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Chapter 11: Functions
# 11.5 Logarithmic Functions
Logarithms come from a rich history, extending from the Babylonians around 1500–2000 BC, through the Indian mathematician Virasena around 700–800 AD, and later rapidly growing and expanding in European science from the mid-1500s and on. Logarithms were developed to reduce multiplication and division to correspond to adding and subtracting numbers on a number line. Quite simply, logarithms reduced the complexity of these functions and retained significance until the advent of the computer. Even so, logarithms are still in use today in many functions. This topic is taught here, since the logarithmic function is the inverse to the exponential function (shown in 11.4). We will use this feature to solve both exponential and logarithmic functions.
In general, a logarithm is the exponent to which the base must be raised to get the number that you are taking the logarithm of. Using logarithms and exponents together, we can start to identify useful relations.
Consider 23. 23 = 2 × 2 × 2 = 8, when written using logarithmic functions, will look like $\log_{2} 8 = 3$. You read this as the log base 2 of 8 equals 3. This means that, if you are using the base 2 and are looking to find the exponent that yields 8, the power needed on the base is 3. You can quantify this relation in either the one of the two equations: $x = a^y$ or $y = \log_{a} x$. Writing this relationship in either form is illustrated in the following examples.
Example 11.5.1
Write the logarithmic equation for each given exponential relation.
1. $x^3 = 12\hspace{0.5in} \text{In a logarithmic equation looks like}\hspace{0.5in} \log_{x}12 = 3$
2. $4^3 = 64\hspace{0.5in}\text{In a logarithmic equation looks like}\hspace{0.5in} \log_{4}64 = 3$
3. $7^2 = 49\hspace{0.5in}\text{In a logarithmic equation looks like}\hspace{0.5in} \log_{7}49 = 2$
4. $y^6 = 102\hspace{0.5in}\text{In a logarithmic equation looks like}\hspace{0.5in} \log_{y}102 = 6$
Example 11.5.2
Write the exponential relation for each given logarithmic equation.
1. $\log_{x}42=5 \hspace{0.54in} \text{In exponential form looks like }\hspace{0.5in}x^5=42$
2. $\log_{4}624 = 5 \hspace{0.5in}\text{In exponential form looks like }\hspace{0.5in}4^5 = 624$
3. $\log_{3}18 = 2\hspace{0.54in}\text{In exponential form looks like }\hspace{0.5in}3^2 = 18$
4. $\log_{y}12=4\hspace{0.54in}\text{In exponential form looks like }\hspace{0.5in}y^4=12$
A further illustration of this relationship is shown below for the exponents and logarithms for the common base values of 2 and 10.
## Examples of Exponents and Logarithms for Base 2 and 10
$\begin{array}{llll} 2^0=1&\hspace{0.5in}\log_{2}1=0&\hspace{0.5in}10^0=1&\hspace{0.5in}\log_{10}1=0 \\ 2^1=2&\hspace{0.5in}\log_{2}2=1&\hspace{0.5in}10^1=10&\hspace{0.5in}\log_{10}10=1 \\ 2^2=4&\hspace{0.5in}\log_{2}4=2&\hspace{0.5in}10^2=100&\hspace{0.5in}\log_{10}100=2 \\ 2^3=8&\hspace{0.5in}\log_{2}8=3&\hspace{0.5in}10^3=1000&\hspace{0.5in}\log_{10}1000=3 \\ 2^4=16&\hspace{0.5in}\log_{2}16=4&\hspace{0.5in}10^4=10000&\hspace{0.5in}\log_{10}10000=4 \\ 2^5=32&\hspace{0.5in}\log_{2}32=5&\hspace{0.5in}10^5=100000&\hspace{0.5in}\log_{10}100000=5 \\ 2^6=64&\hspace{0.5in}\log_{2}64=6&\hspace{0.5in}10^6=1000000&\hspace{0.5in}\log_{10}1000000=6 \\ 2^7=128&\hspace{0.5in}\log_{2}128=7&\hspace{0.5in}10^7=10000000&\hspace{0.5in}\log_{10}10000000=7 \\ 2^8=256&\hspace{0.5in}\log_{2}256=8&\hspace{0.5in}10^8=100000000&\hspace{0.5in}\log_{10}100000000=8 \end{array}$
In the following examples, we evaluate logarithmic functions by converting the logarithms to exponential form.
Example 11.5.3
Evaluate the logarithmic equation $\log_{2}64 = x$.
The exponential form of this logarithm is $2^x = 64$.
Since 64 equals 26, rewrite this as $2^x = 2^6$, which means that $x = 6$.
Often, you are asked to evaluate a logarithm that is not in the form of an equation; rather, it is given as a simple logarithm. For this type of question, set the logarithm to equal $x$ and then solve as we did above.
Example 11.5.4
Evaluate the logarithmic equation $\log_{125}5$.
First, we set this logarithm to equal $x$, so $\log_{125}5 = x$.
The exponential form of this logarithm is $5^x = 125$.
Since 125 equals 53, we rewrite this as $5^x = 5^3$, which means that $x = 3$.
Logarithmic equations that appear more complicated are solved using a somewhat similar strategy as above, except that you often employ algebraic methods. For instance:
Example 11.5.5
Evaluate the logarithmic equation $\log_{2}(3x + 5) = 4$.
The exponential form of this logarithm is $2^4 = 3x + 4$.
This now becomes an algebraic equation to solve:
$\begin{array}{rrrrr} 16&=&3x&+&4 \\ -4&&&-&4 \\ \hline \dfrac{12}{3}&=&\dfrac{3x}{3}&& \\ \\ x&=&4&& \end{array}$
The most common form of logarithm uses base 10. This can be compared to the most common radical of the square root. When encountering base 10 logarithms, they are often written without the base 10 shown. To solve these, rewrite in exponential form using the base 10.
Example 11.5.6
Evaluate the logarithmic equation $\log x = -2$.
The exponential form of this logarithm is $10^{-2} = x$.
Since $10^{-2} = \dfrac{1}{100}$, this means that $x = \dfrac{1}{100}$.
# Questions
Rewrite each equation in exponential form.
1. $\log_{9}81=2$
2. $\log_{b}a=-16$
3. $\log_{7}\dfrac{1}{49}=-2$
4. $\log_{16}256=2$
5. $\log_{13}169=2$
6. $\log_{11}1=0$
Rewrite each equation in logarithmic form.
1. $8^0=1$
2. $17^{-2}=\dfrac{1}{289}$
3. $15^2=225$
4. $144^{\frac{1}{2}}=12$
5. $64^{\frac{1}{6}}=2$
6. $19^2=361$
Evaluate each expression.
1. $\log_{125}5$
2. $\log_{5}125$
3. $\log_{343}\dfrac{1}{7}$
4. $\log_{7}1$
5. $\log_{4}16$
6. $\log_{4}\dfrac{1}{64}$
7. $\log_{6}36$
8. $\log_{36}6$
9. $\log_{2}64$
10. $\log_{3}243$
Solve each equation.
1. $\log_{5}x=1$
2. $\log_{8}k=3$
3. $\log_{2}x=-2$
4. $\log n=3$
5. $\log_{11}k=2$
6. $\log_{4}p=4$
7. $\log_{9}(n+9)=4$
8. $\log_{11}(x-4)=-1$
9. $\log_{5}(-3m)=3$
10. $\log_{2}-8r=1$
11. $\log_{11}(x+5)=-1$
12. $\log_{7}-3n=4$
13. $\log_{4}(6b+4)=0$
14. $\log_{11}(10v+1)=-1$
15. $\log_{5}(-10x+4)=4$
16. $\log_{9}(7-6x)=-2$
17. $\log_{2}(10-5a)=3$
18. $\log_{8}(3k-1)=1$
Answer Key 11.5
## License
Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. |
# Why do we miss 8 in the decimal expansion of 1/81, and 98 in the decimal expansion of 1/9801?
Why do we miss $8$ in the decimal expansion of $1/81$, and $98$ in the decimal expansion of $1/9801$? I’ve seen this happen that when you divide in a fraction using the square of any number with only nines in the denominator. Like in
and in
the decimals go on predictably when suddenly in the first one you miss $8$, and in the second you miss $98$ and it keeps going on forever. How does this happen? Why do we miss numbers like $8$ in the decimal representation of $\frac{1}{9^2},$ or like $98$ in the decimal of $\frac{1}{99^2},$ or $998$ in $\frac{1}{999^2},$ or $9998$ in $\frac{1}{9999^2}\;$?
For $\frac{1}{81}$, there was an $8$, but it got bumped up. We can write $\frac{1}{81}$ as this sum:
This kind of effect of “carrying the $1$” when the nine digit flips to a ten is the thing that is causing the behavior in all of the fractions you are describing.
To follow-up, this should provide a bit more insight as to why interesting patterns appear in the decimal representations of fractions with a power of $9$ or $11$ as the denominator, and see why we can write those numbers like $\frac{1}{81}$ as that sum. First note that
so if we were to consider $\frac{1}{81}$ like before, we would have
Then if we were to want to know the value of the ten-thousandth’s decimal place of $\frac{1}{81}$, we would just have to find the numerator of the term in the expansion of this square with a denominator of $10\,000$, which we can readily see is
So it looks like these evident patterns that appear in the decimal expansions of fraction with a multiple of $9$ in the denominator is due at least partly to the fact that this infinite sum representation of $\frac{1}{9}$ consists entirely of terms with a numerator of $1$, so multiplying this sum into things may result in “predictable” behavior that results in a pattern.
As for why having a multiple of $11$ in the denominator makes similar patterns, note that
So again we have an infinite sum of terms each with a numerator of $1$ (just alternating sign this time) that will result in certain “predictable” patterns when multiplied by things. |
# Rational expression solver
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Best of all, Rational expression solver is free to use, so there's no sense not to give it a try! If you're solving for x with logs, then you're likely only interested in how things are changing over time. This is why we can use logs to calculate percent change. To do this, we first need to transform the data into a proportional format. For example, if we have data in the form of \$x = y and want to know the change in \$x over time, we would take the log of both sides: log(x) = log(y) + log(1/y). Then, we can just plot all of these points on a graph and look for trends. Next, let's say that we have data in the form of \$x = y and want to know the percent change in \$x over time. In this case, instead of taking the log of both sides, we would simply divide by 1: frac{log{\$x} - log{\$y}}{ ext{log}}. Then, we can again plot all of these points on a graph and look for trends.
Solving exponential functions can be a bit tricky because of the tricky constant that appears at the end of the equation. But don’t worry! There are a few ways to solve exponential functions. Let’s start with the easiest way: plugging in values. When your function has a non-zero constant at the end, you can use that constant to find your answer. For example, let’s say our function is y = 2x^3 + 2 and we want to solve for x using this method. First, plug in 2 for x by putting x=2 into our function. Then, multiply both sides by 3 on the left to get x=6. Finally, add 2 to both sides to get x=8. If you were able to do this, then your answer is 8! When you can’t use this method, there are two other ways to solve an exponential equation: tangent or logarithmic. Tangent means “slope”, and it is used when you know the slope of your graph at one point in time (such as when it starts) and want to find out where it ends up at another point in time (such as when it ends). Logarithmic means “log base number”, and it is used when you want to find out how quickly something grows over
Arithmetic math problems are a staple in every grade. They help kids practice basic math facts and develop their ability to count and add numbers. With so much emphasis on arithmetic in school, there are plenty of arithmetic math problems to choose from. Here are some of the best: Here are some tips for solving arithmetic math problems: 1) Keep track of the problem steps. If you’re unsure about how to proceed, write down each step as you go. 2) Be careful with your answer choices. There are two types of answers that students can choose from: right and wrong. Don’t be afraid to pick a right answer if it makes sense, but don’t be too quick to pick the wrong options either. 3) Break down problems into smaller parts. This will help you keep track of all the steps needed to complete the problem and make sure you don’t miss anything along the way. 4) Look for patterns in the problem steps. If you see a pattern repeating itself over and over again, you can use that information to help solve the problem more quickly.
In mathematics, solving a system of equations is the process of turning an equation into a true statement that can be solved for any unknown value. The equation is converted into a set of linear equations using the same variable names as the original equation. Each equation becomes a row in a matrix or array and then the unknown value can be found by solving each row. This example shows how to solve systems of equations. Each row represents an equation. The first column represents the variable on the left side of the equation and the second column represents the variable on the right side of the equation. The last column represents the sum of all other columns. The values in this matrix represent all possible values for each variable. When solving systems of equations, you start by writing down every possible combination of variables that could take place in your problem and then adding up all those numbers to find out what your solution should be. In addition, it is important to work carefully with multiple operations when working with systems of equations. For example, if two different operations are performed on two different sets of equations, one set may become more difficult to solve than another set.
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# EXAMPLE 2 Find the area of a regular polygon DECORATING You are decorating the top of a table by covering it with small ceramic tiles. The table top is.
## Presentation on theme: "EXAMPLE 2 Find the area of a regular polygon DECORATING You are decorating the top of a table by covering it with small ceramic tiles. The table top is."— Presentation transcript:
EXAMPLE 2 Find the area of a regular polygon DECORATING You are decorating the top of a table by covering it with small ceramic tiles. The table top is a regular octagon with 15 inch sides and a radius of about 19.6 inches. What is the area you are covering? SOLUTION STEP 1 Find the perimeter P of the table top. An octagon has 8 sides, so P = 8(15) = 120 inches.
EXAMPLE 2 STEP 2 So, QS = (QP) = (15) = 7.5 inches. 1 2 1 2 To find RS, use the Pythagorean Theorem for RQS. a = RS 19.6 2 – 7.5 2 = 327.91 18.108 Find the apothem a. The apothem is height RS ofPQR. BecausePQR is isosceles, altitude RS bisects QP. Find the area of a regular polygon
EXAMPLE 2 STEP 3 Find the area A of the table top. 1 2 A = aP Formula for area of regular polygon (18.108)(120) 1 2 Substitute. 1086.5 Simplify. Find the area of a regular polygon So, the area you are covering with tiles is about 1086.5 square inches. ANSWER
EXAMPLE 3 Find the perimeter and area of a regular polygon A regular nonagon is inscribed in a circle with radius 4 units. Find the perimeter and area of the nonagon. SOLUTION 360° The measure of central JLK is, or 40°. Apothem LM bisects the central angle, so m KLM is 20°. To find the lengths of the legs, use trigonometric ratios for right KLM. 9
EXAMPLE 3 sin 20° = MK LK sin 20° = MK 4 4 sin 20° = MK cos 20° = LM LK cos 20° = LM 4 4 cos 20° = LM The regular nonagon has side length s = 2MK = 2(4 sin 20°) = 8 sin 20° and apothem a = LM = 4 cos 20°. Find the perimeter and area of a regular polygon
EXAMPLE 3 Find the perimeter and area of a regular polygon So, the perimeter is P = 9s = 9(8 sin 20°) = 72 sin 20° 24.6 units, and the area is A = aP = (4 cos 20°)(72 sin 20°) 46.3 square units. 1 2 1 2 ANSWER
GUIDED PRACTICE for Examples 2 and 3 3. Find the perimeter and the area of the regular polygon. about 46.6 units, about 151.5 units 2 ANSWER
GUIDED PRACTICE for Examples 2 and 3 4.4. Find the perimeter and the area of the regular polygon. 70 units, about 377.0 units 2 ANSWER
GUIDED PRACTICE for Examples 2 and 3 5.5. 30 3 52.0 units, about 129.9 units 2 ANSWER Find the perimeter and the area of the regular polygon.
GUIDED PRACTICE for Examples 2 and 3 6. Which of Exercises 3–5 above can be solved using special right triangles? Exercise 5ANSWER
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# How do you factor 125x 3?
## How do you factor 125x 3?
Rewrite 125 as 53 . Since both terms are perfect cubes, factor using the difference of cubes formula, a3−b3=(a−b)(a2+ab+b2) a 3 – b 3 = ( a – b ) ( a 2 + a b + b 2 ) where a=5 and b=x . Raise 5 to the power of 2 .
How do you factor 125×3 27?
1. 125×3−27 is a difference of cubes.
2. 125×3=(5x)3⇒a=5x.
3. 27=(3)3⇒b=3.
What is the complete factorization of 3x 4 48?
3×4 – 48 = 3(x2 + 4)(x – 2)(x + 2). There are 2 terms and there is a common factor of 2. Then the expression in the brackets is the difference of 2 cubes. Remember how to factor a3 – b3.
### How do you factor 125?
Factors of 125
1. Factors of 125: 1, 5, 25, 125.
2. Prime Factorization of 125: 5 × 5 × 5.
How do you factor 8x 3 y 3?
Algebra Examples Rewrite 8×3 8 x 3 as (2x)3 ( 2 x ) 3 . Since both terms are perfect cubes, factor using the difference of cubes formula, a3−b3=(a−b)(a2+ab+b2) a 3 – b 3 = ( a – b ) ( a 2 + a b + b 2 ) where a=2x a = 2 x and b=y . Simplify. Apply the product rule to 2x 2 x .
How do you factor 49×2 9?
Algebra Examples Rewrite 49×2 49 x 2 as (7x)2 ( 7 x ) 2 . Rewrite 9 as 32 . Since both terms are perfect squares, factor using the difference of squares formula, a2−b2=(a+b)(a−b) a 2 – b 2 = ( a + b ) ( a – b ) where a=7x a = 7 x and b=3 .
#### How to write factor 125 as 53 5 3?
Rewrite 125 125 as 53 5 3. Since both terms are perfect cubes, factor using the difference of cubes formula, a3 −b3 = (a−b)(a2 + ab+b2) a 3 – b 3 = ( a – b) ( a 2 + a b + b 2) where a = x a = x and b = 5 b = 5. Simplify.
How to factor a third power trinomial?
Factor a third power trinomial (a polynomial with three terms) such as x 3+5x 2+6x. Think of a monomial that is a factor of each of the terms in the equation. In x 3+5x 2+6x, x is a common factor for each of the terms.
What do you call a third power polynomial?
A third power polynomial, also called a cubic polynomial, includes at least one monomial or term that is cubed, or raised to the third power. |
Euclid Division Lemma Real Numbers Class Ten Mathematics
# Real Number
## Theorem (Euclid’s Division Lemma):
For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such that
a=bq+r, where 0≤r<b
Explanation:
Thus, for any pair of two positive integers a and b; the relation
a=bq+r, where 0≤r<b
will be true where q is some integer.
Example:
(a) 20, 8
Let 20 = a and 8 = b
Therefore, by applying the relation a=bq+r, where 0≤r<b we get 20=8xx2+4 (In this q=2 and r=4)
(b) 17, 5
Let 17 = a and 5 = b
Therefore, by applying the relation a=bq+r, where 0≤r<b we get 17=5xx3+2 (In this q=3 and r=2
### Exercise 1.1(NCERT Book)
Question - 1: Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Solution: (i) 135 and 225
In the given problem, let 225 = a, and 135 = b
Therefore, by applying the relation a=bq+r, where 0≤r<b we get
225=135xx1+90 (Here r=90)
Since, r (remainder) is not equal to zero (0). Thus, by applying the Euclid’s division algorithm, by taking 135 = a, and 90 = b we get
135=90xx2+45 (Here r=45)
Since, in this step also, r is not equal to zero(0). Thus by continuing the Euclid’s division algorithm, by taking this time, 90 = a, and 45 = b we get
90=45xx2+0 (In this step we get r=0)
Therefore, 45 is the HCF of given pair 225 and 135
Solution: (ii) 196 and 38220
In the given pair, 38220 > 196, thus let 38220 = a and 196 = b
Now by applying the Euclid’s division algorithm, we get
38220=196xx195+0 (Here q=195 and r=0)
Since, in the above equation we get, r = 0, therefore, 196 is the HCF of the given pair 196 and 38220.
(iii) 867 and 255
Solution: Let a = 867 and b = 255, thus after applying the Euclid’s division algorithm we get
867 = 255 xx 3 + 102 ( Where r = 102)
Since, r ≠ 0, therefore, by taking 255 and 102 as a and b respectively we get
255 = 102 xx 2 + 51
Similarly, 102 = 51 xx 2 + 0 (Where, r = 0)
Since, in this term r = 0, thus HCF of the given pair 867 and 255 is equal to 51
Question - 2 – Show that any positive odd integer is of the form 6q + 1 or, 6q +3 or, 6q + 5, where q is some integer.
Solution: Let ‘a’ be any positive odd integer and ‘b = 6’.
Therefore,
a=6q+r where 0≤r<6
Now, by placing r = 0, we get, a = 6q + 0 = 6q
By placing r = 1, we get, a = 6q +1
By placing, r = 2, we get, a = 6q + 2
By placing, r = 3, we get, a = 6q + 3
By placing, r = 4, we get, a = 6q + 4
By placing, r = 5, we get, a = 6q +5
Thus, a = 6q or, 6q +1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q +5
But here, 6q, 6q + 2, 6q +4 are the even integers
Therefore, 6q + 1 or, 6q + 3 or, 6q + 5 are the forms of any positive odd integers.
Alternate Method: Assume any value for q; like 1, 2, 3, ………….n
If q = 1;
Then;
6q+1=6xx1+1=7
If q = 2; Then;
6q+1=6xx2+1=13
6q+3=6xx2+3=15
6q+5=6xx2+5=17
Similarly, we can go on substitiuting different values for q and the result would always be a positive odd integer.
Another rationale for this can be as follows:
Since 6 is an even number and product of any number with an even number is always an even number. Moreover, if an odd number is added to any even number, we get an odd number as result.
Question 3: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution: The required number of column is obtained by the HCF of 616 and 32
Let, a = 616 and b = 32, therefore, by applying Euclid’s division algorithm, we get
616 = 32 xx 19 + 8, since, here r = 8 and p≠0, thus, by continuing the process, we get
32 = 8 xx 2 + 0, here r = 0
Thus, the HCF of 616 and 32 is equal to 8
Thus, the required maximum number of column = 8
Question 4: Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3q, 3q +1 or 3q + 2. Now square each of these and show that they can rewritten in the form 3m or 3m +1]
Solution: Let ‘n’ is any positive integer, then it is of the form of 3q or, 3q + 1 or, 3q + 2
If, n=3q
By squaring both sides we get
n^2=(3q)^2=9q^2=3(3q^2)
Or, n^2=3m, where m=3q^2
By squaring both sides we get
n^2=(3q+1)^2
=9q^2+6q+1
=3(3q^2+2q)+1
Or, n^2=3m+1
Where m=(3q^2+2q)
If, n =3q+2
Then b squaring both sides we get
n^2=(3q+2)^2
=9q^2+12q+4
=9q^2+12q+3+1
=3(q^2+4q+1)+1
Or, n^2=3m+1
Where, m=q^2+4q+1
Therefore, square of any positive integer is either of the form of 3m or 3m + 1
Alternate method:
4=3xx1+1, i.e. 3m+1
Let us take the next square number, i.e. 9
9=3xx3, i.e. 3m
Let us take the next square number, i.e. 16
16=3xx5+1, i.e. 3m+1
Therefore, square of any positive integer is either of the form of 3m or 3m + 1
Question 5: Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution: Let ‘a’ is a positive integer and b = 4
Therefore, according Euclid’s division lemma
a=3q+r, where 0≤r&let;4
Therefore, if r=0 then
a=3q
Or, a^3=(3q)^3=27q^3
Or, a^3=9(3q^3)
Or, a^3=9m, where m=3q^3
If r=1 then
a=3q+1
Or, a^3=(3q+1)^3=27q^3+1+27q^2+9q
(Since (a+b)^3=a^3+b^3+3a^2b+3ab^2)
Or, a^3=27q^3+27q^2+9q+1
Or, a^3=9(3q^3+3q^2+q)+1
Or, a^3=9m+1, where m=3q^3+3q^2+q
If r=2 then
a=3q+2
Or, a^3=(3q+2)^3
Or, a^3=27q^3+54q^2+36q+8
Or, a^3=9(3q^3+6q^2+4q)+8)
Or, a^3=9m, where m=3q^3+6q^2+4q
If r=3 then
a=3q+3
Or, a^3=(3q+3)^3
Or, a^3=27q^3+81q^2+81q+27
Or, a^3=9(3q^3+9q^2+9q+3)
Or, a^3=9m, where m=3q^3+9q^2+9q+3
Thus, cube of any positive integer is in the form of 9m, 9m +1 or 9m +8
Alternate method:
8=9xx0, i.e. 9m+8
Let us take the next cube number, i.e. 27
27=9xx3, i.e. 9m
Let us take the next cube number, i.e. 64
64=9xx7+1, i.e. 9m+1
Thus, cube of any positive integer is in the form of 9m, 9m +1 or 9m +8` |
New Zealand
Level 8 - NCEA Level 3
# Signed area function
Lesson
As we've seen, we need to be mindful whether the area we're calculating is above or below the x-axis as this influences how we set up or finalise our calculation.
Relating back to what we saw on the Fundamental Theorem of Calculus, we can write our area function as $F(x)$F(x). Calculating the integral of $f(x)$f(x) will give us the expression for $F(x)$F(x).
But if the area we want is bellow the x-axis, then we'll need to write our area function as $-F(x)$F(x).
Let's take a look at some examples.
##### Example 1:
Determine an expression for the area function that can be used to find the area shaded on the graph.
We begin by finding the equation of the function graphed.
Since the area is above the $x$x-axis, our area function will simply be the integral of $f(x)$f(x).
Note here that adding our constant of integration is optional. Since we'll be using this function to calculate area, when we subtract the lower limit from the upper limit that $c$c value will cancel out. So you don't actually need it!
##### Example 2:
Graph the function $f(x)=(x-1)(x+2)$f(x)=(x1)(x+2) and determine an expression for the area function that can be used to calculate the area bounded by the curve and the $x$x-axis between the $x$x-intercepts.
First we graph our function and shade the area we're interested in.
Since the area is below the $x$x-axis we'll need to introduce a negative into our area function.
#### Worked Examples
##### Question 1
Consider the function $f\left(x\right)=8-4x$f(x)=84x.
1. Graph the function.
2. What is the $x$x-coordinate of the $x$x-intercept?
3. Let $F\left(x\right)$F(x) be the area function representing the area bound by $f\left(x\right)$f(x) and the $x$x-axis for $x<2$x<2.
Integrate $f\left(x\right)$f(x) to find an expression for $F\left(x\right)$F(x).
Use $C$C as the constant of integration.
##### Question 2
Consider the function $f\left(x\right)=\left(x+2\right)\left(x-4\right)$f(x)=(x+2)(x4).
1. Graph the function.
2. What are the $x$x-coordinates of the $x$x-intercepts?
Write both values on the same line, separated by a comma.
3. Let $F\left(x\right)$F(x) be the area function representing the area bound by $f\left(x\right)$f(x) and the $x$x-axis for $-22<x<4. Integrate$f\left(x\right)$f(x) to find an expression for$F\left(x\right)$F(x). Use$C\$C as the constant of integration.
### Outcomes
#### M8-11
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
#### 91579
Apply integration methods in solving problems |
# Combination of resistors in series and parallel connection
Suppose you have two carbon resistors of 2 ohms each, but you need 1 ohm resistance to connect in a circuit. What will you do if there are no other sources to collect a resistor of resistance 1 ohm? Again, what will you do if you need 4 ohms to use? Yes, we need to use a combination of the available resistors. There are two types of combinations of resistors – Series combination and Parallel combination. In this article, we are going to explain the combination of resistors in series and parallel circuits. Also, we will derive the equivalent resistance in series and parallel connections.
1. Series Combination of Resistors
2. Equivalent resistance of the combination of resistors in series
3. Parallel Combination of Resistors
4. Equivalent resistance for the combination of resistors in parallel
5. Numerical problems on the combinations of resistors
## Combination of Resistorsin series with circuit diagram
The neat circuit diagram for the series connection of resistors is shown above. In a series combination of resistors, we need to connect resistors one by one in a row. That means each resistor will be connected at the ending point of the previous one. The series combination of resistors has a significant role in electrical circuits. Using a series combination of resistors one can increase the effective resistance to a higher value. The connection of a battery across this combination will provide the current flow through the resistors.
## Formula of equivalent resistance for a series combination of resistors
Let, three resistors of resistances R1, R2 and R3 are connected in series and the battery across the series connection supplies the electric current through the circuit. Let, I be the electric current flowing through the circuit. Since, there is only one loop, then the same amount of current will flow through each resistance. Here, the battery supplies voltage across the resistors. If V1, V2 and V3 be the voltage drops across the resistors R1, R2 and R3, then
V = ( V1 + V2 + V3 )
or, IReq = ( IR1 + IR2 + IR3 ) [ Using Ohm’s Law]
or, Req = ( R1 + R2 + R3 ) ………………….(1)
If we connect N number of resistors in series then the equivalent resistance of the series combination of resistors will be
Req = ( R1 + R2 + R3 + ……. + RN ) ……………..(2)
Equation-(2) gives the formula for equivalent resistance in series combination of resistors.
This equation shows that the effective resistance increases if we connect resistors in series. Series equivalent resistance is greater than every individual resistance in that combination.
## Explain the Parallel combination of resistors with a neat diagram
The neat circuit diagram for parallel connection of resistors is shown above. In a parallel combination of resistors, we need to connect resistors in a column between the same two points. The parallel combination of resistors also has a significant role in electric circuits. Using the parallel combination of larger resistors one can decrease the effective resistance to a lower value than individual resistors. The external battery will supply current through the resistors.
## Derive the formula of equivalent resistance for the parallel combination of resistors
Let, three resistors of resistances R1, R2 and R3 are connected in parallel and the battery across the parallel connection supplies the electric current through the circuit. As the resistors are connected parallelly across the same two points, then the voltage difference across all resistances is the same and is equal to the voltage of the external battery. Let, I be the electric current emitting from the battery. This current will divide into three parts to flow through three resistances. If I1, I2 and I3 be the current through the resistors R1, R2 and R3 then
I = ( I1 + I2 + I3 )
or, $\small \frac{V}{R_{eq}}=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}$
or, $\small \frac{1}{R_{eq}}=(\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}})$ ……………(3)
If we connect N number of resistors in parallel combination then the formula for equivalent resistance in parallel combination is
$\small \frac{1}{R_{eq}}=(\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+.........+\frac{1}{R_{N}})$ ………………(4)
From equation-(4), one can see that the equivalent resistance in parallel combination is smaller than individual resistances of the combination. So, in parallel combination, the effective resistance decreases.
## Facts related to the combination of resistors
1. The series connection of resistors increases the effective resistance and the parallel connection of resistors decreases the equivalent resistance.
2. Equivalent resistance of the series combination is greater than the individual resistances used in the combination. Thus, if we connect more resistances in series in a circuit, the amount of current flow decreases if the external voltage remains the same. Since there is only one path in a series connection, the amount of current flow through each resistor will be the same. But, the voltage drop across the resistors will be different.
3. The equivalent resistance of parallel combination is smaller than the individual resistances used in the combination. Thus, if we connect more resistances in parallel in a circuit, the battery will provide more current if the voltage remains the same. This amount of current divides into the branches of resistors. That means the current will not be the same through each resistor. As the resistors in parallel are connected between two same points, the voltage drop across all the resistors will be equal.
## Numeral problems on combination of resistances
1. Two equal resistors are connected once in series and once in parallel combination. Find the ratio of equivalent resistances in these cases.
2. How can you get 3 ohm resistance by using three 2 ohm resistances?
3. Find the minimum effective resistance from the combination of four 2 ohm resistors.
This is all from this article on the combination of resistors in series and parallel and their derivation. If you have any doubt on this topic you can ask me in the comment section.
Thank you!
Related posts:
1. Ohm’s law of current and voltage
2. Carbon resistor Color code
3. Ohm’s law practical
4. Kirchhoff’s law
5. Drift velocity of electrons
6. Mobility of electrons
7. Impedance of L, C, LC, RLC circuit |
Home » How to Perform a Two Proportion Z-Test in Excel
# How to Perform a Two Proportion Z-Test in Excel
A two proportion z-test is used to test for a difference between two population proportions.
For example, suppose a superintendent of a school district claims that the percentage of students who prefer chocolate milk over regular milk in school cafeterias is the same for school 1 and school 2.
To test this claim, an independent researcher obtains a simple random sample of 100 students from each school and surveys them about their preferences. He finds that 70% of students prefer chocolate milk in school 1 and 68% of students prefer chocolate milk in school 2.
We can use a two proportion z-test to test whether or not the percentage of students who prefer chocolate milk over regular milk is the same for both schools.
## Steps to Perform a Two Sample Z-Test
We can use the following steps to perform the two proportion z-test:
Step 1. State the hypotheses.Â
The null hypothesis (H0):Â P1 = P2
The alternative hypothesis: (Ha): P1 ≠P2
Step 2. Find the test statistic and the corresponding p-value.
First, find the pooled sample proportion p:
p = (p1Â * n1Â + p2Â * n2) / (n1Â + n2)
p = (.70*100 + .68*100) / (100 + 100) = .69
Then use p in the following formula to find the test statistic z:
z = (p1-p2) / √p * (1-p) * [ (1/n1) + (1/n2)]
z = (.70-.68) / √.69 * (1-.69) * [ (1/100) + (1/100)] = .02 / .0654 = .306
Use the Z Score to P Value Calculator with a z score of .306 and a two-tailed test to find that the p-value = 0.759.
Step 3. Reject or fail to reject the null hypothesis.
First, we need to choose a significance level to use for the test. Common choices are 0.01, 0.05, and 0.10. For this example, let’s use 0.05. Since the p-value is not less than our significance level of .05, we fail to reject the null hypothesis.
Thus, we do not have sufficient evidence to say that the percentage of students who prefer chocolate milk is different for school 1 and school 2.
## How to Perform a Two Sample Z-Test in Excel
The following examples illustrate how to perform a two sample z-test in Excel.
### Two Sample Z Test (Two-tailed)
A superintendent of a school district claims that the percentage of students who prefer chocolate milk over regular milk in school cafeterias is the same for school 1 and school 2.
To test this claim, an independent researcher obtains a simple random sample of 100 students from each school and surveys them about their preferences. He finds that 70% of students prefer chocolate milk in school 1 and 68% of students prefer chocolate milk in school 2.
Based on these results, can we reject the superintendent’s claim that the percentage of students who prefer chocolate milk is the same for school 1 and school 2? Use a .05 level of significance.Â
The following screenshot shows how to perform a two-tailed two sample z test in Excel, along with the formulas used:
You need to fill in the values for cells B1:B4. Then, the values for cells B6:B8 are automatically calculated using the formulas shown in cells C6:C8.
Note that the formulas shown do the following:
• Formula in cell C6: This calculates the pooled sample proportion using the formula p = (p1 * n1 + p2 * n2) / (n1 + n2)
• Formula in cell C7: This calculates the test statistic z using the formula z = (p1-p2) / √p * (1-p) * [ (1/n1) + (1/n2)] where p is the pooled sample proportion.
• Formula in cell C8: This calculates the p-value associated with the test statistic calculated in cell B7 using the Excel function NORM.S.DIST, which returns the cumulative probability for the normal distribution with mean = 0 and standard deviation = 1. We multiply this value by two since this is a two-tailed test.
Since the p-value (0.759) is not less than our chosen significance level of 0.05, we fail to reject the null hypothesis. Thus, we do not have sufficient evidence to say that the percentage of students who prefer chocolate milk is different for school 1 and school 2.
### Two Sample Z Test (One-tailed)
A superintendent of a school district claims that the percentage of students who prefer chocolate milk over regular milk in school 1 is less than or equal to the percentage in school 2.
To test this claim, an independent researcher obtains a simple random sample of 100 students from each school and surveys them about their preferences. He finds that 70% of students prefer chocolate milk in school 1 and 68% of students prefer chocolate milk in school 2.
Based on these results, can we reject the superintendent’s claim that the percentage of students who prefer chocolate milk in school 1 is less than or equal to the percentage in school 2? Use a .05 level of significance.Â
The following screenshot shows how to perform a one-tailed two sample z test in Excel, along with the formulas used:
You need to fill in the values for cells B1:B4. Then, the values for cells B6:B8 are automatically calculated using the formulas shown in cells C6:C8.
Note that the formulas shown do the following:
• Formula in cell C6: This calculates the pooled sample proportion using the formula p = (p1 * n1 + p2 * n2) / (n1 + n2)
• Formula in cell C7: This calculates the test statistic z using the formula z = (p1-p2) / √p * (1-p) * [ (1/n1) + (1/n2)] where p is the pooled sample proportion.
• Formula in cell C8: This calculates the p-value associated with the test statistic calculated in cell B7 using the Excel function NORM.S.DIST, which returns the cumulative probability for the normal distribution with mean = 0 and standard deviation = 1.
Since the p-value (0.379) is not less than our chosen significance level of 0.05, we fail to reject the null hypothesis. Thus, we do not have sufficient evidence to say that the percentage of students who prefer chocolate milk in school 2 is greater than that of school 1. |
Time-saving math heuristics
# Learn the Assumption Method
### Assumption vs Guess and Check – Which method is better for my child?
Before we answer that question, let’s first learn about the differences between these two approaches to problem solving.
Guessing and checking works for every kind of child. As long as a child has the time and patience, he/she will be able to guess the correct answer to the problem eventually. The number of guesses they need to make really depends on how strong their logic is and how lucky they are .
Well, since this method only requires a child to make an assumption at the start, they won’t need to spend any unnecessary time guessing. All the time and energy saved can then be spent working their way towards the right solution.
The assumption method is definitely the faster and better choice, especially for children who are studying in upper primary (e.g Primary 4, 5 or 6). However, it may be a bit difficult for some children with a shaky Math foundation.
Fear not though, in the next section, Practicle is going to help you learn the assumption method easily!
### How does the assumption method work?
As mentioned earlier, we start by making an assumption about the problem. Most of the time, we are given two types of items and we want to make sure that we assume all the items to be one of the types.
For example, if we have apples and oranges, we’ll assume that all the fruits are apples or all of them are oranges. Either of these works.
Next, we’ll continue to make some deduction with what we have and compare our assumption with what we are given in the problem. Most of the time, it won’t be the same and we need to be able to come up with a logical explanation to help us find the answer.
Lost? Don’t worry! We’ll guide you through this step-by-step with a real word problem below.
Before we learn how to do the Assumption Method, let’s learn how to identify the kind of questions that we can solve with this method.
### Examples of assumption method questions
Here are an example of a lower primary and upper primary Math problem sum questions that we can solve with the Assumption Method:
Primary 3 and 4 Math:
There are 15 mosquitoes and pest busters somewhere in Singapore. There are 54 legs altogether.
How many mosquitoes are there?
Primary 5 and 6 Math:
How many masks did he make?
### How do we identify questions that use the Assumption Method?
Check if your Math question contains:
1. 2 types of items with a common property
2. the total number of items and the total number of the common property
3. an unknown number of each item
If it does, chances are, you are looking at an assumption question.
Now, time to learn how to do the assumption method!
### How to do the assumption method
Let’s use the Primary 4 Math question as an example.
The two types of items we have are mosquitoes and pest busters. The common property they have is legs.
The total number of items is 15. We have 15 mosquitoes and pest busters. What about the total number of legs? We are given that it is 54.. Since this is a Primary 4 Math problem, the P4 students are expected to be able to deduce that a mosquito has 6 legs while a pest buster has 2.
AS for the unknown number of each item, we are not given the exact number of mosquitoes, nor are we given the exact number of pest busters.
### Steps for using the Assumption Method
#### Step 1:
ASSUME everything to be of the same type
Let’s assume that there are only pest busters.
#### Step 2:
MULTIPLY to find the total value
Next, calculate the total number of legs that we have.
No. of legs of 1 pest buster x No. of pest busters in our assumption
= 2 x 15
= 30
Looks like we have 30 legs.
However, when we look at the question, we are supposed to have 54 legs.
#### Step 3:
Find the DIFFERENCE (The gap between what we have in our assumption and what’s given in the problem)
54 – 30 = 24
Subtract the total number of legs in the problem from the total number of legs in our assumption.
Looks like we are short of 24 legs. What shall we do to get more legs then?
Remember the mosquitoes and pest busters that were mentioned in the question?
Since we have been working with only pest busters, it’s time to bring in the mosquitoes!
#### Step 4:
Find the EFFECT of replacing 1 item with the other
Subtract the number of legs of 1 pest busters from the number of legs of 1 mosquito.
6 – 2 = 4
This means that replacing 1 pest buster with 1 mosquito is going to increase the total number of legs by 4.
#### Step 5:
REPLACE subjects until we close the gap
Find the number of replacements we need to make so that we can close the gap of 24 legs.
24 / 4 = 6.
This tells us that we need to replace 6 pest busters with 6 mosquitoes.
That’s how we know we have 6 mosquitoes.
### And that’s how you do the Assumption Method!
The next time you see Guess and Check questions for Primary 4, try using the Assumption Method instead of the Guess and Check method and see how much faster it takes to get the answer.
### Need more examples?
Check out how we can use the assumption method for this p5 question!
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# Fractions as Decimals
## Dividing the denominator into the numerator
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Practice Fractions as Decimals
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Decimal Notation
Which is greater, 1899\begin{align*}\frac{18}{99}\end{align*} or 1580\begin{align*}\frac{15}{80}\end{align*}? How can you use decimals to help you with this problem?
### Guidance
To change a fraction to a number in decimal form, the numerator must be divided by the denominator. You can use long division or a calculator for this calculation. If, after the division, the numbers after the decimal point end, it is a terminating decimal. If, after the division, the numbers after the decimal point repeat in a pattern forever, it is a periodic decimal (also known as a repeating decimal).
• 34=3÷4=0.75\begin{align*}\frac{3}{4}=3\div 4=0.75\end{align*}. This is a terminating decimal.
• 313=3÷13=0.230769230769...\begin{align*}\frac{3}{13}=3\div 13=0.230769230769...\end{align*}. This is a periodic or repeating decimal. The period is 230769 because this is the set of numbers that repeats.
Keep in mind that a rational number is any number that can be written in the form ab\begin{align*}\frac{a}{b}\end{align*} where b0\begin{align*}b \ne 0\end{align*}. Therefore, periodic decimals and terminating decimals are both rational numbers.
#### Example A
What fraction is equal to 0.45454545... ?
Solution: This is a periodic or repeating decimal. The period has a length of two because the pattern that is repeating consists of 2 digits. To express the number as a fraction, follow these steps:
Step 1: Let x=0.45454545\begin{align*}x=0.45454545\end{align*}
Step 2: The repeating digit is 45. Place the repeating digit to the left of the decimal point by moving the decimal point 2 places to the right.
45.454545
To move the decimal point two places to the right, the decimal number was multiplied by 100. If you go back to step 1, you will see the equation x=0.45454545\begin{align*}x=0.45454545\end{align*}. If you multiply one side of the equation by 100, you must multiply the other side by 100. You now have 100x=45.454545\begin{align*}100x=45.454545\end{align*}
Step 3: Subtract the two equations and solve for x\begin{align*}x\end{align*}.
100x=45.454545x=0.4545454599x99=4599 x=4599=511
#### Example B
What fraction is equal to 0.125?
Solution: This number appears to be a terminating decimal number. The steps to follow to express 0.125 as a fraction are:
Step 1: Express the number as a whole number by moving the decimal point to the right. In this case, the decimal must be moved three places to the right.
Step 2: 0.125=125\begin{align*}0.125=125\end{align*}
Express 125 as a fraction with a denominator of 1 and three zeros. The three zeros represent the number of places that the decimal point was moved.
1251000\begin{align*}\frac{125}{1000}\end{align*}
Step 3: If possible, simplify the fraction. If you are not sure of the simplified form, your graphing calculator can do the calculations.
Therefore, the decimal number of 0.125 is equivalent to the fraction 18\begin{align*}\frac{1}{8}\end{align*}.
The method shown above is one that can be used if you can’t remember the place value associated with the decimal numbers. If you remember the place values, you can simply write the decimal as a fraction and simplify that fraction.
#### Example C
Are the following decimal numbers terminating or periodic? If they are periodic, what is the period and what is its length?
i) 0.318181818...
ii) 0.375
iii) 0.3125
iv) 0.1211221112...
Solution:
i) A periodic decimal with a period of 18. The length of the period is 2.
ii) 0.375 A terminating decimal.
iii) 0.3125 A terminating decimal.
iv) 0.1211221112 This decimal is not a terminating decimal nor is it a periodic decimal. Therefore, the decimal is not a rational number. Decimals that are non-periodic belong to the set of irrational numbers.
#### Concept Problem Revisited
You can convert both fractions to decimals in order to figure out which is greater.
1899=.1818...\begin{align*}\frac{18}{99}=.1818...\end{align*}
1580=.1875\begin{align*}\frac{15}{80}=.1875\end{align*}
You can see that 1580\begin{align*}\frac{15}{80}\end{align*} is greater.
### Vocabulary
Irrational Numbers
An irrational number is the set of non-periodic decimal numbers. Some examples of irrational numbers are 3,2\begin{align*}\sqrt{3},\sqrt{2}\end{align*} and π\begin{align*}\pi\end{align*}.
Periodic Decimal
A periodic decimal is a decimal number that has a pattern of digits that repeat. The decimal number 0.1465325325..., is a periodic decimal.
Rational Numbers
A rational number is any number that be written in the form ab\begin{align*}\frac{a}{b}\end{align*} where b0\begin{align*}b \ne 0\end{align*}. Therefore, periodic decimal numbers and terminating decimal numbers are rational numbers.
Terminating Decimal
A terminating decimal is a decimal number that ends. The process of dividing the fraction ends when the remainder is zero. The decimal number 0.25 is a terminating decimal.
### Guided Practice
1. Express 2.018181818 in the form ab\begin{align*}\frac{a}{b}\end{align*}.
2. Express 1511\begin{align*}\frac{15}{11}\end{align*} in decimal form.
3. If one tablet of micro K contains 0.5 grams of potassium, how much is contained in 234\begin{align*}2\frac{3}{4}\end{align*} tablets?
1. Let x=2.018181818\begin{align*}x=2.018181818\end{align*} The period is 18.
1000x=2018.181818\begin{align*}1000x=2018.181818\end{align*}
10x=20.18181818\begin{align*}10x=20.18181818\end{align*}
1000x=2018.181818These are the two equations that must be subtracted.10x=20.18181818990x990=1998990 Solve for x. x=1998990
Use your calculator to simplify the fraction.
x=1998990\begin{align*}x=\frac{1998}{990}\end{align*}
x=11155\begin{align*}x=\frac{111}{55}\end{align*}
2. \begin{align*}\frac{15}{11}=1.3636...\end{align*}
3. The number of tablets is given as a mixed number. \begin{align*} 2 \frac{3}{4}=2.75\end{align*}. \begin{align*}2.75 \times 0.5=1.375 \ grams\end{align*}.
### Practice
Express the following fractions in decimal form.
1. \begin{align*}\frac{1}{12}\end{align*}
2. \begin{align*}\frac{6}{11}\end{align*}
3. \begin{align*}\frac{3}{20}\end{align*}
4. \begin{align*}\frac{1}{13}\end{align*}
5. \begin{align*}\frac{3}{8}\end{align*}
Express the following numbers in the form \begin{align*}\frac{a}{b}\end{align*}.
1. 0.325
2. 3.72727272...
3. 0.245454545...
4. 0.618
5. 0.36363636...
Complete the following table.
Fraction Decimal
11. \begin{align*}\frac{5}{64}\end{align*}
12. \begin{align*}\frac{11}{32}\end{align*}
13. \begin{align*}\frac{1}{20}\end{align*}
14. \begin{align*}0.0703125\end{align*}
15. \begin{align*}0.1875\end{align*}
### Vocabulary Language: English
Irrational Number
Irrational Number
An irrational number is a number that can not be expressed exactly as the quotient of two integers.
Periodic Decimal
Periodic Decimal
A periodic decimal is a decimal number that has a pattern of digits that repeat. The decimal number 0.146532532532..., is a periodic decimal.
Place Value
Place Value
The value of given a digit in a multi-digit number that is indicated by the place or position of the digit.
rational number
rational number
A rational number is a number that can be expressed as the quotient of two integers, with the denominator not equal to zero. |
# Arithmetic and Geometric progression in 3 numbers
Suppose
$$a,b,c \textrm{ is an arithmetic progression}$$ and
$$a^2,b^2,c^2\textrm{ is a geometric progression}$$ $$a+b+c = \frac{3}{2}.$$
From these equations I get
$$2b=a+c \textrm{, from A.P.}$$
$$b^4=a^2c^2 \textrm{, from G.P.}$$
and finally $$a=b=c=\frac{1}{2}.$$
Can there be any such triplet such that $a<b<c$ ?
$2b=a+c$ and $a+b+c=\frac{3}{2}$ gives $b=\frac{1}{2}$ and $a+c=1$.
Also we have $b^4=a^2c^2$, which gives $$a^2c^2=\frac{1}{16}.$$ Thus, $$a^2(1-a)^2=\frac{1}{16},$$ which gives $a(1-a)=\frac{1}{4}$ and $a=c=\frac{1}{2}$ or $$a(1-a)=-\frac{1}{4},$$ which gives $$(a,c)\in\left\{\left(\frac{1+\sqrt2}{2},\frac{1-\sqrt2}{2}\right),\left(\frac{1-\sqrt2}{2},\frac{1+\sqrt2}{2}\right)\right\}.$$
For $a<b<c$ we have the following unique solution: $$(a,b,c)=\left(\frac{1-\sqrt2}{2},\frac{1}{2},\frac{1+\sqrt2}{2}\right)$$
$2b=a+c,\; a+b+c=\dfrac{3}{2}$
give $3b=\dfrac{3}{2}\to b=\dfrac{1}{2}$
and $a+c=1\to c=1-a$
Substitute in $b^4=a^2c^2$ and get
$a^2(1-a)^2=\dfrac{1}{16}$
$a(1-a)=\pm \dfrac{1}{4}$
$a (1-a)=\dfrac{1}{4}$ gives $a=b=c=\dfrac{1}{2}$ you have already found
$a(1-a)=-\dfrac{1}{4}$ gives
$\color{red}{a=\dfrac{1}{2} \left(1-\sqrt{2}\right),\;b=\dfrac{1}{2};\;c=\dfrac{1}{2} \left(1+\sqrt{2}\right)}$ which is the solution you was looking for $a<b<c$
the other one $a=\dfrac{1}{2} \left(1+\sqrt{2}\right),b=\dfrac{1}{2},c=\dfrac{1}{2} \left(1-\sqrt{2}\right)$ does not satisfy the request
Hope this is useful
$b=qa , c=q^2 a$ so we have $$2q=1+q^2$$ hence $q=1.$ Therefore $a=b=c=0.5$ is a unique solution.
If $a^2,b^2,c^2$are in geometric progression,
Then $$\Big(\frac{a}{b}\Big)^2=\Big(\frac{b}{c}\Big)^2$$ $$\color{red}{1. }$$ $$\frac{a}{b}=\frac{b}{c}$$
Hence $a,b,c$ are also in GP.
Let $a,b,c$ be $f,fg,fg^2$
For $a,b,c$ to be in AP, $$f+fg^2=2fg$$ $$g^2+1=2g$$ $$(g-1)^2=0$$ $$g=1$$ If g=1 then $$a=b=c$$ hence there is no possibility for$$a\neq b\neq c$$
$$\color{red}{2. }$$ $$\frac{a}{b}=-\frac{b}{c}$$ $$b^2=a(-c)$$ Hence a,b,-c are in GP, Let them be $f,fg,fg^2$
So now a,b,c are in AP, $$a+c=2b$$ $$f-fg^2=2fg$$ $$g^2+2g-1=0$$ $$(g+1)^2=2$$ $$g=\sqrt{2} - 1 \ , or\, -(\sqrt{2} + 1)$$ Hence $a,b,-c$ are, $$f,(\sqrt{2}-1)f,(3-2\sqrt{2})f$$ Or $$f,-(\sqrt{2}+1)f,(3+2\sqrt{2})f$$ For $a,b,c$ to be in AP $$f-3f+2\sqrt{2}f=2(\sqrt{2}-1)f$$ $$2(\sqrt{2}-1)f=2(\sqrt{2}-1)f$$ Or $$f-3f-\sqrt{2}f=-2(\sqrt{2}+1)f$$ $$-2(\sqrt{2}f+1)f=-2(\sqrt{2}+1)f$$ Which is true, hence such pairs to exist
In you case, alternate solutions are, $$f+(\sqrt{2}-1)f-3f+2\sqrt{2}f=\frac{3}{2}$$ $$f(3\sqrt{2}-3)=\frac{3}{2}$$ $$f=\frac{\sqrt{2}+1}{2}$$ Hence one of the possible outcomes studying your equations is, $$\frac{\sqrt{2}+1}{2},\frac{1}{2},\frac{1-\sqrt{2}}{2}$$ |
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# Solving Two-Step Equations
Solving Two-Step Equations. Teacher: Dr. Epps. What is a Two-Step Equation?. An equation written in the form Ax + B = C. Examples of Two-Step Equations. 3x – 5 = 16 y/4 + 3 = 12 5n + 4 = 6 n/2 – 6 = 4. Steps for Solving Two-Step Equations.
## Solving Two-Step Equations
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### Presentation Transcript
1. Solving Two-Step Equations Teacher: Dr. Epps
2. What is a Two-Step Equation? An equation written in the form Ax + B = C
3. Examples of Two-Step Equations • 3x – 5 = 16 • y/4 + 3 = 12 • 5n + 4 = 6 • n/2 – 6 = 4
4. Steps for Solving Two-Step Equations • Solve for any Addition or Subtraction on the variable side of equation by “undoing” the operation from both sides of the equation. • Solve any Multiplication or Division from variable side of equation by “undoing” the operation from both sides of the equation.
5. Addition Subtraction Opposite Operations Multiplication Division
6. Helpful Hints? • Identify what operations are on the variable side. (Add, Sub, Mult, Div) • “Undo” the operation by using opposite operations. • Whatever you do to one side, you must do to the other side to keep equation balanced.
7. Ex. 1: Solve 4x – 5 = 11 4x – 5 = 15 +5 +5 (Add 5 to both sides) 4x = 20 (Simplify) 4 4 (Divide both sides by 4) x = 5 (Simplify)
8. Try These Examples • 2x – 5 = 17 • 3y + 7 = 25 • 5n – 2 = 38 • 12b + 4 = 28
9. Check your answers!!! • x = 11 • y = 6 • n = 8 • b = 2
11. Ex. 2: Solve x/3 + 4 = 9 x/3 + 4 = 9 - 4 - 4 (Subt. 4 from both sides) x/3 = 5 (Simplify) (x/3) 3 = 5 3 (Mult. by 3 on both sides) x = 15 (Simplify)
12. Try these examples! • x/5 – 3 = 8 • c/7 + 4 = 9 • r/3 – 6 = 2 • d/9 + 4 = 5
13. Check your answers!!! • x = 55 • c = 35 • r = 24 • d = 9
14. Time to Review! • Make sure your equation is in the form Ax + B = C • Keep the equation balanced. • Use opposite operations to “undo” • Follow the rules: • Undo Addition or Subtaction • Undo Multiplication or Division
15. That’s All! Good Bye
More Related |
# Problem Set: Number Lines
15 teachers like this lesson
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## Objective
SWBAT draw number lines with attention to scale.
