance, 2 × 3 = 6
Second, define:
Let n be the number of loads
Let c be the cost.
Write the equation:
c = 3 × n
Example #2:
Write a function rule for the following situation:
Input 1 2 3 4 5 Output 5 6 7 8 9
First, show the relationship:
Output is 4 + input.
Second, define:
Let I = input
Let O = output.
Write the equation:
O = 4 + I
Example #3:
Number of items sold Revenue 1 6 2 10 3 14 4 18 5 22
First, show the relationship:
It is not that easy to see what the relation is in this case.
1 × 6 = 6. However, 2 × 6 is not equal to 10.
1 + 5 = 6. However, 2 + 5 is not equal to 10.
The function rule could be a combination of addition and multiplication.
You could pick a revenue say 10 for instance and ask yourself "What math should I do with the 2 to get this revenue?"
Trial and error can help.
My strategy of trial and error is shown below:
2 × 1 + 8 = 10. However, 3 × 1 + 8 is not equal to 14. This rule does not work
2 × 2 + 6 = 10. However, 3 × 2 + 6 is not equal to 14. This rule does not work
2 × 3 + 4 = 10. However, 3 × 3 + 4 is not equal to 14. This rule does not work
2 × 4 + 2 = 10. Moreover, 3 × 4 + 2 is also equal to 14. This rule seems work! In fact, it will work
The relationship is then revenue is number of items × 4 plus two.
Second, define:
Let x be the number of items
Let R be the revenue
Write the equation:
R = x × 4 + 2
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# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.5 | Set 2
### Question 11. Find the mean, median, and mode of the following data:
Solution:
Let mean (A) = 175
Find Median:
Here N = 25, 5/3 = 25/2 = 12.5 or 13, it lies in the class interval = 50-200.
l = 150, F = 10, f = 6, h = 50
Using median formula, we get
= 150 + 20.83
= 170.83
Find Mean:
Using mean formula we get
= 175 – 6
= 169
Find Mode:
Using mode formula we get
= 150 + 25
= 175
### Question 12. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data.
Solution:
From the given table we conclude that
Modal class = 40-50 (it has maximum frequency)
Also,
l = 40, f = 20, f1 = 12, f2 = 11 and h = 10
By using mode formula, we get
= 40 + 4.70
= 44.7
### Question 13. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean, and mode of the data and compare them:
Solution:
Let mean (A) = 135
Find Median:
Here, N = 34
N/2 = 34,
Class interval = 25-145
Also,
l = 125, F = 22, f = 20 and h = 20
By using the median formula, we get
= 125 + 12
= 137 units
Find Mean:
By using the mean formula, we get
Mean =
= 135 + 2.05
= 137.05 units
Find Mode:
By using the mode formula, we get
= 125 + 10.76
= 135.76 units
### Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, And the modal size of the surnames.
Solution:
Let mean (A) = 8.5
Find Median:
Here, N = 100
So, N/2 = 50
Class interval = 7-10
l = 7, F = 36, f = 40 and h =3
By using the median formula, we get
= 7 + 1.05
= 8.05
Find Mean:
By using the mean formula, we get
Mean =
= 8.5 + 0.18
= 8.32
Find Mode:
We have,
N = 100
N/2 = 100/2 = 50
Here, the cumulative frequency is just greater than N/2 = 76,
Hence, the median class = 7 – 10
l = 7, h = 10 – 7 = 3, f = 40, F = 36
By using the mode formula, we get
Mode = l +
= 7 +
= 7 + 30/34
= 7 + 0.88
= 7.88
### Question 15. Find the mean, median, and mode of the following data:
Solution:
Find Mean:
By using the mean formula, we get
Mean =
Find Median:
We have,
N = 50
Then, N/2 = 50/2 = 25
Here, the cumulative frequency just greater than N/2 = 36
Hence, the median class = 60 – 80
l = 60, h = 80 – 60 = 20, f = 12, F = 24
By using the median formula, we get
Median = l +
= 60 +
= 60 + 20/12
= 60 + 1.67
= 61.67
Find Mode:
We have,
The maximum frequency = 12
Model class = 60 – 80
l = 60, h = 80 – 60 = 20, f = 12, f1 = 10, f2 = 6
By using the mode formula, we get
Mode = l +
= 60 +
= 60 + 40/8
= 65
### Question 16. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
Solution:
From the given table we conclude that
The maximum class frequency = 40
So, modal class = 1500 – 2000
l = 1500, f = 40, h = 500, f1 = 24, f2 = 33
By using the mode formula, we get
Mode = l +
= 1500 +
= 1500 +
= 1500 + 347.826
= 1847.826 ≈ 1847.83
Hence, the modal monthly expenditure = Rs. 1847.83
Now we will find class marks as
Class mark =
Class size (h) of given data = 500
Let mean(a) = 2750, now we are going to calculate diui as follows:
From the table we conclude that
∑fi = 200
∑fidi = -35
Mean = a +
= 2750 +
= 2750 – 87.5
= 2662.5
Hence, the mean monthly expenditure = Rs. 2662.5
### Find the mode of the data
Solution:
From the given table we conclude that
The maximum class frequency = 18
So, modal class = 4000 – 5000
and
l = 4000, f = 18, h = 1000, f1 = 4, f2 = 9
By using the mode formula, we get
Mode = l +
= 4000 +
= 4000 + (14000/23)
= 4000 + 608.695
= 4608.695
Hence, the mode of given data = 4608.7 runs.
### Find the modal agriculture holdings of the village.
Solution:
From the given table we conclude that
The maximum class frequency = 80,
So, the modal class = 5-7
and
l = 5, f0 = 45, h = 2, f1 = 80, f2 = 55
By using the mode formula, we get
Mode = l +
= 5 +
= 5 +
= 5 +
= 5 + 1.2
= 6.2
So, the modal agricultural holdings of the village = 6.2 hectares.
### Calculate the modal income.
Solution:
From the given table we conclude that
The maximum class frequency = 41,
So, modal class = 10000-15000.
Here, l = 10000, f1 = 41, f0 = 26, f2 = 16 and h = 5000
Therefore, by using the mode formula, we get
Mode = l +
= 10000 +
= 10000 +
= 10000 +
= 10000 + 15 × 125
= 10000 + 1875
= 11875
So, the modal income = Rs. 11875.
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# A line segment has endpoints at (5 ,2 ) and (3 ,4 ). If the line segment is rotated about the origin by (3 pi)/2 , translated vertically by 4 , and reflected about the y-axis, what will the line segment's new endpoints be?
Apr 30, 2017
The new end points are $\left(- 2 , - 5\right)$ and $\left(- 4 , - 3\right)$
#### Explanation:
We are going to use matrices.
The matrix of rotation of $\frac{3}{2} \pi$ about the origin is
$\left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right)$
The matrix of reflection in the $y$-axis is
$\left(\begin{matrix}- 1 & 0 \\ 0 & 1\end{matrix}\right)$
The combination of the 2 operations is
$\left(\begin{matrix}- 1 & 0 \\ 0 & 1\end{matrix}\right) \cdot \left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right) = \left(\begin{matrix}0 & - 1 \\ - 1 & 0\end{matrix}\right)$
The new end points are
$\left(\begin{matrix}0 & - 1 \\ - 1 & 0\end{matrix}\right) \left(\begin{matrix}5 \\ 2\end{matrix}\right) = \left(\begin{matrix}- 2 \\ - 5\end{matrix}\right)$
and
$\left(\begin{matrix}0 & - 1 \\ - 1 & 0\end{matrix}\right) \cdot \left(\begin{matrix}3 \\ 4\end{matrix}\right) = \left(\begin{matrix}- 4 \\ - 3\end{matrix}\right)$
|
### Ch11Stu
```Review of Probability Distributions
Probability distribution is a theoretical frequency
distribution.
Example 1.
If you throw a fair die (numbered 1 through 6).
What is the probability that you get a 1? or a 5?
Example 2.
If you throw a fair coin twice. What is the
probability that you get two tails?
Slide 1
Discrete vs. Continuous distributions
A variable can be discrete or continuous
A variable is discrete if it takes on a limited number
of values, which can be listed.
Example: Poisson distribution
Other examples:
A variable is continuous if it can take any value
within a given range.
Example: Exponential distribution.
Other examples:
Slide 2
Poisson Distribution
A Poisson distribution is a discrete distribution that
can take an integer value > 0 (i.e., 0, 1, 2, 3, ….)
Formula
• P(x) = (lx e –l)/x! (where e = natural logarithm or
2.718, and x! = x factorial)
Example
l= 3
• What is P(x = 0)?
• What is P(x = 2)?
Slide 3
Exponential Distribution
An exponential distribution is a continuous random
variable that can take on any positive value.
Formula: f(x) = l e (-lx) ; F(x) = P(X < x) = 1- e (-lx)
for l > 0, and 0 < x < infinity.
Example: l = 3
f(x=5) =
F(x=5)
Slide 4
Relationship between Poisson distribution and
Exponential distribution
Poisson distribution and exponential distribution are
used to describe the same random process.
Poisson distribution describes the probability that there
is/are x occurrence/s per given time period.
Exponential distribution describes the probability that the
time between two consecutive occurrence is within a certain
number x.
Example
If the arrival rate of customers are Poisson distributed
and, say, 6 per hour, then the time between arrivals of
customers are exponentially distributed with a mean of
(1/6) hour or 10 minutes.
Slide 5
Class Exercise
Suppose the arrival rate of customers is 10 per hour,
Poisson distributed
What is the probability that 2 customers are arrival in
one hour?
What is the average inter-arrival time of customers?
What is the probability that the inter-arrival time of
customers is exactly 3 minutes?
What is the probability that the inter-arrival time of
customers is less than or equal to 3 minutes?
Slide 6
Class Exercise
Suppose the arrival rate of customers is 10 per hour,
Poisson distributed
What is the probability that 2 customers are arrival in
one hour?
What is the average inter-arrival time of customers?
What is the probability that the inter-arrival time of
customers is exactly 3 minutes?
What is the probability that the inter-arrival time of
customers is less than or equal to 3 minutes?
Slide 7
Chapter 11: Waiting Line Models
Structure of a Waiting Line System
Queuing Systems
Queuing System Input Characteristics
Queuing System Operating Characteristics
Analytical Formulas
Single-Channel Waiting Line Model with Poisson
Arrivals and Exponential Service Times
Single-Channel Waiting Line Model with Poisson
Arrivals and Constant Service Times
Multiple-Channel Waiting Line Model with Poisson
Arrivals and Exponential Service Times
Economic Analysis of Waiting Lines
Slide 8
Structure of a Waiting Line System
Queuing theory is the study of waiting lines.
Four characteristics of a queuing system are:
• the manner in which customers arrive
• the time required for service
• the priority determining the order of service
• the number and configuration of servers in the
system.
Slide 9
Structure of a Waiting Line System
Distribution of Arrivals
• Generally, the arrival of customers into the system is
a random event.
• Frequently the arrival pattern is modeled as a
Poisson process
Distribution of Service Times
• Service time is also usually a random variable.
• A distribution commonly used to describe service
time is the exponential distribution.
Queue Discipline
• Most common queue discipline is first come, first
served (FCFS).
• What is the queue discipline in elevators?
Slide 10
Structure of a Waiting Line System
Single Service Channel
Customer
arrives
Waiting line
Multiple Service Channels
System
S1
Customer
leaves
System
S1
Customer
arrives
Waiting line
S2
Customer
leaves
S3
Slide 11
When a business like a restaurant opens in the
morning, no customers are in the restaurant.
state.
The beginning or start-up period is referred to as the
transient period.
The transient period ends when the system reaches
Waiting line/Queueing models describe the steadystate operating characteristics of a waiting line.
Slide 12
Queuing Systems
A three part code of the form A/B/k is used to
describe various queuing systems.
A identifies the arrival distribution, B the service
(departure) distribution, and k the number of
identical servers for the system.
Symbols used for the arrival and service processes
are: M - Markov distributions
(Poisson/exponential), D - Deterministic (constant)
and G - General distribution (with a known mean
and variance).
For example, M/M/k refers to a system in which
arrivals occur according to a Poisson distribution,
service times follow an exponential distribution
and there are k servers working at identical service
rates.
Slide 13
Analytical Formulas
When the queue discipline is FCFS, analytical formulas
have been derived for several different queuing models
including the following:
• M/M/1
• M/D/1
• M/M/k
Analytical formulas are not available for all possible
queuing systems. In this event, insights may be gained
through a simulation of the system.
Slide 14
Queuing Systems Assumptions
The arrival rate is l and arrival process is Poisson
There is one line/channel
The service rate, m, is per server (even for M/M/K).
The queue discipline is FCFS
Unlimited maximum queue length
Infinite calling population
Once the customers arrive they do not leave the
system until they are served
Slide 15
Queuing System Input Characteristics
l
1/l
µ
1/µ
=
=
=
=
=
the arrival rate
the average time between arrivals
the service rate for each server
the average service time
the standard deviation of the service time
Suppose the arrival rate, l, is 6 per hour.
What is the average time between arrivals?
Slide 16
Relationship between L and Lq and W and Wq.
Single Service Channel
Customer
arrives
System
S1
Customer
leaves
How many customers are waiting in the queue?
How many customers are in the system?
Suppose a customer waits for 10 minutes before she is
served and the service time takes another 5 minutes.
What is the waiting time in the queue?
What is the waiting time in the system?
Slide 17
Queuing System Operating Characteristics
P0 = probability the service facility is idle
Pn = probability of n units in the system
Pw = probability an arriving unit must wait for
service
Lq = average number of units in the queue
awaiting service
L = average number of units in the system
Wq = average time a unit spends in the queue
awaiting service
W = average time a unit spends in the system
Slide 18
M/M/1 Operating Characteristics
P0 = 1 – l/m
Pn = (l/m)n P0 = (l/m)n (1 – l/m)
Pw = l/m
Lq = l2 /{m(m – l)}
L = Lq + l/m = l /(m – l)
Wq = Lq/l = l /{m(m – l)}
W = Wq + 1/m = 1 /(m – l)
Slide 19
Some General Relationships for Waiting Line
Models (M/M/1, M/D/1, and M/M/K)
Little's flow equations are:
L = lW and Lq = lWq
Little’s flow equations show how operating
characteristics L, Lq, W, and Wq are related in any
waiting line system. Arrivals and service times do
not have to follow specific probability distributions
for the flow equations to be applicable.
Slide 20
Single-Channel Waiting Line Model
M/M/1 queuing system
Number of channels =
Arrival process =
Service-time distribution =
Queue length =
Calling population =
Customer leave the system without service?
Examples:
• Single-window theatre ticket sales booth
• Single-scanner airport security station
Slide 21
Example: SJJT, Inc. (A)
M/M/1 Queuing System
Joe Ferris is a stock trader on the floor of the New
York Stock Exchange for the firm of Smith, Jones,
Johnson, and Thomas, Inc. Daily stock transactions
arrive at Joe’s desk at a rate of 20 per hour, Poisson
distributed. Each order received by Joe requires an
average of two minutes to process, exponentially
distributed. Joe processes these transactions in FCFS
order.
Slide 22
Example: SJJT, Inc. (A)
What is the probability that an arriving order does
not have to wait to be processed?
What percentage of the time is Joe processing orders?
Slide 23
Example: SJJT, Inc. (A)
What is the probability that Joe has exactly 3 orders
waiting to be processed?
What is the probability that Joe has at least 2 orders in
the system?
Slide 24
Example: SJJT, Inc. (A)
What is the average time an order must wait from the
time Joe receives the order until it is finished being
processed (i.e. its turnaround time)?
What is the average time an order must wait from
before Joe starts processing it?
Slide 25
Example: SJJT, Inc. (A)
What is the average number of orders Joe has waiting
to be processed?
What is the average number of orders in the system?
Slide 26
Single-Channel Waiting Line Model with
Poisson Arrivals and Constant Service Times
M/D/1 queuing system
Single channel
Poisson arrival-rate distribution
Constant service time
Unlimited maximum queue length
Infinite calling population
Examples:
• Single-booth automatic car wash
• Coffee vending machine
Slide 27
M/D/1 Operating Characteristics
P0 = 1 – l/m
Pw = l/m
Lq = l2 /{2m(m – l)}
L = Lq + l/m
Wq = Lq/l = l /{2m(m – l)}
W = Wq + 1/m
Slide 28
Example: SJJT, Inc. (B)
M/D/1 Queuing System
The New York Stock Exchange the firm of Smith,
Jones, Johnson, and Thomas, Inc. now has an
opportunity to purchase a new machine that can
process the transactions in exactly 2 minutes. Instead
of using Joe, the company would like to evaluate the
impact of using the new machine. Daily stock
transactions still arrive at a rate of 20 per hour,
Poisson distributed.
Slide 29
Example: SJJT, Inc. (B)
What is the average time an order must wait from the
time the order arrives until it is finished being
processed (i.e. its turnaround time)?
What is the average time an order must wait from
before machine starts processing it?
Slide 30
Example: SJJT, Inc. (B)
What is the average number of orders waiting to be
processed?
What is the average number of orders in the system?
Slide 31
Improving the Waiting Line Operation
Waiting line models often indicate when improvements
in operating characteristics are desirable.
To make improvements in the waiting line operation,
analysts often focus on ways to improve the service
rate by:
- Increasing the service rate by making a creative
design change or by using new technology.
- Adding one or more service channels so that more
customers can be served simultaneously.
Slide 32
Multiple-Channel Waiting Line Model with
Poisson Arrivals and Exponential Service Times
M/M/k queuing system
Multiple channels (with one central waiting line)
Poisson arrival-rate distribution
Exponential service-time distribution
Unlimited maximum queue length
Infinite calling population
Examples:
• Four-teller transaction counter in bank
• Two-clerk returns counter in retail store
Slide 33
M/M/k Example: SJJT, Inc. (C)
M/M/2 Queuing System
Smith, Jones, Johnson, and Thomas, Inc. has
begun a major advertising campaign which it
believes will increase its business 50%. To handle the
increased volume, the company has hired an
the same speed as Joe Ferris.
Note that the new arrival rate of orders, l , is
50% higher than that of problem (A). Thus, l =
1.5(20) = 30 per hour.
Slide 34
M/M/k Example: SJJT, Inc. (C)
Sufficient Service Rate: l > km
Question
Will Joe Ferris alone not be able to handle the
increase in orders?
Since Joe Ferris processes orders at a mean rate of
µ = 30 per hour, then l = µ = 30 and the utilization
factor is 1.
This implies the queue of orders will grow
infinitely large. Hence, Joe alone cannot handle this
increase in demand.
Slide 35
M/M/k Example: SJJT, Inc. (C)
Probability of No Units in System (continued)
P0 =
1
n
k 1 (l
/ m)
n= 0
n!
(l / m )
k!
k
(
km
km l
)
Given that l = 30, µ = 30, k = 2 and (l /µ) = 1, the
probability that neither Joe nor Fred will be working is:
=
What is the probability that neither Joe nor Fred will be
working on an order at any point in time?
Slide 36
M/M/k Example: SJJT, Inc. (C)
Probability of n Units in System
Pn =
Pn =
(l / m )
n
P0 _ for _ n k
n!
(l / m )
k! k
n
( nk )
P0 _ for _ n > k
Slide 37
Example: SJJT, Inc. (C)
Average Length of the Queue
Lq =
lm (l m )
k
( k 1)!( k m l )
2
( P0 ) =
(30)(30)(30 30)
(1!)(2(30) 30)
2
2
(1/3) =
1
3
The average number of orders waiting to be filled
with both Joe and Fred working is 1/3.
Average Length of the system
L = Lq + (l /µ) =
Slide 38
Example: SJJT, Inc. (C)
Average Time in Queue
Wq = Lq /l =
Average Time in System
W = L/l =
Question
What is the average turnaround time for an order
with both Joe and Fred working?
Slide 39
Example: SJJT, Inc. (C)
Economic Analysis of Queuing Systems
The advertising campaign of Smith, Jones,
Johnson and Thomas, Inc. (see problems (A) and (B))
was so successful that business actually doubled.
The mean rate of stock orders arriving at the
exchange is now 40 per hour and the company must
decide how many floor traders to employ. Each floor
trader hired can process an order in an average time
of 2 minutes.
Slide 40
Example: SJJT, Inc. (C)
Economic Analysis of Queuing Systems
Based on a number of factors the brokerage firm
has determined the average waiting cost per minute
for an order to be \$.50. Floor traders hired will earn
\$20 per hour in wages and benefits. Using this
information compare the total hourly cost of hiring 2
Slide 41
Economic Analysis of Waiting Lines
The total cost model includes the cost of waiting and
the cost of service.
TC = cwL csk
where:
cw = the waiting cost per time period for each unit
L = the average number of units in the system
cs = the service cost per time period for each channel
k = the number of channels
TC = the total cost per time period
Slide 42
Example: SJJT, Inc. (C)
Economic Analysis of Waiting Lines
Total Hourly Cost
= (Total hourly cost for orders in the system)
+ (Total salary cost per hour)
= (\$30 waiting cost per hour)
x (Average number of orders in the system)
= 30L + 20k
Thus, L must be determined for k = 2 traders and
for k = 3 traders with l = 40/hr. and m = 30/hr. (since
the average service time is 2 minutes (1/30 hr.).
Slide 43
Example: SJJT, Inc. (C)
Cost of Two Servers
P0 =
1
n
k 1 (l
/ m)
n= 0
n!
(l / m )
k!
k
(
km
km l
)
P0 = 1 / [1+(1/1!)(40/30)]+[(1/2!)(40/30)2(60/(60-40))]
= 1 / [1 + (4/3) + (8/3)]
= 1/5
Slide 44
Example: SJJT, Inc. (C)
Cost of Two Servers (continued)
Thus,
Lq =
lm (l m )
k
( k 1)!( k m l )
2
( P0 ) =
(40)(30)(40 30)
(1!)(2(30) 40)
2
2
(1/5) =
16
15
L = Lq + (l /µ) = 16/15 + 4/3 = 2.40
Total Cost = 30(2.40) + (20)(2) =
\$112.00 per hour
Slide 45
Example: SJJT, Inc. (C)
Cost of Three Servers
P0 =
1
n
k 1 (l
/ m)
n= 0
n!
(l / m )
k!
k
(
km
km l
)
P0 =
Slide 46
Example: SJJT, Inc. (C)
Cost of Three Servers (continued)
Lq =
lm (l m )
k
( k 1)!( k m l )
2
( P0 ) =
(30)(40)(40 30)
(2!)(3(30) 40)
3
2
(15/59) = .1446
Thus, L = .1446 + 40/30 = 1.4780
Total Cost = 30(1.4780) + (20)(3) = \$104.35 per hour
Slide 47
Example: SJJT, Inc. (C)
System Cost Comparison
Waiting
Cost/Hr
\$82.00
44.35
Wage
Cost/Hr
\$40.00
60.00
Total
Cost/Hr
\$112.00
104.35
Thus, the cost of having 3 traders is less than that
|
# APEX Calculus
## Section6.4Trigonometric Substitution
In Section 5.2 we defined the definite integral as the “signed area under the curve.” In that section we had not yet learned the Fundamental Theorem of Calculus, so we only evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate
$$\int_{-3}^3\sqrt{9-x^2}\, dx = \frac{9\pi}{2}\tag{6.4.1}$$
as we recognized that $$f(x) = \sqrt{9-x^2}$$ described the upper half of a circle with radius 3.
We have since learned a number of integration techniques, including Substitution and Integration by Parts, yet we are still unable to evaluate the above integral without resorting to a geometric interpretation. This section introduces Trigonometric Substitution, a method of integration that fills this gap in our integration skill. This technique works on the same principle as Substitution as found in Section 6.1, though it can feel “backward.” In Section 6.1, we set $$u=f(x)\text{,}$$ for some function $$f\text{,}$$ and replaced $$f(x)$$ with $$u\text{.}$$ In this section, we will set $$x=f(\theta)\text{,}$$ where $$f$$ is a trigonometric function, then replace $$x$$ with $$f(\theta)\text{.}$$
We start by demonstrating this method in evaluating the integral in (6.4.1). After the example, we will generalize the method and give more examples.
### Example6.4.2.Using Trigonometric Substitution.
Evaluate $$\ds \int_{-3}^3\sqrt{9-x^2}\, dx\text{.}$$
Solution 1. Video solution
Solution 2.
We begin by noting that $$9\left(\sin^2(\theta) + \cos^2(\theta)\right) = 9\text{,}$$ and hence $$9\cos^2(\theta) = 9-9\sin^2(\theta)\text{.}$$ If we let $$x=3\sin(\theta)\text{,}$$ then $$9-x^2 = 9-9\sin^2(\theta) = 9\cos^2(\theta)\text{.}$$
Setting $$x=3\sin(\theta)$$ gives $$dx = 3\cos(\theta) \, d\theta\text{.}$$ We are almost ready to substitute. We also wish to change our bounds of integration. The bound $$x=-3$$ corresponds to $$\theta = -\pi/2$$ (for when $$\theta = -\pi/2\text{,}$$ $$x=3\sin(\theta) = -3$$). Likewise, the bound of $$x=3$$ is replaced by the bound $$\theta = \pi/2\text{.}$$ Thus
\begin{align*} \int_{-3}^3\sqrt{9-x^2}\, dx \amp = \int_{-\pi/2}^{\pi/2} \sqrt{9-9\sin^2(\theta) }\,(3\cos(\theta) )\, d\theta\\ \amp = \int_{-\pi/2}^{\pi/2} 3\sqrt{9\cos^2(\theta) } \cos(\theta) \, d\theta\\ \amp =\int_{-\pi/2}^{\pi/2} 3\abs{3\cos(\theta) } \cos(\theta) \, d\theta\text{.} \end{align*}
On $$[-\pi/2,\pi/2]\text{,}$$ $$\cos(\theta)$$ is always positive, so we can drop the absolute value bars, then employ a power-reducing formula:
\begin{align*} \int_{-3}^3\sqrt{9-x^2}\, dx \amp = \int_{-\pi/2}^{\pi/2} 9\cos^2(\theta) \, d\theta\\ \amp = \int_{-\pi/2}^{\pi/2} \frac{9}{2}\big(1+\cos(2\theta)\big)\, d\theta\\ \amp = \left.\frac92 \big(\theta +\frac12\sin(2\theta)\big)\right|_{-\pi/2}^{\pi/2}\\ \amp = \frac92\pi\text{.} \end{align*}
This matches our answer from before.
We now describe in detail Trigonometric Substitution. This method excels when dealing with integrands that contain $$\sqrt{a^2-x^2}\text{,}$$ $$\sqrt{x^2-a^2}$$ and $$\sqrt{x^2+a^2}\text{.}$$ The following Key Idea outlines the procedure for each case, followed by more examples. Each right triangle acts as a reference to help us understand the relationships between $$x$$ and $$\theta\text{.}$$
### Key Idea6.4.3.Trigonometric Substitution.
1. Integrands containing $$\sqrt{a^2-x^2}$$.
Let $$x=a\sin(\theta)\text{,}$$ $$dx = a\cos(\theta) \, d\theta\text{.}$$ Thus $$\theta = \sin^{-1}(x/a)\text{,}$$ for $$-\pi/2\leq \theta\leq \pi/2\text{.}$$ On this interval, $$\cos(\theta) \geq 0\text{,}$$ so $$\sqrt{a^2-x^2} = a\cos(\theta)\text{.}$$
2. Integrands containing $$\sqrt{x^2+a^2}$$.
Let $$x=a\tan(\theta)\text{,}$$ $$dx = a\sec^2(\theta) \, d\theta\text{.}$$ Thus $$\theta = \tan^{-1}(x/a)\text{,}$$ for $$-\pi/2 \lt \theta \lt \pi/2\text{.}$$ On this interval, $$\sec(\theta) \gt 0\text{,}$$ so $$\sqrt{x^2+a^2} = a\sec(\theta)\text{.}$$
3. Integrands containing $$\sqrt{x^2-a^2}$$.
Let $$x=a\sec(\theta)\text{,}$$ $$dx = a\sec(\theta) \tan(\theta) \, d\theta\text{.}$$ Thus $$\theta = \sec^{-1}(x/a)\text{.}$$ If $$x/a\geq 1\text{,}$$ then $$0\leq\theta\lt \pi/2\text{;}$$ if $$x/a \leq -1\text{,}$$ then $$\pi/2\lt \theta\leq \pi\text{.}$$ We restrict our work to where $$x\geq a\text{,}$$ so $$x/a\geq 1\text{,}$$ and $$0\leq\theta\lt \pi/2\text{.}$$ On this interval, $$\tan(\theta) \geq 0\text{,}$$ so $$\sqrt{x^2-a^2} = a\tan(\theta)\text{.}$$
### Example6.4.7.Using Trigonometric Substitution.
Evaluate $$\ds \int \frac{1}{\sqrt{5+x^2}}\, dx\text{.}$$
Solution 1. Video solution
Solution 2.