#### Big Idea
A number line gives us a way to visualize order of operations, bridging the gap between abstract and quantitative reasoning . As we begin, students pay close attention to scale.
## Opener: Order of Operations
3 minutes
Today's Order of Ops opener consists of two arithmetic expressions that students are asked to evaluate. The first one is from last night's homework, and includes a fraction in front of some parentheses; kids always have questions about this one. The second example is heavy on grouping symbols, and I like to give the class a chance to try to make sense of it. Today's exit slip includes an expression similar to the second example.
As students give these a try, I walk around to glance at the progress students have made on last night's homework. If they have questions about the homework (whose content was the same as this opener), I can spend a few minutes addressing these after going over the opener.
I say that I want students to think about keeping great notes. After giving them a few minutes to grapple, I show them how to simplify the expression one step at a time; this comprises the day's notes. I'm touching on order of operations here, because it's important for students to grasp it, but I'm not dwelling on it for long. If we spend a little time each day, and we begin to use these skills in context, I hope to see that students are confident in what they know and can do.
## Number Line Problems
35 minutes
Today, students begin the first problem set of the year. It is due at the start of class tomorrow, so whatever is not done today in class serves as tonight’s homework. The problem set is called Number Line Problems.
When it comes time to graph functions and to create statistical representations later in this class, an initial roadblock that many students face is their inability to properly construct and scale their axes. Rather than knowing this is coming and then bemoaning it, the CCS gives us a chance to preempt this by including the standards 8.NS.2 and N-Q.1, which I combine into the learning target
SLT 1.2: I can create a number line with attention to scale, and I can locate the exact and approximate locations of all sorts of numbers on a number line.
I find that taking the time to really think about number lines now - both with this problem set, and with a project that begins next week - pays great dividends later in the class.
What is a Problem Set?
To get started, I distribute the handout, and tell students that they’re going to learn about another new class structure now. First, I give the class some notes about problem sets. I post the document on the board, I fill in the blanks, and ask students to do the same on their handouts. Here are the notes: Notes on Problem Sets. With all of that in mind, I explain what it means to complete a problem set.
Everyone gets a sheet of graph paper. I make a really big deal of writing the heading correctly. I tell students that by starting with attention to details like this, they set themselves up to learn more and be better organized later.
Next we read the learning targets. Mathematical Practice #1 will be on every problem set. Practice #2 we’ve seen before, and number lines will help us make some connections between quantitative and abstract reasoning. Then we read SLT 1.2. I ask for a volunteer to read the learning target aloud, then I invite students to shout out the key words in the SLT. I really want to draw their attention to the word “scale”. I show them a sketch of what I call a "screwed up ruler," which has the 1 and 2 inch marks very close together, and the 3 inch mark way beyond the 2. I ask, "What’s wrong with this?" We consider the idea that a well-defined scale gives equal space to equivalent values on a number line.
How Work Time Works
As everyone gets started on the work, I say that I want students to collaborate on this assignment. "If you have a question," I say, "try talking to each other before you talk to me." I emphasize conversation, and I hope that the structure of this problem set allows it to occur naturally. I emphasize hard work. When I see students who are off task or taking their time to get started, I provide little leeway. I counter this by demonstrating endless patience for mathematical questions of any sort.
What I Glean and What Students Learn / Common Conversations
Moving forward, I'll use this problem set to identify what my students know and to plan my next instructional steps with each student. Although a lot this work will happen after I collect the problem sets tomorrow, it begins as soon as kids get to work.
The first problem on the set says to draw a number line that goes from 0 to 10, and it students have any reaction to this, it's just to say, "Oh this is easy!" Then they get to the second problem, and say, "Wait you want me to count up to 500?" This is where I say, "Well, there aren't 500 squares on your graph paper. And if I ask you to count to 500, would you really want to count by 1's?" I start counting, with a little wink. For some students, that's enough to get them going, but if I see they're hesitating, I continue, "What would you like to count by?" Most say 100's, others say 10's or 50's. Whatever the case, I tell them to give it a try. This conversation continues on each problem.
The placement of zero is another big deal, and in every class, there are always students who place zero closer to 1 than 1 is to 2 (or in the case of the second problem, 100 closer to zero than to 200). This is exactly why it's worth our time to complete this set. I move from table to table making sure that they scale is impeccable on the first four number lines that all students draw. I often use the line, "I know this is going to sound really picky, but it's going to be very important later..." when I open a conversation about the location of 0.
As the problem set continues into problems that span both the negative and positive ends of the number line, the question of 0's location can puzzle even more students, and again, that's exactly why we're doing this work.
We also have the opportunity to begin to clear up some misunderstandings about the relationship between the values of numbers in decimal form. Which is greater, for example: 0.5 or 0.25? And once we get that figured out, how do -0.5 and -0.25 look on the number line? Each problem here gives me a chance to figure out what my students know, and for them to shore up some knowledge that I took for granted earlier in my career. Even for my brightest students, this assignment is an introduction to the kind of attention to detail that I'll look for as the year continues.
So with all that in mind, here are snapshots of some student work. What "sort-of picky" advice would you give to these kids?
## Exit Slip: Order of Operations
5 minutes
With five minutes left in class, I distribute today's exit slip. I took a quick glance to see that students are doing homework, but I usually don't collect and grade textbook homework. Instead, I give this quick slip to get a look at what my students know and can with the order of operations.
I say, "Show me perfect steps, because I'm not asking for 20 problems right now. I'm giving you 5 minutes to give me perfect work on these two problems." I'll use this exit slip to continue learning what level of knowledge my new students bring to class.
The document is made to print six to a page, which serves two purposes: it saves paper, and it makes for a small stack that I can stick in my pocket and flip through whenever I have a free moment. |
# Reverse distributive property simplification
• Oct 17th 2012, 04:10 PM
xXplosionZz
Reverse distributive property simplification
Ok, so i have a generally clear idea on what then distributive property is. I understand that if a number is outside of a pair of brackets then you multiply that number by all the entities in the brackets although I have no idea how to solve this question. Expand, using the distributive property. Simplify. (5 - 4 times the square root of 3)(-2 + the square root of 3)
(I am only grade 10, so please try to avoid using any complicated (higher grade) math short cuts without clearly describing them) The answer is -22 + 13 square root of 3.
I would greatly appreciate any answer displaying the steps on how to solve this, and what you did each step.(Happy)
• Oct 17th 2012, 05:42 PM
johnsomeone
Re: Reverse distributive property simplification
$\displaystyle (5 - 4 \sqrt{3})(-2 + \sqrt{3})$
$\displaystyle = (5 - 4 \sqrt{3})(-2) + (5 - 4 \sqrt{3})(\sqrt{3})$
$\displaystyle \text{ (That's using the distributive law: } (x)(-2 + \sqrt{3}) = (x)(-2) + (x)(\sqrt{3}).\text{ )}$
$\displaystyle = [ \ (5)(-2) + (- 4 \sqrt{3})(-2) \ ] + [ \ (5)(\sqrt{3}) + (- 4 \sqrt{3})(\sqrt{3}) \ ]$
$\displaystyle \text{ (That's using the distributive law twice again - once for each bracket [ ]. )}$
$\displaystyle = (-10) + (- 4)(-2)\sqrt{3} + (5 \sqrt{3}) + (- 4)(\sqrt{3})^2$
$\displaystyle = -10 + 8\sqrt{3} + 5\sqrt{3} - (4)(3)$
$\displaystyle = -10 + (8+5)(\sqrt{3}) - 12$
$\displaystyle \text{ (That's using the distributive law again: } 8x + 5x = (8 + 5)x \text{. )}$
$\displaystyle = -22 + (13)(\sqrt{3})$
$\displaystyle = -22 + 13 \sqrt{3}$
• Oct 17th 2012, 05:53 PM
xXplosionZz
Re: Reverse distributive property simplification
$\displaystyle = (5 - 4 \sqrt{3})(-2) + (5 - 4 \sqrt{3})(\sqrt{3})$
$\displaystyle \text{ (That's using the distributive law: } (x)(-2 + \sqrt{3}) = (x)(-2) + (x)(\sqrt{3}).\text{ )}$
I don't really understand this first part? how did you pull a -2 out of the first pair of parenthesis and a sqrt{3} out of the second pair?
• Oct 17th 2012, 06:10 PM
xXplosionZz
Re: Reverse distributive property simplification
I apologize as after a decently long pondering on how you have done that I realized you applied the brackets to each other, Would an easy was of saying this law be (x)(a+b) = (ax + bx) ? Thank you so much for you time and effort. :)
My brain sort of throws basic thought out of the window when so many applications of one law start to intermingle. Do you have any advice on how to to view these problems with greater ease in, say, a test situation?
• Oct 17th 2012, 06:58 PM
pickslides
Re: Reverse distributive property simplification
$\displaystyle (a+b)(c+d)= ac+ad+bc+bd$
• Oct 18th 2012, 06:46 AM
HallsofIvy
Re: Reverse distributive property simplification
Quote:
Originally Posted by xXplosionZz
I apologize as after a decently long pondering on how you have done that I realized you applied the brackets to each other, Would an easy was of saying this law be (x)(a+b) = (ax + bx) ? Thank you so much for you time and effort. :)
Yes, that is precisely the "distributive law" which you said you understood.
Perhaps you learned the distributive law as (a+ b)c= ac+ bc. Because multiplication of numbers is commutative, the left side is the same as c(a+ b) and the right side is the same as ca+ cb. So c(a+ b)= ca+ cb.
Was that what you meant by "reverse" distributive property? I had thought you mean going from ab+ ac to a(b+ c)!
My brain sort of throws basic thought out of the window when so many applications of one law start to intermingle. Do you have any advice on how to to view these problems with greater ease in, say, a test situation?[/QUOTE]
• Oct 18th 2012, 02:49 PM
xXplosionZz
Re: Reverse distributive property simplification
Ahhh yes just today I learned that We can use the Foil law. I apologize as I had convinced myself the only way to somehow perform the second step, was to factor. Thank you guys for clearly describing everything :) |
# How to solve multi step inequalities
This can help the student to understand the problem and How to solve multi step inequalities. We will also look at some example problems and how to approach them.
## How can we solve multi step inequalities
We will explore How to solve multi step inequalities can help students understand and learn algebra. Differential equations are mathematical equations that describe how a function changes over time. In many cases, these equations can't be solved analytically, meaning that we can't find a closed-form solution for the function. In these cases, we must use numerical methods to approximates the solution. One popular numerical method for solving differential equations is the differential solver. This method works by discretizing the differential equation, meaning that we break it up into a finite number of small
There are a few different methods that can be used to solve equations, and the substitution method is one of them. With this method, you essentially solve one equation for one of the variables, and then plug that value into the other equation. This can be a helpful way to approach solving equations, especially if the equations are fairly simple.
I love doing word searches, and I especially love finding words that are hidden in plain sight within mathematical problems. It's like a little game for me, and it's a great way to spend some time when I'm feeling bored or mentally exhausted. I find that the challenge of spotting the words among all the numbers and symbols really helps to wake me up and get my brain working again. Plus, it's just a lot of fun!
To solve an absolute value equation, you need to isolate the absolute value term on one side of the equation. Once you have done this, you can then solve the equation as you would any other equation. Remember that the solutions to an absolute value equation are always the same, regardless of which side of the equation the absolute value term is on.
A math answer generator is a tool that can be used to generate answers to math problems. This can be useful for students who are struggling with a particular problem, or for teachers who want to check their students' work. There are a variety of different math answer generators available online, and they can be customized to suit the needs of the user. |
• July 5, 2022
### How Do You Calculate One Standard Deviation From The Mean?
How do you calculate one standard deviation from the mean? Answer: The value of standard deviation, away from mean is calculated by the formula, X = µ ± Zσ The standard deviation can be considered as the average difference (positive difference) between an observation and the mean.
## What does 1 standard deviation away from the mean mean?
Specifically, if a set of data is normally (randomly, for our purposes) distributed about its mean, then about 2/3 of the data values will lie within 1 standard deviation of the mean value, and about 95/100 of the data values will lie within 2 standard deviations of the mean value.
## What percent is one standard deviation from the mean?
Under this rule, 68% of the data falls within one standard deviation, 95% percent within two standard deviations, and 99.7% within three standard deviations from the mean.
## How do you find the deviation from the mean?
Calculating the mean average helps you determine the deviation from the mean by calculating the difference between the mean and each value. Next, divide the sum of all previously calculated values by the number of deviations added together and the result is the average deviation from the mean.
## What value is 1 standard deviation above the mean?
For instance, if we say that a given score is one standard deviation above the mean, what does that tell us? Perhaps the easiest way to begin thinking about this is in terms of percentiles. Roughly speaking, in a normal distribution, a score that is 1 s.d. above the mean is equivalent to the 84th percentile.
## Related advise for How Do You Calculate One Standard Deviation From The Mean?
### Is 1 a good standard deviation?
For an approximate answer, please estimate your coefficient of variation (CV=standard deviation / mean). As a rule of thumb, a CV >= 1 indicates a relatively high variation, while a CV < 1 can be considered low. A "good" SD depends if you expect your distribution to be centered or spread out around the mean.
### What is 1 standard deviation on a normal curve?
For an approximately normal data set, the values within one standard deviation of the mean account for about 68% of the set; while within two standard deviations account for about 95%; and within three standard deviations account for about 99.7%.
### What percentage of cases fall 1 standard deviation above the mean?
In other words, we know that approximately 34 percent of our data will fall between the mean and one standard deviation above the mean. We can also say that a given observation has a 34 percent chance of falling between the mean and one standard deviation above the mean.
### What percentage of cases fall between 1 standard deviation above the mean and 1 standard deviation below the mean?
In normally distributed data, about 34% of the values lie between the mean and one standard deviation below the mean, and 34% between the mean and one standard deviation above the mean. In addition, 13.5% of the values lie between the first and second standard deviations above the mean.
### How do I find the deviation from the individual?
• First, determine n, which is the number of data values.
• Second, calculate the arithmetic mean, which is the sum of scores divided by n.
• Then, subtract the mean from each individual score to find the individual deviations.
• Then, square the individual deviations.
• ### How do you find the deviation from the mean for each data item?
• Calculate the mean or average of each data set.
• Subtract the deviance of each piece of data by subtracting the mean from each number.
• Square each of the deviations.
• Add up all of the squared deviations.
• ### Is mean deviation and standard deviation the same?
The average deviation, or mean absolute deviation, is calculated similarly to standard deviation, but it uses absolute values instead of squares to circumvent the issue of negative differences between the data points and their means. To calculate the average deviation: Calculate the mean of all data points.
### What is 1 SD and 2SD?
A smaller SD represents data where the results are very close in value to the mean. In fact, 68% of all data points will be within ±1SD from the mean, 95% of all data points will be within + 2SD from the mean, and 99% of all data points will be within ±3SD.
### What is 2 standard deviations from the mean?
Standard deviation tells you how spread out the data is. It is a measure of how far each observed value is from the mean. In any distribution, about 95% of values will be within 2 standard deviations of the mean.
### Why is the mean 0 and the standard deviation 1?
The mean of 0 and standard deviation of 1 usually applies to the standard normal distribution, often called the bell curve. The most likely value is the mean and it falls off as you get farther away. If you have a truly flat distribution then there is no value more likely than another.
### How do you find standard deviation of 1 sigma?
• Calculate the mean of the data set (μ)
• Subtract the mean from each value in the data set.
• Square the differences found in step 2.
• Add up the squared differences found in step 3.
• Divide the total from step 4 by N (for population data).
• ### What is the value of 1 sigma?
One standard deviation, or one sigma, plotted above or below the average value on that normal distribution curve, would define a region that includes 68 percent of all the data points. Two sigmas above or below would include about 95 percent of the data, and three sigmas would include 99.7 percent.
### What is the relationship between standard deviation and mean?
Standard deviation is basically used for the variability of data and frequently use to know the volatility of the stock. A mean is basically the average of a set of two or more numbers. Mean is basically the simple average of data. Standard deviation is used to measure the volatility of a stock.
### What percentage of scores in a normal distribution is between +1 and 1 standard deviation of the mean?
In a normal curve, the percentage of scores which fall between -1 and +1 standard deviations (SD) is 68%.
### What percentage of the data in a normal distribution is between 1 standard deviation below the mean and 2 standard deviations above the mean?
The Empirical Rule. You have already learned that 68% of the data in a normal distribution lies within 1 standard deviation of the mean, 95% of the data lies within 2 standard deviations of the mean, and 99.7% of the data lies within 3 standard deviations of the mean.
### What percentage is 1.5 standard deviations from the mean?
For a normal curve, how much of the area lies within 1.5 standard deviations of the mean? I already know about the 68–95–99.7 rule, and see that it should be between 68% and 95%. I also know that it should be closer to 95%, so I estimate it to be around 80%.
### What percentage of Presidents ages fall within one standard deviation of the mean round to 1 decimal?
In a perfect normal distribution, 68.26% of the data should fall within one standard deviation of the mean. The presidential ages are an approximately normal distribution, so you should have arrived at approximately 68.26% in your last answer. Below is a normal curve, showing the distribution of the data.
### How do you find the percentage of one standard deviation?
It is expressed in percent and is obtained by multiplying the standard deviation by 100 and dividing this product by the average. Example: Here are 4 measurements: 51.3, 55.6, 49.9 and 52.0.
### How do I calculate standard deviation?
To calculate the standard deviation, first, calculate the difference between each data point and the mean. The differences are then squared, summed, and averaged to produce the variance. The standard deviation, then, is the square root of the variance, which brings it back to the original unit of measure.
### How do you find the sum of deviations from the mean?
• Step 1: Calculate the Sample Mean.
• Step 2: Subtract the Mean From the Individual Values.
• Step 3: Square the Individual Variations.
• Step 4: Add the the Squares of the Deviations. |
## Thursday, December 22, 2016
### Advanced Math Solutions – Vector Calculator, Simple Vector Arithmetic
Vectors are used to represent anything that has a direction and magnitude, length. The most popular example of a vector is velocity. Given a car’s velocity of 50 miles per hour going north from the origin, we can draw a vector. In this blog post, we will focus on the simpler aspects of vectors. We won’t talk about how to graph vectors.
Position vectors are vectors that give the position of a point from the origin. The vector is denoted as \vec{v}=<a_1,\:a_2,\:a_3> and starts at point A=(0, 0, 0) and ends at point B=(a_1,\:a_2,\:a_3).
This brings us to how to find a vector given an initial and final point. Given two points A=(a_1,\:a_2,\:a_3) and B=(b_1,\:b_2,\:b_3), the vector \vec{AB}, which goes from point A to B, is \vec{v}=<b_1-a_1,\:b_2-a_2,\:b_3-a_3>.
Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and \vec{b}=<b_1,\:b_2,\:b_3> , \vec{a}+\vec{b}=<a_1+b_1,\:a_2+b_2,\:a_3+b_3>
Subtracting vectors is just as simple
Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and \vec{b}=<b_1,\:b_2,\:b_3> , \vec{a}-\vec{b}=<a_1-b_1,\:a_2-b_2,\:a_3-b_3>
Scalar multiplication is used to lengthen or shorten a vector
Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and any number c, c\vec{a}=<ca_1,\:ca_2\:ca_3>
Every vector has a magnitude and a direction. The direction is where its arrow is pointed and the magnitude is the length of the vector. If the magnitude of a vector is 1, then we call that vector a unit vector.
Magnitude is denoted as |\vec{a}| or ||\vec{a}||.
We will use ||\vec{a}||, so we don't get confused with absolute values.
||\vec{a}||=\sqrt{a_x^2+a_y^2}
A unit vector \hat{u\:}, is a vector with length 1.
\hat{u\:}=\frac{u}{||u||}
Here’s an example of finding the unit vector of a vector (click here):
Here are some properties to memorize about basic vector arithmetic:
The topics we covered in this blog are simple. I recommend practicing a few examples and memorizing the formulas, and you should be good to go. We are going to cover some of the heavier vector topics in next blog.
Until next time,
Leah
## Tuesday, December 13, 2016
### High School Math Solutions – Inequalities Calculator, Exponential Inequalities
Last post, we talked about how to solve logarithmic inequalities. This post, we will learn how to solve exponential inequalities. The method of solving exponential inequalities is very similar to solving logarithmic inequalities. In these problems, the variable is in the exponent and our goal is to isolate the variable, which means getting it out of the exponent. We will see some these problems in the forms:
e^x>a
a^x<b
Let’s see how to solve these problems step by step.
1. Use algebraic manipulation to move everything that is not in the exponential expression to one side
2. Isolate the variable by getting rid of the exponential expression
Example:
e^x>a
e^x>e^{\ln(a)}
x>\ln(a)
Example:
a^x<b
a^x<a^{\log_a(b)}
x<\log_a(b)
3. Solve the inequality
These problems are a little simpler than solving logarithmic inequalities. Let’s see how to solve an example step by step.
3^{x+1}+1<2
Step 1: Use algebraic manipulation to move everything that is not in the exponential expression to one side
3^{2x+1}+1<2
3^{2x+1}<1
Step 2: Isolate the variable by getting rid of the exponential expression
3^{2x+1}<1
3^{2x+1}<3^{\log_3(1)}
2x+1<\log_3(1)
Step 3: Solve the inequality
2x+1<\log_3(1)
2x+1<0
x<\frac{-1}{2}
That was pretty simple. Let’s see some more examples.
Solving exponential inequalities are simpler than solving logarithmic inequalities. However, it can still get a little tricky when solving these inequalities with more parts. For more help and practice on solving exponential inequalities, visit Symbolab’s practice.
Until next time,
Leah
## Tuesday, November 29, 2016
### High School Math Solutions – Inequalities Calculator, Logarithmic Inequalities
Last post, we talked about radical inequalities. In this post, we will talk about how to solve logarithmic inequalities. We’ll see logarithmic inequalities in forms such as \log_b(f(x))<a or \ln(f(x))<a. In order to solve these inequalities, the goal will be to isolate the variable, just as in any inequality, and we will do this by getting rid of the log function. Let’s dive in and see how to solve logarithmic inequalities.
Steps to solve logarithmic inequalities:
1. Use algebraic manipulation to move anything that is not in the logarithmic expression to one side
2. Combine logarithmic expressions
3. Isolate the variable by getting rid of the logarithmic expression
Ex: \log_b(f(x))<a
b^(\log_b(f(x))) <b^a
f(x)<b^a
Ex: \ln(f(x))<a
\ln(f(x))<\ln(e^a )
f(x)<e^a
4. Solve inequality
5. Get the range for the expression in the original log function
Ex: \log_b(f(x))
Range: f(x)>0
6. Combine ranges
Let’s do an example step by step now.
\log_4(x+3)-\log_4(x+2)\ge\frac{3}{2}
Step 1: Use algebraic manipulation to move anything that is not in the logarithmic expression to one side
There’s nothing to move, so we can skip this step.
Step 2: Combine logarithmic expressions
\log_4(x+3)-\log_4(x+2)\ge\frac{3}{2}
\log_4(\frac{x+3}{x+2})≥\frac{3}{2}
Step 3: Isolate the variable by getting rid of the logarithmic expression
\log_4(\frac{x+3}{x+2})≥\frac{3}{2}
4^(\log_4(\frac{x+3}{x+2})) ≥4^(\frac{3}{2})
\frac{x+3}{x+2}≥8
Step 4: Solve inequality
\frac{x+3}{x+2}≥8
We can see that this is now a rational inequality. We won’t solve this step by step; I will show the answer after solving this inequality. If you are struggling with solving this inequality, visit the blog post on rational inequalities.
-2<x≤\frac{-13}{7}
Step 5: Get the range for the expression in the original log function
\log_4(x+3)-\log_4(x+2)\ge\frac{3}{2}
x+3>0 x+2>0
x>-3 x>-2
Step 6: Combine ranges
-2<x≤\frac{-13}{7}, x>-2, x>-3
-2<x≤\frac{-13}{7}
That wasn’t too bad! Let’s see some more examples.
Solving logarithmic inequalities is not too difficult. Just remember to get the ranges inside the logarithmic expressions and to double check your work. For more help and practice on this topic visit Symbolab’s practice.
Until next time,
Leah
## Tuesday, November 22, 2016
### High School Math Solutions – Inequalities Calculator, Radical Inequalities
Last post, we went over how to solve absolute value inequalities. For today’s post, we will talk about how to solve radical inequalities. Solving radical inequalities is easier than solving absolute value inequalities and require fewer steps. Let’s see the steps on how to solve these inequalities.
Steps:
1. Isolate the square root
2. Check that the inequality is true (i.e. not less than 0)
3. Find the real region for the square root, (i.e. see when the expression inside square root igreater than or equal to 0)
4. Simplify and compute the inequality
5. Combine the ranges
Let’s see how to do one example step by step.
\sqrt{5+x}-1<3
Step 1: Isolate the square root
\sqrt{5+x}<4
Step 2: Check that the inequality is true
Yes, this inequality is true the radical is not less than 0.
Step 3: Find the real region for the square root
5+x≥0
x≥-5
Step 4: Simplify and compute the inequality
\sqrt{5+x}<4
(\sqrt{5+x})^2<4^2
5+x<16
x<11
Step 5: Combine the ranges
x≥-5 and x<11
-5≤x<11
That wasn’t too difficult. Let’s see some more examples.
Solving radical inequalities isn’t too difficult, however, they require practice. For more practice examples check out Symbolab’s practice.
Until next time,
Leah
## Thursday, November 3, 2016
### High School Math Solutions – Polynomials Calculator, Dividing Polynomials (Long Division)
Last post, we talked dividing polynomials using factoring and splitting up the fraction. In this post, we will talk about another method for dividing polynomials, long division. Long division with polynomials is similar to the basic numerical long division, except we are dividing variables. This is where it gets tricky. I will talk about the steps to dividing polynomials using long division to help make the process easier and go into detail.
Steps for polynomial long division:
1. Organize each polynomial by higher order
• We want to make sure that each polynomial is written in order of the variable with the highest exponent to the variable with the lowest exponent
• you can skip this step if they are already in high order
2. Set up in long division form
• The denominator becomes the divisor and the numerator becomes the dividend
3. Write 0 as the coefficient for missing terms in the dividend
• Since we’ve put in order the terms based on their exponent, we can see which terms are missing (i.e. x^4+x^2 we can see we are missing an x^3 term so we will add that in to its proper spot and make the coefficient 0)
• This will help you with step 6, so you don’t subtract the wrong terms
• You can skip this step if there are no missing terms
4. Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
• This is allows us to see what we need to multiply the divisor by to get rid of the first term of the dividend
5. Multiply the divisor by that term
• Write the term down on top of line where the term that is getting eliminated is
6. Subtract this from the dividend
• This gives you a new polynomial to work with
7. Repeat steps 4-6 until you get a remainder
• When you repeat step 4, move onto the newest first term from step 6
8. Put the remainder over the divisor to create a fraction and add it to the new polynomial
This may seem a bit confusing, so we will go through two examples step by step to understand better how to solve these problems.
\frac{(x^4+6x^2+2)}{(x^2+5)}
1. Organize each polynomial by high order
We can skip this step because the polynomials are already in high order
2. Set up in long division form
3. Write 0 as the coefficient for missing terms in the dividend
4. Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
\frac{x^4}{x^2}= x^2
5. Multiply the divisor by that term
x^2∙(x^2+5)=x^4+5x^2
6. Subtract this from the dividend
7. Repeat step 4-6 until you get a remainder
\frac{x^2}{x^2} =1
8. Put the remainder over the divisor to create a fraction and add it to the new polynomial
x^2+1+\frac{(-3)}{(x^2+5)}
\frac{(2x^2-18+5x)}{(x+4)}
1. Organize each polynomial by high order
\frac{(2x^2+5x-18)}{(x+4)}
2. Set up in long division form
3. Write 0 as the coefficient for missing terms in the dividend
We can skip this step because there are no missing terms.
4. Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
\frac{2x^2}{x}=2x
5. Multiply the divisor by that term
2x(x+4)=2x^2+8x
6. Subtract this from the dividend
7. Repeat steps 4-6 until you get the remainder
\frac{(-3x)}{x}=-3
8. Put the remainder over the divisor to create a fraction and add it to the new polynomial
2x-3+\frac{(-6)}{(x+4)}
Dividing polynomials using long division is very tricky. It is so easy to skip an exponent, have an algebraic error, and forget a step. This is why practicing this type of problem is so important. The only way to get better at it is to keep practicing it. Check out Symbolab’s Practice for practice problems and quizzes.
Until next time,
Leah.
## Wednesday, October 26, 2016
### High School Math Solutions – Polynomials Calculator, Dividing Polynomials
In the last post, we talked about how to multiply polynomials. In this post, we will talk about to divide polynomials. When we divide polynomials, we can write these problems in the form of a fraction. This allows us to reduce the fraction and get our answer. Although this method doesn’t work every time we divide polynomials, when it does, it is a quick, simple method.
Once written in the form of a fraction, there are two ways to reduce the fraction. One method is to split the fraction up, so there is one term per fraction, having the denominator be the same. The other method is to factor the numerator and denominator and cancel out terms that are the same in the numerator and denominator.
Let’s see some examples to better understand how to solve these problems.
We will use the splitting method first for our first example.
(32x^4-56x^2)\div 8x
1. Rewrite the problem in the form of a fraction
\frac{32x^4-56x^2}{8x}
2. Split the fraction
\frac{32x^4}{8x}-\frac{56x^2}{8x}
3. Reduce the fractions
\frac{8\cdot 4x^4}{8x}-\frac{8\cdot7x^2}{8x}
4x^3-7x
Next example:
We will use our same problem above, but use factoring to solve the problem.
(32x^4-56x^2)\div 8x
1. Rewrite the problem in the form of a fraction
\frac{32x^4-56x^2}{8x}
2. Factor the numerator and denominator
\frac{8x^2(4x^2-7)}{8x}
3. Cancel out common factors
4. Simplify
x(4x^2-7)
4x^3-7x
(2x^3+11x^2+5x)\div(2x^2+x)
1. Rewrite the problem in the form of a fraction
\frac{2x^3+11x^2+5x}{2x^2+x}
2. Factor the numerator and denominator
\frac{(2x+1)(x+5)x}{(2x+1)x}
3. Cancel out common factors
4. Simplify
x+5
As you can see, factoring is a big part of dividing polynomials. Before attempting these problems, make sure you’ve mastered factoring. Dividing polynomials by these two methods is pretty simple. However, these methods may not work for certain division problems. Next blog post, we will talk about another method to use, when these methods don’t work. For more practice and help on this topic, checkout Symbolab’s Practice.
Until next time,
Leah
### Middle School Math Solutions – Polynomials Calculator, Factoring Quadratics
Just like numbers have factors (2×3=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Factoring is the process of finding factors of a number or expression, where we find what multiplies together to make the expression or number. In today’s blog post, we will talk about how to factor simple expressions and quadratics.
Factoring simple expressions –
Given a simple expression, ax+b, pull out the greatest common factor from the expression. Pretty simple!
Factor\:2x+6
2x+2\cdot3
2(x+3)
Quadratics have the form: (ax)^2+bx+c, where a, b, and c are numbers.
Here are the steps for factoring quadratics:
1. Find u and v such that u∙v=a∙c and u+v=b
This means u and v are factors of a∙c that when added together equal b
2. Rewrite the expression as (ax^2+ux)+(vx+c)
3. Factor out what you can from each parentheses
4. Factor out a common term
5. Check by multiplying the factors together (FOIL)
Factor\:x^2-5x+6
1. Find u and v such that u∙v=a∙c and u+v=b
a∙c=1∙6=6
6 can be written as the product of 1 and 6, -1 and -6, 3 and 2, or of -3 and -2. We need to pick the factors of 6 that equal -5.
-3+(-2)=-5
-3∙-2=6
u=-3,\:v=-2
2. Rewrite the expression as (ax^2+ux)+(vx+c)
(x^2+(-3x))+(-2x+6)
(x^2-3x)+(-2x+6)
3. Factor out what you can from each parentheses
(x^2-3x)+(-2x+6)
x(x-3)+2(-x+3)
x(x-3)-2(x-3)
We factored out a -1 from (-x+3) on the last step because we want the expressions inside the parentheses to be the same for step 4.
4. Factor out a common term
x(x-3)-2(x-3)
(x-3)(x-2)
5 .Check by multiplying the factors together
(x-3)(x-2)=x^2-2x-3x+6=x^2-5x+6
Factor\:2x^2+x-6
1. Find u and v such that u∙v=a∙c and u+v=b
a∙c=2∙-6=-12
Factors of -12: -12 and 1, 12 and -1, -6 and 2, 6 and -2, -4 and 3, 4 and -3
4-3=1
4∙-3=-12
u=-4,\:v=3
2. Rewrite the expression as (ax^2+ux)+(vx+c)
(2x^2-4x)+(3x-6)
3. Factor out what you can from each parentheses
(2x^2-4x)+(3x-6)
2x(x-2)+3(x-2)
4. Factor out a common term
(x-2)(2x+3)
5. Check by multiplying the factors together
(x-2)(2x+3)=2x^2+3x-4x-6=(2x)^2-x-6
The best advice for factoring quadratics is to practice factoring as many quadratics as you can. The more you practice factoring, the faster you will get. Soon, you’ll be able to skip the steps and factor it all in your head within seconds. For more help or practice on the topic, check out Symbolab’s Practice.
Until next time,
Leah
### Symbolab Study Groups…groups that work
stud•y group
noun
plural noun: study groups
a group of people who meet to study a particular subject and then report their findings or recommendations.
Study groups are great; more brainpower, boost motivation, support system. But let’s be honest, it’s not always the most effective way to learn. Sessions can turn into social events, schedule doesn’t work for everyone, we’re not always prepared for the sessions…
Symbolab Groups is a game changer. There’s no better way to connect, share notes, and work through difficult problems together. Symbolab Groups is a stress free study environment. You get all the benefits of a study group minus the distractions.
Symbolab helps you stay focused. No need to worry about missing out on group meetings, or not taking notes. Stuck on a problem? Just ask, your friends can help you instantly. Share problems, exercises and graphs, start a discussion, answer questions.
Symbolab Groups are there for you, always.
Cheers,
Michal
## Wednesday, September 28, 2016
### Middle School Math Solutions – Polynomials Calculator, Multiplying Polynomials
Multiplying polynomials can be tricky because you have to pay attention to every term, not to mention it can be very messy. There are a few ways of multiplying polynomials, depending on how many terms are in each polynomial. In this post, we will focus on how to multiply two term polynomials and how to multiply two or more term polynomials.
Multiply two term polynomials
When multiplying polynomials with two terms, you use the FOIL method. The FOIL method only works for multiplying two term polynomials. FOIL stands for first, outer, inner, last. This lets you know the order of how to distribute and multiply the terms. Let’s see how it works.
After FOILing, multiply the terms, group like terms, and add like terms if there are any.
Here is another helpful identity to use when multiplying two term polynomials:
(a+b)(a-b)=a^2-b^2
Multiplying these polynomials is pretty simple because if you memorize these identities then you just plug in the values and have an answer.
Multiplying multiple term polynomials
You cannot use the FOIL method to multiply these polynomials. Instead, you have to multiply each term in one polynomial by each term in the other. You can do this by multiplying each term of one polynomial by the other polynomial. This can be tricky because it is easy to miss one term. When we do examples of this, it will become easier to understand how to solve them.
When multiplying polynomials, you may come across multiplying variables with exponents by variables with exponents. In this case, we use this exponent rule:
x^n\cdot x^m=x^(n+m)
For this rule, the base or variable must be the same. When multiplying variables with exponents, you add the exponents together.
Let’s see some examples to understand how to multiply polynomials.
(2x-1)(5x-6)
We will use the FOIL method to solve this.
1. Use FOIL identity
(2x-1)(5x-6)
2x\cdot 5x+2x\cdot -6+(-1)\cdot 5x+(-1)\cdot -6
2. Multiply terms
10x^2-12x-5x+6
3. Group like terms
10x^2-12x-5x+6
(Luckily, everything was already grouped together)
10x^2-17x+6
(2x^2+6)(2x^2-6)
Here, we can use another one of the identities for multiplying two term polynomials.
1. Use (a+b)(a-b)=a^2-b^2
(2x^2+6)(2x^2-6)
(2x^2 )^2-6^2
2. Simplify
4x^4-6^2
4x^4-36
(x^2+2x-1)(2x^2-3x+6)
1. Multiply each term in one polynomial by the other polynomial
x^2 (2x^2-3x+6)+2x(2x^2-3x+6)-1(2x^2-3x+6)
2. Distribute and multiply
2x^2\cdot x^2-3x\cdot x^2+6\cdot x^2+2x^2\cdot 2x-3x\cdot 2x+6\cdot 2x+2x^2\cdot -1-3x\cdot -1+6\cdot -1
2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6
3. Group like terms
2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6
2x^4-3x^3+4x^3+6x^2-6x^2-2x^2+12x+3x-6
2x^4+x^3+6x^2-6x^2-2x^2+12x+3x-6
2x^4+x^3-2x^2+12x+3x-6
2x^4+x^3-2x^2+15x-6
Multiplying polynomials looks intimidating, but as long as you keep your work neat and double check your work, it should be pretty easy. Practice will be one of the biggest things that will help you. The more you practice, the easier multiplying polynomials will be because you will get the hang and flow of how to multiply them. Check out Symbolab’s Practice for more help and practice.
Until next time,
Leah
## Tuesday, September 20, 2016
### Middle School Math Solutions – Polynomials Calculator, Subtracting Polynomials
In the previous post, we talked about how to add polynomials. In this post, we will talk about subtracting polynomials. The key to subtracting polynomials is to make sure that you distribute the minus sign to the expression in the parentheses. Imagine that instead of – there is a -1. This should help you visualize why you are distributing the minus sign. Once the minus sign is distributed, the minus sign will turn into a plus sign. Then, you’ll be able to add the polynomials together. Let’s see the steps for subtracting polynomials.
Steps:
1. Distribute the negative sign
• Include this step only if there is a minus sign in front of a polynomial in parentheses
2. Remove the parentheses
• Include this step if there are polynomials in parentheses
3. Group like terms
• Put together and order like terms (terms with the same variables and the same exponent)
• Add the coefficients of the like terms
Let’s see some examples to better understand distributing the minus sign.
(x^2+2x-1)-(2x^2-3x+6)
1. Distribute the negative sign
(x^2+2x-1)-(2x^2-3x+6)
(x^2+2x-1)-1(2x^2-3x+6)
(x^2+2x-1)+(-2x^2+3x-6)
When we imagine that there is a -1 instead of – outside the parentheses, it is easy to see and remind ourselves that we have to distribute the negative. Distributing the negative sign allows that minus sign to turn into a plus sign.
2. Remove the parentheses
x^2+2x-1+-2x^2+3x-6
x^2+2x-1-2x^2+3x-6
3. Group like terms
x^2+2x-1-2x^2+3x-6
x^2-2x^2+2x+3x-1-6
x^2-2x^2+2x+3x-1-6
-x^2+5x-7
(2x^3+2x-1)-(2x^2-5x-6)
1. Distribute the negative sign
(2x^3+2x-1)-(2x^2-5x-6)
(2x^3+2x-1)-1(2x^2-5x-6)
(2x^3+2x-1)+(-2x^2+5x+6)
2. Remove the parentheses
2x^3+2x-1+-2x^2+5x+6
2x^3+2x-1-2x^2+5x+6
3. Group like terms
2x^3+2x-1-2x^2+5x+6
2x^3-2x^2+2x+5x-1+6
2x^3-2x^2+7x-1+6
2x^3-2x^2+7x+5
(4x^3-x^2+x-2)-(-x^2+3)
1. Distribute the negative sign
(4x^3-x^2+x-2)-(-x^2+3)
(4x^3-x^2+x-2)-1(-x^2+3)
(4x^3-x^2+x-2)+(x^2-3)
2. Remove the parentheses
4x^3-x^2+x-2+x^2-3
3. Group like terms
4x^3-x^2+x-2+x^2-3
4x^3-x^2+x^2+x-2-3
4x^3+x-2-3
4x^3+x-5
Subtracting polynomials takes some practice to get the hang of distributing and remembering to distribute. Once you’ve mastered adding polynomials, subtracting them should be simple. For more practice, check out Symbolab’s Practice.
Until next time,
Leah
## Monday, August 15, 2016
### Middle School Math Solutions – Polynomials Calculator, Adding Polynomials
A polynomial is an expression of two or more algebraic terms, often having different exponents. Adding polynomials is pretty simple. The only tricky part is that there are different terms. The key is to add the like terms. Like terms are terms that have the same variable with the same exponent, even if they have different coefficients. There are three steps to adding polynomials:
1. Remove parentheses - only include this step if the polynomials are in parentheses
2. Group like terms - put together and order like terms (terms with the same variable and same exponent)
3. Add like terms - add the coefficients of the like terms
Let’s check out some examples step by step.
(x^2+2x-1)+(2x^2-3x+6)
1. Remove parentheses
x^2+2x-1+2x^2-3x+6
2. Group like terms
x^2+2x^2+2x-3x-1+6
When I group like terms, I like to group and put the terms in order from the largest exponent to the smallest exponent.
3x^2+2x-3x-1+6
3x^2-x-1+6
3x^2-x+5
(2x^3+4x+7)+(2x^2+5x-6)
1. Remove parentheses
2x^3+4x+7+2x^2+5x-6
2. Group like terms
2x^3+2x^2+4x+5x+7-6
2x^3+2x^2+9x+1
4x+(4x-2)+(x^2-3)
1. Remove parentheses
4x+4x-2+x^2-3
2. Group like terms
x^2+4x+4x-2-3
x^2+8x-2-3
x^2+8x-5
Adding polynomials is pretty simple and requires just a few steps. The trickiest part about adding polynomials is making sure that you pay attention the exponents. It is easy to overlook and confuse a 2 for a 3. For more practice on this topic, check out Symbolab’s Practice.
Until next time,
Leah
## Thursday, June 9, 2016
### High School Math Solutions – Inequalities Calculator, Absolute Value Inequalities Part II
Last post we talked about absolute value inequalities with one absolute value expression. In this post, we will learn how to solve absolute value inequalities with two absolute value expressions. To do this, we will use some concepts from a previous post on rational functions. This can get a little tricky and confusing, so take time to read everything carefully.
I’ll show you the steps to solving these inequalities and then I will go through one example step by step.
Steps:
1. For each absolute value expression, figure out their negative and positive ranges.
2. Combine ranges if needed
3. For each range, evaluate the absolute value inequality
4. Combine the ranges
\frac{|3x+2|}{|x-1|} >2
Step 1: For each absolute value expression, figure out their negative and positive ranges
3x+2 x-1
Positive (≥0) 3x+2≥0
Range:x\ge-\frac{2}{3}
Positive absolute value expression:|3x+2|=3x+2
x-1≥0
Range:x≥1
Positive absolute value expression:|x-1|=x-1
Negative (<0) 3x+2<0
Range:x\le-\frac{2}{3}
Positive absolute value expression:|3x+2|=-(3x+2)
x-1<0
Range:x<1
Positive absolute value expression:|x-1|=-(x-1)
Step 2: Combine ranges if needed
Step 3: For each range, evaluate the absolute value inequality
For x<-\frac{2}{3}:
\frac{-(3x+2)}{-(x-1)}>2 We use the negative absolute value expressions because for both expressions the range is less than their negative range. \frac{3x+2}{x-1}>2 \frac{x+4}{x-1}>0 Now, we see that this is just a rational inequality. So we solve the rational inequality x<-4\:or\:x>1 The steps aren’t shown to get the answer. You can refer back to the rational inequalities blog if you forget how to solve it
For -\frac{2}{3}≤x<1:
\frac{3x+2}{-(x-1)}>2 We use the negative absolute value expression for the |x-1| because the range is less than its range. \frac{5x}{1-x}>0 0
For x≥1:
\frac{3x+2}{x-1}>2 \frac{x+4}{x-1}>2 x<-4\:or\:x>1
Step 4: Combine the ranges
(x<-\frac{2}{3}\:and\:x<-4\:or\:x>1)\:or\:(-\frac{2}{3}≤x<1\:and\:0<x<1)\:or\:(x≥1\:and\:x<-4\:or\:x>1)
Answer: x<-4 or x>1 or 0<x<1
Hopefully, that wasn’t too complicated. Now let’s see one more example.
As you can see, this topic can be very confusing and tricky. The more you practice, the more natural it will become. For more practice examples check out Symbolab’s practice.
Until next time,
Leah
### High School Math Solutions – Inequalities Calculator, Absolute Value Inequalities Part I
Last post, we learned how to solve rational inequalities. In this post, we will learn how to solve absolute value inequalities. There are two ways to solve these types of inequalities depending on what the inequality sign is. Let’s see how to solve them . . .
Case I (< or ≤):
Given |x|<a, we read this as “x is less than a units for 0.” On a number that looks like:
The red line is the solution for x. Remember because it is “less than,” a is not included in the points (this is why the circle is not filled in).
So we rewrite
The solution is in this form: -a<x<a.
The same goes for ≤. Just substitute ≤ for < and fill in the circles in the number line.
Case II(> or ≥):
Given |x|>a, we read this as “ x is more than a units from 0.” On a number line, that looks like:
The red line is the solution for x. Remember because it is “greater than.” A is not included in the points.
So we rewrite
The solution is 2 inequalities, NOT one. It is in the form: x<-a or x>a.
The same goes for ≥. Just substitute ≥ for > and fill in the circles in the number line.
NOTE: Remember to carry out any algebraic manipulations outside of the absolute value before continuing, for example:
2|x|-2>4
|x|>3
ANOTHER NOTE: Make sure a is positive or else there is no solution.
Let’s see some examples . . .
Notice that the algebraic manipulations to be done are inside the absolute value, so that is done last.
That wasn’t too complicated. Next post, we will learn how to solve absolute value inequalities with two absolute value expressions in it. Make sure you practice because next post the problems will become trickier.
Until next time,
Leah
### High School Math Solutions – Inequalities Calculator, Rational Inequalities
Last post, we talked about solving quadratic inequalities. In this post, we will talk about rational inequalities. Let’s recall rational functions are an algebraic fraction that contain polynomials in the numerator and denominator. Solving rational inequalities is a little different than solving quadratic inequalities. Both share the concept of a table and testing values.
Let’s see how to solve rational inequalities.
Here the steps:
1. Move everything to one side of the inequality sign
2. Simplify the rational function
3. Find the zeros from the numerator and undefined points in the denominator
4. Derive intervals
5. Find the sign of the rational function on each interval
6. Select the proper inequality
Let’s go through our first example step by step to understand the concept better.
\frac{x}{x-3}<4
Step 1: Move everything to one side of the inequality sign
\frac{x}{x-3}<4
\frac{x}{x-3}-4<0
\frac{-3x+12}{x-3}<0
Make sure you combine everything into one rational function.
Step 2: Simplify the rational function
\frac{-3x+12}{x-3}<0
\frac{3(-x+4)}{x-3}<0
It’s already simplified. Nothing to cancel out.
Step 3: Find the zeros from the numerator and undefined points in the denominator
-3x+12=0 x-3=0
x=4 x=3
Zero Undefined point
Step 4: Derive intervals
Step 5: Find the sign of the rational function on each interval
\frac{-3x+12}{x-3} \frac{-3(0)+12}{(0)-3}=-4 \frac{-3(3.5)+12}{(3.5)-3}=3 \frac{-3(5)+12}{(5)-3}=-\frac{3}{2}
We changed the format of the intervals to inequalities. We pick a number in the interval, plug the number in the rational function and see what sign the answer is (negative or positive).
Step 6: Select the proper inequality
\frac{-3x+12}{x-3}<0
We refer back to the original inequality and see which inequality satisfies the original inequality. We are looking for a number that produces a negative number. We refer back to the table and see that x<3 and x>4 satisfy this.
Alright, that was a mouthful. Let’s see some more examples now…
Make sure you double check your work for calculation errors because that is where it’s very easy to make a mistake. For more practice, check out Symbolab’s practice.
Until next time,
Leah
### High School Math Solutions – Inequalities Calculator, Quadratic Inequalities
We’ve learned how to solve linear inequalities. Now, it’s time to learn how to solve quadratic inequalities. Solving quadratic inequalities is a little harder than solving linear inequalities. Let’s see how to solve them.
There are a couple ways to solve quadratic inequalities depending on the inequality. I’ll focus on explaining the more complicated version.
We’re given the quadratic inequality: x^2+2x-8\le0
Here are the steps to solving it:
1. Move everything to one side of the inequality sign
2. Set the inequality sign to an equal sign and solve for x
3. Create three intervals
4. Pick a number in each inequality and see if it satisfies the original inequality
5. Select the proper inequality
Now we will go through this example step by step to understand a little better how to solve it.
x^2+2x-8\le0
Looks good!
Step 2: Set the inequality sign to an equal sign and solve for x
x^2+2x-8=0
(x-2)(x+4)=0
x=2\:and\:x=-4
Step 3: Create three intervals
We are able to pick three intervals from looking at the number and seeing where the function crosses the x-axis (i.e. where the function is equal to 0).
Step 4: Pick a number in each inequality and see if it satisfies the original inequality
x<-4 -4<x<2 x>2
x^2+2x-8 (-5)^2+2(-5)-8=7 (0)^2+2(0)-8=-8 (3)^2+2(3)-8=7
We’ve turned the intervals into inequalities. Then, we picked a number in each inequality to see if it satisfied the original inequality, x^2+2x-8\le0.
Step 5: Select the proper inequality
-4<x<2
-4\lex\le2
We’ve selected -4<x<2 because it satisfies the original inequality because the quadratic is negative when x is between -4 and 2. However, don’t forget that the original equality contains the ≤ symbol so that means x can equal 0 too. We change the inequality signs because we know -4 and 2 are the zeros of the quadratic.
Let’s see some more examples…
Hopefully, that wasn’t too hard! Solving quadratic inequalities sometimes require patience to write everything out. For more help check out Symbolab’s practice.
Until next time,
Leah
### High School Math Solutions – Inequalities Calculator, Compound Inequalities
In the previous post, we talked about solving linear inequalities. In today’s post we will focus on compound inequalities, which are just a little more complicated than linear inequalities.
A compound inequality is an equation of two or more inequalities joined together by “and” or “or”. Sometimes you see compound inequalities joined by “and” written like -3<2x-1<5. Compound inequalities have three parts: the left, the middle, and then right. Just like compound inequalities, our goal is to isolate the variable. However, we want the variable to be isolated in the middle part.
How do we get that answer? The key is to break the compound inequalities.
Here are the steps for solving compound inequalities:
1. Break the inequality into two parts
2. Solve each linear inequality by isolating the variable
3. Combine the inequalities
Simple enough? Let’s break down an example step by step to understand the concept.
-17<3+10x\le33
1. Break the inequality into two parts
-17<3+10x 3+10x\le33
Based on the definition of a compound inequality, we know that this compound is made of two inequalities. We break the compound inequality down into its two inequalities.
2. Solve each linear inequality by isolating the variable
-17<3+10x 3+10x\le33
-20<10x 10x\le30
-2<x x\le3
Using algebra, we can isolate x and get two simplified inequalities.
3. Combine the inequalities
-2<x x\le3
-2<x\le3
Make sure after combining the inequalities that the simplified compound inequality makes logical sense.
Let’s see two more examples . . . |
Lesson: M9 ========== The Law of Sines ---------------- .Definition. ************* sin asciimath:[alpha=frac{text{secant}}{text{diameter}}]. ************* For this definition to make sense, we need the .Peripheral Angle Theorem. **************** Two peripheral angles subtending the same secant of a circle are equal or supplementary and therefore have the same sine. **************** 'Proof.' Let asciimath:[A] be the center of a circle. First we consider the case where one side is a diameter. Let asciimath:[DB] be a diameter. Then for a point asciimath:[C] on the circle, asciimath:[/_BAC] is an exterior angle of the isoceles △asciimath:[DAC]. [frame="none",cols="^",valign="middle",grid="all"] |=================== |pass:[] |*Click image to view in KSEG.* | |=================== By the Exterior Angle Theorem of Euclid, we have [frame="none",cols="^",valign="middle",grid="none"] |=================== |asciimath:[/_BAC=/_ADC+/_DCA=alpha+alpha=2 alpha]. |=================== So the peripheral angle is half the central angle to the same secant asciimath:[CB]. Recall the Law of Sines from trigonometry: .Law of Sines. **************** [frame="none",cols="^",valign="middle",grid="none"] |=================== |asciimath:[frac{sin\ alpha}{a}=frac{sin\ beta}{b}=frac{sin\ gamma}{c}]. |=================== **************** The Cross Ratio --------------- .Definition. ************* Given four collinear points asciimath:[ABCD] (in any order), the 'cross ratio' asciimath:[CR(ABCD)] is [frame="none",cols=">,^,>",valign="middle",grid="none"] |=================== ||asciimath:[frac{A-B}{A-D}\ frac{C-D}{C-B}\ =\ frac{sin/_BPA}{sin/_DPA}\ frac{sin/_DPC}{sin/_BPC}]|asciimath:[(**)], |=================== where asciimath:[P] is any point not on the line. ************* [frame="none",cols="^",valign="middle",grid="all"] |=================== |pass:[] |*Click image to view in KSEG.* | |=================== 'Proof of asciimath:[(**)].' Apply the Law of Sines to △asciimath:[ABP] then [frame="none",cols="^",valign="middle",grid="none"] |=================== |asciimath:[frac{A-B}{sin/_BPA}=frac{a}{sin/_B}=frac{b}{sin/_A}], |=================== where asciimath:[A-B] is the signed length asciimath:[|A-B|] and asciimath:[sin/_B=sin/_PBA=sin/_CBP] since supplementary angles have the same sine. Now let's calculate [frame="none",cols="^",valign="middle",grid="none"] |=================== |asciimath:[frac{A-B}{A-D}\ frac{C-D}{C-B}\ =\ frac{A-B}{sin/_BPA}\ frac{C-D}{sin/_DPC}\ =\ frac{P-A}{sin/_B}\ frac{P-C}{sin/_D}]. |=================== Similarly, [frame="none",cols="^",valign="middle",grid="none"] |=================== |asciimath:[frac{sin/_BPA}{sin/_DPA}\ frac{sin/_DPC}{sin/_BPC}\ =\ frac{A-D}{sin/_DPA}\ frac{C-B}{sin/_BPC}\ =\ frac{P-A}{sin/_D}\ frac{P-C}{sin/_B}]. |=================== So asciimath:[frac{A-B}{A-D}\ frac{C-D}{C-B}\ =\ frac{sin/_BPA}{sin/_DPA}\ frac{sin/_DPC}{sin/_BPC}\ \ square]. Perspective Rulers ------------------ .Definition. ************* A 'ruler' is a line together with a copy of the real numbers on it. ************* A 'perspective ruler' on a line with a finite vanishing point is obtained with the following procedure: given asciimath:[l] with vanishing point asciimath:[V], - Choose a point asciimath:[P] not on asciimath:[l] on the extension of asciimath:[l] past asciimath:[V]. - Choose a line asciimath:[s\ ||\ (PV)]. - On asciimath:[s] choose a Euclidean ruler. - Transfer this scale through the lens asciimath:[P] to asciimath:[l]. [frame="none",cols="^",valign="middle",grid="all"] |=================== |pass:[] |*Click image to view in KSEG.* | |=================== |
# Methods of Proof — Induction
In this final post on the basic four methods of proof (but perhaps not our last post on proof methods), we consider the proof by induction.
## Proving Statements About All Natural Numbers
Induction comes in many flavors, but the goal never changes. We use induction when we want to prove something is true about all natural numbers. These statements will look something like this:
For all natural numbers n, $1 + 2 + \dots + n = n(n+1)/2$.
Of course, there are many ways to prove this fact, but induction relies on one key idea: if we know the statement is true for some specific number $n$, then that gives us information about whether the statement is true for $n+1$. If this isn’t true about the problem, then proof by induction is hopeless.
Let’s see how we can apply it to the italicized statement above (though we haven’t yet said what induction is, this example will pave the way for a formal description of the technique). The first thing we notice is that indeed, if we know something about the first $n$ numbers, then we can just add $n+1$ to it to get the sum of the first $n+1$ numbers. That is,
$\displaystyle 1 + \dots + n + (n+1) = (1 + \dots + n) + (n+1)$
Reiterating our key point, this is how we know induction is a valid strategy: the statement written for a fixed $n$ translates naturally into the statement about $n+1$. Now suppose we know the theorem is true for $n$. Then we can rewrite the above sum as follows:
$\displaystyle 1 + \dots + n + (n+1) = \frac{n(n+1)}{2} + (n+1)$
With some algebra, we can write the left-hand side as a single fraction:
$\displaystyle 1 + \dots + (n+1) = \frac{n(n+1) + 2(n+1)}{2}$
and factoring the numerator gives
$\displaystyle 1 + \dots + (n+1) = \frac{(n+1)(n+2)}{2}$
Indeed, this is precisely what we’re looking for! It’s what happens when you replace $n$ by $n+1$ in the original statement of the problem.
At this point we’re very close to being finished. We proved that if the statement is true for $n$, then it’s true for $n+1$. And by the same reasoning, it will be true for $n+2, n+3,$ and all numbers after $n$. But this raises the obvious question: what’s the smallest number that it’s true for?
For this problem, it’s easy to see the answer is $n=1$. A mathematician would say: the statement is trivially true for $n=1$ (here trivial means there is no thinking required to show it: you just plug in $n=1$ and verify). And so by our reasoning, the statement is true for $n=2, n=3,$ and so on forever. This is the spirit of mathematical induction.
## Formal Nonsense
Now that we’ve got a taste of how to use induction in practice, let’s formally write down the rules for induction. Let’s have a statement which depends on a number $n$, and call it $p(n)$. This is written as a function because it actually is one (naively). It’s a function from the set of natural numbers to the set of all mathematical statements. In our example above, $p(n)$ was the statement that the equality $1 + \dots + n = n(n+1)/2$ holds.
We can plug in various numbers into this function, such as $p(1)$ being the statement “$1 = 1(1+1)/2$ holds,” or $p(n+1)$ being “$1 + \dots + (n+1) = (n+1)(n+1+1)/2$ holds.”
The basic recipe for induction is then very simple. Say you want to prove that $p(n)$ is true for all $n \geq 1$. First you prove that $p(1)$ is true (this is called the base case), and then you prove the implication $p(n) \to p(n+1)$ for an arbitrary $n$. Each of these pieces can be proved with any method one wishes (direct, contrapositive, contradiction, etc.). Once they are proven, we get that $p(n)$ is true for all $n$ automatically.
Indeed, we can even prove it. A rigorous proof requires a bit of extra knowledge, but we can easily understand the argument: it’s just a proof by contradiction. Here’s a quick sketch. Let $X$ be the set of all natural numbers $n$ for which $p(n)$ is false. Suppose to the contrary that $X$ is not empty. Every nonempty set of natural numbers has a smallest element, so let’s call $m$ the smallest number for which $p(m)$ is false. Now $m-1 < m$, so $p(m-1)$ must be true. But we proved that whenever $p(n)$ is true then so is $p(n+1)$, so $p(m-1 + 1) = p(m)$ is true, a contradiction.
Besides practicing proof by induction, that’s all there is to it. One more caveat is that the base case can be some number other than 1. For instance, it is true that $n! > 2^n$, but only for $n \geq 4$. In this case, we prove $p(4)$ is true, and $p(n) \to p(n+1)$ with the extra assumption that $n \geq 4$. The same inductive result applies.
Here are some exercises the reader can practice with, and afterward we will explore some variants of induction.
1. Prove that $n! > 2^n$ for all $n \geq 4$.
2. Prove that for all $n \geq 1$ the following equality holds: $1/(1 \cdot 2) + 1/(2 \cdot 3) + \dots + 1/(n \cdot (n+1)) = n/(n+1)$.
3. Prove that for every natural number $n$, a set of $n$ elements has $2^n$ subsets (including the empty subset).
This last exercise gives a hint that induction can prove more than arithmetic formulas. Indeed, if we have any way to associate a mathematical object with a number, we can potentially use induction to prove things about those objects. Unfortunately, we don’t have any mathematical objects to work with (except for sets and functions), and so we will stick primarily to proving facts about numbers.
One interesting observation about proof by induction is very relevant to programmers: it’s just recursion. That is, if we want to prove a statement $p(n)$, it suffices to prove it for $p(n-1)$ and do some “extra computation” to arrive at the statement for $p(n)$. And of course, we want to make sure the recursion terminates, so we add in the known result for $p(1)$.
## Variations on Induction, and Connections to Dynamic Programming
The first variation of induction is simultaneous induction on multiple quantities. That is, we can formulate a statement $p(n,m)$ which depends on two natural numbers independently of one another. The base case is a bit trickier, but paralleling the above remark about recursion, double-induction follows the same pattern as a two-dimensional dynamic programming algorithm. The base cases would consist of all $p(1,m)$ and all $p(n,1)$, and the inductive step to get $p(n,m)$ requires $p(n-1,m)$ and $p(n,m-1)$ (and potentially $p(n-1, m-1)$ or others; it depends on the problem).
Unfortunately, natural instances where double induction is useful (or anywhere close to necessary) are rare. Here is an example of a (tricky) but elementary example. Call
$\displaystyle C(m,n) = \frac{(2m)!(2n)!}{m!n!(m+n)!}$,
where the exclamation point denotes the factorial function. We will outline a proof that $C(m,n)$ is always an integer for all $m, n \geq 0$. If we look at the base cases, $C(0,n), C(m,0)$ (recalling that 0! = 1), we get $(2n!)/(n! n!)$, and this happens to be in the form of a binomial coefficient (here, the number of ways to choose $n!$ objects from a collection of $(2n)!$ objects), and binomial coefficients are known to be integers. Now the inductive step relies on the fact that $C(m,n-1)$ and $C(m+1, n-1)$ are both integers. If this is true then
$\displaystyle C(m,n) = 4C(m,n-1) - C(m+1, n-1)$,
which is obviously again an integer.
If we write these values in a table, we can see how the induction progresses in a “dynamic programming” fashion:
In order to fill the values in the next $n$ column (prove the statement for those values of $n$), we need to fill the entire $n-1$ column (for indeed, we rely on the inductive hypothesis for both the $m+1$ and $m$ row). But since our base case was the entire $n=0$ column, we can fill the entire table. In fact, we have just described a dynamic programming algorithm for computing the value of $C(m,n)$ in $mn$ steps. The correctness of the algorithm is indeed an inductive proof of the theorem.
Perhaps uninterestingly, this is asymptotically slower than the naive algorithm of computing $C(m,n)$ directly by computing $(2n)!, (2m)!, (n+m)!$ directly; this would take a linear number of steps in $n$, assuming $n > m$. In passing, this author wonders if, when the numbers are really large, the lack of division and multiplication in the dynamic program (multiplying by 4 using bit shifting instead) would overtake the naive algorithm. But $C(m,n)$ is certainly not interesting enough in its own right for anyone to care 🙂
At this point, we have noticed that we sometimes use induction and assume that many smaller instances of the statement are true. Indeed, why not inductively assume that the statement holds for all smaller $n$. This would certainly give the prover more tools to work with. Indeed, this technique is sometimes called strong induction, in the sense that we assume a stronger inductive hypothesis when we’re trying to prove $p(n+1)$. It may not be entirely obvious (especially to one well versed in the minutiae of formal logic) that this is actually equivalent to normal induction, but it is. In fact, the concept of “strong” induction is entirely pedagogical in nature. Most working mathematicians will not mention the difference in their proofs.
The last variant we’ll mention about induction is that of transfinite induction. The concept is that if you have any set $X$ which is well-ordered (essentially this means: allowing one to prove induction applies as we did earlier in the post), then we can perform induction its elements. In this way, we can “parameterize” statements by elements of an arbitrary well-ordered set, so that instead of $p(n)$ being a function from natural numbers to mathematical statements, it’s a function from $X$ to mathematical statements. One somewhat common example of when $X$ is something besides natural numbers is when we use the so-called cardinal numbers. We have already seen two distinct infinite cardinal numbers in this series: the cardinality of the integers and the cardinality of the real numbers (indeed, “cardinal number” just means a number which is the cardinality of a set). It turns out that there are many more kinds of cardinal numbers, and you can do induction on them, but this rarely shows up outside of mathematical logic.
And of course, we should close this post on an example of when induction goes wrong. For this we point the reader to our proof gallery, and the false proof that all horses are the same color. It’s quite an amusing joke, and hopefully it will stimulate the reader’s mind to recognize the pitfalls that can occur in a proof by induction.
So those are the basic four proof techniques! Fortunately for the reader, pretty much all proofs presented on this blog follow one of these four techniques. I imagine many of my readers skip over the proofs entirely (if only I could put proofs in animated gifs, with claims illustrated by grumpy cats!). So hopefully, if you have been intimidated or confused by the structure of the proofs on this blog, this will aid you in your future mathematical endeavors. Butchering an old phrase for the sake of a bad pun, the eating of the pudding is in the proof. 🙂
Until next time!
## 2 thoughts on “Methods of Proof — Induction”
1. Justin Reynen
Hello! I’m confused by where the 4 comes in this statement. Can you expand a bit on how you got here? I understand the base cases being proven because of the binomial coefficient, but I don’t get how you get to this :
C(m,n) = 4C(m,n-1) – C(m+1, n-1)
Great blog BTW!! I’m reading through all the primers I can and then plan on going through all your other articles once I have a bit of the basics under my belt. The previous 3 articles in your proof series were very easy to follow, great writing!
Liked by 1 person
• This problem is from the 1972 International Mathematical Olympiad, so how “I” found that recurrence relation is 1. I looked it up while I was looking for good examples of double-induction, and 2. Really creative people thought really hard and found the identity when the problem was first posed.
The important part is the double induction. I don’t believe how to find the identity is obvious or easy.
Liked by 1 person |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 5.7: Real-World Applications of Systems of Inequalities
Difficulty Level: At Grade Created by: CK-12
## The Vertex Theorem for Feasible Regions
Introduction
In this lesson you will learn about the vertex theorem for feasible regions and how to apply this theorem to real-world problems. You will learn to write a system of linear inequalities to model the real-world problem. This system will then be graphed to determine the solution set for the system of inequalities. Using the vertex theorem, you will then answer the real-world problem.
Objectives
The lesson objectives for the Vertex Theorem for Feasible Regions are:
• Understanding the vertex theorem.
• Writing a system of inequalities for a real-world problem.
• Solving the system of inequalities by graphing
• Determining the vertices algebraically by solving the linear inequalities.
• Using the vertex theorem to determine the answer to the real-world problem.
Introduction
A system of linear inequalities is often used to determine the best solution to a problem. This solution could be as simple as determining how many of a product should be produced to maximize a profit or as complicated as determining the correct combination of drugs to give a patient. Regardless of the problem, there is a theorem in mathematics that is used, with a system of linear inequalities, to determine the best solution to the problem.
Guidance
The following diagram shows a feasible region that is within a polygonal region.
The linear function \begin{align*}z=2x+3y\end{align*} will now be evaluated for each of the vertices of the polygon.
To evaluate the value of ‘\begin{align*}z\end{align*}’ substitute the coordinates of the point into the expression for ‘\begin{align*}x\end{align*}’ and ‘\begin{align*}y\end{align*}’.
\begin{align*}& (0, 0) && z=2x+3y \rightarrow z=2(0)+3(0) \rightarrow z=0+0 \rightarrow z=0\\ & && \text{Therefore} \ 2x+3y=0\\ \\ & (0, 4) && z=2x+3y \rightarrow z=2(0)+3(4) \rightarrow z=0+12 \rightarrow z=12\\ & && \text{Therefore} \ 2x+3y=12\\ \\ & (6, 0) && z=2x+3y \rightarrow z=2(6)+3(0) \rightarrow z=12+0 \rightarrow z=12\\ & && \text{Therefore} \ 2x+3y=12\\ \\ & (3, 6) && z=2x+3y \rightarrow z=2(3)+3(6) \rightarrow z=6+18 \rightarrow z=24\\ & && \text{Therefore} \ 2x+3y=24\\ \\ & (9, 4) && z=2x+3y \rightarrow z=2(9)+3(4) \rightarrow z=18+12 \rightarrow z=30\\ & && \text{Therefore} \ 2x+3y=30\end{align*}
The value of \begin{align*}z=2x+3y\end{align*}, for each of the vertices, remains constant along any line with a slope of \begin{align*}-\frac{2}{3}\end{align*}. This is obvious on the following graph.
As the line moved away from the origin, the value of \begin{align*}z=2x+3y\end{align*} increased. The maximum value for the shaded region occurred at the vertex (9, 4) while the minimum value occurred at the vertex (0, 0). These statements confirm the vertex theorem for a feasible region:
If a linear expression \begin{align*}\boxed{z=ax+by+c}\end{align*} is to be evaluated for all points of a convex, polygonal region, then the maximum value of \begin{align*}z\end{align*}, if one exists, will occur at one of the vertices of the feasible region. Also, the minimum value of \begin{align*}z\end{align*}, if one exists, will occur at one of the vertices of the feasible region.
Example A
Evaluate the expression \begin{align*}z=3x+4y\end{align*} for the given feasible region, to determine the point at which ‘\begin{align*}z\end{align*}’ has a maximum value and the point at which ‘\begin{align*}z\end{align*}’ has a minimum value.
\begin{align*}&(-4, 0) && z=3x+4y \rightarrow z=3(-4)+4(0) \rightarrow z=-12+0 \rightarrow z=-12\\ & && \text{Therefore} \ 3x+4y=-12\\ \\ & (5, 0) && z=3x+4y \rightarrow z=3(5)+4(0) \rightarrow z=15+0 \rightarrow z=15\\ & && \text{Therefore} \ 3x+4y=15\\ \\ & (6, 3) && z=3x+4y \rightarrow z=3(6)+4(3) \rightarrow z=18+12 \rightarrow z=30\\ & && \text{Therefore} \ 3x+4y=30\\ \\ & (8, -4) && z=3x+4y \rightarrow z=3(8)+4(-4) \rightarrow z=24-16 \rightarrow z=8\\ & && \text{Therefore} \ 3x+4y=8\end{align*}
The maximum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (6, 3).
The minimum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (-4, 0).
Using the vertices of the feasible region to determine the maximum or the minimum value is the branch of mathematics known as linear programming. Linear programming is a technique used by business to solve problems. The types of problems that usually employ linear programming are those where the profit is to be maximized and those where the expenses are to be minimized. However, linear programming can also be used to solve other types of problems. The solution provides the business with a program to follow to obtain the best results for the company. The following examples will demonstrate different types of real-world problems which use linear programming to obtain the solution.
Example B
A company that produces flags makes two flags for Nova Scotia-the traditional blue flag and the green flag for Cape Breton. To produce each flag, two types of material, nylon and cotton, are used. The company has 450 units of nylon in stock and 300 units of cotton. The traditional blue flag requires 6 units of nylon and 3 units of cotton. The Cape Breton flag requires 5 units of nylon and 3 units of cotton.
Each blue flag that is made realizes a profit of $12 for the company, whereas each Cape Breton flag realizes a profit of$15. For the nylon and cotton that the company currently has in stock, how many of each flag should the company make to maximize their profit?
Let ‘\begin{align*}x\end{align*}’ represent the number of blue flags.
Let ‘\begin{align*}y\end{align*}’ represent the number of green flags.
Step 1:
Transfer the information presented in the problem to a table.
Units Required per Blue Flag Units Required Per Green Flag Units Available
Nylon 6 5 450
Cotton 3 5 300
Profit(per flag) $12$15
The information presented in the problem identifies the restrictions or conditions on the production of the flags. These restrictions are known as constraints and are written as inequalities to represent the information presented in the problem.
Step 2:
From the information (now in the table), list the constraints.
The number of blue flags that are produced must be either zero or greater than zero. Therefore, the constraint is \begin{align*}\boxed{x \ge 0}\end{align*}.
The number of green flags that are produced must be either zero or greater than zero. Therefore, the constraint is \begin{align*}\boxed{y \ge 0}\end{align*}.
The total number of units of nylon required to make both types of flags cannot exceed 450. Therefore, the constraint is \begin{align*}\boxed{6x+5y \le 450}\end{align*}.
The total number of units of cotton required to make both types of flags cannot exceed 300. Therefore, the constraint is \begin{align*}\boxed{3x+5y \le 300}\end{align*}.
Step 3:
Write an equation to identify the profit. \begin{align*}\boxed{P=12x+15y}\end{align*}
Step 4:
Graph the listed constraints to identify the feasible region.
The feasible region is the area shaded in teal blue.
Step 5:
Algebraically, determine the exact point of intersection between the constraints. Also, the \begin{align*}x-\end{align*}intercept of the feasible region must be calculated.
Write the constraints as linear equations and solve the system by elimination.
\begin{align*}& 6x+5y=450 \quad \rightarrow && \qquad 6x+5y=450 \quad \rightarrow \qquad 6x+\cancel{5y}=450 && 6x+5y=450\\ & 3x+5y=300 && -1(3x+5y=300) \ \rightarrow \quad \underline{-3x-\cancel{5y}=-300} && 6({\color{red}50})+5y=450\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ 3x=150 \quad \rightarrow && {\color{red}300}+5y=450\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \frac{\cancel{3}x}{\cancel{3}}=\frac{\overset{{\color{red}50}}{\cancel{150}}}{\cancel{3}} && 300{\color{red}-300}+5y=450{\color{red}-300}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x=50 && 5y=150\\ & && && \frac{\cancel{5}y}{\cancel{5}}=\frac{\overset{{\color{red}30}}{\cancel{150}}}{\cancel{5}}\\ & && && y=30\end{align*}
\begin{align*}\boxed{l_1 \cap l_2 @ (50, 30)}\end{align*}
The \begin{align*}x-\end{align*}intercept for the inequality \begin{align*}6x+5y \le 450\end{align*} must be calculated.
Write the inequality as a linear equation. Set ‘\begin{align*}y\end{align*}’ equal to zero and solve the equation for ‘\begin{align*}x\end{align*}’.
\begin{align*}6x+5y&=450\\ 6x+5({\color{red}0})&=450\\ 6x&=450\\ \frac{\cancel{6}x}{\cancel{6}}&=\frac{\overset{{\color{red}75}}{\cancel{450}}}{\cancel{6}}\\ x&=75\end{align*}
The \begin{align*}x-\end{align*}intercept of the feasible region is (75, 0).
The \begin{align*}y-\end{align*}intercept is (0, 60). This point was plotted when the inequalities were put into slope-intercept form for graphing.
The following graph shows the vertices of the polygon than encloses the feasible region.
Step 6:
Calculate the profit, using the profit equation, for each vertex of the feasible region:
\begin{align*}& (0, 0) && P=12x+15y \rightarrow P=12(0)+15(0) \rightarrow P=0+0 \rightarrow P=0\\ & && \text{Therefore} \ 12x+15y=\ 0\\ \\ & (0, 60) && P=12x+15y \rightarrow P=12(0)+15(60) \rightarrow P=0+900 \rightarrow P=900\\ & && \text{Therefore} \ 12x+15y=\ 900\\ \\ & (50, 30) && P=12x+15y \rightarrow P=12(50)+15(30) \rightarrow P=600+450 \rightarrow P=1050\\ & && \text{Therefore} \ 12x+15y=\ 1050\\ \\ & (75, 0) && P=12x+15y \rightarrow P=12(75)+15(0) \rightarrow P=900+0 \rightarrow P=900\\ & && \text{Therefore} \ 12x+15y=\ 900\end{align*}
The maximum profit occurred at the vertex (50, 30). This means, with the supplies in stock, the company should make 50 blue flags and 30 green flags to maximize their profit.
Example C
A local smelting company is able to provide its customers with iron, lead and copper by melting down either of two ores, A or B. The ores arrive at the company in railroad cars. Each railroad car of ore A contains 3 tons of iron, 3 tons of lead and I ton of copper. Each railroad car of ore B contains 1 ton of iron, 4 tons of lead and 3 tons of copper. The smelting receives an order for 7 tons of iron, 19 tons of lead and 8 tons of copper. The cost to purchase and process a carload of ore A is $7000 while the cost for ore B is$6000. If the company wants to fill the order at a minimum cost, how many carloads of each ore must be bought?
Let ‘\begin{align*}x\end{align*}’ represent the number of carloads of ore A to purchase.
Let ‘\begin{align*}y\end{align*}’ represent the number of carloads of ore B to purchase.
Step 1:
Transfer the information presented in the problem to a table.
One Carload of ore A One Carload of ore B Number of tons to fill the order
Tons of Iron 3 1 7
Tons of Lead 3 4 19
Tons of Copper 1 3 8
Step 2:
From the information, list the constraints.
The number of carloads of ore A that must be bought is either zero or greater than zero. Therefore, the constraint is \begin{align*}\boxed{x \ge 0}\end{align*}.
The number of carloads of ore B that must be bought is either zero or greater than zero. Therefore, the constraint is \begin{align*}\boxed{y \ge 0}\end{align*}.
The total number of tons of iron from ore A and ore B must be greater than or equal to the 7 tons needed to fill the order. Therefore, the constraint is \begin{align*}\boxed{3x+y \ge 7}\end{align*}.
The total number of tons of lead from ore A and ore B must be greater than or equal to the 20 tons needed to fill the order. Therefore, the constraint is \begin{align*}\boxed{3x +4y \ge 19}\end{align*}.
The total number of tons of copper from ore A and ore B must be greater than or equal to the 8 tons needed to fill the order. Therefore, the constraint is \begin{align*}\boxed{x+3y \ge 8}\end{align*}.
Step 3:
Write an equation to represent the cost in dollars of \begin{align*}x\end{align*} carloads of ore A and y carloads of ore B. \begin{align*}\boxed{c=7000x+6000y}\end{align*}
Step 4:
Graph the listed constraints to identify the feasible region.
The feasible region shows that there are an infinite number of ways to fill the order. The feasible region is the large shaded that is sitting above the graphed lines.
Step 5:
Algebraically, determine the exact point of intersection between the constraints. Also, the \begin{align*}x-\end{align*}intercept of the feasible region must be calculated.
Write the constraints as linear equations and solve the system by elimination.
\begin{align*}& \ 3x+y=7 \quad \rightarrow && -1(3x+y=7) \quad \rightarrow \quad -\cancel{3x}-y=-7 && 3x+4y=19\\ & 3x+4y=19 && \quad \ \ 3x+4y=19 \quad \rightarrow \quad \ \underline{\cancel{3x}+4y=19} && 3x+4({\color{red}4})=19\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 3y=12 \quad \rightarrow && 3x+{\color{red}16}=19\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \frac{\cancel{3}y}{\cancel{3}}=\frac{\overset{{\color{red}4}}{\cancel{12}}}{\cancel{3}} && 3x+{\color{red}16-16}=19{\color{red}-16}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad y=4 && 3x=3\\ & && && \frac{\cancel{3}x}{\cancel{3}}=\frac{\overset{{\color{red}1}}{\cancel{3}}}{\cancel{3}}\\ & && && x=1\end{align*}
\begin{align*}\boxed{l_1 \cap l_2 @ (1,4)}\end{align*}
\begin{align*}&3x+4y=19 \quad \rightarrow && \quad \ 3x+4y=19 \quad \rightarrow \quad \cancel{3x}+4y=19 && 3x+4y=19\\ & \ x+3y=8 && -3(x+3y=8) \quad \rightarrow \ \ \underline{\cancel{3x}-9y=-24} && 3x+4({\color{red}1})=19\\ & && \qquad \qquad \qquad \qquad \qquad \quad \ -5y=-5 \quad \rightarrow && 3x+{\color{red}4}=19\\ & && \qquad \qquad \qquad \qquad \qquad \quad \ \ \frac{\cancel{-5}y}{\cancel{-5}}=\frac{\overset{{\color{red}1}}{\cancel{-5}}}{\cancel{-5}} && 3x+{\color{red}4-4}=19{\color{red}-4}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \quad y=1 && 3x=15\\ & && && \frac{\cancel{3}x}{\cancel{3}}=\frac{\overset{{\color{red}5}}{\cancel{15}}}{\cancel{3}}\\ & && && x=5\end{align*}
\begin{align*}\boxed{l_2 \cap l_3 @ (5,1)}\end{align*}
The \begin{align*}x-\end{align*}intercept for the inequality \begin{align*}x+3y \ge 8\end{align*} must be calculated.
Write the inequality as a linear equation. Set ‘\begin{align*}y\end{align*}’ equal to zero and solve the equation for ‘\begin{align*}x\end{align*}’.
\begin{align*}x+3y &=8\\ x+3({\color{red}0})&=8\\ x &=8\end{align*}
The \begin{align*}x-\end{align*}intercept of the feasible region is (8, 0).
The \begin{align*}y-\end{align*}intercept is (0, 7). This point was plotted when the inequalities were put into slope-intercept form for graphing.
The following graph shows the vertices of the region borders the feasible region.
Step 6:
Calculate the cost, using the cost equation, for each vertex of the feasible region:
\begin{align*}& (0, 7) && c=7000x+6000y \rightarrow c=7000(0)+6000(7) \rightarrow c=0+42000 \rightarrow P=42000\\ & && \text{Therefore} \ 7000x+6000y=\ 42000\\ \\ & (1, 4) && c=7000x+6000y \rightarrow c=7000(1)+6000(4) \rightarrow c=7000+24000 \rightarrow P=31000\\ & && \text{Therefore} \ 7000x+6000y=\ 31000\\ \\ & (5, 1) && c=7000x+6000y \rightarrow c=7000(5)+6000(1) \rightarrow c=35000+6000 \rightarrow P=41000\\ & && \text{Therefore} \ 7000x+6000y=\ 41000\\ \\ & (8, 0) && c=7000x+6000y \rightarrow c=7000(8)+6000(0) \rightarrow c=56000+0 \rightarrow P=56000\\ & && \text{Therefore} \ 7000x+6000y=\ 56000\end{align*}
The minimum cost is located at the vertex (1, 4). Therefore the company should buy one carload of ore A and four carloads of ore B.
Vocabulary
Constraint
A constraint is a restriction or condition presented in a real-world problem. The constraints are written as inequalities and are used to solve the problem.
Linear Programming
Linear programming is a branch of mathematics that uses systems of linear inequalities to solve real-world problems. The vertex theorem of regions is applied to the vertices to determine the best solution to the problem.
Vertex Theorem for Regions
The vertex theorem for regions states:
If a linear expression \begin{align*}z=ax+by+c\end{align*} is to be evaluated for all points of a convex, polygonal region, then the maximum value of \begin{align*}z\end{align*}, if one exists, will occur at one of the vertices of the feasible region. Also, the minimum value of \begin{align*}z\end{align*}, if one exists, will occur at one of the vertices of the feasible region.
Guided Practice
1. For the following graphed region and the expression \begin{align*}z=5x+7y-1\end{align*}, find a point where ‘\begin{align*}z\end{align*}’ has a maximum value and a point where ‘\begin{align*}z\end{align*}’ has a minimum value.
2. The following table shows the time required on three machines for a company to produce Super 1 and Super 2 coffee percolators. The table also shows the amount of time that each machine is available during a one hour period. The company is trying to determine how many of each must be made to maximize a profit if they make $30 on each Super 1 model and$35 on each Super 2 model. List the constraints and write a profit statement to represent the information.
Super 1 Super 2 Time Available
Machine A 1 minute 3minutes 24 minutes
Machine B 3 minutes 2minutes 36 minutes
Machine C 3 minutes 4 minutes 44 minutes
3. A local paint company has created two new paint colors. The company has 28 units of yellow tint and 22 units of red tint and intends to mix as many quarts as possible of color X and color Y. Each quart of color X requires 4 units of yellow tint and 1 unit of red tint. Each quart of color Y requires 1 unit of yellow tint and 4 units of red tint. How many quarts of each color can be mixed with the units of tint that the company has available? List the constraints, complete the graph and determine the solution using linear programming.
1. The vertices of the polygonal region are (-7, -1); (2, 5); (6, 1); and (0, -4).
\begin{align*}& (-7, -1) && z=5x+7y-1 \rightarrow z=5(-7)+7(-1)-1 \rightarrow z=-35-7-1 \rightarrow z=-43\\ & && \text{Therefore} \ 5x+7y-1=-43\\ \\ & (2, 5) && z=5x+7y-1 \rightarrow z=5(2)+7(5)-1 \rightarrow z=10+35-1 \rightarrow z=44\\ & && \text{Therefore} \ 5x+7y-1=44\\ \\ & (6, 1) && z=5x+7y-1 \rightarrow z=5(6)+7(1)-1 \rightarrow z=30+7-1 \rightarrow z=36\\ & && \text{Therefore} \ 5x+7y-1=36\\ \\ & (0, -4) && z=5x+7y-1 \rightarrow z=5(0)+7(-4)-1 \rightarrow z=0-28-1 \rightarrow z=-29\\ & && \text{Therefore} \ 5x+7y-1=-29\end{align*}
The maximum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (2, 5).
The minimum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (-7, -1).
2. Let ‘\begin{align*}x\end{align*}’ represent the number of Super 1 coffee percolators.
Let ‘\begin{align*}y\end{align*}’ represent the number of Super 2 coffee percolators.
The number of Super 1 coffee percolators that are made must be either zero or greater than zero. Therefore, the constraint is \begin{align*}\boxed{x \ge 0}\end{align*}.
The number of Super 2 coffee percolators that are made must be either zero or greater than zero. Therefore, the constraint is \begin{align*}\boxed{y \ge 0}\end{align*}.
The total amount of time that both a Super1 and a Super 2 model can be processed on Machine A is less than or equal to 24 minutes. Therefore, the constraint is \begin{align*}\boxed{x+3y \le 24}\end{align*}.
The total amount of time that both a Super1 and a Super 2 model can be processed on Machine B is less than or equal to 36 minutes. Therefore, the constraint is \begin{align*}\boxed{3x+2y \le 36}\end{align*}.
The total amount of time that both a Super1 and a Super 2 model can be processed on Machine C is less than or equal to 44 minutes. Therefore, the constraint is \begin{align*}\boxed{3x+4y \le 44}\end{align*}.
The profit equation is \begin{align*}\boxed{P=30x+35y}\end{align*}
3. Table:
Color X Color Y Units Available
Yellow Tint 4 units 1 unit 28
Red Tint 1 unit 4 units 22
Constraints:
Let ‘\begin{align*}x\end{align*}’ represent the number of quarts of Color X paint to be made.
Let ‘\begin{align*}y\end{align*}’ represent the number of quarts of Color Y paint to be made.
The number of quarts of Color X paint that are mixed must be either zero or greater than zero. Therefore, the constraint is \begin{align*}\boxed{x \ge 0}\end{align*}.
The number of quarts of Color Y paint that are mixed must be either zero or greater than zero. Therefore, the constraint is \begin{align*}\boxed{y \ge 0}\end{align*}.
The total amount of yellow tint that is used to mix Color X and Color Y must be less than or equal to 28. Therefore, the constraint is \begin{align*}\boxed{4x+y \le 28}\end{align*}.
The total amount of red tint that is used to mix Color X and Color Y must be less than or equal to 22. Therefore, the constraint is \begin{align*}\boxed{x+4y \le 22}\end{align*}.
Graph:
Vertices:
\begin{align*}& 4x+y =28 \quad \rightarrow && \qquad 4x+y=28 \quad \rightarrow \qquad \ \ \cancel{4x}+y=28 && 4x+y=28\\ & x+4y=22 && -4(x+4y=22) \ \rightarrow \quad -\underline{\cancel{4x}-16y=-88} && 4x+({\color{red}0})=28\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \ \ -15y=-60 \quad \rightarrow && 4x=28\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \quad \frac{\cancel{-15}y}{\cancel{-15}} = \frac{\overset{{\color{red}4}}{\cancel{-60}}}{\cancel{-15}} && \frac{\cancel{4}x}{\cancel{4}} = \frac{\overset{{\color{red}7}}{\cancel{28}}}{\cancel{4}}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad y=4 && x=7\end{align*}
\begin{align*}\boxed{l_1 \cap l_2 @ (7,4)}\end{align*}
The company wants to mix as many quarts as possible of Color X and of Color Y paint. Therefore, the \begin{align*}x\end{align*} and \begin{align*}y-\end{align*}intercepts have no meaning in this problem.
The company should mix 7 quarts of Color X paint and 4 quarts of Color Y paint.
Summary
In this lesson you have learned that a system of linear inequalities can be used to solve real- world problems. The solution to the problem is determined by graphing the system and using the vertices of the feasible region to calculate the best answer. The vertices were used since the Vertex Theorem for regions states that the maximum and minimum values, if they exist, occur at a vertex.
To ensure that the vertices were accurate, the inequalities of the intersecting lines were solved algebraically. The method that is used to solve the system of inequalities is a personal choice. Regardless of the method used, the results will not differ. In many problems it was also necessary to determine the \begin{align*}x\end{align*} and \begin{align*}y-\end{align*}intercepts for the feasible region.
Problem Set
For each graphed region and corresponding equation, find a point at which ‘\begin{align*}z\end{align*}’ has a maximum value and a point at which ‘\begin{align*}z\end{align*}’ has a minimum value.
1. \begin{align*}\boxed{z=7x-2y}\end{align*}
2. \begin{align*}\boxed{z=3y-4x}\end{align*}
3. \begin{align*}\boxed{z=4x-2y}\end{align*}
4. \begin{align*}\boxed{z=10x+20y}\end{align*}
5. \begin{align*}\boxed{z=20x-15y+4}\end{align*}
For each of the following problems, list the constraints, complete the graph and determine the maximum profit. Show all of your work.
1. A small manufacturing company makes $125 on each DVD player it produces and$100 profit on each color TV set it makes. Each DVD player and each TV must be processed by a cutting machine (A), a fitting machine (B) and a polishing machine (C). Each DVD player must be processed on Machine A for one hour, on Machine B for one hour and on Machine C for four hours. Each TV set must be processed on Machine A for two hours, on Machine B for one hour and on Machine C for one hour. Machines A, B, and C are available for 16, 9, and 24 hours per day respectively. How many DVD players and TV sets must be made each day to maximize the profit?
2. April has a small business during the winter months making hats and scarves. A hat requires 2 hours on Machine A, 4 hours on Machine B and 2 hours on Machine C. A scarf requires 3 hours on Machine A, 3 hours on Machine B and 1 hour on Machine C. Machine A is available 36 hours each week, Machine B is available 42 hours each week and Machine C is available 20 hours each week. The profit on a hat is $7.00 and the profit on a scarf is$4.00. How many of each should be made each week to maximize the profit?
3. Beth is knitting mittens and gloves. Each pair must be processed on three machines. Each pair of mittens requires 2 hours on Machine A, 2 hours on Machine B and 4 hours on Machine C. Each pair of gloves requires 4 hours on Machine A, 2 hours on Machine B and 1 hour on Machine C. Machine A, B, and C are available 32, 18 and 24 minutes each day respectively. The profit on a pair of mittens is $8.00 and on a pair of gloves is$10.00. How many pairs of each should be made each day to maximize the profit?
4. A patient is prescribed a pill that contains vitamins A, B and C. These vitamins are available in two different brands of pills. The first type is called Brand X and the second type is called Brand Y. The following table shows the amount of each vitamin that a Brand X and a Brand Y pill contain. The table also shows the minimum daily requirement needed by the patient. Each Brand X pill costs 32¢ and each Brand Y pill costs 29¢. How many pills of each brand should the patient take each day to minimize the cost?
Brand X Brand Y Minimum Daily Requirement
Vitamin A 2mg 1mg 5mg
Vitamin B 3mg 3mg 12mg
Vitamin C 25mg 50mg 125mg
1. A local smelting company is able to provide its customers with lead, copper and iron by melting down either of two ores, X or Y. The ores arrive at the company in railroad cars. Each railroad car of ore X contains 5 tons of lead, 1 ton of copper and I ton of iron. Each railroad car of ore Y contains 1 ton of lead, 1 ton of copper and 1 ton of iron. The smelting receives an order for 20 tons of lead, 12 tons of copper and 20 tons of iron. The cost to purchase and process a carload of ore X is $6000 while the cost for ore Y is$5000. If the company wants to fill the order at a minimum cost, how many carloads of each ore must be bought?
1. The vertices of the polygonal region are (0, 0); (0, 5); (6, 7); (7, 4) and (6, 0).
\begin{align*}& (0, 0) && z=7x-2y \rightarrow z=7(0)-2(0) \rightarrow z=0-0 \rightarrow z=0\\ & && \text{Therefore} \ 7x-2y=0\\ \\ & (0, 5) && z=7x-2y \rightarrow z=7(0)-2(5) \rightarrow z=0-10 \rightarrow z=-10\\ & && \text{Therefore} \ 7x-2y=-10\\ \\ & (6, 7) && z=7x-2y \rightarrow z=7(6)-2(7) \rightarrow z=42-14 \rightarrow z=28\\ & && \text{Therefore} \ 7x-2y=28\\ \\ & (7, 4) && z=7x-2y \rightarrow z=7(7)-2(4) \rightarrow z=49-8 \rightarrow z=41\\ & && \text{Therefore} \ 7x-2y=41\\ \\ & (6, 0) && z=7x-2y \rightarrow z=7(6)-2(0) \rightarrow z=42-0 \rightarrow z=42\\ & && \text{Therefore} \ 7x-2y=42\end{align*}
The maximum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (6, 0).
The minimum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (0, 5).
1. The vertices of the polygonal region are (-3, 2); (5, 4) and (1, -5).
\begin{align*}&(-3, 2) && z=4x-2y \rightarrow z=4(-3)-2(2) \rightarrow z=12-4 \rightarrow z=-16\\ & && \text{Therefore} \ 4x-2y=-16\\ \\ & (5, 4) && z=4x-2y \rightarrow z=4(5)-2(4) \rightarrow z=20-8 \rightarrow z=12\\ & && \text{Therefore} \ 4x-2y=-12\\ \\ &(1, -5)&& z=4x-2y \rightarrow z=4(-1)-2(5) \rightarrow z=-4+10 \rightarrow z=14\\ & && \text{Therefore} \ 4x-2y=-14\end{align*}
The maximum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (1, -5).
The minimum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (-3, 2).
1. The vertices of the polygonal region are (-7, 0); (-5, 5); (0, 0); (3, 4); (6, 0) and (0, -5).
\begin{align*}& (-7, 0) && z=20x-15y+4 \rightarrow z=20(-7)-15(0)+4 \rightarrow z=-140-0+4 \rightarrow z=-136\\ & && \text{Therefore} \ 20x-15y+4=-136\\ \\ & (-5, 5) && z=20x-15y+4 \rightarrow z=20(-5)-15(5)+4 \rightarrow z=-100-75+4 \rightarrow z=-171\\ & && \text{Therefore} \ 20x-15y+4=-171\\ \\ & (0, 0) && z=20x-15y+4 \rightarrow z=20(0)-15(0)+4 \rightarrow z=0-0+4 \rightarrow z=4\\ & && \text{Therefore} \ 20x-15y+4=4\\ \\ & (3, 4) && z=20x-15y+4 \rightarrow z=20(3)-15(4)+4 \rightarrow z=60-60+4 \rightarrow z=4\\ & && \text{Therefore} \ 20x-15y+4=4\\ \\ & (6, 0) && z=20x-15y+4 \rightarrow z=20(6)-15(0)+4 \rightarrow z=120-0+4 \rightarrow z=124\\ & && \text{Therefore} \ 20x-15y+4=124\\ \\ & (0, -5) && z=20x-15y+4 \rightarrow z=20(0)-15(-5)+4 \rightarrow z=0+75+4 \rightarrow z=79\\ & && \text{Therefore} \ 20x-15y+4=79\end{align*}
The maximum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (6, 0).
The minimum value of ‘\begin{align*}z\end{align*}’ occurred at the vertex (-5, 5).
1. 1. Let ‘\begin{align*}x\end{align*}’ represent the number of DVDs to be made. Let ‘\begin{align*}y\end{align*}’ represent the number of TVs to be made.
Table:
DVD TV Time Available hr/ day
Machine A 1hr 2hr 16
Machine B 1hr 1hr 9
Machine C 4hr 1hr 24
Constraints:
\begin{align*}\begin{Bmatrix} x+2y \le 16\\ x+y \le 9\\ 4x+y \le 24\\ x \ge 0\\ y \ge 0 \end{Bmatrix}\end{align*}
Profit equation: \begin{align*}\boxed{P=125x+100y}\end{align*}
Graph:
The feasible region is the area shaded in yellow.
Vertices:
\begin{align*}& x+2y=16 \quad \rightarrow && \quad \ \ x+2y=16 \quad \rightarrow \qquad \cancel{x}+2y=16 && x+y=9\\ & \ \ x+y=9 && -1(x+y=9) \quad \rightarrow \quad \ \underline{-\cancel{1x}-y=-9} && x+({\color{red}7})=9\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ y=7 \rightarrow && x+7{\color{red}-7}=9{\color{red}-7}\\ & && && x={\color{red}2}\\ & && && x=2\end{align*}
\begin{align*}\boxed{l_1 \cap l_2 @ (2,7)}\end{align*}
\begin{align*}& \ x+y=9 \quad \rightarrow && \qquad \ \ x+y=9 \qquad \rightarrow \qquad x+\cancel{y}=9 && x+y=9\\ & 4x+y=24 && -1(4x+y=24) \quad \rightarrow \quad \underline{-4x-\cancel{y}=-24} && (5)+y=9\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \ \ -3x=-15 \quad \rightarrow && 5{\color{red}-5}+y=9{\color{red}-5}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \ \ \frac{\cancel{-3}x}{\cancel{-3}}=\frac{\overset{{\color{red}5}}{\cancel{-15}}}{\cancel{-3}} && y={\color{red}4}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x=5 && y=4\end{align*}
\begin{align*}\boxed{l_2 \cap l_3 @ (5,4)}\end{align*}
The \begin{align*}y-\end{align*}intercept of \begin{align*}x+2y \le 16\end{align*} is (0, 8).
\begin{align*}4x+y&=24\\ 4x+({\color{red}0})&=24\\ 4x&=24\\ \frac{\cancel{4}x}{\cancel{4}} &= \frac{\overset{{\color{red}6}}{\cancel{24}}}{\cancel{4}}\\ x&=6\end{align*}
The \begin{align*}x-\end{align*}intercept of \begin{align*}4x+y \le 24\end{align*} is (6, 0)
The vertices of the feasible region are (0, 0); (0, 8); (2, 7); (5, 4) and (6, 0).
Profit:
\begin{align*}& (0, 0) && P=125x+100y \rightarrow P=125(0)+100(0) \rightarrow P=0+0 \rightarrow P=0\\ & && \text{Therefore} \ 125x+100y=\ 0\\ \\ & (0, 8) && P=125x+100y \rightarrow P=125(0)+100(8) \rightarrow P=0+800 \rightarrow P=800\\ & && \text{Therefore} \ 125x+100y=\ 800\\ \\ & (2, 7) && P=125x+100y \rightarrow P=125(2)+100(7) \rightarrow P=250+700 \rightarrow P=950\\ & && \text{Therefore} \ 125x+100y=\ 950\\ \\ & (5, 4) && P=125x+100y \rightarrow P=125(5)+100(4) \rightarrow P=625+400 \rightarrow P=1025\\ & && \text{Therefore} \ 125x+100y=\ 1025\\ \\ & (6, 0) && P=125x+100y \rightarrow P=125(6)+100(0) \rightarrow P=750+0 \rightarrow P=750\\ & && \text{Therefore} \ 125x+100y=\ 750\end{align*}
To maximize their profit, the company should make five DVDs and 4 TVs each day.
1. Let ‘\begin{align*}x\end{align*}’ represent the number of Brand X pills to be used. Let ‘\begin{align*}y\end{align*}’ represent the number of Brand Y pills to be used.
Table:
Brand X Brand Y Minimum Daily Requirement
Vitamin A 2mg 1mg 5mg
Vitamin B 3mg 3mg 12mg
Vitamin C 25mg 50mg 125mg
Constraints:
\begin{align*}\begin{Bmatrix} 2x+y \ge 5\\ 3x+3y \ge 12\\ 25x+50y \ge 125\\ x \ge 0\\ y \ge 0 \end{Bmatrix}\end{align*}
Cost equation: \begin{align*}\boxed{c=.32x+.29y}\end{align*}
Graph:
The feasible region is the area shaded in pink.
Vertices:
\begin{align*}& \ \ 2x+y=5 \quad \rightarrow && -3(2x+y=5) \quad \rightarrow \quad -6x-\cancel{3y}=-15 && 2x+y=5\\ & 3x+3y=12 && \quad \ 3x+3y=12 \quad \rightarrow \qquad \underline{3x +\cancel{3y}=12} && 2({\color{red}1})+y=5\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \ \ -3x=-3 \quad \rightarrow && 2{\color{red}-2}+y=5{\color{red}-2}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \ \ \frac{\cancel{-3}x}{\cancel{-3}}=\frac{\overset{{\color{red}1}}{\cancel{-3}}}{\cancel{-3}} && y={\color{red}3}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad x=1 && y=3\end{align*}
\begin{align*}\boxed{l_2 \cap l_3 @ (1,3)}\end{align*}
\begin{align*}& \quad 3x+3y=12 \quad \rightarrow && \quad \ \ 25(3x+3y=12) \quad \rightarrow \qquad \ \cancel{75x}+75y=300 && 3x+3y=12\\ & 25x+50y=125 && -3(25x+50y=125) \ \ \rightarrow \quad \underline{-\cancel{75x}-150y=-375} && 3x+3({\color{red}1})=12\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ -75y=-75 && 3x{\color{red}+3}=12\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \frac{\cancel{-75}y}{\cancel{-75}}=\frac{\overset{{\color{red}1}}{\cancel{-75}}}{\cancel{-75}} \quad \rightarrow && 3x+3{\color{red}-3}=12{\color{red}-3}\\ & && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad y=1 && 3x={\color{red}9}\\ & && && \frac{\cancel{3}x}{\cancel{3}}=\frac{\overset{{\color{red}3}}{\cancel{9}}}{\cancel{3}}\\ & && && x=3\end{align*}
\begin{align*}\boxed{l_2 \cap l_3 @ (3,1)}\end{align*}
\begin{align*}25x+50y&=125\\ 25(x)+({\color{red}0})&=125\\ 25x &= 125\\ \frac{\cancel{25}x}{\cancel{25}}&=\frac{\overset{{\color{red}5}}{\cancel{125}}}{\cancel{25}}\\ x &= 5\end{align*}
The \begin{align*}x-\end{align*}intercept of \begin{align*}25x+50y \ge 125\end{align*} is (5, 0)
The \begin{align*}y-\end{align*}intercept of \begin{align*}2x+y \ge 5\end{align*} is (0, 5)
The vertices of the feasible region are (0, 5); (1, 3); (3, 1) and (5, 0).
Cost:
\begin{align*}& (0, 5) && c=.32x+.29y \rightarrow c=.32(0)+.29(5) \rightarrow c=0+1.45 \rightarrow P=1.45\\ & && \text{Therefore} \ .32x+.29y=\ 1.45\\ \\ & (1, 3) && c=.32x+.29y \rightarrow c=.32(1)+.29(3) \rightarrow c=.32+.87 \rightarrow P=1.19\\ & && \text{Therefore} \ .32x+.29y=\ 1.19\\ \\ & (3, 1) && c=.32x+.29y \rightarrow c=.32(3)+.29(1) \rightarrow c=.96+.29 \rightarrow P=1.25\\ & && \text{Therefore} \ .32x+.29y=\ 1.25\\ \\ & (5, 0) && c=.32x+.29y \rightarrow c=.32(5)+.29(0) \rightarrow c=1.60+0 \rightarrow P=1.60\\ & && \text{Therefore} \ .32x+.29y=\ 1.60\end{align*}
To minimize the cost of the pills, the patient should take one Brand X pill and three Brand Y pills daily.
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# Calculate: x^2+3x-18=0
## Expression: ${x}^{2}+3x-18=0$
Identify the coefficients $a$, $b$ and $c$ of the quadratic equation
$\begin{array} { l }a=1,& b=3,& c=-18\end{array}$
Substitute $a=1$, $b=3$ and $c=-18$ into the quadratic formula $x=\frac{ -b\pm\sqrt{ {b}^{2}-4ac } }{ 2a }$
$x=\frac{ -3\pm\sqrt{ {3}^{2}-4 \times 1 \times \left( -18 \right) } }{ 2 \times 1 }$
Any expression multiplied by $1$ remains the same
$x=\frac{ -3\pm\sqrt{ {3}^{2}-4 \times \left( -18 \right) } }{ 2 \times 1 }$
Any expression multiplied by $1$ remains the same
$x=\frac{ -3\pm\sqrt{ {3}^{2}-4 \times \left( -18 \right) } }{ 2 }$
Evaluate the power
$x=\frac{ -3\pm\sqrt{ 9-4 \times \left( -18 \right) } }{ 2 }$
Multiply the numbers
$x=\frac{ -3\pm\sqrt{ 9+72 } }{ 2 }$
$x=\frac{ -3\pm\sqrt{ 81 } }{ 2 }$
Evaluate the square root
$x=\frac{ -3\pm9 }{ 2 }$
Write the solutions, one with a $+$ sign and one with a $-$ sign
$\begin{array} { l }x=\frac{ -3+9 }{ 2 },\\x=\frac{ -3-9 }{ 2 }\end{array}$
Simplify the expression
$\begin{array} { l }x=3,\\x=\frac{ -3-9 }{ 2 }\end{array}$
Simplify the expression
$\begin{array} { l }x=3,\\x=-6\end{array}$
The equation has $2$ solutions
$\begin{array} { l }x_1=-6,& x_2=3\end{array}$
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5/6 as a Decimal: A Simple Guide to Converting Fractions
Published:
If you’ve ever wondered how to convert the fraction 5/6 as a decimal, you’ve come to the right place. Fractions and decimals are fundamental concepts in mathematics, and understanding how to convert between them is a valuable skill. In this article, we’ll walk you through the step-by-step process of converting 5/6 into its decimal equivalent. By the end, you’ll not only know the answer but also understand the underlying concept.
1. Introduction: The Significance of Converting Fractions to Decimals
2. Understanding the Basics: What Is 5/6?
3. Method 1: Long Division
4. Method 2: Using a Calculator
5. Decimal Equivalents of Common Fractions
6. Why Is Converting Fractions to Decimals Important?
7. Real-Life Applications
8. Fractions and Decimals in Daily Life
9. The Role of Fractions in Mathematics
10. Converting Fractions in Other Bases
11. Converting Decimals to Fractions
12. Challenges and Pitfalls
13. Practice Problems
14. Tips for Mastery
15. Conclusion
1. Introduction: The Significance of Converting Fractions to Decimals
Understanding the relationship between fractions and decimals is crucial in various aspects of life, from everyday calculations to more complex mathematical problems. Converting fractions to decimals is a fundamental skill that empowers you to work with numbers more flexibly and efficiently.
2. Understanding the Basics: What Is 5/6?
Before we delve into the conversion methods, let’s ensure we’re clear on what 5/6 represents. In simple terms, 5/6 is a fraction where the numerator (5) represents the part we have, and the denominator (6) represents the whole. To convert it to a decimal, we need to find its equivalent in decimal form.
3. Method 1: Long Division
One common method to convert fractions to decimals is long division. Here’s how you can do it for 5/6:
1. Divide 5 by 6: 5 ÷ 6 = 0 with a remainder of 5.
2. Add a decimal point after the 0 to make it 0.0.
3. Bring down a 0 to the remainder, making it 50.
4. Divide 50 by 6: 50 ÷ 6 = 8 with a remainder of 2.
5. Continue this process as far as you need for accuracy.
So, 5/6 as a decimal using long division is approximately 0.8333 (repeating).
4. Method 2: Using a Calculator
In our modern age, calculators have made mathematical tasks significantly easier. To convert 5/6 to a decimal using a calculator, simply input 5 ÷ 6, and you’ll get the result: 0.8333 (repeating).
5. Decimal Equivalents of Common Fractions
Understanding the decimal equivalents of common fractions can save you time in everyday calculations. Here are some other popular fractions and their decimal counterparts:
• 1/2 = 0.5
• 1/4 = 0.25
• 3/4 = 0.75
• 1/3 = 0.3333 (repeating)
• 2/3 = 0.6666 (repeating)
6. Why Is Converting Fractions to Decimals Important?
Converting fractions to decimals is essential because it allows for easier comparisons and calculations. It’s particularly valuable in fields like science, engineering, and finance, where precise numerical analysis is required.
7. Real-Life Applications
In real-life scenarios, you might need to convert fractions to decimals when dealing with recipes, measurements, or financial calculations. Being able to quickly convert between the two forms can make your tasks more efficient.
8. Fractions and Decimals in Daily Life
Even in everyday life, fractions and decimals play a significant role. From dividing a pizza into equal slices to calculating discounts during sales, these concepts are all around us.
9. The Role of Fractions in Mathematics
Fractions are a fundamental part of mathematics, appearing in equations, proportions, and more. Mastering their conversion to decimals is a stepping stone to mathematical proficiency.
10. Converting Fractions in Other Bases
In addition to base 10 (decimal), fractions can be converted in other bases, such as binary or hexadecimal. These conversions are essential in computer science and digital systems.
11. Converting Decimals to Fractions
While this article focuses on converting fractions to decimals, the reverse process is equally important. Converting decimals to fractions can be handy for precise representations.
12. Challenges and Pitfalls
Some fractions result in repeating decimals, like 1/3, which becomes 0.3333 (repeating). Understanding these patterns is key to mastering conversions.