Using Item 2Key Idea 6.4.3, we recognize $$a=\sqrt{5}$$ and set $$x= \sqrt{5}\tan(\theta)\text{.}$$ This makes $$dx = \sqrt{5}\sec^2(\theta) \, d\theta\text{.}$$ We will use the fact that $$\sqrt{5+x^2} = \sqrt{5+5\tan^2(\theta) } = \sqrt{5\sec^2(\theta) } = \sqrt{5}\sec(\theta)\text{.}$$ Substituting, we have:
\begin{align*} \int \frac{1}{\sqrt{5+x^2}}\, dx \amp = \int \frac{1}{\sqrt{5+5\tan^2(\theta) }}\sqrt{5}\sec^2(\theta) \, d\theta\\ \amp = \int \frac{\sqrt{5}\sec^2(\theta) }{\sqrt{5}\sec(\theta) } \, d\theta\\ \amp = \int \sec(\theta) \, d\theta\\ \amp = \ln\abs{\sec(\theta) +\tan(\theta) }+C\text{.} \end{align*}
While the integration steps are over, we are not yet done. The original problem was stated in terms of $$x\text{,}$$ whereas our answer is given in terms of $$\theta\text{.}$$ We must convert back to $$x\text{.}$$
The reference triangle given in Key Idea 6.4.3(b) helps. With $$x=\sqrt{5}\tan(\theta)\text{,}$$ we have
\begin{equation*} \tan(\theta) = \frac x{\sqrt{5}} \text{ and } \sec(\theta) = \frac{\sqrt{x^2+5}}{\sqrt{5}}\text{.} \end{equation*}
This gives
\begin{align*} \int \frac{1}{\sqrt{5+x^2}}\, dx \amp = \ln\abs{\sec(\theta) +\tan(\theta) }+C\\ \amp = \ln\abs{\frac{\sqrt{x^2+5}}{\sqrt{5}}+ \frac x{\sqrt{5}}}+C\text{.} \end{align*}
We can leave this answer as is, or we can use a logarithmic identity to simplify it. Note:
\begin{align*} \ln\abs{\frac{\sqrt{x^2+5}}{\sqrt{5}}+ \frac x{\sqrt{5}}}+C \amp = \ln\abs{\frac{1}{\sqrt{5}}\big(\sqrt{x^2+5}+ x\big)}+C\\ \amp = \ln\abs{\frac{1}{\sqrt{5}}} + \ln\abs{\sqrt{x^2+5}+ x}+C\\ \amp = \ln\abs{\sqrt{x^2+5}+ x}+C\text{,} \end{align*}
where the $$\ln\big(1/\sqrt{5}\big)$$ term is absorbed into the constant $$C\text{.}$$ (In Section 6.6 we will learn another way of approaching this problem.)
### Example6.4.8.Using Trigonometric Substitution.
Evaluate $$\ds \int \sqrt{4x^2-1}\, dx\text{.}$$
Solution 1. Video solution
Solution 2.
We start by rewriting the integrand so that it looks like $$\sqrt{x^2-a^2}$$ for some value of $$a\text{:}$$
\begin{align*} \sqrt{4x^2-1} \amp = \sqrt{4\left(x^2-\frac14\right)}\\ \amp = 2\sqrt{x^2-\left(\frac12\right)^2}\text{.} \end{align*}
So we have $$a=1/2\text{,}$$ and following Key Idea 6.4.3(c), we set $$x= \frac12\sec(\theta)\text{,}$$ and hence $$dx = \frac12\sec(\theta) \tan(\theta) \, d\theta\text{.}$$ We now rewrite the integral with these substitutions:
\begin{align*} \int \sqrt{4x^2-1}\, dx \amp = \int 2\sqrt{x^2-\left(\frac12\right)^2}\, dx\\ \amp = \int 2\sqrt{\frac14\sec^2(\theta) - \frac14}\left(\frac12\sec(\theta) \tan(\theta) \right)\, d\theta\\ \amp =\int \sqrt{\frac14(\sec^2(\theta) -1)}\Big(\sec(\theta) \tan(\theta) \Big)\, d\theta\\ \amp =\int\sqrt{\frac14\tan^2(\theta) }\Big(\sec(\theta) \tan(\theta) \Big)\, d\theta\\ \amp =\int \frac12\tan^2(\theta) \sec(\theta) \, d\theta\\ \amp =\frac12\int \Big(\sec^2(\theta) -1\Big)\sec(\theta) \, d\theta\\ \amp =\frac12\int \big(\sec^3(\theta) - \sec(\theta) \big)\, d\theta\text{.} \end{align*}
We integrated $$\sec^3(\theta)$$ in Example 6.3.11, finding its antiderivatives to be
\begin{equation*} \int \sec^3(\theta) \, d\theta = \frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C\text{.} \end{equation*}
Thus
\begin{align*} \amp \int \sqrt{4x^2-1}\, dx =\frac12\int \big(\sec^3(\theta) - \sec(\theta) \big)\, d\theta\\ \amp = \frac12\left(\frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big) -\ln\abs{\sec(\theta) + \tan(\theta) }\right) + C\\ \amp = \frac14\left(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\right)+C\text{.} \end{align*}
We are not yet done. Our original integral is given in terms of $$x\text{,}$$ whereas our final answer, as given, is in terms of $$\theta\text{.}$$ We need to rewrite our answer in terms of $$x\text{.}$$ With $$a=1/2\text{,}$$ and $$x=\frac12\sec(\theta)\text{,}$$ the reference triangle in Key Idea 6.4.3(c) shows that
\begin{equation*} \tan(\theta) = \sqrt{x^2-1/4}\Big/(1/2) = 2\sqrt{x^2-1/4} \text{ and } \sec(\theta) = 2x\text{.} \end{equation*}
Thus
\begin{align*} \amp \frac14\Big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C\\ \amp\quad = \frac14\Big(2x\cdot 2\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C\\ \amp\quad = \frac14\Big(4x\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C\text{.} \end{align*}
The final answer is given in the last line above, repeated here:
\begin{equation*} \int \sqrt{4x^2-1}\, dx = \frac14\Big(4x\sqrt{x^2-1/4} - \ln\abs{2x + 2\sqrt{x^2-1/4}}\Big)+C\text{.} \end{equation*}
### Example6.4.9.Using Trigonometric Substitution.
Evaluate $$\ds \int \frac{\sqrt{4-x^2}}{x^2}\, dx\text{.}$$
Solution 1. Video solution
Solution 2.
We use Key Idea 6.4.3(a) with $$a=2\text{,}$$ $$x=2\sin(\theta)\text{,}$$ $$dx = 2\cos(\theta)$$ and hence $$\sqrt{4-x^2} = 2\cos(\theta)\text{.}$$ This gives
\begin{align*} \int \frac{\sqrt{4-x^2}}{x^2}\, dx \amp = \int \frac{2\cos(\theta) }{4\sin^2(\theta) }(2\cos(\theta) )\, d\theta\\ \amp = \int \cot^2(\theta) \, d\theta\\ \amp = \int (\csc^2(\theta) -1)\, d\theta\\ \amp = -\cot(\theta) -\theta + C\text{.} \end{align*}
We need to rewrite our answer in terms of $$x\text{.}$$ Using the reference triangle found in Key Idea 6.4.3(a), we have $$\cot(\theta) = \sqrt{4-x^2}/x$$ and $$\theta = \sin^{-1}(x/2)\text{.}$$ Thus
\begin{equation*} \int \frac{\sqrt{4-x^2}}{x^2}\, dx = -\frac{\sqrt{4-x^2}}x-\sin^{-1}\left(\frac x2\right) + C\text{.} \end{equation*}
Trigonometric Substitution can be applied in many situations, even those not of the form $$\sqrt{a^2-x^2}\text{,}$$ $$\sqrt{x^2-a^2}$$ or $$\sqrt{x^2+a^2}\text{.}$$ In the following example, we apply it to an integral we already know how to handle.
### Example6.4.10.Using Trigonometric Substitution.
Evaluate $$\ds \int\frac1{x^2+1}\, dx\text{.}$$
Solution.
We know the answer already as $$\tan^{-1}(x) +C\text{.}$$ We apply Trigonometric Substitution here to show that we get the same answer without inherently relying on knowledge of the derivative of the arctangent function.
Using Key Idea 6.4.3(b), let $$x=\tan(\theta)\text{,}$$ $$dx=\sec^2(\theta) \, d\theta$$ and note that $$x^2+1 = \tan^2(\theta) +1 = \sec^2(\theta)\text{.}$$ Thus
\begin{align*} \int \frac1{x^2+1}\, dx \amp = \int \frac{1}{\sec^2(\theta) }\sec^2(\theta) \, d\theta\\ \amp = \int 1\, d\theta\\ \amp = \theta + C\text{.} \end{align*}
Since $$x=\tan(\theta)\text{,}$$ $$\theta = \tan^{-1}(x)\text{,}$$ and we conclude that $$\ds \int\frac1{x^2+1}\, dx = \tan^{-1}(x) +C\text{.}$$
The next example is similar to the previous one in that it does not involve a square-root. It shows how several techniques and identities can be combined to obtain a solution.
### Example6.4.11.Using Trigonometric Substitution.
Evaluate $$\ds\int\frac1{(x^2+6x+10)^2}\, dx\text{.}$$
Solution 1. Video solution
Solution 2.
We start by completing the square, then make the substitution $$u=x+3\text{,}$$ followed by the trigonometric substitution of $$u=\tan(\theta)\text{:}$$
\begin{align} \int \frac1{(x^2+6x+10)^2}\, dx =\int \frac1{\big((x+3)^2+1\big)^2}\, dx\amp = \int \frac1{(u^2+1)^2}\,du.\notag\\ \end{align}
Now make the substitution $$u=\tan(\theta)\text{,}$$ $$du=\sec^2(\theta) \, d\theta\text{:}$$
\begin{align} \amp = \int \frac1{(\tan^2(\theta) +1)^2}\sec^2(\theta) \, d\theta\notag\\ \amp = \int\frac 1{(\sec^2(\theta) )^2}\sec^2(\theta) \, d\theta\notag\\ \amp = \int \cos^2(\theta) \, d\theta.\notag\\ \end{align}
Applying a power reducing formula, we have
\begin{align} \amp = \int \left(\frac12 +\frac12\cos(2\theta)\right)\, d\theta\notag\\ \amp = \frac12\theta + \frac14\sin(2\theta) + C\text{.}\tag{6.4.2} \end{align}
We need to return to the variable $$x\text{.}$$ As $$u=\tan(\theta)\text{,}$$ $$\theta = \tan^{-1}(u)\text{.}$$ Using the identity $$\sin(2\theta) = 2\sin(\theta) \cos(\theta)$$ and using the reference triangle found in Key Idea 6.4.3(b), we have
\begin{equation*} \frac14\sin(2\theta) = \frac12\frac u{\sqrt{u^2+1}}\cdot\frac 1{\sqrt{u^2+1}} = \frac12\frac u{u^2+1}\text{.} \end{equation*}
Finally, we return to $$x$$ with the substitution $$u=x+3\text{.}$$ We start with the expression in Equation (6.4.2):
\begin{align*} \frac12\theta + \frac14\sin(2\theta) + C \amp = \frac12\tan^{-1}(u) + \frac12\frac{u}{u^2+1}+C\\ \amp = \frac12\tan^{-1}(x+3) + \frac{x+3}{2(x^2+6x+10)}+C\text{.} \end{align*}
Stating our final result in one line,
\begin{equation*} \int\frac1{(x^2+6x+10)^2}\, dx=\frac12\tan^{-1}(x+3) + \frac{x+3}{2(x^2+6x+10)}+C\text{.} \end{equation*}
Our last example returns us to definite integrals, as seen in our first example. Given a definite integral that can be evaluated using Trigonometric Substitution, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms of $$x$$ to one in terms of $$\theta\text{,}$$ then converting back to $$x$$) and then evaluate using the original bounds. It is much more straightforward, though, to change the bounds as we substitute.
### Example6.4.12.Definite integration and Trigonometric Substitution.
Evaluate $$\ds\int_0^5\frac{x^2}{\sqrt{x^2+25}}\, dx\text{.}$$
Solution 1. Video solution
Solution 2.
Using Key Idea 6.4.3(b), we set $$x=5\tan(\theta)\text{,}$$ $$dx = 5\sec^2(\theta) \, d\theta\text{,}$$ and note that $$\sqrt{x^2+25} = 5\sec(\theta)\text{.}$$ As we substitute, we can also change the bounds of integration.
The lower bound of the original integral is $$x=0\text{.}$$ As $$x=5\tan(\theta)\text{,}$$ we solve for $$\theta$$ and find $$\theta = \tan^{-1}(x/5)\text{.}$$ Thus the new lower bound is $$\theta = \tan^{-1}(0) = 0\text{.}$$ The original upper bound is $$x=5\text{,}$$ thus the new upper bound is $$\theta = \tan^{-1}(5/5) = \pi/4\text{.}$$
Thus we have
\begin{align*} \int_0^5\frac{x^2}{\sqrt{x^2+25}}\, dx \amp = \int_0^{\pi/4} \frac{25\tan^2(\theta) }{5\sec(\theta) }5\sec^2(\theta) \, d\theta\\ \amp = 25\int_0^{\pi/4} \tan^2(\theta) \sec(\theta) \, d\theta\text{.} \end{align*}
We encountered this indefinite integral in Example 6.4.8 where we found
\begin{equation*} \int \tan^2(\theta) \sec(\theta) \, d\theta = \frac12\big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\big)\text{.} \end{equation*}
So
\begin{align*} 25\int_0^{\pi/4} \tan^2(\theta) \sec(\theta) \, d\theta \amp = \left.\frac{25}2\big(\sec(\theta) \tan(\theta) -\ln\abs{\sec(\theta) +\tan(\theta) }\big)\right|_0^{\pi/4}\\ \amp = \frac{25}2\big(\sqrt2-\ln(\sqrt2+1)\big)\\ \amp \approx 6.661\text{.} \end{align*}
The following equalities are very useful when evaluating integrals using Trigonometric Substitution.
### Key Idea6.4.13.Useful Equalities with Trigonometric Substitution.
1. $$\displaystyle \sin(2\theta) = 2\sin(\theta) \cos(\theta)$$
2. $$\displaystyle \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) -1 = 1-2\sin^2(\theta)$$
3. $$\displaystyle \ds \int \sec^3(\theta) \, d\theta = \frac12\Big(\sec(\theta) \tan(\theta) + \ln\abs{\sec(\theta) +\tan(\theta) }\Big)+C$$
4. $$\ds \int \cos^2(\theta) \, d\theta = \int \frac12\big(1+\cos(2\theta)\big)\, d\theta = \frac12\big(\theta+\sin(\theta) \cos(\theta) \big)+C\text{.}$$
The next section introduces Partial Fraction Decomposition, which is an algebraic technique that turns “complicated” fractions into sums of “simpler” fractions, making integration easier.
### ExercisesExercises
#### Terms and Concepts
##### 1.
Trigonometric Substitution works on the same principles as Integration by Substitution, though it can feel “ ”.
##### 2.
If one uses Trigonometric Substitution on an integrand containing $${\sqrt{16-x^{2}}}\text{,}$$ then one should set $$x={}$$.
##### 3.
Consider the Pythagorean Identity $$\sin^2(\theta) +\cos^2(\theta) = 1\text{.}$$
1. What identity is obtained when both sides are divided by $$\cos^2(\theta)\text{?}$$
2. Use the new identity to simplify $$9\tan^2(\theta) + 9\text{.}$$
##### 4.
Why does Key Idea 6.4.3(a) state that $$\sqrt{a^2-x^2} = a\cos(\theta)\text{,}$$ and not $$\abs{a\cos(\theta) }\text{?}$$
#### Problems
##### Exercise Group.
Apply Trigonometric Substitution to evaluate the indefinite integral.
###### 5.
$$\ds \int \sqrt{x^2+1}\, dx$$
###### 6.
$$\ds \int \sqrt{x^2+4}\, dx$$
###### 7.
$$\ds \int \sqrt{1-x^2}\, dx$$
###### 8.
$$\ds \int \sqrt{9-x^2}\, dx$$
###### 9.
$$\ds \int \sqrt{x^2-1}\, dx$$
###### 10.
$$\ds \int \sqrt{x^2-16}\, dx$$
###### 11.
$$\ds \int {\sqrt{16x^{2}+1}}\, dx$$
###### 12.
$$\ds \int {\sqrt{1-25x^{2}}}\, dx$$
###### 13.
$$\ds \int {\sqrt{36x^{2}-1}}\, dx$$
###### 14.
$$\ds \int {\frac{6}{\sqrt{x^{2}+2}}}\, dx$$
###### 15.
$$\ds \int {\frac{7}{\sqrt{13-x^{2}}}}\, dx$$
###### 16.
$$\ds \int {\frac{8}{\sqrt{x^{2}-6}}}\, dx$$
##### Exercise Group.
Evaluate the indefinite integral. Trigonometric Substitution may not be required.
###### 17.
$$\ds \int {\frac{\sqrt{x^{2}-15}}{x}}\, dx$$
###### 18.
$$\ds \int \frac {1}{(x^2+1)^2}\, dx$$
###### 19.
$$\ds \int {\frac{x}{\sqrt{x^{2}-3}}}\, dx$$
###### 20.
$$\ds \int x^2\sqrt{1-x^2}\, dx$$
###### 21.
$$\ds \int {\frac{x}{\left(x^{2}+25\right)^{\left({\frac{3}{2}}\right)}}}\, dx$$
###### 22.
$$\ds \int {\frac{6x^{2}}{\sqrt{x^{2}-5}}}\, dx$$
###### 23.
$$\ds \int {\frac{1}{\left(x^{2}-4x+85\right)^{2}}}\, dx$$
###### 24.
$$\ds \int x^2(1-x^2)^{-3/2}\, dx$$
###### 25.
$$\ds \int {\frac{\sqrt{5-x^{2}}}{8x^{2}}}\, dx$$
###### 26.
$$\ds \int {\frac{x^{2}}{\sqrt{x^{2}+15}}}\, dx$$
##### Exercise Group.
Evaluate the definite integral by making the proper trigonometric substitution and changing the bounds of integration. (Note: the corresponding indefinite integrals appeared previously in the Section 6.4 exercises.)
###### 27.
$$\ds \int_{-1}^1 \sqrt{1-x^2}\, dx$$
###### 28.
$$\ds \int_{4}^{6} \sqrt{x^2-16}\, dx$$
###### 29.
$$\ds \int_{0}^{3} \sqrt{x^2+4}\, dx$$
###### 30.
$$\ds \int_{-4}^{4} \frac1{(x^2+1)^2}\, dx$$
###### 31.
$$\ds \int_{-2}^{2} \sqrt{9-x^2}\, dx$$
###### 32.
$$\ds \int_{-1}^1 x^2\sqrt{1-x^2}\, dx$$
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# Video:How to Make Basic Algebraic Word Problems
Want to learn how to make basic algebraic word problems? Here, see tips and tricks for the best ways to compose them for students to practice.See Transcript
## Transcript:How to Make Basic Algebraic Word Problems
Hi, my name is Bassem Saad. I'm an associate math instructor and math Ph.D. candidate. I'm here today for About.com to show you how to make a basic algebra word problem. So a well-written word problem should model reality. At every step of the way of your creation of the word problem, you should ask yourself if the problem or the solution is reasonable.
### Steps for Making Basic Algebraic Word Problems
Let's begin with step one: Begin with a mathematical expression. For example, 12, minus two x, equals ten. Once we have our mathematical expression, we can move on to step two, using key words to write out the expression. So for this example, we'll say the difference of 12 and twice x, equals ten. Now we're ready for step three where you try to build a plausible story around the expression.
### Example of a Basic Algebraic Word Problem
For example, our mathematical expression has turned into the following word problem: Bill has 12 bananas and x apples. The difference between the number of bananas and twice the number of apples is ten. How many apples does Bill have? Notice that we've included the formula in the second sentence; but for this problem to make sense, apples has to be a positive whole number, which leads us to step four: We solve the problem and verify for consistency. So from the word problem we should get back our expression, and then we can solve our expression so that we see we have one apple. One apple makes sense because it's a positive whole number.
### Another Example of a Basic Algebraic Word Problem
So let's do one more example. We'll begin with the expression: 20x equals 450. Then we'll move on to step two: re-writing the expression in words using key words. The product of 20 and x, equals 450. Now we're ready to move on to a plausible story. Sam has a rectangular garden with an area of 450 square feet. The length of the garden is 20 feet and the width of the garden is x feet. Remember that the area of a rectangle is the product of the length and the width. Find the width of Sam's garden. So now we're ready to solve and verify. From our word problem, we retrieve our original mathematical expression; that is 20x equals 450. From this mathematical expression, we can divide 20 into both sides to find out that x equals 22-and-a-half feet. So now we've seen how to construct basic algebraic word problems.
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Definitions
# trigonometry
[trig-uh-nom-i-tree]
trigonometry [Gr.,=measurement of triangles], a specialized area of geometry concerned with the properties of and relations among the parts of a triangle. Spherical trigonometry is concerned with the study of triangles on the surface of a sphere rather than in the plane; it is of considerable importance in surveying, navigation, and astronomy.
## The Basic Trigonometric Functions
Trigomometry originated as the study of certain mathematical relations originally defined in terms of the angles and sides of a right triangle, i.e., one containing a right angle (90°). Six basic relations, or trigonometric functions, are defined.If A, B, and C are the measures of the angles of a right triangle (C=90°) and a, b, and c are the lengths of the respective sides opposite these angles, then the six functions are expressed for one of the acute angles, say A, as various ratios of the opposite side (a), the adjacent side (b), and the hypotenuse (c), as set out in the table.Although the actual lengths of the sides of a right triangle may have any values, the ratios of the lengths will be the same for all similar right triangles, large or small; these ratios depend only on the angles and not on the actual lengths. The functions occur in pairs—sine and cosine, tangent and cotangent, secant and cosecant—called cofunctions. In equations they are usually represented as sin, cos, tan, cot, sec, and csc. Since in ordinary (Euclidean) plane geometry the sum of the angles of a triangle is 180°, angles A and B must add up to 90° and therefore are complementary angles. From the definitions of the functions, it may be seen that sin B=cos A, cos B=sin A, tan B=cot A, and sec B=csc A; in general, the function of an angle is equal to the cofunction of its complement. Since the hypotenuse (c), is always the longest side of a right triangle, the values of the sine and cosine are always between zero and one, the values of the secant and cosecant are always equal to or greater than one, and the values of the tangent and cotangent are unbounded, increasing from zero without limit.
For certain special right triangles the values of the functions may be calculated easily; e.g., in a right triangle whose acute angles are 30° and 60° the sides are in the ratio 1 :3 : 2, so that sin 30°=cos 60°=1/2, cos 30°=sin 60°=3/2, tan 30°=cot 60°=1/3, cot 30°=tan 60°=3, sec 30°=csc 60°=2/3, and csc 30°=sec 60°=2. For other angles, the values of the trigonometric functions are usually found from a set of tables or a scientific calculator. For the limiting values of 0° and 90°, the length of one side of the triangle approaches zero while the other approaches that of the hypotenuse, resulting in the values sin 0°=cos 90°=0, cos 0°=sin 90°=1, tan 0°=cot 90°=0, and sec 0°=csc 90°=1; since division by zero is undefined, cot 0°, tan 90°, csc 0°, and sec 90° are all undefined, having infinitely large values.
A general triangle, not necessarily containing a right angle, can also be analyzed by means of trigonometry, and various relationships are found to exist between the sides and angles of the general triangle. For example, in any plane triangle a/sin A=b/sin B=c/sin C. This relationship is known as the Law of Sines. The related Law of Cosines holds that a2=b2+c2-2bc cosA and the Law of Tangents holds that (a-b)/(a+b)=[tan 1/2(A-B)]/[tan 1/2(A+B)]. Each of the trigonometric functions can be represented by an infinite series.
## Extension of the Trigonometric Functions
The notion of the trigonometric functions can be extended beyond 90° by defining the functions with respect to Cartesian coordinates. Let r be a line of unit length from the origin to the point P (x,y), and let θ be the angle r makes with the positive x-axis. The six functions become sin θ =y/r=y, cos θ=x/r=x, tan θ=y/x, cot θ=x/y, sec θ=r/x=1/x, and csc θ=r/y=1/y. As θ increases beyond 90°, the point P crosses the y-axis and x becomes negative; in quadrant II the functions are negative except for sin θ and csc θ. Beyond θ=180°, P is in quadrant III, y is also negative, and only tan θ and cot θ are positive, while beyond θ=270° P moves into quadrant IV, x becomes positive again, and cos θ and sec θ are positive.Since the positions of r for angles of 360° or more coincide with those already taken by r as θ increased from 0°, the values of the functions repeat those taken between 0° and 360° for angles greater than 360°, repeating again after 720°, and so on.This repeating, or periodic, nature of the trigonometric functions leads to important applications in the study of such periodic phenomena as light and electricity.
Mathematical discipline dealing with the relationships between the sides and angles of triangles. Literally, it means triangle measurement, though its applications extend far beyond geometry. It emerged as a rigorous discipline in the 15th century, when the demand for accurate surveying techniques and navigational methods led to its use for the “solution” of right triangles, or the calculation of the lengths of two sides of a right triangle given one of its acute angles and the length of one side. The solution can be found by using ratios in the form of the trigonometric functions.
Trigonometry (from Greek trigōnon "triangle" + metron "measure") is a branch of mathematics that deals with triangles, particularly those plane triangles in which one angle has 90 degrees (right triangles). Trigonometry deals with relationships between the sides and the angles of triangles and with the trigonometric functions, which describe those relationships.
Trigonometry has applications in both pure mathematics and in applied mathematics, where it is essential in many branches of science and technology. It is usually taught in secondary schools either as a separate course or as part of a precalculus course. Trigonometry is informally called "trig".
A branch of trigonometry, called spherical trigonometry, studies triangles on spheres, and is important in astronomy and navigation.
## History
Trigonometry was probably developed for use in sailing as a navigation method used with astronomy. The origins of trigonometry can be traced to the civilizations of ancient Egypt, Mesopotamia and the Indus Valley (India), more than 4000 years ago. The common practice of measuring angles in degrees, minutes and seconds comes from the Babylonian's base sixty system of numeration.
The first recorded use of trigonometry came from the Hellenistic mathematician Hipparchus circa 150 BC, who compiled a trigonometric table using the sine for solving triangles. Ptolemy further developed trigonometric calculations circa 100 AD.
The ancient Sinhalese in Sri Lanka, when constructing reservoirs in the Anuradhapura kingdom, used trigonometry to calculate the gradient of the water flow. Archeological research also provides evidence of trigonometry used in other unique hydrological structures dating back to 4 BC.
The Indian mathematician Aryabhata in 499, gave tables of half chords which are now known as sine tables, along with cosine tables. He used zya for sine, kotizya for cosine, and otkram zya for inverse sine, and also introduced the versine. Another Indian mathematician, Brahmagupta in 628, used an interpolation formula to compute values of sines, up to the second order of the Newton-Stirling interpolation formula.
In the 10th century, the Persian mathematician and astronomer Abul Wáfa introduced the tangent function and improved methods of calculating trigonometry tables. He established the angle addition identities, e.g. sin (a + b), and discovered the sine formula for spherical geometry:
$frac\left\{sin A\right\}\left\{sin a\right\} = frac\left\{sin B\right\}\left\{sin b\right\} = frac\left\{sin C\right\}\left\{sin c\right\}.$
Also in the late 10th and early 11th centuries, the Egyptian astronomer Ibn Yunus performed many careful trigonometric calculations and demonstrated the formula
$cos a cos b = frac\left\{cos\left(a+b\right) + cos\left(a-b\right)\right\}\left\{2\right\}.$
Persian mathematician Omar Khayyám (1048-1131) combined trigonometry and approximation theory to provide methods of solving algebraic equations by geometrical means. Khayyam solved the cubic equation $x^3 + 200 x = 20 x^2 + 2000$ and found a positive root of this cubic by considering the intersection of a rectangular hyperbola and a circle. An approximate numerical solution was then found by interpolation in trigonometric tables.
Detailed methods for constructing a table of sines for any angle were given by the Indian mathematician Bhaskara in 1150, along with some sine and cosine formulae. Bhaskara also developed spherical trigonometry.
The 13th century Persian mathematician Nasir al-Din Tusi, along with Bhaskara, was probably the first to treat trigonometry as a distinct mathematical discipline. Nasir al-Din Tusi in his Treatise on the Quadrilateral was the first to list the six distinct cases of a right angled triangle in spherical trigonometry.
In the 14th century, Persian mathematician al-Kashi and Timurid mathematician Ulugh Beg (grandson of Timur) produced tables of trigonometric functions as part of their studies of astronomy.
The mathematician Bartholemaeus Pitiscus published an influential work on trigonometry in 1595 which may have coined the word "trigonometry" itself.
## Overview
If one angle of a triangle is 90 degrees and one of the other angles is known, the third is thereby fixed, because the three angles of any triangle add up to 180 degrees. The two acute angles therefore add up to 90 degrees: they are complementary angles. The shape of a right triangle is completely determined, up to similarity, by the angles. This means that once one of the other angles is known, the ratios of the various sides are always the same regardless of the overall size of the triangle. These ratios are given by the following trigonometric functions of the known angle A, where a, b and c refer to the lengths of the sides in the accompanying figure:
• The sine function (sin), defined as the ratio of the side opposite the angle to the hypotenuse.
$sin A=frac\left\{textrm\left\{opposite\right\}\right\}\left\{textrm\left\{hypotenuse\right\}\right\}=frac\left\{a\right\}\left\{,c,\right\},.$
• The cosine function (cos), defined as the ratio of the adjacent leg to the hypotenuse.
$cos A=frac\left\{textrm\left\{adjacent\right\}\right\}\left\{textrm\left\{hypotenuse\right\}\right\}=frac\left\{b\right\}\left\{,c,\right\},.$
• The tangent function (tan), defined as the ratio of the opposite leg to the adjacent leg.
$tan A=frac\left\{textrm\left\{opposite\right\}\right\}\left\{textrm\left\{adjacent\right\}\right\}=frac\left\{a\right\}\left\{,b,\right\}=frac\left\{sin A\right\}\left\{cos A\right\},.$
The hypotenuse is the side opposite to the 90 degree angle in a right triangle; it is the longest side of the triangle, and one of the two sides adjacent to angle A. The adjacent leg is the other side that is adjacent to angle A. The opposite side is the side that is opposite to angle A. The terms perpendicular and base are sometimes used for the opposite and adjacent sides respectively. Many people find it easy to remember what sides of the right triangle are equal to sine, cosine, or tangent, by memorizing the word SOH-CAH-TOA (see below under Mnemonics).
The reciprocals of these functions are named the cosecant (csc or cosec), secant (sec) and cotangent (cot), respectively. The inverse functions are called the arcsine, arccosine, and arctangent, respectively. There are arithmetic relations between these functions, which are known as trigonometric identities.
With these functions one can answer virtually all questions about arbitrary triangles by using the law of sines and the law of cosines. These laws can be used to compute the remaining angles and sides of any triangle as soon as two sides and an angle or two angles and a side or three sides are known. These laws are useful in all branches of geometry, since every polygon may be described as a finite combination of triangles.
### Extending the definitions
The above definitions apply to angles between 0 and 90 degrees (0 and π/2 radians) only. Using the unit circle, one can extend them to all positive and negative arguments (see trigonometric function). The trigonometric functions are periodic, with a period of 360 degrees or 2π radians. That means their values repeat at those intervals.
The trigonometric functions can be defined in other ways besides the geometrical definitions above, using tools from calculus and infinite series. With these definitions the trigonometric functions can be defined for complex numbers. The complex function cis is particularly useful
$operatorname\left\{cis\right\},x = cos x + isin x ! = e^\left\{ix\right\}.$
See Euler's and De Moivre's formulas.
### Mnemonics
Students often use mnemonics to remember facts and relationships in trigonometry. For example, the sine, cosine, and tangent ratios in a right triangle can be remembered by representing them as strings of letters, as in SOH-CAH-TOA.
Sine = Opposite ÷ Hypotenuse
Alternatively, one can devise sentences which consist of words beginning with the letters to be remembered. For example, to recall that Tan = Opposite/Adjacent, the letters T-O-A must be remembered. Any memorable phrase constructed of words beginning with the letters T-O-A will serve.
It is of ethnographic interest to note that the mnemonic TOA-CAH-SOH can be translated in the local Singaporean Hokkien dialect to 'big-legged woman', serving as an additional learning aid for students in Singapore.
Another type of mnemonic describes facts in a simple, memorable way, such as "Plus to the right, minus to the left; positive height, negative depth," which refers to trigonometric functions generated by a revolving line.
### Calculating trigonometric functions
Trigonometric functions were among the earliest uses for mathematical tables. Such tables were incorporated into mathematics textbooks and students were taught to look up values and how to interpolate between the values listed to get higher accuracy. Slide rules had special scales for trigonometric functions.
Today scientific calculators have buttons for calculating the main trigonometric functions (sin, cos, tan and sometimes cis) and their inverses. Most allow a choice of angle measurement methods: degrees, radians and, sometimes, Grad. Most computer programming languages provide function libraries that include the trigonometric functions. The floating point unit hardware incorporated into the microprocessor chips used in most personal computers have built in instructions for calculating trigonometric functions.
## Applications of trigonometry
There are an enormous number of applications of trigonometry and trigonometric functions. For instance, the technique of triangulation is used in astronomy to measure the distance to nearby stars, in geography to measure distances between landmarks, and in satellite navigation systems. The sine and cosine functions are fundamental to the theory of periodic functions such as those that describe sound and light waves.
Fields which make use of trigonometry or trigonometric functions include astronomy (especially, for locating the apparent positions of celestial objects, in which spherical trigonometry is essential) and hence navigation (on the oceans, in aircraft, and in space), music theory, acoustics, optics, analysis of financial markets, electronics, probability theory, statistics, biology, medical imaging (CAT scans and ultrasound), pharmacy, chemistry, number theory (and hence cryptology), seismology, meteorology, oceanography, many physical sciences, land surveying and geodesy, architecture, phonetics, economics, electrical engineering, mechanical engineering, civil engineering, computer graphics, cartography, crystallography and game development.
## Common formulae
Certain equations involving trigonometric functions are true for all angles and are known as trigonometric identities. Many express important geometric relationships. For example, the Pythagorean identities are an expression of the Pythagorean Theorem. Here are some of the more commonly used identities, as well as the most important formulae connecting angles and sides of an arbitrary triangle. For more identities see trigonometric identity.
### Trigonometric identities
#### Pythagorean identities
begin\left\{align\right\}
sin^2 alpha + cos^2 alpha = 1 tan^2 alpha + 1 = sec^2 alpha 1+cot^2 alpha = csc^2 alpha end{align}
#### Sum and product identities
##### Sum to product
begin\left\{align\right\}
sin alpha pm sin beta &= 2sin left(frac{alpha pm beta}{2}right)cos left(frac{alpha mp beta}{2} right) cos alpha + cos beta &= 2cos left(frac{alpha + beta}{2} right)cos left(frac{alpha - beta}{2}right) cos alpha - cos beta &= -2sin left(frac{alpha + beta}{2} right) sin left(frac{alpha - beta}{2}right) end{align}
##### Product to sum
begin\left\{align\right\}
cos alpha ,cos beta &= frac{1}{2}[cos(alpha - beta) + cos (alpha + beta)] sin alpha ,sin beta &= frac{1}{2}[cos(alpha - beta) - cos (alpha + beta)] cos alpha ,sin beta &= frac{1}{2}[sin(alpha + beta) - sin (alpha - beta)] sin alpha ,cos beta &= frac{1}{2}[sin(alpha + beta) + sin (alpha - beta)] end{align}
##### Sine, cosine, and tangent of a sum
Detailed, diagramed proofs of the first two of these formulas are available
begin\left\{align\right\}
sin(alpha pm beta) &= sin alpha cos beta pm cos alpha sin beta cos(alpha pm beta) &= cos alpha cos beta mp sin alpha sin beta tan(alpha pm beta) &= frac{tan alpha pm tan beta}{1 mp tan alpha tan beta} end{align}
#### Half-angle identities
Note that $pm$ is correct, it means it may be either one, depending on the value of A/2.
begin\left\{align\right\}
sin frac{A}{2} &= pm sqrt{frac{1-cos A}{2}} cos frac{A}{2} &= pm sqrt{frac{1+cos A}{2}} tan frac{A}{2} &= pm sqrt{frac{1-cos A}{1+cos A}} = frac {sin A}{1+cos A} = frac {1-cos A}{sin A} end{align}
#### Stereographic (or parametric) identities
begin\left\{align\right\}
sin alpha &= frac{2T}{1+T^2} cos alpha &= frac{1-T^2}{1+T^2} end{align}
where $T=tan frac\left\{alpha\right\}\left\{2\right\}$.
### Triangle identities
In the following identities, A, B and C are the angles of a triangle and a, b and c are the lengths of sides of the triangle opposite the respective angles.
#### Law of sines
The law of sines (also know as the "sine rule") for an arbitrary triangle states:
$frac\left\{a\right\}\left\{sin A\right\} = frac\left\{b\right\}\left\{sin B\right\} = frac\left\{c\right\}\left\{sin C\right\} = 2R,$
where R is the radius of the circumcircle of the triangle.
#### Law of cosines
The law of cosines (also known as the cosine formula, or the "cos rule") is an extension of the Pythagorean theorem to arbitrary triangles:
$c^2=a^2+b^2-2abcos C ,,$
or equivalently:
$cos C=frac\left\{a^2+b^2-c^2\right\}\left\{2ab\right\}.,$
#### Law of tangents
The law of tangents:
$frac\left\{a+b\right\}\left\{a-b\right\}=frac\left\{tanleft\left[tfrac\left\{1\right\}\left\{2\right\}\left(A+B\right)right\right]\right\}\left\{tanleft\left[tfrac\left\{1\right\}\left\{2\right\}\left(A-B\right)right\right]\right\}$
## References
Christopher M. Linton (2004). From Eudoxus to Einstein: A History of Mathematical Astronomy . Cambridge University Press.
Weisstein, Eric W. "Trigonometric Addition Formulas". Wolfram MathWorld.
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Dividing Monomials
Monomials can be divided in the same manner as numbers.
To divide monomials, we use properties of fractions and properties of exponents.
Let's start from a simple example, involving only one variable.
Example 1. Simplify ${\left({14}{{x}}^{{5}}\right)}\div{\left({7}{{x}}^{{2}}\right)}$.
Since fraction denotes division, then it is better (for visual perception) to rewrite ${\left({14}{{x}}^{{5}}\right)}\div{\left({7}{{x}}^{{2}}\right)}$ as $\frac{{{14}{{x}}^{{5}}}}{{{7}{{x}}^{{2}}}}$.
$\frac{{{14}{{x}}^{{5}}}}{{{7}{{x}}^{{2}}}}=$
$=\frac{{14}}{{7}}\cdot\frac{{{{x}}^{{5}}}}{{{{x}}^{{2}}}}=$ (multiplication of fractions in reverse direction: $\frac{{{a}\cdot{c}}}{{{b}\cdot{d}}}=\frac{{a}}{{b}}\cdot\frac{{c}}{{d}}$)
$={2}\frac{{{{x}}^{{5}}}}{{{{x}}^{{2}}}}=$ (simplify coefficient)
$={2}{{x}}^{{{5}-{2}}}=$ (rule for subtracting exponents)
$={2}{{x}}^{{3}}$ (simplify).
Answer: ${\left({14}{{x}}^{{5}}\right)}{\left({7}{{x}}^{{2}}\right)}={2}{{x}}^{{3}}$.
Let's see how to multiply, if there are more than one variable (in fact technique is same).
Also, following examples, shows how to handle negative exponents.
Example 2. Simplify $\frac{{\frac{{1}}{{2}}{{x}}^{{4}}{{y}}^{{5}}{z}}}{{-{7}{{z}}^{{3}}{x}{{y}}^{{3}}}}$.
$\frac{{\frac{{1}}{{2}}{{x}}^{{4}}{{y}}^{{5}}{z}}}{{-{7}{{z}}^{{3}}{x}{{y}}^{{3}}}}=$
$=\frac{{\frac{{1}}{{2}}}}{{-{7}}}\cdot\frac{{{{x}}^{{4}}}}{{{x}}}\cdot\frac{{{{y}}^{{5}}}}{{{{y}}^{{3}}}}\cdot{\left(\frac{{z}}{{{z}}^{{3}}}\right)}$ (break down fraction)
$=-\frac{{1}}{{14}}\cdot{{x}}^{{{4}-{1}}}\cdot{{y}}^{{{5}-{2}}}\cdot{{z}}^{{{1}-{3}}}=$ (rule for subtracting exponents).
$=-\frac{{1}}{{14}}{{x}}^{{3}}{{y}}^{{3}}{{z}}^{{-{2}}}$ (simplify).
You can leave answer as it is, but, in most cases, teachers don't allow negative exponents, so you need to get rid of negative exponents.
$=-\frac{{1}}{{14}}{{x}}^{{3}}{{y}}^{{3}}\frac{{1}}{{{{z}}^{{2}}}}=$ (get rid of negative exponent)
$=-\frac{{{{x}}^{{3}}{{y}}^{{3}}}}{{{14}{{z}}^{{2}}}}$ (write more compactly, using rule for multiplying fractions)
Answer: ${\left(\frac{{1}}{{2}}{{x}}^{{4}}{{y}}^{{5}}{z}\right)}\div{\left(-{7}{{z}}^{{3}}{x}{{y}}^{{3}}\right)}=-\frac{{{{x}}^{{3}}{{y}}^{{3}}}}{{{14}{{z}}^{{2}}}}$.
Note: last example shows, that result of division monomial by monomial is not always monomial.
We can even divide more than two monomials!
Example 3. Divide ${\left({4}{{x}}^{{3}}{{y}}^{{2}}\right)}\div{\left({3}{{z}}^{{5}}{y}{{x}}^{{3}}\right)}\div{\left(-{2}{{y}}^{{2}}{z}{x}\right)}$.
You need to do it step by step.
First divide ${\left({4}{{x}}^{{3}}{{y}}^{{2}}\right)}\div{\left({3}{{z}}^{{5}}{y}{{x}}^{{3}}\right)}$: $\frac{{{4}{{x}}^{{3}}{{y}}^{{2}}}}{{{3}{{z}}^{{5}}{y}{{x}}^{{3}}}}=\frac{{{4}{y}}}{{{3}{{z}}^{{5}}}}$.
Now, divide resulting fraction by third monomial (using rule for dividing fractions): $\frac{{{4}{y}}}{{{3}{{z}}^{{5}}}}\div{\left(-{2}{{y}}^{{2}}{z}{x}\right)}=\frac{{\frac{{{4}{y}}}{{{3}{{z}}^{{5}}}}}}{{-{2}{{y}}^{{2}}{z}{x}}}=\frac{{{4}{y}}}{{{3}{{z}}^{{5}}\cdot{\left(-{2}{{y}}^{{2}}{z}{x}\right)}}}=\frac{{2}}{{{3}{x}{y}{{z}}^{{6}}}}$.
Answer: ${\left({4}{{x}}^{{3}}{{y}}^{{2}}\right)}\div{\left({3}{{z}}^{{5}}{y}{{x}}^{{3}}\right)}\div{\left(-{2}{{y}}^{{2}}{z}{x}\right)}=\frac{{2}}{{{3}{x}{y}{{z}}^{{6}}}}$.
Note: after some practice, you will want to skip some steps and just divide "like" variables immediately, making commutations and splittings in your head.
Now, it is time to exercise.
Exercise 1. Divide $\frac{{{7}{{a}}^{{7}}}}{{{3}{{a}}^{{2}}}}$.
Answer: $\frac{{7}}{{3}}{{a}}^{{5}}$.
Exercise 2. Simplify ${\left(-\frac{{1}}{{3}}{{x}}^{{3}}{{y}}^{{2}}{{z}}^{{7}}\right)}\div{\left(-\frac{{1}}{{6}}{{y}}^{{7}}{{z}}^{{3}}{{x}}^{{4}}\right)}$.
Answer: $\frac{{{2}{{z}}^{{4}}}}{{{x}{{y}}^{{5}}}}$.
Exercise 3. Divide the following: ${\left(-\frac{{3}}{{5}}{{c}}^{{2}}{{a}}^{{3}}{{b}}^{{7}}\right)}\div{\left(-\frac{{2}}{{7}}{{b}}^{{4}}{a}{{c}}^{{3}}\right)}\div{\left(-{a}{b}{c}\right)}$.
Answer: $-\frac{{{21}{a}{{b}}^{{2}}}}{{{10}{{c}}^{{2}}}}$.
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# 1.1 Introduction to Whole Numbers
The topics covered in this section are:
## 1.1.1 Identify Counting Numbers and Whole Numbers
Learning algebra is similar to learning a language. You start with a basic vocabulary and then add to it as you go along. You need to practice often until the vocabulary becomes easy to you. The more you use the vocabulary, the more familiar it becomes.
Algebra uses numbers and symbols to represent words and ideas. Let’s look at the numbers first. The most basic numbers used in algebra are those we use to count objects: $1,2,3,4,5,…$ and so on. These are called the counting numbers. The notation “…” is called an ellipsis, which is another way to show “and so on”, or that the pattern continues endlessly. Counting numbers are also called natural numbers.
### Counting Numbers
The counting numbers start with $1$ and continue.
$1, 2, 3, 4, 5 …$$1, 2, 3, 4, 5…$
Counting numbers and whole numbers can be visualized on a number line as shown in Figure 1.2:
The point labeled origin. The points are equally spaced to the right of $0$ and labeled with the counting numbers. When a number is paired with a point, it is called the coordinate of the point.
The discovery of the number zero was a big step in the history of mathematics. Including zero with the counting numbers gives a new set of numbers called the whole numbers.
### Whole Numbers
The whole numbers are the counting numbers and zero.
$0, 1 ,2 ,3 ,4 ,5…$
We stopped at $5$ when listing the first few counting numbers and whole numbers. We could have written more numbers if they were needed to make the patterns clear.
#### Example 1
Which of the following are a. counting numbers? b. whole numbers?
$0, \frac{1}{4}, 3, 5.2, 15, 105$
Solution
a. The counting numbers start at $1$ so $0$ is not a counting number. The numbers $3, 15$ and $105$ are all counting numbers.
b. Whole numbers are counting numbers and $0$. The numbers $0, 3, 15$ and $105$ are whole numbers.
The numbers $\frac{1}{4}$ and $5.2$ are neither counting numbers nor whole numbers. We will discuss these numbers later.
## 1.1.2 Model Whole Numbers
Our number system is called a place value system because the value of a digit depends on its position, or place, in a number. The number $537$ has a different value than the number $735$. Even though they use the same digits, their value is different because of the different placement of the $7$ and the $5$.
Money gives us a familiar model of place value. Suppose a wallet contains three $\$100$bills, seven$\$10$ bills, and four $\$1$bills. The amounts are summarized in Figure 1.3. How much money is in the wallet? Find the total value of each kind of bill, and then add to find the total. The wallet contains$\$374$
Base-10 blocks provide another way to model place value, as shown in Figure 1.4. The blocks can be used to represent hundreds, tens, and ones. Notice that the tens rod is made up of $10$ ones, and the hundreds square is made of $10$ tens, or $100$ ones.
Figure 1.5 shows the number $138$ modeled with base-$10$ blocks.
Digit Place Value Number Value Total Value $1$ hundreds $1$ $100$ $100$ $3$ tens $3$ $10$ $30$ $8$ ones $8$ $1$ $8$ Sum $=138$
#### Example 2
Use place value notation to find the value of the number modeled by the base-$10$ blocks shown.
Solution
There are $2$ hundreds squares, which is $200$.
There is $1$ tens rod, which is $10$.
There are $5$ ones blocks, which is $5$.
Digit Place Value Number Value Total Value $2$ hundreds $2$ $100$ $200$ $1$ tens $1$ $10$ $10$ $5$ ones $5$ $1$ $+5$ $215$
The base-$10$ blocks model the number $215$.
## 1.1.3 Identify the Place Value of a Digit
By looking at money and base-$10$ blocks, we saw that each place in a number has a different value. A place value chart is a useful way to summarize this information. The place values are separated into groups of three, called periods. The periods are ones, thousands, millions, billions, trillions, and so on. In a written number, commas separate the periods.
Just as with the base-$10$ blocks, where the value of the tens rod is ten times the value of the ones block and the value of the hundreds square is ten times the tens rod, the value of each place in the place-value chart is ten times the value of the place to the right of it.
Figure 1.6 shows how the number $5,278,194$ is written in a place value chart.
• The digit $5$ is in the millions place. Its value is $5,000,000$.
• The digit $2$ is in the hundred thousands place. Its value is $200,000$.
• The digit $7$ is in the ten thousands place. Its value is $70,000$.
• The digit $8$ is in the thousands place. Its value is $8,000$.
• The digit $1$ is in the hundreds place. Its value is $100$.
• The digit $9$ is in the tens place. Its value is $90$.
• The digit $4$ is in the ones place. Its value is $4$.
#### Example 3
In the number $63,407,218$ find the place value of each of the following digits:
• $7$
• $0$
• $1$
• $6$
• $3$
Solution
Write the number in a place value chart, starting at the right.
• The $7$ is in the thousands place.
• The $0$ is in the ten thousands place.
• The $1$ is in the tens place.
• The $6$ is in the ten millions place.
• The $3$ is in the millions place.
## 1.1.4 Use Place Value to Name Whole Numbers
When you write a check, you write out the number in words as well as in digits. To write a number in words, write the number in each period followed by the name of the period without the ‘s’ at the end. Start with the digit at the left, which has the largest place value. The commas separate the periods, so wherever there is a comma in the number, write a comma between the words. The ones period, which has the smallest place value, is not named.
So the number $37,519,248$ is written thirty-seven million, five hundred nineteen thousand, two hundred forty-eight.
Notice that the word and is not used when naming a whole number.
### How To: Name a Whole Number in Words.
1. Starting at the digit on the left, name the number in each period, followed by the period name. Do not include the period name for the ones.
2. Use commas in the number to separate the periods.
#### Example 4
Name the number $8,165,432,098,710$ in words.
Solution
Putting all of the words together, we write $8,165,432,098,710$ as eight trillion, one hundred sixty-five billion, four hundred thirty-two million, ninety-eight thousand, seven hundred ten.
#### Example 5
A student conducted research and found that the number of mobile phone users in the United States during one month in $2014$ was $327,577,529$. Name that number in words.
Solution
Identify the periods associated with the number.
Name the number in each period, followed by the period name. Put the commas in to separate the periods.
Millions period: three hundred twenty-seven million
Thousands period: five hundred seventy-seven thousand
Ones period: five hundred twenty-nine
So the number of mobile phone users in the Unites States during the month of April was three hundred twenty-seven million, five hundred seventy-seven thousand, five hundred twenty-nine.
## 1.1.5 Use Place Value to Write Whole Numbers
We will now reverse the process and write a number given in words as digits.
### How to: Use Place Value to Write a Whole Number
1. Identify the words that indicate periods. (Remember the ones period is never named.)
2. Draw three blanks to indicate the number of places needed in each period. Separate the periods by commas.
3. Name the number in each period and place the digits in the correct place value position.
#### Example 6
Write the following numbers using digits.
• fifty-three million, four hundred one thousand, seven hundred forty-two
• nine billion, two hundred forty-six million, seventy-three thousand, one hundred eighty-nine
Solution
• Identify the words that indicate periods.
Except for the first period, all other periods must have three places. Draw three blanks to indicate the number of places needed in each period. Separate the periods by commas.
Then write the digits in each period.
Put the numbers together, including the commas. The number is $53,401,742$.
• Identify the words that indicate periods.
Except for the first period, all other periods must have three places. Draw three blanks to indicate the number of places needed in each period. Separate the periods by commas.
Then write the digits in each period.
• The number is $9,246,073,189$.
Notice that in part 2, a zero was needed as a place-holder in the hundred thousands place. Be sure to write zeros as needed to make sure that each period, except possibly the first, has three places.
#### Example 7
A state budget was about $\$77$billion. Write the budget in standard form. Solution Identify the periods. In this case, only two digits are given and they are in the billions period. To write the entire number, write zeros for all of the other periods. So the budget was about$\$77,000,000,000$.
## 1.1.7 Round Whole Numbers
In $2013$, the U.S. Census Bureau reported the population of the state of New York as $19,651,127$ people. It might be enough to say that the population is approximately $20$ million. The word approximately means that $20$ million is not the exact population, but is close to the exact value.
The process of approximating a number is called rounding. Numbers are rounded to a specific place value depending on how much accuracy is needed. $20$ million was achieved by rounding to the millions place. Had we rounded to the one hundred thousands place, we would have $19,700,000$ as a result. Had we rounded to the ten thousands place, we would have $19,650,000$ as a result, and so on. The place value to which we round to depends on how we need to use the number.
Using the number line can help you visualize and understand the rounding process. Look at the number line in Figure 1.7. Suppose we want to round the number $76$ to the nearest ten. Is $76$ closer to $70$ or $80$ on the number line?
Now consider the number $72$. Find $72$ in Figure 1.8.
How do we round $75$ to the nearest ten? Find $75$ in Figure 1.9.
So that everyone rounds the same way in cases like this, mathematicians have agreed to round to the higher number, $80$. So $75$ rounded to the nearest ten is $80$.
Now that we have looked at this process on the number line, we can introduce a more general procedure. To round a number to a specific place, look at the number to the right of that place. If the number is less than $5$, round down. If it is greater than or equal to $5$, round up.
So, for example, to round $76$ to the nearest ten, we look at the digit in the ones place.
The digit in the ones place is a $6$. Because $6$ is greater than or equal to $5$, we increase the digit in the tens place by one. So the $7$ in the tens place becomes an $8$. Now, replace any digits to the right of the $8$ with zeros. So, $76$ rounds to $80$.
Let’s look again at rounding $72$ to the nearest ten. Again, we look to the ones place.
The digit in the ones place is $2$. Because $2$ is less than $5$, we keep the digit in the tens place the same and replace the digits to the right of it with zero. So $72$ rounded to the nearest ten is $70$.
### HOW TO: Round a whole number to a specific place value.
1. Locate the given place value. All digits to the left of that place value do not change unless the digit immediately to the left is $9$, in which case it may. (See Step 3.)
2. Underline the digit to the right of the given place value.
3. Determine if this digit is greater than or equal to $5$.
• Yes—add $1$ to the digit in the given place value. If that digit is $9$, replace it with $0$ and add $1$ to the digit immediately to its left. If that digit is also a $9$, repeat.
• No—do not change the digit in the given place value.
4. Replace all digits to the right of the given place value with zeros.
#### Example 8
Round $843$ to the nearest ten.
Solution
#### Example 9
Round each number to the nearest hundred:
• $23,658$
• $3,978$
Solution
Part 1
Part 2
#### Example 10
Round each number to the nearest thousand:
• $147,032$
• $29,504$
Solution
Part 1
Part 2
Notice that in part 1, when we add $1$ thousand to the $9$ thousands, the total is $10$ thousands. We regroup this as $1$ ten thousand and $0$ thousands. We add the $1$ ten thousand to the $2$ ten thousands and put a $0$ in the thousands place.
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This week, we will take up some questions on co-ordinate geometry. Let me re-cap the relations we discussed in the last post.
Say, the equations of 2 lines are:
ax + by + c = 0
and
mx + ny + p = 0
1. A single point of intersection between two lines: a/m ? b/n
2. Distinct parallel lines: a/m = b/n ? c/p
3. The same line: a/m = b/n = c/p
4. Perpendicular lines: am = -bn
Question 1: A given line L has an equation 3x+4y=5. Which of the following is the equation of line which does not intersect the above line?
(A) 4x + 3y = 5
(B) 3x + 4y = 10
(C) 3x + 5y = 5
(D) 3x + 5y = 3
(E) 3x – 4y = 5
Solution: A line that does not intersect with L, is a distinct line parallel to L. The relation of coefficients between distinct parallel lines is a/m = b/n ? c/p
Equation of L is 3x + 4y – 5 = 0.
a = 3
b = 4
c = -5
For option (B), m = 3, n = 4 and p = -10
We see that a/m (3/3) = b/n (4/4) ? c/p (-5/-10)
Question 2: What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?
(A) 0
(B) 1/4
(C) 1/2
(D) ?2/2
(E) ?2/4
Solution:
Any two lines in the xy plane will be either parallel or intersecting. If the lines intersect, the shortest distance between them will be 0.
The two given lines are:
x + y = 3 (shown by the red line)
2x + 2y = 8 which is same as x + y = 4 (shown by the blue line)
We notice that a/m (1/1) = b/n (1/1) ? c/p (3/4). Hence, the lines are parallel.
They intersect the x axis at x = 3 and x = 4 and the y axis at y = 3 and y = 4 (as shown in the figure)
Now there are many ways of getting the distance between them. The first method I will discuss is using a formula. The second method will use the properties of right triangles.
Would I advise you to learn up the formula? No. GMAT requires you to know only the very basic formulas. This is certainly not one of them. Remember it only if you have already come across it sometime during the course of your study and seeing it here is enough for you to recall it during the exam ( if need be). If you are seeing this formula for the first time, don’t worry about adding it to your list. There will always be other, more intuitive ways of getting to the answer.
First Method: Using the formula
If the equations of two parallel lines are: y = mx + b and y = mx + c (note that they have the same slope, m, but different y intercepts, b and c)
Distance between them = |b-c|/?(m^2 + 1)
Here the parallel lines are: y = -x + 3 and y = -x + 4
Distance between them = |4-3|/?((-1)^2 + 1) = 1/?2 = ?2/2
Second Method: Using right triangles
Use the little triangle ABC. Co-ordinates of A are (3, 0) and of C are (4, 0).
When you draw the red line, you notice that its x and y intercepts are the same.
i.e. x + y = 3 intersects x axis at 3 and y axis also at 3. So it forms an isosceles triangle.
Similarly, x + y = 4 intersects x axis at 4 and y axis also at 4. It also forms an isosceles triangle so angle BCA is 45 degrees.
AB is dropped perpendicular to the blue line. This is the distance between the two parallel lines. Since angle ABC is 90 degrees, angle BAC will also be 45 degrees (to make the sum 180). So AB = AC.
In an isosceles right triangle, the ratio of the sides is 1:1: ?2 where ?2 is the hypotenuse. Since we know that the hypotenuse is actually 1 (= 4 – 3), the measure of equal sides (AB and BC) will be 1/?2 each. Multiply and divide this by ?2 to get ?2/2.
Hence, the distance between the two lines, AB = ?2/2.
Hope the application of the concept discussed is clear. We will continue working on co-ordinate geometry next week. Till then, keep practicing!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep in Detroit, and regularly participates in content development projects such as this blog!
### 4 Responses
1. Krishna says:
2nd qn – I too used the second method visualizing :-)
• Karishma says:
Great! When I write it down to explain, it seems long and cumbersome. But when we visualize it on our own, it happens in a blink!
2. Mike says:
Hi Karishma,
In Question 2, what if the line did not intercept the x-axis and y-axis at 45 degree (for example, the line intercept the x-axis at 67 degree)? Does that mean you can NOT use the Second Method: Right Triangles property? If the lines do not lintercept the x-axis in a 45 degree, can you still use the Formula that you posted as the “First Method”? Thanks.
• Karishma says:
First method can be used in any case in which lines are parallel. The second method can be used in case it is a 45-45-90 or 30-60-90 triangle since we know the ratio of sides in that case. Remember, GMAT doesn’t expect you to memorize these formulas so the questions you will get will be special cases – i.e. cases in which there are other methods of getting the answer.
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JEE > Single Correct MCQs: Inverse Trignometric Functions
# Single Correct MCQs: Inverse Trignometric Functions - Notes | Study All Types of Questions for JEE - JEE
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Q.1. The principal value of tan-1(tan 3π/5) is
(a) 2π/5
(b) -2π/5
(c) 3π/5
(d) -3π/5
tan-1 (tan 3π/5)
This can be written as:
tan-1 (tan 3π/5) = tan-1 (tan[π – 2π/5])
= tan-1 (- tan 2π/5) {since tan(π – x) = -tan x}
= –tan-1 (tan 2π/2)
= –2π/5
Q.2. sin[π/3 – sin-1(-½)] is equal to:
(a) ½
(b) ⅓
(c) -1
(d) 1
sin[π/3 – sin-1(-½)]
= sin[π/3 – sin-1[sin (-π/6))]
sin[π/3 – (-π/6)]
= sin(π/3 + π/6)
= sin π/2
= 1
Q.3. The domain of sin–1(2x) is
(a) [0, 1]
(b) [– 1, 1]
(c) [-1/2, 1/2]
(d) [–2, 2]
Let sin–1(2x) = θ.