13. Practice Problems
To enhance your skills, we’ve included some practice problems at the end of this article. Test your knowledge and become more confident in converting fractions to decimals.
14. Tips for Mastery
We provide tips and tricks to make the conversion process smoother. Whether you’re a student or a professional, these insights can boost your math skills.
15. Conclusion
Converting fractions to decimals is a valuable skill with practical applications in various fields. It empowers you to work with numbers more efficiently and make informed decisions. So, the next time you encounter a fraction like 5/6, you’ll know how to effortlessly convert it to a decimal.
FAQs
1. What is the quickest way to convert fractions to decimals?
• The quickest way is to use a calculator, especially for complex fractions.
2. Why do some fractions result in repeating decimals?
• This happens when the denominator cannot be evenly divided by the numerator.
3. Can I convert decimals back to fractions?
• Yes, decimals can be converted back to fractions using specific methods.
4. Are there shortcuts for converting common fractions to decimals?
• Yes, there are shortcuts for fractions like 1/2, 1/4, and 3/4.
5. Where can I practice more fraction-to-decimal conversions?
• You can find practice problems online or in math textbooks to enhance your skills.
Now that you’ve learned how to convert 5/6 to a decimal and gained insights into the importance of this skill, you’re better equipped to tackle various mathematical challenges in your daily life. Mathematics is all about building skills step by step, and this is undoubtedly a valuable one. So, embrace the world of fractions and decimals, and keep calculating! |
# Trigonometric Integrals
In this topic, we will study how to integrate certain combinations involving products and powers of trigonometric functions.
We consider 8 cases.
## 1. Integrals of the form $$\int {\cos ax\cos bxdx} ,$$ $$\int {\sin ax\cos bxdx} ,$$ $$\int {\sin ax\sin bxdx}$$
To evaluate integrals of products of sine and cosine with different arguments, we apply the identities
$\cos ax\cos bx = \frac{1}{2}\left[ {\cos \left( {ax + bx} \right) + \cos \left( {ax - bx} \right)} \right];$
$\sin ax\cos bx = \frac{1}{2}\left[ {\sin \left( {ax + bx} \right) + \sin \left( {ax - bx} \right)} \right];$
$\sin ax\sin bx = - \frac{1}{2}\left[ {\cos \left( {ax + bx} \right) - \cos \left( {ax - bx} \right)} \right].$
## 2. Integrals of the form $$\int {{\sin^m}x\,{\cos^n}xdx}$$
We assume here that the powers $$m$$ and $$n$$ are non-negative integers.
To find an integral of this form, use the following substitutions:
1. If $$m$$ (the power of sine) is odd, we use the $$u-$$substitution
$u = \cos x,\;\; du = - \sin xdx$
and the identity
${\sin ^2}x + {\cos ^2}x = 1$
to express the remaining even power of sine in $$u-$$terms.
2. If $$n$$ (the power of cosine) is odd, we use the $$u-$$substitution
$u = \sin x,\;\; du = \cos xdx$
and the identity
${\sin ^2}x + {\cos ^2}x = 1$
to express the remaining even power of cosine in $$u-$$terms.
3. If both powers $$m$$ and $$n$$ are even, we reduce the powers using the half-angle formulas
${\sin ^2}x = \frac{{1 - \cos 2x}}{2},\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2}.$
The integrals of type $$\int {{{\sin }^n}xdx}$$ and $$\int {{{\cos }^n}xdx}$$ can be evaluated by reduction formulas
$\int {{{\sin }^n}xdx} = - \frac{{{{\sin }^{n - 1}}x\cos x}}{n} + \frac{{n - 1}}{n}\int {{{\sin }^{n - 2}}xdx} ,$
$\int {{{\cos }^n}xdx} = \frac{{{{\cos }^{n - 1}}x\sin x}}{n} + \frac{{n - 1}}{n}\int {{{\cos }^{n - 2}}xdx} .$
## 3. Integrals of the form $$\int {{\tan^n}xdx}$$
The power of the integrand can be reduced using the trigonometric identity
$1 + {\tan ^2}x = {\sec ^2}x$
and the reduction formula
$\int {{\tan^n}xdx} = \int {{{\tan }^{n - 2}}x\,{{\tan }^2}xdx} = \int {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)dx} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}xdx} .$
## 4. Integrals of the form $$\int {{\cot^n}xdx}$$
The power of the integrand can be reduced using the trigonometric identity
$1 + {\cot^2}x = {\csc^2}x$
and the reduction formula
$\int {{{\cot }^n}xdx} = \int {{{\cot }^{n - 2}}x\,{{\cot }^2}xdx} = \int {{{\cot }^{n - 2}}x\left( {{{\csc }^2}x - 1} \right)dx} = - \frac{{{{\cot }^{n - 1}}x}}{{n - 1}} - \int {{{\cot }^{n - 2}}xdx} .$
## 5. Integrals of the form $$\int {{\sec^n}xdx}$$
This type of integrals can be simplified with help of the reduction formula:
$\int {{\sec^n}xdx} = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{\sec^{n - 2}}xdx} .$
## 6. Integrals of the form $$\int {{\csc^n}xdx}$$
Similarly to the previous examples, this type of integrals can be simplified by the formula
$\int {{\csc^n}xdx} = - \frac{{{\csc^{n - 2}}x \cot x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{\csc^{n - 2}}xdx} .$
## 7. Integrals of the form $$\int {{\tan^m}x\,{\sec^n}xdx}$$
1. If the power of the secant $$n$$ is even, then using the identity
$1 + {\tan ^2}x = {\sec ^2}x$
the secant function is expressed as the tangent function. The factor $${\sec ^2}x$$ is separated and used for transformation of the differential. As a result, the entire integral (including differential) is expressed in terms of the function $$\tan x.$$
2. If both the powers $$n$$ and $$m$$ are odd, then the factor $$\sec x \tan x,$$ which is necessary to transform the differential, is separated. Then the entire integral is expressed in terms of $$\sec x.$$
3. If the power of the secant $$n$$ is odd, and the power of the tangent $$m$$ is even, then the tangent is expressed in terms of the secant using the identity
$1 + {\tan ^2}x = {\sec ^2}x.$
After this substitution, you can calculate the integrals of the secant.
## 8. Integrals of the form $$\int {{\cot^m}x\,{\csc^n}xdx}$$
1. If the power of the cosecant $$n$$ is even, then using the identity
$1 + {\cot^2}x = {\csc ^2}x$
the cosecant function is expressed as the cotangent function. The factor $${\csc^2}x$$ is separated and used for transformation of the differential. As a result, the integrand and differential are expressed in terms of $$\cot x.$$
2. If both the powers $$n$$ and $$m$$ are odd, then the factor $$\cot x \csc x,$$ which is necessary to transform the differential, is separated. Then the integral is expressed in terms of $$\csc x.$$
3. If the power of the cosecant $$n$$ is odd, and the power of the cotangent $$m$$ is even, then the cotangent is expressed in terms of the cosecant using the identity
$1 + {\cot^2}x = {\csc ^2}x.$
After this substitution, you can find the integrals of the cosecant.
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
Calculate the integral $\int {{\sin^3}xdx}.$
### Example 2
Evaluate the integral $\int {{\cos^5}xdx}.$
### Example 3
Find the integral $\int {{\sin^6}xdx}.$
### Example 4
Find the integral $\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.$
### Example 5
Calculate the integral $\int {{{\sin }^2}x\,{{\cos }^4}xdx}.$
### Example 6
Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^4}xdx}.$
### Example 7
Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^5}xdx}.$
### Example 8
Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^3}xdx}.$
### Example 1.
Calculate the integral $\int {{\sin^3}xdx}.$
Solution.
Let $$u = \cos x,$$ $$du = -\sin xdx.$$ Then
$\int {{\sin^3}xdx} = \int {{\sin^2}x\sin xdx} = \int {\left( {1 - {\cos^2}x} \right)\sin xdx} = - \int {\left( {1 - {u^2}} \right)du} = \int {\left( {{u^2} - 1} \right)du} = \frac{{{u^3}}}{3} - u + C = \frac{{{{\cos }^3}x}}{3} - \cos x + C.$
### Example 2.
Evaluate the integral $\int {{\cos^5}xdx}.$
Solution.
Making the substitution $$u = \sin x,$$ $$du = \cos xdx$$ and using the identity
${\cos ^2}x = 1 - {\sin ^2}x,$
we obtain:
$\int {{\cos^5}xdx} = \int {{{\left( {{\cos^2}x} \right)}^2}\cos xdx} = \int {{{\left( {1 - {{\sin }^2}x} \right)}^2}\cos x dx} = \int {{{\left( {1 - {u^2}} \right)}^2}du} = \int {\left( {1 - 2{u^2} + {u^4}} \right)du} = u - \frac{{2{u^3}}}{3} + \frac{{{u^5}}}{5} + C = \sin x - \frac{{2{{\sin }^3}x}}{3} + \frac{{{{\sin }^5}x}}{5} + C.$
### Example 3.
Find the integral $\int {{\sin^6}xdx}.$
Solution.
Using the identities
${\sin ^2}x = \frac{{1 - \cos 2x}}{2}\;\;\text{and}\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2},$
we can write:
$I = \int {{\sin^6}xdx} = \int {{{\left( {{\sin^2}x} \right)}^3}dx} = \frac{1}{8}\int {{{\left( {1 - \cos 2x} \right)}^3}dx} = \frac{1}{8}\int {\left( {1 - 3\cos 2x + 3\,{{\cos }^2}2x - {{\cos }^3}2x} \right)dx} = \frac{x}{8} - \frac{3}{8} \cdot \frac{{\sin 2x}}{2} + \frac{3}{8}\int {{\cos^2}2xdx} - \frac{3}{8}\int {{\cos^3}2xdx}.$
Calculate the integrals in the latter expression.
$\int {{\cos^2}2xdx} = \int {\frac{{1 + \cos 4x}}{2}dx} = \frac{1}{2}\int {\left( {1 + \cos 4x} \right)dx} = \frac{1}{2}\left( {x + \frac{{\sin 4x}}{4}} \right) = \frac{x}{2} + \frac{{\sin 4x}}{8}.$
To find the integral $$\int {{\cos^3}2xdx},$$ we make the substitution $$u = \sin 2x,$$ $$du =$$ $$2\cos 2xdx.$$ Then
$u = \sin 2x,\;du = 2\cos 2xdx.$
Then
$\int {{\cos^3}2xdx} = \frac{1}{2}\int {2{{\cos }^2}2x\cos 2xdx} = \frac{1}{2}\int {2\left( {1 - {{\sin }^2}2x} \right)\cos 2xdx} = \frac{1}{2}\int {\left( {1 - {u^2}} \right)du} = \frac{u}{2} - \frac{{{u^3}}}{6} = \frac{{\sin 2x}}{2} - \frac{{{{\sin }^3}2x}}{6}.$
Hence, the initial integral is
$I = \frac{x}{8} - \frac{{3\sin 2x}}{{16}} + \frac{3}{8}\left( {\frac{x}{2} + \frac{{\sin 4x}}{8}} \right) - \frac{1}{8}\left( {\frac{{\sin 2x}}{2} - \frac{{{{\sin }^3}2x}}{6}} \right) + C = \frac{{5x}}{{16}} - \frac{{\sin 2x}}{4} + \frac{{3\sin 4x}}{{64}} + \frac{{{{\sin }^3}2x}}{{48}} + C.$
### Example 4.
Find the integral $\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.$
Solution.
The power of cosine is odd, so we make the substitution
$u = \sin x,\;\;du = \cos xdx.$
We rewrite the integral in terms of $$\sin x$$ to obtain:
$\int {{{\sin }^2}x\,{{\cos }^3}xdx} = \int {{{\sin }^2}x\,{{{\cos }^2}x}\cos xdx} = \int {{{\sin }^2}x{\left( {1 - {{\sin }^2}x} \right)}\cos xdx} = \int {{u^2}\left( {1 - {u^2}} \right)du} = \int {\left( {{u^2} - {u^4}} \right)du} = \frac{{{u^3}}}{3} - \frac{{{u^5}}}{5} + C = \frac{{{{\sin }^3}x}}{3} - \frac{{{{\sin }^5}x}}{5} + C.$
### Example 5.
Calculate the integral $\int {{{\sin }^2}x\,{{\cos }^4}xdx}.$
Solution.
We can write:
$I = \int {{{\sin }^2}x\,{{\cos }^4}xdx} = \int {{{\left( {\sin x\cos x} \right)}^2}{{\cos }^2}xdx} .$
We convert the integrand using the identities
$\sin x\cos x = \frac{{\sin 2x}}{2},\;\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2},\;\;\;{\sin ^2}x = \frac{{1 - \cos 2x}}{2}.$
This yields
$I = \int {{{\left( {\frac{{\sin 2x}}{2}} \right)}^2}\frac{{1 + \cos 2x}}{2}dx} = \frac{1}{8}\int {{{\sin }^2}2x\left( {1 + \cos 2x} \right)dx} = \frac{1}{8}\int {{{\sin }^2}2xdx} + \frac{1}{8}\int {{{\sin }^2}2x\cos 2xdx} = \frac{1}{8}\int {\frac{{1 - \cos 4x}}{2}dx} + \frac{1}{{16}}\int {2{{\sin }^2}2x\cos 2xdx} = \frac{1}{{16}}\int {\left( {1 - \cos 4x} \right)dx} + \frac{1}{{16}}\int {{{\sin }^2}2x\,d\left( {\sin 2x} \right)} = \frac{1}{{16}}\left( {x - \frac{{\sin 4x}}{4}} \right) + \frac{1}{{16}} \cdot \frac{{{{\sin }^3}2x}}{3} + C = \frac{x}{{16}} - \frac{{\sin 4x}}{{64}} + \frac{{{{\sin }^3}2x}}{{48}} + C.$
### Example 6.
Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^4}xdx}.$
Solution.
As the power of sine is odd, we use the substitution
$u = \cos x,\;\;du = - \sin xdx.$
The integral is written as
$I = \int {{{\sin }^3}x\,{{\cos }^4}xdx} = \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} .$
By the Pythagorean identity,
${\sin ^2}x = 1 - {\cos ^2}x.$
Hence
$I = \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^4}x\sin xdx} = - \int {\left( {1 - {u^2}} \right){u^4}du} = \int {\left( {{u^6} - {u^4}} \right)du} = \frac{{{u^7}}}{7} - \frac{{{u^5}}}{5} + C = \frac{{{{\cos }^7}x}}{7} - \frac{{{{\cos }^5}x}}{5} + C.$
### Example 7.
Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^5}xdx}.$
Solution.
We see that both powers are odd, so we can substitute either $$u = \sin x$$ or $$u = \cos x.$$ Choosing the least exponent, we have
$u = \cos x,\;\;du = - \sin xdx.$
The integral takes the form
$I = \int {{{\sin }^3}x\,{{\cos }^5}xdx} = \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} .$
Using the Pythagorean identity
${\sin ^2}x = 1 - {\cos ^2}x,$
we can write
$I = \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^5}x\sin xdx} = - \int {\left( {1 - {u^2}} \right){u^5}du} = \int {\left( {{u^7} - {u^5}} \right)du} = \frac{{{u^8}}}{8} - \frac{{{u^6}}}{6} + C = \frac{{{{\cos }^8}x}}{8} - \frac{{{{\cos }^6}x}}{6} + C.$
### Example 8.
Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^3}xdx}.$
Solution.
The powers of both sine and cosine are odd. Hence we can use the substitution $$u = \sin x$$ or $$u = \cos x.$$ Let's apply the substitution $$u = \sin x.$$ Then $$du = \cos x dx,$$ and the integral becomes
$I = \int {{{\sin }^3}x\,{{\cos }^3}xdx} = \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} .$
By the Pythagorean identity,
${\sin ^2}x = 1 - {\cos ^2}x,$
so we obtain
$I = \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} = \int {{{\sin }^3}x\left( {1 - {{\sin }^2}x} \right)\cos xdx} = \int {{u^3}\left( {1 - {u^2}} \right)du} = \int {\left( {{u^3} - {u^5}} \right)du} = \frac{{{u^4}}}{4} - \frac{{{u^6}}}{6} + C = \frac{{{{\sin }^4}x}}{4} - \frac{{{{\sin }^6}x}}{6} + C.$
See more problems on Page 2. |
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# Statistics Homework
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Statistics 112: Homework 1 Answer to Question 1:
Variable that is dependent:
Position on women's rights,
Age, gender, and degree of education are independent variables.
Variable\sType\sScale
Attitudes concerning women's rights
Numerical\sInterval
Respondent's age in years
Numerical\sRatio
Degree of Education Categorical Variable Gender Categorical Variable
To be utilized plot
Attitudes concerning women's rights
Histogram and boxplot of respondent's age in years
Histogram,\sGender
Pie and bar graphs
Education Level
Graph with bars
To demonstrate the relationship in question, we can utilize a side by side boxplot.
To demonstrate the relationship in question, we can utilize a side by side bar chart.We can use line chart to exhibit the relationship in question.
Solution to Question Two:
The command in “R” to read the csv file:
Figure 1: Boxplot of Ethnicity
Here is the code and output for the problem:
The code for drawing histogram:
Histogram:
The code for drawing boxplot:
Figure 2: Boxplot for Education Level
e. The required codes are as follows:
Solution to Question Three:
a) Given data: 3, 5, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 10, 11, 13, 15, 44
Mean = = = 10
The data are arranged in ascending order and there are even number of elements (20 data). So, the average of 10th and 11th data will give the median.
Median = = = 8
The first quartile, Q1 will be the average of the 5th and 6th element, which is equal to = 7.
The third quartile, Q3 will be the average of the 15th and 16th element, which is equal to = 9.5.
Inter Quartile Range (IQR) = (9.5–7) = 2.5.
Variance = = …
= 67.5
Standard Deviation = √variance = √ 67.5 = 8.215
(b) Here, Q3 + 1.5 IQR = 9.5 + 1.5×2.5 = 13.5 and Q1–1.5IQR = 3.25
We have 3 values i.e. 3, 15, and 44, which falls outside this range and are outliers.
(c) The modified data without outliers is: 5, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 10, 11, 13
Mean = = = 8.11
The data are arranged in ascending order and there are odd number of elements (17 data). So, the 9th data will give the median.
Median = 8
The first quartile, Q1 will be the 5th element = 7.
The third quartile, Q3 will be the 13th , which is equal to = 9
Inter Quartile Range (IQR) = (9–7) = 2
Variance = = …
= 7.05
In the absence of the outliers, the statistical parameters can predict the central tendency more accurately i.e. they are truer representative of their actual data.
Solution to Question Four:
(a) Z-score = = -1.25; the value of critical value of z from the z-table is = 0.11.
Which means John falls within the 10% percentile.
(b) The required R code to estimate percentile for John:
The percentile value for John is 0.1, which means that John falls within the 10% of the population, who have IQ higher than or equal to 80.
Solution to Question Five:
(a) The required R code and output is given below:
(b) 8.5 Kg
Solution to Question Six:
The R code for plotting histogram is given below:
(b) The required R code:
(c) The required R code is given below:
Solution to Question Seven:
(a) Watching educational short film 🡪 Independent Variable
Attitude of policemen and policewomen toward treating minority 🡪 Dependent variable.
(b) Showing or not showing short film, attitude of policemen and policewomen toward treating minority
(c) He should use a survey based research design
(d) Experimental group 🡪 Those who watched film, Control Group 🡪 Those who did not.
(e) Yes.
(f) Whether education short film has influence on the attitude of policemen and policewomen.
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# [Maths Class Notes] on Vertical Line Pdf for Exam
The line is long but not wide. A line is a two-dimensional geometric figure that can move in either direction. A line is made up of an infinite number of points. It is infinite and has no beginnings or endings on either side. A line has only one dimension. Ancient mathematicians introduced the concept of line or straight line in geometry to represent straight objects with negligible width and depth. It’s frequently described in terms of two points.
In the concept of analytic geometry, a line in the plane is often defined as the set of points whose coordinates satisfy a given linear equation, whereas in the concept of incidence geometry, a line may be an independent object distinct from the set of points which lie on it.
### A Coordinate Plane
A coordinate plane is a plane where two lines intersect each other at 90o. The horizontal line is known as the x-axis and the vertical line is known as the y-axis. Together the lines are called axes and the point where the two lines intersect each other is known as the origin. There are four quadrants (I, II, III, IV). These quadrants are formed due to axes.
Parallel Lines in the Coordinate Plane:
Parallel lines are those lines which never intersect each other as shown in the figure.
Perpendicular Lines in the Coordinate Plane:
Perpendicular lines are those lines which intersect each other at a right angle.
### Graph:
The graph is a design that shows the relation between the variables, quantities, etc. The graph is a collection of points called vertices, and lines between those points are called edges.
### Vertical Lines:
The vertical line in coordinate geometry is a line that goes up to down. In other words, the line that is parallel to the y-axis and has the same x coordinate point is vertical. A vertical line can only be drawn when the line has the same x coordinate.
The slope of the vertical line is zero.
The equation of the vertical line is given as
x = a
Where,
x = coordinates of any point on the line
a = point where the line crosses the x-axis
Hence, it can be seen that the equation is independent of y.
### Equation of a Vertical Line
The equation of a vertical line is given by x = a, or x = -a,
Here,
x is the x coordinate of any point on the line
a is where the line crosses the x-intercept.
### Vertical Line Properties
• A vertical line in the graph that is parallel to the y-axis has the equation x = a.
• A vertical line’s slope is infinity or undefined because it lacks a y-intercept and has a zero denominator in the slope formula.
• In mathematics, we use a vertical line to determine whether the relationship is a function. It is a function if all vertical lines intersect it at least once. This is also referred to as a vertical line test. If a graph has only one output y for each input x, it is considered a function. As a result, a vertical line cannot serve as a function.
### Conclusion
The vertical line is a perpendicular to the surface or another line that serves as the base. Vertical lines in coordinate geometry are parallel to the y-axis and generally perpendicular to horizontal lines. A vertical line is always a straight line that runs from top to bottom or top to bottom. Vertical lines are also referred to as standing lines. Vertical lines are typically drawn by connecting the bases of a square or a rectangle. |
# Factors of 2310 in pair
Factors of 2310 in pair are (1, 2310) , (2, 1155) , (3, 770) , (5, 462) , (6, 385) , (7, 330) , (10, 231) , (11, 210) , (14, 165) , (15, 154) , (21, 110) , (22, 105) , (30, 77) , (33, 70) , (35, 66) and (42, 55)
#### How to find factors of a number in pair
1. Steps to find factors of 2310 in pair 2. What is factors of a number in pair? 3. What are Factors? 4. Frequently Asked Questions 5. Examples of factors in pair
### Example: Find factors of 2310 in pair
Factor Pair Pair Factorization
1 and 2310 1 x 2310 = 2310
2 and 1155 2 x 1155 = 2310
3 and 770 3 x 770 = 2310
5 and 462 5 x 462 = 2310
6 and 385 6 x 385 = 2310
7 and 330 7 x 330 = 2310
10 and 231 10 x 231 = 2310
11 and 210 11 x 210 = 2310
14 and 165 14 x 165 = 2310
15 and 154 15 x 154 = 2310
21 and 110 21 x 110 = 2310
22 and 105 22 x 105 = 2310
30 and 77 30 x 77 = 2310
33 and 70 33 x 70 = 2310
35 and 66 35 x 66 = 2310
42 and 55 42 x 55 = 2310
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 2310. They are called negative pair factors.
Hence, the negative pairs of 2310 would be ( -1 , -2310 ) .
#### Definition of factor pairs?
In mathematics, factor pair are often given as pair of numbers which when multiplied together give the original number. Every natural number is a product of atleast one factor pair. Eg- Factors of 2310 are 1 , 2 , 3 , 5 , 6 , 7 , 10 , 11 , 14 , 15 , 21 , 22 , 30 , 33 , 35 , 42 , 55 , 66 , 70 , 77 , 105 , 110 , 154 , 165 , 210 , 231 , 330 , 385 , 462 , 770 , 1155 , 2310. So, factors of 2310 in pair are (1,2310), (2,1155), (3,770), (5,462), (6,385), (7,330), (10,231), (11,210), (14,165), (15,154), (21,110), (22,105), (30,77), (33,70), (35,66), (42,55).
#### What are factors?
In mathematics, a factor is that number which divides into another number exactly, without leaving a remainder. A factor of a number can be positive or negative.
#### Properties of Factors
• Each number is a factor of itself. Eg. 2310 is a factor of itself.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, 2310 are exact divisors of 2310.
• 1 is a factor of every number. Eg. 1 is a factor of 2310.
• Every number is a factor of zero (0), since 2310 x 0 = 0.
#### Steps to find Factors of 2310
• Step 1. Find all the numbers that would divide 2310 without leaving any remainder. Starting with the number 1 upto 1155 (half of 2310). The number 1 and the number itself are always factors of the given number.
2310 ÷ 1 : Remainder = 0
2310 ÷ 2 : Remainder = 0
2310 ÷ 3 : Remainder = 0
2310 ÷ 5 : Remainder = 0
2310 ÷ 6 : Remainder = 0
2310 ÷ 7 : Remainder = 0
2310 ÷ 10 : Remainder = 0
2310 ÷ 11 : Remainder = 0
2310 ÷ 14 : Remainder = 0
2310 ÷ 15 : Remainder = 0
2310 ÷ 21 : Remainder = 0
2310 ÷ 22 : Remainder = 0
2310 ÷ 30 : Remainder = 0
2310 ÷ 33 : Remainder = 0
2310 ÷ 35 : Remainder = 0
2310 ÷ 42 : Remainder = 0
2310 ÷ 55 : Remainder = 0
2310 ÷ 66 : Remainder = 0
2310 ÷ 70 : Remainder = 0
2310 ÷ 77 : Remainder = 0
2310 ÷ 105 : Remainder = 0
2310 ÷ 110 : Remainder = 0
2310 ÷ 154 : Remainder = 0
2310 ÷ 165 : Remainder = 0
2310 ÷ 210 : Remainder = 0
2310 ÷ 231 : Remainder = 0
2310 ÷ 330 : Remainder = 0
2310 ÷ 385 : Remainder = 0
2310 ÷ 462 : Remainder = 0
2310 ÷ 770 : Remainder = 0
2310 ÷ 1155 : Remainder = 0
2310 ÷ 2310 : Remainder = 0
Hence, Factors of 2310 are 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, and 2310
• Is 2310 a composite number?
Yes 2310 is a composite number.
• Is 2310 a prime number?
No 2310 is not a prime number.
• What is the mean of factors of 2310?
Factors of 2310 are 1 , 2 , 3 , 5 , 6 , 7 , 10 , 11 , 14 , 15 , 21 , 22 , 30 , 33 , 35 , 42 , 55 , 66 , 70 , 77 , 105 , 110 , 154 , 165 , 210 , 231 , 330 , 385 , 462 , 770 , 1155 , 2310. therefore mean of factors of 2310 is (1 + 2 + 3 + 5 + 6 + 7 + 10 + 11 + 14 + 15 + 21 + 22 + 30 + 33 + 35 + 42 + 55 + 66 + 70 + 77 + 105 + 110 + 154 + 165 + 210 + 231 + 330 + 385 + 462 + 770 + 1155 + 2310) / 32 = 216.00.
• What do you mean by proper divisors?
A number x is said to be the proper divisor of y if it divides y completely, given that x is smaller than y.
• What do you mean by improper divisors?
An improper divisor of a number is the number itself apart from this any divisor of a given number is a proper divisor.
#### Examples of Factors
Can you help Rubel in arranging 2310 blocks in order to form a rectangle in all possible ways? For arranging 2310 blocks in order to form a rectangle in all possible ways we need to find factors of 2310 in pair.
Factors of 2310 are 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, 2310. So, factors of 2310 in pair are (1,2310), (2,1155), (3,770), (5,462), (6,385), (7,330), (10,231), (11,210), (14,165), (15,154), (21,110), (22,105), (30,77), (33,70), (35,66), (42,55).
Ariel has been asked to write all factor pairs of 2310 but she is finding it difficult. Can you help her find out?
Factors of 2310 are 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, 2310. So, factors of 2310 in pair are (1,2310), (2,1155), (3,770), (5,462), (6,385), (7,330), (10,231), (11,210), (14,165), (15,154), (21,110), (22,105), (30,77), (33,70), (35,66), (42,55).
How many total number of factors of 2310 in pair are possible?
Factors of 2310 are 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, 2310. Hence, the factors of 2310 in pair are (1,2310), (2,1155), (3,770), (5,462), (6,385), (7,330), (10,231), (11,210), (14,165), (15,154), (21,110), (22,105), (30,77), (33,70), (35,66), (42,55). Therefore, in total 16 pairs of factors are possible.
Sammy wants to write all the negative factors of 2310 in pair, but don't know how to start. Help Sammy in writing all the factor pairs.
Negative factors of 2310 are -1, -2, -3, -5, -6, -7, -10, -11, -14, -15, -21, -22, -30, -33, -35, -42, -55, -66, -70, -77, -105, -110, -154, -165, -210, -231, -330, -385, -462, -770, -1155, -2310. Hence, factors of 2310 in pair are (-1,-2310), (-2,-1155), (-3,-770), (-5,-462), (-6,-385), (-7,-330), (-10,-231), (-11,-210), (-14,-165), (-15,-154), (-21,-110), (-22,-105), (-30,-77), (-33,-70), (-35,-66), (-42,-55).
Help Deep in writing the positive factors of 2310 in pair and negative factor of 2310 in pair.
Factors of 2310 are 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, 2310. Positive factors of 2310 in pair are (1,2310), (2,1155), (3,770), (5,462), (6,385), (7,330), (10,231), (11,210), (14,165), (15,154), (21,110), (22,105), (30,77), (33,70), (35,66), (42,55). Negative factors of 2310 in pair are (-1,-2310), (-2,-1155), (-3,-770), (-5,-462), (-6,-385), (-7,-330), (-10,-231), (-11,-210), (-14,-165), (-15,-154), (-21,-110), (-22,-105), (-30,-77), (-33,-70), (-35,-66), (-42,-55).
Find the product of all factors of 2310.
Factors of 2310 are 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, 2310. So the product of all factors of 2310 would be 1 x 2 x 3 x 5 x 6 x 7 x 10 x 11 x 14 x 15 x 21 x 22 x 30 x 33 x 35 x 42 x 55 x 66 x 70 x 77 x 105 x 110 x 154 x 165 x 210 x 231 x 330 x 385 x 462 x 770 x 1155 x 2310 = 6.573424486790346e+53.
A student has been assigned the following tasks by the teacher:
- Finding out all positive factors of 2310.
- Writing all prime factors of 2310.
- Writing all the possible factors of 2310 in pair.
Help him in writing all these.
Positive factors of 2310 are 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, 2310.
Prime factors of 2310 are 2, 3, 5, 7, 11.
Factors of 2310 in pair are (1,2310), (2,1155), (3,770), (5,462), (6,385), (7,330), (10,231), (11,210), (14,165), (15,154), (21,110), (22,105), (30,77), (33,70), (35,66), (42,55).
Can you help Sammy list the factors of 2310 and also find the factor pairs?
Factors of 2310 are 1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 21, 22, 30, 33, 35, 42, 55, 66, 70, 77, 105, 110, 154, 165, 210, 231, 330, 385, 462, 770, 1155, 2310.
Factors of 2310 in pair are (1,2310), (2,1155), (3,770), (5,462), (6,385), (7,330), (10,231), (11,210), (14,165), (15,154), (21,110), (22,105), (30,77), (33,70), (35,66), (42,55). |
# Class 8 NCERT Solutions – Chapter 3 Understanding Quadrilaterals – Exercise 3.2
• Last Updated : 15 Dec, 2020
### Question 1. Find x in the following figures.
#### Solution:
As we know , the sum of the measures of the external angles of any polygon is 360°.
(a) 125° + 125° + x = 360° ⇒ 250° + x = 360° ⇒ x = 110°
(a) 70° + 60° + x + 90° + 90° = 360° ⇒ 310° + x = 360° ⇒ x = 50°
### (i) 9 sides
#### Solution:
Measure of angles = 360°/ 9 = 40°
As in a regular polygon all interior angle are equal so all the exterior angles will also be equal.
### (ii) 15 sides
#### Solution:
Measure of angles = 360°/ 15 = 24°
### Question 3. How many sides does a regular polygon have if the measure of an exterior angle is 24°?
#### Solution:
Given measure of exterior angle = 24°
(no. of sides) x (measure of exterior angle) = 360° ⇒ no. of sides = 360°/24° = 15
### Question 4. How many sides does a regular polygon have if each of its interior angles is 165°?
#### Solution:
Given measure of interior angle = 165°
measure of exterior angle = 180° – 165° = 15°
(no. of sides) x (measure of exterior angle) = 360° ⇒ no. of sides = 360°/15° = 24
### (b) Can it be an interior angle of a regular polygon? Why?
#### Solution:
(a) given here, measure of exterior angle = 22°
As we know, (no. of sides) x (measure of exterior angle) = 360°
putting here value, we get
No. of sides = 360°/22° ≈ 16.36 (approx), which is never possible .
Number of sides can never be fractional.
(b) No. It cannot be an interior angle of a regular polygon.
In this case , measure of exterior angle will be = 158°
again we will get fractional no. of sides. Hence not possible.
### (b) What is the maximum exterior angle possible for a regular polygon?
#### Solution:
(a) An Equilateral Triangle is a regular polygon with minimum number of sides because all sides are equal in it. We know that each angle of an equilateral triangle measures 60° . Hence, 60° is the minimum possible value for internal angle of a regular polygon.
(b) Each exterior angle of an equilateral triangle is 120° and hence this the maximum possible value of exterior angle of a regular polygon. This can also be proved by another principle; which states that each exterior angle of a regular polygon is equal to 360° divided by number of sides in the polygon. If 360° is divided by 3, we get 120° .
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# Quadratic equation class 10 formulas
## Quadratic equation class 10 formulas
a, b, c \in \mathbb{R} and a \ne 0 ,
ax^2 + bx + c = 0
Equations of the form are called quadratic equations with one unknown.
• x is the unknown of the equation, a , b and c are the coefficients of the equation.
• Quadratic equations can contain only the second ( x^2 ) and first order ( x ) powers of x .
• a is the leading number of the equation, c is the constant term.
• If $a = 0$the equation becomes a first-degree equation, so a cannot be zero.
• The expression $ax^2 + bx + c$is also a quadratic polynomial.
There are two methods we can use to find the solution set of quadratic equations.
• Factorization method: In this method, all terms are added on one side and the expression is factored (if it is separable). The values x that make these factors zero are solutions to the equation.
• Discriminant method: In this method, real or complex roots of all quadratic equations can be found with the discriminant formula.
• Graph method: If the graph of a function of the form f(x) = ax^2 + bx + c is given, the abscissa values of the points where this graph intersects the x axis are also ax^2 + bx + c = 0 are a root of the quadratic equation of the form.
### Factoring Methods
Example:
x^2 - 25 = 0
x^2 - 5^2 = 0
(x - 5)(x + 5) = 0
Solution set: x \in \{ -5, 5 \}
## Discriminant (Delta) of Quadratic Equations
Formula: \Large x_{1, 2} = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Example:
Find the roots of the equation x^2 - 20x + 99 = 0.
x_{1, 2} = \frac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 99}}{2 \cdot 1}
= \dfrac{20 \pm \sqrt{400 - 396}}{2}
= 10 \pm 1
Solution set: x \in \{9, 11\}
### Note on Discriminant Roots of Quadratic Equations
• In the case of \Delta \gt 0 , \sqrt{\Delta} is also a positive real number, so two roots of different real numbers are formed.
• In case of \Delta = 0, \sqrt{\Delta} would also be zero, hence a single real number root.
• In the case of \Delta \lt 0 the result of \sqrt{\Delta} is a complex number, not a real number, so two complex roots are conjugated to each other. |
# Understanding Whole Numbers Lesson 1-1. Vocabulary standard form – a number is written using digits and place value (the regular way to write numbers).
## Presentation on theme: "Understanding Whole Numbers Lesson 1-1. Vocabulary standard form – a number is written using digits and place value (the regular way to write numbers)."— Presentation transcript:
Understanding Whole Numbers Lesson 1-1
Vocabulary standard form – a number is written using digits and place value (the regular way to write numbers). expanded form – a number is written as a sum using the place and value of each digit.
Place Value Chart
How To Read a Large Number Numbers are grouped in sets of three (each set is called a period). Only read three numbers at a time. Say the name of the period that the numbers are in. Say “and” for the decimal, but do not say “and” if there isn’t a decimal.
Example 4,658,089 Millions periodThousands periodOnes period Four million, six hundred fifty-eight thousand, eighty-nine.
Number Lines Numbers towards the right on a number line are larger. As you move to the left on a number line, the numbers get smaller. What’s Bigger? 1 or -2? 1 is larger because it is to the right of the -2. What numbers are smaller than -2? -3 and -4 are both smaller than -2 because they are to the left of -2.
Comparing Numbers Line up the numbers vertically (up and down) by the ones place (or the decimal, if there is one). Start at the left and compare the digits. Move towards the right until you find a difference.
Just a Reminder… < means “less than.” > means “greater than.” = means “equal to.”
Example 45,312 45,321 45,312 45,321 1 is less than 2 <
Example 2 – Put the numbers in order from least to greatest. 321; 345; 354; 29; 1,013; 312; 332 321 345 354 29 1013 312 332 largest smallest 29 1,013 312321332 345354 <<< <<
Homework Time Commonly Misspelled Numbers: hundred thousand eight forty ninety
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# Question Video: Finding the Equations of Two Tangents to a Given Circle in Which They Make a Given Angle with a Coordinate Axis Mathematics
Find the equations of the two tangents to the circle π₯Β² + π¦Β² = 125 that are inclined to the positive π₯-axis by an angle whose tangent is 2.
06:32
### Video Transcript
Find the equations of the two tangents to the circle π₯ squared plus π¦ squared equals 125 that are inclined to the positive π₯-axis by an angle whose tangent is two.
Letβs think about what we actually know here. The circle with equation π₯ squared plus π¦ squared equals 125 has a center at the origin zero, zero. Whatβs not entirely necessary to know its radius is equal to the square root of 125, or five root five units. What weβre actually interested in is finding the equations of the two tangents to the circle inclined to the positive π₯-axis by an angle whose tangent is two. Now these might look a little something like this.
But what does it mean if theyβre inclined to the positive π₯-axis by an angle whose tangent is two? Well, letβs suppose the angle is equal to π. Thinking about the right triangle, we know that the tangent ratio links the opposite side in this triangle to the adjacent. Specifically, tan π is equal to opposite over adjacent. This means the ratio of the opposite to the adjacent side is equal to two. We could then say that the opposite side is equal to two length units and the adjacent side is equal to one. Then the length of the opposite divided by the adjacent side is two as we required.
Of course, we could have had any multiple of these values, but weβll see why this is the best option in a moment. Now, since the two tangents are straight lines, we know we can use the formula π¦ minus π¦ one equals π times π₯ minus π₯ one to find their equations. Now here π is the slope of the line, whilst π₯ one, π¦ one is a point that they pass through. But of course, π, the slope, can be found by calculating rise over run or, equivalently, change in π¦ divided by change in π₯. Now, in this case, we can see that the change in π¦ is two and the change in π₯ is one. So the slope of the line is two divided by one which is equal to two.
Now, itβs no coincidence that the slope of our tangents are equal to the value of the tangent of the angle at which theyβre inclined by. In general, the slope of line is equivalent to the tangent of the angle of the slope. So we actually have the slope of both of our lines; theyβre two. But weβre going to need to find the points that they pass through. And so weβre going to go back to the equation of the circle; that is, π₯ squared plus π¦ squared equals 125.
We know that given an equation of a curve, we can find the slope of the curve at a given point by thinking about the derivative. So what weβre going to do is find the derivative function for the equation of our circle and then set this equal to two. This will tell us the points at which our tangents touch the circle. So how do we differentiate the equation π₯ squared plus π¦ squared equals 125? Weβre going to do it in fact term by term. First, letβs differentiate π₯ squared with respect to π₯. Using the general power rule for differentiation, we multiply by the exponent and then reduce that exponent by one. So its derivative is simply two π₯.
But what about the derivative of π¦ squared with respect to π₯? We use implicit differentiation here. This is a special version of the chain rule. Essentially, we differentiate with respect to π¦ and then multiply by dπ¦ by dπ₯. The derivative of π¦ squared with respect to π¦ is two π¦. So when we differentiate π¦ squared with respect to π₯, we get two π¦ times dπ¦ by dπ₯. Then, on the right-hand side, the derivative of 125 with respect to π₯ is equal to zero. So our expression for the derivative is two π₯ plus two π¦ dπ¦ by dπ₯ equals zero.
Letβs divide through by two and then make dπ¦ by dπ₯ the subject. We can subtract π₯ from both sides, and then we divide through by π¦. And that gives us that dπ¦ by dπ₯ is equal to negative π₯ over π¦. We want to find the points where the slope is equal to two because thatβs the slope of the tangents weβre interested in. So we set dπ¦ by dπ₯ equal to two. So two is negative π₯ over π¦. And if we rearrange to make π₯ the subject, multiplying by π¦ and then multiplying by negative one, we get negative two π¦ equals π₯.
This shows us the relationship between the π₯- and π¦-values at the two points where our lines meet the circle. This means then that we can substitute this into the equation for the circle and weβll find the two points where they meet. Replacing π₯ with negative two π¦ and our equation becomes negative two π¦ squared plus π¦ squared equals 125. Negative two π¦ squared is four π¦ squared, so the left-hand side becomes five π¦ squared. Then, we divide through by five. So π¦ squared is equal to 25.
Our final step will be to take the square root. Now, we need to think about both the positive and negative square root of 25. Since the square root of 25 is five, we can say that π¦ must be equal to positive or negative five. Then, we can substitute this back into the equation π₯ equals negative two π¦ to find the corresponding π₯-values. When we do, we find that π₯ is equal to negative or positive 10. In other words, our two tangents pass through the points 10, negative five and negative 10, five.
We now have the two points through which our two tangents pass and the value of their slope. So we can substitute all of this into our equation for a straight line. Beginning with the point 10, negative five β that will be this point here β we get π¦ minus negative five equals two times π₯ minus 10. Distributing the parentheses, this becomes π¦ plus five equals two π₯ minus 20. And then subtracting two π₯ and adding 20, we find the equation of our first line is π¦ minus two π₯ plus 25 equals zero.
Letβs now repeat this with the point negative 10, five; thatβs this one. Substituting in, we get π¦ minus five equals two times π₯ minus negative 10. That gives us π¦ minus five equals two π₯ plus 20. And rearranging once again and we find the equation of this line is π¦ minus two π₯ minus 25 equals zero. And so our equations are π¦ minus two π₯ plus 25 equals zero and π¦ minus two π₯ minus 25 equals zero. |
# Foirmlean cur-ris
Nuair a bhios sinn a' cur-ris no a' toirt-air-falbh cheàrnan canaidh sinn ceàrn dà-fhillte ris an toradh. Mar eisimpleir, 's e ceàrn dà-fhillte a th' ann an $$30^\circ + 120^\circ$$. Le àireamhair, obraichidh sinn a-mach:
$\sin (30^\circ + 120^\circ ) = \sin (210^\circ ) = - 0.5$
$\sin (30^\circ ) + \sin (120^\circ ) = 1.366\,(gu\,3\,id.)$
Tha seo a' sealltainn nach eil $$\sin (A + B)$$ co-ionann ri $$\sin A + \sin B$$. An àite sin, faodaidh sinn na co-ionannachdan a leanas a chleachdadh:
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
$\sin (A - B) = \sin A\cos B - \cos A\sin B$
$\cos (A + B) = \cos A\cos B - \sin A\sin B$
$\cos (A - B) = \cos A\cos B + \sin A\sin B$
• Bidh foirmlean airson cur-ris air an sgrìobhadh ann an riochd goirid:
• $\sin (A \pm B) = \sin A\cos B \pm \cos A\sin B$
• $\cos (A \pm B) = \cos A\cos B \pm \sin A\sin B$
Seo mar a bhios sinn a' dol an sàs sa cheist seo sa bheil dà phàirt:
Question
1. Sgrìobh $$75^\circ = 45^\circ + 30^\circ$$ agus obraich a-mach an luach mionaideach aig $$\sin 75^\circ$$
2. Obraich a-mach an luach mionaideach aig $$\cos \left( {\frac{{7\pi }}{{12}}} \right)$$
1. $$\sin 75^\circ = \sin (45 + 30)^\circ$$
A' cleachdadh an fhoirmle $$\sin (A + B)$$
$= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$
A' cleachdadh luachan mionaideach air am bu chòir fios a bhith agad:
$= \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} \times \frac{1}{2}$
$= \frac{{\sqrt 3 }}{{2\sqrt 2 }} + \frac{1}{{2\sqrt 2 }}$
$= \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
2. Bhon a tha $$\frac{{7\pi }}{{12}} = \frac{\pi }{3} + \frac{\pi }{4}$$ tha:
$\cos \left( {\frac{{7\pi }}{{12}}} \right) = \cos \left( {\frac{\pi }{3} + \frac{\pi }{4}} \right)$
A' cleachdadh an fhoirmle airson $$\cos (A + B)$$
$= \cos \frac{\pi }{3}\cos \frac{\pi }{4} - \sin \frac{\pi }{3}\sin \frac{\pi }{4}$
A' cleachdadh luachan mionaideach air am bu chòir fios a bhith agad:
$= \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}$
$= \frac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$ |
Question
# If four times the number of permutations of $n$ things taking $3$ at a time is equal to five times the number of permutations of $n - 1$ things taking $3$ at a time, find the value of $n$.
Hint: Use formula of permutation, $^n{P_r} = \frac{{n!}}{{(n - r)!}}.$
The number of permutations of $n$ things taking $3$ at a time $= {{\text{ }}^n}{P_3}.$
Similarly, the number of permutations of $n - 1$ things taking $3$ at a time $= {{\text{ }}^{n - 1}}{P_3}.$
$\Rightarrow 4{ \times ^n}{P_3} = 5{ \times ^{n - 1}}{P_3}$
We know that, $^n{P_r} = \frac{{n!}}{{(n - r)!}}$ using this we’ll get:
$\Rightarrow 4 \times \frac{{n!}}{{(n - 3)!}} = 5 \times \frac{{(n - 1)!}}{{(n - 4)!}} \\ \Rightarrow \frac{{4n(n - 1)!}}{{(n - 3)(n - 4)!}} = \frac{{5(n - 1)!}}{{(n - 4)!}} \\ \Rightarrow \frac{{4n}}{{n - 3}} = 5 \\ \Rightarrow 4n = 5n - 15 \\ \Rightarrow n = 15 \\$
Therefore, the required value of $n$ is $15$ |
# RD Sharma Chapter 13 Class 9 Maths Exercise 13.4 Solutions
RD Sharma Chapter 13 Class 9 Maths Exercise 13.4 Solutions is about the Equations of lines parallel to the x-axis and y-axis. Any point in the pattern(x, 0), where x is the real number, lies on the x-axis because the y-coordinate of every spot on the x-axis is zero. The equation of the x-axis is y= 0. Likewise, the equation of the y-axis is x= 0 as the x-coordinate of every spot on the y-axis is zero. Practicing questions based on the graph is comparatively from the other problems. Also, students can obtain the marks more easily as these questions get solved quickly. It is effortless to score well in the exam with the help of problems on graphs.
We have attached the RD Sharma Chapter 13 Class 9 Maths Exercise 13.4 Solutions PDF to practice with the various types of questions related to the x-axis and the y-axis. The PDF is prepared by our experts for the students in which the solutions of problems are provided in a stepwise and easy manner.
Learn about RD Sharma Chapter 13 (Linear Equations In Two Variables) Class 9
## Download RD Sharma Chapter 13 Class 9 Maths Exercise 13.4 Solutions PDF
Solutions for Class 9 Maths Chapter 13 Linear Equations in Two Variables Exercise 13.4
## Important Definitions RD Sharma Chapter 13 Class 9 Maths Exercise 13.4 Solutions
In the following points, we will discuss the Equation of a Line Parallel to the x-axis and the y-axis separately with definitions and examples.
### Equation of a Line Parallel to the x-Axis
To obtain the equation of the x-axis and a line parallel to the x-axis-
Consider ‘AB’ to be a straight line parallel to the x-axis at the distance ‘b’ units from it. Then, all points on a line ‘AB’ have the corresponding ordinate ‘b’. Thus, ‘AB’ can be recognized as the locus of a point at a distance ‘b’ from the x-axis, and all points on a line ‘AB’ meet the condition y = b.
If P(x, y) in any position on ‘AB’, then y = b.
Consequently, the equation of a straight line parallels to the x-axis at a distance ‘b’ from it is y = b.
The equation of the x-axis is y = 0 since the x-axis is parallel to itself at a distance of ‘0’ from it.
Or
Let P(x,y) be any position on the x-axis. Simply, for all positions of ‘P’, we shall have the same ordinate 0 or y = 0.
Therefore, an equation of the x-axis is y = 0.
If a straight line is parallel and down to the x-axis at a distance ‘b,’ then the equation is y = -b.
### Equation of a Line Parallel to the y-Axis
Here, we will study how to obtain the equation of the y-axis and an equation of a line parallel to the y-axis.
Consider ‘AB’ to be a straight line parallel to the y-axis at a distance ‘a’ units from it. Then, all points on a line AB have the same abscissa ‘a’. So, ‘AB’ can be recognized as the locus of the point at a distance ‘a’ from the y-axis, and all points on a line ‘AB’ meet the condition x = a.
If P(x, y) in any position on ‘AB’, then x = a.
Consequently, the equation of a straight line parallels to the y-axis at the distance ‘a’ from it is x = a.
The equation of the y-axis is x = 0 since the y-axis is the parallel to itself at a distance of ‘0’ from it.
Or
Let P (x, y) be any position on the y-axis. Then simply, for all positions of ‘P,’ we shall have the same abscissa 0 or, x = 0.
Accordingly, the equation of the y-axis is x = 0.
If a straight line is parallel and to the left of the x-axis at a distance ‘a’, then the equation is x = -a.
### Examples related to the Equations of lines parallel to the x-axis and y-axis of RD Sharma Chapter 13 Class 9 Maths Exercise 13.4 Solutions
Ques 1- Find an equation of a straight line parallels to the x-axis at the distance of 10 units above the x-axis.
Ans- We know that an equation of a straight line parallel to the x-axis at a distance ‘b’ from it is y = b. Hence, an equation of a straight line parallels to the x-axis at a distance of 10 (ten) units above the x-axis is y = 10.
Ques 2- Find an equation of a straight line parallels to the y-axis at the distance of 3 units on the left-hand side of the y-axis.
Ans- We know that an equation of a straight line is parallel, and to the left of the x-axis at a distance ‘a’, then the equation is x = -a. Hence, an equation of the straight line parallel to the y-axis at the distance of 3 units on the left-hand side of the y-axis is x = -3. |
In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.2 Pdf, These solutions are solved subject experts from the latest edition books.
## MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Let the number of boys = x and Number of girls = y
∴ x + y = 10 …. (1)
Also, Number of girls = [Number of boys] + 4
∴ y = x + 4 ⇒ x – y = – 4 … (2)
Now, from the equation (1), we have :
l1 : x + y = 10 ⇒ y = 10 – x
And from the equation (2), we have
l2 : x – y = -4 ⇒ y = x + 4
The lines l1 and l2 intersects at the point (3, 7)
∴ The solution of the pair of linear equations is : x = 3, y = 7
⇒ Number of boys = 3
and number of girls = 7
Let the cost of a pencil = ₹ x
and cost of a pen = ₹ y
Since, cost of 5 pencils + Cost of 7 pens = ₹ 50
5 x + 7y = 50 … (1)
Also, cost of 7 pencils + cost of 5 pens = ₹ 46
7x + 5y = 46 …. (2)
Now, from equation (1), we have
And from equation (2), we have
Plotting the points (10, 0), (3, 5) and (-4, 10), we get a straight line l1 and plotting the points (8, -2), (3, 5) and (0, 9.2) we get a straight line l2.
These two straight lines intersects at (3, 5).
∴ Cost of a pencil = ₹ 3 and cost of a pen = ₹ 5.
Question 2.
On comparing the ratios $$\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}$$ and $$\frac{c_{1}}{c_{2}}$$ find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
Comparing the given equations with
a1x + b1y + C1 = 0, a2x + b2y + c2 = 0, we have
(i) For, 5x – 4y + 8 = 0, 7x + 6y – 9 = 0,
a1 = 5, b1 = -4, C1 = 8; a2 = 7, b2 = 6, c2 = -9
So, the lines are intersecting, i.e., they intersect at a point.
(ii) For, 9x + 3y + 12 = 0,18x + 6y + 24 = 0,
a1 = 9, b1 = 3, C1 = 12; a2 = 18, b2 = 6, c2 = 24
So, the lines are coincident.
(iii) For, 6x – 3y +10 = 0, 2x – y + 9 = 0
So, the lines are parallel.
Question 3.
On comparing the ratios $$\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}$$ and $$\frac{c_{1}}{c_{2}}$$, find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y = 8; 4x – 6y = 9
(iii) $$\frac{3}{2} x+\frac{5}{3} y$$ = 7; 9x-10y = 14
(iv) 5x – 3y = 11; -10x + 6y = -22
Solution:
Comparing the given equations with
So, lines are intersecting i.e., they intersect at a point.
∴ It is consistent pair of equations.
(ii) For 2x – 3y = 8, 4x – 6y = 9, 2x – 3y – 8 = 0 and 4x – 6y – 9 = 0
So, lines are parallel i.e., the given pair of linear equations has no solution.
∴ It is inconsistent pair of equations.
(iii)
⇒ The given pair of linear equations has exactly one solution.
∴ It is a consistent pair of equations.
(iv)
So, lines are coincident.
⇒ The given pair of linear equations has infinitely many solutions.
Thus, it is consistent pair of equations
(v)
So, the lines are coincident.
⇒ The given pair of linear equations have infinitely many solutions.
Thus it is consistent pair of equations.
Question 4.
Which of the following pairs of linear equations are consistent / inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5; 2x + 2y = 10
(ii) x – y = 8; 3x – 3y = 16
(iii) 2x + y – 6 = 0; 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
Solution:
(i) For, x + y = 5, 2x + 2y = 10 ⇒ x + y – 5 = 0 and 2x + 2y – 10 = 0
So, lines l1 and l2 are coinciding. i.e., They have infinitely many solutions.
(ii) For, x – y = 8,
3x – 3y = 16
⇒ x – y – 8 = 0
and 3x – 3y = 16
so, line are parallel
∴ The pair of linear equations are inconsistent
(iii) For 2x + y – 6 = 0, 4x – 2y – 4 = 0
The pair of linear equations are inconsistent
Question 5.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let the width of the garden = x m
and the length of the garden = y m
According to question,
The lines l1 and l2 intersect at (16, 20).