Thus, 2x = sin θ.
As we know, – 1 ≤ sin θ ≤ 1
We can write this as– 1 ≤ 2x ≤ 1, which gives -1/2 ≤ x ≤ 1/2.
Therefore, the domain of sin-1(2x) is [-½, ½].
Q.4. If sin–1x + sin–1y = π/2, then value of cos–1x + cos–1y is
(a) π/2
(b) π
(c) 0
(d) 2π/3
Given,
sin–1 x + sin–1 y = π/2
[(π/2) – cos-1x] + [(π/2) – cos-1y] = π/2
(π/2) + (π/2) – (π/2) = cos-1x + cos-1y
Therefore, cos–1x + cos–1y = π/2.
Q.5. Which of the following is the principal value branch of cos–1x?
(a) [–π/2, π/2]
(b) (0, π)
(c) [0, π]
(d) (0, π) – {π/2}
The principal value branch of cos–1x is [0, π].
Q.6. The value of the expression sin [cot–1 (cos (tan–1 1))] is
(a) 0
(b) 1
(c) 1/√3
(d) √(2/3)
sin [cot–1 (cos (tan–1 1))]
= sin[cot-1 {cos (tan-1 (tan π/4))}] {since tan π/4 = 1}
= sin[cot-1 (cos π/4)]
= sin[cot-1(1/√2)]
= sin [sin-1(√(⅔))] {by Pythagoras theorem}
= √(⅔)
Q.7. The domain of y = cos–1 (x2 – 4) is
(a) [3, 5]
(b) [0, π]
(c) [-√5, -√3] ∩ [-√5, √3]
(d) [-√5, -√3] ∪ [√3, √5]
Given,
y = cos–1 (x2 – 4 )
⇒ cos y = x2 – 4
As we know, –1 ≤ cos y ≤ 1
So, – 1 ≤ x2 – 4 ≤ 1
Adding 4 on both sides, we get;
⇒ 3 ≤ x2 ≤ 5
Taking square root on both sides, we get;
⇒ √3 ≤ x ≤ √5
⇒ x∈ [-√5, -√3] ∪ [√3, √5]
Q.8. If α ≤ 2 sin–1x + cos–1x ≤ β, then
(a) α = -π/2, β = π/2
(b) α = 0, β = π
(c) α = -π/2, β = 3π/2
(d) α = 0, β = 2π
Given,
α ≤ 2 sin–1x + cos–1x ≤ β
We know that,
-π/2 ≤ sin–1 x ≤ π/2
⇒ (-π/2) + (π/2) ≤ sin–1x + (π/2) ≤ (π/2) + (π/2)
⇒ 0 ≤ sin–1x + (sin–1x + cos–1x) ≤ π
⇒ 0 ≤ 2 sin–1x + cos–1x ≤ π
By comparing with α ≤ 2 sin–1x + cos–1x ≤ β, we get α = 0, β = π.
Q.9. The value of sin (2 tan–1 (.75)) is equal to
(a) .75
(b) 1.5
(c) .96
(d) sin 1.5
sin (2tan–1 (.75))
Let, tan–1 (.75) = θ
tan θ = 0.75
tan θ = 3/4
Thus by Pythagoras theorem, we get;
sin θ = 3/5 and cos θ = 4/5.
Now,
sin (2tan–1 (.75)) = sin 2θ {as tan-1(.75) = θ}
= 2 sin θ cos θ
= 2 × (3/5) × (4/5)
= 24/25
= 0.96
Therefore, sin (2tan–1 (.75)) = .96.
Q.10. sin(tan-1 x), where |x| < 1, is equal to:
(a) x/√(1 – x2)
(b) 1/√(1 – x2)
(c) 1/√(1 + x2)
(d) x/√(1 + x2)
Let tan-1x = θ.
So, tan θ = x = x/1
From this, we can write the sin θ and cos θ values as:
sin θ = x/√(1 + x2)
cos θ = 1/√(1 + x2)
Now,
sin(tan-1 x) = sin θ = x/√(1 + x2).
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Inequalities | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: Algebra II Go to the latest version.
# 3.1: Inequalities
Difficulty Level: At Grade Created by: CK-12
Name: __________________
1. $4<6$ (I think we can all agree on that, yes?)
1. Add $4$ to both sides of the equation. ___________ Is it still true?
2. Add $-4$ to both sides of the (original) equation. ___________ Is it still true?
3. Subtract $10$ from both sides of the (original) equation. ___________ Is it still true?
4. Multiply both sides of the (original) equation by $4$. ___________ Is it still true?
5. Divide both sides of the (original) equation by $2$. ___________ Is it still true?
6. Multiply both sides of the (original) equation by $-3$. ___________ Is it still true?
7. Divide both sides of the (original) equation by $-2$. ___________ Is it still true?
8. In general: what operations, when performed on an inequality, reverse the inequality?
2. $2x+3 < 7$
1. Solve for $x$.
2. Draw a number line below, and show where the solution set to this problem is.
3. Pick an $x-$value which, according to your drawing, is inside the solution set. Plug it into the original inequality $2x+3<7$. Does the inequality hold true?
4. Pick an $x-$value which, according to your drawing, is outside the solution set. Plug it into the original inequality $2x+3<7$. Does the inequality hold true?
3. $10-x \ge 4$
1. Solve for $x$. Your first step should be adding $x$ to both sides, so in your final equation, $x$ is on the right side.
2. Solve for $x$ again from the original equation. This time, leave $x$ on the left side.
3. Did your two answers come out the same?
4. Draw a number line below, and show where the solution set to this problem is.
5. Pick an $x-$value which, according to your drawing, is inside the solution set. Plug it into the original inequality $10-x \ge 4$. Does the inequality hold true?
6. Pick an $x-$value which, according to your drawing, is outside the solution set. Plug it into the original inequality $10-x \ge 4$. Does the inequality hold true?
4. $x = \pm 4$
1. Rewrite this statement as two different statements, joined by “and” or “or.”
2. Draw a number line below, and show where the solution set to this problem is.
5. $-3 < x \le 6$
1. Rewrite this statement as two different statements, joined by “and” or “or.”
2. Draw a number line below, and show where the solution set to this problem is.
6. $x > 7$ or $x < -3$
1. Draw a number line below, and show where the solution set to this problem is.
7. $x > 7$ and $x < -3$
1. Draw a number line below, and show where the solution set to this problem is.
8. $x < 7$ or $x > -3$
1. Draw a number line below, and show where the solution set to this problem is.
9. $x > \pm 4$
1. Rewrite this statement as two different statements, joined by “and” or “or.”
2. Draw a number line below, and show where the solution set to this problem is.
Name: __________________
Homework: Inequalities
1. $2x+7 \le 4x+4$
a. Solve for $x$.
b. Draw a number line below, and show where the solution set to this problem is.
c. Pick an $x-$value which, according to your drawing, is inside the solution set. Plug it into the original inequality $2x+7 \le 4x+4$. Does the inequality hold true?
d. Pick an $x-$value which, according to your drawing, is outside the solution set. Plug it into the original inequality $2x+7 \le 4x+4$. Does the inequality hold true?
2. $14-2x < 20$
a. Solve for $x$.
b. Draw a number line below, and show where the solution set to this problem is.
c. Pick an $x-$value which, according to your drawing, is inside the solution set. Plug it into the original inequality $14-2x<20$. Does the inequality hold true?
d. Pick an $x-$value which, according to your drawing, is outside the solution set. Plug it into the original inequality $14-2x<20$. Does the inequality hold true?
3. $-10 < 3x+2 \le 5$
a. Solve for $x$.
b. Draw a number line below, and show where the solution set to this problem is.
c. Pick an $x-$value which, according to your drawing, is inside the solution set. Plug it into the original inequality $-10<3x+2 \le 5$. Does the inequality hold true?
d. Pick an $x-$value which, according to your drawing, is outside the solution set. Plug it into the original inequality $-10<3x+2 \le 5$. Does the inequality hold true?
4. $x < 3$ and $x < 7$. Draw a number line below, and show where the solution set to this problem is.
5. $x < 3$ or $x < 7$. Draw a number line below, and show where the solution set to this problem is.
6. $x-2y \ge 4$
a. Solve for $y$.
b. Now—for the moment—let’s pretend that your equation said equals instead of “greater than” or “less than.” Then it would be the equation for a line. Find the slope and the $y-$intercept of that line, and graph it.
Slope: _________
$y-$Intercept: _________
c. Now, pick any point $(x,y)$ that is above that line. Plug the $x$ and $y$ coordinates into your inequality from part $(a)$. Does this point fit the inequality? (Show your work...)
d. Now, pick any point $(x,y)$ that is below that line. Plug the $x$ and $y$ coordinates into your inequality from part $(a)$. Does this point fit the inequality? (Show your work...)
e. So, is the solution to the inequality the points below or above the line? Shade the appropriate region on your graph.
7. Using a similar technique, draw the graph of $y \ge x^2$. (If you don’t remember what the graph of $y=x^2$ looks like, try plotting a few points!)
Name: ____________________
Inequality Word Problems
1. Jacob is giving a party. $20$ people showed up, but he only ordered $4$ pizzas! Fortunately, Jacob hasn’t cut the pizzas yet. He is going to cut each pizza into $n$ slices, and he needs to make sure there are enough slices for everyone at the party to get at least one. Write an inequality or set that describes what $n$ has to be.
2. Whitney wants to drive to Seattle. She needs $100\;\mathrm{gallons}$ of gas to make the trip, but she has only $\80$ allocated for gas. Her strategy is to wait until the price of gas is low enough that she can make the trip. Write an inequality or set that describes what the price of gas has to be for Whitney to be able to reach Seattle. Be sure to clearly define your variable$(s)$!
3. Your evil math teacher, who shall go nameless, is only giving two tests for the whole grading period. They count equally—your grade will be the average of the two. Your first test was a $90$. Write an inequality or set that describes what your second test grade has to be, in order for you to bring home an A on your report card. (“A” means $93$ or above.) Be sure to clearly define your variable$(s)$!
4. Laura $L$ is going to build a movie theater with $n$ screens. At each screen, there will be $200$ seats for the audience to watch that movie. (So maximum capacity is $200$ audience members per screen.) In addition to audience members, there are $20$ employees on the premises at any given time (selling tickets and popcorn and so on). According to code (which I am making up), she must have at least one bathroom for each $100$ people in the building. (Of course, it’s fine to build more bathrooms than that, if she wants!)
1. Write a function (this will be an equation) relating the number of screens $(n)$ to the total number of people who can possibly be in the building $(p)$. Which one is dependent? Which one is independent?
2. Write an inequality relating the total number of people who can possibly be in the building $(p)$ to the number of bathrooms $(b)$.
3. Now write a composite inequality (I just made that word up) that tells Laura: if you build this many screens, here is how many bathrooms you need.
5. Make up your own word problem for which the solution is an inequality, and solve it. The topic should be breakfast.
Name:____________________
## Absolute Value Equations
1. $|4| =$
2. $|-5| =$
3. $|0| =$
4. OK, now, I’m thinking of a number. All I will tell you is that the absolute value of my number is $7$.
a. Rewrite my question as a math equation instead of a word problem.
b. What can my number be?
5. I’m thinking of a different number. This time, the absolute value of my number is $0$.
a. Rewrite my question as a math equation instead of a word problem.
b. What can my number be?
6. I’m thinking of a different number. This time, the absolute value of my number is $-4$.
a. Rewrite my question as a math equation instead of a word problem.
b. What can my number be?
7. I’m thinking of a different number. This time, the absolute value of my number is less than $7$.
a. Rewrite my question as a math inequality instead of a word problem.
b. Does $8$ work?
c. Does $6$ work?
d. Does $-8$ work?
e. Does $-6$ work?
f. Write an inequality that describes all possible values for my number.
8. I’m thinking of a different number. This time, the absolute value of my number is greater than $4$.
a. Rewrite my question as a math inequality instead of a word problem.
b. Write an inequality that describes all possible values for my number. (Try a few numbers, as we did in $^\#7$.)
9. I’m thinking of a different number. This time, the absolute value of my number is greater than $-4$.
a. Rewrite my question as a math inequality instead of a word problem.
b. Write an inequality that describes all possible values for my number.
Stop at this point and check your answers with me before going on to the next side.
10. $|x+3|=7$
a. First, forget that it says, $x+3''$ and just think of it as “a number.” The absolute value of this number is $7$. So what can this number be?
b. Now, remember that “this number” is $x+3$. So write an equation that says that $x+3$ can be <your answer(s) in part a>.
c. Solve the equation(s) to find what $x$ can be.
d. Plug your answer(s) back into the original equation $|x+3|=7$ and see if they work.
11. $4|3x-2|+5=17$
a. This time, because the absolute value is not alone, we’re going to start with some algebra. Leave $|3x-2|$ alone, but get rid of everything around it, so you end up with $|3x-2|$ alone on the left side, and some other number on the right.
b. Now, remember that “some number” is $3x-2$. So write an equation that says that $3x-2$ can be <your answer(s) in part a>.
c. Solve the equation(s) to find what $x$ can be.
d. Plug your answer(s) back into the original equation $4|3x-2|+5=17$ and see if they work.
12. $|x-3|+5=4$
a. Solve, by analogy to the way you solved the last two problems.
b. Plug your answer(s) back into the original equation $|x-3|+5=4$ and see if they work.
13. $|x-2|=2x-10$.
a. Solve, by analogy to the way you solved the last two problems.
b. Plug your answer(s) back into the original equation $|x-2|=2x-10$ and see if they work.
Name: __________________
Homework: Absolute Value Equations
1. $|x| = 5$
1. Solve for $x$
2. Check your answer(s) in the original equation.
2. $|x| = 0$
1. Solve for $x$
2. Check your answer(s) in the original equation.
3. $|x| = -2$
1. Solve for $x$
2. Check your answer(s) in the original equation.
4. $10|x| = 5$
1. Solve for $x$
2. Check your answer(s) in the original equation.
5. $|x+3| = 1$
1. Solve for $x$
2. Check your answer(s) in the original equation.
6. $\frac{4|x-2|}{3} = 2$
1. Solve for $x$
2. Check your answer(s) in the original equation.
7. $7|2x+3| - 4 = 4$
1. Solve for $x$
2. Check your answer(s) in the original equation.
8. $|2x-3| = x$
1. Solve for $x$
2. Check your answer(s) in the original equation.
9. $|2x+2| = x$
1. Solve for $x$
2. Check your answer(s) in the original equation.
10. $|x-5| = 2x-7$
1. Solve for $x$
2. Check your answer(s) in the original equation.
Check-yourself hint: for numbers $8, 9$, and $10$, one of them has no valid solutions, one has one valid solution, and one has two valid solutions.
Name: __________________
## Absolute Value Inequalities
1. $|x| \le 7$
1. Solve.
2. Graph your solution on a number line
3. Choose a point that is in your solution set, and test it in the original inequality. Does it work?
4. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
2. $|2x+3| \le 7$
1. Write down the solution for what $2x+3$ has to be. This should look exactly like your answer to number $1$, except with a $(2x+3)$ instead of an $(x)$.
2. Now, solve that inequality for $x$.
3. Graph your solution on a number line
4. Choose a point that is in your solution set, and test it in the original inequality. Does it work?
5. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
3. $4|3x-6|+7 > 19$
1. Solve for $|3x-6|$. (That is, leave the $|3x-6|$ part alone, but get rid of all the stuff around it.)
2. Write down the inequality for what $(3x-6)$ has to be.
3. Now, solve that inequality for $x$.
4. Graph your solution on a number line
5. Choose a point that is in your solution set, and test it in the original inequality. Does it work?
6. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
4. $\frac{|3x-4|}{2}+6 < 3$
1. Solve for $x$. (You know the drill by now!)
2. Graph your solution on a number line
3. Choose a point that is in your solution set, and test it in the original inequality. Does it work?
4. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
5. $6|2x^2-17x-85|+5 \ge 3$
1. Solve for $x$.
2. Graph your solution on a number line
3. hoose a point that is in your solution set, and test it in the original inequality. Does it work?
4. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
Name: __________________
Homework—Absolute Value Inequalities
1. $|4+3x| = 2+5x$ (...OK, this isn’t an inequality, but I figured you could use a bit more practice at these)
2. $|x| = x-1$
3. $4|2x-3|-5 \ge 3$
1. Solve for $x$.
2. Graph your solution on a number line
3. Choose a point that is in your solution set, and test it in the original inequality. Does it work?
4. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
4. $3|x-5|+2 < 17$
1. Solve for $x$.
2. Graph your solution on a number line
3. Choose a point that is in your solution set, and test it in the original inequality. Does it work?
4. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
5. $-3|x-5|+2 < 17$
1. Solve for $x$.
2. Graph your solution on a number line
3. Choose a point that is in your solution set, and test it in the original inequality. Does it work?
4. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
6. $2|x+2|+6 < 6$
1. Solve for $x$.
2. Graph your solution on a number line
3. Choose a point that is in your solution set, and test it in the original inequality. Does it work?
4. Choose a point that is not in your solution set, and test it in the original inequality. Does it work?
Name: __________________
## Graphing Inequalities and Absolute Values
1. $9x+3y \le 6$
1. Put into a sort of $y=mx+b$ format, except that it will be an inequality.
2. Now, ignore the fact that it is an inequality—pretend it is a line, and graph that line.
3. Now, to graph the inequality, shade in the area either above the line, or below the line, as appropriate. (Hint: does $y$ have to be less than the values on the line, or greater than them?)
4. Test your answer. Choose a point (any point) in the region you shaded, and test it in the inequality. Does the inequality work? (Show your work.)
5. Choose a point (any point) in the region you did not shade, and test it in the inequality. Does the inequality work? (Show your work.)
2. $4x \le 2y+5$
1. Graph the inequality, using the same steps as above.
2. Test your answer by choosing one point in the shaded region, and one point that is not in the shaded region. Do they give you the answers they should? (Show your work.)
3. $y = |x|$
1. Create a table of points. Your table should include at least two positive $x-$values, two negative $x-$values, and $x=0$.
2. Graph those points, and then draw the function.
4. $y = |x|+3$. Graph this without a table of points, by remembering what “adding 3” does to any graph. (In other words, what will these $y-$values be like compared to your $y-$values in $^\#3$?)
5. $y = -|x|$. Graph this without a table of points, by remembering what “multiplying by $-1$” does to any graph. (In other words, what will these $y-$values be like compared to your $y-$values in $^\#3$?)
6. Now, let’s put it all together!!!
1. Graph it. $y = -|x|+2$
2. Graph $y \ge -|x|+2$. Your answer will either be a shaded region on a $2-$dimensional graph, or on a number line.
3. Test your answer by choosing one point in the shaded region, and one point that is not in the shaded region. Do they give you the answers they should? (Show your work.)
4. Graph $-|x|+2<0$. Your answer will either be a shaded region on a $2-$dimensional graph, or on a number line.
5. Test your answer by choosing one point in the shaded region, and one point that is not in the shaded region. Do they give you the answers they should? (Show your work.)
7. Extra for experts: $y \ge 3|x+4|$
1. Graph it. Think hard about what that $+4$ and that $3$ will do. Generate a few points if it will help you!
2. Test your answer by choosing one point in the shaded region, and one point that is not in the shaded region. Do they give you the answers they should? (Show your work.)
Name: __________________
Homework: Graphing Inequalities and Absolute Values
1. The famous detectives Guy Noir and Nick Danger are having a contest to see who is better at catching bad guys. At 8:00 in the evening, they start prowling the streets of the city. They have twelve hours. Each of them gets $10$ points for every purse-snatcher he catches, and $15$ points for every cat-burglar. At 8:00 the next morning, they meet in one of their dingy offices to compare notes. “I got $100$ points,” brags Nick. If Guy gets enough snatchers and burglars, he will win the contest.
a. Label and clearly describe the relevant variables.
b. Write an inequality relating the variables you listed in part (a). I should be able to read it as, “If this inequality is true, then Guy wins the contest.”
c. Graph the inequality from part (b).
2. The graph below shows the function $y=f(x)$.
a. Graph $y \le f(x)$. Your answer will either be a shaded region on a $2-$dimensional graph, or on a number line.
b. Graph $f(x) < 0$. Your answer will either be a shaded region on a $2-$dimensional graph, or on a number line.
3. $x-2y > 4$
a. Graph.
b. Pick a point in your shaded region, and plug it back into our original equation $x-2y > 4$. Does the inequality work? (Show your work!)
c. Pick a point which is not in your shaded region, and plug it into our original equation $x-2y>4$. Does the inequality work? (Show your work!)
4. $|x|-y \ge 2$
a. Graph.
b. Pick a point in your shaded region, and plug it back into our original equation $|x|-y \ge 2$. Does the inequality work? (Show your work!)
c. Pick a point which is not in your shaded region, and plug it into our original equation $|x|-y \ge 2$. Does the inequality work? (Show your work!)
5. $y>x^3$
a. Graph. (Plot points to get the shape.)
b. Pick a point in your shaded region, and plug it back into our original equation $y>x^3$. Does the inequality work? (Show your work!)
c. Pick a point which is not in your shaded region, and plug it into our original equation $y>x^3$. Does the inequality work? (Show your work!)
6. Graph: $y+|x| < -|x|$. Think hard—you can do it!
Name: __________________
Sample Test: Inequalities and Absolute Values
1. $1 < 4-3x \le 10$
1. Solve for $x$.
2. Draw a number line below, and show where the solution set to this problem is.
3. Pick an $x-$value which, according to your drawing, is inside the solution set. Plug it into the original inequality $1<4-3x \le 10$. Does the inequality hold true? (Show your work!)
4. Pick an $x-$value which, according to your drawing, is outside the solution set. Plug it into the original inequality $1<4-3x \le 10$. Does the inequality hold true? (Show your work!)
2. Find the $x$ value(s) that make this equation true: $4|2x+5|-3=17$
3. Find the $x$ value(s) that make this equation true: $|5x-23| = 21-6x$
4. $\frac{|2x-3|}{3} +7 > 9$
1. Solve for $x$.
2. Show graphically where the solution set to this problem is.
5. $-3|x+4|+7 \ge 7$
1. Solve for $x$.
2. Show graphically where the solution set to this problem is.
6. Make up and solve an inequality word problem having to do with hair.
1. Describe the scenario in words.
2. Label and clearly describe the variable or variables.
3. Write the inequality. (Your answer here should be completely determined by your answers to (a) and (b) - I should know exactly what you’re going to write. If it is not, you probably did not give enough information in your scenario.)
7. Graph $y \ge -|x|+2$
8. $|2y|-|x|>6$
1. Rewrite this as an inequality with no absolute values, for the fourth quadrant (lower-right-hand corner of the graph).
2. Graph what this looks like, in the fourth quadrant only.
9. Graph $y=x-|x|$
## Date Created:
Feb 23, 2012
Apr 29, 2014
Files can only be attached to the latest version of None
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## Search This Blog
### Odds against/in favor of an event
"Odds against" an event is the ratio of the probability of not happening of that event to the probability of happening of that event.
For example, odds against getting a particular number, say 5, on rolling a number cube is the ratio of the probability of not getting 5 to the probability of getting 5 on rolling that number cube.
"Odds in favor of" an event is the ratio of the probability of happening of that event to the probability of not happening of that event.
For example, odds in favor of getting the number 5 on rolling a number cube is the ratio of the probability of getting the number 5 to the probability of not getting the number 5.
Thus, we conclude that the odds against an event are the opposite of the odds in favor of the event.
Let us understand the meaning of "odds" with an example. This will also lead us on to the formula written below it.
#### What are the odds in favor of getting the number 5 on rolling a number cube?
First, calculate the probability of getting the number 5 on rolling a number cube. We use the formula for theoretical probability to calculate it.
P(5) = "Number of favorable outcomes"/"Total number of outcomes"
P(5) = 1/6
Now calculate the probability of not getting the number 5 on rolling the number cube. This is the complement of the probability of getting 5, thus,
P("not 5") = 1 - P(5) = 1 - 1/6 = 5/6
Now, odds in favor of getting the number 5 on rolling a number cube are given by the ratio of the above probabilities.
"Odds in favor of getting 5" = (P(5))/(P("not 5"))
"Odds in favor of getting 5" = (1/6)/(5/6) = 1/5"
Thus, the odds in favor of getting the number 5 on rolling a number cube are 1/5.
From the above example, we can conclude that the formula for odds in favor of an event can be written as follows:
"Odds in favor of an event E" = (P("E"))/(P("not E"))
Similarly, since odds against an event is the ratio of the probability of not happening of that event to the probability of happening of that event, therefore its formula is
"Odds against of an event E" = (P("not E"))/(P("E"))
Thus, as stated previously, we can say that the odds against an event are the complete opposite of the odds in favor of it. In other words, the odds against an event are the reciprocal of the odds in favor of an event. For example, if the odds against an event are 1/2, then the odds in favor of that event are 2/1.
The above example leads us to another conclusion: Since the odds against or in favor of an event are the ratio of probabilities and not the probabilities themselves, therefore they can be greater than 1as opposed to probability of an event (recall that the probability of an event can not be greater than 1).
Further, if we use the formula for theoretical probability, we can write
P("E") = "Number of favorable outcomes"/"Total number of outcomes"
and,
P("not E") = "Number of outcomes that are not favorable"/"Total number of outcomes"
If we place the above two formulas for P(E) and P(Not E) in the formula for odds in favor of an event E, we get
"Odds in favor of event E" = (P("E"))/(P("Not E"))
"Odds in favor of event E" = ("Number of favorable outcomes"/"Total number of outcomes")/("Number of outcomes that are not favorable"/"Total number of outcomes")
Simplifying, we get,
"Odds in favor event E" = "Number of favorable outcomes"/"Number of outcomes that are not favorable"
The above formula is extremely useful in calculating the odds in favor of an event. It helps you calculate the odds in favor of an event without calculating the probability of its happening or not happening.
Since the odds against an event are the reciprocal of the odds in favor of it, thus, we can also write the formula for odds against an event as follows:
"Odds against an event E" = "Number of outcomes that are not favorable"/"Number of favorable outcomes"
### Solved Examples
#### 1. There are five red, four blue and three white marbles in a bag. What are the odds against and in favor of getting a red marble on drawing one marble from the bag?
There are five red marbles in the bag, therefore the number of favorable outcomes for drawing a red marble is 5. The other 7 marbles are not red. Thus the number of outcomes that are not favorable are 7. Thus,
"Odds against getting a red marble" = "Number of outcomes that are not favorable"/"Total number of outcomes"
"Odds against getting a red marble" = 7/5
Since the odds in favor of an event are the opposite of the odds against it, therefore,
"Odds in favor of getting a red marble" = 1/"Odds against it" = 5/7
#### 2. A number cube is rolled. What are the odds against getting a number greater than 4 on it? What are the odds in favor?
There are two numbers, 5 and 6, greater than 4 on a number cube. Thus, the number of favorable outcomes is 2. The other four numbers on the number cube (1 to 4) are not greater than 4. Thus the number of outcomes that are not favorable is 4.
"Odds against getting a number greater than 4" = "Number of outcomes that are not favorable"/"Number of outcomes that are favorable"
"Odds against getting a number greater than 4" = 4/2 = 2/1
Since the odds in favor of an event are the opposite of the odds against an event, therefore
"Odds in favor of getting a number greater than 4 = 1/2"
#### 3. One card is drawn from a deck of fifty two playing cards. What are the odds in favor of getting a red King?
There are two red Kings in a standard deck of fifty two cards. Thus the number of favorable of favorable outcomes is 2. The other 50 cards are not red Kings. Thus the number of outcomes that are not favorable are 50.
"Odds in favor of getting a red King" = "Number of favorable outcomes"/"Total number of outcomes"
"Odds in favor of getting a red King" = 2/50= 1/25
#### 4. It rained on four out of five consecutive days in a week. What are the odds against raining on the sixth day?
Since it rained on four days, number of favorable outcomes for raining = 4
Since it did not rain on one day, therefore number of non-favorable outcomes = 1
Applying the formula for "odds against",
"Odds against raining" = "Number of non-favorable outcomes"/"Number of favorable outcomes" = 1/4
#### 5. Team A won three of the four matches against Team B. What are the odds in favor of team A winning the fifth match?
Since team A won three matches, therefore number of favorable events for wining a match of team A = 3
Since team A did not win one match out of four, therefore number of non-favorable outcomes for winning a match of team A = 1
Applying the formula for "odds in favor",
"Odds in favor of team A winning the match" = "Number of favorable outcomes"/"Number of non-favorable outcomes" = 3/1 = 3
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# Can we divide a vector by another vector? How about this: $a = vdv/dx?$
The statement $$a = v (dv/dx)$$ only holds in that form for one-dimensional motion, where the quantities $$v$$ and $$x$$ are just numbers rather than vectors. It follows from the chain rule, if we view $$v$$ as a function of $$x$$ instead of as a function of $$t$$: $$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v.$$
If you're doing 2-D or 3-D motion, you can still do something similar, but you have to let $$\vec{v}$$ be a function of $$x$$, $$y$$, and $$z$$, since $$\vec{v}$$ can change as each of these quantities changes. This means that you need to use multi-variable calculus to write out an equivalent statement. For example, we have $$a_x = \frac{dv_x}{dt} = \frac{\partial v_x}{\partial x} \frac{dx}{dt} + \frac{\partial v_x}{\partial y} \frac{dy}{dt} + \frac{\partial v_x}{\partial z} \frac{dz}{dt} \\= \frac{\partial v_x}{\partial x} v_x + \frac{\partial v_x}{\partial y} v_y + \frac{\partial v_x}{\partial z} v_z.$$ As you can see, we're never "dividing by" the entire vector $$\vec{x}$$ when we take these derivatives; we only ever "divide by" its components $$x$$, $$y$$, or $$z$$, which is a valid mathematical operation.
Your expression doesn't make sense for (non-trivial) vectors, only for scalars (which are one-dimensional, a.k.a. "trivial" vectors).
The expression $$a=v\frac{dv}{dx}$$ only makes sense if $$v$$ and $$x$$ are both functions from one dimension to one dimension. You will never see the expression $$\vec{a}=\vec{v}\frac{d\vec{v}}{d\vec{x}}$$, because such an operation is not defined.
In general, multiplication of vectors is only well-defined in specific circumstances. There are three commonly-used multiplications of vectors: the dot product (which returns a scalar), the cross product (which returns a vector), and the tensor product (which returns a matrix). The cross product, in addition, only makes sense for three-dimensional and seven-dimensional vectors (the reason for this has to do with the existence of certain extensions of complex numbers called the quaternions and the octonions).
Division of two vectors is in general not defined, because it's not possible to undo either type of vector multiplication. For example, for any given vector, there are an infinite number of other vectors whose dot product with that vector will be zero (namely, all vectors that are perpendicular to it), and similarly, there are an infinite number of other vectors whose cross product with that vector will be zero (namely, all vectors that are parallel to it). Defining an inverse would require a one-to-one correspondence between the input and output of this multiplication, so defining division is not strictly possible.
That said, defining differentiation of vectors is different, namely because differentiation does not involve dividing two vectors. For a vector which is a function of space $$\vec{v}(\vec{x})$$, there are four ways to define differentiation:
• differentiation by a scalar parameter $$t$$, which is defined for an $$n$$-dimensional vector as:
$$\frac{d\vec{v}(\vec{x},t)}{dt}=\left\langle \frac{d v_1(\vec{x},t)}{d t},...,\frac{d v_n(\vec{x},t)}{d t}\right\rangle$$
• the divergence, which is defined for an $$n$$-dimensional vector as:
$$\nabla \cdot \vec{v}(\vec{x})=\sum_{i=0}^n \frac{\partial v_i(\vec{x})}{\partial x_i}$$
• the curl, which is defined for three-dimensional and seven-dimensional vectors, and, in three dimensions, in Cartesian coordinates, is:
$$\nabla \times\vec{v}(\vec{x})=\left\langle\frac{\partial v_y(\vec{x})}{\partial z}-\frac{\partial v_z(\vec{x})}{\partial y},\frac{\partial v_z(\vec{x})}{\partial x}-\frac{\partial v_x(\vec{x})}{\partial z},\frac{\partial v_x(\vec{x})}{\partial y}-\frac{\partial v_y(\vec{x})}{\partial x}\right\rangle$$
• the Jacobian matrix $$\frac{d\vec{v}(\vec{x})}{d\vec{x}}=J$$, whose entries $$J_{ij}$$ are given by
$$J_{ij}=\frac{\partial v_i(\vec{x})}{\partial x_j}$$
As you can see, none of these involve dividing a vector by a vector; in fact, all of them involve multiplying a vector by another vector, namely, the vector of partial derivative operators $$\nabla$$. For the divergence, this is a dot product; for the curl, this is a cross product; and for the Jacobian matrix, this is a tensor product.
So how do we rework the one-dimensional expression for acceleration so that it makes sense in any number of dimensions? The key is to start with the basic definition, assuming, as you did, that there is no explicit time dependence for $$\vec{v}$$:
\begin{align} \vec{a}(\vec{x})&=\frac{d\vec{v}(\vec{x})}{dt}\\ &=\left\langle \sum_{i=1}^n \frac{\partial v_1(\vec{x})}{\partial x_i}\frac{\partial x_i}{\partial t},...,\sum_{i=1}^n\frac{\partial v_n(\vec{x})}{\partial x_i}\frac{\partial x_i}{\partial t}\right\rangle\\ &=\left\langle \sum_{i=1}^n \frac{\partial v_1(\vec{x})}{\partial x_i}v_i,...,\sum_{i=1}^n\frac{\partial v_n(\vec{x})}{\partial x_i}v_i\right\rangle\\ &=(\vec{v}\cdot\nabla)\vec{v} \end{align}
There is nothing wrong with differentiating a vector, since it doesn't involve division by a vector.
Well, we know how to multiply vectors, via the dot product $$\vec a \cdot \vec b$$. So we could define a division through:
$$\frac{\vec a}{\vec b} \doteqdot \vec a \cdot \frac{1}{\vec b}.$$
Now we need to define what is meant by $$\frac{1}{\vec b}$$. Well, what about:
$$\frac{1}{\vec b} \doteqdot \frac{\vec b}{|\vec b|^2}.$$
This vector inversion gives a new vector with same direction, but reciprocal magnitude. In fact it's how we define an inversion for other things like the special conformal transformation. Seems like as good a definition as any. Then division of vectors becomes:
$$\frac{\vec a}{\vec b} = \frac{\vec a \cdot \vec b}{|\vec b|^2} = \frac{|a|}{|b|}\cos\theta.$$
This division satisfies what you would expect:
$$\frac{\vec a}{\vec a} = 1,$$
but different from usual division, the inverse of $$\vec a$$ isn't unique.
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Shared15 - Taylor Series Assignment / Taylor Series Notes.sagewsOpen in CoCalc
This material was developed by Aaron Tresham at the University of Hawaii at Hilo and is
### Prerequisites:
• Intro to Sage
• Series
# Taylor Series
Definition: A power series centered at $x=a$ is a series of the form $\sum_{n=0}^{\infty}\left[c_n\cdot(x-a)^n\right]=c_0+c_1\cdot(x-a)+c_2\cdot(x-a)^2+\cdots$
where $a$ and all the $c_i$ are constants.
Notice that a power series is a function of $x$. The domain is all values of $x$ for which the series converges. There is always at least one number in the domain, namely $a$.
You can think of a power series as a polynomial with (possibly) infinite degree.
One of the nice things about power series is that it's easy to differentiate and integrate them. Just like regular polynomials, you can take the derivative or integral of each term first, and then add up the results.
$\frac{d}{dx}\sum_{n=0}^{\infty}\left[c_n\cdot(x-a)^n\right]=\sum_{n=0}^{\infty}\frac{d}{dx}\left[c_n\cdot(x-a)^n\right]$
$\int\sum_{n=0}^{\infty}\left[c_n\cdot(x-a)^n\right]\,dx=\sum_{n=0}^{\infty}\int\left[c_n\cdot(x-a)^n\right]\,dx$
## Example 1
A geometric series with $x$ as the common ratio is a simple example of a power series ($a=0$ and $c_i=c$ for all $i$):
$\sum_{n=0}^{\infty}c\cdot x^n$
We know when this series converges and what it coverges to:
$\sum_{n=0}^{\infty}c\cdot x^n=\frac{c}{1-x}$
provided that $-1 < x < 1$.
So $\displaystyle\sum_{n=0}^{\infty}c\cdot x^n$ is a function with domain $(-1,1)$. On this domain, this power series function happens to be equal to the function $\displaystyle f(x)=\frac{c}{1-x}$.
## Taylor Series
This raises the question of whether other functions are equal to the sum of a power series (at least for part of their domains).
In other words, given a function $f(x)$, can we find $c_i$ so that $f(x)=\sum_{n=0}^{\infty}\left[c_n\cdot(x-a)^n\right]$
for some interval of x-values?
This process of representing a function by a power series is called "expanding" the function into a series. The power series you get is called a Taylor series expansion of $f(x)$, after mathematician Brook Taylor (1685-1731).
Expanding functions into Taylor series and differentiating and integrating the series had a number of applications back then. For example, you can use Taylor Series to approximate the values of numbers like $\pi$ and $e$. Or consider the logarithmic and trigonometric functions. These are often difficult to calculate, but if you expand these into Taylor series, then you can approximate values of these functions using only polynomials (and polynomials only require arithmetic to calculate).
Fortunately, finding the right power series to represent a function is fairly straightforward, as long as the function is repeatedly differentiable. The secret is to find derivatives of every order and evaluate them at $x=a$.
Suppose $\displaystyle f(x)=\sum_{n=0}^{\infty}\left[c_n\cdot(x-a)^n\right]$, take the $m^{th}$ derivative, and plug in $x=a$:
$f^{(m)}(a)=\sum_{n=0}^{\infty}\left.\frac{d^m}{dx^m}\left[c_n\cdot(x-a)^n\right]\right|_{x=a}$
For all $n < m$, $\displaystyle \frac{d^m}{dx^m}\left[c_n\cdot(x-a)^n\right]=0$, so when you plug in $x=a$ it's still 0.
For all $n > m$, $\displaystyle \frac{d^m}{dx^m}\left[c_n\cdot(x-a)^n\right]=\frac{n!}{(n-m)!}\cdot c_n\cdot(x-a)^{n-m}$, with $n-m > 0$. So when you plug in $x=a$, you get $\displaystyle \frac{n!}{(n-m)!}\cdot c_n\cdot(a-a)^{n-m}=0$.
Finally, when $n=m$, $\displaystyle \frac{d^m}{dx^m}\left[c_n\cdot(x-a)^n\right]=n!\cdot c_n\cdot (x-a)^0=n!\cdot c_n=m!\cdot c_m$.
Putting this all together, we see that $\displaystyle f^{(m)}(a)=m!\cdot c_m$.
Therefore, $\displaystyle c_m=\frac{f^{(m)}(a)}{m!}$.
Conclusion: If a function $f(x)$ with derivatives of every order may be represented by a power series centered at $x=a$ on some interval $I$, then that power series is $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}\cdot(x-a)^n,$
where the series converges on the interval $I$.
Notice the "if" in the last sentence. There are functions that are not equal to the sum of their Taylor series, even if the series converges. We are not going to deal with such functions in this lab.
## Example 2
Find the Taylor series of $f(x)=e^x$ centered at $x=0$ (Taylor series centered at 0 are also called Maclaurin series).
We know $f^{(n)}(x)=f(x)=e^x$ in this case. Since $e^0=1$, we have $f^{(n)}(0)=1$ for all $n$.
Thus, $\displaystyle c_n=\frac{1}{n!}$.
If $e^x$ equals the sum of its Taylor series, then $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$. For this particular function, the Taylor series converges for all $x$, and $e^x$ does equal the sum of the series (take my word for it).
## Taylor Polynomials
The partial sums of a Taylor series are actual polynomials, called Taylor polynomials. In other words, the Taylor polynomial of degree $m$ is $\displaystyle\sum_{n=0}^m\frac{f^{(n)}(a)}{n!}(x-a)^n$.
We can approximate a Taylor series to whatever level of accuracy we want by using a Taylor polynomial of high enough degree.
Note: If $f^{(m)}(a)=0$, then this polynomial will actually have degree less than $m$.
Notice that the Taylor polynomial of degree 1 is $\displaystyle\sum_{n=0}^1\frac{f^{(n)}(a)}{n!}(x-a)^n=\frac{f^{(0)}(a)}{0!}(x-a)^0+\frac{f^{(1)}(a)}{1!}\cdot(x-a)^1=f(a)+f'(a)(x-a)$.
Does this look familiar? It should! This is an equation for the tangent line to $f$ at $a$. In other words, Taylor polynomials are generalizations of the tangent line to higher degree polynomials.
Also note that the Taylor polynomial of degree 2 is $\displaystyle\sum_{n=0}^2\frac{f^{(n)}(a)}{n!}(x-a)^n=$ $\frac{f^{(0)}(a)}{0!}(x-a)^0+\frac{f^{(1)}(a)}{1!}\cdot(x-a)^1+\frac{f^{(2)}(a)}{2!}\cdot(x-a)^2=$
$f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2$
This is the formula we used for Quadratic Approximation in Calc 1.
## Example 3
We will use Taylor polynomials to approximate the value of $e$. We saw above that $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$.
The Taylor polynomial of degree $m$ is $\displaystyle\sum_{n=0}^{m}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots+\frac{x^m}{m!}$.
So $\displaystyle e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots+\frac{x^m}{m!}$. The approximation improves as $m$ increases.
If we plug in $x=1$, we get $\displaystyle e\approx 1+1+\frac{1}{2}+\frac{1}{6}+\cdots+\frac{1}{m!}$.
Let's approximate the value of $e$.
%auto
%var n
@interact
def _(m=input_box(1,label='Degree of Taylor polynomial',width=20)):
print "Approximation =",N(sum(1/factorial(n),n,0,m),digits=30)
print "Actual value =",N(e,digits=30)
How big must $m$ be so that our approximation is correct for all the decimal places shown?
Let's look at some graphs.
$f(x)=e^x$ is plotted in black, along with three Taylor polynomials (blue = degree 2, green = degree 3, red = degree 4)
%var n
plot(sum(x^n/factorial(n),n,0,2),xmin=-1, xmax=1,color='blue',legend_label='degree 2')+plot(sum(x^n/factorial(n),n,0,3),xmin=-1, xmax=1,color='green',legend_label='degree 3')+plot(sum(x^n/factorial(n),n,0,4),xmin=-1, xmax=1,color='red',legend_label='degree 4')+plot(e^x,xmin=-1, xmax=1,ymax=e,color='black',legend_label='$e^x$')
Notice that the higher the degree the better the approximation.
Now try xmin=-10; xmax=10; ymax=1000
%var n
plot(sum(x^n/factorial(n),n,0,2),xmin=-10,xmax=10,color='blue',legend_label='degree 2')+plot(sum(x^n/factorial(n),n,0,3),xmin=-10, xmax=10,color='green',legend_label='degree 3')+plot(sum(x^n/factorial(n),n,0,4),xmin=-10, xmax=10,color='red',legend_label='degree 4')+plot(e^x,xmin=-10, xmax=10,ymax=1000,color='black',legend_label='$e^x$')
Notice that the Taylor polynomials are better approximations the closer you get to the center of the expansion (in this case $x=0$), just like the tangent line is a better approximation the closer you get to the point of tangency.
In the animation below, we see taylor polynomials of increasing degree (in blue) plotted with $f(x)=e^x$ (in black).
%auto
s=[]
taylorpoly=1
p=plot(e^x,-5,15,ymin=-100,ymax=1500,color='black')
for n in [1..10]:
taylorpoly+=x^n/factorial(n)
s+=[p+plot(taylorpoly,-5,15)]
a=animate(s)
show(a,delay=50)
We can use Sage to calculate Taylor polynomials using the "taylor" command. This command takes four arguments: the expression or function to expand, the variable of expansion, the center of the expansion, and the degree of polynomial you want.
## Example 4
Use Sage to find the 10th-degree Taylor polynomial centered at $x=0$ for $f(x)=e^x$.
show(taylor(e^x,x,0,10))
$\displaystyle \frac{1}{3628800} \, x^{10} + \frac{1}{362880} \, x^{9} + \frac{1}{40320} \, x^{8} + \frac{1}{5040} \, x^{7} + \frac{1}{720} \, x^{6} + \frac{1}{120} \, x^{5} + \frac{1}{24} \, x^{4} + \frac{1}{6} \, x^{3} + \frac{1}{2} \, x^{2} + x + 1$
## Example 5
Find the 15th-degree Taylor polynomials of $\sin(x)$ centered at $x=0$, $x=-\pi$, and $x=\pi$.
show(taylor(sin(x),x,0,15))
$\displaystyle -\frac{1}{1307674368000} \, x^{15} + \frac{1}{6227020800} \, x^{13} - \frac{1}{39916800} \, x^{11} + \frac{1}{362880} \, x^{9} - \frac{1}{5040} \, x^{7} + \frac{1}{120} \, x^{5} - \frac{1}{6} \, x^{3} + x$
show(taylor(sin(x),x,-pi,15))
$\displaystyle -\pi + \frac{1}{1307674368000} \, {\left(\pi + x\right)}^{15} - \frac{1}{6227020800} \, {\left(\pi + x\right)}^{13} + \frac{1}{39916800} \, {\left(\pi + x\right)}^{11} - \frac{1}{362880} \, {\left(\pi + x\right)}^{9} + \frac{1}{5040} \, {\left(\pi + x\right)}^{7} - \frac{1}{120} \, {\left(\pi + x\right)}^{5} + \frac{1}{6} \, {\left(\pi + x\right)}^{3} - x$
show(taylor(sin(x),x,pi,15))
$\displaystyle \pi - \frac{1}{1307674368000} \, {\left(\pi - x\right)}^{15} + \frac{1}{6227020800} \, {\left(\pi - x\right)}^{13} - \frac{1}{39916800} \, {\left(\pi - x\right)}^{11} + \frac{1}{362880} \, {\left(\pi - x\right)}^{9} - \frac{1}{5040} \, {\left(\pi - x\right)}^{7} + \frac{1}{120} \, {\left(\pi - x\right)}^{5} - \frac{1}{6} \, {\left(\pi - x\right)}^{3} - x$
Notice that some of the coefficients (the $c_i$) are 0 in this example.
## Example 6
Find the Taylor polynomials of $\cos(x)$ centered at $x=0$ of degrees 5, 10, and 15.
show(taylor(cos(x),x,0,5))
$\frac{1}{24} \, x^{4} - \frac{1}{2} \, x^{2} + 1$
show(taylor(cos(x),x,0,10))
$-\frac{1}{3628800} \, x^{10} + \frac{1}{40320} \, x^{8} - \frac{1}{720} \, x^{6} + \frac{1}{24} \, x^{4} - \frac{1}{2} \, x^{2} + 1$
show(taylor(cos(x),x,0,15))
$-\frac{1}{87178291200} \, x^{14} + \frac{1}{479001600} \, x^{12} - \frac{1}{3628800} \, x^{10} + \frac{1}{40320} \, x^{8} - \frac{1}{720} \, x^{6} + \frac{1}{24} \, x^{4} - \frac{1}{2} \, x^{2} + 1$
## Example 7
Find the 10th-degree Taylor polynomial centered at $x=0$ of $\displaystyle f(x)=\frac{c}{1-x}$.
%var c
show(taylor(c/(1-x),x,0,10))
$\displaystyle c x^{10} + c x^{9} + c x^{8} + c x^{7} + c x^{6} + c x^{5} + c x^{4} + c x^{3} + c x^{2} + c x + c$
Here's a graph of this one for $c=1$.
plot(taylor(1/(1-x),x,0,10))+plot(1/(1-x),color='black',ymax=6,ymin=0)
|
# Introduction To The Practice Of Statistics
Introduction of the statistics should be the formal science that of generating the efficient usage in numerical terms. These are having the introduction of median, mean, range and mode. This introduction can alsohave interpretation and analysis of the numerical terms. In this article we are going to learn about example and practice problems with the introduction of statistics.
Please express your views of this topic Assumptions of Chi Square Test by commenting on blog.
## Example problems for statistics:
Here we are going to learn few example problems in statistics.
Example 1:
Find the mode of the numbers 2,6,5,1,2,8,9,2.
Solution:
Given numbers are, 2,6,5,1,2,8,9,2.
Ascending order of numbers = 1,2,2,2,5,6,8,9
Here, The number '2' is occurring 3 times.
So, Mode value of the given numbers is 2.
Mode = 2
Example 2:
What is the median of the given series?
94,42,61,58,39
Solution:
Given series is, 94,42,61,58,39
The sorting list of given series is, 39,42,58,61,94
Here, total numbers = 5. This is the odd number.
Therefore, Median = Middle value in sorted list
= 58
Example 3:
Find the range of the numbers 57,3,61,23,85.
Solution:
Given numbers are, 57,3,61,23,85.
The largest number = 85
The smallest number = 3
Thus, Range = Largest number - Smallest number
= 85 - 3
= 82
Between, if you have problem on these topics Categorical Data Definition, please browse expert math related websites for more help on sample paper for class 11 cbse .
## Few more problems for statistics:
Example 4:
The weights of 6 students in a class are, 45,58,47,51,39,55. Find the mean of the marks?
Solution:
Given weights of 6 students in class are,
45,58,47,51,39,55
The formula for finding mean is,
Sum of all the marks = 45+58+47+51+39+55
= 295
Total number of students = 6
Therefore, mean = `(295)/(6)`
= 49.17
Example 5:
What is the median of the given numbers, 87,12,19,28,47,36,44,50.
Solution:
Given, data sets are, 87,12,19,28,47,36,44,50
First step is finding the sorting list of numbers. That is, 12,19,28,36,44,47,50,87
Here, total numbers is even = 8
So, Median = `(m_(1)+m_(2))/(2)`
Here, m1 and m2 are the two middle terms = 36 and 44
Therefore, Median = `(36+44)/(2)`
= `(80)/(2)`
= 40
Practice problems:
Practice problem 1:
Find the mean of the numbers 13,21,9,8,55
Practice problem 2:
Find the range of the numbers 95,41,13,74,62,15.
|
Informative line
### Application Of Derivative
Find the equation of the tangent line and Normal to the curve, the derivative of the function at the given point. Practice to find the point on the curve at which tangent is horizontal (parallel to x axis).
# Finding the Derivatives at Particular Values of x
When we are asked to find $$f'(a)$$, we first find the derivative of $$f$$ with respect to $$x$$ and then put $$x=a$$. It will not mean the derivative of the value $$f(a)$$ because this value will always be 0, since $$f(a)$$ is a constant.
Note:
Do not simplify the expression of derivative, just put the value of $$x$$ as soon as differentiation step is complete.
#### If $$f(t)=\dfrac{t^2-5t-1}{t^3}$$, find the value of $$f'(2)$$.
A $$\dfrac{25}{18}$$
B $$\dfrac{19}{16}$$
C $$\dfrac{4}{3}$$
D $$\dfrac{-7}{8}$$
×
$$f(t)=\dfrac{t^2-5t-1}{t^3}$$
$$\Rightarrow$$ $$f'(t)$$ $$=\underbrace{\dfrac{t^3\dfrac{d}{dt}(t^2-5t-1)-(t^2-5t-1)\dfrac{d}{dt}t^3}{(t^3)^2}}_{Use \,Quotient\,Rule}$$
$$=\dfrac{t^3×(2t-5)-(t^2-5t-1)×3t^2}{t^6}$$
$$f'(2)$$$$\dfrac{8×(4-5)-(4-10-1)×3×4}{2^6}$$
$$=\dfrac{-8+84}{64}$$
$$=\dfrac{76}{64}$$
$$=\dfrac{19}{16}$$
### If $$f(t)=\dfrac{t^2-5t-1}{t^3}$$, find the value of $$f'(2)$$.
A
$$\dfrac{25}{18}$$
.
B
$$\dfrac{19}{16}$$
C
$$\dfrac{4}{3}$$
D
$$\dfrac{-7}{8}$$
Option B is Correct
# Equation of Tangent to a Curve at a given Point
Let $$y=f(x)$$ be a curve and $$(a,\,f(a))$$ be any point on it. The equation of tangent at this point is given by.
$$y-f(a)=$$ $$f'(a)$$ $$(x-a)$$
#### Find the equation of tangent to the curve $$y=\dfrac{1+2x}{3-4x}$$ at the point $$(1,\,-1)$$ on it.
A $$5x+y+17=0$$
B $$x+2y+18=0$$
C $$2x-y-3=0$$
D $$4x+11y-7=0$$
×
$$y=\dfrac{1+2x}{3-4x}$$
$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{(3-4x)×2-(1+2x)×-4}{(3-4x)^2}$$
$$=\dfrac{6-8x-4+8x}{(3-4x)^2}$$
$$=\dfrac{2}{(3-4x)^2}$$
$$\Rightarrow$$ $$y'$$ at $$(1,\,-1)$$ $$=\dfrac{2}{(3-4)^2}=2$$
Equation of tangent is $$y-(-1)=$$ $$y'$$$$(x-1)$$
$$\Rightarrow\,y+1=2(x-1)$$
$$\Rightarrow\,2x-y-3=0$$
### Find the equation of tangent to the curve $$y=\dfrac{1+2x}{3-4x}$$ at the point $$(1,\,-1)$$ on it.
A
$$5x+y+17=0$$
.
B
$$x+2y+18=0$$
C
$$2x-y-3=0$$
D
$$4x+11y-7=0$$
Option C is Correct
# Finding the Point on the Curve when Slope of Tangent is given
• Given the slope, equate the derivative of the function to that slope and solve for $$x$$, hence find the point.
#### Find the point(s) on the curve $$y=\dfrac{x+1}{x-1}$$ at which the slope of tangent is –2.
A $$(0,\,-1)\,\;or\;\;(2,\,3)$$
B $$(0,\,1)\,\;or\;\;(3,\,5)$$
C $$\left(5,\,\dfrac{3}{2}\right)\,\;or\;\;(1,\,0)$$
D $$(2,\,3)\,\;or\;\;(-1,\,0)$$
×
$$y=\dfrac{x+1}{x-1}$$
$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{(x-1)×1-(x+1)×1}{(x-1)^2}$$
$$\therefore\,\dfrac{dy}{dx}=\dfrac{-2}{(x-1)^2}$$
Now
$$\dfrac{dy}{dx}=-2$$
$$\Rightarrow\,\dfrac{-2}{(x-1)^2}=-2$$
$$\Rightarrow\,(x-1)^2=1$$
$$\Rightarrow\,x-1=\pm1$$
$$\Rightarrow\,x=2\;or\;x=0$$
$$\therefore\,$$ required points are $$(0,\,-1)\;or\;(2,\,3)$$
### Find the point(s) on the curve $$y=\dfrac{x+1}{x-1}$$ at which the slope of tangent is –2.
A
$$(0,\,-1)\,\;or\;\;(2,\,3)$$
.
B
$$(0,\,1)\,\;or\;\;(3,\,5)$$
C
$$\left(5,\,\dfrac{3}{2}\right)\,\;or\;\;(1,\,0)$$
D
$$(2,\,3)\,\;or\;\;(-1,\,0)$$
Option A is Correct
# Evaluating Derivatives of Mixed Function at Particular Values of x
If we are given a mixed function and its derivative is desired at particular values of $$x$$, then use the derivative of component function given at that $$x$$ and appropriate rule of differentiation.
#### If $$f$$ and $$g$$ are differentiable functions such that $$f'(3)$$$$=4$$, $$f(3)=5$$, $$g'(3)$$$$=7$$, $$g(3)=-2$$ then the value of $$\left(\dfrac{f}{g}\right)' at\, x=3\, is$$
A $$\dfrac{-43}{4}$$
B $$\dfrac{76}{3}$$
C $$\dfrac{18}{5}$$
D –19
×
$$\left(\dfrac{f}{g}\right)'=\dfrac{gf'-fg'}{g^2}$$
(Quotient Rule)
$$\left(\dfrac{f}{g}\right)'\Bigg|_{x=3}=\dfrac{g(3)f'(3)-f(3)g'(3)}{(g(3))^2}$$
$$=\dfrac{-2×4-5×7}{(-2)^2}$$
$$=\dfrac{-8-35}{4}$$
$$=\dfrac{-43}{4}$$
### If $$f$$ and $$g$$ are differentiable functions such that $$f'(3)$$$$=4$$, $$f(3)=5$$, $$g'(3)$$$$=7$$, $$g(3)=-2$$ then the value of $$\left(\dfrac{f}{g}\right)' at\, x=3\, is$$
A
$$\dfrac{-43}{4}$$
.
B
$$\dfrac{76}{3}$$
C
$$\dfrac{18}{5}$$
D
–19
Option A is Correct
# Equation of Normal to a Curve at a given Point
If two lines are perpendicular to each other then the product of their slopes is –1
i.e. $$m_1m_2=-1$$ if $$m_1,\,m_2$$ are slopes of $$\bot$$ lines.
Let $$y=f(x)$$ be a curve and $$(a,\,f(a))$$ be any point on it, then the equation of normal at this point is
$$y-f(a)={\dfrac{-1}{f'(a)}}(x-a)$$
#### Find the equation of normal to the curve $$y=3x-2\sqrt x$$ at the point $$(4,\,8)$$ on it.
A $$x+7y-84=0$$
B $$3x+8y-1=0$$
C $$2x+5y-48=0$$
D $$11x+2y+7=0$$
×
$$y=3x-2\sqrt x$$
$$\Rightarrow$$ $$y'$$ $$=3-\dfrac{2}{2\sqrt x}$$
$$\Rightarrow$$$$y'$$$$=3-\dfrac{1}{\sqrt x}$$
$$\Rightarrow$$ $$y'$$ at $$(4,\,8)=3-\dfrac{1}{2}=\dfrac{5}{2}$$
Equation of normal is $$y-8=\dfrac{-1}{y'}(x-4)$$
$$\Rightarrow\,y-8=\dfrac{-1}{\dfrac{5}{2}}(x-4)$$
$$\Rightarrow\,y-8=\dfrac{-2}{5}(x-4)$$
$$\Rightarrow\,5y-40=-2x+8$$
$$\Rightarrow\,2x+5y-48=0$$
### Find the equation of normal to the curve $$y=3x-2\sqrt x$$ at the point $$(4,\,8)$$ on it.
A
$$x+7y-84=0$$
.
B
$$3x+8y-1=0$$
C
$$2x+5y-48=0$$
D
$$11x+2y+7=0$$
Option C is Correct
# Finding the Point on the Curve at which Tangent is Horizontal(Parallel to x axis)
To find the point where tangent is horizontal or parallel to $$x$$ axis.
1. Take the derivative, find $$\dfrac{dy}{dx}$$
2. Put $$\dfrac{dy}{dx}=0$$
3. Solve for $$x$$.
#### For what value of $$x$$ will the graph of $$y=2x^3-15x^2+36x+1$$ have a horizontal tangent ?
A $$x=2\;or\;x=3$$
B $$x=8\;or\;x=1$$
C $$x=-7\;or\;x=5$$
D $$x=17\;or\;x=-18$$
×
For horizontal tangent $$\to \dfrac{dy}{dx}=0$$
$$y=2x^3-15x^2+36x+1$$
$$\Rightarrow\,\dfrac{dy}{dx}=6x^2-30x+36$$
then $$\dfrac{dy}{dx}=0$$
$$\Rightarrow\,6x^2-30x+36=0$$
$$\Rightarrow\,x^2-5x+6=0$$
$$\Rightarrow\,(x-2)\,(x-3)=0$$
$$\Rightarrow\,x=2\;or\;x=3$$
### For what value of $$x$$ will the graph of $$y=2x^3-15x^2+36x+1$$ have a horizontal tangent ?
A
$$x=2\;or\;x=3$$
.