∴ x = 16 and y = 20
So, Length = 20m, and Width = 16m
Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Let the pair of linear equations be 2x + 3y – 8 = 0
⇒ a1 = 2, b1 = 3 and C1 = -8 and a2x + b2y + c2 = 0.
(i) For intersecting lines, we have:
∴ We can have a2 = 3, b2 = 2 and c2 = – 7
∴ The required equation can be 3x + 2y – 7 = 0
(ii) For parallel lines, $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
∴ Any line parallel to 2x + 3y – 8 = 0, can be taken as 2x + 3y -12 = 0
(iii) For coincident lines, $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
∴ Any line parallel to 2x + 3y – 8 = 0 can be 2(2x + 3y – 8 = 0)
⇒ 4x + 6y – 16 = 0
Question 7.
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
Since, we have
The lines l1 and l2 intersect at (2, 3). Thus, co-ordinates of the vertices of the shaded triangular region are (4, 0), (-1, 0) and (2, 3). |
# How to Figure the Number of Octagon and Square Tiles Needed
Octagon/Square Tile Calculator
Tile Side Length
Tile Spacing
(between octagon tiles)
Project Width
Project Length
Although you cannot tile a surface using only regular octagons, you can create a seamless repeating pattern using octagons and squares. This is called a truncated square tiling, which is an example of a semi-regular tessellation. An octagon-and-square tiling pattern is a popular choice for indoor and outdoor flooring projects because it breaks up the monotony of a simple square pattern.
One of the most commonly asked questions when installing octagon and square tiles is how many tiles are needed. In such a pattern, one regular octagon is used for every square, so the total number of squares and octagons will be equal. To figure out what this number is, you can either follow the math steps below or use the calculator on the left.
To use the calculator on the left, you must enter the length of the side of a tile, the spacing between adjacent octagon tiles, and the total width and length of the space to be tiled. Since the octagon and square tiles are assumed to have the same side length, the spacing between two adjacent octagons will be slightly wider than the spacing between octagon and square. See the diagrams below for more detail.
Step 1: First compute the approximate amount of 2-dimensional space that each octagon-and-square unit occupy in the tile arrangement. If the side length of one tile is T inches and the spacing between adjacent octagon tiles is S inches, then the spacing between an octagon and square tile is S/√2. Using the formulas for the area of a square and regular octagon, the total amount of space they take up is
(2 + 2√2)[T + S/√2]² + [T + S/√2]² square inches
= (3 + 2√2)[T + S/√2]² square inches.
The expression above can be approximated by the simpler formula
Tile Area = 5.8(T + 0.7S)² in².
Step 2: Compute the area of the space you are going to tile, in square inches. This is done by calculating the square footage and then multiplying that number by 12*12 = 144. (The conversion factor for sq.ft to sq.in.) For a rectangular surface with a length of L feet and a width of W feet, the area is
144LW square inches.
Step 3: Divide the number you obtained in Step 2 by the number you obtained in Step 1. This is the exact number of octagon-square tile units you would need to cover the area, including any tiles that need to be cut, without any wasted tiles.
This does not give you the most realistic estimate for number of octagon and square tiles needed, since inevitably you will end up wasting some material. For a better estimate, add 5%-7% to this number.
Alternatively, you can add two tile side lengths to both the width and length in Step 2. For example, suppose the area to be tiled has a width of 15'4" and a length of 20'7", and suppose the side length of a tile is 5". Then the adjusted values of L and W are
L = 20'7" + 5" + 5" = 21'5"
W = 15'4" + 5" + 5" = 16'2"
This method is employed by the calculator above.
Example: You are tiling a floor space that is 14' wide and 22' long. You are using regular octagon and square tiles that have a side length of 3" and a spacing of 1/4" (0.25") between adjacent octagon tiles.
First, you use Step 1 to figure the amount of space each tile needs. This is approximately:
5.8(3 + 0.7*0.25)² = 58.46 in²
Next, the total square inches of the floor space to be tiled is 144*14*22 = 44352 in².
Now divide the first number into the second to compute the exact number of octagon-square tile units needed: 44352/58.46 = 759.
Finally, add about 6% more to this figure: 759 + 0.06*759 = 805. This means you need 805 square tiles and 805 octagon tiles, for a total of 1610 tiles. |
These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. This video is all about Solving Routine and Non- Routine Problems Involving Factors, Multiples and Divisibility Rules for 2, 3, 4, 5, 6, 8, 9, 10, 11, and 12 These division story problems deal with only whole divisions (quotients without remainders.) In the given number 504, twice the digit in one's place is, The number formed by the digits except the digit in one's place is, The difference between twice the digit in one's place and the number formed by the other digits is. This partitioning operation is the key concept in many division story problems, and students may need to have an example explained either on paper or with manipulatives before they're ready to tackle a worksheet independently. These worksheets will produce ten problems per worksheet. Problems Involving Divisibility Rules: Difficult Problems with Solutions. Questions and their detailed solutions on divisibility are presented. Now, it is clear that 41295 is divisible by both 3 and 5. Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems. Word Problem Lesson The following video is about writing an equation and finding the dimensions of a rectangle knowing the perimeter and some information about the about the length and width. The divisibility rule for 3 is equivalent to choosing . The division of integers by integers could be written in different forms. After how many days both salesman will be again in New York on same day? Divisibility Rules For 2, 3, 4, 5, 6, 8, 9, 10, And 12. Welcome to the Prime numbers, factors and multiples section at Tutorialspoint.com.On this page, you will find worksheets on divisibility of numbers, even and odd numbers, prime numbers, factors, greatest common factors and least common multiples of two or more numbers using prime factorization method, division method and so on. Normal. A number is divisible by 2 if its last digit is even or the last digit is 0,2,4,6,or 8. Test skills with the revision worksheets. Next lesson. Fun maths practice! De Morgan's Laws ... and some of the most beautiful and elegant proofs bring apparently unrelated parts of mathematics together to solve a problem. C. Assessment. Logical Operations; 2. Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. So, the given number 504 is divisible by 7. Problem 1. Use the data below. In the number 328, the number formed by last two digits is 28 which is divisible by 4. Divisibility This is a complete lesson with explanations and exercises about the concept of divisibility, and about factors, divisors, and multiples, meant for fourth grade math. Divisibility rules and examples showing how to use the rules. According to the rule, if the sum of the digits in a number is a multiple of 3, then it is divisible by 3. Divisibility rules help us work out whether a number is exactly divisible by other numbers (i.e. Multiplying or dividing to solve word problems involving multiplication comparison; Interpreting remainders in division word problems; Solving word problems with four operations; Solving two-step problems; inding factors for whole numbers between 1 and 100; Identifying factors and multiples; Identifying prime numbers; Prime factorization 1. The rules are shortcuts for finding out whether numbers are exactly divisible without doing division calculations. Now, it is clear that 5832 is divisible by both 2 and 3. Create a problem involving factors, multiples and divisibility rules for each set of . Using Prime Factors to Find the Least Common Multiple of Two Numbers A salesman goes to New York every 15 days for one day and another every 24 days, also for one day. Because 18 is a multiple of 3, 5832 is divisible by 3. This lesson also lets students explore the concept of divisibility by 3, … One step equation word problems Because 5832 is an even number, it is divisible by 2. This is a great first step to recognizing the keywords that signal you are solving a division word problem. According to the rule, if a number is divisible by both 2 and 3, then it is divisible by 6. It comprises a divisibility rules chart for divisors 2-12 and adequate exercises to apply these rules involving single and multiple divisors. Division word problems can be some of the more confusing problems for students to understand. Directions: Use the rules of divisibility to answer the following word problems. Master the art of dividing lengthy numbers in a jiffy with this array of printable worksheets on divisibility tests for children of grade 3 through grade 6. Solving linear equations using elimination method, Solving linear equations using substitution method, Solving linear equations using cross multiplication method, Solving quadratic equations by quadratic formula, Solving quadratic equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic equations, Complementary and supplementary worksheet, Complementary and supplementary word problems worksheet, Sum of the angles in a triangle is 180 degree worksheet, Special line segments in triangles worksheet, Proving trigonometric identities worksheet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributive property of multiplication worksheet - II, Writing and evaluating expressions worksheet, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is proportional worksheet, Trigonometric ratios of some specific angles, Trigonometric ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric ratios of 180 degree plus theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios of 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonometric ratios of angles greater than or equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions, Domain and range of inverse trigonometric functions, Sum of the angle in a triangle is 180 degree, Different forms equations of straight lines, Word problems on direct variation and inverse variation, Complementary and supplementary angles word problems, Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Volume and Surface Area of Composite Solids Worksheet, Example Problems on Surface Area with Combined Solids. Solution :According to the rule, in a number, if the sum of the digits in odd places and sum of the digits in even places are equal or they differ by a number divisible by 11, then the number is divisible by 11. Two or more step word problem involving multiplication of fractions. a. Now, it is clear that 8520 is divisible by both 3 and 4. Word Problems Involving Addition, Subtraction, Multiplication, and Division: Word Problems with Multiplication and Division: Divisibility Rules for 2, 3, 5, 9, and 10: Division by 10 and 100: Dividing Negative Numbers: Dividing Decimals by Powers of Ten: Word Problems Involving Quotients and Differences: For example integer n divided by integer m yields a remainder r is written as: n = q m + r n is the dividend, q is the quotient, m is the divisor and r is the remainder. Geometry Word Problems Involving Area Girl Scouts . If a division is exact (there is no remainder), then we say a number is divisible by another. A number a is divisible by another number b if the division a ÷ b is exact (no remainder). Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. In this algebra lesson, students work in groups as they solve division problems. According to the rule, if a number is divisible by both 3 and 5, then it is divisible by 15. The lesson also reviews the divisibility rules for 2, and 5, and 10. View Module 8 Solving Word Problem Involving Division of Fractions.pdf from MATH 2014 at University of the Philippines Diliman. Factors and multiples. So, the given number 4328 is divisible by 8. Create some word problems involving factors and multiples. Divisibility rule 2 for 7 says that for , . For instance, 8596742 is divisible by 2 because the last digit is 2. Divisibility rules worksheets. As with the other word problem worksheets here, these division worksheets start with very basic problems to help surmount this challenge. Now, it is clear that 1458 is divisible by both 2 and 9. The divisibility rule for 11 is equivalent to choosing . 1. Quantifiers; 3. Use the work space provided below. Challenge students to answer these questions involving multiplying decimals, finding patterns, and understanding divisibility rules. Create a word problem involving multiplication of fractions for each of the following. According to the rule, if the last digit of a number is either 0 or 5, then it is divisible by 5. Some rules give us quick guidance as to whether one number is divisible by another. Customary units word problems - length 2.MD.B.5 - Use addition and subtraction within 100 to solve word problems involving lengths that are given in the same units. The why of the 9 divisibility rule. Today, both are in New York. But, the number formed by the last three digits is 328 which is divisible by 8. According to the divisibility rule for 7, in a number, if the difference between twice the digit in one's place and the number formed by other digits is either zero or a multiple of 7, then the number is divisible by 7. According to the rule, if the sum of the digits in a number is a multiple of 9, then it is divisible by 9. The why of the 3 divisibility rule. In 8520, the number formed by the last two digits is 20 which is divisible by 4. They have 342 guests that plan to attend. In this divisibility activity, students use the rules for divisibility to help them solve the division word problems. Our mission is to provide a free, world-class education to anyone, anywhere. This is the currently selected item. This Divisibility: Problem Solving Worksheet is suitable for 5th - 6th Grade. Question 1: 2.2 Divisibility [Jump to exercises] Collapse menu 1 Logic. According to the rule, if the last two digits of a number are zeroes or the number formed by the last two digits is a multiple of 25, then the number is divisible by 25. Because 15 is a multiple of 3, 8520 is divisible by 3. Multi-Step All Operations Word Problems These Word Problems worksheets will produce word problems involving all basic operations. Improve your math knowledge with free questions in "Divisibility rules: word problems" and thousands of other math skills. Which of the following numbers is divisible by both 2 and 3? You may choose the format of the answers. According to the rule, in a number, if the digit in one's place is 0, then it is divisible by 10. Students complete four word problems. information given below. Many division problems will use words like 'share among' or 'give to each' or similar phraseology to imply that a total amount is to be split evenly into groups. number of tray. Problems Involving Divisibility Rules. Rule # 2: divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. Word problems on fractions. ... Students solve problems using the divisibility rule. Acceptable answers: yes, no. Problem 2. Word problems on mixed fractrions. According to the divisibility rule for 6, in a number, if the last three digits are zeros or the number formed by the last 3 digits is divisible by 8, then the number is divisible by 8. Division word problems can be some of the more confusing problems for students to understand. According to the rule, if a number is divisible by both 3 and 4, then it is divisible by 12. John and Susan are getting married! Once these division concepts are well understood, proceeding to worksheets that mix multiplication and division, and even addition and subtraction, can provide excellent practice to make sure students understand how each operation needs to be chosen appropriately based on the problem description. Is 215496 divisible by 6? In 1782, the sum of the digits in odd places : In 1782, the sum of the digits in even places : In 1782, the sum of the digits in odd places and sum of the digits in even places are equal. put in a small tray. Trigonometry word problems. According to the rule, if a number is divisible by both 2 and 9, then it is divisible by 18. there is no remainder). Double facts word problems. Directions: Using the rules of divisibility, choose one digit a … Name Monthly Salary Monthly Savings Randy Php 12,000 Ryan Php 15, 000 63. Free, printable worksheets from K5 Learning. Many division problems will use words like 'share among' or 'give to each' or similar phraseology to imply that a total amount is to be split evenly into groups. Because 1458 is an even number, it is divisible by 2. Because 21 is a multiple of 3, 41295 is divisible by 3. Practice: Divisibility tests. PROBLEMS ON DIVISIBILITY RULES Problem 1 : Check whether 18 is divisible by 2. Leap year is only every 4 years! According to the rule, if a number is an even number, then it is divisible by 2. Because 18 is a multiple of 9, 1458 is a divisible by 9. These rules can also be found under the appropriate conditions in number bases other than 10. Improve your skills with free problems in 'Divisibility rules: word problems' and thousands of other practice lessons. A lesson with explanations and exercises about the concept of divisibility, for third grade. Learn the divisibility rules for 2, 3, 4, 5, 6, … Easy. Divisibility rules lesson plans and worksheets from thousands of teacher-reviewed resources to help you inspire students learning. Apply Your Skills! In the given number 4328, the last three digits are not zeroes. 30 bananas, 60 chicos, and 120pears. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. 2. Rule #1: divisibility by 2. 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Question Video: Finding the Length of a Projection of a Right-Angled Triangle Side on the Straight Line Carrying Another Side Using Pythagoras’s Theorem | Nagwa Question Video: Finding the Length of a Projection of a Right-Angled Triangle Side on the Straight Line Carrying Another Side Using Pythagoras’s Theorem | Nagwa
# Question Video: Finding the Length of a Projection of a Right-Angled Triangle Side on the Straight Line Carrying Another Side Using Pythagorasβs Theorem Mathematics • Second Year of Preparatory School
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Given that π΄π΅ = 29, πΆπ΅ = 20, and πΆπ· = 35, calculate the length of the projection of line segment πΆπ· on line π΄π·.
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### Video Transcript
Given that π΄π΅ equals 29, πΆπ΅ equals 20, and πΆπ· equals 35, calculate the length of the projection of line segment πΆπ· on line π΄π·.
Before we can calculate the length of this projection, weβll need to see which line segment this projection is. Line π΄π· is our target line. Thatβs the line upon which the projection will fall. And πΆπ· is the line segment that weβll use to create our projection. But in order to create a projection, weβll need perpendicular lines to our target line. Because we know that line π΅πΆ and line π΄π· are parallel, the angle π΅πΆπ΄ is an alternate interior angle to the angle πΆπ΄π·, which means the line segment π΄πΆ is perpendicular to the line segment π΄π·.
Once we imagine the light source as the set of all the perpendicular lines to our target line, we can find the endpoints of our projection. Starting at πΆ, the projection of point πΆ onto π΄π· would be the point π΄ and the projection of point π· onto π΄π· would be itself. And that means the projection of πΆπ· is going to be equal to π΄π·. This is the value we want to calculate the length of. To do that, letβs clear out our light source lines and think about what we know about right triangles.
First of all, we know that π΄π΅ has a measure of 29 and πΆπ΅ has a measure of 20 and πΆπ· has a measure of 35. What we have in our figure is two separate right triangles that make up this quadrilateral. And so we remember that we can find side lengths in right triangles using the Pythagorean theorem, where π and π represent the two smaller sides and π represents the hypotenuse. π squared plus π squared equals π squared. To use the Pythagorean theorem, you need at least two of the lengths. Since we donβt know the length of π΄πΆ, we canβt find the length of π΄π·. However, we can use the information about the smaller triangle π΄π΅πΆ to solve for the side length π΄πΆ first.
When we plug in what we know, we get 20 squared plus π΄πΆ squared equals 29 squared. 400 plus π΄πΆ squared equals 841. So weβll subtract 400 from both sides of the equation, and weβll get π΄πΆ squared equals 441. After that, weβll take the square root of both sides. Weβre only interested in the positive square root since weβre dealing with distance, and so we see that π΄πΆ equals 21. Now that we know that π΄πΆ is 21, we know two distances in our right triangle and weβll be able to find the side length of our third distance π΄π·. But weβll need to set up the Pythagorean theorem for a second time.
This time, weβll have 21 squared plus π΄π· squared equals 35 squared. 441 plus 80 squared equals 1225. And so we subtract 441 from both sides, and weβll get π΄π· squared equals 784. So we take the square root of both sides. Again, weβre only interested in the positive square root of 784, which is 28. The line segment π΄π· will be equal to 28. The line segment π΄π· is the projection of the line segment πΆπ· onto line π΄π·, and it has a measure of 28.
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# a poster is 25cm taller than it is wide. it is mounted on a piece of cardboard so that there is a 5cm border on all sides. if the area of the border alone 1350cm^2, what are the dimensions of the poster?
It is almost a must to draw a diagram but I can't do that on the computer. It will help you to do so.
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height of inside poster = w+25
width of inside poster = w
area of inside poster =w*(w+25)
Border is 5 cm on each side; therefore, its height is w+25+top5cm+bottom 5cm=w+35
Border width = w + left 5 cm + right 5 cm = w + 10 cm.
area of entire unit is (w+35)*(w+10)
area of inside poster is w*(w+25)
The border is 1350 cm2. Set this into an equation.
area entire unit-1350=area inside unit.
(w+35)*(w+10)-1350 = w(w+25)
Solve for w, which is the width of the inside poster. Add 25 to that to obtain the height. Then add 10 to each to obtain the dimensions of the entire frame. Post your work if you get stuck.
## To find the dimensions of the poster, we can follow these steps:
1. Let's assume the width of the poster is 'w' cm.
2. Since the poster is 25 cm taller than its width, the height of the poster would be (w + 25) cm.
3. The border on each side is 5 cm, so the actual width of the poster, including the border, would be (w + 2*5) cm, which simplifies to (w + 10) cm.
4. Similarly, the actual height of the poster, including the border, would be ((w + 25) + 2*5) cm, which simplifies to (w + 35) cm.
5. The area of the border alone is given as 1350 cm².
6. We can set up an equation for the area of the entire unit (including the border) and subtract the area of the poster to get the area of the border alone: (w + 35) * (w + 10) - w * (w + 25) = 1350 cm².
- The first term, (w + 35) * (w + 10), represents the area of the entire unit (including the poster and the border).
- The second term, w * (w + 25), represents the area of the poster.
7. Simplify and solve the equation for 'w' to find the width of the poster.
8. Add 25 to the width to find the height of the poster.
9. Add 10 to the width and 35 to the height to find the dimensions of the entire frame.
By following these steps, you should be able to determine the dimensions of the poster. |
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# Kids Math
## Glossary and Terms: Measurement
Area - The measurement of the surface of an area is defined in square units. The area is the number of squares that fit inside the region. The formula for determining the area of a rectangle is length x width.
Example: Area = 5 x 6 = 30 square units, where the length = 5 units and the width = 6 units
Capacity - The capacity is the measurement of how much something will hold. It is also the volume of an object.
Cubed - Cubed is a three dimensional measurement of a cube with all three sides the same size. It is represented by a small 3 next to the number like X³.
Example: 5 cubed = 5³ = 5 x 5 x 5 = 125
Degree - A unit of measure used for angles.
Example: A circle is divided up into 360 degrees. A right angle is 90 degrees.
Dimension - In measurement, the dimensions are represented by length, width, and height.
Distance - The length between two points as drawn by a straight line.
Example: It is 5 miles from my house to my school. The distance is 5 miles.
Estimate - When guessing a measurement close to the exact measurement, this is called an estimate.
Height - The measurement of the vertical distance.
Example: That girl is 5 feet tall. Her height is 5 feet.
Length - A measurement of distance. It is usually the longer of two distances of an object, the other being the width.
Example: A football field is 100 yards long. Its length is 100 yards.
Measurement - Determining the physical quantity of something such as length, time, temperature, or volume in terms of a unit of measurement such as feet, seconds, or degrees.
Perimeter - The perimeter is the path that surrounds an area. The distance of the path is the measurement of the perimeter.
Example: The perimeter of this polygon is 3 + 7 + 5 + 4 + 6 + 4 = 29.
Rounding - Rounding a number or measurement is when you write the number in a shorter or simpler form that is an approximation of the exact number or measurement.
Example:
671.98 rounded to the nearest 1s place is 672.
5,213 rounded to the nearest 100s place is 5,200.
Ruler - A ruler is a tool used to measure distance. It can also help to draw straight lines.
Surface Area - The total surface area of three dimensional object is the added areas of all the objects surfaces (for a cube this would be 6 sides).
Example: The total surface area of a cube with side length 4 would be 4x4 = 16 (for each side). Then multiply by 6 for the six sides: 16x6 = 96.
Temperature - The measurement of an object that expresses its hotness or coldness. The units of measurement are degrees Fahrenheit or degrees Celsius.
Example: The oven was hot, it was 350 degrees F.
Thermometer - A device for measuring temperature.
Time - A measurement of the dimension that represents intervals of the past, present, and future. Units of time include seconds, minutes, and hours.
Unit - A unit of measurement is a standard by which measurements can be compared. Different types of measurements use different units.
Example: Inches are a unit of measurement used to measure distance. Other units of measurement include seconds (time), degrees (temperature), and grams (mass).
Volume - The quantity of space that an object takes up. It is measured in cubic units.
Example: The volume of a box is equal to the length x width x height.
Weight - The measurement of the pull of gravity on an object. Typical units include pounds, ounces, and tons.
Width - The measurement of the distance of a side of an object. Usually this is the shorter side while the length is the longer side.
More Math Glossaries and Terms
Algebra glossary
Angles glossary
Figures and Shapes glossary
Fractions glossary
Graphs and lines glossary
Measurements glossary
Mathematical operations glossary
Probability and statistics glossary
Types of numbers glossary
Units of measurements glossary
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# Problem of the Week Problem D and Solution Mangoes and Oranges
## Problem
At POTW’s Supermarket, Livio stocks mangoes and Dhruv stocks oranges. One day they noticed that an equal number of mangoes and oranges were rotten. Also, $$\frac{2}{3}$$ of the mangoes were rotten and $$\frac{3}{4}$$ of the oranges were rotten. What fraction of the total number of mangoes and oranges was rotten?
## Solution
Solution 1:
Let the total number of mangoes be represented by $$a$$ and the total number of oranges be represented be $$b$$. Since there were an equal number of rotten mangoes and rotten oranges, then $$\frac{2}{3}a=\frac{3}{4}b$$, so $$b=\frac{4}{3}(\frac{2}{3}a)=\frac{8}{9}a$$.
Therefore, there were a total of $$a+b=a+\frac{8}{9}a=\frac{17}{9}a$$ mangoes and oranges.
Also, the total amount of rotten fruit was $$2(\frac{2}{3}a)=\frac{4}{3}a$$.
Therefore, $$\frac{\frac{4}{3}a}{\frac{17}{9}a}=\frac{4}{3} \left( \frac{9}{17}\right) = \frac{12}{17}$$ of the total number of mangoes and oranges was rotten.
Solution 2:
Since $$\frac{2}{3}$$ of the mangoes were rotten, $$\frac{3}{4}$$ of the oranges were rotten, and the number of rotten mangoes equaled the number of rotten oranges, suppose there were $$6$$ rotten mangoes. (We choose $$6$$ as it is a multiple of the numerator of each fraction.) Then the number of rotten oranges will also be $$6$$.
If there were 6 rotten mangoes, then there were a total of $$6 \div \frac{2}{3}= 6\left( \frac{3}{2}\right)=9$$ mangoes.
If there were 6 rotten oranges, then there were a total of $$6 \div \frac{3}{4}= 6\left( \frac{4}{3}\right)=8$$ oranges.
Therefore, there were $$9+8 =17$$ pieces if fruit in total, of which $$6+6 = 12$$ were rotten.
Thus, $$\frac{12}{17}$$ of the total number of mangoes and oranges was rotten.
Note: In Solution 2 we could have used any multiple $$6$$ for the number of rotten mangoes and thus the number of rotten oranges. The final fraction would always reduce to $$\frac{12}{17}$$. We will show this in general in Solution 3.
Solution 3:
According to the problem, there were an equal number of rotten mangoes and rotten oranges.
Let the number of rotten mangoes and rotten oranges each be $$6x$$, for some positive integer $$x$$.
The total number of mangoes was thus $$6x \div \frac{2}{3} = 9x$$.
The total number of oranges was thus $$6x \div \frac{3}{4} = 8x$$.
Therefore, the total number of mangoes and oranges was $$9x + 8x = 17x.$$
Also, the total number of rotten mangoes and rotten oranges was $$6x + 6x = 12x$$.
Therefore, $$\frac{12x}{17x}= \frac{12}{17}$$ of the total number of mangoes and oranges was rotten. |
# How do you factor completely 12x^3y + 6x^2y^2 - 9xy^3?
Feb 19, 2017
$12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3} = \frac{3}{4} x y \left(4 x + \left(1 - \sqrt{13}\right) y\right) \left(4 x + \left(1 + \sqrt{13}\right) y\right)$
#### Explanation:
Given:
$12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3}$
Note that all of the terms are of degree $4$ and all are divisible by $3 x y$:
$12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3} = 3 x y \left(4 {x}^{2} + 2 x y - 3 {y}^{2}\right)$
We can factor the remaining quadratic by completing the square as follows:
$4 {x}^{2} + 2 x y - 3 {y}^{2} = \frac{1}{4} \left(16 {x}^{2} + 8 x y - 12 {y}^{2}\right)$
$\textcolor{w h i t e}{4 {x}^{2} + 2 x y - 3 {y}^{2}} = \frac{1}{4} \left({\left(4 x\right)}^{2} + 2 \left(4 x\right) \left(y\right) + {y}^{2} - 13 {y}^{2}\right)$
$\textcolor{w h i t e}{4 {x}^{2} + 2 x y - 3 {y}^{2}} = \frac{1}{4} \left({\left(4 x + y\right)}^{2} - {\left(\sqrt{13} y\right)}^{2}\right)$
$\textcolor{w h i t e}{4 {x}^{2} + 2 x y - 3 {y}^{2}} = \frac{1}{4} \left(\left(4 x + y\right) - \sqrt{13} y\right) \left(\left(4 x + y\right) + \sqrt{13} y\right)$
$\textcolor{w h i t e}{4 {x}^{2} + 2 x y - 3 {y}^{2}} = \frac{1}{4} \left(4 x + \left(1 - \sqrt{13}\right) y\right) \left(4 x + \left(1 + \sqrt{13}\right) y\right)$
Putting it all together:
$12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3} = \frac{3}{4} x y \left(4 x + \left(1 - \sqrt{13}\right) y\right) \left(4 x + \left(1 + \sqrt{13}\right) y\right)$
Feb 19, 2017
${3}^{2} x y \left(y - \frac{x}{3} \left(1 - \sqrt{13}\right)\right) \left(y - \frac{x}{3} \left(1 + \sqrt{13}\right)\right)$
#### Explanation:
$f \left(x , y\right) = 12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3}$ is a homogeneous function
Making $y = \lambda x$ and substituting
$f \left(x , \lambda\right) = 3 {x}^{4} \lambda \left(4 + \left(2 - 3 \lambda\right) \lambda\right)$
Solving $\lambda \left(4 + \left(2 - 3 \lambda\right) \lambda\right) = 0$ we have
$\lambda = \left\{0 , \frac{1}{3} \left(1 - \sqrt{13}\right) , \frac{1}{3} \left(1 + \sqrt{13}\right)\right\}$ or
$f \left(x , \lambda\right) = {3}^{2} {x}^{4} \lambda \left(\lambda - \frac{1}{3} \left(1 - \sqrt{13}\right)\right) \left(\lambda - \frac{1}{3} \left(1 + \sqrt{13}\right)\right)$ so
$f \left(x , y\right) = {3}^{2} x y \left(y - \frac{x}{3} \left(1 - \sqrt{13}\right)\right) \left(y - \frac{x}{3} \left(1 + \sqrt{13}\right)\right)$ |
# Quantitative Comparison and Manipulation
Many quantitative questions have variables in both columns. While your first instinct may be to work algebraically, this strategy is not always best. Often the fastest way to a solution is by plugging in different values to see which column is greater.
Developing a sense of when to plug in and when to solve algebraically takes practice. Here are a few helpful guidelines when trying to determine which approach to use.
## Work algebraically if the question is a polynomial
If you are dealing with a polynomial, simplify. Plugging in may require too much calculation. Instead, work with the familiar algebraic forms shown below:
Now let’s take a look at a question.
Column A Column B
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
In Quantitative Comparison we can make it so each side is equal. Then we can balance the equation, adding and subtracting, multiplying and dividing, where necessary.
First, note that can be factored into . Now we can set both columns equal to each other:
At this point we have to be careful. While algebra tells us to divide both sides by (x – 2), we need to be aware of the following: if x is between 0 and 2, (x-2) yields a negative. However, (x + 2) yields a positive, meaning (x – 2)(x + 2) gives us a negative. In this case (A) would be bigger.
However, if we divide each side by (x – 2) and solve we get the following:
DIVIDE EACH SIDE BY (x – 2)
SUBTRACT ‘X’ FROM BOTH SIDES
This hardly looks like a solution (in fact it looks like I forgot to go to grade school!). However, what this yields is the important insight: Column B is now 4 greater than Column A. Therefore, the answer is (D). Alternatively you could pick numbers, such as ‘0’ and ‘4’. Each gives us different answers, leading to the same conclusion: Answer (D).
## If the question has variables, but there no polynomials, plug in values.
This advice pertains to variables that are not in polynomial form, as seen above. Here come up with easy numbers to plug in to see which values the columns yield.
0 > x > y > z > -1
Column A Column B
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
.
It would be very nice if we could just stop here and choose answer (B). However, things are not so simple. When plugging in one set of numbers we will inevitably come up with one outcome. Once we’ve plugged in and come up with one answer, whether it is (A), (B), or (C), our job is to disprove that answer.
Can we make Column A larger than Column B? Well what if we plug in a values for x and y that are very close to one another. . Now, for z we can plug in a value close to -1, say -8/9. This gives us . Now you can see (B) is much smaller. Therefore the answer is (D).
## Takeaway
You should be adept at both algebra and plugging-in to efficiently—and accurately—answer a quantitative comparison question that contains variables. Typically it is best to simply check for polynomials before plugging in. To test out your skills, try these GRE math questions for practice!
## Author
• Chris graduated from UCLA with a BA in Psychology and has 20 years of experience in the test prep industry. He’s been quoted as a subject expert in many publications, including US News, GMAC, and Business Because. |
250 is an also number and need to be a element of 2. Looking in ~ the number in the units place, i.e. 0, us can likewise find the 5 and 10 room its factors. In this lesson, let us discover to uncover the other factors the 250. When the list may appear long, it"s relatively simple come remember particularly when you identify the pair determinants of 250 or look at the aspect tree that 250.
You are watching: What are the factors of 250
Factors the 250: 1, 2, 5, 10, 25, 50, 125 and 250Prime factorization of 250: 2 × 5 × 5 × 5
1 What are the factors of 250? 2 How to Calculate determinants of 250? 3 Factors the 250 by prime Factorization 4 Factors the 250 in Pairs 5 FAQs-on-Factors-of-250
Factors the 250 are all the number which top top multiplying provide the value 250. 250 is an even composite number.
250 is a many of 2, 5 and also 10. For this reason 2, 5, 10 room the determinants of 250. Let"s uncover which every numbers multiply with each other to form 250.
The components of 250 are integers that divide 250 without any kind of remainder.
Let"s learn just how to calculate determinants of 250Let"s start dividing 250 v the natural numbers, starting with 1. 250 ÷ 1 = 250, Remainder = 0250 ÷ 2 = 125, Remainder = 0250 ÷ 5 = 50, Remainder = 0250 ÷ 10 = 25, Remainder = 0All the divisors and also the quotients thus acquired on division form the determinants of 250. They are 1, 2, 5, 10, 25, 50, 125 and 250.
Explore components using illustrations and also interactive examples
Prime factorization way to express a composite number together the product the its prime factors.
To acquire the prime factorization of 250, we divide it by its smallest prime aspect which is 2250 ÷ 2 = 125Now divide 125 through its the smallest prime factor.125 ÷ 5 = 25This process goes ~ above till we get the quotient as 1.25 ÷ 5 = 5 and also 5 ÷ 5 = 1The element factorization of 245 is shown below in 2 ways:
We stop the aspect tree as we can not branch more because 2 and 5 room the element numbers. Therefore, the prime factorization that 250 is 2 x 5 x 5 x 5.
These 4 prime factors can be multiplied in pairs in every the possible ways to get the factors of 250.
## Factors the 250 in Pairs
Now that we are familiar with the factors of 250, let"s write determinants of 250 in pairs.
The factor-pairs the 250 are (1, 250), (2, 125), (5, 50), (25, 10).
It is feasible to have an unfavorable pair factors as well because the product of two negative numbers additionally gives a hopeful number.
So, we have the right to have an adverse factor-pairs that 250 as (-1,-250), (-2,-125), (-5,-50), (-25, -10)
Annie is offered 2 numbers, 10 and also 250. She is listed with a few cards numbered 1, 2, 5, 10, 25, and 50. Can you help her in listing the typical factors the 10 and also 250?
Solution:
The factors of 10 room 1, 2, 5, and 10.
The factors of 250 space 1, 2, 5, 10, 25, 50, 125 and also 250.
The typical factors that 50 and also 250 space 1, 2, 5, and also 10.
Example 2:Jane bought a rectangle-shaped carpet because that the life room of she house. The area the the carpet is around 250 square feet. What room the possible dimensions that the carpet?
Solution:
The area of a rectangle is length × breadth sq units.
Given area = 250 square feet
So the possible length and breadth space the element pairs the give the product 250.
Length the carpet (in ft)Breadth that carpet (in ft)
1250
2125
550
1025
2510
505
2501
Therefore, there room 7 feasible dimensions of the carpet.
### What room the determinants of 250?
The components of 250 space 1, 2, 5, 10, 25, 50, 125 and 250.
### What space all the components of 250 and also 300?
The factors of 250 are 1, 2, 5, 10, 25, 50, 125 and 250.
The factors of 300 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150 and 300.
### Is 250 a perfect cube?
The cube root of 250 is not an integer, and therefore 250 is not a perfect cube.
See more: Telling People What They Want To Hear Or Tell Them The Truth?
### How many factors that 250 room there?
The factors of 250 space 1, 2, 5, 10, 25, 50, 125 and 250. We have actually 8 determinants of 250. |
# COMPOUND INTEREST PROBLEMS
Problem 1 :
\$800 is invested in compound interest where the rate of interest is 20% per year. If interest is compounded half yearly, what will be the accumulated value and compound interest after 2 years?
Solution :
The formula to find accumulated value in compound interest is
A = P(1 + r/n)nt
A ----> Accumulated value (final value)
P ----> Principal (initial value of an investment)
r ----> Annual interest rate (in decimal)
n ----> Number of times interest compounded per year
t ----> Time (in years)
Here,
P = 800
r = 20% = 0.2
n = 2
t = 2
Then, the accumulated value is
A = 800(1 + 0.1)4
A = 800(1.1)4
A = 800 1.4641
A = \$1171.28
Compound interest is
= A - P
= 1171.28 - 800
= \$371.28
Problem 2 :
A sum of money placed at compound interest doubles itself in 3 years. In how many years will it amount to four times itself?
Solution :
Let P be the amount invested initially.
From the given information,
P becomes 2P in 3 years
Because the investment is in compound interest, the principal in the 4th year will be 2P.
And 2P becomes 4P (it doubles itself) in the next 3 years.
Therefore, at the end of 6 years accumulated value will be 4P.
So, the amount deposited will amount to 4 times itself in 6 years.
Problem 3 :
The compound interest and simple interest on a certain sum for 2 years is \$ 1230 and \$ 1200 respectively. The rate of interest is same for both compound interest and simple interest and it is compounded annually. What is the principle?
Solution :
Simple interest for two years is 1200 and interest for one year is 600.
So, compound interest for 1st year is 600 and for 2nd year is 630.
(Since it is compounded annually, simple interest and compound interest for 1st year would be same)
When we compare the compound interest for 1st year and 2nd year, it is clear that the interest earned in 2nd year is 30 more than the first year.
Because, interest 600 earned in 1st year earned this 30 in 2nd year.
It can be considered as simple interest for one year.
That is, principal = 600, interest = 30.
I = Prt/100
30 = (600 ⋅ r ⋅ 1)/100
30 = 6r
5% = r
In the given problem, simple interest earned in two years is 1200.
I = Prt/100
1200 = (P 5 2)/100
1200 = P/10
Multiply each side by 10.
12000 = P
So, the principal is \$ 12,000.
Problem 4 :
Mr. David borrowed \$15,000 at 12% per year compounded annually. He repaid \$7000 at the end of 1st year. What amount should he pay at the end of second year to completely discharge the load?
Solution :
The formula to find accumulated value in compound interest is
A = P(1 + r/n)nt
To find the accumulated value for the first year,
Substitute P = 15000, r = 0.12, n = 1 and t = 1 in the above formula.
A = 15000(1 + 0.12/1)1x1
= 15000(1 + 0.12)1
= 15000(1.12)
= 16800
Amount paid at the end of 1st year is 7000.
Balance to be repaid :
= 16800 - 7000
= 9800
This 9800 is going to be the principal for the 2nd year.
Now we need the accumulated value for the principal 9800 in one year. (That is, at the end of 2nd year)
A = 9800(1 + 0.12/1)1x1
= 9800(1 + 0.12)1
= 9800 1.12
= 10976
So, to completely discharge the loan, at the end of 2nd year, Mr. David has to pay \$ 10,976.
Problem 5 :
There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of \$12,000 after 3 years at the same rate?
Solution :
Let the principal in simple interest be \$100.
Since there is 60% increase, simple interest = 60
We already know the formula for S.I.
That is,
I = Prt/100
Here, I = 60, P = 100, t = 6.
60 = (100 r 6)/100
60 = 6r
Divide each side by 6.
10% = r
Because the rate of interest is same for both S.I and C.I, we can use the rate of interest 10% in C.I.
To know compound interest for 3 years,
Substitute P = 12000, r = 0.1, n = 1 and t = 3 in the formula of C.I.
A = 12000(1 + 0.1/1)1x3
= 12000(1 + 0.1)3
= 12000(1.1)3
= 12000(1.331)
= 15972
Compound interest = A - P
= 15972 - 12000
= 3972
So, the compound interest after 3 years at the same rate of interest is \$3972.
Kindly mail your feedback to v4formath@gmail.com
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# Assessment for Learning Explain how you solved this problem. equipment to show your solution? STRAND CODE Year 2 Objectives Year 2 Year 2
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## Transcription
3 Understanding Shape Visualise common -D shapes and 3- D solids; identify shapes from pictures of them in different positions and orientations; sort, make and describe shapes, referring to their properties How do you know that this shape is a square? What is special about it? Two of these shapes are not hexagons. Which are they? Here are five identical triangles. Use some or all of the triangles to make a bigger triangle Is there another way to do it? Geometry shapes 1 identify and describe the properties of -D shapes, including the number of sides identify and describe the properties of 3-D shapes, including the number of edges, vertices and faces 3 identify -D shapes on the surface of 3-D shapes, for example a circle on a cylinder and a triangle on a pyramid 4 compare and sort common -D and 3-D shapes and everyday objects. Geometry position and direction 1.order and arrange combinations of mathematical objects in patterns.use mathematical vocabulary to describe position, direction and movement including distinguishing between rotation as a turn and in terms of right angles for quarter, half and threequarter turns (clockwise and anti-clockwise), and movement in a straight line.
6 Read the numbered divisions on a scale, and interpret the divisions between them (e.g. on a scale from to 5 with intervals of 1 shown but only the divisions 0, 5, 10, 15 and 0 numbered); use a ruler to draw and measure lines to the nearest centimetre Follow and give instructions involving position, direction and movement. [Point to 65 cm on a metre stick marked in centimetres and numbered in tens.] What measurement is this? [Point to half a litre on a 1 litre measuring jug.] What measurement is this? The tick is in square B5. Follow my instructions. Draw a cross in square D. Draw a circle in square E4. Draw a triangle in square A5. 1.choose and use appropriate standard units to estimate and measure length/height in any direction (m/cm); mass (kg/g); temperature ( C); capacity (litres/ml) to the nearest appropriate unit, using rulers, scales, thermometers and measuring vessels Geometry Position and direction 1.order and arrange combinations of mathematical objects in patterns Shape Now tell me where to put a cross, a circle and a triangle. use mathematical vocabulary to describe position, direction and movement including distinguishing between rotation as a turn and in terms of right angles for quarter, half and three-quarter turns (clockwise and anti-clockwise), and movement in a straight line.
8 Calculating Represent repeated addition and arrays as multiplication, and sharing and repeated subtraction (grouping) as division; use practical and informal written methods and related vocabulary to support multiplication and division, including calculations with remainders Look at these jumps on a number line. What does it show? How could you record that? Is there another way that you could record t? Show me on a number line how you could do: Show me on a number line how you could do: Look at these diagrams: 4.Solve problems involving multiplication and division, using materials, arrays, repeated addition, mental methods, and multiplication and division facts, including problems in contexts. Use the symbols +,,, and = to record and interpret number sentences involving all four operations; calculate the value of an unknown in a number sentence (e.g. = 6, Is 4 the same as 4? How do you know? Look at these problems. What number sentences could you write to record them? How many tens make 80? Jo s box is 5 cm wide. Mary s box is twice as wide as Jo s box. How wide is Mary s box? How many wheels are there on 3 cars?.calculate mathematical statements for multiplication and division within the multiplication tables and write them using the multiplication ( ), division ( ) and equals (=) 30 = 4)
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# App to solve math word problems
One tool that can be used is App to solve math word problems. We can help me with math work.
## The Best App to solve math word problems
This App to solve math word problems provides step-by-step instructions for solving all math problems. One step equations word problems can be solved by using the addition, subtraction, multiplication, or division operations. In order to solve a one step equation, you must first identify the operation that is being used. Next, you will need to solve the equation by using the inverse operation. For example, if the equation is 4x + 2 = 10, then you would use division to solve for x. This is because division is the inverse of multiplication. Therefore, you would divide both sides of the equation by 4 in order to solve for x. Once you have solved for x, you can then plug the value back into the original equation to check your work. One step equations word problems can be tricky, but with a little practice, you will be able to solve them with ease!
A factorial is a mathematical operation that multiplies a given number by all the numbers below it. For example, the factorial of 5 would be 5x4x3x2x1, which equals 120. Factorials are typically denoted using an exclamation mark, so the factorial of 5 would be written as 5!. Factorials are commonly used in statistics and probability theory. They can also be used to calculate permutations and combinations. To solve a factorial, you simply need to multiply the given number by all the numbers below it. So, to solve thefactorial of 5, you would multiply 5 by 4, 3, 2, and 1. This would give you the answer of 120.
To solve inequality equations, you need to first understand what they are. Inequality equations are mathematical equations that involve two variables which are not equal to each other. The inequalities can be either greater than or less than. To solve these equations, you need to find the value of the variable that makes the two sides of the equation equal. This can be done by using the properties of inequality. For example, if the equation is x+5>9, then you can subtract 5 from both sides to get x>4. This means that the solutions to this inequality are all values of x that are greater than 4. Solve inequality equations by using the properties of inequality to find the value of the variable that makes the two sides of the equation equal.
Solving inequality equations requires a different approach than solving regular equations. Inequality equations involve two variables that are not equal, so they cannot be solved using the same methods as regular equations. Instead, solving inequality equations requires using inverse operations to isolate the variable, and then using test points to determine the solution set. Inverse operations are operations that undo each other, such as multiplication and division or addition and subtraction. To solve an inequality equation, you must use inverse operations on both sides of the equation until the variable is isolated on one side. Once the variable is isolated, you can use test points to determine the solution set. To do this, you substitute values for the other variable into the equation and see if the equation is true or false. If the equation is true, then the point is part of the solution set. If the equation is false, then the point is not part of the solution set. By testing multiple points, you can determine the full solution set for an inequality equation.
Math can be a difficult subject for many students, but there are simple solutions that can help to make it easier. One way to make math less daunting is to approach it in small steps. When solving a problem, take the time to break it down into smaller pieces. This will make it easier to understand and will prevent you from feeling overwhelmed. Additionally, it can be helpful to practice regularly. Just as with any other skill, the more you practice math, the better you will become at it. There are many resources available that can help you to find problems to solve, so there is no excuse not to keep your skills sharp. With a little effort, anyone can improve their math skills and find success in the subject.
## Help with math
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# Lesson 10: Comparing Situations by Examining Ratios
Let’s use ratios to compare situations.
Mai and Jada each ran on a treadmill. The treadmill display shows the distance, in miles, each person ran and the amount of time it took them, in minutes and seconds.
1. What is the same about their workouts? What is different about their workouts?
2. If each person ran at a constant speed the entire time, who was running faster? Explain your reasoning.
## 10.2: Concert Tickets
Diego paid \$47 for 3 tickets to a concert. Andre paid \$141 for 9 tickets to a concert. Did they pay at the same rate? Explain your reasoning.
GeoGebra Applet ycJSQpXT
## 10.3: Sparkling Orange Juice
Lin makes sparkling orange juice by mixing 3 liters of orange juice with 4 liters of soda water. Noah makes sparkling orange juice by mixing 4 liters of orange juice with 5 liters of soda water. How do the two mixtures compare in taste? Explain your reasoning. If you get stuck, you can draw double number line diagrams to represent each situation.
GeoGebra Applet ycJSQpXT
GeoGebra Applet ycJSQpXT
## Summary
Sometimes we want to know whether two situations are described by the same rate. To do that, we can write an equivalent ratio for one or both situations so that one part of their ratios has the same value. Then we can compare the other part of the ratios.
For example, do these two paint mixtures make the same shade of orange?
• Kiran mixes 9 teaspoons of red paint with 15 teaspoons of yellow paint.
• Tyler mixes 7 teaspoons of red paint with 10 teaspoons of yellow paint.
Here is a double number line that represents Kiran's paint mixture. The ratio $9:15$ is equivalent to the ratios $3:5$ and $6:10$.
For 10 teaspoons of yellow paint, Kiran would mix in 6 teaspoons of red paint. This is less red paint than Tyler mixes with 10 teaspoons of yellow paint. The ratios $6:10$ and $7:10$ are not equivalent, so these two paint mixtures would not be the same shade of orange.
When we talk about two things happening at the same rate, we mean that the ratios of the quantities in the two situations are equivalent. There is also something specific about the situation that is the same.
• If two ladybugs are moving at the same rate, then they are traveling at the same constant speed.
• If two bags of apples are selling for the same rate, then they have the same unit price.
• If we mix two kinds of juice at the same rate, then the mixtures have the same taste.
• If we mix two colors of paint at the same rate, then the mixtures have the same shade.
## Glossary
same rate
#### same rate
In two situations involving ratios of the same two quantities, if the ratio of the quantities in one situation is equivalent to the ratio of the quantities in the other situation then we say the two situations involve the same rate. |
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Studying for a test? Prepare with these 10 lessons on Rational relationships.
See 10 lessons
# Simplifying rational expressions: common binomial factors
Video transcript
Given a rectangle with length a-squared plus 6a minus 27 and a width a-squared minus 9, Write the ratio of the width of the rectangle to its length as a simplified rational expression. So we want the ratio of the width to the length to the length of the rectangle, and they give us the expressions for each of these. The width—the expression for the width of the rectangle—is a-squared minus 9. So, the width —let me do it in this pink—is a-squared minus 9, and we want the ratio of that to the length; the ratio of the width of the rectangle to its length. The length is given right over there; it is a-squared plus 6a minus 27, they want us to simplify this. And so the best way to simplify this, whether we're dealing with expressions in the numerator or denominator, or just numbers, is we want to factor them and see if they have common factors and if they do we might be able to cancel them out. So if we factor this top expression over here—which was the expression for the width—this is of the form a-squared minus b-squared, where b-squared is 9. So this is going to be the same thing as a plus the square root of 9 times a minus the square root of 9. So this is a plus 3 times a minus 3 and I just recognized that from just the pattern; if you ever see something a-squared minus b-squared, it's a plus b times a minus b and you can verify that for yourself; multiply this out, you'll get a-squared minus b-squared. So, the width can be factored into a plus 3 times a minus 3. Let's see if we can do something for the denominator. So here, if we want to factor this out, we have to think of two numbers that when we add them, I get positive 6, and when I take their product, I get negative 27. Let's see, if I have positive 9 and negative 3, that would work. So, this could be factored as a plus 9 and a minus 3. 9 times a is 9a, a times negative 3 is negative 3a, when you add those two middle terms together, you'll get 6a, just like that and then, 9 times negative 3 is negative 27—of course the a times a is a squared. So I've factored the two expressions and let's see if we can simplify it. And before we simplify it, because when we simplify it we lose information, let's just remember what are allowable a's here, so we don't lose that information. Are there any a values here that will make this expression undefined? Well, any a value that makes the denominator zero will make this undefined. So a cannot be equal to negative 9 or 3, 'cause if a was negative 9 or 3, then the denominator would be zero; this expression would be undefined. So we have to remember this, this is part of the expression; we don't want to change this domain; we don't want to allow things that weren't allowable to begin with, so let's just remember this right over here. Now with that said, now that we've made this constraint, we can simplify it more; we can say, look we have an a minus 3 in the numerator and we have an a minus 3 in the denominator and we're assuming that a is not going to be equal to 3, so it's not like we're dividing zero over zero. So a will not be equal to 3, any other number; this will be an actual number, you divide the numerator and denominator by that same value, and we are left with a plus 3 over a plus 9 and the constraint here—we don't want to forget the constraints on our domain— a cannot equal negative 9 or 3. And it's important that we write this here, because over here we lost the information that a could not be equal to 3, but in order for this to really be the same thing as this thing over here, when a was equal to 3 it wasn't defined so in order for this to be the same thing, we have to constrain the domain right over there; a cannot be equal to 3. Hopefully you found that useful. |
DOC
# In a resistive circuit Ohm's Law states that voltage is equal to
By Harold Washington,2014-05-06 17:33
9 views 0
In a resistive circuit Ohm's Law states that voltage is equal to
Electricity Practice Problems June 17, 2010
Ohm’s Law Practice Problems
In a resistive circuit, Ohm's Law states that: voltage is equal to current times resistance.
Ohm's Law: V = I x R
where:
V = voltage (volts)
I = current (amps)
R = resistance (ohms)
1) A circuit consists of a 10 ohm resistor and carries a 5 amp current. What is the
voltage in the circuit (in volts)?
2) One day, Olive has what she thinks is an ingenious idea to connect her dad‟s
ammeter to a flashlight which uses a 6 volt battery. (An ammeter is a device that
measures the current flow, in amps, in an electric circuit.) She connects the
ammeter and sees that there is a current of 3 amps flowing in the circuit. What is
the flashlight‟s resistance?
3) Olive built the circuit below with a variable-voltage power supply (the circuit‟s 1leftmost circuit element) and a resistor defined by the diagram below.
resistor
Using her ammeter (shown by the circuit element with an encircled „A‟), she took measurements of current passing through the resistor. She also used a voltmeter (as shown by the circuit element in the encircled „V‟) to measure the voltage across the resistor. She recorded these numerical values in a table, with the results shown below:
1
Voltage Current
0.66 V 0.22 A
1.42 V 0.47 A
2.54 V 0.84 A
3.16 V 1.03 A
4.51 V 1.50 A
5.41 V 1.80 A
5.99 V 2.00 A
7.49 V 2.51 A
Plot the figures on the graph.
What mathematical relationship would you tell Olive that you see between voltage and
current in this simple circuit?
4) A circuit is comprised of a battery and variable resistor, known as a rheostat (see
figure). What happens to the current if resistance is doubled?
a) Current doubles resistor
b) Current halves
c) Current remains the same
d) The current goes to zero
5) Which of the following statements is NOT true?
a) Voltage is inversely proportional to the value of Resistance
b) Voltage is directly proportional to current
c) Current is inversely proportional to Resistance
d) Voltage and Resistance are always proportional
2
6) Olive‟s family car wouldn‟t start and she helped her father install a new 12-volt
battery, but that did not fix the problem. Olive‟s dad then had the car towed to the
auto mechanic‟s (Olive got to ride in the tow truck!), where the mechanic told her
father that the car would need a new 600 amp start motor. Olive, wanting to show
that she understood about such things, speaks up, and what does she proudly tell
her father and the mechanic that she has figured out about the start motor?
1) V = I x R, or (5 amps) x (10 ohms) = 50 volts
2) V = I x R which can be converted to R = V ? I, so:
(6 volts) ? (3 amps) = (2 ohms)
3) Current is linearly proportional to voltage, or equivalently, that the numerical
value of the voltage is always three times that of the current, or equivalently,
that the resistor has a value of three ohms which makes the current always one-
third of the voltage.
4) b
5) a
6) 0.02 ohm
3
Electrical Power Law Practice Problems
Power is the energy being used at a given instant in time and is measured in watts. In an electric circuit, power can be calculated by multiplying the circuit‟s voltage times
its amperage:
P = I x V (Power Law)
P - Power consumed or produced by a circuit
(watts, or equivalently: amp-volts)
7) How much power is consumed by
a) The resistor in problem #1 above?
b) The light bulb in problem #2?
c) The start motor in problem #6?
8) If voltage in household circuits is 120 volts, what is the current flow through a
a) 60 watt incandescent light bulb?
b) 12 watt CFL light bulb (which produces light approximately equal to a 60
watt incandescent bulb)?
c) 1500 watt hair drier?