B
$$x=8\;or\;x=1$$
C
$$x=-7\;or\;x=5$$
D
$$x=17\;or\;x=-18$$
Option A is Correct
# Finding the Equation of Tangent to a Curve using the given Slope
Sometimes equation of a tangent is desired whose contact point is not given but its slope is given, either directly or indirectly.
• First we find the contact point, and then find equation of tangent.
#### Find the equation of tangent line to $$y=x^3+x-2$$ which is parallel to straight line $$y=4x-1$$
A $$4x-y-4=0 \;\;\&\;\;4x-y=0$$
B $$2x+3y+4=0 \;\;\&\;\;x-y=0$$
C $$4x+y+7=0 \;\;\&\;\;x+2y+8=0$$
D $$x+3y=0 \;\;\&\;\;2x+y+7=0$$
×
If two lines are parallel then there slopes are equal. So the required tangent has slope = 4.
(Same as slope of $$y=4x-1$$)
$$\therefore\,\dfrac{dy}{dx}=4$$
$$\Rightarrow\,3x^2+1=4$$
$$\Rightarrow\,3x^2=3$$
$$\Rightarrow\,x=\pm1$$
$$\therefore$$ The point at which tangent have slope 4 are $$(1,\,0)$$ and $$(-1,\,-4)$$.
$$\therefore$$ There are two tangents -
(1) $$y-0=4(x-1)$$
$$\Rightarrow\,4x-y-4=0$$
(2) $$y+4=4(x+1)$$
$$\Rightarrow\,4x-y=0$$
(There can be more then one tangent to a curve of same slope).
### Find the equation of tangent line to $$y=x^3+x-2$$ which is parallel to straight line $$y=4x-1$$
A
$$4x-y-4=0 \;\;\&\;\;4x-y=0$$
.
B
$$2x+3y+4=0 \;\;\&\;\;x-y=0$$
C
$$4x+y+7=0 \;\;\&\;\;x+2y+8=0$$
D
$$x+3y=0 \;\;\&\;\;2x+y+7=0$$
Option A is Correct
# Identifying the Parameters of a Curve
If slope of tangent at some point of a graph is given, equate it to the derivative at that point. If the function contains some parameter, they can be solved.
#### A curve has the equation of the form $$y=ax^2+bx+c$$. It has slope 9 at $$x=1$$, slope 13 at $$x=2$$ and passes through $$(0,\,3)$$, find the curve.
A $$y=2x^2+5x+3$$
B $$y=5x^2+8x+7$$
C $$y=7x^2=x-9$$
D $$y=3x^2+8x+18$$
×
Slope = 9 at $$x=1$$
$$\Rightarrow\,\dfrac{dy}{dx}\Bigg|_{x=1}=9$$
$$\Rightarrow\,2ax+b\Bigg|_{x=1}=9$$
$$\Rightarrow\,2a+b=9\to$$ (1)
Slope = 13 at $$x=2$$
$$\Rightarrow\,\dfrac{dy}{dx}\Bigg|_{x=2}=13$$
$$\Rightarrow\,4a+b=13\to$$ (2)
Passes through $$(0,\,3)$$
$$\Rightarrow3=c$$
Solving equation (1) and (2) for $$'a'$$ and $$'b'$$ we get
$$a=2,\,\;b=5$$
$$\therefore$$ the required curve is $$y=2x^2+5x+3$$.
### A curve has the equation of the form $$y=ax^2+bx+c$$. It has slope 9 at $$x=1$$, slope 13 at $$x=2$$ and passes through $$(0,\,3)$$, find the curve.
A
$$y=2x^2+5x+3$$
.
B
$$y=5x^2+8x+7$$
C
$$y=7x^2=x-9$$
D
$$y=3x^2+8x+18$$
Option A is Correct
|
# Eureka Math Precalculus Module 4 End of Module Assessment Answer Key
## Engage NY Eureka Math Precalculus Module 4 End of Module Assessment Answer Key
### Eureka Math Precalculus Module 4 End of Module Assessment Task Answer Key
Question 1.
a. In the following diagram, triangle XYZ has side lengths a, b, and c as shown. The angle α indicated is acute. Show that the area A of the triangle is given by A=$$\frac{1}{2}$$ ab sin(α).
Draw an altitude as shown. Call its length h.
We have sin (α ) = $$\frac{h}{b}$$ , so h = b sin (α) .
The area A of the triangle is given as “half base times height.” So
A = $$\frac{1}{2}$$ × a × b sin (α ) = $$\frac{1}{2}$$ ab sin (α ) .
b. In the following diagram, triangle PQR has side lengths p, q, and r as shown. The angle β indicated is obtuse. Show that the area A of the triangle is given by A=$$\frac{1}{2}$$ pq sin(β).
Draw in an altitude as shown. Call its length h .
We have sin (π – β ) = $$\frac{h}{p}$$, so h = p sin (π – β) . Since sin (π – β ) = sin (β) , this can be rewritten h = p sin (β) .
The area A of the triangle is thus $$\frac{1}{2}$$ × q × p sin (β ) = $$\frac{1}{2}$$ pq sin (β) .
c. To one decimal place, what is the area of the triangle with sides of lengths 10 cm, 17 cm, and 21 cm? Explain how you obtain your answer.
Let θ be the measure of the angle between the sides of lengths 10 cm and 17 cm . By the law of cosines, we have 212 = 102 + 172 – 2 ∙ 10 ∙ 17 cos (θ) . This gives
cos (θ ) = $$\frac{52}{340}$$ = $$\frac{13}{85}$$, and so θ = cos–1 ($$\frac{13}{85}$$) ≈ 1.42 radians.
$$\frac{1}{2}$$ ∙ 10 ∙ 17 sin (θ ) ≈ 85 sin (1.42) ≈ 84.0
The area of the triangle is 84.0 square centimeters.
Question 2.
Triangle ABC with side lengths a, b, and c as shown is circumscribed by a circle with diameter d.
a. Show that $$\frac{a}{\sin (A)}$$ =d.
Consider the pointA’ on the circle with $$\overline{\mathrm{A}^{\prime} B}$$ a diameter of the circle.
By Thales’ theorem (an inscribed angle that intercepts a semi-circle is a right angle),∠A’CB is a right angle. Thus,sin (A’ ) = $$\frac{a}{d}$$.
But by the inscribed angle theorem (angles intercepting the same arc are congruent), the inscribed angle atA’ has the same measure as the inscribed angle atA . So,sin (A’ ) = sin (A) , and our equation readssin (A ) = $$\frac{a}{d}$$. Rearranging gives $$\frac{a}{\sin (A)}$$ = d .
b. The law of sines states that $$\frac{a}{\sin (A)}$$=$$\frac{b}{\sin (B)}$$=$$\frac{c}{\sin (C)}$$ for any triangle ABC with side lengths a, b, and c (with the side of length a opposite vertex A, the side of length b opposite vertex B, and the side of length c opposite vertex C). Explain why the law of sines holds for all triangles.
The relationship between a, sin(A), and d holds for any side of the triangle. So we also have $$\frac{b}{\sin (B)}$$ = d and $$\frac{c}{\sin (C)}$$ = d . This shows that $$\frac{a}{\sin (A)}$$ = $$\frac{b}{\sin (B)}$$ = $$\frac{c}{\sin (C)}$$ for a triangle circumscribed by a circle.
As every triangle can be circumscribed by a circle, the law of sines,
$$\frac{a}{\sin (A)}$$ = $$\frac{b}{\sin (B)}$$ = $$\frac{c}{\sin (C)}$$, thus holds for all triangles.
c. Prove that c2=a2+b2-2ab cos(C) for the triangle shown in the original diagram.
Draw an altitude as shown for the triangle, and identify the three lengths x , y , and h as shown.
Now x = b cos (C ),h = b sin (C) , and
y = a – x = a – b cos (C) .
Applying the Pythagorean theorem to the right triangle on the right, we have
y2 + h2 = c2
(a – b cos (C))2 + (b sin (C))2 = c2
a2 – 2ab cos (C ) + b2 cos2 (C ) + b2 sin2 (C ) = c2
Using cos2 (C ) + sin2 (C ) = 1 this reads
a2 – 2ab cos (C ) + b2 = c2
or
c2 = a2 + b2 – 2ab cos (C) .
Question 3.
Beatrice is standing 20 meters directly east of Ari, and Cece is standing 15 meters directly northeast of Beatrice.
a. To one decimal place, what is the distance between Ari and Cece?
The following diagram depicts the situation described:
By the law of cosines
|AC |2 = 202 + 152 – 2 ∙ 15 ∙ 20 ∙ cos ($$\frac{3\pi}{4}$$)
|AC |2 = 400 + 225 – 600 (– $$\frac{1}{\sqrt{2}}$$)
|AC |2 = 625 + 300 $$\sqrt{2}$$
|AC | = $$\sqrt{625+300 \sqrt{2}}$$ ≈ 32.4
Thus, the distance between Ari and Cece is approximately 32.4 meters.
b. To one decimal place, what is the measure of the smallest angle in the triangle formed by Ari, Beatrice, and Cece?
The angle of the smallest measure in a triangle lies opposite the shortest side of the triangle. Thus, we seek the measure of the angle at Ari’s position.
By the law of sines
$$\frac{\sin (A)}{15}=\frac{\sin \left(\frac{3 \pi}{4}\right)}{|A C|}$$
giving
sin (A ) = $$\frac{15}{\sqrt{2}|A C|}$$ ≈ $$\frac{1.5}{\sqrt{2} \cdot 32.4}$$ ≈ 0.33.
Thus, m ∠A ≈ sin-1 (0.33 ) ≈ 0.33 radian. (This is about 19° .)
Question 4.
a. Is it possible to construct an inverse to the sine function if the domain of the sine function is restricted to the set of real values between $$\frac{\pi}{2}$$ and $$\frac{3\pi}{2}$$? If so, what is the value of sin-1 ($$\frac{1}{2}$$) for this inverse function? Explain how you reach your conclusions.
We see, when restricted to inputs between $$\frac{\pi}{2}$$ and $$\frac{3\pi}{2}$$, the graph of y = sin (x) is strictly decreasing.
Thus for each value between –1 and 1 , there is an input value x within this range for which sin (x) has this value. That is, there is indeed an inverse function for sine in this restricted domain.
We know that sin ($$\frac{\pi}{6}$$) = $$\frac{1}{2}$$. So it follows that sin (π + $$\frac{\pi}{6}$$) = – $$\frac{1}{2}$$. For our inverse function we have
sin-1 (– $$\frac{1}{2}$$) = π + $$\frac{\pi}{6}$$ = $$7\frac{\pi}{6}$$ .
b. Is it possible to construct an inverse to the cosine function if the domain of the cosine function is restricted to the set of real values between $$\frac{\pi}{2}$$ and $$\frac{3\pi}{2}$$? If so, what is the value of cos-1 ($$\frac{1}{2}$$) for this inverse function? Explain how you reach your conclusions.
The value of the cosine function is neither strictly increasing nor strictly decreasing on that restricted domain.
There are distinct inputs from the restricted domain that give the same cosine values, and so it is not possible to construct an inverse to the cosine for this domain.
c. Is it possible to construct an inverse to the tangent function if the domain of the tangent function is restricted to the set of real values between $$\frac{\pi}{2}$$ and $$\frac{3\pi}{2}$$? If so, what is the value of tan-1 (-1) for this inverse function? Explain how you reach your conclusions.
The graph of the tangent function is strictly increasing on the restricted domain:
For each real number y , there is indeed a unique real number x in this restricted domain with tan (x ) = y . We can thus construct an inverse function.
We know that tan ($$\frac{\pi}{4}$$ ) = 1 , and so we see that tan (π + $$\frac{\pi}{4}$$) = –1 . Thus for our inverse function, tan–1 (–1 ) = $$\frac{5\pi}{4}$$.
Question 5.
The diagram shows part of a rugby union football field. The goal line (marked) passes through two goal posts (marked as black circles) set 5.6 meters apart.
According to the rules of the game, an attempt at a conversion must be taken at a point on a line through the point of touchdown and perpendicular to the goal line. If a touchdown occurred 5 meters to one side of a goal post on the goal line, for example, the dashed line in the diagram indicates the line on which the conversion must be attempted.
Suppose the conversion is attempted at a distance of x meters from the goal line. Let w be the angle (measured in radians) indicated subtended by the goal posts.
a. Using inverse trigonometric functions, write an expression for w in terms of the distance x.
Label the angle y as shown:
We have tan (y ) = $$\frac{5}{x}$$ and tan (y + w ) = $$\frac{10.6}{x}$$ and so
w = (y + w ) – y = tan–1 ($$\frac{10.6}{x}$$) – tan–1 ($$\frac{5}{x}$$).
b. Using a graphing calculator or mathematics software, sketch a copy of the graph of the angle measure w as a function of x on the axes below. Indicate on your sketch the value of x that maximizes w. What is that maximal angle measure? (Give all your answers to two decimal places.)
We see
At x = 7.28 , the angle w has a measure of 0.37 radian. (This is about 21° .)
c. In the original diagram, we see that the angle of measure w is one of three angles in an obtuse triangle. To two decimal places, what is the measure of the obtuse angle in that triangle when w has its maximal possible measure?
Label the obtuse angle a and the length L as shown.
For x = 7.28 and w = 0.37 , we have
L = $$\sqrt{10.6^{2}+7.28^{2}}$$ ≈ 12.86 , in meters.
By the law of sines,
$$\frac{\sin (a)}{L}=\frac{\sin (w)}{5.6}$$
So sin (a ) = $$\frac{L \sin (w)}{5.6}$$ ≈ $$\frac{12.86 \times \sin (0.37)}{5.6}$$ ≈ 0.83 .
Thus, a = sin–1 (0.83 ) = 0.98 or π – 0.98 .
Since we are working with an obtuse angle, we must have a = π – 0.98 ≈ 2.16 radians. (This is about 124° .)
Question 6.
While riding her bicycle, Anu looks down for an instant to notice a reflector attached to the front wheel near its rim. As the bicycle moves, the wheel rotates and the position of the reflector relative to the frame of the bicycle changes. Consequently, the angle down from the horizontal that Anu needs to look in order to see the reflector changes with time.
Anu also notices the air valve on the rim of the front wheel tire and observes that the valve and the reflector mark off about one-sixth of the perimeter of the front wheel.
As Anu rides along a straight path, she knows that there will be a moment in time when the reflector, the valve, and her eye will be in line. She wonders what the angle between the horizontal from her eye and the line from her eye to the reflector passing through the valve is at this special moment.
She estimates that the reflector and the valve are each 1.5 feet from the center of the front wheel, that her eye is 6 feet away from the center of that wheel, and that the line between her eye and the wheel center is 45° down from the horizontal.
According to these estimates, what is the measure, to one decimal place in radians, of the angle Anu seeks?
In this diagram, O represents the center of the front wheel, E the location of Anu’s eye, and R and V the positions of the reflector and valve, respectively, at the instant R, V, and E are collinear.
We are told that the length of the arc between V and R is one-sixth of the perimeter of the wheel. Thus m ∠VOR = $$\frac{1}{6}$$ ∙ 2π = $$\frac{\pi}{3}$$ radian. Consequently, m ∠EVO = $$\frac{2\pi}{3}$$.
Looking at triangle EVO , the law of sines gives $$\frac{6}{\sin \left(\frac{2 \pi}{3}\right)}$$ = $$\frac{1.5}{\sin (m \angle V E O)}$$, and so sin (m∠VEO ) = $$\frac{\sin \left(\frac{2 \pi}{3}\right)}{4}$$ =$$\frac{\sqrt{3}}{8}$$. Thus, m ∠VEO = sin–1 ($$\frac{\sqrt{3}}{8}$$) radians. Since ∠EVO is obtuse, ∠VEO is acute, and we must work with the value of sin–1 ($$\frac{\sqrt{3}}{8}$$) that corresponds to the measure of an acute angle.
It follows that m ∠a = $$\frac{\pi}{3}$$ – sin–1 ($$\frac{\sqrt{3}}{8}$$) ≈ 0.6 radians (which is about 34° ).
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# Bending Moment MCQ Quiz - Objective Question with Answer for Bending Moment - Download Free PDF
Last updated on Apr 23, 2024
## Latest Bending Moment MCQ Objective Questions
#### Bending Moment Question 1:
For cantilever beam, bending moment will be zero at
1. 1/4th from free end
2. Fixed end
3. Center of beam
4. Free end
Option 4 : Free end
#### Bending Moment Question 1 Detailed Solution
Explanation:
For a cantilever beam, the bending moment will be zero at the free end. This is because the free end is not supported, and there is no force acting on it to cause bending.
The bending moment will be maximum at the fixed end of the beam. This is because the fixed end is where the beam is attached to the support, and the support prevents the beam from rotating.
The bending moment will also be zero at the center of the beam if the beam is symmetrically loaded. This is because the forces acting on the beam will be balanced, and there will be no net force causing bending.
Therefore, the only correct option is that the bending moment will be zero at the free end for a cantilever beam.
#### Bending Moment Question 2:
An overhanging beam CADEBF is shown in the figure below. Calculate the sum of the bending moment values at A and B. Ignore the sign conventions.
1. 13.50 kN-m
2. 10.00 kN-m
3. 12.50 kN-m
4. 16.78 kN-m
Option 1 : 13.50 kN-m
#### Bending Moment Question 2 Detailed Solution
Explanation:
Concept:
• It is the algebraic sum of moments acting on either side of a section along the length of the beam.
Calculation:
The bending moment at A ( from left) = 9×1.5×$$\frac{1.5}{2}$$ = 10.125 kN-m
The bending moment at B (from right) = 3×1.5×$$\frac{1.5}{2}$$ = 3.375 kN-m
the Thus sum of moments = 10.125 +3.375 = 13.50 kN-m
#### Bending Moment Question 3:
Find the bending moment at a distance L/4 from end A of the simply supported beam as shown below.
1. Zero
2. $$\rm \frac{3.wL^2}{32}$$
3. $$\rm \frac{wL^2}{8}$$
4. $$\rm \frac{wL^2}{32}$$
Option 2 : $$\rm \frac{3.wL^2}{32}$$
#### Bending Moment Question 3 Detailed Solution
Concept:
The bending moment and shear force diagram for a simply supported beam with UDL is shown below.
Calculation:
Calculating the support reactions, due to loading symmetry, will be equal at A and B.
So, $$R_A=R_B=\frac{wL}{2}$$
The bending moment at a distance L/4 from end A of the simply supported beam is given by
M = Moment due to RA from L/4 distance - Moment due to load w from L/4 distance
$$M=\frac{wL}{2}\times\frac{L}{4}-w\times\frac{L}{4}\times(\frac{1}{2}\times\frac{L}{4})$$
$$M=\frac{3wL^2}{32}$$
#### Bending Moment Question 4:
The value of slope at fixed support of a cantilever beam of length- L, flexural rigidity- EI, subjected to a point load ‘P’ at its mid span, and another point load ‘P’ at its free end, is:
1. $$\rm \frac{PL}{2.El}$$
2. $$\rm \frac{3.PL}{4.El}$$
3. $$\rm \frac{PL}{4.El}$$
4. zero
Option 4 : zero
#### Bending Moment Question 4 Detailed Solution
For a cantilever and beam fixed beam, the slope and deflection at deflection at the fixed support for 0, as fixed support are restrained against the rotation and translation motion both
Important PointsWe can prove the same, by assuming any loading condition, as below
Explanation:
Cantilever Beam:
From the moment-curvature relationship, we know that, $${\rm{EI}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{\;}} - {{\rm{M}}_{\rm{x}}}$$
where, E = Modulus of elasticity, I = Moment of Inertia, y = deflection, Mx = Bending moment at a distance x
and $$\;\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}\;$$ is the curvature of the beam.
A cantilever beam with BMD and SFD is shown in the following figure below:
Bending moment at a distance x from the free end is given by, Mx = - P × x.
$$∴ {\rm{EI}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{\;}} - {{\rm{M}}_{\rm{x}}} = {\rm{Px}}$$
$$∴ {\rm{\;}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \frac{{\rm{P}}}{{{\rm{EI}}}}{\rm{x\;}}$$
Now double integrating the above equation we get,
$$∴ \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\rm{P}}}{{{\rm{EI}}}}\smallint {\rm{x\;dx}}$$
$$⇒ {\rm{\;}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\rm{P}}}{{2{\rm{EI}}}}{{\rm{x}}^2} + {{\rm{C}}_1}$$
$$⇒ {\rm{y}} = \smallint \left( {\frac{{\rm{P}}}{{2{\rm{EI}}}}{{\rm{x}}^2} + {{\rm{C}}_1}} \right){\rm{dx}}$$
$$⇒ {\rm{y}} = {\rm{\;}}\frac{{{\rm{P}}{{\rm{x}}^3}}}{{6{\rm{EI}}}} + {{\rm{C}}_1}{\rm{x}} + {{\rm{C}}_2}$$
Putting boundary condition, at x = L, y = 0 and at x = L, dy/dx=0
$${\rm{So}},{\rm{\;}}{{\rm{C}}_1} = - \frac{{{\rm{P}}{{\rm{L}}^2}}}{{2{\rm{EI}}}}{\rm{\;\;and}},{\rm{\;}}{{\rm{C}}_2} = \frac{{{\rm{P}}{{\rm{L}}^3}}}{{3{\rm{EI}}}}{\rm{\;}}$$
So, deflection at any point at distance x from free end is given by,
$${\rm{y}}\left( {\rm{x}} \right) = \frac{{{\rm{P}}{{\rm{x}}^3}}}{{6{\rm{EI}}}} - \frac{{{\rm{P}}{{\rm{L}}^2}}}{{2{\rm{EI}}}}{\rm{x}} + \frac{{{\rm{P}}{{\rm{L}}^3}}}{{3{\rm{EI}}}}$$
by differentiating w.r.t.
$$\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{P}}{{\rm{x}}^2}}}{{2{\rm{EI}}}} - \frac{{{\rm{P}}{{\rm{L}}^2}}}{{2{\rm{EI}}}} + ~0$$
Now, at x = 0 ⇒ $$\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = ~0$$
and we know that, slope of any curve can be represents as
$$tan~\theta ~=\frac{{{\rm{dy}}}}{{{\rm{dx}}}} \Rightarrow 0$$
∴ the slope of the curve at fixed end is zero degree.
#### Bending Moment Question 5:
A cantilever beam carries a uniformly distributed load over a span of 1 m as shown in the figure below. The reactive moment at point A is _____.
1. 30 kN-m
2. 0 kN-m
3. 10 kN-m
4. 5 kN-m
Option 4 : 5 kN-m
#### Bending Moment Question 5 Detailed Solution
Concept:
Calculation:
Given:
w = 10 kN/m, span = 1 m
∴ Load = 10 × 1 = 10 kN which will act at 0.5 m from point A.
∴ Moment at A = Reactive moment at point A = 10 × 0.5 = 5 kN-m
## Top Bending Moment MCQ Objective Questions
#### Bending Moment Question 6
A cantilever 9 m long has uniformly distributed load over the entire length. The maximum bending moment is 8100 N-m, the rate of loading is:
1. 200 N/m
2. 100 N/m
3. 400 N/m
4. 900 N/m
Option 1 : 200 N/m
#### Bending Moment Question 6 Detailed Solution
Explanation:
Cantilever beam with uniformly distributed load:
So, the cantilever beam has a maximum bending moment at the fixed end. and it is given as, $$M=\frac{wL^2}{2}$$
Calculation:
Given:
M = 8100 N-m, L = 9 m
$$8100=\frac{w~\times~ 9^2}{2}$$
w = 200 N/m
#### Bending Moment Question 7
A bending moment causing concavity upward will be taken as _____ and called as ______ bending moment.
1. positive, sagging
2. positive, hogging
3. negative, sagging
4. negative, hogging
Option 1 : positive, sagging
#### Bending Moment Question 7 Detailed Solution
Explanation:
Let us assume a simply supported beam.
The uniformly distributed load (UDL) of w/length is acted on the beam.
Due to downward load, the beam is sagging.
We also know that when a simply supported beam is subjected to UDL the bending moment will be positive.
Sagging Or Positive Bending Moment
We take bending moment at a section as positive if
• Force tends to bend the beam at that considered point.
• This bending forms to curvature having concavity at the top
• Concavity at the top indicates compression in the top fibers of the beam.
• Hence bottom fibers of the beam would have tension.
• Here due to UDL simply supported beam will give us the positive value of bending moment that indicates sagging.
• Due to this, the upper layer fibers are getting compressive stresses and the bottom layer fibers are getting tensile stresses.
Hogginng or Negative Bending Moment
We take bending moment at a section as positive if
• Force tends to bend the beam at that considered point.
• This bending forms to curvature having convexity at the top
• Convexity at the top indicates tension in the top fibers of the beam.
• Hence top fibers of the beam would have compression.
Points of zero bending moment
• The points of contra flexure (or inflection) are points of zero bending moment, i.e. where the beam changes its curvature from hogging to sagging.
• In a bending beam, a point of contra flexure is a location where the bending moment is zero (changes its sign).
• In a bending moment diagram, it is the point at which the bending moment curve intersects with the zero lines.
for example:
Points A and B are contraflexure points.
#### Bending Moment Question 8
The BM diagram of the beam shown in figure is:
1. A rectangle
2. A triangle
3. A trapezium
4. A parabola
Option 2 : A triangle
#### Bending Moment Question 8 Detailed Solution
Concept:
Find the Reaction force at A and B and draw the bending moment diagram
$${\rm{\Sigma }}{{\rm{F}}_{\rm{x}}} = 0,\;\;{\rm{\Sigma }}{{\rm{F}}_y} = 0,\;\;{\rm{\Sigma M}}_{\left( {{\rm{about\;a\;point}}} \right)} = 0$$
Calculation:
Given:
$$\sum {F_y} = 0\;$$
RA + RB = 0
$$\sum {M_A} = 0$$
M - R× L = 0
$${R_B} = \frac{M}{L}$$ and $${R_A} = - \frac{M}{L}$$
Shear force:
$${\left( {S.F} \right)_B} ={\left( {S.F} \right)_A} = - \frac{M}{L}$$
Bending movement:
$${\left( {B.M} \right)_{X - X}} = M - \frac{M}{L}x$$ (x taken from the left side)
Clockwise bending moment -ve, Anticlockwise bending moment +ve
$$% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaqadaWdaeaaieWapeGaa8Nqaiaac6cacaWFnbaacaGLOaGaayzk % aaWdamaaBaaaleaapeGaa8hwaiabgkHiTiaa-Hfaa8aabeaak8qacq % GHDisTcaWFybaaaa!3F7A! {\left( {B.M} \right)_{X - X}} \propto X% MathType!End!2!1!$$ (Bending moment varies linearly)
$${\left( {B.M} \right)_A} = M$$
$${\left( {B.M} \right)_B} = M - \frac{M}{L} \times L = 0\;$$
∴ bending moment diagram will be a TRINGLE.
Important Points
• If SFD is constant throughout the span of the beam then BMD will be linear.
• If at a point a couple is acting then there will be a sudden jump in the BMD.
#### Bending Moment Question 9
In the figure given below, the beam will be stable only if -
1. x = √2y
2. x = 2y
3. √2x = y
4. 2x = y
Option 3 : √2x = y
#### Bending Moment Question 9 Detailed Solution
Concept-
For a system to be in stable condition, it should be in stable equilibrium.
So to be in stable equilibrium the moment about the support point should be zero.
∑ MB = 0
$$⇒ 2W \times x \times \dfrac{x}{2} = W \times y \times \dfrac{y}{2}$$
$$⇒ x^2 = \dfrac{y^2}{2}$$
⇒ y2 = 2x2
⇒ y = √2 x
#### Bending Moment Question 10
In a simply supported beam of 10 m span having udl of 8 kN/m, the maximum Bending Moment shall be
1. 80 kNm
2. 100 kNm
3. 0.8 kNm
4. 10 kNm
Option 2 : 100 kNm
#### Bending Moment Question 10 Detailed Solution
Concept:
The maximum bending moment for a simply supported beam with a uniformly distributed load W per unit length is wL2/8.
Calculation:
Given:
w = 8 kN/m
L = 10 m
Maximum bending moment = $$\frac{{w{L^2}}}{8} = \frac{{8 \times \left( {{{10}^2}} \right)}}{8} = 100kNm$$
#### Bending Moment Question 11
Consider the following statements:
If a simply supported beam of uniform cross-section is subjected to a clockwise moment at the left support and an equal anticlockwise moment at the right support, then the:
1. B.M.D. will be in the shape of a rectangle
2. S.F.D. will be a straight line coinciding with the base
3. Deflection curve will be in the shape of a circular arc
Of these statements
1. 1, 2 and 3 are correct
2. 1 and 2 are correct
3. 1 and 3 are correct
4. 2 and 3 are correct
Option 1 : 1, 2 and 3 are correct
#### Bending Moment Question 11 Detailed Solution
Explanation:
As given in the question that the simply supported beam is subjected to a clockwise moment on the left side and an anti-clockwise moment of the same magnitude on the right side.
∴ Net moments on the beam = +M - M = 0
Reaction of the beam = $$\frac{Net~moments~on~the~beam}{Span~of~the~beam}$$
So, RA = RB = 0.
∴ Shear force = 0
⇒ The sheer force diagram will be a straight line coinciding with the base.
Concentrated moment at A = M (Clockwise)
The concentrated moment at B = M (Anti-Clockwise)
And these concentrated moments are shown as a vertical straight line in the bending moment diagram.
B.M.D. will be in the shape of a rectangle.
δA = δB = 0
And Maximum deflection at the midsection,
δmax = $$\frac{ML^2}{8EI_{NA}}$$
So, the deflection curve will be in the shape of a circular arc.