9) A circuit breaker is a safety device that protects a household electric circuit, such as all the electric outlets in a particular room in a house, from having too much electricity running through it. A breaker‟s function is to stop electric current flow
when the current exceeds some preset maximum as given by the breaker‟s “rating”. What do Olive and her twin sister Penny have to do when they both dry their hair at the same time (assume that all of their house‟s bedroom circuit breakers are rated at
15 amps and all bathroom circuit breakers are rated at 20 amps)?
4
7 a) 250 watts (50 volts x 5 amps)
b) 18 watts (6 volts x 3 amps)
c) 7200 watts (12 volts x 600 amps)
8) a) 0.5 amp (P = I x V ; I = P ? V)
b) 0.1 amp
c) 12.5 amps
9) Since hair driers draw 12.5 amps each (from problem #8), only one hair drier at a time can operate from any single bedroom or bathroom circuit in the house without tripping a circuit breaker. Therefore, if the sisters need to both dry their hair at the same time they either use the same hair drier or they dry their hair in separate rooms.
5
Electric “Power Company Law” Practice Problems
Power consumption, which is also known as work, is defined as the total energy used or generated during a given period of time. In electrical circuits it is measured in kilowatt-hours (1000 watts of power lasting for one hour of time).
W = P x t
P - Power consumed by a circuit (typically measured in kilowatts)
t - Amount of time that electric circuit operates
Typically measured in hours for household use.
W - Work, or equivalently: “total power used”
Typically measured in kilowatt-hours (KWH),
but also can be measured in watt-hours, where
1 x KWH = 1000 x watt-hours.
Assume 1 x KWH costs 20? on your monthly electric bill.
Based on the above, you can calculate the cost of energy use for various electrical appliances and equipment if you have these three pieces of information:
The rated power of the appliance, usually given in watts;
The length of operating time,
And the cost of electricity.
10) a) How much power does a 100 watt light bulb use in a year if it is on a timer that
turns on at 6:00 PM every night and off at midnight? (Assume there are
approximately 9,000 hours in a year.)
b) What is the cost of the electricity the bulb consumes in the course of a year?
c) How much would be saved by switching the bulb to a 20 watt CFL bulb (which
produces the same amount of light as the 100 watt bulb)?
6
11) a) When their dad pulls out the kitchen refrigerator to clean behind it, Olive sees a
small metal plate on the back of the refrigerator that says it consumes 200 watts of
power. During the cleaning, she notices that the refrigerator comes on for about
four minutes every ten minutes or so. Later that day she tells her twin sister Penny
what she has found out about the refrigerator, and after some figuring, what can
Penny tell Olive about how much it costs to run the refrigerator for a year?
b) Penny sees an on-line ad for a brand new “energy efficient” refrigerator that
costs \$1000. The ad claims the refrigerator costs only a dollar a day to run. Should
Penny advise her family to replace their old refrigerator?
12) Penny and Olive have a pet hamster named Max. The twins love to watch Max exercise by running in the spinner in his cage. For a science fair project, Penny got the idea to hook up Max‟s spinner to a fan blade that she disconnected from a handheld battery-operated personal electric fan. After she did so, she thought that when Max was running his very fastest that he was able to turn the fan blade about as fast as the blade had originally spun when attached to the electric fan.
a) The fan could also be plugged into the wall using a power-supply that came with the fan. To help her sister with the science fair project, Olive connected the ammeter and voltmeter to the power supply while the fan was turning the blade and found that the fan operated at three volts and 0.5 amps. How much power did Max expend to turn the fan blade when it was attached to his spinner?
For living things and for nutritional purposes, energy use is usually measured in
Calories, with one nutritional Calorie being approximately equal to 0.001 KWH, or
equivalently, there are approximately 1000 Calories per KWH.
b) If Max always exercised after his dinner every day by jogging for six minutes on
his spinner and assuming he was using one watt of power per second while doing
so, how many Calories did he use to turn the blade during his after-dinner jog?
c) What would be the equivalent electricity cost Max was capable of producing if he
jogged for an hour? How much would this add up to for a year‟s worth of daily
post-dinner jogging?
7
10 a) 225 KWH
W = P x t
? 24 hrs/day) x (9,000 hours/year)] = 225,000 watt-hours (100 watts) x [(6 hrs/day
= 225 kilowatt-hours
b) \$45 (225 KWH per year) x (\$0.20/KWH)
c) \$9 - the CFL consumes one-fifth the power of the 100 watt bulb, so its cost is 1/5.
11 a) This problem is similar to 10a, but with the time usage in minutes for a ten-minute
(200 watts) x (4 minutes on ? 10 minutes) x (9,000 hrs/yr) = 720,000 watt-hours
= 720 KWH
720 KWH x \$0.20/KWH = \$142 per year of electricity usage.
b) At a “dollar per day” that is probably about \$365 per year, so the “energy efficient”
refrigerator is costing over twice as much per year as the existing one, probably not a very
good switch unless the new one is much, much larger or has some other major advantage.
12 a) 1.5 watts (P = I x V)
b) Max used one watt of power for six minutes, which is equal to
1 w X 0.1 hr = 0.1 w-hr = 0.0001 KWH (1w-hr = 0.001 KWH);
this means he used 0.1 Calories (= 0.0001 KWH X 1000 Calories/KWH).
c) Based on the answer to problem b above,
0.0001 KWH/day X 365 days/yr = 0.365 KWH/yr
At \$0.20 / KWH, this means Max could generate the equivalent of
approximately 7? worth of electricity in a year‟s worth of jogging.
8
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# 4/6 Simplified, Simplify 4/6 in its Simplest Form
2/3 is the simplified reduced form of the fraction 4/6. In this post, we will learn to simplify 4/6 in its simplest form and as a result, we will get 4/6 simplified to lowest terms.
## What is 4/6 in Simplest Form
Solution:
[Method 1: Division Method]
To simplify the fraction 4/6, we will divide both 4 and 6 by their common divisors. As 4 and 6 are even numbers, they will be divisible by 2. Thus, dividing them by 2, we get that
$\dfrac{4}{6}$ = $\dfrac{4 \div 2}{6 \div 2}$ = $\dfrac{2}{3}$.
Observe that there are no common divisors of 2 and 3 other than 1, so we cannot simplify it.
Therefore, 2/3 is the simplified form of 4/6 to lowest terms.
[Method 2: Prime Factorization Method]
To simplify the fraction 4/6 by the prime factorization method, we need to write the prime factorizations of 4 and 6.
Note that 4 = 2 × 2
6 = 2 × 3
So
$\dfrac{4}{6}$ = $\dfrac{2 \times 2}{2 \times 3}$
= $\dfrac{\cancel{2} \times 2}{\cancel{2} \times 3}$, cancelling the common terms.
= $\dfrac{2}{3}$
So 2/3 is the simplest reduced form of 4/6.
Have You Read These Fractions?
8/12 simplified
14/21 simplified
15/21 simplified
15/60 simplified
## FAQs
### Q1: What is 4/6 in simplest form?
Answer: 4/6 in simplest form is equal to 2/3.
### Q2: What is 4/6 simplified as a fraction?
Answer: 4/6 simplified as a fraction is equal to 2/3. |
1. Evaluating this infinite sum
Is it possible to evaluate this infinite sum:
$\sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
2. Originally Posted by acevipa
Is it possible to evaluate this infinite sum:
$\sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
This way seems messy, but it works..
Skip proving convergence.... it converges.
$\displaystyle L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}\left(\frac{1}{3}\right)$
$\displaystyle 3L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}$
$\displaystyle 2L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^i$
This form is "cleaner." Write some terms
$2L = \dfrac{2}{3} + 2\left(\dfrac{4}{9}\right) + 3\left(\dfrac{8}{27}\right) + 4\left(\dfrac{16}{81}\right) + \dots$
Consider what happens when we multiply by 3,
$6L = 2 + \dfrac{2\cdot4}{3} + \dfrac{3\cdot8}{9} + \dfrac{4\cdot16}{27} + \dots$
Add the two together, matching up terms with common denominators.
$8L = 2 + \dfrac{2+2\cdot4}{3} + \dfrac{2\cdot4+3\cdot8}{9} + \dfrac{3\cdot8+4\cdot16}{27} + \dots$
$= 2 + 5\left(\dfrac{2}{3}\right) + 8\left(\dfrac{2}{3}\right)^2 + 11\left(\dfrac{2}{3}\right)^3 + \dots$
So
$\displaystyle 8L = 2 + \sum_{i=1}^{\infty}(3i+2)\left(\frac{2}{3}\right)^ i$
$\displaystyle = 2 + \sum_{i=1}^{\infty}3i\left(\frac{2}{3}\right)^i + 2\sum_{i=1}^{\infty}\left(\frac{2}{3}\right)^i$
Simplifying,
$8L = 2 + 3(2L) + 4$
$2L = 6$
$L = 3$
3. For $|x|<1$ is...
$\displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ (1)
... so that is...
$\displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}}$ (2)
Now from (2) follows...
$\displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3$ (3)
Kind regards
$\chi$ $\sigma$
4. Originally Posted by chisigma
For $|x|<1$ is...
$\displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ (1)
... so that is...
$\displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}}$ (2)
Now from (2) follows...
$\displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3$ (3)
Kind regards
$\chi$ $\sigma$
Great method. Thanks
5. Originally Posted by acevipa
Great method. Thanks
If you can follow chisigma's solution you have posted this in the wrong forum and wasted the time of those helpers who have given or may have considered giving solutions that do not use calculus.
CB
6. Originally Posted by acevipa
Is it possible to evaluate this infinite sum:
$\sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
$\displaystyle\ S_{\infty}=\frac{1}{3}+\left(\frac{2}{3}\right)\le ft(\frac{2}{3}\right)+\left(\frac{3}{3}\right)\lef t(\frac{2}{3}\right)^2+\left(\frac{4}{3}\right)\le ft(\frac{2}{3}\right)^3+.....$
$3\displaystyle\ S_{\infty}-1=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig ht)^2+4\left(\frac{2}{3}\right)^3+....$
$s=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig ht)^2+.....$
$\frac{2}{3}s=2\left(\frac{2}{3}\right)^2+3\left(\f rac{2}{3}\right)^3+4\left(\frac{2}{3}\right)^4....$
$s-\frac{2}{3}s=\frac{s}{3}=2\left(\frac{2}{3}\right) +\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\righ t)^3+.....$
$\frac{s-2}{3}=\left(\frac{2}{3}\right)+\left(\frac{2}{3}\r ight)^2+\left(\frac{2}{3}\right)^3+....$
which is purely a geometric series.
Then use the result of this to calculate the original sum.
i will use
$\Delta f(x)=f(x+1)-f(x)$ for the difference operator
and $\sum\limits_x f(x) = g(x)$ if and only if $\Delta g(x) =f(x)$ (the indefinite summation)
we have the summation by parts method
$\sum\limits_{x} g(x)\Delta
f(x)=f(x).g(x)-\sum\limits_{x} f(x+1).\Delta g(x)$
(its similar to the integration by parts)
if we apply this method "n" times we get
$$\sum\limits_{x} g(x).\Delta f(x)=$$
$\sum\limits^{n}_{k=0}\Delta^{k}g(x).(-1)^{k}\Delta^{-k}E^{k}f(x)+(-1)^{n+1}\sum\limits_{x}\Delta^{n+1}g(x)\Delta^{-n}E^{n+1}f(x)$
( where $E^k f(x)=f(x+k)$)
if, g(x) has degree n , (so $\Delta ^{n+1} g(x)=0$ )and $|a|<1$ we can use it on the summation by parts to find
$\sum\limits^{\infty}_{x=c} g(x)a^{x}=a^{c-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}g(x)|_{x=c} }{(b-1)^{k+1}}$
where $b= \frac{1}{a}$
if $g(x)=x^{n}$ and e c=0
$\sum\limits^{\infty}_{x=0} x^{n}a^{x}=a^{-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}x^{n}|_{x=0 }}{(b-1)^{k+1}}$
but $\Delta^{k}x^{n}|_{x=0}=k!\bigg\{^{n}_{k}\bigg\}$ (here $\bigg\{^{n}_{k}\bigg\}$ it's the stirling number of the second kind, knuth's notation )
so
$\sum\limits^{\infty}_{x=0} x^{n}a^{x}= a^{-1}\sum\limits^{n}_{k=0}\frac{k! \bigg\{^{n}_{k}\bigg\}}{(b-1)^{k+1}}.$
8. (other way , the general case)
we can define the operator xD, (xD) f(x) =x [D f(x)] ( where "D" is the derivative )
then we can show
$(xD)^{n}g(x)=\sum\limits^{n}_{k=0}\bigg\{^{n}_{k}\ bigg\}x^{k}D^{k}g(x)$
(again the stirling numbers of the second kind)
take the series
$\sum\limits^{\infty}_{n=0}x^{n}=\frac{1}{1-x}$
apply $(xD)^p$ in the both sides
$\sum\limits^{\infty}_{n=0}n^{p}x^{n}=(xD)^p\frac{1 }{1-x}$
using the identity
$(xD)^{p}g(x)=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\ bigg\}x^{k}D^{k}g(x)$
we have
$(xD)^{p}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}x^{k }D^{k}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}\fra c{k!x^{k}}{(1-x)^{k+1}}$
so
$\sum\limits^{\infty}_{n=0}n^{p}x^{n}=\sum\limits^{ p}_{k=0}\bigg\{^{p}_{k}\bigg\}\frac{k!x^{k}}{(1-x)^{k+1}}.$ |
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# Venn diagrams
Updated: Jun 22, 2021
Dear Secondary Math students, in our previous math article, we have covered set language and notations. It is necessary to understand the different set language and notations as we will be using them a lot in this chapter.
Please make sure that you understand and master it before proceeding to Venn Diagrams. Here is the link to our Set Language and Notations article.
By mathematical definition, Venn diagrams are “diagrams representing mathematical or logical sets pictorially as circles or closed curves within an enclosing rectangle (the universal set), common elements of the sets being represented by intersections of the circles.”
Let’s take a look at the diagram shown below:
As you can see, sets A and B are denoted by two large circles in the diagram. The universal set, on the other hand, is denoted by the white rectangle containing sets A and B.
Set A is the large yellow circle while Set B is the green circle, which in between is the intersection between the sets, shown as the blue portion.
However, the diagram above is just for illustration purposes. In actual examinations, the circles will be white in color, and you will be asked to shade portions that represents the notation shown in the question.
*IMPORTANT: General rule of thumb when solving questions involving Venn Diagrams is to solve the question part by part.
What this means is to look at each of the sets from the question individually, then shade in a different direction for every part you solve, and depending on the notation, the resulting portion which crosses will be your answer! Remember to erase the portions which are not part of the answer!
Look at the example provided below:
Another thing to note is that this does not work for every scenario, so do read the question carefully and approach the question flexibly!
Examples of some types of Venn diagrams and how they are shaded:
The Venn diagram shown above represents the notation, “AB”, which is basically the set that includes all the elements from both A and B, hence both the circles must be shaded fully.
The Venn diagram shown above represents the notation, “AB”, which is basically only the elements that exists in both set A and B, hence only the intersection between the circles must be shaded.
The Venn diagram shown above represents the notation, “A’B”, which is basically only the intersection between the complement of A and set B, hence only strictly set B is being shaded, with the exclusion of the intersection between the two sets as well.
The Venn diagram shown above represents the notation, “(A B)’”, which is basically the complement of the union of set A and set B, hence the whole rectangle is shaded except for the union of set A and set B.
And that’s all for today, students! Math Lobby hopes that after this article, you have a clear understanding and is equipped with the skills to deal with questions involving Venn diagrams!
If you have any pending questions, please do go on to our Facebook page, Instagram or contact us directly at Math Lobby! We have certified mathematics tutors to aid you in your journey to becoming a better student!
As always: Work hard, stay motivated and we wish all students a successful and enjoyable journey with Math Lobby!
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# Top Reasons Why Arrays Are Essential for Teaching Multiplication
What is an array? Why should we use arrays? An array is a way of arranging objects in groups or columns. In math, we generally see arrays with rows and columns of dots, like this: However, arrays appear in real life too! Take a look at these pictures below:
### Why are they useful?
Arrays allow us to see a visual representation of repeated addition or multiplication. For example, in the egg carton above, we see 2 rows of 5. We can think of this as 5+5 or 2×5. In the muffin tin, we see 3 groups of 4. We can think of this as 4+4+4 or 3×4. Arrays also help us see that the order of the factors doesn’t matter when solving a multiplication problem. For example, 3×8 will have the same product as 8×3. To visualize this, we can have our students rotate the arrays. The picture below shows a 4×5 array that has been rotated to show 5×4. This shows us that there are the same number of objects in 5 rows of 4 as there are in 4 rows of 5.
### When should they be used?
Arrays can be used for addition in the early grades. Students might use blocks, counters, or bingo chips to create arrays. Then a repeated addition sentence can be created. For example, 6 rows of 3 can be written as 3+3+3+3+3+3=18. Arrays are essential as a tool for multiplication. We can begin by making equal groups using blocks or counters. Then we can arrange those objects into an array.
### Tips for Using Arrays To Teach Multiplication Concepts:
• Have students work with rotating the array to show the commutative property of multiplication (the order of the factors doesn’t change the product). For example, a 4 by 3 array has the same number of objects as a 3 by 4 array.
• Have students show equal groups using an array. For example, a 4 by 3 array represents 4 groups of 3. This can be written like 4×3.
• Have students break up the array to show the distributive property. For example, a 4×5 array can be broken up into a 3×5 piece and a 1×5 piece. This proves that 4×5 is equal to (3×5)+(1×5).
• We know that our students move through three stages in terms of number – concrete, representational, and finally abstract. Use arrays as the “concrete” part. Then have students represent the array in a drawing (representational). Lastly, have students write an equation to represent the array (abstract). Remember that lots of practice in the concrete and representational stages will help once we expect students to work in the abstract stage!
How do you use arrays in your classroom? I’d love to hear in the comments below! |
# Eureka Math Grade 8 Module 4 Lesson 26 Answer Key
## Engage NY Eureka Math 8th Grade Module 4 Lesson 26 Answer Key
### Eureka Math Grade 8 Module 4 Lesson 26 Exercise Answer Key
Exercises
Exercise 1.
Sketch the graphs of the system.
y = $$\frac{2}{3}$$ x + 4
y = $$\frac{4}{6}$$ x – 3
Answer:
a. Identify the slope of each equation. What do you notice?
Answer:
The slope of the first equation is $$\frac{2}{3}$$, and the slope of the second equation is $$\frac{4}{6}$$. The slopes are equal.
b. Identify the y – intercept point of each equation. Are the y – intercept points the same or different?
Answer:
The y – intercept points are (0,4) and (0, – 3). The y – intercept points are different.
Exercise 2.
Sketch the graphs of the system.
y = – $$\frac{5}{4}$$ x + 7
y = – $$\frac{5}{4}$$ x + 2
Answer:
a. Identify the slope of each equation. What do you notice?
Answer:
The slope of both equations is – $$\frac{5}{4}$$. The slopes are equal.
b. Identify the y – intercept point of each equation. Are the y – intercept points the same or different?
Answer:
The y – intercept points are (0,7) and (0,2). The y – intercept points are different.
Exercise 3.
Sketch the graphs of the system.
y = 2x – 5
y = 2x – 1)
Answer:
a. Identify the slope of each equation. What do you notice?
Answer:
The slope of both equations is 2. The slopes are equal.
b. Identify the y – intercept point of each equation. Are the y – intercept points the same or different?
Answer:
The y – intercept points are (0, – 5) and (0, – 1). The y – intercept points are different.
Exercise 4.
Write a system of equations that has no solution.
Answer:
Answers will vary. Verify that the system that has been written has equations that have the same slope and unique y – intercept points. Sample student solution:
Exercise 5.
Write a system of equations that has (2,1) as a solution.
Answer:
Answers will vary. Verify that students have written a system where (2,1) is a solution to each equation. Sample student solution:
Exercise 6.
How can you tell if a system of equations has a solution or not?
Answer:
If the slopes of the equations are different, the lines will intersect at some point, and there will be a solution to the system. If the slopes of the equations are the same, and the y – intercept points are different, then the equations will graph as parallel lines, which means the system will not have a solution.
Exercise 7.
Does the system of linear equations shown below have a solution? Explain.
6x – 2y = 5
4x – 3y = 5
Answer:
Yes, this system does have a solution. The slope of the first equation is 3, and the slope of the second equation is $$\frac{4}{3}$$. Since the slopes are different, these equations will graph as nonparallel lines, which means they will intersect at some point.
Exercise 8.
Does the system of linear equations shown below have a solution? Explain.
– 2x + 8y = 14
x = 4y + 1
Answer:
No, this system does not have a solution. The slope of the first equation is $$\frac{2}{8}$$ = $$\frac{1}{4}$$, and the slope of the second equation is $$\frac{1}{4}$$. Since the slopes are the same, but the lines are distinct, these equations will graph as parallel lines. Parallel lines never intersect, which means this system has no solution.
Exercise 9.
Does the system of linear equations shown below have a solution? Explain.
12x + 3y = – 2
4x + y = 7
Answer:
No, this system does not have a solution. The slope of the first equation is – $$\frac{12}{3}$$ = – 4, and the slope of the second equation is – 4. Since the slopes are the same, but the lines are distinct, these equations will graph as parallel lines. Parallel lines never intersect, which means this system has no solution.
Exercise 10.
Genny babysits for two different families. One family pays her $6 each hour and a bonus of$20 at the end of the night. The other family pays her $3 every half hour and a bonus of$25 at the end of the night. Write and solve the system of equations that represents this situation. At what number of hours do the two families pay the same for babysitting services from Genny?
Answer:
Let y represent the total amount Genny is paid for babysitting x hours. The first family pays y = 6x + 20. Since the other family pays by the half hour, 3∙2 would represent the amount Genny is paid each hour. So, the other family pays y = (3∙2)x + 25, which is the same as y = 6x + 25.
y = 6x + 20
y = 6x + 25
Since the equations in the system have the same slope and different y – intercept points, there will not be a point of intersection. That means that there will not be a number of hours for when Genny is paid the same amount by both families. The second family will always pay her \$5 more than the first family.
### Eureka Math Grade 8 Module 4 Lesson 26 Problem Set Answer Key
Answer Problems 1–5 without graphing the equations.
Question 1.
Does the system of linear equations shown below have a solution? Explain.
2x + 5y = 9
– 4x – 10y = 4
Answer:
No, this system does not have a solution. The slope of the first equation is – $$\frac{2}{5}$$, and the slope of the second equation is – $$\frac{4}{10}$$, which is equivalent to – $$\frac{2}{5}$$. Since the slopes are the same, but the lines are distinct, these equations will graph as parallel lines. Parallel lines never intersect, which means this system has no solution.
Question 2.
Does the system of linear equations shown below have a solution? Explain.
$$\frac{3}{4}$$ x – 3 = y
4x – 3y = 5
Answer:
Yes, this system does have a solution. The slope of the first equation is $$\frac{3}{4}$$, and the slope of the second equation is $$\frac{4}{3}$$. Since the slopes are different, these equations will graph as nonparallel lines, which means they will intersect at some point.
Question 3.
Does the system of linear equations shown below have a solution? Explain.
x + 7y = 8
7x – y = – 2
Answer:
Yes, this system does have a solution. The slope of the first equation is – $$\frac{1}{7}$$, and the slope of the second equation is 7. Since the slopes are different, these equations will graph as nonparallel lines, which means they will intersect at some point.
Question 4.
Does the system of linear equations shown below have a solution? Explain.
y = 5x + 12
10x – 2y = 1
Answer:
No, this system does not have a solution. The slope of the first equation is 5, and the slope of the second equation is $$\frac{10}{2}$$, which is equivalent to 5. Since the slopes are the same, but the lines are distinct, these equations will graph as parallel lines. Parallel lines never intersect, which means this system has no solution.
Question 5.
Does the system of linear equations shown below have a solution? Explain.
y = $$\frac{5}{3}$$ x + 15
5x – 3y = 6
Answer:
No, this system does not have a solution. The slope of the first equation is $$\frac{5}{3}$$, and the slope of the second equation is $$\frac{5}{3}$$. Since the slopes are the same, but the lines are distinct, these equations will graph as parallel lines. Parallel lines never intersect, which means this system has no solution.
Question 6.
Given the graphs of a system of linear equations below, is there a solution to the system that we cannot see on this portion of the coordinate plane? That is, will the lines intersect somewhere on the plane not represented in the picture? Explain.
Answer:
The slope of l1 is $$\frac{4}{7}$$, and the slope of l2 is $$\frac{6}{7}$$. Since the slopes are different, these lines are nonparallel lines, which means they will intersect at some point. Therefore, the system of linear equations whose graphs are the given lines will have a solution.
Question 7.
Given the graphs of a system of linear equations below, is there a solution to the system that we cannot see on this portion of the coordinate plane? That is, will the lines intersect somewhere on the plane not represented in the picture? Explain.
Answer:
The slope of l1 is – $$\frac{3}{8}$$, and the slope of l2 is – $$\frac{1}{2}$$. Since the slopes are different, these lines are nonparallel lines, which means they will intersect at some point. Therefore, the system of linear equations whose graphs are the given lines will have a solution.
Question 8.
Given the graphs of a system of linear equations below, is there a solution to the system that we cannot see on this portion of the coordinate plane? That is, will the lines intersect somewhere on the plane not represented in the picture? Explain.
Answer:
The slope of l1 is – 1, and the slope of l2 is – 1. Since the slopes are the same the lines are parallel lines, which means they will not intersect. Therefore, the system of linear equations whose graphs are the given lines will have no solution.
Question 9.
Given the graphs of a system of linear equations below, is there a solution to the system that we cannot see on this portion of the coordinate plane? That is, will the lines intersect somewhere on the plane not represented in the picture? Explain.
Answer:
The slope of l1 is $$\frac{1}{7}$$, and the slope of l2 is $$\frac{2}{11}$$. Since the slopes are different, these lines are nonparallel lines, which means they will intersect at some point. Therefore, the system of linear equations whose graphs are the given lines will have a solution.
Question 10.
Given the graphs of a system of linear equations below, is there a solution to the system that we cannot see on this portion of the coordinate plane? That is, will the lines intersect somewhere on the plane not represented in the picture? Explain.
Answer:
Lines l1 and l1 are horizontal lines. That means that they are both parallel to the x – axis and, thus, are parallel to one another. Therefore, the system of linear equations whose graphs are the given lines will have no solution.
### Eureka Math Grade 8 Module 4 Lesson 26 Exit Ticket Answer Key
Does each system of linear equations have a solution? Explain your answer.
Question 1.
y = $$\frac{5}{4}$$ x – 3
y + 2 = $$\frac{5}{4}$$ x
Answer:
No, this system does not have a solution. The slope of the first equation is $$\frac{5}{4}$$ , and the slope of the second equation is $$\frac{5}{4}$$. Since the slopes are the same, and they are distinct lines, these equations will graph as parallel lines. Parallel lines never intersect; therefore, this system has no solution.
Question 2.
y = $$\frac{2}{3}$$ x – 5
4x – 8y = 11
Answer:
Yes, this system does have a solution. The slope of the first equation is $$\frac{2}{3}$$, and the slope of the second equation is $$\frac{1}{2}$$. Since the slopes are different, these equations will graph as nonparallel lines, which means they will intersect at some point.
Question 3.
$$\frac{1}{3}$$ x + y = 8
x + 3y = 12
Answer:
No, this system does not have a solution. The slope of the first equation is – $$\frac{1}{3}$$, and the slope of the second equation is – $$\frac{1}{3}$$. Since the slopes are the same, and they are distinct lines, these equations will graph as parallel lines. Parallel lines never intersect; therefore, this system has no solution.
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## Probability Calculator
A Probability Calculator is a tool used to calculate the probability of an event or outcome occurring in a given situation.
Desktop
Desktop
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# Exploring Probability: Your Guide to Understanding and Using Probability Calculators
## Introduction: Unraveling the World of Probability Calculators
Imagine being able to easily anticipate results, comprehend uncertainty, and make wise judgments with the aid of a tool. Greetings from the probability calculator universe! These calculators are an excellent resource for anybody interested in statistics, dice probability in games, or seeking to understand concepts such as conditional probability or implied chances.
## What is a Probability Calculator? Understanding the Basics
A probability calculator is fundamentally your mathematical crystal ball. It's a computer program that calculates the probability of an event occurring under specific circumstances. Do you want to know the likelihood of rolling a particular number on the dice? A dice probability calculator can be useful in this situation.
## Probability Formulas and Definitions
The probability of an event happening can be calculated using the following formula:
$$P(A) = \frac{n(A)}{n(S)}$$
Where:
• $$P(A)$$ is the probability of event $$A$$.
• $$n(A)$$ is the number of favorable outcomes for event $$A$$.
• $$n(S)$$ is the total number of possible outcomes in the sample space.
For example, when rolling a fair six-sided die, the probability of getting a 4 would be:
$$P(4) = \frac{1}{6}$$
Another important concept is conditional probability, which is the probability of an event occurring given that another event has already occurred. It is calculated using the formula:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
Where:
• $$P(A|B)$$ is the probability of event $$A$$ occurring given that event $$B$$ has occurred.
• $$P(A \cap B)$$ is the probability of both events $$A$$ and $$B$$ occurring.
• $$P(B)$$ is the probability of event $$B$$ occurring.
This formula helps in understanding the likelihood of an event happening under certain conditions. For instance, the probability of drawing two aces in a deck of cards given that the first card drawn was an ace can be calculated using conditional probability.
## how to calculate probability?
Probability measures the likelihood of an event happening. It's calculated as:
$$\text{Probability} (P) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}}$$
Here's a step-by-step guide to calculating probability:
1. Determine the Event: Identify the specific event you want to calculate the probability for.
2. Count Favorable Outcomes: Determine the number of outcomes favorable to the event.
3. Count Total Possible Outcomes: Find the total number of outcomes in the sample space.
4. Apply the Formula: Use the formula $$P = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}}$$ to compute the probability.
5. Express as a Fraction or Decimal: The probability can be represented as a fraction, decimal, or percentage.
For example, when rolling a fair six-sided die:
• Favorable Outcomes: The number of outcomes resulting in a 4 is 1.
• Total Possible Outcomes: There are 6 possible outcomes (numbers 1 through 6) on the die.
Therefore, the probability of rolling a 4 is:
$$P(4) = \frac{1}{6} \approx 0.167 \text{ (or } 16.7\% \text{)}$$
Remember, probability ranges from 0 (impossible event) to 1 (certain event). It's a powerful tool for analyzing uncertainties and predicting outcomes.
## Probability Calculator Examples and Solutions
### Example 1: Coin Toss Probability
Consider a fair coin. What is the probability of getting heads?
Solution:
$$P(\text{Heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{2} = 0.5$$
### Example 2: Dice Probability
If you roll a six-sided die, what is the probability of rolling an even number?
Solution:
$$P(\text{Even number}) = \frac{\text{Number of even outcomes}}{\text{Total possible outcomes}} = \frac{3}{6} = \frac{1}{2} = 0.5$$
### Example 3: Card Probability
You draw a single card from a standard deck. What is the probability of drawing a heart?
Solution:
$$P(\text{Heart}) = \frac{\text{Number of hearts}}{\text{Total cards}} = \frac{13}{52} = \frac{1}{4} = 0.25$$
### Example 4: Bag Probability
In a bag, there are 4 red, 3 blue, and 5 green balls. What is the probability of drawing a blue ball?
Solution:
$$P(\text{Blue ball}) = \frac{\text{Number of blue balls}}{\text{Total balls}} = \frac{3}{12} = \frac{1}{4} = 0.25$$
## Types of Probability Calculator
There are several types of probability calculators designed to address different scenarios and calculations. Here's an explanation of some common types:
### 1. Basic Probability Calculator:
• Calculates the probability of a single event occurring based on the ratio of favorable outcomes to the total possible outcomes.
### 2. Dice Probability Calculator:
• Specifically designed for dice-based scenarios, determining probabilities of various outcomes when rolling one or more dice.
### 3. Card Probability Calculator:
• Calculates probabilities related to decks of cards, such as the chance of drawing specific cards from a deck.
### 4. Conditional Probability Calculator:
• Computes the probability of an event occurring given that another event has already occurred, dealing with dependent events.
### 5. Implied Probability Calculator:
• Commonly used in betting scenarios, it calculates the implied probability from given betting odds.
### 6. Multivariate Probability Calculator:
• Handles situations involving multiple variables or events and computes the probability of their combined occurrence.
### 7. Labor Probability Calculator:
• Helps estimate the probability of labor starting within a certain timeframe during pregnancy.
### 8. Probability Calculator with Mean and Standard Deviation:
• Utilizes statistical measures like mean and standard deviation to calculate probabilities in datasets.
### 9. Joint Probability Calculator:
• Focuses on computing probabilities of multiple events occurring simultaneously.
### 10. Bayesian Probability Calculator:
• Incorporates Bayes' theorem to update probabilities based on new evidence or information.
These calculators cater to diverse scenarios, offering users the ability to compute probabilities relevant to their specific situations or fields of interest.
## Dice Probability Calculator: Rolling the Odds in Your Favor
Have you ever pondered how often it is to roll two dice and get a double? These intricate dice rolls may be reduced to easily understood odds with a dice probability calculator. It's like having a board game cheat code that makes it easy to plan and predict results.
## Navigating Z-Score Probability Calculator: Decoding Standard Deviation
Dive into the world of statistics with a z-score probability calculator. It's your guide to understanding deviations from the mean and figuring out the probability of a score occurring within a standard distribution. This tool unlocks insights, making statistical analysis less daunting.
Moving beyond the basics, advanced concepts such as the Z-score probability calculator, conditional probability calculator, and implied probability calculator delve into nuanced scenarios. The Z-score probability calculator, for instance, evaluates probabilities concerning standard deviations from the mean in a dataset.
## Multi-Event Probability Calculation
When dealing with multiple events, a probability calculator for three events becomes handy. In specific domains like healthcare, a labor probability calculator aids in assessing childbirth outcomes based on various factors.
## Conditional Probability Calculator: When Events Depend on Each Other
Probability isn't always straightforward. Sometimes, one event's outcome hinges on another. Enter the conditional probability calculator. It's like peeking into the future by considering the relationship between different events, making predictions more accurate.
## Labor Probability Calculator: Navigating Pregnancy Probabilities
Expecting parents often seek insights into labor probabilities. This specialized calculator estimates the likelihood of labor starting within a certain timeframe, offering a comforting sense of preparedness.
## Probability Calculator with Mean and Standard Deviation
Understanding Data Spread When dealing with datasets, understanding the spread of values is crucial. A probability calculator incorporating mean and standard deviation aids in comprehending how data points are dispersed around the average.
## Probability with Statistical Parameters
Integrating statistical parameters like mean and standard deviation into a probability calculator elevates its functionality. This advanced feature empowers users to compute probabilities within a specified range, offering a more nuanced analysis.
## Real-World Applications of Probability Calculators
Probability calculators find widespread applications across diverse fields. In business and finance, these tools aid in risk assessment, decision-making, and forecasting. Moreover, in scientific research, they assist in hypothesis testing and predicting experimental outcomes.
## Conclusion
Probability calculators serve as indispensable tools, offering insights and predictions crucial for decision-making across industries. Embracing their functionality enables users to navigate uncertainty with informed calculations and strategic foresight.
How do probability calculators work?
An online application called a probability calculator is used to figure out the chances of an event or result happening. It aids in comprehending the likelihood of various events occurring.
How should I operate the probability calculator?
The Probability Calculator is easy to use. When finished, click the "Calculate" button after entering the total number of outcomes and the number of favourable circumstances. The likelihood of the event will be shown by the calculator immediately.
Can the Probability Calculator handle compound events?
Yes, the Probability Calculator can handle compound events, where two or more independent events are involved. It calculates the combined probability of these events.
Can I use the calculator for conditional probability?
Yes, the Probability Calculator can calculate conditional probability. Simply input the number of outcomes and favorable events for each scenario to find the conditional probability. |
# Thread: Proof By Mathematical Induction
1. ## Proof By Mathematical Induction
The sequence (an) is defined inductively (or recursively) by an1 = 6 and an+1 = (sqrt)(4an + 5) for n >= 1.
(a) Use the Principle of Mathematical Induction to prove that the sequence (an) bounded above by 6.
(b) Use the Principle of Mathematical Induction to prove that the sequence (an) is bounded below by 5.
(c) Use the Principle of Mathematical Induction to prove that the sequence (an) is decreasing.
I have done it already.. but it kinda seems too simple, so I wanna check with the solutions of other people.
2. Hello, ah-bee!
I think you're right . . . It is rather simple.
The sequence $\displaystyle a_n$ is defined recursively by: .$\displaystyle a_1 = 6\,\text{ and }\,a_{n+1} = \sqrt{4a_n + 5}\,\text{ for }n \geq 1$
(a) Use Induction to prove that the sequence is bounded above by 6.
Verify $\displaystyle S(1)\!:\;\;a_1\,=\,6\,\leq 6$ . . . True!
Assume $\displaystyle S(k)\!:\;\;a_k\:\leq\:6$
. . Multiply by 4: .$\displaystyle 4a_k \:\leq \:24$
. . Add 5:. . $\displaystyle 4a_k + 5 \:\leq \:29$
. . $\displaystyle \text{Take the square root: }\;\underbrace{\sqrt{4a_k + 5}}_{\text{This is }a_{k+1}} \:\leq \:\sqrt{29} \:<\:6$
We have shown that: .$\displaystyle a_{k+1}\:\leq\:6$
The inductive proof is compete.
3. yeah thats what i got... seems a bit too simple. thanks for verification |
# Modes, Medians and Means: A Unifying Perspective
March 22, 2013
By
(This article was first published on John Myles White » Statistics, and kindly contributed to R-bloggers)
### Introduction / Warning
Any traditional introductory statistics course will teach students the definitions of modes, medians and means. But, because introductory courses can’t assume that students have much mathematical maturity, the close relationship between these three summary statistics can’t be made clear. This post tries to remedy that situation by making it clear that all three concepts arise as specific parameterizations of a more general problem.
To do so, I’ll need to introduce one non-standard definition that may trouble some readers. In order to simplify my exposition, let’s all agree to assume that $0^0 = 0$. In particular, we’ll want to assume that $|0|^0 = 0$, even though $|\epsilon|^0 = 1$ for all $\epsilon > 0$. This definition is non-standard, but it greatly simplifies what follows and emphasizes the conceptual unity of modes, medians and means.
### Constructing a Summary Statistic
To see how modes, medians and means arise, let’s assume that we have a list of numbers, $(x_1, x_2, \ldots, x_n)$, that we want to summarize. We want our summary to be a single number, which we’ll call $s$. How should we select $s$ so that it summarizes the numbers, $(x_1, x_2, \ldots, x_n)$, effectively?
To answer that, we’ll assume that $s$ is an effective summary of the entire list if the typical discrepancy between $s$ and each of the $x_i$ is small. With that assumption in place, we only need to do two things: (1) define the notion of discrepancy between two numbers, $x_i$ and $s$; and (2) define the notion of a typical discrepancy. Because each number $x_i$ produces its own discrepancy, we’ll need to introduce a method for aggregating the individual discrepancies to order to say something about the typical discrepancy.
### Defining a Discrepancy
We could define the discrepancy between a number $x_i$ and another number $s$ in many ways. For now, we’ll consider only three possibilities. All of these three options satisfies a basic intuition we have about the notion of discrepancy: we expect that the discrepancy between $x_i$ and $s$ should be $0$ if $|x_i – s| = 0$ and that the discrepancy should be greater than $0$ if $|x_i – s| > 0$. That leaves us with one obvious question: how much greater should the discrepancy be when $|x_i – s| > 0$?
To answer that question, let’s consider three definitions of the discrepancy, $d_i$:
1. $d_i = |x_i – s|^0$
2. $d_i = |x_i – s|^1$
3. $d_i = |x_i – s|^2$
How should we think about these three possible definitions?
The first definition, $d_i = |x_i – s|^0$, says that the discrepancy is $1$ if $x_i \neq s$ and is $0$ only when $x_i = s$. This notion of discrepancy is typically called zero-one loss in machine learning. Note that this definition implies that anything other than exact equality produces a constant measure of discrepancy. Summarizing $x_i = 2$ with $s = 0$ is no better nor worse than using $s = 1$. In other words, the discrepancy does not increase at all as $s$ gets further and further from $x_i$. You can see this reflected in the far-left column of the image below:
The second definition, $d_i = |x_i – s|^1$, says that the discrepancy is equal to the distance between $x_i$ and $s$. This is often called an absolute deviation in machine learning. Note that this definition implies that the discrepancy should increase linearly as $s$ gets further and further from $x_i$. This is reflected in the center column of the image above.
The third definition, $d_i = |x_i – s|^2$, says that the discrepancy is the squared distance between $x_i$ and $s$. This is often called a squared error in machine learning. Note that this definition implies that the discrepancy should increase super-linearly as $s$ gets further and further from $x_i$. For example, if $x_i = 1$ and $s = 0$, then the discrepancy is $1$. But if $x_i = 2$ and $s = 0$, then the discrepancy is $4$. This is reflected in the far right column of the image above.
When we consider a list with a single element, $(x_1)$, these definitions all suggest that we should choose the same number: namely, $s = x_1$.
### Aggregating Discrepancies
Although these definitions do not differ for a list with a single element, they suggest using very different summaries of a list with more than one number in it. To see why, let’s first assume that we’ll aggregate the discrepancy between $x_i$ and $s$ for each of the $x_i$ into a single summary of the quality of a proposed value of $s$. To perform this aggregation, we’ll sum up the discrepancies over each of the $x_i$ and call the result $E$.
In that case, our three definitions give three interestingly different possible definitions of the typical discrepancy, which we’ll call $E$ for error:
$$E_0 = \sum_{i} |x_i – s|^0.$$
$$E_1 = \sum_{i} |x_i – s|^1.$$
$$E_2 = \sum_{i} |x_i – s|^2.$$
When we write down these expressions in isolation, they don’t look very different. But if we select $s$ to minimize each of these three types of errors, we get very different numbers. And, surprisingly, each of these three numbers will be very familiar to us.
### Minimizing Aggregate Discrepancies
For example, suppose that we try to find $s_0$ that minimizes the zero-one loss definition of the error of a single number summary. In that case, we require that,
$$s_0 = \arg \min_{s} \sum_{i} |x_i – s|^0.$$
What value should $s_0$ take on? If you give this some extended thought, you’ll discover two things: (1) there is not necessarily a single best value of $s_0$, but potentially many different values; and (2) each of these best values is one of the modes of the $x_i$.
In other words, the best single number summary of a set of numbers, when you use exact equality as your metric of error, is one of the modes of that set of numbers.
What happens if we consider some of the other definitions? Let’s start by considering $s_1$:
$$s_1 = \arg \min_{s} \sum_{i} |x_i – s|^1.$$
Unlike $s_0$, $s_1$ is a unique number: it is the median of the $x_i$. That is, the best summary of a set of numbers, when you use absolute differences as your metric of error, is the median of that set of numbers.
Since we’ve just found that the mode and the median appear naturally, we might wonder if other familiar basic statistics will appear. Luckily, they will. If we look for,
$$s_2 = \arg \min_{s} \sum_{i} |x_i – s|^2,$$
we’ll find that, like $s_1$, $s_2$ is again a unique number. Moreover, $s_2$ is the mean of the $x_i$. That is, the best summary of a set of numbers, when you use squared differences as your metric of error, is the mean of that set of numbers.
To sum up, we’ve just seen that the three most famous single number summaries of a data set are very closely related: they all minimize the average discrepancy between $s$ and the numbers being summarized. They only differ in the type of discrepancy being considered:
1. The mode minimizes the number of times that one of the numbers in our summarized list is not equal to the summary that we use.
2. The median minimizes the average distance between each number and our summary.
3. The mean minimizes the average squared distance between each number and our summary.
In equations,
1. $\text{The mode of } x_i = \arg \min_{s} \sum_{i} |x_i – s|^0$
2. $\text{The median of } x_i = \arg \min_{s} \sum_{i} |x_i – s|^1$
3. $\text{The mean of } x_i = \arg \min_{s} \sum_{i} |x_i – s|^2$
### Summary
We’ve just seen that the mode, median and mean all arise from a simple parametric process in which we try to minimize the average discrepancy between a single number $s$ and a list of numbers, $x_1, x_2, \ldots, x_n$ that we try to summarize using $s$. In a future blog post, I’ll describe how the ideas we’ve just introduced relate to the concept of $L_p$ norms. Thinking about minimizing $L_p$ norms is a generalization of taking modes, medians and means that leads to almost every important linear method in statistics — ranging from linear regression to the SVD.
### Thanks
Thanks to Sean Taylor for reading a draft of this post and commenting on it. |
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# 垂直关系的性质 - PowerPoint PPT Presentation
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（A）3对（B）4对（C）5对（D）6对
1、PA⊥平面ABCD
4、BC⊥平面PAB
α
a
β
L
1、当平面α⊥平面β时，平面α里面的任意一条直线a和平面β之间会存在些什么样的位置关系？
2、平面α里哪些直线是垂直于平面β的
α
A
L
β
B
M
AM⊥β
∵α⊥β
∴∠AMB=90。
∴AM⊥BM
∴AM⊥β
1
：如图，平面
AED
⊥平面
ABCD
，⊿
AED
a
2
a
ABCD
,AB=
,
(1)
EA
CD
(2)
EC
ABCD
E
∴CD⊥平面AED
∵AE在平面AED内
∴CD⊥EA
D
C
∵平面AED⊥平面ABCD
B
M
∴EM⊥平面ABCD
A
∴∠AMC即为直线EC与平面ABCD所成的角
A
D
P
B
Q
C
∴DC⊥平面ABC
∵AB在面ABC内
∴DC⊥AB
AC∩CD=C，
∵AB⊥平面ACD
A
D
C
D
C
N
A
B
M
E
E
B
AB与β内两条相交直线垂直即可，引导学生找出在
β内适合题意的直线是解决本题的关键。
∵α⊥β，∴AB⊥BE．
∴AB⊥β．
1.
2. 已知
1、两个平面垂直的性质定理
2、“转化思想” |
# SAT II Math I : Analyzing Figures
## Example Questions
### Example Question #1 : Analyzing Figures
The sides of a triangle have lengths 6 yards, 18 feet, and 216 inches. Which of the following is true about this triangle?
This triangle is right and isosceles, but not equilateral.
This triangle is acute and isosceles, but not equilateral.
The triangle is acute and equilateral.
This triangle is acute and scalene.
This triangle is right and scalene.
The triangle is acute and equilateral.
Explanation:
One yard is equal to 3 feet; it is also equal to 36 inches. Therefore:
18 feet is equal to yards,
and
216 feet is equal to yards.
The three sides are congruent, making the triangle equilateral - and all equilateral triangles are acute.
### Example Question #1 : Analyzing Figures
Figures not drawn to scale
# The triangles above are similar. Given the measurements above, what is the length of side c?
inches
inches
inches
inches
inches
inches
Explanation:
You can find the length of c by first finding the length of the hypotenuse of the larger similar triangle and then setting up a ratio to find the hypotenuse of the smaller similar triangle.
You also could have found 10 by recognizing this triangle is a form of a 3-4-5 triangle.
### The hypotenuse of the bigger triangle is 10 inches.
Now that we know the length of the hypotenuse for the larger triangle, we can set up a ratio equation to find the hypotenuse of the smaller triangle.
cross multiply
### Example Question #2 : Analyzing Figures
If line 2 and line 3 eventually intersect when extended to the left which of the following could be true?
Cannot be determined
I, II, and III
I and II
I and III
I only
I only
Explanation:
Read the question carefully and notice that the image is deceptive: these lines are not parallel. So we cannot apply any of our rules about parallel lines. So we cannot infer II or III, those are only true if the lines are parallel. If we sketch line 2 and line 3 meeting we will form a triangle and it is possible to make a = e. One such solution is to make a and e 60 degrees.
### Example Question #3 : Analyzing Figures
What is the maximum number of distinct regions that can be created with 4 intersecting circles on a plane?
Explanation:
Try sketching it out.
Start with one circle and then keep adding circles like a venn diagram and start counting. A region is any portion of the figure that can be defined and has a boundary with another portion. Don't forget that the exterior (labeled 14) is a region that does not have exterior boundaries.
### Example Question #1 : Analyzing Figures
Note: Figure may not be drawn to scale
In rectangle has length and width and respectively. Point lies on line segment and point lies on line segment . Triangle has area , in terms of and what is the possible range of values for ?
cannot be determined
Explanation:
Notice that the figure may not be to scale, and points and could lie anywhere on line segments and respectively.
Next, recall the formula for the area of a triangle:
To find the minimum area we need the smallest possible values for and .
To make smaller we can shift points and all the way to point . This will make triangle have a height of :
is the minimum possible value for the area.
To find the maximum value we need the largest possible values for and . If we shift point all the way to point then the base of the triangle is and the height is , which we can plug into the formula for the area of a triangle:
which is the maximum possible area of triangle |
## Cute inequality
Here is a problem that appeared in Romanian Mathematical Olympiad (Junior Team Selection Test 2002).
If $a,b,c\in (0,1)$ prove that
$\displaystyle \sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1$
First Solution: Use Cauchy-Schwarz inequality to obtain:
$\displaystyle \left(\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}\right)^2 \le \left[ab+(1-a)(1-b)\right]\left(c+(1-c)\right)=$
$\displaystyle =2ab-a-b+1\le a^2+b^2-a-b+1=(a^2-a)+(b^2-b)+1<1$
Here we used the fact that $2ab\le a^2+b^2$ (which follows from $(a-b)^2\ge 0$), $a^2 and $b^2 (which follows from $a,b\in (0,1)$). Taking square root of both sides, we obtain the desired inequality.
Second Solution: Since $a,b,c\in (0,1)$, there exists $x,y,z\in (0,\pi/2)$ such that
$\displaystyle a=\sin^2(x) \ \ \ \ b=\sin^2(y) \ \ \ \ c=\sin^2(z)$
Thus, the problem now reads as
$\displaystyle \sin(x)\sin(y)\sin(z)+\sqrt{(1-\sin^2(x))(1-\sin^2(y))(1-\sin^2(z))} < 1$
or equivalently,
$\displaystyle \sin(x)\sin(y)\sin(z)+\cos(x)\cos(y)\cos(z) < 1$
We can prove this inequality as follows
$\displaystyle \sin(x)\sin(y)\sin(z)+\cos(x)\cos(y)\cos(z) <\sin(x)\sin(y)+\cos(x)\cos(y)=\cos(x-y)\le 1$
Done! |
# How do you solve 8^x=1000?
Nov 28, 2016
$x \approx 3.322$
#### Explanation:
Convert the exponential form to a logarithmic form ${a}^{x} = b \to x = {\log}_{a} b$
${8}^{x} = 1000$
$x = {\log}_{8} 1000$
You can use the 'change of base law' to calculate it.
${\log}_{a} b = \frac{{\log}_{c} b}{{\log}_{c} a} \text{ }$ (c is usually 10)
$x = {\log}_{10} 1000 / {\log}_{10} 8$
$x \approx 3.322$
Nov 29, 2016
x≈3.322
#### Explanation:
Use the $\textcolor{b l u e}{\text{law of logarithms}}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\log {x}^{n} = n \log x} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
Applies to logarithms to any base.
Take the ln ( natural log) of both sides.
$\Rightarrow \ln {8}^{x} = \ln 1000$
Using the above law.
$\Rightarrow x \ln 8 = \ln 1000$
divide both sides by ln8
$\frac{x \cancel{\ln 8}}{\cancel{\ln 8}} = \ln \frac{1000}{\ln} 8$
rArrx≈3.322" to 3 decimal places" |
$$\require{cancel}$$
# 23.2: Reactance, Inductive and Capacitive
Learning Objectives
By the end of this section, you will be able to:
• Sketch voltage and current versus time in simple inductive, capacitive, and resistive circuits.
• Calculate inductive and capacitive reactance.
• Calculate current and/or voltage in simple inductive, capacitive, and resistive circuits.
Many circuits also contain capacitors and inductors, in addition to resistors and an AC voltage source. We have seen how capacitors and inductors respond to DC voltage when it is switched on and off. We will now explore how inductors and capacitors react to sinusoidal AC voltage.
## Inductors and Inductive Reactance
Suppose an inductor is connected directly to an AC voltage source, as shown in Figure. It is reasonable to assume negligible resistance, since in practice we can make the resistance of an inductor so small that it has a negligible effect on the circuit. Also shown is a graph of voltage and current as functions of time.
The graph in Figure(b) starts with voltage at a maximum. Note that the current starts at zero and rises to its peak after the voltage that drives it, just as was the case when DC voltage was switched on in the preceding section. When the voltage becomes negative at point a, the current begins to decrease; it becomes zero at point b, where voltage is its most negative. The current then becomes negative, again following the voltage. The voltage becomes positive at point c and begins to make the current less negative. At point d, the current goes through zero just as the voltage reaches its positive peak to start another cycle. This behavior is summarized as follows:
AC Voltage in an Inductor
When a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle, or by a $$90^o$$ phase angle.
Current lags behind voltage, since inductors oppose change in current. Changing current induces a back emf $$V = -L(\Delta I/\Delta t)$$. This is considered to be an effective resistance of the inductor to AC. The rms current $$I$$ through an inductor $$L$$ s given by a version of Ohm’s law:vo
$I = \dfrac{V}{X_L},$ where $$V$$ is the rms voltage across the inductor and $$X_L$$ is defined to be $X_L = 2\pi fL,$ with $$f$$ the frequency of the AC voltage source in hertz (An analysis of the circuit using Kirchhoff’s loop rule and calculus actually produces this expression). $$X_L$$ is called the inductive reactance, because the inductor reacts to impede the current. $$X_L$$ has units of ohms ($$1 \, H = 1 \, \Omega \cdot s$$, so that frequency times inductance has units of (cycles/s)$$(\Omega \cdot s) = \Omega$$), consistent with its role as an effective resistance. It makes sense that $$X_L$$ is proportional to $$L$$, since the greater the induction the greater its resistance to change. It is also reasonable that $$X_L$$ is proportional to frequency $$f$$, since greater frequency means greater change in current. That is, $$\Delta I/\Delta t$$ is large for large frequencies (large $$f$$, small $$\Delta t$$). The greater the change, the greater the opposition of an inductor.
Example $$\PageIndex{1}$$: Calculating Inductive Reactance and then Current
(a) Calculate the inductive reactance of a 3.00 mH inductor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current at each frequency if the applied rms voltage is 120 V?
Strategy
The inductive reactance is found directly from the expression $$X_L = 2\pi fL$$. Once $$X_L$$ has been found at each frequency, Ohm’s law as stated in the Equation $$I = V/X_L$$ can be used to find the current at each frequency.
Solution for (a)
Entering the frequency and inductance into Equation $$X_L = 2\pi fL$$ gives
$X_L = 2\pi fL = 6.28(60.0/s)(3.00 \, mH) = 1.13 \, \Omega \, at \, 60 \, Hz.$
Similarly, at 10 kHz,
$X_L = 2\pi fL = 6.28(1.00 \times 10^4/s)(3.00 \, mH) = 188 \, \Omega \, at \, 10 \, kHz.$
Solution for (b)
The rms current is now found using the version of Ohm’s law in Equation $$I = V/X_L$$, given the applied rms voltage is 120 V. For the first frequency, this yields
$I = \dfrac{V}{X_L} = \dfrac{120 \, V}{1.13 \, \Omega} = 106 \, A \, at \, 60 \, Hz.$
Similarly, at 10 kHz,
$I = \dfrac{V}{X_L} = \dfrac{120 \, V}{188 \, \Omega} = 0.637 \, A \, at \, 10 \, kHz.$
Discussion
The inductor reacts very differently at the two different frequencies. At the higher frequency, its reactance is large and the current is small, consistent with how an inductor impedes rapid change. Thus high frequencies are impeded the most. Inductors can be used to filter out high frequencies; for example, a large inductor can be put in series with a sound reproduction system or in series with your home computer to reduce high-frequency sound output from your speakers or high-frequency power spikes into your computer.
Note that although the resistance in the circuit considered is negligible, the AC current is not extremely large because inductive reactance impedes its flow. With AC, there is no time for the current to become extremely large.
## Capacitors and Capacitive Reactance
Consider the capacitor connected directly to an AC voltage source as shown in Figure. The resistance of a circuit like this can be made so small that it has a negligible effect compared with the capacitor, and so we can assume negligible resistance. Voltage across the capacitor and current are graphed as functions of time in the figure.
The graph in Figure starts with voltage across the capacitor at a maximum. The current is zero at this point, because the capacitor is fully charged and halts the flow. Then voltage drops and the current becomes negative as the capacitor discharges. At point a, the capacitor has fully discharged ($$Q = 0$$ on it) and the voltage across it is zero. The current remains negative between points a and b, causing the voltage on the capacitor to reverse. This is complete at point b, where the current is zero and the voltage has its most negative value. The current becomes positive after point b, neutralizing the charge on the capacitor and bringing the voltage to zero at point c, which allows the current to reach its maximum. Between points c and d, the current drops to zero as the voltage rises to its peak, and the process starts to repeat. Throughout the cycle, the voltage follows what the current is doing by one-fourth of a cycle:
AC Voltage in a Capacitor
When a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a phase angle.
The capacitor is affecting the current, having the ability to stop it altogether when fully charged. Since an AC voltage is applied, there is an rms current, but it is limited by the capacitor. This is considered to be an effective resistance of the capacitor to AC, and so the rms current in the circuit containing only a capacitor $$I$$ is given by another version of Ohm’s law to be
$I = \dfrac{V}{X_C},$ where $$V$$ is the rms voltage and $$X_C$$ is defined (As with $$X_L$$, this expression for $$X_C$$ results from an analysis of the circuit using Kirchhoff’s rules and calculus) to be
$X_C = \dfrac{1}{2\pi fC},$ where $$X_C$$ is called the capacitive reactance, because the capacitor reacts to impede the current. $$X_C$$ has units of ohms (verification left as an exercise for the reader). $$X_C$$ is inversely proportional to the capacitance $$C$$, the larger the capacitor, the greater the charge it can store and the greater the current that can flow. It is also inversely proportional to the frequency $$f$$, the greater the frequency, the less time there is to fully charge the capacitor, and so it impedes current less.
Example $$\PageIndex{2}$$: Calculating Capacitive Reactance and then Current
(a) Calculate the capacitive reactance of a 5.00 mF capacitor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current if the applied rms voltage is 120 V?
Strategy
The capacitive reactance is found directly from the expression in $$X_C = \frac{1}{2\pi fC}$$. Once $$X_C$$ has been found at each frequency, Ohm’s law stated as $$I = V/X_C$$
can be used to find the current at each frequency.
Solution for (a)
Entering the frequency and capacitance into $$X_C = \frac{1}{2\pi fC}$$ gives
$X_C = \dfrac{1}{2\pi fC}$
$= \dfrac{1}{6.28(60.0/s)(5.00 \, \mu F)} = 531 \, \Omega \, at \, 60 \, Hz.$
Similarly, at 10 kHz,
$X_C = \dfrac{1}{2\pi fC} = \dfrac{1}{6.28(1.00 \times 10^4/s)(5.00 \, \mu F)}.$
$= 3.18 \, \Omega \, at \, 10 \, kHz$
Solution for (b)
The rms current is now found using the version of Ohm’s law in $$I = V/X_C$$, given the applied rms voltage is 120 V. For the first frequency, this yields
$I = \dfrac{V}{X_C} = \dfrac{120 \, V}{531 \, \Omega} = 0.226 \, A \, at \, 60 \, Hz.$
Similarly, at 10 kHz,
$I = \dfrac{V}{X_C} = \dfrac{120 \, V}{3.18 \, \Omega} = 37.7 \, A \, at \, 10 \, kHz.$
Discussion
The capacitor reacts very differently at the two different frequencies, and in exactly the opposite way an inductor reacts. At the higher frequency, its reactance is small and the current is large. Capacitors favor change, whereas inductors oppose change. Capacitors impede low frequencies the most, since low frequency allows them time to become charged and stop the current. Capacitors can be used to filter out low frequencies. For example, a capacitor in series with a sound reproduction system rids it of the 60 Hz hum.
Although a capacitor is basically an open circuit, there is an rms current in a circuit with an AC voltage applied to a capacitor. This is because the voltage is continually reversing, charging and discharging the capacitor. If the frequency goes to zero (DC), $$X_C$$ tends to infinity, and the current is zero once the capacitor is charged. At very high frequencies, the capacitor’s reactance tends to zero—it has a negligible reactance and does not impede the current (it acts like a simple wire). Capacitors have the opposite effect on AC circuits that inductors have.
## Resistors in an AC Circuit
Just as a reminder, consider Figure, which shows an AC voltage applied to a resistor and a graph of voltage and current versus time. The voltage and current are exactly in phase in a resistor. There is no frequency dependence to the behavior of plain resistance in a circuit:
AC Voltage in a Resistor
When a sinusoidal voltage is applied to a resistor, the voltage is exactly in phase with the current—they have a $$0^o$$ phase angle.
## Summary
• For inductors in AC circuits, we find that when a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle, or by a $$90^o$$ phase angle.
• The opposition of an inductor to a change in current is expressed as a type of AC resistance.
• Ohm’s law for an inductor is $I = \dfrac{V}{X_L},$ where $$V$$ is the rms voltage across the inductor.
• $$X_L$$ is defined to be the inductive reactance, given by $X_L = 2\pi fL,$ with $$f$$ the frequency of the AC voltage source in hertz.
• Inductive reactance $$X_L$$ has units of ohms and is greatest at high frequencies.
• For capacitors, we find that when a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a $$90^o$$ phase angle.
• Since a capacitor can stop current when fully charged, it limits current and offers another form of AC resistance; Ohm’s law for a capacitor is $I = \dfrac{V}{X_C},$ where $$V$$ is the rms voltage across the capacitor.
• $$X_C$$ is defined to be the capacitive reactance, given by $X_C = \dfrac{1}{2\pi fC}.$
• $$X_C$$ has units of ohms and is greatest at low frequencies.
## Glossary
inductive reactance
the opposition of an inductor to a change in current; calculated by $$X_L = 2\pi fL$$
capacitive reactance
the opposition of a capacitor to a change in current; calculated by $$X_C = \frac{1}{2\pi fC}$$
Contributors
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). |
Wazeesupperclub.com Blog What steps do you take to ensure your work is correct?
What steps do you take to ensure your work is correct?
What steps do you take to ensure your work is correct?
How to Make Fewer Mistakes at Work and Boost Productivity:
3. Always clarify all your doubts.
4. Avoid all kinds of distractions.
7. Get a second set of eyes.
8. Stop procrastinating.
Use these six proofreading tips when you think you’re too busy to do it right:
2. Use a spell checker, but know its limits.
5. Go slow to go fast.
6. Accept the fact that someone will have to do it.
How can I check my own work?
Here are some tips for checking your work:
1. Slow down. Sometimes you just need to slow down.
2. Practice, practice and practice more.
3. Create checklists.
4. Do not proofread until finished.
5. Sentence by sentence.
6. Facts, dates, tables and references.
7. Spellcheck.
What are the steps to solving a linear equation?
1. Step 1: Simplify each side, if needed.
2. Step 2: Use Add./Sub. Properties to move the variable term to one side and all other terms to the other side.
3. Step 3: Use Mult./Div.
5. I find this is the quickest and easiest way to approach linear equations.
6. Example 6: Solve for the variable.
How do you solve an equation with 2 variables?
In a two-variable problem rewrite the equations so that when the equations are added, one of the variables is eliminated, and then solve for the remaining variable. Step 1: Multiply equation (1) by -5 and add it to equation (2) to form equation (3) with just one variable.
How do you solve linear equations with constants?
General strategy for solving linear equations
1. Simplify each side of the equation as much as possible.
2. Collect all the variable terms to one side of the equation.
3. Collect all the constant terms to the other side of the equation.
4. Make the coefficient of the variable term to equal to 1.
5. Check the solution.
How do you double check in math?
If you had to add or multiply a series of numbers when you answered the problem, redo the calculation, but do it in reverse order. For example, if the problem required that you add 7+when you did the problem, add the numbers in order 7 when you double check your work.
How do you know if an equation is correct?
In a math class, verifying that you arrived at the correct solution is very good practice. We check a solution to an equation by replacing the variable in the equation with the value of the solution. A solution should result in a true statement when simplified.
How do you check your work in algebra?
Actually plugging in your answer requires you to go through the algebra manipulations in the problem. You add, subtract, multiply, and divide to see if you get a true statement using your answer. For example, you need to perform the operations in the following equation after plugging in your answer for the variable x.
How do you check the answer of a linear equation?
Add the x terms and the constants on the left side of the equation. The answer is x=3. Check the solution by substituting 3 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
How do I check my math answers?
There are a couple of ways to check your math answer. The easiest way is to plug your solution back into the problem and work backwards to see if you get the right numbers that way. You can also take a look at your answer and see if it makes sense.
The method works like this: For any addition or multiplication problem, take the digits of each number you’re adding or multiplying and add them together. So for instance, if we were adding 218 and 435, we would add 2+1+8=11, and 4+3+5=12. Then, if we get a two-digit answer we’d repeat the process.
1. Do the division problem.
2. Multiply the quotient times the divisor.
3. If there is a remainder, add it to the multiplication product.
4. Compare this answer to the dividend. They should be the same number (630 = 630).
Tips for Proofreading for Yourself and Others
1. Slow down. Proofread line by line and focus on each line.
2. Know your own weaknesses. Make a list of common errors and check every document for those errors, one at a time.
3. Do not proof for every type of mistake at once.
5. Eliminate distractions.
6. Make a hard copy.
7. Sleep on it.
8. Don’t be afraid to cut.
How do you solve and verify equations?
The process of making sure a solution is correct by making sure it satisfies any and all equations and/or inequalities in a problem. Example: Verify that x = 3 is a solution of the equation x2 – 5x + 6 = 0. To do this, substitute x = 3 into the equation.
What is the constant in a linear equation?
In the equation y = a + bx, the constant a is called as the y-intercept. Graphically, the y-intercept is the y coordinate of the point where the graph of the line crosses the y axis. At this point x = 0. The slope of a line is a value that describes the rate of change between the independent and dependent variables. |
# Factorial simplification
Can somebody please explain to me this simplification and how it's done?
$$\frac{n!}{(2n)!}$$ = $$\frac{1}{(2n)(2n-1)...(n+1)}$$
Thanks a lot.
Homework Helper
The original denominator is
$$(2n)! = (2n)(2n-1) \cdots (n+1) n!$$
so things simply cancel.
The original denominator is
$$(2n)! = (2n)(2n-1) \cdots (n+1) n!$$
so things simply cancel.
Thanks for the help. I still don't understand the (n + 1), where does it come from? I've tried to search the net, and my textbooks but I never found examples of (xn)!, only n! = n(n-1)!.
Think about the meaning of $$(2n)!$$. It contains all the integers from $$2n$$ down to $$1$$. When you write out the entire factorial you must write down each one of those integers, and $$n + 1$$ is one of them.
As a specific (but small enough to write down) example, look what happens for $$n = 4$$. This clearly means $$n+1 = 5$$, which is the number I've placed in a box.
\begin{align*} \frac{n!}{(2n)!} & =\frac{4!}{8!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot \boxed{5} \cdot 4 \cdot 3 \cdot 2 \cdot 1}\\ & = \frac{1}{8 \cdot 7 \cdot 6 \cdot \boxed{5}}= \frac{1}{(2n)\cdots (n+1)} \end{align*}
Basically, when you write out the factorials in numerator and denominator, the final $$n$$ factors cancel. Hope this helps. |
Multiplying Decimals By Whole Numbers For Kids Multiplying decimals by whole numbers can be made easier with a step-by-step approach. Here’s a detailed explanation with examples. Steps to Multiply Decimals by Whole Numbers Ignore the Decimal Point Temporarily: First, treat the decimal number as if it were a whole number. This simplifies the multiplication process. […]
What are 2D shapes ? 2D shapes, or two-dimensional shapes, are flat shapes that have only two dimensions, length and width. They do not have depth or height. These shapes can be drawn on a flat surface, like a piece of paper, and they include various types of polygons and circles. Here’s a detailed explanation […]
Law of Exponents In this chapter, we will study about law of exponents. What is an Exponent? When a number repeatedly multiplied by itself, multiple times is known as Exponent. For example: p raised to q means that p is multiplied by itself q times. = pq = p x p x p x p […]
Commutative Property of Natural Numbers With Examples The Commutative Property is a fundamental principle in mathematics that applies to addition and multiplication. This property states that the order in which two numbers are added or multiplied does not change the result. Let’s break this down with detailed explanations and examples. Commutative Property of Addition The […]
Associative Property of Whole Numbers With Examples The Associative Property is a fundamental concept in mathematics, especially useful in simplifying calculations and understanding the structure of arithmetic operations. This property applies to addition and multiplication but not to subtraction or division. Let’s break it down with examples for better clarity. Associative Property of Addition The […]
Identity Property of Whole Numbers With Examples The Identity Property of Whole Numbers is a fundamental concept in mathematics, particularly in arithmetic. This property comes in two forms: The Identity Property of Addition and the Identity Property of Multiplication. Let’s explore each in detail with examples. 1. Identity Property of Addition Definition: The Identity Property […]
Properties of Fractional Division Let’s dive into the properties of fractional division. Understanding how to divide fractions involves grasping a few fundamental concepts and properties. Here are the key properties and steps: 1. Reciprocal Property Definition: The reciprocal (or multiplicative inverse) of a fraction is obtained by flipping the numerator and the denominator. Example: The […] |
2011 AIME I Problems/Problem 1
Problem 1
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is $k\%$ acid. From jar C, $\frac{m}{n}$ liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$.
Solution
There are $\frac{45}{100}(4)=\frac{9}{5}$ L of acid in Jar A. There are $\frac{48}{100}(5)=\frac{12}{5}$ L of acid in Jar B. And there are $\frac{k}{100}$ L of acid in Jar C. After transfering the solutions from jar C, there will be
$4+\frac{m}{n}$ L of solution in Jar A and $\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}$ L of acid in Jar A.
$6-\frac{m}{n}$ L of solution in Jar B and $\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}\right$ (Error compiling LaTeX. ! Missing delimiter (. inserted).) of acid in Jar B.
Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.
$$\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}$$ $$\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}$$ Add the equations to get $$\frac{42}{5}+\frac{k}{50}=10$$ Solving gives $k=80$.
If we substitute back in the original equation we get $\frac{m}{n}=\frac{2}{3}$ so $3m=2n$. Since $m$ and $n$ are relatively prime, $m=2$ and $n=3$. Thus $k+m+n=80+2+3=\boxed{085}$. |
# Introduction
Equations involving derivatives are known as differential equations.
Examples: (dy/dx)+ 3y=5
(d2y/dx2)+3(dy/dx)+5x=0
## Order of a differential equation
Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation.
Consider the following equations:
(dy/dx)+3y=0 ………………… (1)
(d2y/dx2)+3(dy/dx)+4y=7 …………………. (2)
(d3y/dx3)2+(d2y/dx2)5-3(dy/dx)=1 ………… (3)
Order of (1) = 1
Order of (2) = 2
Order of (3) = 3
## Degree of a differential equation
The degree of a differential equation is the power of highest order derivative.
Degree of (1) above =1
Degree of (2) above = 1
Degree of (3) above = 2
1) Find the order and degree of the following:
a) (d2y/dx2)3+(dy/dx)2+sin(dy/dx)+1=0
Order = 2
Degree = not defined [since sin (dy/dx) is not defined]
b) 2x2(d2y/dx2)-3(dy/dx)+y=0
Order = 2
Degree=1
## General and Particular solutions of a differential equation
Here an equation and a differential equation will be given; we have to verify whether the given equation is a solution of the given differential equation.
Example: Verify that the function y=x2+2x+C is a solution of the differential equation y'-2x-2=0
y=x2+2x+C
y'=2x+2
y'-2x-2=0
Hence the given equation is a solution of the differential equation
## Formation of a differential equation whose general solution is given
Given the equation of a family of curves having arbitrary constants, we have to differentiate and eliminate the constants (arbitrary). If there is one constant differentiate once, if two constants are there then differentiate two times and eliminate the arbitrary constants to form the differential equation.
Example: Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Let P denote the family of above said parabolas and let (0,a) be the focus, where 'a' is arbitrary constant. Hence the equation of P is x2=4ay ……. (1)
Differentiating both sides, 2x=4ay'
a=x/[2y']
Hence (1) becomes x2=4[x/(2y')]
xy'-2y=0
Therefore the differential equation is xy'-2y=0.
## Methods of solving first order, first degree differential equations
### Differential equations with variables separable
A first order-first degree differential equation is of the form (dy/dx)= F(x,y).
If F(x,y) can be expressed as a product g(x) h(y) where g(x) is a function of 'x' and h(y) is a function of 'y' then it is said to be variable separable type.
After separating, the next step is to integrate to obtain the solution of the differential equation.
Example: Solve (dy/dx) + y=1 (y≠1)
(dy/dx)=1-y
dy/(1-y) = dx
∫dy/(1-y) = ∫dx
-log(1-y)=x+c
x+log(1-y)+c=0, which is the required solution.
Now try it yourself! Should you still need any help, click here to schedule live online session with e Tutor! |
# 2015 AMC 8 Problems/Problem 23
Tom has twelve slips of paper which he wants to put into five cups labeled $A$, $B$, $C$, $D$, $E$. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E$. The numbers on the papers are $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,$ and $4.5$. If a slip with $2$ goes into cup $E$ and a slip with $3$ goes into cup $B$, then the slip with $3.5$ must go into what cup?
$\textbf{(A) } A \qquad \textbf{(B) } B \qquad \textbf{(C) } C \qquad \textbf{(D) } D \qquad \textbf{(E) } E$
## Solution
The numbers have a sum of $6+5+12+4+8=35$, which averages to $7$, which means $A, B, C, D, E$ will have values $5, 6, 7, 8, 9$, respectively. Now it's process of elimination: Cup $A$ will have a sum of $5$, so putting a $3.5$ slip in the cup will leave $5-3.5=1.5$; However, all of our slips are bigger than $1.5$, so this is impossible. Cup $B$ has a sum of $6$, but we are told that it already has a $3$ slip, leaving $6-3=3$, which is too small for the $3.5$ slip. Cup $C$ is a little bit trickier, but still manageable. It must have a value of $7$, so adding the $3.5$ slip leaves room for $7-3.5=3.5$. This looks good at first, as we do have slips smaller than that, but upon closer inspection, we see that no slip fits exactly, and the smallest sum of two slips is $2+2=4$, which is too big, so this case is also impossible. Cup $E$ has a sum of $9$, but we are told it already has a $2$ slip, so we are left with $9-2=7$, which is identical to the Cup C case, and thus also impossible. With all other choices removed, we are left with the answer: Cup $\boxed{\textbf{(D)}~D}$
~savannahsolver |
# How to obtain the squares of the number ending in 05?
This is an interesting find. While teaching students on obtaining the square of any number, I observed an amazing patterns in the squares of the number which ends with the digits 05 like 105, 205, 1105 etc.
You already know the shortcut, how to multiply any number ending with 5. You can use the same shortcut here as well. However, you will surprised to know how the square pattern look, if the number ends with the digits 05.
Consider an example. Let’s say, we want to calculate the square of 205 i.e., 2052 . The steps can be illustrated in the following figure.
Step 1 : Break the number such that 05 is on the right and the rest of the digits ( i.e, 2 in this case) is on the left.
Step 2 : Obtain the square of 05 on the right to get 025. This forms the second part of the answer. Please note, a number ending with 05 will always have 025 as the last part of the answer when you take its square.
Step 3 : Take the rest of the digits(2) and take its square followed by the digit itself which forms the LHS part of the answer. So, 2052 = 22/2/025 = 42025.
Important : The square of any number which ends with 05 would look like a52/a/025.
Let’s take another example to make it clear. Can you guess, what is the square of 505 ?
So, 5052 = 52/5/025 = 255025
Very easy right
Let’s consider another example : What is 12052 ?
Notice the difference from the previous example. In this case, we have 2-digits on the left. So, we need to consider the carry – over.
So, 12052 = 122 / 12 /025 = 144+1/2/025 = 1452025
On the similar lines, can you mentally calculate?
• 3052
• 6052
• 14052
• 15052 |
# 3.2: Equations and Mass Relationships
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Consider the balanced chemical equation (i.e., catalytic oxidation of ammonia) such as
$4 \text{ N} \text{H}_{3} (g) + 5 \text{O}_{2} (g) \rightarrow 4 \text{ N} \text{O} (g) + 6 \text{ H}_{2} \text{O} (g) \label{1}$
not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation $$\ref{1}$$ (represented molecularly by the image below it) says that 4 NH3 molecules can react with 5 O2 molecules to give 4 NO molecules and 6 H2O molecules. It also says that 4 mol NH3 would react with 5 mol O2 yielding 4 mol NO and 6 mol H2O.
The balanced equation does more than this, though. It also tells us that $$2 \cdot4 = 8 \text{mol NH}_3$$ will react with $$2 \cdot5 = 10 \text{mol O}_2$$, and that $$\small\frac{1}{2} \cdot4 = 2 \text{mol NH}_3$$ requires only $$\small\frac{1}{2} \cdot5 = 2.5 \text{mol O}_2$$. In other words, the equation indicates that exactly 5 mol O2 must react for every 4 mol NH3 consumed. For the purpose of calculating how much O2 is required to react with a certain amount of NH3 therefore, the significant information contained in Equation $$\ref{1}$$ is the ratio
$\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{\text{3}}}\label{2}$
We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Equation $$\ref{1}$$,
$\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{NH}_{\text{3}}} \right)=\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{\text{3}}} \label{3}$
The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.
##### Example $$\PageIndex{1}$$: Stoichiometric Ratios
Derive all possible stoichiometric ratios from Equation $$\ref{1}$$.
###### Solution
Any ratio of amounts of substance given by coefficients in the equation may be used:
\begin{align*} &\text{S}\left(\frac{\ce{NH3}}{\ce{O2}}\right) = \frac{\text{4 mol NH}_3}{\text{5 mol O}_2} &\text{S}\left(\frac{\ce{O2}}{\ce{NO}}\right) &= \frac{\text{5 mol O}_2}{\text{4 mol NO}} \\ { } \\ &\text{S}\left(\frac{\ce{NH3}}{\ce{NO}}\right) = \frac{\text{4 mol NH}_3}{\text{4 mol NO}} &\space\text{S}\left(\frac{\ce{O2}}{\ce{H2O}}\right) &= \frac{\text{5 mol O}_2}{\text{6 mol }\ce{H2O}} \\ { } \\ &\text{S}\left(\frac{\ce{NH3}}{\ce{H2O}}\right) = \frac{\text{4 mol NH}_3}{\text{6 mol }\ce{H2O}} &\space\text{S}\left(\frac{\ce{NO}}{\ce{H2O}}\right) &= \frac{\text{4 mol NO}}{\text{6 mol }\ce{H2O}} \end{align*} \nonumber
There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Equation $$\ref{3}$$ gives one of them.]
When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Equation $$\ref{2}$$ as an example, this means that the ratio of the amount of O2 consumed to the amount of NH3 consumed must be the stoichiometric ratio S(O2/NH3):
$\frac{n_{\text{O}_{\text{2}}\text{ consumed}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} =\text{S} \left(\frac{\text{O}_2}{\text{NH}_3}\right) = \frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{3}}\label{9}$
Similarly, the ratio of the amount of H2O produced to the amount of NH3 consumed must be S(H2O/NH3):
$\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} =\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{3}} \right) = \frac{\text{6 mol H}_{\text{2}}\text{O}}{\text{4 mol NH}_{3}} \label{10}$
In general we can say that
$\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}} \label{11}$
or, in symbols,
$\text{S}\left( \frac{\text{X}}{\text{Y}} \right)= \frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}} \label{12}$
Note that in the word Equation $$\ref{11}$$ and the symbolic Equation $$\ref{12}$$, $$X$$ and $$Y$$ may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.
##### Example $$\PageIndex{2}$$: Ratio of Water
Find the amount of water produced when 3.68 mol NH3 is consumed according to Equation $$\ref{10}$$.
###### Solution
The amount of water produced must be in the stoichiometric ratio S(H2O/NH3) to the amount of ammonia consumed:
$\text{S}\left( \dfrac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{\text{3}}} \right)=\dfrac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} \nonumber$
Multiplying both sides nNH3 consumed, by we have
\begin{align} n_{\text{H}_{\text{2}}\text{O produced}} &= n_{\text{NH}_{\text{3}}\text{ consumed}} \normalsize \cdot\text{S}\left( \frac{\ce{H2O}}{\ce{NH3}} \right) \\ { } \\ & =\text{3.68 mol NH}_3 \cdot\frac{\text{6 mol }\ce{H2O}}{\text{4 mol NH}_3} \\ & =\text{5.52 mol }\ce{H2O} \end{align} \nonumber
This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example $$\PageIndex{2}$$ is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example $$\PageIndex{2}$$ is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Equation $$\ref{9}$$ when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form
$\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \\ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced} \nonumber$
or symbolically.
$n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}} \nonumber$
When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol NH3 cancels 1 mol NH3 but does not cancel 1 mol H2O.
The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.
##### Example $$\PageIndex{3}$$: Mass Produced
Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is reacted with FeS2 according to the equation
$\ce{4FeS2 + 11O2 -> 2Fe2O3 + 8SO2} \nonumber$
###### Solution
The problem asks that we calculate the mass of SO2 produced. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of SO2 to the mass of SO2. Therefore this problem in effect is asking that we calculate the amount of SO2 produced from the amount of O2 consumed. This is the same problem as in Example 2. It requires the stoichiometric ratio:
$$\text{S}\left( \frac{\text{SO}_{\text{2}}}{\text{O}_{\text{2}}} \right)=\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}$$
The amount of SO2 produced is then:
\begin{align*} n_{\ce{SO2}\text{ produced}} & = n_{\ce{O2}\text{ consumed}}\text{ }\normalsize\cdot\text{ conversion factor} \\ & =\text{3.84 mol O}_2\cdot\frac{\text{8 mol SO}_2}{\text{11 mol O}_2} \\ & =\text{2.79 mol SO}_2 \end{align*} \nonumber
The mass of SO2 is:
\begin{align*}\text{m}_{\text{SO}_{\text{2}}} & =\text{2.79 mol SO}_2\cdot\frac{\text{64.06 g SO}_2}{\text{1 mol SO}_2} \\& =\text{179 g SO}_2 \end{align*} \nonumber
With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O2 to moles of SO2 and the molar mass will convert moles of SO2 to grams of SO2. A schematic road map for the one-step calculation can be written as:
$n_{\text{O}_{\text{2}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/O}_{\text{2}}\text{)}}\text{ }n_{\text{SO}_{\text{2}}}\text{ }\xrightarrow{M_{\text{SO}_{\text{2}}}}\text{ }m_{\text{SO}_{\text{2}}} \nonumber$
Thus:
$\text{m}_{\text{SO}_{\text{2}}}=\text{3}\text{.84 mol O}_{\text{2}}\cdot\text{ }\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}\normalsize\text{ }\cdot\text{ }\frac{\text{64}\text{.06 g}}{\text{1 mol SO}_{\text{2}}}=\normalsize\text{179 g} \nonumber$
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations.
Calculations
$$4 \text{ FeS}_{2}$$ $$+ 11 \text{ O}_{2}$$ $$\rightarrow 2 \text{Fe}_2 \text{O}_3$$ $$+ 8 \text{SO}_2$$
m (g) 168 123 111 179
M (g/mol) 120.0 32.0 159.7 64.06
n (mol) 1.40 3.84 0.698 2.79
The chemical reaction in this example is of environmental interest. Iron pyrite (FeS2) is often an impurity in coal, and so burning this fuel in a power plant produces sulfur dioxide (SO2), a major air pollutant. Our next example also involves burning a fuel and its effect on the atmosphere.
##### Example $$\PageIndex{4}$$: Mass of Oxygen
What mass of oxygen would be consumed when 3.3 × 1015 g, 3.3 Pg (petagrams), of octane (C8H18) is burned to produce CO2 and H2O?
###### Solution
First, write a balanced equation
$\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O} \nonumber$
The problem gives the mass of C8H18 burned and asks for the mass of O2 required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass of O2 permits calculation of the mass of O2. Symbolically
$m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}} \nonumber$
\begin{align} m_{\text{O}_{\text{2}}} & =\text{3}\text{.3 }\cdot\text{ 10}^{\text{15}}\text{ g }\cdot\text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\cdot\text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\cdot \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}} \\ & =\text{1}\text{.2 }\cdot\text{ 10}^{\text{16}}\text{ g } \end{align*} \nonumber
Thus 12 Pg (petagrams) of O2 would be needed.
The large mass of oxygen obtained in this example is an estimate of how much O2 is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O2. Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen.
This page titled 3.2: Equations and Mass Relationships is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn. |
## Trout Pond Population
Trout Pond
9-12
1
This investigation illustrates the use of iteration, recursion and algebra to model and analyze a changing fish population. Graphs, equations, tables, and technological tools are used to investigate the effect of varying parameters on the long-term population.
In this unit, students will be investigating the numeric, graphical, and symbolic representations of a recursive function. Specifically, students will explore a scenario in which a trout pond loses a portion of its population to natural causes, but the pond is restocked with fish each year.
To begin the lesson, have students think about the following situation:
Each spring, a trout pond is restocked with fish. That is, the population decreases each year due to natural causes, but at the end of each year, more fish are added. Here’s what you need to know.
• There are currently 3000 trout in the pond.
• Due to fishing, natural death, and other causes, the population decreases by 20% each year, regardless of restocking.
• At the end of each year, 1000 trout are added to the pond.
Allow students time to think about this situation. It might be advantageous for students to work in pairs to discuss their findings. When students have had ample opportunity to investigate the situation, ask the following questions:
1. Do you think the population will grow without bound, level off, oscillate, or die out? Explain why you think your conjecture about long-term population is reasonable.
2. Let the word NEXT represent the population next year, and NOW represent the population this year. Write an equation using NEXT and NOW that represents the assumptions given above.
To allow students to focus solely on the mathematics of this activity, you may wish to have them use the Trout Pond Exploration activity sheet to structure their investigation.
Trout Pond Exploration Activity Sheet
With a chart to tally data and a list of guiding questions, this sheet will help students focus on the concept of recursion rather than on the skill of organizing data. However, if you wish to have students devise their own methods for organizing information, this activity sheet may provide too much guidance.
Extensions
How do you think the population will change over time if the parameters are changed? That is, what will happen if a change is made to the initial number of fish, the rate at which the population decreases, or the number that is restocked each year?
Questions for Students
1. What will happen to the population of trout in the pond? Will it increase, decrease, or level off?
Teacher Reflection
• Did you use the activity sheet to guide students' investigation? If so, did it provide more guidance than was necessary? If not, would using the sheet help the next time you have students explore this situation?
• How did you organize this open-ended exploration to avoid classroom management problems?
### Trout Pond: Numerical Analysis
9-12
This investigation illustrates the use of iteration, recursion and algebra to model and analyze a changing fish population. Graphs, equations, tables, and technological tools are used to investigate the effect of varying parameters on the long-term population.
### Graphical Analysis
9-12
This investigation illustrates the use of iteration, recursion and algebra to model and analyze a changing fish population. Graphs, equations, tables, and technological tools are used to investigate the effect of varying parameters on the long-term population.
### Symbolic Analysis
9-12
This investigation illustrates the use of iteration, recursion and algebra to model and analyze a changing fish population. Graphs, equations, tables, and technological tools are used to investigate the effect of varying parameters on the long-term population.
### Learning Objectives
Students will:
• Use iteration, recursion and algebra to model and analyze a changing fish population
• Use graphs, equations, tables, and technology to investigate the effect of varying parameters on the long-term population |
### Area of a Triangle Practice
#### Question 2
Given a triangle with the area of 20in2. Find the height of this triangle given that its base is 5cm.
A. 8 in
B. 8 in2
C. 7 in
D. 7 in2
### Step by Step Solution
• #### Step 1
The picture below shows the triangle with the base of 5cm and the height of h.
• #### Step 2
Since the value of the area is given as 20in2, we can substitute A with 30. Similarly, since the base is given as 5cm, we can substitute b with 5. After doing so, we can simplify the equation as shown below:
• #### Step 3
Now, we have 5h = 40. To find h, we need to remove 5. We can do so by dividing both sides of the equation with 5. This is shown below:
• #### Step 4
Now, the calculated number 8, has no meaning unless we include the unit for it. Since the area is in inch2, h will be in inch. Hence:
h = 8 in |
NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3
# NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3
NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3 are the part of NCERT Solutions for Class 6 Maths (Rationalised Contents). Here you can find the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3.
## NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.3 (Rationalised Contents)
### Ex 7.3 Class 6 Maths Question 1.
Write the fractions. Are all these fractions equivalent?
Solution:
After converting all the fractions in their simplest form, they are not equal.
They are not equivalent fractions.
### Ex 7.3 Class 6 Maths Question 2.
Write the fractions and pair up the equivalent fractions from each row.
Solution:
The following pairs of fractions represent the equivalent fractions.
(a) ↔ (ii) = 1/2
(b) ↔ (iv) = 2/3
(c) ↔ (i) = 1/3
(d) ↔ (v) = 1/4
(e) ↔ (iii) = 3/4
### Ex 7.3 Class 6 Maths Question 3.
Replace boxes in each of the following by the correct number:
Solution:
### Ex 7.3 Class 6 Maths Question 4.
Find the equivalent fraction of 3/5 having
(a) denominator 20
(b) numerator 9
(c) denominator 30
(d) numerator 27
Solution:
(a) Here, we need denominator 20.
Let N be the numerator of the fraction.
The required equivalent fraction is 12/20.
(b) Here, we need numerator 9.
Let D be the denominator of the fraction.
The required equivalent fraction is 9/15.
(c) Here, we need denominator 30.
Let N be the numerator of the fraction.
The required equivalent fraction is 18/30.
(d) Here, we need numerator 27.
Let D be the denominator of the fraction.
The required equivalent fraction is 27/45.
### Ex 7.3 Class 6 Maths Question 5.
Find the equivalent fraction of 36/48 with
(a) numerator 9
(b) denominator 4
Solution:
(a) Given: Numerator of the equivalent fraction = 9
Let D be the denominator of the required fraction.
So, the equivalent fraction is 9/12.
(b) Given: Denominator of the equivalent fraction = 4
Let N be the numerator of the required fraction.
N/4 = 36/48 N × 48 = 4 × 36
N = (4 × 36)/48 = 3
So, the equivalent fraction is 3/4.
### Ex 7.3 Class 6 Maths Question 6.
Check whether the given fractions are equivalent:
Solution:
(a) 5/9 and 30/54
Using cross-multiplication, 5 × 54 = 270
and 9 × 30 = 270
Here, 5 × 54 = 9 × 30
5/9 and 30/54 are equivalent fractions.
(b) 3/10 and 12/50
Using cross-multiplication, 3 × 50 = 150
and 10 × 12 = 120
Here, 3 × 50 ≠ 10 × 12
3/10 and 12/50 are not equivalent fractions.
(c) 7/13 and 5/11
Using cross-multiplication, 7 × 11 = 77 and 5 × 13 = 65
Here, 7 × 11 ≠ 5 × 13
7/13 and 5/11 are not equivalent fractions.
### Ex 7.3 Class 6 Maths Question 7.
Reduce the following fractions to simplest form:
Solution:
### Ex 7.3 Class 6 Maths Question 8.
Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils.
Solution:
Ramesh used up 10 pencils out of 20 pencils.
Sheelu used up 25 pencils out of 50 pencils.
Jamaal used up 40 pencils out of 80 pencils.
Yes, each has used up an equal fraction of his/her pencils, i.e., 1/2.
### Ex 7.3 Class 6 Maths Question 9.
Match the equivalent fractions and write two more for each.
Solution:
Two more examples of equivalent fractions are:
Two more examples of equivalent fractions are:
Two more examples of equivalent fractions are:
Two more examples of equivalent fractions are:
Two more examples of equivalent fractions are:
Related Links:
NCERT Solutions for Maths Class 7
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12
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The Learning Point > Mathematics >
### Linear Algebra - Matrices Part I - A Tutorial with Examples
------------xxxx------------
Linear Algebra: Matrices Part 1
## Linear Algebra - Matrices Part I - Outline of Contents:
Introduction to Matrices. Theory, definitions. What a Matrix is, order of a matrix, equality of matrices, different kind of matrices: row matrix, column matrix, square matrix, diagonal, identity and triangular matrices. Definitions of Trace, Minor, Cofactors, Adjoint, Inverse, Transpose of a matrix. Addition, subtraction, scalar multiplication, multiplication of matrices. Defining special types of matrices like Symmetric, Skew Symmetric, Idempotent, Involuntary, Nil-potent, Singular, Non-Singular, Unitary matrices.
Matrices
This tutorial contains an introduction to various terms and definitions used while dealing with Matrices
A Quick Introduction (Covered in greater detail in the tutorial document at the end)
A rectangular arrangement of numbers (which may be real or complex numbers) in rows and columns, is called a matrix. This arrangement is enclosed by small ( ) or big [ ] brackets. The numbers are called the elements of the matrix or entries in the matrix. Example:-
1 2 3
4 5 6
7 8 9
A matrix having m rows and n columns is called a matrix of order m×n matrix (read as an m by n matrix). We will frequently use this notation A=[ ]m×n represents the element in the i-th row and the j-th column in a matrix of order m×n.
Two matrices A and B are said to be equal matrix if they are of same order and their
corresponding elements are equal. Example:
A =
1 2 3
4 5 6
7 8 9
B =
1 2 3
4 5 6
7 8 9
#### 1. Row matrix:
The matrix has only one row and any number of columns. Example
[ 1 2 3 4 ]
#### 2. Column matrix:
The matrix has only one columns and any number of rows. Example
1
2
3
#### 3. Singleton matrix:
If a matrix has only one element. Example [2]
#### 4. Null or Zero matrix:
All the elements are zero in such a matrix. Example:
0 0 0
0 0 0
0 0 0
#### 5. Square matrix:
If the number of rows and columns in a matrix are equal, then it is
called a square matrix. Thus A=[ ] m×n is a square matrix if m=n
#### 6. Trace of a matrix:
The sum of the diagonal elements of a square matrix. A is called the trace
of matrix A, which is denoted by tr A=a11+a22+…….ann
#### 7. Diagonal Matrix:
All elements not on the principal diagonal are zero
Example:
5 0 0
0 1 0
0 0 4
#### 8. Identical Matrix:
A diagonal matrix in which all elements on the principal diagonal are set to one.
#### 9. Scalar Matrix:
A diagonal matrix in which all elements on the principal diagonal are equal.
## Other concepts introduced in the tutorial :
Triangular Matrices: Upper and Lower triangular matrices
Addition, subtraction and scalar multiplication of matrices
Multiplication of matrices
How to compute the minors, cofactors, adjoint, transpose and inverse of a matrix. Properties of the adjoint, inverse and transpose of a matrix
Special types of matrices: Symmetric matrix, skew-symmetric matrix, singular matrix, non-singular matrix, orthogonal matrix, idempotent matrix, involuntary matrix, nilpotent matrix, unitary matrix, periodic matrix, hermitian matrix, skew-hermitian matrix, conjugate of a matrix
## Tutorial :
Matrices: Brief summary of things we read in the above tutorial.
#### Definition:
A rectangular arrangement of numbers (which may be real or complex numbers) in rows and columns, is called a matrix. This arrangement is enclosed by small ( ) or big [ ] brackets. The numbers are called the elements of the matrix or entries in the matrix.
#### Order of a matrix:
A matrix having m rows and n columns is called a matrix of order m×n matrix (read as an m by n matrix).
#### Equality of Matrices
Two matrices A and B are said to be equal matrix if they are of same order and their corresponding elements are equal.
#### Types of Matrices:
1. Row matrix: The matrix has only one row and any number of columns.
2. Column matrix: The matrix has only one columns and any number of rows.
3. Singleton matrix: If a matrix has only one element.
4. Null or Zero matrix: All the elements are zero in such a matrix.
5. Square matrix: If the number of rows and columns in a matrix are equal, then it is called a square matrix.
#### Trace of a matrix
The sum of the diagonal elements of a square matrix. A is called the trace of matrix A
#### Diagonal Matrix
If all elements except the principal diagonal in a square matrix are zero, it is called a diagonal matrix.
#### Identity Matrix
A square matrix in which elements in the main diagonal are all 1 and rest are all zero
#### Scalar matrix:
A square matrix whose all non-diagonal elements are zero and diagonal elements are equal is called a scalar matrix.
#### Triangular Matrix
A square matrix [ ] is said to be triangular matrix if each element above or below the principal diagonal is zero
We will add the corresponding elements.
#### Subtraction of matrices:
We will subtract the corresponding elements.
#### Scalar multiplication of matrices
We will multiply the corresponding elements with the given scalar. (Explained in the tutorial above)
#### Multiplication of matrices
Two matrices A and B are conformable for the product AB if the number of columns in A (pre multiplier) is same as the number of rows in B(post multiplier).Thus, if A=[ ]m×n and B=[ ]n×p are two matrices of order m×n and n×p respectively then there product is of order m×p (Explained with examples in the tutorial above).
#### Recap of the basic properties of matrix addition
If A,B,C are three matrices such that the product is defined, then
(i) AB≠BA
(ii) (AB)C=A(BC)
(iii)IA=A=AI
(iv) A(B+C)=AB+AC
(v) If AB=AC
(vi) If AB=0
#### Transpose of a Matrix:
The matrix obtained from a given matrix A by changing its rows into columns or columns into rows is called transpose of Matrix A and is denoted by AT
Special Types of Matrices:
1. Symmetric Matrix
2. Skew-Symmetric Matrix
3. Singular Matrix:
4. Non Singular Matrix:
5. Orthogonal Matrix:
6. Idempotent matrix:
7. Involuntary matrix:
8. Nilpotent matrix:
9. Unitary matrix:
10. Periodic matrix: A matrix is called a periodic matrix if Ak+1=A where k is a positive
integer. If however k is the least positive integer for which Ak+1=A then k is said to
be the period of A
11. Hermitian matrix
12. Skew Hermitian matrix
13. Conjugate of matrix: The matrix obtained from any given matrix A containing complex number as its elements, on replacing its elements by the corresponding
conjugate complex numbers is called conjugate of A.
14. Submatrix
15. Transpose of a conjugate of a matrix
### You might like to take a look at some of our other Linear Algebra tutorials :
Introduction to Matrices - Part I Introduction to Matrices. Theory, definitions. What a Matrix is, order of a matrix, equality of matrices, different kind of matrices: row matrix, column matrix, square matrix, diagonal, identity and triangular matrices. Definitions of Trace, Minor, Cofactors, Adjoint, Inverse, Transpose of a matrix. Addition, subtraction, scalar multiplication, multiplication of matrices. Defining special types of matrices like Symmetric, Skew Symmetric, Idempotent, Involuntary, Nil-potent, Singular, Non-Singular, Unitary matrices. Introduction to Matrices - Part II Problems and solved examples based on the sub-topics mentioned above. Some of the problems in this part demonstrate finding the rank, inverse or characteristic equations of matrices. Representing real life problems in matrix form. Determinants Introduction to determinants. Second and third order determinants, minors and co-factors. Properties of determinants and how it remains altered or unaltered based on simple transformations is matrices. Expanding the determinant. Solved problems related to determinants. Simultaneous linear equations in multiple variablesRepresenting a system of linear equations in multiple variables in matrix form. Using determinants to solve these systems of equations. Meaning of consistent, homogeneous and non-homogeneous systems of equations. Theorems relating to consistency of systems of equations. Application of Cramer rule. Solved problems demonstrating how to solve linear equations using matrix and determinant related methods. Basic concepts in Linear Algebra and Vector spacesTheory and definitions. Closure, commutative, associative, distributive laws. Defining Vector space, subspaces, linear dependence, dimension and bias. A few introductory problems proving certain sets to be vector spaces. Introductory problems related to Vector Spaces - Problems demonstrating the concepts introduced in the previous tutorial. Checking or proving something to be a sub-space, demonstrating that something is not a sub-space of something else, verifying linear independence; problems relating to dimension and basis; inverting matrices and echelon matrices. More concepts related to Vector SpacesDefining and explaining the norm of a vector, inner product, Graham-Schmidt process, co-ordinate vectors, linear transformation and its kernel. Introductory problems related to these. Problems related to linear transformation, linear maps and operators - Solved examples and problems related to linear transformation, linear maps and operators and other concepts discussed theoretically in the previous tutorial. Definitions of Rank, Eigen Values, Eigen Vectors, Cayley Hamilton Theorem Eigenvalues, eigenvectors, Cayley Hamilton Theorem More Problems related to Simultaneous Equations; problems related to eigenvalues and eigenvectors Demonstrating the Crammer rule, using eigenvalue methods to solve vector space problems, verifying Cayley Hamilton Theorem, advanced problems related to systems of equations. Solving a system of differential equations . A few closing problems in Linear AlgebraSolving a recurrence relation, some more of system of equations. |
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RS Aggarwal Solutions: Number System Exercise - 1F Notes | EduRev
Class 6 : RS Aggarwal Solutions: Number System Exercise - 1F Notes | EduRev
The document RS Aggarwal Solutions: Number System Exercise - 1F Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
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Find the estimated quotient for each of the following :
Q.1. 87/28.
Ans. 87/28
87 is estimated to the nearest ten = 90
28 is estimated to the nearest ten = 30
∴ 90/30 = 3
Q.2. 83/17.
Ans. 83/17
83 is estimated to the nearest ten = 80
17 is estimated to the nearest ten = 20
∴ 80/20 = 4
Q.3. 75/23.
Ans. 75/23
75 is estimated to the nearest ten = 80
23 is estimated to the nearest ten = 20
∴ 80/20 = 4
Q.4. 193/24.
Ans.193/24
193 is estimated to the nearest ten = 200
24 is estimated to the nearest ten = 20
∴ 200/20 = 10
Q.5. 725/23.
Ans. 725/23
725 is estimated to the nearest hundred = 700
23 is estimated to the nearest ten = 20
∴ 700/20 = 35
Q.6. 275/25.
Ans. 275/25
275 is estimated to the nearest hundred = 300
25 is estimated to the nearest ten = 30
∴ 300/30 = 10
Q.7. 633/33.
Sol. 633/33
633 is estimated to the nearest hundred = 600
33 is estimated to the nearest ten = 30
∴ 600/30 = 20
Q. 8. 729/29.
Ans. 729/29
729 is estimated to the nearest hundred = 700
29 is estimated to the nearest ten = 30
∴ 700/30 = 70/3 = 23 (approximately)
Q.9. 858/39.
Ans. 858/39
858 is estimated to the nearest hundred = 900
39 is estimated to the nearest ten = 40
∴ 900/40 = 90/4 = 23 (approximately)
Q.10. 868/38.
Ans. 868/38
868 is estimated to the nearest hundred = 900
38 is estimated to the nearest ten = 40
∴ 900/40 = 90/4 = 23 (approximately)
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Mathematics (Maths) Class 6
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There are four suits in a deck of cards.
# Complete the Sum 1
Practise your multiplication skills in this Complete the Sum quiz.
In this series of 11-plus verbal reasoning quizzes, you have to find the number that will complete the sum correctly. The only operations that will be used are multiply, divide, add and subtract. Choose the correct answer from the four choices available.
Sometimes you have to work out the sum to find the last number, and sometimes you have to find a number in the middle of the sum. An example has been done for you (see below and read it carefully so you understand before playing the quiz).
Speed is important in these, as they are usually easy marks to pick up, so you need to work quickly in order to leave more time for the harder questions on your actual test.
Did you know...
You can play all the teacher-written quizzes on our site for just £9.95 per month. Click the button to sign up or read more.
Example:
26 + 7 = 18 + ?
13
14
15
16
This is because 26 + 7 = 33. 18 + ? = 33. This makes 15 the only possible answer.
1. Find the number that will complete the sum below correctly.
12 x 4 = 52 - ?
This is because 12 x 4 = 48 and 52 - 4 is also 48
2. Find the number that will complete the sum below correctly.
35 - ? = 4 x 4
This is because 4 x 4 = 16. This is equal to 35 - 19. When we multiply a number by itself, the product is known as the square of that number. So 16 is the square of 4. Another example is 25 which is the square of 5
3. Find the number that will complete the sum below correctly.
? + 26 = 30 + 4.
This is because 30 + 4 = 34. To turn ? + 26 into 34 we need to add 8. Here's an interesting fact for you: a number is always divisible by 8 if its last three digits are also divisible by 8, for example 6,240
4. Find the number that will complete the sum below correctly.
19 x 4 = ? + 52
This is because 24 + 52 = 76. 19 x 4 also gives the answer of 76. Many things are arranged in fours. A deck of cards has four suits, there are four points of a compass and, if you're lucky, there are four leaves on a clover!
5. Find the number that will complete the sum below correctly.
14 ÷ 2 = 63 ÷ ?
This is because 14 ÷ 2 = 7 and 63 ÷ 9 is also equal to 7. Can you remember what a prime number is? Good. Well, 2 is the only prime number which is also an even number
6. Find the number that will complete the sum below correctly.
18 - 5 = 78 ÷ ?
This is because 18 - 5 = 13. To get from 78 to 13 we need to divide it by 6
7. Find the number that will complete the sum below correctly.
17 + ? = 49 - 13.
This is because 49 - 13 is 36. The answer must be 19 as 17 + 19 = 36. 19 is a prime number. A prime number is a number which is only divisible by itself and 1
8. Find the number that will complete the sum below correctly.
56 x 2 = 127 - ?
This is because 127 - 15 = 112 which is equal to 56 x 2
9. Find the number that will complete the sum below correctly.
? ÷ 9 = 15 - 7
This is because 15 - 7 = 8 and 72 ÷ 9 = 8
10. Find the number that will complete the sum below correctly.
45 - ? = 13 x 3
This is because 45 - 6 = 39, which is the same as = 13 x 3. Some people consider 13 to be an unlucky number and Friday 13th is thought to be an unlucky date. Any month's 13th day will fall on a Friday if the month begins with a Sunday
Quiz yourself clever - 3 free quizzes in every section |
What is a BODMAS Rule and How It Is Applied in Mathematics
What is BODMAS in math? In mathematics, BODMAS or Bodmas rule identifies the order of operations to follow when carrying out calculations. A child needs to know in which order to perform operations to get the right answer. This guide will help you understand how to apply the rule to math problems.
What is BODMAS?
What does BODMAS stand for? BODMAS stands for Brackets, Orders (powers and roots), Division, Multiplication, Addition, and Subtraction. The interpretation of the rule is the following: you have to solve brackets first, then orders (powers, square roots), division and multiplication, and lastly addition and subtraction.
The BODMAS order of operations is important because we all may interpret arithmetic expressions differently. One person might add first, and another one might multiply first. The use of BODMAS makes sure that everyone makes calculations in a given mathematical expression following the same pattern.
The BODMAS helps us know the correct order of operations to solve mathematical questions. Also, the rule tells us that we must work out the answer to any sum in a certain order. Here is the summary of how we can arrive at the correct answer when dealing with BODMAS:
Numbers
Numbers are the objects which allow us to count things, compare amounts, and make calculations. We do not give numbers priority in BODMAS. We treat them as equals, whether they are small or large.
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Operators
Operators are symbols that perform mathematical operations on a value. They are applicable in expressions, which are groups of symbols that evaluate a single value. Also, mathematical operators are symbols that represent an action, such as addition or multiplication. Operators do specific actions when used in mathematical equations. These are:
BODMAS Example
The order of operations is always important. For example, let’s take the expression
3 + 2 x 5 =
Since there are no brackets and orders, you must perform multiplication before you add the numbers. This is because multiplication has a higher precedence than addition.
BODMAS Meaning
BODMAS full form is Brackets, Orders, Division, Multiplication, Addition, and Subtraction. You need to follow this order when you complete a calculation that involves more than one operation. If you fail to follow it, you will get the wrong answer.
How to Use BODMAS Rule
BODMAS rule in math is:
• If there are brackets, then first solve the Bracket.
• If no brackets are there, then first solve Orders.
• If no orders are evident, then solve Multiplication and Division.
• If no multiplication and division are there, then solve Addition and Subtraction.
• Add and subtract in order from left to right.
The order in which you perform operations is important because it determines how to calculate the final answer. For instance, you have to calculate the answer to this equation:
2 + 3 x 4
You would get an answer of 20 if you performed the addition first, and then the multiplication. However, if you performed the multiplication first and the addition then, you would get an answer of 14. In other words, the order in which you perform operations can change the final answer.
BODMAS vs. PEMDAS
BODMAS and PEMDAS are both acronyms designed to help you remember the order of operations in math. They can also be used when solving equations as both stand for order of operations.
The difference between these two rules is that they work differently. We use BODMAS when there are multiple operations within parentheses or brackets. We apply PEMDAS in case of multiple operations outside of parentheses or brackets.
PEMDAS stands for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction, whereas BODMAS stands for Brackets, Orders (or exponents), Division, Multiplication, Addition, and Subtraction.
The two methods have the same steps to use when solving a problem. The difference is in how you apply them. In BODMAS, operation orders follow left to right, while PEMDAS applies them from right to left.
How and When Is BODMAS Taught?
BODMAS should be taught after children have learned the four operations with numbers: addition, subtraction, multiplication, and division. The BODMAS rule is usually studied in elementary school by students aged 9-12 and used to perform mathematical operations in the proper order.
Conclusion
When you are unsure about the sequence of operations, use BODMAS to help guide your choices.
Once a teacher gets convinced that children are grasping the concept of BODMAS and are able to apply it for simple problems, they may proceed using the rule in more complex problems.