#### Bending Moment Question 12
A uniformly loaded propped cantilever beam and its free body diagram are shown below. The reactions are
1. $${R_1} = \frac{{5ql}}{8},{R_2} = \frac{{3ql}}{8},M = \frac{{q{l^2}}}{8}$$
2. $${R_1} = \frac{{3ql}}{8},{R_2} = \frac{{5ql}}{8},M = \frac{{q{l^2}}}{8}$$
3. $${R_1} = \frac{{5ql}}{8},{R_2} = \frac{{3ql}}{8},M = 0$$
4. $${R_1} = \frac{{3ql}}{8},{R_2} = \frac{{5ql}}{8},M = 0$$
Option 1 : $${R_1} = \frac{{5ql}}{8},{R_2} = \frac{{3ql}}{8},M = \frac{{q{l^2}}}{8}$$
#### Bending Moment Question 12 Detailed Solution
The given propped cantilever beam can be assumed to be consisting of two types of loads. First part is simple cantilever with UDL and the second part is cantilever beam with point load of R2 at end.
Balancing the forces we get
R1 + R2 = qL ----------(1)
Balancing the deflection at end point as net deflection at the end is zero.
Deflection at B due to the UDL alone
$${\delta _B} = \frac{{q{L^4}}}{{8EI}}$$
Deflection at B due to R2 alone,
$$\delta _B^{'} = \frac{{{R_2}{L^3}}}{{3EI}}$$
$$\therefore {\delta _B} = \delta _B^{'}$$
$$\Rightarrow \frac{{q{L^4}}}{{8EI}} = \frac{{{R_2}{L^3}}}{{3EI}} \Rightarrow {R_2} = \frac{{3qL}}{8}$$
From 1)
$${R_1} = qL - {R_2}$$
$$= qL - \frac{{3qL}}{8} = \frac{{5qL}}{8}$$
$$Moment,\;M = {R_2}L - qL \times \frac{L}{2}$$
$$= \frac{{3q{L^2}}}{8} - \frac{{q{L^2}}}{2} = - \frac{{q{L^2}}}{8}$$
#### Bending Moment Question 13
A propped cantilever beam of span L when loaded with a uniformly distributed load of intensity p will have a bending moment M at the fixed end, where M is:
1. wL2/12
2. wL2/16
3. wL2/24
4. wL2/8
Option 4 : wL2/8
#### Bending Moment Question 13 Detailed Solution
Concept:
Using super position principle:
From compatibility condition:
δB = 0
δ, due to case (I), $${\delta _{B,I}} = \frac{{W{L^4}}}{{8\;EI}}\;\left( \downarrow \right)$$
due to case (II), $${\delta _{B,\;II}} = \frac{{{R_B}\; \times \;{L^3}}}{{3\;EI}}\;\left( \uparrow \right)$$
so, $$\Rightarrow \frac{{W{L^4}}}{{8\;EI}} = \frac{{{R_B}\; \times \;{L^3}}}{{3\;EI}}$$
$$\Rightarrow {R_B} = \frac{{3\;WL}}{8}$$
So, bending moment at fixed end,
$${M_A} = \frac{{3w\ell }}{8} \times \ell - \frac{{w\; \times \;{\ell ^2}}}{2}$$
$$= \frac{{3w{\ell ^2}}}{8} - \frac{{w{\ell ^2}}}{2}$$
$$= \frac{{ - w{\ell ^2}}}{8}$$ [Sagging moment)
So, $$M = \frac{{w{\ell ^2}}}{8}$$
#### Bending Moment Question 14
A simply supported beam of length 3 m carries a concentrated load of 15 kN at distance of 1 m from left support. The maximum bending moment in the beam:
1. 15 kNm
2. 5 kNm
3. 10 kNm
4. 40 kNm
Option 3 : 10 kNm
#### Bending Moment Question 14 Detailed Solution
Concept:
The bending moment due to point load P at a distance of x is
= P × x
Calculation:
Given:
Load at point C = 15 kN
Distance of application of load from left support A = 1 m
Let’s assume reaction at the supports A and B as RA and RB respectively.
From the equilibrium of forces in y-direction
ΣFy = 0
RA + RB = 15 kN …(1)
Taking moment about point A and using the equilibrium condition
ΣMA = 0
-(15 × 1) + (RB × 3) = 0
B = 5 kN …(2)
From eq. (1), we get
RA = 10 kN …(3)
The bending moment diagram is shown as follow
The bending moment is maximum at point C with a value of 10 kNm
• Be careful while taking the sign conventions for shear force and bending moment diagrams.
#### Bending Moment Question 15
A massless beam has a loading pattern shown in figure. Find the bending moment at mid span?
1. 1 kN-m
2. 3 kN-m
3. 2 kN-m
4. 0.0 kN-m
Option 1 : 1 kN-m
#### Bending Moment Question 15 Detailed Solution
Concept:
Equilibrium of a beam is satisfied by following there equations
∑fx = 0
∑fy = 0
∑M = 0
Total load = wL, where w is the distributed load/unit length, and L is the length of that section of the beam where distributed load is acting
And acts at the centroid of the section in which distributed load is acting.
Calculation:
Total load adding on the beam $$=4~\left( \frac{kN}{m} \right)\times 1\left( m \right)=~4\text{ }\!\!~\!\!\text{ kN}$$
From Vertical Equilibrium
RA + RC = 4 kN
Now from moment equilibrium
∑MA = 0
RC × 2 – 4 × 0.5 = 0
RC = 1 kN
RA = 3 kN
Now, the moment about point B = RC × 1 = 1 kN × 1 m = 1 kN⋅m
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# Factors of 12
Last Updated: August 16, 2024
## Factors of 12
The factors of 12 is a fundamental concept in mathematics. Factors are numbers that can be multiplied together to produce another number. For 12, the factors are pairs of numbers that, when multiplied, equal 12. Identifying these factors helps in various mathematical applications, such as simplifying fractions, finding least common multiples, and solving algebraic equations. The factors of 12 include 1, 2, 3, 4, 6, and 12, as these numbers can be evenly divided into 12. This article will explore these factors in detail, providing clear examples and explanations to enhance your mathematical knowledge.
## What are the Factors of 12?
The factors of 12 are the numbers that can be multiplied together to yield the product of 12. These factors include 1, 2, 3, 4, 6, and 12. To determine these factors, we look for all pairs of whole numbers that, when multiplied, result in 12. Starting with 1 and 12 (since 1 × 12 = 12), we then find 2 and 6 (since 2 × 6 = 12), and finally, 3 and 4 (since 3 × 4 = 12). These pairs represent all the possible combinations of whole numbers that can be multiplied to give 12. Understanding these factors is useful in various mathematical operations, including simplifying fractions and finding common denominators.
## Factors Pairs of 12
The factor pairs of 12 are (1, 12), (2, 6), and (3, 4).
In multiplication, 1 times 12 equals 12, forming the factor pair (1, 12).
In multiplication, 2 times 6 equals 12, forming the factor pair (2, 6).
In multiplication, 3 times 4 equals 12, forming the factor pair (3, 4).
The factor pairs of 12 are (1, 12), (2, 6), and (3, 4).
## How to Calculate Prime Factors of 12?
Calculating the prime factors of 12 involves breaking down the number into the prime numbers that multiply together to give the original number. Here are the steps to determine the prime factors of 12:
### Step 1: Identify the Number
Start with the number you want to factor, which in this case is 12.
### Step 2: Divide by the Smallest Prime Number
Begin by dividing 12 by the smallest prime number, which is 2. 12÷2=6
### Step 3: Continue Dividing by Prime Numbers
Next, take the result (6) and continue dividing by the smallest prime numbers. 6÷2=3
### Step 4: Check for Remaining Prime Factors
Finally, take the last result (3) and check if it is a prime number. Since 3 is a prime number, it cannot be divided further except by 1 and itself.
### Step 5: List All Prime Factors
Combine all the prime numbers obtained from the division steps: 2,2,3
Therefore, the prime factors of 12 are 2, 2, and 3, or written as 22×3.
## Factors of 12 : Examples
### Example 1: Listing All Factors
List all the factors of 12.
The factors of 12 are the numbers that can be multiplied in pairs to give 12.
These pairs are: 1×12=12,2×6=12,3×4=12 So, the factors of 12 are 1, 2, 3, 4, 6, and 12.
### Example 2: Prime Factorization
Find the prime factorization of 12.
1. Divide 12 by the smallest prime number (2): 12÷2=6
2. Divide 6 by 2: 6÷2=3
3. Since 3 is a prime number, we stop here. The prime factors of 12 are: 2×2×3 Or 22×3
### Example 3: Using Factors to Simplify a Fraction
Simplify the fraction 12/18.
1. Find the greatest common divisor (GCD) of 12 and 18. The factors of 12 are 1, 2, 3, 4, 6, and 12. The factors of 18 are 1, 2, 3, 6, 9, and 18. The GCD is 6.
2. Divide both the numerator and the denominator by the GCD: 12÷6/18÷6=2/3 So, 12/18 simplifies to 2/3.
### Example 4: Finding Common Factors
Find the common factors of 12 and 15.
• Factors of 12: 1, 2, 3, 4, 6, 12
• Factors of 15: 1, 3, 5, 15 The common factors are 1 and 3.
### Example 5: Using Factors to Find Multiples
Determine if 36 is a multiple of 12. Solution:
1. Divide 36 by 12: 36÷12=3
2. Since the result is a whole number (3), 36 is a multiple of 12.
## Factors of 12 : Tips
Understanding the factors of 12 is a basic but essential skill in mathematics. Here are some tips to help you work with the factors of 12 effectively:
1. Identify the Number: Start by understanding that factors are numbers that divide evenly into 12 without leaving a remainder.
2. Begin with the Smallest Factor: Always start with 1, as it is a factor of every number.
3. Use Division: Check divisibility by dividing 12 by smaller numbers (2, 3, 4, etc.) to see which ones are factors.
4. Pair Up Factors: For each factor, find its corresponding pair that multiplies to 12 (e.g., 2 and 6).
5. Prime Factorization: Break down 12 into its prime factors (2 and 3) to understand its basic building blocks.
6. Factor List: Memorize the complete list of factors of 12: 1, 2, 3, 4, 6, and 12.
7. Practice with Examples: Regularly solve problems involving the factors of 12 to strengthen your understanding.
8. Apply in Fractions: Use factors to simplify fractions by finding common factors between the numerator and the denominator.
## Is 12 a prime number?
No, 12 is not a prime number. A prime number has exactly two distinct positive divisors: 1 and itself. 12 has six positive divisors: 1, 2, 3, 4, 6, and 12.
## Are the factors of 12 and 24 related?
Yes, the factors of 12 and 24 are related. Since 24 is a multiple of 12, all factors of 12 are also factors of 24. The factors of 12 (1, 2, 3, 4, 6, 12) are included in the factors of 24.
## What is the sum of the factors of 12?
The sum of the factors of 12 is the total when all the factors are added together: 1+2+3+4+6+12=28
## How do you use the factors of 12 to find the greatest common divisor (GCD)?
To find the GCD of two numbers, list the factors of each number and identify the largest number that appears in both lists. For example, the GCD of 12 and 18 is 6 because 6 is the largest number that appears in both lists of factors.
## Write down the positive and negative pair factors of 12?
The positive pair factors of 12 are (1, 12), (2, 6), and (3, 4). The negative pair factors of 12 are (-1, -12), (-2, -6), and (-3, -4).
## What are the Prime Divisors of 12?
The prime divisors of 12 are 2 and 3. The number 12 can be broken down into its prime factors as 22×3, showing that the prime numbers 2 and 3 divide 12.
## How do factors of 12 relate to the concept of multiples?
Factors of 12 are directly related to its multiples. A multiple of 12 is any number that can be expressed as 12 times an integer. Knowing the factors helps in identifying these multiples and understanding their properties.
## How can factors of 12 be used to simplify fractions?
To simplify a fraction with 12 in the numerator or denominator, you can divide both the numerator and the denominator by any common factors they share with 12. For example, to simplify 12/18, divide both by their GCD, which is 6, resulting in 2/3.
Text prompt
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# Question #6f780
Jan 19, 2018
$\text{please have a look at the fallowing details. (option c)}$
#### Explanation:
1- In order to understand the solution, we need to remember the following important points.
• An electric field is formed around the electric charges.
• Electric field is a vectorial quantity.
• The direction of the electric field generated by the electric charge with positive sign is directed outward from the charge.
• The direction of the electric field generated by the electric charge in the negative sign is directed from the outside to the charge.
• We calculate the electric field intensity at a point far from an electric charge by using the following formula...
$E = k \cdot \frac{q}{r} ^ 2$
• The growth of r (away from load) causes the intensity of the electric field to decrease.
• In the given diagram space is colored by dividing into three regions.
• Two vectors representing the electric field in each region are drawn.
• Note the directions of the vectors in the regions.
• The point we are looking for can not be in the yellow zone (because the vectors are in the same direction).
• In order for the electric field to be zero, the magnitudes of the vectors must be equal and opposite.
• Which of the green or blue regions may be zero? we need to answer the question.
• To make it easier to understand, let's take the distance between points equally (indicated by x).
1. Now we can solve the problem.
2. Let's calculate the electric field at point K.
${\vec{E}}_{K} = k \cdot \frac{+ q}{x} ^ 2 + k \cdot \frac{- 3 q}{2 x} ^ 2 = \frac{k q}{x} ^ 2 - \frac{3 k q}{4 {x}^{2}}$
• Let's calculate the electric field at point R.
${\vec{E}}_{R} = k \cdot \frac{+ q}{4 x} ^ 2 + k \cdot \frac{- 3 q}{x} ^ 2 = \frac{k q}{16 {x}^{2}} - \frac{3 k q}{x} ^ 2$
• Electric field intensity is proportional to the magnitude of the load and is inversely proportional to the distance.
$\frac{3 k q}{x} ^ 2 > \frac{k q}{16 {x}^{2}}$
• The point we're looking for is not in the blue zone.
• The intensity of the electric field can be zero in the green-painted space zone.
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# NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Written by Team Trustudies
Updated at 2021-05-07
## NCERT solutions for class 6 Maths Chapter 10 Mensuration Exercise 10.1
Q.1 Find the perimeter of each of the following figures:
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Perimeter = Sum of all the sides
= 1 + 2 + 4 + 5
= 12 cm
(b) Perimeter = Sum of all the sides
= 23 + 35 + 35 + 40
= 133 cm
(c) Perimeter = Sum of all the sides
= 15 + 15 + 15 + 15
= 60 cm
(d) Perimeter = Sum of all the sides
= 4 + 4 + 4 + 4 + 4
=20 cm
(e) Perimeter = Sum of all the sides
= 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4
= 15 cm
(f) Perimeter = Sum of all the sides
= 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3
= 52 cm
Q.2 The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Length of required tape = Perimeter of rectangle
= 2 (40 + 10)
= 2 × (50) = 100 cm
Q.3 A table top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table top?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 (Length + Breadth)
= 2 (2.25 + 1.50)
= 2 (3.75)
= 2 × 3.75
= 7.5 m
Q.4 What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Required length of wooden strip = Perimeter of photograph
= 2 (32 + 21)
= 2 (53)
= 2 × 53
= 106 cm
Q.5 A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Perimeter of the field = 2 (Length + Breadth)
= 2 (0.7 + 0.5)
= 2 (1.2)
= 2 × 1.2
= 2.4 km
Q.6 Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm
(b) An equilateral triangle of side 9 cm
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Perimeter of triangle = 3 + 4 + 5
= 12 cm
(b) Perimeter of an equilateral triangle = 3 × side
= 3 × 9
= 27 cm
(c) Perimeter of isosceles triangle = 8 + 8 + 6
= 22 cm
Q.7 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Perimeter of triangle = 10 + 14 + 15
= 39 cm
Q.8 Find the perimeter of a regular hexagon with each side measuring 8 m.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Perimeter of hexagon = 6 × 8 = 48 m
Q.9 Find the side of the square whose perimeter is 20 m.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Perimeter of square = 4 × side
20 = 4 × side
Side =$\frac{20}{4}$
Side = 5 m
Q.10 The perimeter of a regular pentagon is 100 cm. How long is its each side?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Perimeter of regular pentagon = 100 cm
5 × side = 100 cm
Side = $\frac{100}{5}$
Side = 20 cm
Q.11 A piece of strings is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Perimeter of square = 30 cm
4 × side = 30
Side = $\frac{30}{4}$
Side = 7.5 cm
(b) Perimeter of an equilateral triangle = 30 cm
3 × side = 30
Side = $\frac{30}{10}$
Side = 10 cm
(c) Perimeter of a regular hexagon = 30 cm
6 × side = 30
Side = $\frac{30}{6}$
Side = 5 cm
Q.12 Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Let a cm be the third side
Perimeter of triangle = 36 cm
12 + 14 + a = 36
26 + a = 36
a = 36 – 26
a = 10 cm
Q.13 Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Side of square = 250 m
Perimeter of square = 4 × side
= 4 × 250
= 1000 m
Cost of fencing = Rs 20 per m
Cost of fencing for 1000 m = Rs 20 × 1000
= Rs 20,000
Q.14 Find the cost of fencing a rectangular park of length 175 cm and breadth 125 m at the rate of ? 12 per metre.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Length = 175 cm
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (175 + 125)
= 2 × (300)
= 2 × 300
= 600 m
Cost of fencing = 12 × 600 = Rs 7200
Q.15 Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Perimeter of square = 4 × side
= 4 × 75
= 300 m
$?$Distance covered by Sweety is 300 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (60 + 45)
= 2 × 105
= 210 m
Hence, Bulbul covers less distance than Sweety.
Q.16 What is the perimeter of each of the each of the following figures? What do you infer from the the answers?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Perimeter of square = 4 × side
= 4 × 25
= 100 cm
(b) Perimeter of rectangle = 2 (40 + 10)
= 2 × 50
= 100 cm
(c) Perimeter of rectangle = 2 (Length + Breadth)
= 2 (30 + 20)
= 2 (50)
= 2 × 50
= 100 cm
(d) Perimeter of triangle = 30 + 30 + 40
= 100 cm
All the figures have same perimeter.
Q.17 Avneet buys 9 square paving slabs, each with a side of 1 / 2 m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [fig 10.7(i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig 10.7 (ii)]?
(c) Which has greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e they cannot be broken.)
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Side of square = 3 × side
= 3 × $\frac{1}{2}$
= 3 / 2 m
Perimeter of Square = 4 × $\frac{3}{2}$
= 2 × 3
= 6 m
(b) Perimeter = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
= 10 m
(c) The arrangement in the form of cross has greater perimeter
(d) Perimeters greater than 10 m cannot be determined.
## NCERT solutions for class 6 Maths Chapter 10 Mensuration Exercise 10.2
Q.1 Find the areas of the following figures by counting square:
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Number of full squares = 9
Area of 1 square = 1 sq unit
$?$Area of 9 squares = 9 x 1 sq unit
= 9 sq units.
So, the area of the portion covered by 9 squares = 9 sq units
(b) Number of full squares = 5
$?$ Area of the figure = 5 x 1 sq unit = 5 sq units
(c) Number of full squares = 2
Number of half squares = 4
$?$ Area of the covered figure
= 2 x 1 + 4 x $\frac{1}{2}$ = 2 + 2 = 4 sq units
(d) Number of full squares = 8
$?$ Area of the covered portion of the figure
= 8 x 1 sq unit = 8 sq units.
(e) Number of full squares = 10
Area covered by the figure = 10 x 1 sq unit = 10 sq units.
(f) Number of full squares = 2
Number of half squares = 4
$?$ Area of the covered figure
= (2 x 1 + 4 x $\frac{1}{2}$)= (2 + 2) sq units = 4 sq units.
(g) Number of full squares = 4
Number of half squares = 4
$?$ Area of the covered figure = (4 x 1 + 4 x 12)
= (4 + 2) sq units = 6 sq units.
(h) Number of full squares = 5
$?$ Area of the covered figure
= 5 x 1 sq unit = 5 sq units.
(i) Number of full squares = 9
$?$ Area of the covered figure
= 9 x 1 sq units = 9 sq units.
(j) Number of full squares = 2
Number of half squares = 4
$?$ Area of the covered figure
=(2 x 1 + 4 x $\frac{1}{2}$) sq units
= (2 + 2) sq units = 4 sq units.
(k) Number of full squares = 4
Number of half squares = 2
$?$ Area of the covered figure
= (4 x 1 + 2 x $\frac{1}{2}$)sq units
= (4 + 1)sq units = 5sq units
(l) Number of full squares = 4
Number of squares more than half = 3
Number of half squares = 2
$?$ Area of the covered figure
= (4 x 1 + 3 x 1 + 2 x $\frac{1}{2}$ sq units
= (4 + 3 + 1) sq units = 8 sq units.
(m) Number of full squares = 6
Number of more than half squares = 8
Area of the covered figure = (6 x 1 + 8 x 1) sq units
= (6 + 8) sq units = 14 sq units.
(n) Number of full squares = 9
Number of more than half squares = 9
$?$ Area of the covered figure
= (9 x 1 + 9 x 1) sq units = (9 + 9) sq units = 18 sq units.
## NCERT solutions for class 6 Maths Chapter 10 Mensuration Exercise 10.3
Q.1 Find the area of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Area of rectangle = Length × Breadth
(a) l = 3 cm and b = 4 cm
Area = l × b = 3 × 4 = 12 $c{m}^{2}$
(b) l = 12 m and b = 21 m
Area = l × b = 12 × 21
= 252 ${m}^{2}$
(c) l = 2 km and b = 3 km
Area = l × b = 2 × 3
= 6 $k{m}^{2}$
(d) l = 2 m and b = 70 cm = 0.70 m
Area = l × b = 2 × 0.70
= 1.40 ${m}^{2}$
Q.2 Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Area of square = $sid{e}^{2}$
= ${10}^{2}=100c{m}^{2}$
(b) Area of square =$sid{e}^{2}$ $={14}^{2}=196c{m}^{2}$
(c) Area of square = $sid{e}^{2}$
$={5}^{2}=25c{m}^{2}$
Q.3 The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Area of rectangle = l × b
$=9×6=54{m}^{2}$
(b) Area of rectangle = l × b
$=17×3=51{m}^{2}$
(c) Area of rectangle = l × b
$=4×14=56{m}^{2}$
Area of rectangle 56 ${m}^{2}$ i.e (c) is the largest area and area of rectangle 51 ${m}^{2}$ i.e (b) is the smallest area
Q.4 The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Area of rectangle = length × width
300 = 50 × width
width = 300 / 50
width = 6 m
Q.5 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m.?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Area of land = length × breadth
$=500×200=1,00,000{m}^{2}$
Cost of tiling 1,00,000 sq m of land = (8 × 1,00,000) / 100
= Rs 8000
Q.6 A table top measures 2 m by 1 m 50 cm. What is its area in square metres?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
l = 2m
b = 1m 50 cm = 1.50 m
Area = l × b = 2 × 1.50
$=3{m}^{2}$
Q.7 A room is 4 m long and 3 m 50 cm wide. Howe many square metres of carpet is needed to cover the floor of the room?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
l = 4m
b = 3 m 50 cm = 3.50 m
Area = l × b = 4 × 3.50
=14 ${m}^{2}$
Q.8 A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Area of floor = l × b = 5 × 4
$=20{m}^{2}$
Area of square carpet = 3 × 3
$=9{m}^{2}$
Area of floor that is not carpeted = 20 – 9
$=11{m}^{2}$
Q.9 Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Area of flower square bed = 1 × 1
$=1{m}^{2}$
Area of 5 square bed = 1 × 5
$=5{m}^{2}$
Area of land = 5 × 4
$=20{m}^{2}$
Remaining part of the land = Area of land – Area of 5 square bed
$=20–5=15{m}^{2}$
Q.10 By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Area of HKLM = 3 x 3 =$9c{m}^{2}$
Area of IJGH = 1 x 2 = 2 $c{m}^{2}$
Area of FEDG = 3 x 3 = 9 $c{m}^{2}$
Area of ABCD = 2 x 4 = 8 $c{m}^{2}$
Total area of the figure = 9 + 2 + 9 + 8 = 28 $c{m}^{2}$
(b) Area of ABCD = 3 x 1 = 3 $c{m}^{2}$
Area of BDEF = 3 x 1 = 3 $c{m}^{2}$
Area of FGHI = 3 x 1 = 3 $c{m}^{2}$
Total area of the figure = 3 + 3 + 3 = 9 $c{m}^{2}$
Q.11 Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Area of rectangle ABCD = 2 x 10 = 20 $c{m}^{2}$
Area of rectangle DEFG = 10 x 2 = 20 $c{m}^{2}$
Total area of the figure = 20 + 20 = 40 $c{m}^{2}$
(b) There are 5 squares each of side 7 cm.
Area of one square = 7 x 7 = 49$c{m}^{2}$
Area of 5 squares = 49 x 5 = 245 $c{m}^{2}$
(c) Area of rectangle ABCD = 5 x 1 = 5 $c{m}^{2}$
Area of rectangle EFGH = 4 x 1 = 4 $c{m}^{2}$
Total area of the figure = 5 + 4 $c{m}^{2}$
Q.12 How many tiles whose length and breadth are 12 cm and 5 cm, respectively will be needed to fit in a rectangular region whose length and breadth are respectively?
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
(a) Area of rectangle = 100 × 144
= 14400 cm
Area of one tile = 5 × 12
$=60c{m}^{2}$
Number of tiles = $\frac{\left(Areaofrectangle\right)}{\left(Areaofonetile\right)}$
$=\frac{14400}{60}$ = 240
Hence, 240 tiles are needed
(b) Area of rectangle = 70 × 36
$=2520c{m}^{2}$
Area of one tile = 5 × 12
$=60c{m}^{2}$
Number of tiles = $\frac{\left(Areaofrectangle\right)}{\left(Areaofonetile\right)}$
=$\frac{2520}{60}=42$
##### FAQs Related to NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
There are total 30 questions present in ncert solutions for class 6 maths chapter 10 mensuration
There are total 1 long question/answers in ncert solutions for class 6 maths chapter 10 mensuration
There are total 3 exercise present in ncert solutions for class 6 maths chapter 10 mensuration
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# Multiples of 23
Created by: Team Maths - Examples.com, Last Updated: May 23, 2024
## Multiples of 23
Multiples of 23 are numbers obtained by multiplying 23 by any integer. In mathematics, a multiple is a product of a number and an integer. The sequence of multiples of 23 includes 23, 46, 69, and so on. These numbers are significant in number theory as they illustrate the relationship between multiplication, factors, and divisors. Understanding multiples helps in identifying common factors and solving problems related to divisibility and integer operations.
## What are Multiples of 23?
Multiples of 23 are the products obtained when 23 is multiplied by integers (e.g., 23, 46, 69, 92). They form an arithmetic sequence where each term increases by 23. These multiples are essential in understanding factors, divisors, and various mathematical operations involving integers.
Prime factorization of 23: 23 = 1 × 23 First 10 multiples of 23: 23, 46, 69, 92, 115, 138, 161, 184, 207, 230.
Table of 23
## Important Notes
• Definition: A multiple of 23 is any number that can be expressed as 23×n, where n is an integer.
• Sequence: The sequence of multiples of 23 starts at 23 and increases by 23 each time: 23, 46, 69, 92, 115, etc.
• Divisibility: For a number to be a multiple of 23, it must be divisible by 23 without leaving a remainder.
• Prime Number: Since 23 is a prime number, its only factors are 1 and 23.
• Applications: Understanding multiples of 23 is useful in number theory, arithmetic operations, and finding common multiples in problems involving division and multiplication.
## Examples on Multiples of 23
### Example 1: Calculating the 5th Multiple of 23
• Step 1: Identify the position of the multiple.
• Step 2: Multiply 23 by 5.23×5 = 115
• Result: The 5th multiple of 23 is 115.
### Example 2: Checking if 184 is a Multiple of 23
• Step 1: Divide 184 by 23.184÷23 = 8
• Step 2: Check if the quotient is an integer without a remainder.
• Result: Since the quotient is 8 (an integer), 184 is a multiple of 23.
### Example 3: Finding the Next Multiple of 23 After 69
• Step 1: Identify the last known multiple (69).
• Step 2: Add 23 to the last multiple.69+23 = 92
• Result: The next multiple of 23 after 69 is 92.
### Example 4: Determining if 250 is a Multiple of 23
• Step 1: Divide 250 by 23.250÷23 = 10.87
• Step 2: Check the remainder.250÷23 = 10
• Result: Since the quotient is not an integer and there is a remainder of 20, 250 is not a multiple of 23.
### Example 5: Calculating the 12th Multiple of 23
• Step 1: Identify the position of the multiple.
• Step 2: Multiply 23 by 12.23×12 = 276
• Result: The 12th multiple of 23 is 276.
## What are multiples of 23?
Multiples of 23 are numbers that can be expressed as 23×n, where n is an integer. Examples include 23, 46, 69, 92, and so on.
## How do you find multiples of 23?
To find multiples of 23, multiply 23 by any integer. For example, 23×2 = 46 and 23×3 = 69.
## Is 46 a multiple of 23?
Yes, 46 is a multiple of 23 because 23×2 = 46.
## How can you verify if a number is a multiple of 23?
To verify if a number is a multiple of 23, divide the number by 23. If the quotient is an integer and there is no remainder, it is a multiple of 23.
## Are multiples of 23 always even?
No, multiples of 23 are not always even. For example, 23 itself is an odd number.
## What is the 10th multiple of 23?
The 10th multiple of 23 is 23×10 = 230.
## Is 115 a multiple of 23?
Yes, 115 is a multiple of 23 because 23×5 = 115.
## What is the next multiple of 23 after 138?
The next multiple of 23 after 138 is 138+23 = 161.
## Why is 69 considered a multiple of 23?
69 is considered a multiple of 23 because 23×3 = 69.
## Can a negative number be a multiple of 23?
Yes, negative numbers can be multiples of 23. For example, -23 and -46 are multiples of 23 because 23×−1 = −23 and 23×−2 = −46.
Text prompt
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# Multiples of 130 – Find & Write All Multiples
If you are looking for the solution to the maths query, “Find Multiples of 130” or “Write all Multiples of 130” then you have visited the right place.
We here at multiplesof.com provide you the full detailed guide to the Multiples of 130.
## What Are The Multiples of 130?
The result/outcome from multiplying 130 with any natural number is known as multiples of 130.
In other simple words, the multiples of 130 are the numbers that, when divided by 130, have a remainder value of 0.
The multiples of 130 are : 130, 260, 390, 520, 650, 780, 910, 1040, 1170, 1300, 1430, 1560, 1690, 1820, 1950, 2080, 2210, 2340, 2470, 2600, 2730, 2860, 2990, 3120, 3250, 3380, 3510, 3640, 3770, 3900, 4030, 4160, 4290, 4420, 4550, 4680 ……..
## List of Multiples of 130
The following table below shows the list of Multiples of 130.
## What is the 5th Multiple of 130?
The 5th Multiple of 130 is 650.
## What is the Smallest Multiple of 130?
The smallest multiple of 130 is 130.
## How to Find Multiples of 130?
Multiply the number by any natural number to find multiples of 130. For example, 130 x 5 = 650.
It states that 650 is the fifth multiple of 130.
By using the above steps you may be able to find all multiples of 130.
Every multiple of 130 number is greater than 130 or equal to 130 itself.
### What Does Multiple of 130 Mean?
the multiples of 130 are the numbers or digits, when the number is divided by 130 and leave remainder value 0.
All numbers that are divided by 130 give the remainder 0 are the multiples of number 130.
Examples of multiples of 130 are 260, 390, 520 ,650, and so on.
The result of multiplying 5 by 130 is 650, The number 650 is a multiple of 130.
### What is the Sum of First Five Multiples of 130?
The sum of the first five multiples of 130 is 1950.
## How Many Multiples Does 130 Have?
There are infinite multiples of 130. The Multiples of 130 are follows as: 130, 260, 390, 520, 650, 780, 910, 1040, 1170, 1300, 1430, 1560, 1690, 1820, 1950, 2080, 2210, 2340, 2470, 2600, 2730, 2860, 2990, 3120, 3250, 3380, 3510, 3640, 3770, 3900, 4030, 4160, 4290, 4420, 4550, 4680…… to infinity.
## What are the First 3 Multiples of 130?
The first three multiples of 130 are: 130, 260, 390.
## Are Multiples of 130 Always Odd?
No, It is not compulsory that the multiples of 130, 260, 390 are always odd.
For example, multiples of 130 that are even numbers include 1560.
The result of multiplying 130 by an even number will always give you the even number.
## Write Down the First 10 Multiples of 130
The first 10 multiples of 130, written as : 130, 260, 390, 520, 650, 780, 910, 1040, 1170, 1300.
• 130 x 1 = 130
• 130 x 2 = 260
• 130 x 3 = 390
• 130 x 4 = 520
• 130 x 5 = 650
• 130 x 6 = 780
• 130 x 7 = 910
• 130 x 8 = 1040
• 130 x 9 = 1170
• 130 x 10 = 1300
### Is 1 a Multiple of 130?
No, the number 1 is a multiple of 1 itself. 1 is not a multiple of 130. Thus, 1 is a factor of 130 not a multiple.
## What are the Multiples of 130?
The multiples of 130 are : 130, 260, 390, 520, 650, 780, 910, 1040, 1170, 1300, 1430, 1560, 1690, 1820, 1950, 2080, 2210, 2340, 2470, 2600, 2730, 2860, 2990, 3120, 3250, 3380, 3510, 3640, 3770, 3900, 4030, 4160, 4290, 4420, 4550, 4680 ……………… .
## What are the First 10 Multiples of 130?
The first 10 multiples of 130 are 130, 260, 390, 520, 650, 780, 910, 1040, 1170, 1300.
Hope from the above table, you might be able to find the first 10 multiples of 130.
## Can 130 Be A Multiple Of Itself?
Yes, because 130 = 130×1.
You will find that every number is a multiple of itself.
### Write all Multiples of 130 Number less than 100
Multiple of 130 less then 100 are : Not Available.
### Write the Multiples of 130 Between 40 and 50
The multiples of 130 between 40 and 50 are Not Available.
### Write the Multiples of 130 between 1 and 30
The multiples of 130 between 1 and 30 are Not Available.
### What is the Fifth Multiple of 130?
The fifth multiple of 130 is 650.
### What is the 11th Multiple of 130?
The 11th Multiple of 130 is 1430.
### What is the 8th Multiple of 130?
The 8th Multiple of 130 is 1040.
## What is the Average of First Five Multiples of 130?
The average of First Five Multiples of 130 is 390.
The average of first five , 130’s multiples calculated as the sum of first 5 multiples of 130 divided by the total number.
= Sum of 130/Total Numbers
Sum of First Five Multiples of = 130+260+390+520+650 = 1950
Total Numbers = 5
= 1950/5
= 390
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# Change Of Coordinate Transformations
Change of Coordinate Transformations
Field: Other
Image Created By: [[Author:| ]]
Change of Coordinate Transformations
Image displaying various forms of the transformations // temporary
# Basic Description
A Change Of Coordinate Transformation is a transformation that converts coordinates from one coordinate system to another coordinate system. Transformations such as scaling, rotating, and translating are usually looked upon as changing or manipulating the geometry itself. However, with change of coordinate transformations, it is important to realize that the coordinate representation of the geometry is modified, rather than the geometry itself.
The Change of Coordinate Systems in general is also common for converting coordinates from system to another, such as from Cartesian coordinates to Cylindrical coordinates.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Linear Algebra
Change of coordinate transformations are different for vectors and points.
### Vectors
[...]
Change of coordinate transformations are different for vectors and points.
### Vectors
Transformation of vector $\boldsymbol{\vec{p}}$ from coordinate system A to coordinate system B.
Consider a coordinate system A, and a vector $\boldsymbol{\vec{p}}$. The coordinates of $\boldsymbol{\vec{p}}$ relative to coordinate system A is $\boldsymbol{\vec{p}}_A = (x, y)$. It is also apparent that
$\boldsymbol{\vec{p}} = x\boldsymbol{\hat{u}} + y\boldsymbol{\hat{v}}$
In which $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$ are unit vectors along the x and y-axes of coordinate system A. Now consider a second coordinate system, B. In coordinate system B,
$\boldsymbol{\vec{p}}_B = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B$
More generally, given $\boldsymbol{\vec{p}}_A = (x, y)$ along with $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$, $\boldsymbol{\vec{p}}_B = (x', y')$ may be found using the formula above.
In 3-dimensional space, given $\boldsymbol{\vec{p}}_A = (x, y, z)$ then
$\boldsymbol{\vec{p}}_B = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B + z\boldsymbol{\hat{w}}_B$
In which $\boldsymbol{\hat{u}}$, $\boldsymbol{\hat{v}}$, and $\boldsymbol{\hat{w}}$ are unit vectors along the x, y, and z-axes of coordinate system A.
Given:
$\boldsymbol{\vec{p}}_A = \begin{bmatrix} 5 & 7 & 13 \end{bmatrix}$
$\boldsymbol{\hat{u}}_B = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$
$\boldsymbol{\hat{v}}_B = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$
### Points
Transformation of point$\boldsymbol{p}$ from coordinate system A to coordinate system B.
Consider a coordinate system A, and a point $\boldsymbol{q}$. Point $\boldsymbol{q}$ may be expressed as:
$\boldsymbol{q} = x\boldsymbol{\hat{u}} + y\boldsymbol{\hat{v}} + \boldsymbol{O}$
In which $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$ are unit vectors along the x and y-axes of coordinate system A, and $\boldsymbol{O}$ is the origin of coordinate system A. Now consider a second coordinate system, B. In coordinate system B,
$\boldsymbol{q}_B = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B + \boldsymbol{O}_B$
More generally, given $\boldsymbol{q}_A = (x, y)$ along with $\boldsymbol{\hat{u}}$, $\boldsymbol{\hat{v}}$, and $\boldsymbol{O}$ relative to coordinate system B, then $\boldsymbol{q}_B = (x', y')$ may be found using the above formula.
In 3-dimensional space, given $\boldsymbol{q}_A = (x, y, z)$ then
$\boldsymbol{q}_B = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B + z\boldsymbol{\hat{w}}_B + \boldsymbol{O}_B$
In which $\boldsymbol{\hat{u}}$, $\boldsymbol{\hat{v}}$, and $\boldsymbol{\hat{w}}$ are unit vectors along the x, y, and z-axes of coordinate system A, and $\boldsymbol{O}$ is the origin of coordinate system A.
### Matrix Representation
The change of coordinate transformation varies for points and vectors and thus results in two different equations. However, by using homogeneous coordinates, both cases may be handled with the following equation:
$(x', y', z', w) = x\boldsymbol{u}_B + y\boldsymbol{v}_B + z\boldsymbol{w}_B + w\boldsymbol{O}_B$
When $w = 1$, the equation handles the change of coordinate transformation for points; when $w = 0$, the equation handles the transformation for vectors. As long as the $w$-coordinate is set correctly, there is no need to keep track of two different equations. Thus the change of coordinate matrix may be defined as:
$\begin{bmatrix} x' & y' & z' & w \end{bmatrix} = \begin{bmatrix} x & y & z & w \end{bmatrix} \begin{bmatrix} u_x & u_y & u_z & 0 \\ v_x & v_y & v_z & 0 \\ w_x & w_y & w_z & 0 \\ O_x & O_y & O_z & 1 \\ \end{bmatrix} = x\boldsymbol{\hat{u}}_B + y\boldsymbol{\hat{v}}_B + z\boldsymbol{\hat{w}}_B + w\boldsymbol{O}_B$
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PYTHAGOREAN THEOREM PROOF
Updated 08/04/2010
There are literally dozens of proofs for the Pythagorean Theorem. The proof shown here is probably the clearest and easiest to understand.
The Pythagorean Theorem states that for any right triangle the square of the hypotenuse equals the sum of the squares of the other 2 sides.
If we draw a right triangle having sides 'a' 'b' and 'c' (with 'c' being the hypotenuse)
then according to the theorem, the length of c² = a² + b²
In order to prove the theorem, we construct squares on each of the sides of the triangle.
It is important to realize that squaring the length of side a is exactly the same thing as determining the area of the green square. (The same applies to side b with the red square and side c with the blue square and this is the important concept of this proof). Basically, if we can show that the area of the green square plus the area of the red square equals the area of the blue square, we have proven the Pythagorean Theorem.
Now let's construct those same squares around the remaining three sides of the blue square.
Gee, that diagram sure looks confusing doesn't it? However, you can see that those eight squares have "drawn" a square around the blue square. When we remove the 6 squares we just added, we'll have a diagram very similar to the first one except now the blue square is surrounded by a square with sides of a length 'a' plus 'b'. The area of this new square would be (a + b)² and the new diagram would look like the one drawn below.
Area of green square = a² Area of red square = b²
The area of the larger square surrounding the blue square equals (a+b)² which equals a² + 2ab + b².
Note that the blue square is surrounded by 4 right triangles, the area of each being ½ (a•b) making the total area of all 4 triangles equal 2•a•b.
So, the area of the blue square = area of the surrounding square minus the area of the 4 triangles.
Area of blue square = a² + 2ab + b² minus 2ab
Blue Square Area = c² = a² + b²
We have just proven the Pythagorean Theorem.
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How is the addition of vectors, it is not always clear to students. Children do not represent what is hidden behind them. You just have to memorize the rules, and not ponder the essence. Therefore, it is about the principles of addition and subtraction of vector quantities that a lot of knowledge is required.
As a result of the addition of two or more vectors, one is always obtained. Moreover, it will always be the same, regardless of the reception of its location.
Most often in the school course of geometry is considered the addition of two vectors. It can be performed by the rule of a triangle or parallelogram. These figures look different, but the result of the action is the same.
## How is the addition of the rule of the triangle?
It is applied when vectors are noncollinear. That is, do not lie on one line or on parallel lines.
In this case, from some arbitrary point you need to postpone the first vector. From its end it is required to conduct a parallel and equal to the second. The result will be a vector originating from the beginning of the first and ending at the end of the second. Figure resembles a triangle. Hence the name of the rule.
If the vectors are collinear, then this rule can also be applied. Only the drawing will be located along one line.
## How is the addition according to the parallelogram rule?
Yet again? applies only to non-collinear vectors. The construction is carried out according to another principle. Although the beginning is the same. It is necessary to postpone the first vector. And from its beginning - the second. Based on them, complete the parallelogram and draw a diagonal from the beginning of both vectors. She will be the result. This is the way to add vectors according to the parallelogram rule.
So far, there have been two. And what if there are 3 or 10? Use the following trick.
## How and when does the polygon rule apply?
If you want to perform the addition of vectors, the number of which is more than two, you should not be afraid. It is enough to postpone them all sequentially and connect the beginning of the chain with its end. This vector will be the desired sum.
## What properties are valid for actions with vectors?
About zero vector. Which asserts that when added with it, the original is obtained.
On the opposite vector. That is, one that has the opposite direction and is equal in magnitude to the value. Their sum will be equal to zero.
On commutativity of addition. What is known since elementary school. Changing the places of the items does not change the result. In other words, no matter which vector to put off first. The answer will still be true and unique.
On the associativity of addition. This law allows you to add in pairs any vectors from a triple and add a third one to them. If you write it with the help of signs, you get the following:
first + (second + third) = second + (first + third) = third + (first + second).
## What is known about the difference of vectors?
A separate subtraction operation does not exist. This is due to the fact that it is, in fact, an addition. Only the second of them is given the opposite direction. And then everything is done as if the addition of vectors were considered. Therefore, they practically do not speak about their differences.
In order to simplify the work with their subtraction, the triangle rule has been modified. Now (when subtracting) the second vector must be postponed from the beginning of the first. The answer will be the one that connects the end point of the deductible with it. Although it is possible to postpone as described earlier, simply by changing the direction of the second.
## How to find the sum and difference of vectors in coordinates?
The problem gives the coordinates of the vectors and you want to know their values for the final. In this construction is not necessary to perform. That is, you can use simple formulas that describe the rule of addition of vectors. They look like this:
a (x, y, z) + in (k, l, m) = c (x + k, y + l, z + m);
a (x, y, z) -c (k, l, m) = c (xk, y-l, z-m).
It is easy to notice that the coordinates you just need to add or subtract, depending on the specific task.
## The first example with the solution
Condition. Given a rectangle AVSD. Its sides are 6 and 8 cm. The intersection point of the diagonals is denoted by the letter O. It is required to calculate the difference of the vectors AO and VO.
Decision. First you need to draw these vectors. They are directed from the vertices of the rectangle to the intersection point of the diagonals.
If you look closely at the drawing, you can see that the vectors are already aligned so that the second of them is in contact with the end of the first. That's just his direction is wrong. It must start from this point. This is if the vectors are added up, and in the problem - subtraction. Stop. This action means that you need to add an oppositely directed vector. This means that VO needs to be replaced with OB. And it turns out that two vectors have already formed a pair of sides from the rule of a triangle. Therefore, the result of their addition, that is, the desired difference, is the vector AB.
And it coincides with the side of the rectangle. In order to record a numerical answer, the following will be required. Draw a rectangle along so that the big side goes horizontally. The numbering of the vertices start from the bottom left and go counterclockwise. Then the length of the vector AB will be equal to 8 cm.
Answer. The difference between AO and VO is 8 cm.
## The second example and its detailed solution
Condition. The rhombus AVSD diagonal is 12 and 16 cm. The point of their intersection is denoted by the letter O. Calculate the length of the vector formed by the difference of the vectors of AO and VO.
Decision. Let the designation of the vertices of a rhombus be the same as in the previous problem. Similarly to the solution of the first example, it turns out that the desired difference is equal to the vector AB. And its length is unknown. The solution of the problem was reduced to calculating one of the sides of the rhombus.
For this purpose, you need to consider the triangle ABO. It is rectangular, because the diagonal of the rhombus intersects at an angle of 90 degrees. And his legs are equal to half the diagonals. That is, 6 and 8 cm. The side sought in the problem coincides with the hypotenuse in this triangle.
To find it, we need the Pythagorean theorem. The square of the hypotenuse will be equal to the sum of the numbers 6 2 and 8 2. After squaring, the values will be 36 and 64. Their sum is 100. It follows that the hypotenuse is 10 cm.
Answer. The difference between the vectors of AO and HE is 10 cm.
## The third example with a detailed solution
Condition. Calculate the difference and the sum of two vectors. Their coordinates are known: in the first - 1 and 2, in the second - 4 and 8.
Decision. To find the amount you will need to add in pairs the first and second coordinates. The result will be the numbers 5 and 10. The answer will be a vector with coordinates (5; 10).
For the difference you need to perform the subtraction of coordinates. After performing this action, you get the numbers -3 and -6. They will be the coordinates of the desired vector.
Answer. The sum of vectors is (5; 10), their difference is (-3; -6).
## Fourth example
Condition. The length of the vector AB is 6 cm, BC - 8 cm. The second one is plotted from the end of the first at an angle of 90 degrees. Calculate: a) the difference of the modules of the vectors BA and BC and the module of the difference BA and BC; b) the sum of the same modules and the modulus of the sum.
Solution: a) The lengths of the vectors are already given in the problem. Therefore, to calculate their difference is not difficult. 6 - 8 = -2. The situation with the difference module is somewhat more complicated. First you need to know which vector will be the result of subtraction. For this purpose, you should postpone the vector BA, which is directed in the opposite direction AB. Then from its end to hold the vector of the sun, directing it in the direction opposite to the original. The result of the subtraction is the vector CA. Its module can be calculated by the Pythagorean theorem. Simple calculations lead to a value of 10 cm.
b) The sum of the modules of the vectors is 14 cm. To search for the second answer, some conversion is required. Vector BA is oppositely directed to that given by - AB. Both vectors are directed from one point. In this situation, you can use the parallelogram rule. The result of the addition will be a diagonal, and not just a parallelogram, but a rectangle. Its diagonals are equal, which means that the modulus of the sum is the same as in the previous paragraph.
Answer: a) -2 and 10 cm; b) 14 and 10 cm.
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# Study Guide And Review Math Worksheet With Answer Key - Chapter 4
The range of a relation is the domain of the inverse
of the relation. The statement is false.
Study Guide and Review - Chapter 4
false, domain
State whether each sentence is true or false . If
Write an equation of a line in slope-intercept
form with the given slope and y-intercept. Then
false , replace the underlined term to make a
graph the equation.
true sentence.
10. slope: 3, y-intercept: 5
2. The process of using a linear equation to make
predictions about values that are beyond the range of
SOLUTION:
the data is called linear regression.
The slope-intercept form of a line is y = mx + b,
SOLUTION:
where m is the slope, and b is the y-intercept.
The statement is false. The process of using a linear
equation to make predictions about values that are
beyond the range of the data is called linear
extrapolation. Linear regression is an algorithm to
find a precise line of fit for a set of data.
To graph the equation, plot the y-intercept (0, 5).
Then move down 3 units and left 1 unit. Plot the
point. Draw a line through the two points.
false, linear extrapolation
4. The correlation coefficient describes whether the
correlation between the variables is positive or
negative and how closely the regression equation is
modeling the data.
SOLUTION:
The correlation coefficient describes whether the
correlation between the variables is positive or
negative and how closely the equation is modeling the
data. So, the statement is true.
true
y = 3x + 5
6. Lines that intersect at acute angles are called
perpendicular lines.
SOLUTION:
The statement is false. Lines that intersect at right
angles are called perpendicular lines. Acute angles
have measures less than 90.
false, right
8. The range of a relation is the range of its inverse
function.
12. slope:
, y-intercept: 3
SOLUTION:
The range of a relation is the domain of the inverse
SOLUTION:
of the relation. The statement is false.
The slope-intercept form of a line is y = mx + b,
is the y -intercept.
where m is the slope, and b
false, domain
Write an equation of a line in slope-intercept
form with the given slope and y-intercept. Then
graph the equation.
Page 1
10. slope: 3, y-intercept: 5
To graph the equation, plot the y-intercept (0, 3).
SOLUTION:
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LCM of 4 and also 9 is the the smallest number amongst all typical multiples that 4 and 9. The first couple of multiples of 4 and also 9 space (4, 8, 12, 16, 20, 24, 28, . . . ) and also (9, 18, 27, 36, . . . ) respectively. There space 3 typically used methods to uncover LCM that 4 and 9 - by prime factorization, by listing multiples, and by department method.
You are watching: Least common multiples of 4 and 9
1 LCM the 4 and 9 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM that 4 and 9 is 36.
Explanation:
The LCM of two non-zero integers, x(4) and y(9), is the smallest positive integer m(36) the is divisible by both x(4) and also y(9) without any type of remainder.
The techniques to find the LCM of 4 and also 9 are described below.
By Listing MultiplesBy prime Factorization MethodBy division Method
### LCM that 4 and also 9 by Listing Multiples
To calculate the LCM that 4 and 9 by listing out the typical multiples, we deserve to follow the given listed below steps:
Step 1: perform a few multiples the 4 (4, 8, 12, 16, 20, 24, 28, . . . ) and 9 (9, 18, 27, 36, . . . . )Step 2: The common multiples from the multiples that 4 and 9 are 36, 72, . . .Step 3: The smallest usual multiple that 4 and also 9 is 36.
∴ The least typical multiple the 4 and also 9 = 36.
### LCM the 4 and 9 by prime Factorization
Prime administer of 4 and also 9 is (2 × 2) = 22 and also (3 × 3) = 32 respectively. LCM that 4 and also 9 have the right to be derived by multiplying prime factors raised to their respective highest power, i.e. 22 × 32 = 36.Hence, the LCM the 4 and 9 by element factorization is 36.
### LCM the 4 and 9 by division Method
To calculation the LCM that 4 and also 9 by the department method, we will divide the numbers(4, 9) by their prime determinants (preferably common). The product of these divisors offers the LCM of 4 and 9.
Step 3: continue the procedures until just 1s space left in the last row.
The LCM the 4 and also 9 is the product of all prime number on the left, i.e. LCM(4, 9) by division method = 2 × 2 × 3 × 3 = 36.
☛ also Check:
## FAQs top top LCM the 4 and also 9
### What is the LCM the 4 and also 9?
The LCM that 4 and 9 is 36. To uncover the least typical multiple (LCM) that 4 and 9, we need to discover the multiples of 4 and also 9 (multiples of 4 = 4, 8, 12, 16 . . . . 36; multiples that 9 = 9, 18, 27, 36) and also choose the the smallest multiple the is precisely divisible through 4 and also 9, i.e., 36.
### How to find the LCM of 4 and 9 by prime Factorization?
To discover the LCM the 4 and 9 making use of prime factorization, we will find the prime factors, (4 = 2 × 2) and also (9 = 3 × 3). LCM of 4 and 9 is the product of prime factors raised to your respective highest exponent among the number 4 and 9.⇒ LCM of 4, 9 = 22 × 32 = 36.
### What is the Relation between GCF and LCM of 4, 9?
The complying with equation can be provided to to express the relation in between GCF and LCM of 4 and also 9, i.e. GCF × LCM = 4 × 9.
### If the LCM the 9 and 4 is 36, uncover its GCF.
LCM(9, 4) × GCF(9, 4) = 9 × 4Since the LCM that 9 and 4 = 36⇒ 36 × GCF(9, 4) = 36Therefore, the greatest typical factor (GCF) = 36/36 = 1.
See more: Arctan(-Sqrt(3)) - Evaluate Arctan(( Square Root Of 3)/3)
### What is the the very least Perfect Square Divisible by 4 and 9?
The the very least number divisible by 4 and also 9 = LCM(4, 9)LCM of 4 and 9 = 2 × 2 × 3 × 3 ⇒ the very least perfect square divisible by every 4 and 9 = 36 Therefore, 36 is the required number.
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# 数学代写|线性规划作业代写Linear Programming代考|Motivation-Finding Upper Bounds
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## 数学代写|线性规划作业代写Linear Programming代考|Motivation-Finding Upper Bounds
We begin with an example:
\begin{aligned} \operatorname{maximize} & 4 x_1+x_2+3 x_3 \ \text { subject to } \quad x_1+4 x_2 & \leq 1 \ 3 x_1-x_2+x_3 & \leq 3 \ x_1, x_2, x_3 & \geq 0 . \end{aligned}
Our first observation is that every feasible solution provides a lower bound on the optimal objective function value, $\zeta^$. For example, the solution $\left(x_1, x_2, x_3\right)=(1,0,0)$ tells us that $\zeta^ \geq 4$. Using the feasible solution $\left(x_1, x_2, x_3\right)=(0,0,3)$, we see that $\zeta^* \geq 9$. But how good is this bound? Is it close to the optimal value? To answer, we need to give upper bounds, which we can find as follows. Let’s multiply the first constraint by 2 and add that to 3 times the second constraint:
\begin{aligned} 2\left(x_1+4 x_2 \quad\right) & \leq 2(1) \ +3\left(3 x_1-x_2+x_3\right) & \leq 3(3) \ \hline 11 x_1+5 x_2+3 x_3 & \leq 11 \end{aligned}
Now, since each variable is nonnegative, we can compare the sum against the objective function and notice that
$$4 x_1+x_2+3 x_3 \leq 11 x_1+5 x_2+3 x_3 \leq 11 .$$
## 数学代写|线性规划作业代写Linear Programming代考|The Dual Problem
Given a linear programming problem in standard form,
\begin{aligned} & \operatorname{maximize} \sum_{j=1}^n c_j x_j \ & \text { subject to } \sum_{j=1}^n a_{i j} x_j \leq b_i \quad i=1,2, \ldots, m \ & x_j \geq 0 \quad j=1,2, \ldots, n, \ & \end{aligned}
the associated dual linear program is given by
\begin{aligned} & \operatorname{minimize} \sum_{i=1}^m b_i y_i \ & \text { subject to } \sum_{i=1}^m y_i a_{i j} \geq c_j \quad j=1,2, \ldots, n \ & y_i \geq 0 \quad i=1,2, \ldots, m . \ & \end{aligned}
Since we started with (5.1), it is called the primal problem. Our first order of business is to show that taking the dual of the dual returns us to the primal. To see this, we first must write the dual problem in standard form. That is, we must change the minimization into a maximization and we must change the first set of greater-thanor-equal-to constraints into less-than-or-equal-to. Of course, we must effect these changes without altering the problem. To change a minimization into a maximization, we note that to minimize something it is equivalent to maximize its negative and then negate the answer:
$$\min \sum_{i=1}^m b_i y_i=-\max \left(-\sum_{i=1}^m b_i y_i\right)$$
# 线性规划代写
## 数学代写|线性规划作业代写Linear Programming代考|Motivation-Finding Upper Bounds
\begin{aligned} \operatorname{maximize} & 4 x_1+x_2+3 x_3 \ \text { subject to } \quad x_1+4 x_2 & \leq 1 \ 3 x_1-x_2+x_3 & \leq 3 \ x_1, x_2, x_3 & \geq 0 . \end{aligned}
\begin{aligned} 2\left(x_1+4 x_2 \quad\right) & \leq 2(1) \ +3\left(3 x_1-x_2+x_3\right) & \leq 3(3) \ \hline 11 x_1+5 x_2+3 x_3 & \leq 11 \end{aligned}
$$4 x_1+x_2+3 x_3 \leq 11 x_1+5 x_2+3 x_3 \leq 11 .$$
## 数学代写|线性规划作业代写Linear Programming代考|The Dual Problem
\begin{aligned} & \operatorname{maximize} \sum_{j=1}^n c_j x_j \ & \text { subject to } \sum_{j=1}^n a_{i j} x_j \leq b_i \quad i=1,2, \ldots, m \ & x_j \geq 0 \quad j=1,2, \ldots, n, \ & \end{aligned}
\begin{aligned} & \operatorname{minimize} \sum_{i=1}^m b_i y_i \ & \text { subject to } \sum_{i=1}^m y_i a_{i j} \geq c_j \quad j=1,2, \ldots, n \ & y_i \geq 0 \quad i=1,2, \ldots, m . \ & \end{aligned}
$$\min \sum_{i=1}^m b_i y_i=-\max \left(-\sum_{i=1}^m b_i y_i\right)$$
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## MATLAB代写
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Instruction
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The notion of critical point functions are closely related to the notion of its derivative at that point. Namely, point is called critical if the derivative of a function it does not exist or is zero. Critical points are the internal points of the domain of definition of the function.
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To determine the critical points of this function, you must perform several actions: find the domain of the function, calculate derivative, find the domain of definition of the derivative function, find the point treatment of the derivative to zero, to prove the belonging of the points defining the original function.
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Example 1Определите critical points of the function y = (x - 3)2·(x-2).
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Reinitiate the domain of the function, in this case there are no restrictions: x ∈ (-∞; +∞);Compute the derivative y’. According to the rules of differentiation of product of two functions is: y’ = ((x - 3)2)’·(x - 2) + (x - 3)2·(x - 2)’ = 2·(x - 3)·(x - 2) + (x - 3)2·1. After opening the brackets it turns out the quadratic equation: y’ = 3·x2 – 16·x + 21.
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Find the definition of the derivative of the function: x ∈ (-∞; +∞).Solve the equation 3·x2 – 16·x + 21 = 0 to find which x the derivative becomes zero: 3·x2 – 16·x + 21 = 0.
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D = 256 – 252 = 4x1 = (16 + 2)/6 = 3; x2 = (16 - 2)/6 = 7/3.So, the derivative becomes zero at values of x equal to 3 and 7/3.
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Determine whether found point determining the original function. Since x (-∞; +∞), then both these points are critical.
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Example 2Определите critical points of the function y = x2 – 2/x.
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Rectiability function definition: x ∈ (-∞; 0) ∪ (0; +∞), since x is in the denominator.Calculate the derivative y’ = 2·x + 2/x2.
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The scope of the definition of the derivative function is the same as the original: x ∈ (-∞; 0) ∪ (0; +∞).Solve the equation 2·x + 2/x2 = 0:2·x = -2/x2 → x = -1.
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So the derivative vanishes at x = -1. Done necessary but not sufficient condition for criticality. Since x=-1 falls within the interval (-∞; 0) ∪ (0; +∞), this point is critical.
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