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Title
Properties of Parabolas
Problem Statement
Compare the graphs of quadratic functions in the form: y = ax2 + bx + c
a. What are the coordinates of the y-intercept?
b. What must be true in order for the graph to reflect over the y-axis?
c. What must be true in order for the graph to open down?
d. What must be true in order for the graph to open up?
Problem setup
What happens to the parabola represented by the quadratic equation y = ax
2 + bx + c when the variables are changed from positive to negative?
Plans to Solve/Investigate the Problem
We will use the graphing calculator to observe the quadratic function y = ax2 + bx + c to find the coordinates of the y-intercept. We will try different values for a, b, and c in order to explore the results on the graph.
Investigation/Exploration of the Problem
a.) In the function y=ax2 + bx + c, c is our y intercept and our constant. When we first enter the function, we find that a, b, and c are all one and the parabola opens upward with the vertex at the point (-0.5, 0.75). When we change the value of a to 3 and keep b and c equal to one, the parabola moves to the right and becomes more narrow with the vertex at (-0.234375, 0.93041992).
Blue y = ax2 + bx + c
Green y = 3x2 + bx + c
b.) In order to make the parabola reflect over the y axis, we found that b must be 0. At the point (0,0) the parabola is symmetrical. In other words, all parabolas in the form y = ax2 + bx + c, where b is 0, because b x 0 is 0, leaves the equation y=ax2 + c, and will be reflected symmetrically over the y axis.
c. and d.) When the value of a is a positive number in the equation y = x2 + x + 1, so that y = ax2 + bx + c, the parabola opens upward. When the value of a is changed to a negative number, the parabola reverses and opens downward. In our example, we show the vertex moves to (0.25, 1.125) when a is -2.
Purple
Red
In our original investigations, we tried different values for a, b, and c. When we returned a to one, and changed b to 4, with c remaining at one. The parabola we found had x and y intercepts were both one. Changing b to -4 shifted the parabola’s vertex to the first quadrant and kept the y intercept at one. We wondered if changing b to -4 would move the parabola’s vertex further to the right. In fact, the vertex became (2,-3) and the parabola increased in size. Reversing the value of b to +4, the parabola moved to the mirror image position with the vertex at (-2,-3).
Red
Purple
In an attempt to create another parabola that opens downward, we changed c to a negative number, but found that the parabola remained open upwards, with only the y intercept changing to -2.
Purple
Red
Blue
Extensions of the Problem
If ax2 + bx + c = 0, where is the solution on the graph?
The above graph is the graph 2x2 + x - 3 = y. As we can see, the graph intersects the x axis at x= -1.5 and x = 1 which are the roots of the equation 2x2 + x – 3 = 0. Actually this graph represents the roots of the equation, and the roots of the equation ax2 + bx + c = 0 are where the graph intersects the x axis.
Author & Contact
Jill Jackson and Shirley Crawford.
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# 8.2: Testing a Proportion Hypothesis
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Test a hypothesis about a population proportion by applying the binomial distribution approximation.
• Test a hypothesis about a population proportion using the P\begin{align*}P-\end{align*}value.
## Introduction
In the previous section we studied the test statistic that is used when you are testing hypotheses about the mean of a population and you have a large sample (>30)\begin{align*}(>30)\end{align*}.
Often statisticians are interest in making inferences about a population proportion. For example, when we look at election results we often look at the proportion of people that vote and who this proportion of voters choose. Typically, we call these proportions percentages and we would say something like “Approximately 68 percent of the population voted in this election and 48 percent of these voters voted for Barack Obama.”
So how do we test hypotheses about proportions? We use the same process as we did when testing hypotheses about populations but we must include sample proportions as part of the analysis. This lesson will address how we investigate hypotheses around population proportions and how to construct confidence intervals around our results.
### Hypothesis Testing about Population Proportions by Applying the Binomial Distribution Approximation
We could perform tests of population proportions to answer the following questions:
• What percentage of graduating seniors will attend a 4-year college?
• What proportion of voters will vote for John McCain?
• What percentage of people will choose Diet Pepsi over Diet Coke?
To test questions like these, we make hypotheses about population proportions. For example,
H0:35%\begin{align*}H_0: 35\%\end{align*} of graduating seniors will attend a 4-year college.
H0:42%\begin{align*}H_0:42\%\end{align*} of voters will vote for John McCain.
H0:26%\begin{align*}H_0:26\%\end{align*} of people will choose Diet Pepsi over Diet Coke.
To test these hypotheses we follow a series of steps:
• Hypothesize a value for the population proportion P\begin{align*}P\end{align*} like we did above.
• Randomly select a sample.
• Use the sample proportion p^\begin{align*}\hat{p}\end{align*} to test the stated hypothesis.
To determine the test statistic we need to know the sampling distribution of the sample proportion. We use the binomial distribution which illustrates situations in which two outcomes are possible (for example, voted for a candidate, didn’t vote for a candidate), remembering that when the sample size is relatively large, we can use the normal distribution to approximate the binomial distribution. The test statistic is
zz=sample estimatevalue under the null hypothesisstandard error under the null hypothesis=p^p0p0(1p0)n\begin{align*}z &= \frac{\text{sample estimate}-\text{value under the null hypothesis}}{\text{standard error under the null hypothesis}}\\ z &= \frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\end{align*}
where:
p0\begin{align*}p_0\end{align*} is the hypothesized value of the proportion under the null hypothesis
n\begin{align*}n\end{align*} is the sample size
Example: We want to test a hypothesis that 60 percent of the 400 seniors graduating from a certain California high school will enroll in a two or four-year college upon graduation. What would be our hypotheses and the test statistic?
Since we want to test the proportion of graduating seniors and we think that proportion is around 60 percent, our hypotheses are:
H0:pHa:p=.6.6\begin{align*}H_0: p &= .6\\ H_a: p & \neq .6\end{align*}
The test statistic would be z=p^.6.6(1.6)n\begin{align*}z=\frac{\hat{p}-.6}{\sqrt{\frac{.6(1-.6)}{n}}}\end{align*}. To complete this calculation we would have to have a value for the sample size (n).
### Testing a Proportion Hypothesis
Similar to testing hypotheses dealing with population means, we use a similar set of steps when testing proportion hypotheses.
• Determine and state the null and alternative hypotheses.
• Set the criterion for rejecting the null hypothesis.
• Calculate the test statistic.
• Decide whether to reject or fail to reject the null hypothesis.
• Interpret your decision within the context of the problem.
Example: A congressman is trying to decide on whether to vote for a bill that would legalize gay marriage. He will decide to vote for the bill only if 70 percent of his constituents favor the bill. In a survey of 300 randomly selected voters, 224 (74.6%) indicated that they would favor the bill. Should he or should he not vote for the bill?
First, we develop our null and alternative hypotheses.
H0:pHa:p=.7>.7\begin{align*}H_0: p &=.7\\ H_a: p &> .7\end{align*}
Next, we should set the criterion for rejecting the null hypothesis. Choose α=.05\begin{align*}\alpha=.05\end{align*} and since the null hypothesis is considering p>.7\begin{align*}p > .7\end{align*}, this is a one tailed test. Using a standard z\begin{align*}z\end{align*} table or the TI 83/84 calculator we find the critical value for a one tailed test at an alpha level of .05 to be 1.645.
The test statistic is z=.74.7.7(1.7)3001.51\begin{align*}z=\frac{.74-.7}{\sqrt{\frac{.7(1-.7)}{300}}} \approx1.51 \end{align*}
Since our critical value is 1.645 and our test statistic is 1. 51, we cannot reject the null hypothesis. This means that we cannot conclude that the population proportion is greater than .70 with 95 percent certainty. Given this information, it is not safe to conclude that at least 70 percent of the voters would favor this bill with any degree of certainty. Even though the proportion of voters supporting the bill is over 70 percent, this could be due to chance and is not statistically significant.
Example: Admission staff from a local university is conducting a survey to determine the proportion of incoming freshman that will need financial aid. A survey on housing needs, financial aid and academic interests is collected from 400 of the incoming freshman. Staff hypothesized that 30 percent of freshman will need financial aid and the sample from the survey indicated that 101 (25.3%) would need financial aid. Is this an accurate guess?
First, we develop our null and alternative hypotheses.
H0:pHa:p=.3.3\begin{align*}H_0: p &= .3\\ H_a: p & \neq .3\end{align*}
Next, we should set the criterion for rejecting the null hypothesis. The .05 alpha level is used and for a two tailed test the critical values of the test statistic are 1.96 and -1.96.
To calculate the test statistic:
z=.25.3.3(1.3)4002.18\begin{align*}z=\frac{.25-.3}{\sqrt{\frac{.3(1-.3)}{400}}} \approx -2.18\end{align*}
Since our critical values are ±1.96\begin{align*}\pm 1.96\end{align*} and 2.18<1.96\begin{align*}-2.18 < -1.96\end{align*} we can reject the null hypothesis. This means that we can conclude that the population of freshman needing financial aid is significantly more or less than 30 percent. Since the test statistic is negative, we can conclude with 95% certainty that in the population of incoming freshman, less than 30 percent of the students will need financial aid.
## Lesson Summary
In statistics, we also make inferences about proportions of a population. We use the same process as in testing hypotheses about populations but we must include hypotheses about proportions and the proportions of the sample in the analysis. To calculate the test statistic needed to evaluate the population proportion hypothesis, we must also calculate the standard error of the proportion which is defined as sp=p0(1p0)n\begin{align*}s_p=\sqrt{\frac{p_0(1-p_0)}{n}}\end{align*}
The formula for calculating the test statistic for a population proportion is
z=p^p0p0(1p0)n\begin{align*}z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\end{align*}
where:
p^\begin{align*}\hat{p}\end{align*} is the sample proportion
p0\begin{align*}p_0\end{align*} is the hypothesized population proportion
We can construct something called the confidence interval that specifies the level of confidence that we have in our results. The formula for constructing a confidence interval for the population proportion is p^±zα2(p^(1p^)n)\begin{align*}\hat{p} \pm z_{\frac{\alpha}{2}} \left(\frac{\hat{p}(1-\hat{p})}{n}\right)\end{align*}.
For an explanation on finding the mean and standard deviation of a sampling proportion, p, and normal approximation to binomials (7.0)(9.0)(15.0)(16.0), see American Public University, Sampling Distribution of Sample Proportion (8:24)
For a calculation of the z-statistic and associated P-Value for a 1-proportion test (18.0), see kbower50, Test of 1 Proportion: Worked Example (3:51)
## Review Questions
1. The test statistic helps us determine ___.
2. True or false: In statistics, we are able to study and make inferences about proportions, or percentages, of a population.
3. A state senator cannot decide how to vote on an environmental protection bill. The senator decides to request her own survey and if the proportion of registered voters supporting the bill exceeds 0.60, she will vote for it. A random sample of 750 voters is selected and 495 are found to support the bill.
1. What are the null and alternative hypotheses for this problem?
2. What is the observed value of the sample proportion?
3. What is the standard error of the proportion?
4. What is the test statistic for this scenario?
5. What decision would you make about the null hypothesis if you had an alpha level of .01?
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Question
Draw 'less than' as well as 'more than' ogives for the following data: Weight (in Kg) 30−34 35−39 40−44 45−49 50−54 55−59 60−64 Frequency 3 5 12 18 14 6 2
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Solution
Before proceeding to construct the ogives, we first need to convert the given inclusive series into an exclusive series using the following formula. The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. $\mathrm{Here},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{adjustment}=\frac{35-34}{2}=0.5$ Therefore, we add 0.5 to the upper limit and subtract 0.5 from the lower limit of each class. Weight Frequency 29.5 − 34.5 34.5 − 39.5 39.5 − 44.5 44.5 − 49.5 49.5 − 54.5 54.5 − 59.5 59.5 − 64.5 3 5 12 18 14 6 2 Now for constructing a less than ogive, we convert the frequency distribution into a less than cumulative frequency distribution as follows. Weight Cumulative Frequency Less than 34.5 Less than 39.5 Less than 44.5 Less than 49.5 Less than 54.5 Less than 59.5 Less than 64.5 3 3 + 5 = 8 8 + 12 = 20 20 + 18 = 38 38 + 14 = 52 52 + 6 = 58 58 + 2 = 60 We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive. For constructing a more than ogive, we convert the frequency distribution into a more than cumulative frequency distribution as follows. Weight Cumulative Frequency More than 0 More than 34.5 More than 39.5 More than 44.5 More than 49.5 More than 54.5 More than 59.5 More than 64.5 60 60 − 3 = 57 57 − 5 = 52 52 − 12 = 40 40 − 18 = 22 22 − 14 = 8 8 − 6 = 2 2 − 2 = 0 We now plot the cumulative frequencies against the lower limit of the class intervals. The curve obtained on joining the points so plotted is known as the more than ogive.
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# Solve this
Question:
If $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$, then prove that $A^{2}-4 A-5 /=0$
Solution:
Given : $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
Now.
$A^{2}=A A$
$\Rightarrow A^{2}=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{lll}1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]$
$A^{2}-4 A-5 I$
$\Rightarrow A^{2}-4 A-5 I=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-4\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]-5\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow A^{2}-4 A-5 I=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-\left[\begin{array}{lll}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]$
$\Rightarrow A^{2}-4 A-5 I=\left[\begin{array}{lll}9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5\end{array}\right]$
$\Rightarrow A^{2}-4 A-5 I=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=0$
Hence proved. |
# How to Find the Mean
How to Find the Mean
• The mean is an average.
• We will find the average number of counters in these 3 groups.
• To find the mean, we add up the numbers to find the total.
• 2 + 4 + 3 = 9.
• We now share the counters out equally between the 3 groups.
• 9 ÷ 3 = 3
• The mean number of counters is 3.
To find the mean we add the numbers up and divide this total by how many numbers there are.
• To find the mean of a set of numbers, add the numbers and divide by how many numbers there are.
• Adding the numbers we have: 2 + 4 + 3 = 9.
• There are 3 numbers in total, which are ‘2’, ‘4’ and ‘3’.
• 9 ÷ 3 = 3.
• The mean of the numbers is 3.
# How to Find the Mean
The mean is an average of a set of numbers.
To calculate the mean of a set of numbers, use the following steps:
1. Add the numbers to make a total.
2. Divide this total by how many numbers there are.
First we will look at what the mean is.
Below, we have three groups of counters.
We will share these counters equally between the three groups.
We begin by grouping all of the counters together as a total.
We then divide them equally across the three groups.
There are now 3 counters in each group.
The mean is 3.
When teaching the mean, we can introduce the idea of an average through physical objects like counters. However once the idea is understood, it is best to teach the mean with mathematical steps.
We will look at the same example but this time we will use the steps for finding a mean.
2 + 4 + 3 = 9
Step 2: Divide by how many numbers there are.
There are three numbers (‘2’, ‘4’ and ‘3’), so we divide 9 by 3.
9 ÷ 3 = 3
Therefore, the mean is 3.
Here is an example of finding the mean of the four numbers: 4, 6, 7 and 3.
The first step to find the mean is to add the numbers.
It helps to look for number bonds. For example, we have 4 + 6 and 7 + 3. Both of these pairs of numbers add to make 10.
4 + 6 + 7 + 3 = 20
The second step is to divide this resulting total by how many numbers there are.
There are four numbers: ‘4’, ‘6’, ‘7’ and ‘3’.
We divide 20 by 4.
20 ÷ 4 = 5
The mean of this set of numbers is 5.
In this next example we are calculating the mean of five numbers: 6, 0, 3, 4 and 2.
The first step is to add the numbers.
We can add 6 + 4 to make 10 and then add on the 3 and the 2.
6 + 0 + 3 + 4 + 2 = 15.
The next step is to divide by how many numbers there are. There are five numbers.
A common question is whether we include the zero as one of the numbers. We do include the ‘0’. If we do not include the zero, it will give us a different answer.
15 ÷ 5 = 3
And so, the mean of this set of data is 3.
Now try our lesson on Finding the Median where we learn how to calculate the median.
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What Percentage Is 4 Out of 14
## What Percentage Is 4 Out of 14
The fraction 14/22 is 4/14. What else is 14/22?
### Multiply
via GIPHY
Let's give ourselves a little bit of practice with percentages. So let's ask ourselves, what percent of-- I don't know, let's say what percent of 16 is 4? And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying, what fraction of 16 is 4? And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4/16, which is the same thing as 1/4. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1/4 of 16. But that still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. Percent literally means "per cent." The word "cent" you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator to go from-- I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1/4 is the same thing as 25/100. And another way of saying 25/100 is this is 25 per 100, or 25%. So this is equal to 25%. Now, there's a couple of other ways you could have thought about it. You could have said well, 4/16, this is literally 4 divided by 16. Well, let me just do the division and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 zero times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding zeroes to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some zeroes right over here. The decimal makes sure we keep track of the fact that we are now in the tenths, and in the hundredths, and in the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 two times. 2 times 16 is 32. If you subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80. Let's see, 16 goes into 80 five times. 5 times 16 is 80. You subtract, you have no remainder, and you're done. 4/16 is the same thing as 0.25. Now, 0.25 is the same thing as twenty-five hundredths. Or, this is the same thing as 25/100, which is the same thing as 25%.
Let's give ourselves a little bit of practice with percentages. So let's ask ourselves, what percent of-- I don't know, let's say what percent of 16 is 4? And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying, what fraction of 16 is 4? And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4/16, which is the same thing as 1/4. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1/4 of 16. But that still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. Percent literally means "per cent." The word "cent" you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator to go from-- I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1/4 is the same thing as 25/100. And another way of saying 25/100 is this is 25 per 100, or 25%. So this is equal to 25%. Now, there's a couple of other ways you could have thought about it. You could have said well, 4/16, this is literally 4 divided by 16. Well, let me just do the division and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 zero times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding zeroes to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some zeroes right over here. The decimal makes sure we keep track of the fact that we are now in the tenths, and in the hundredths, and in the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 two times. 2 times 16 is 32. If you subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80. Let's see, 16 goes into 80 five times. 5 times 16 is 80. You subtract, you have no remainder, and you're done. 4/16 is the same thing as 0.25. Now, 0.25 is the same thing as twenty-five hundredths. Or, this is the same thing as 25/100, which is the same thing as 25%. (Source: www.khanacademy.org)
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# How do you simplify 3sqrt98-4sqrt28?
Jun 25, 2018
See a solution process below:
#### Explanation:
First, rewrite the expression as:
$3 \sqrt{49 \cdot 2} - 4 \sqrt{4 \cdot 7}$
$\sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}}$
$3 \sqrt{\textcolor{red}{49} \cdot \textcolor{b l u e}{2}} - 4 \sqrt{\textcolor{red}{4} \cdot \textcolor{b l u e}{7}} \implies$
$3 \sqrt{\textcolor{red}{49}} \sqrt{\textcolor{b l u e}{2}} - 4 \sqrt{\textcolor{red}{4}} \sqrt{\textcolor{b l u e}{7}} \implies$
(3 * 7)sqrt(color(blue)(2)) - (4 * 2))sqrt(color(blue)(7)) =>
$21 \sqrt{2} - 8 \sqrt{7}$
Jun 25, 2018
$21 \sqrt{2} - 8 \sqrt{7}$
#### Explanation:
Let's see if we can factor a perfect square out of the radicals. The key here is that we can leverage the radical property
$\sqrt{a b} = \sqrt{a} \sqrt{b}$
We can rewrite $98$ as $49 \cdot 2$ and $28$ as $4 \cdot 7$. We now have
$3 \sqrt{49 \cdot 2} - 4 \sqrt{4 \cdot 7}$
$\implies 3 \cdot 7 \sqrt{2} - 4 \cdot 2 \sqrt{7}$
We can simplify this to
$21 \sqrt{2} - 8 \sqrt{7}$
Hope this helps! |
# 2015 AMC 10B Problems/Problem 9
## Problem
The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius $3$ and center $(0,0)$ that lies in the first quadrant, the portion of the circle with radius $\tfrac{3}{2}$ and center $(0,\tfrac{3}{2})$ that lies in the first quadrant, and the line segment from $(0,0)$ to $(3,0)$. What is the area of the shark's fin falcata?
$[asy] import cse5;pathpen=black;pointpen=black; size(1.5inch); D(MP("x",(3.5,0),S)--(0,0)--MP("\frac{3}{2}",(0,3/2),W)--MP("y",(0,3.5),W)); path P=(0,0)--MP("3",(3,0),S)..(3*dir(45))..MP("3",(0,3),W)--(0,3)..(3/2,3/2)..cycle; draw(P,linewidth(2)); fill(P,gray); [/asy]$
$\textbf{(A) } \dfrac{4\pi}{5} \qquad\textbf{(B) } \dfrac{9\pi}{8} \qquad\textbf{(C) } \dfrac{4\pi}{3} \qquad\textbf{(D) } \dfrac{7\pi}{5} \qquad\textbf{(E) } \dfrac{3\pi}{2}$
## Solution
The area of the shark's fin falcata is just the area of the quarter-circle minus the area of the semicircle. The quarter-circle has radius $3$ so it has area $\dfrac{9\pi}{4}$. The semicircle has radius $\dfrac{3}{2}$ so it has area $\dfrac{9\pi}{8}$. Thus, the shaded area is $\dfrac{9\pi}{4}-\dfrac{9\pi}{8}=\boxed{\mathbf{(B)}\ \dfrac{9\pi}{8}}$
## Video Solution 1
~Education, the Study of Everything
~savannahsolver |
Tamilnadu State Board New Syllabus Samcheer Kalvi 11th Business Maths Guide Pdf Chapter 8 Descriptive Statistics and Probability Ex 8.1 Text Book Back Questions and Answers, Notes.
## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.1
### Samacheer Kalvi 11th Business Maths Descriptive Statistics and Probability Ex 8.1 Text Book Back Questions and Answers
Question 1.
Find the first quartile and third quartile for the given observations.
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.
Solution:
Given data are arranged in ascending order 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.
Here the number of observations is n = 11
Question 2.
Find Q1, Q3, D8, and P67 of the following data:
Solution:
Question 3.
Find lower quartile, upper quartile, 7th decile, 5th decile, and 60th percentile for the following frequency distribution.
Solution:
Lower quartile, Q1 = size of $$\left(\frac{N}{4}\right)^{\mathrm{th}}$$ value
= $$\left(\frac{120}{4}\right)^{\mathrm{th}}$$
= size of 30th value
Q1 lies in the class (40 – 50) and its corresponding values are L = 40, $$\frac{N}{4}$$ = 30, pcf = 15, f = 21 and C = 10
Q3 lies in the class (60 – 70) and its corresponding values are L = 60, $$\frac{3 \mathrm{N}}{4}$$ = 90, pcf = 79, f = 32 and C = 10.
Thus D7 lies in the class (60 – 70) and its corresponding values are L = 60, $$\frac{7 \mathrm{N}}{10}$$ = 84, pcf = 79, f = 32 and C = 10.
Thus D5 lies in the class (50 – 60) and its corresponding values are L = 50, $$\frac{5 \mathrm{N}}{10}$$ = 60, pcf = 36, f = 43 and C = 10.
Thus P60 lies in the class (50 – 60) and its corresponding values are L = 50, $$\frac{60 \mathrm{N}}{100}$$ = 72, pcf = 36, f = 43 and C = 10.
Question 4.
Calculate GM for the following table gives the weight of 31 persons in the sample survey.
Solution:
= Antilog (2.1540)
= 142.560
= 142.56 lbs
Question 5.
The price of a commodity increased by 5% from 2004 to 2005, 8% from 2005 to 2006, and 77% from 2006 to 2007. Calculate the average increase from 2004 to 2007?
Solution:
In averaging ratios and percentages, the geometric mean is more appropriate.
Let us consider X represents prices at the end of the year.
Here n = 3
= Antilog (2.1008)
= 126.1246
Average rate of increase of price = 126.1246 – 100
= 26.1246
= 26.1%
Question 6.
An aeroplane flies, along the four sides of a square at speeds of 160, 200, 300, and 400 kilometres per hour respectively. What is the average speed of the plane in its flight around the square?
Solution:
Harmonic mean would he suitable. Harmonic mean of n observations is
$$\mathrm{HM}=\frac{n}{\Sigma \frac{1}{x}}$$
Here n = 4
Question 7.
A man travelled by car for 3 days. He covered 480 km each day. On the first day, he drove for 10 hours at 48 km an hour. On the second day, he drove for 12 hours at 40 km an hour, and for the last day, he drove for 15 hours at 32 km. What is his average speed?
Solution:
Total distance covered 480 km.
The first-day distance covered 48 km.
The second-day distance covered 40 km.
Third-day distance covered 32 km.
Number of observations = 3
Average speed = HM
Question 8.
The monthly incomes of 8 families in rupees in a certain locality are given below. Calculate the mean, the geometric mean, and the harmonic mean and confirm that the relations among them hold true. Verify their relationships among averages.
Solution:
Let us first find AM. Here number of observations, n = 8
AM = $$\frac{70+10+50+75+8+25+8+42}{8}$$ = $$\frac{288}{8}$$ = 36
Now we will find the geometric mean. (GM)
Here number of observations is n = 8
GM = Antilog $$$\left(\frac{\Sigma \log X}{n}\right)$$$
= Antilog $$\left(\frac{11.2465}{8}\right)$$
= Antilog 1.4058
GM = 25.4566
Now we will find Harmonic Mean (HM).
Here the number of observations is n = 8.
Thus AM = 36, GM = 25.466, HM = 17.3385
36 > 25.466 > 17.3385
Hence AM > GM > HM
Thus their relations among them is verified for the given data.
Question 9.
Calculate AM, GM, and HM and also verify their relations among them for the following data:
Solution:
Question 10.
Calculate AM, GM, and HM from the following data and also find its relationship:
Solution:
Question 11.
Calculate the quartile deviation and its coefficient from the following data:
Solution:
Question 12.
Calculate quartile deviation and its relative measure from the following data:
Solution:
Thus Q1 lies in the class 20 – 30 and corresponding values are L = 20, $$\frac{\mathrm{N}}{4}$$ = 17, pcf = 15, f = 13, C = 10.
Thus Q3 lies in the class 40 – 50 and corresponding values are L = 40, $$\frac{3 \mathrm{N}}{4}$$ = 51, pcf = 46, f = 14, C = 10
Relative measure, coefficient of QD
Question 13.
Compute mean deviation about median from the following data:
Solution:
Question 14.
Compute the mean deviation about mean from the following data:
Solution:
Question 15.
Find out the coefficient of mean deviation about median in the following series:
Solution:
The class interval corresponding to cumulative frequency 75 is (40 – 50).
So, the corresponding values from the median class are L = 40, pcf = 56, f = 37, C = 10, N = 75.
Now we calculate the mean deviation about the median 45.11 |
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Lesson_8.1_ExploringLogarithms
# Lesson_8.1_ExploringLogarithms - (a y = 5 x(b Example#2...
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Lesson_8.1_ExploringLogarithms.notebook 8.1 Exploring the Logarithmic Function What is the inverse of the exponential function y = a x ? We can rewrite the inverse of the exponential function as a logarithm (also known as a logarithmic function ). The following two statements are equivalent: x = a y y=log a x How do we rewrite exponentials as logarithms? y = log a x base exponent answer is equivalent to x = a y exponent = log base answer Memory Aid: eg. 3 2 = 9 becomes 2 = log 3 9
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Lesson_8.1_ExploringLogarithms.notebook Example #1: Write the equation of the inverse of each in
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Unformatted text preview: (a) y = 5 x (b) Example #2: Write the equation of each logarithmic in exponential form, then write the equation of the inverse. (a) y = log 5 x (b) Lesson_8.1_ExploringLogarithms.notebook Example #3: Evaluate each of the following: (a) log 5 25 (b) log 3 27 (c) Example #4: (a) Sketch the graph of y = 2 x on the grid provided. (b) Sketch the graph of the inverse of y = 2 x on the same grid. Lesson_8.1_ExploringLogarithms.notebook Page 451 # 1ac, 2, 5 - 10 Homework...
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# 2009 Indonesia MO Problems/Problem 2
## Problem
For any real $x$, let $\lfloor x\rfloor$ be the largest integer that is not more than $x$. Given a sequence of positive integers $a_1,a_2,a_3,\ldots$ such that $a_1>1$ and $$\left\lfloor\frac{a_1+1}{a_2}\right\rfloor=\left\lfloor\frac{a_2+1}{a_3}\right\rfloor=\left\lfloor\frac{a_3+1}{a_4}\right\rfloor=\cdots$$ Prove that $$\left\lfloor\frac{a_n+1}{a_{n+1}}\right\rfloor\leq1$$ holds for every positive integer $n$.
## Solution
By assuming $\left\lfloor\frac{a_n+1}{a_{n+1}}\right\rfloor >2$ we can conclude that: $$\left\lfloor\frac{a_1+1}{a_2}\right\rfloor=\left\lfloor\frac{a_2+1}{a_3}\right\rfloor=\left\lfloor\frac{a_3+1}{a_4}\right\rfloor=\cdots >2$$
Since $\left\lfloor\frac{a_1+1}{a_{2}}\right\rfloor >2$ , it is also true that $\frac{a_1+1}{a_{2}} > 2$, implying $a_1 > 2a_2 - 1 > a_2$.
After repeating this same process to all given fractions, we get: $a_1 > a_2 > a_3 > ...$ which is a impossible statement because $a_1, a_2, a_3, ...$ are all positive integers.
Therefore, $\left\lfloor\frac{a_n+1}{a_{n+1}}\right\rfloor\leq1$
~NounZero
## See Also
2009 Indonesia MO (Problems) Preceded byFirst Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed byProblem 3 All Indonesia MO Problems and Solutions |
# What are 5 examples of quadratic equations
Examples of the standard form of a quadratic equation (ax² + bx + c = 0) include:
• 6x² + 11x – 35 = 0.
• 2x² – 4x – 2 = 0.
• -4x² – 7x +12 = 0.
• 20x² -15x – 10 = 0.
• x² -x – 3 = 0.
• 5x² – 2x – 9 = 0.
• 3x² + 4x + 2 = 0.
• -x² +6x + 18 = 0.
Similarly, how do you solve examples of quadratic equations?
What is quadratic equation with examples? In mathematics, we define a quadratic equation as an equation of degree 2, which means that the highest exponent of this function is 2. The standard form of the quadratic is y = ax^2 + bx + c, where a, b, and c are numbers and a cannot be 0. Examples of quadratic equations all include: y = x^2 + 3x + 1 .
What is a quadratic equation in mathematics? Definitions: A quadratic equation has one variable x ax2+bx+c=0 . In a second order polynomial equation . 0 . with . Because it is a second-order polynomial equation, the Fundamental Theorem of Algebra guarantees that it has at least one solution. The solution can be real or complex.
Second how do you solve quadratic equations of square 9?
## What is the quadratic formula of class 9?
• Quadratic Equation: The quadratic equation in the variable x has the form. ax 2 + bx + c = 0 , where a, b, c are real numbers and a 0. • Roots of Quadratic Equation: A real number α is said to be the root of. The quadratic equation ax2 + bx + c = 0, if aα2 + bα + c = 0.
So why do we solve quadratic equations? So why are quadratic functions important? Quadratic functions have a unique place in the school curriculum . They are functions whose values can be easily computed from the input values, so they move a bit over linear functions and provide a significant step away from straight lines by implication. (What are 5 examples of quadratic equations)
The given equation is 3m+8=15. …the highest degree of the equation is 1. However, it is known that the highest degree of a quadratic equation is 2. Should be . Therefore, the given equation is not quadratic.
## What are not examples of quadratic equation?
• bx – 6 = 0 is not a quadratic equation because there is no x . There is no period.
• – x -5 = 0 is not a quadratic equation because one x . Has terms (not allowed in quadratic equations).
Which is a quadratic term? The tea term ax2 is called the quadratic term (hence the name given to the function), the term bx is called the linear term, and the term c is called the constant term. … The graphs of all quadratic functions are parabolas.
What is the quadratic formula of class 10th?
There are polynomial equations of degree 2 in one variable of the type quadratic equation f(x) = ax 2 + bx + c where a, b, c, r and a0.
What is the root of quadratic equation? The roots of a quadratic equation are the values of the variables that satisfy the equation . They are also referred to as the “solution” or “zero” of a quadratic equation. For example, the quadratic equation x . The roots of 2 – 7x + 10 = 0 are x = 2 and x = 5 because they satisfy the equation.
## Which is not a quadratic equation?
bx – 6 = 0 is not a quadratic equation because no x2 . No periods. x 3 – x 2 -5 = 0 is not a quadratic equation because an x3 . Has terms (not allowed in quadratic equations).
How is quadratic equation used in real life?
Answer In daily life we use the quadratic formula to calculate areas, determine the profit of a product or to calculate the speed of an object . Furthermore, quadratic equation refers to an equation that has at least one square variable. (What are 5 examples of quadratic equations)
• military and law enforcement. Quadratic equations are often used to describe the motion of objects flying in the air. ,
• Engineering. Engineers of all kinds use these equations. ,
• science. ,
• Management and clerical work. ,
• agriculture.
Quadratic or not? How to know if the equation is quadratic? We just check the degree of the equation. If the power of the equation is equal to 2, then it is only a quadratic equation.
## What type of equation is 12 4x 0?
Linear equation with one unknown.
How do you solve quadratic equations by factorization?
What are the steps to solve a quadratic equation by factorization?
To solve a quadratic equation using factoring:
1. 1. Using the standard form, transform the equation with one of the sides being zero.
2. 2. Factorise the non-zero side.
3. 3. Set each factor to zero (remember: the product of the factors is zero if and only if one or more factors are zero).
4. 4. Solve each resulting equation.
How many solutions do quadratic equations have? As we have seen, there can be 0, 1, or 2 solutions to a quadratic equation depending on whether the value inside the square root sign, (b is 2 – 4ac), is positive, negative, or zero. This expression has a special name: discriminant. |
# How to Find Square Root and Cube Root of Number
Here you will learn how to find square root and cube root of a number and properties of squares and cubes.
Let’s begin –
## Square Root and Cube Root
### (a) Square and Square Roots
When any number multiplied by itself, it is called as the square of the number.
Thus, 3 $$\times$$ 3 = $$3^2$$ = 9
#### How to find square root of a given number
We will understand this by taking an example. Let 7016 be a given number.
1). Write down the number 7016 as a product of its prime factors,
7016 = $$2 \times 2 \times 2 \times 2 \times 21 \times 21$$
= $$2^4\times 21^2$$
2). The required square root is obtained by having the values of the powers.
Hence. $$sqrt{7016}$$ = $$2^2\times 21^1$$ = 84
#### Properties of Squares
1). When a perfect square is written as a product of its prime factors each prime factor will appear an even number of times.
2). The difference between the square of two consecutive natural numbers is always equal to the sum of the natural numbers. Thus, $$41^2$$ – $$40^2$$ = (40 + 41) = 81
3). The square of a number ending in 1, 5 or 6 also ends in 1, 5 or 6 respectively.
4). The square of any number ending in 5 : The last two digits will always be 25.
5). The square of any number is always non-negative.
### Cubes and Cube Roots
When a number s multiplied with itself two times, we get the cube of the number.
Thus, $$x \times x \times x$$ = $$x^3$$
#### How to find cube root of a given number
In order to find the cube root of a number, first we write it in its standard form and divide all powers by 3.
Thus, the cube root of $$3^6 \times 5^9 \times 17^3 \times 2^6$$ is given by
$$3^2 \times 5^3 \times 17^ \times 2^2$$
#### Properties of Cubes
1). When a perfect cube is written in its standard form the values of the powers on each prime factor will be a multiple of 3.
2). The cubes of all numbers (integers and decimals) greater than 1 are greater than the number itself.
3). The value of the cubes of a number between 0 and 1 is lower than the number itself.
4). The cube of a number between 0 and -1 is greater than the number itself.
5). The cube of any number less than -1, is always lower than the number. |
# Differentiation of sec inverse x
Here you will learn differentiation of sec inverse x or arcsecx x by using chain rule.
Let’s begin –
## Differentiation of sec inverse x or $$sec^{-1}x$$ :
If x $$\in$$ R – [-1, 1] . then the differentiation of $$sec^{-1}x$$ with respect to x is $$1\over | x |\sqrt{x^2 – 1}$$.
i.e. $$d\over dx$$ $$sec^{-1}x$$ = $$1\over | x |\sqrt{x^2 – 1}$$.
## Proof using chain rule :
Let y = $$sec^{-1}x$$. Then,
$$sec(sec^{-1}x)$$ = x
$$\implies$$ sec y = x
Differentiating both sides with respect to x, we get
$$d\over dx$$(sec y) = $$d\over dx$$(x)
$$d\over dx$$ (sec y) = 1
By chain rule,
sec y tan y $$dy\over dx$$ = 1
$$dy\over dx$$ = $$1\over sec y tan y$$
$$dy\over dx$$ = $$1\over | sec y | | tan y |$$
$$dy\over dx$$ = $$1\over | sec y | \sqrt{tan^2 y}$$
$$\implies$$ $$dy\over dx$$ = $$1\over | sec y | \sqrt{sec^2 y – 1}$$
$$\implies$$ $$d\over dx$$ $$sec^{-1}x$$ = $$1\over | x |\sqrt{x^2 – 1}$$
Hence, the differentiation of $$sec^{-1}x$$ with respect to x is $$1\over | x |\sqrt{x^2 – 1}$$.
Example : What is the differentiation of $$sec^{-1} x^2$$ with respect to x ?
Solution : Let y = $$sec^{-1} x^2$$
Differentiating both sides with respect to x and using chain rule, we get
$$dy\over dx$$ = $$d\over dx$$ ($$sec^{-1} x^2$$)
$$dy\over dx$$ = $$1\over | x^2 | \sqrt{x^4 – 1}$$.(2x) = $$2x\over | x^2 | \sqrt{x^4 – 1}$$
Hence, $$d\over dx$$ ($$sec^{-1} x^2$$) = $$2x\over | x^2 | \sqrt{x^4 – 1}$$
Example : What is the differentiation of x + $$sec^{-1} x$$ with respect to x ?
Solution : Let y = x + $$sec^{-1} x$$
Differentiating both sides with respect to x, we get
$$dy\over dx$$ = $$d\over dx$$ (x) + $$d\over dx$$ ($$sec^{-1} x$$)
$$dy\over dx$$ = 1 + $$1\over | x | \sqrt{x^2 – 1}$$
Hence, $$d\over dx$$ (x + $$sec^{-1} x$$) = 1 + $$1\over | x | \sqrt{x^2 – 1}$$
### Related Questions
What is the Differentiation of tan inverse x ?
What is the Differentiation of secx ?
What is the Integration of Sec Inverse x and Cosec Inverse x ? |
# Calculate f'(x) given d/dx(f(3x^4))=6x^4?
• firegoalie33
In summary, to find f'(x) we can use the power rule for derivatives. We divide both sides by 3 and substitute x for 3x^4 to get f'(x) = 2x^3. f'(x) represents the derivative of the function f(x) with respect to x, while d/dx(f(x)) represents the derivative of the entire function f(x). The quotient rule can be used to find f'(x), but in this case, the power rule is more efficient. The chain rule can also be used by multiplying the derivative of the outer function by the derivative of the inner function. The product rule is not applicable in this scenario, as we are dealing with only one function, f
firegoalie33
I am told that d/dx(f(3x^4))=6x^4. I need to calculate f ' (x). i have tried like 3 different methods and have no idea how to do this.
Can you find f'(3x4)?
By the chain rule,
$$\frac{df(3x^4)}{dx}= f'(3x^4)(12x^3)$$
So you must have $f'(3x^4)(12x^3)= 6x^4$.
That let's you find $f'(3x^4)$ per Office Shredder's question. Now, if you let $u= 3x^4$, you can find f'(u).
## 1. How do I find f'(x) given d/dx(f(3x^4))=6x^4?
To find f'(x), we can use the power rule for derivatives. We know that d/dx(x^n) = nx^(n-1), so in this case, we have d/dx(f(3x^4)) = 6x^4. This means that f'(3x^4) = 6x^4. To find f'(x), we simply need to divide both sides by 3 and substitute x for 3x^4, giving us f'(x) = 2x^3.
## 2. What is the difference between f'(x) and d/dx(f(x))?
f'(x) represents the derivative of the function f(x) with respect to x. This means that it tells us the rate of change of f(x) at a specific point, or the slope of the tangent line at that point. On the other hand, d/dx(f(x)) represents the derivative of the entire function f(x), not just at a specific point. It tells us the general rate of change of the function at any point along its curve.
## 3. Can I use the quotient rule to find f'(x) instead of the power rule?
Yes, you can use the quotient rule to find f'(x) as well. However, in this specific case, the power rule is easier and more efficient to use. The quotient rule is typically used when the function is in the form of f(x)/g(x), where f(x) and g(x) are both functions of x.
## 4. How does the chain rule apply in this scenario?
The chain rule states that when finding the derivative of a composite function, we must multiply the derivative of the outer function by the derivative of the inner function. In this case, our composite function is f(3x^4), where the outer function is f(x) and the inner function is 3x^4. Therefore, the chain rule would be used to find f'(x) by multiplying the derivative of f(x) by the derivative of 3x^4, which is 12x^3.
## 5. Can I use the product rule to find f'(x) in this case?
No, the product rule is not applicable in this scenario because we are not dealing with a product of two functions. The product rule is used when finding the derivative of a function that is in the form of f(x)g(x), where f(x) and g(x) are both functions of x. In this case, we only have one function, f(x), and d/dx(f(3x^4)) is simply a notation for the derivative of that function with respect to x.
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# 1.5 | Analyzing Graphs of Functions
• A function $$f$$ is increasing on an interval if, for any $$x_1$$ and $$x_2$$ in the interval, $$x_1 < x_2$$ implies $$f(x_1) < f(x_2)$$.
• A function $$f$$ is decreasing on an interval if, for any $$x_1$$ and $$x_2$$ in the interval, $$x_1 < x_2$$ implies $$f(x_1) > f(x_2)$$.
• A function $$f$$ is constant on an interval if, for any $$x_1$$ and $$x_2$$ in the interval, $$f(x_1) = f(x_2)$$.
A function value $$f(a)$$ is called a relative maximum of $$f$$ if there exists an interval $$(x_1, x_2)$$ that contains $$a$$ such that
$$x_1 < x < x_2 \text{ implies } f(a) \leq f(x).$$ A function value $$f(a)$$ is called a relative maximum of $$f$$ if there exists an interval $$(x_1, x_2)$$ that contains $$a$$ such that
$$x_1 < x < x_2 \text{ implies } f(a) \geq f(x).$$
The average rate of change between any two points $$(x_1, f(x_1))$$ and $$(x_2, f(x_2))$$ is the slope of the line through the two points and is usually denoted $$f_{avg}$$. The line through the two points is called a secant line, and the slope of the line is denoted $$m_{sec}$$ and can be computed using the slope formula. Therefore,
$$f_{avg} = m_{sec} = \frac{f(x_2) -f(x_1)}{x_2-x_1}$$
A function $$y = f(x)$$ is even if, for each $$x$$ in the domain of $$f$$,
$$f(-x) = f(x).$$
A function $$y = f(x)$$ is odd if, for each $$x$$ in the domain of $$f$$,
$$f(-x) = – f(x).$$ |
The sum of
the digits in
the product of
any number and
nine will always
sum to a multiple
of nine.
The difference
of any number and
the sum of its
digits will always
result in a number
that is a multiple
of nine.
The difference
between the larger
and smaller of two
numbers consisting
of exactly the same
digits will always
result in a number
whose digits sum
to a multiple of
nine.
## The First Mysterious Nine Trick
1. Have a friend think of a number between 1 and 9.
2. Then, ask them to multiply this number by 9.
more than one digit.
4. When they let you know they have finished, pretend you are thinking real
hard and then tell them they got nine.
If they have a pencil and paper handy, you can ask them to sum the digits in the
product of any number and nine, until they get one digit. Once they arrive at
that digit, it will always be nine.
Take a look at the nine times table below. Do you see why this works?
1 x 9 = 9..........9 = 9
2 x 9 = 18..... 1 + 8 = 9
3 x 9 = 27..... 2 + 7 = 9
4 x 9 = 36..... 3 + 6 = 9
5 x 9 = 45..... 4 + 5 = 9
6 x 9 = 54..... 5 + 4 = 9
7 x 9 = 63..... 6 + 3 = 9
8 x 9 = 72..... 7 + 2 = 9
9 x 9 = 81..... 9 + 1 = 9
10 x 9 = 90..... 9 + 0 = 9
11 x 9 = 99..... 9 + 9 = 18 and 1 + 8 = 9
12 x 9 = 108..... 1 + 0 + 8 = 9
Now, let's try this with a random number, say 1492.
1492 x 9 = 13,428
Now add the digits 1 + 3 + 4 + 2 + 8 = 18. Since18 is more than two digits,
keep adding across until you get one digit. 1 + 8 = 9!
Try it for yourself.
## The Second Mysterious Nine Trick
1. Have a friend pick a random three or four digit number. Yes, this trick
will work for digits of any length, but don't make it too difficult for
the other person.
3. When they have done this, ask them to subtract this sum from the number.
4. After they subtract, ask them to circle one of the digits in their
answer that is not a nine, and tell you only the numbers they did not circle.
5. To find the digit they circled you must add the digits they did not
circle, and then use the following rules:
• Rule number one: if the sum of the digits is less than or equal to 9,
subtract this sum from 9.
• Rule number two: if the sum of the digits is more than 9, add the
digits again until you arrive at single digit, and then subtract
this number from 9.
Your result will be the number they circled.
Here is an example:
Random number: 7832
Sum the digits: 7 + 8 + 3 + 2 = 20
Subtract this sum from the original number 7832 - 20 = 7812
Circle one of the digits in the answer.
Let's assume your friend circled the 2 in the number 7812. In this case the
other person would tell you that the numbers not circled were 7, 8, and 1.
You add these, and get 7 + 8 + 1 = 16. Since 16 is more than two digits you keep
adding digits to get 1 + 6 = 7. Since 7 is less than 9 follow Rule number one and
subtract this sum from 9. 9-7=2. 2 is the number they circled!
Here is why this works.
For any random number, if you subtract the sum of the random number's digits from
the original number, the sum of the digits in the difference always sums to 9.
Look at the example above.
The difference was 7812, and 7+ 8 + 1 + 2 = 18, and 1 + 8 = 9!
So, if you know that the sum of the digits in the difference is always 9, and
you know all the digits except one, then the number circled plus the ones you
know must sum to nine. What you know plus what you don't know = nine,
therefore, nine minus what you know is what you don't know. Got it?
Let the computer guess
## The Third Mysterious Nine Trick
1. Have a friend pick a random three or four digit number, for example 3428
2. Ask them to scramble the same digits to form a different number, for example 4823
3. Tell them to subtract the smaller from the larger, 4823-3428 = 1395
4. When they have done this, ask them to circle one of the digits in the
answer that is not a nine and tell you the ones not circled.
Let's say that the other person circled the 5, and reported that
the numbers not circled were 1, 3, and 9. To find the number they circled; add
the numbers they did not circle and use one of the following rules to
solve the mystery.
• Rule number one: if the sum of the digits is less than or equal to 9,
subtract this sum from 9.
• Rule number two: if the sum of the digits is more than 9, add the
digits again until you arrive at single digit, and then subtract
this number from 9.
Your result will be the number they circled.
In this case, the sum of the digits is 13. 1 + 3 + 9 = 13 so we would follow Rule
number two, and tell the other person that the number they circled was a 5
after first adding the 1 and 3 in 13 to get 4, and then subtracting this from 9.
Here is why this works.
For any random number, if you rearrange the digits in this number, and then
subtract the smaller from the larger, the sum of the digits in the difference
always sums to 9, if you keep summing until you get only one digit.
Using the example above, the difference was 1395, 1 + 3 + 9 + 5 = 18 and 1 + 8 = 9!
So, if you know that the sum of the digits is always 9, and you also know all
the digits except one, then the sum of the number circled and the ones you were
told must equal nine. What you know plus what you don't know = nine,
therefore, nine minus what you know is what you don't know. Got it?
Would you like the computer to guess? |
# How do you find the domain and range of f(x)= sqrt(4-x^2) /( x-3)?
Sep 16, 2015
Domain $\left\{x : \mathbb{R} , - 2 \le x \le 2\right\}$
Range $\left\{y : \mathbb{R} , - \frac{2}{\sqrt{5}} \le y \le 0\right\}$
#### Explanation:
Domain is quite evident that $x \ne 3$ and $x \le + 2$ and $- 2 \le x$ which essentially boils down to $\left\{x : \mathbb{R} , - 2 \le x \le 2\right\}$
For finding the range, consider the domain from which it can be seen that y would never have any positive value and at end points +2, -2 it is 0, hence $y \le 0$ on the upper side.
Next square both sides, so that it is a quadratic equation in x,
${x}^{2} \left({y}^{2} + 1\right) - 6 {y}^{2} x + 9 {y}^{2} - 4 = 0$. Solve it for x using quadratic formula, x= (3y^2 +- sqrt( -5y^2 +4) )/((y^2+1)
For a function Real to Real $5 {y}^{2} \le 4$ or $\pm y \le \frac{2}{\sqrt{5}}$. Since it is already settled that $y \le 0$ on the upper side, reject $y \le \frac{2}{\sqrt{5}}$' Hence it should be $- y \le \frac{2}{\sqrt{5}}$ or $y \ge - \frac{2}{\sqrt{5}}$.
The range of the function would this be$\left\{y : \mathbb{R} , - \frac{2}{\sqrt{5}} \le y \le 0\right\}$ |
# Question 15 & 16 Exercise 4.5
Solutions of Question 15 & 16 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
A man wishes to save money by setting aside Rs. I the first day, Rs. $2$ the second day, Rs. $4$ the third day, and so on, doubling the amount each day. If this continụed, how much must be set aside on the $15^{\text {th }}$ day? What is the total amoumt saved at the end of thirty days?
Let the money saves on first day $a_1=R s .1$
Saves on second day $a_2=R s .2$
Saves on third day $a_3=R s .4$ and so on.
Hence the sequence is $1,2,4,8, \ldots$ is a geometric sequence, with $a_1=1 . \quad r=2 . \quad n=15$
We know that $a_n=a_1 r^{n-1}$.
The money he saves on $15^{1 / 2}$ day is $$a_{15}=a_1 r^{14}$$ becomes in the given case $$a_{15}=1 .(2)^{1 4}=R s .16384$$ Now the total amount saved at the end of 30 days is
$$S_{30}=\dfrac{a_1(r^{30}-1)}{r-1}$$ putting $r-2$ and $a_1=1$, then
\begin{align}S_{30}&=\dfrac{1[2^{30}-1]}{2-1}=2^{30}-1 \\ \Rightarrow S_{30}&=R s .1073741823 \end{align} $$\text{Rs.}=16384; \text{Rs.}= 1073741823$$
The number of bacteria in a culture increased geometrically from $64000$ to $729000$ in $6$ days. Find the daily rate of increase if the rate is assumed to be constant.
The number of bacteria at the start $a_1=64000$
Let number of bacteria after first day be $a_2$, after second be $a_3$ and so on.
The number of bacteria after $6$th day $a_7=729000$. Since the number is increasing geometrically,
we know that $a_n=a_1 r^{n-1}$, putting $a_1$ and $n=7$, then \begin{align}&729000=64000 r^{7-1} \\ & \Rightarrow r^6=\dfrac{729000}{64000} \\ & \Rightarrow r^6=\dfrac{729}{64}=\dfrac{3^6}{2^6} \\ & \Rightarrow r^6=(\dfrac{3}{2})^6 \\ & \Rightarrow r=\dfrac{3}{2}\end{align} Hence the number of bacteria increasing with $\dfrac{3}{2}$ of the slarting number each day. |
# Triangle constructions
## Lesson details
### Key learning points
1. In this lesson, we will learn how to construct triangles given two sides and an angle.
### Licence
This content is made available by Oak National Academy Limited and its partners and licensed under Oak’s terms & conditions (Collection 1), except where otherwise stated.
## Video
Share with pupils
## Worksheet
Share with pupils
## Starter quiz
Share with pupils
### 5 Questions
Q1.
Complete the sentence: We can draw different triangles that have the same ______ angles using a ruler and a ________.
exterior, compass
exterior, protractor
interior, compass
Q2.
What is the length of the missing side, k, on the larger triangle?
10.4cm
10.8 cm
2.7cm
Q3.
Yasmin and Xiavier are talking about constructing triangles. Yasmin says you need 2 sides lengths to be able to construct a triangle. Xiavier says you need a protractor to construct a triangle. Who is correct?
Both Yasmin and Xiavier
Only Xiavier
Only Yasmin
Q4.
What equipment would you use to construct 95°, 6cm, 55° triangle correctly?
Others
Ruler and pencil
Ruler, pencil and compass
Correct answer: Ruler, pencil and protractor
Q5.
Both angles have the same interior angles of ____, _____ and _______
25°, 25°, 48°
48°, 25°, 73°
48°, 48°, 25°,
## Exit quiz
Share with pupils
### 5 Questions
Q1.
How many can be constructed with the following angle measures?100°, 40°, 40°
Exactly one triangle can be constructed
Correct answer: More than one triangles can be constructed
No triangles can be constructed
Q2.
How many triangles can you draw with three side lengths of 4 cm, 5 cm and 10cm?
Four possibilities
One possibility
Two possibilities
Q3.
Complete the sentence about the image below: You can construct different triangles if you are given two _____ and one _____ angle.
exterior, side
interior, side
side, exterior
Q4.
For the image below, both of the triangles above have side lengths of ______ and 7cm and an interior angle of ______.
4cm and 148° |
Worksheet given in this section will be much useful for the students who would like to practice problems on factoring quadratic expressions.
Before look at the worksheet, if you would like to learn how to factor quadratic expressions,
Question 1 :
Factor :
x2 + 5x + 6
Question 2 :
Factor :
x2 - 2x - 15
Question 3 :
Factor :
x2 - 4x + 3
Question 4 :
Factor :
y2 - 21y - 72
Question 5 :
Factor :
3x2 – 5x – 12
## Factoring Quadratic Expressions Worksheet - Solutions
Question 1 :
Factor :
x2 + 5x + 6
Solution :
In the quadratic expression above, the coefficient of x2 is 1.
Decompose the constant term +6 into two factors such that the product of the two factors is equal to +6 and the addition of two factors is equal to the coefficient of x, that is +5.
Then, the two factors of +6 are
+2 and +3
Factor the given quadratic expression using +2 and +3.
x2 + 5x + 6 = (x + 2)(x + 3)
Therefore, the factors of the given quadratic expression are
(x + 2) and (x + 3)
Question 2 :
Factor :
x2 - 2x - 15
Solution :
In the quadratic expression above, the coefficient of x2 is 1.
Decompose the constant term +6 into two factors such that the product of the two factors is equal to -15 and the addition of two factors is equal to the coefficient of x, that is -2.
Then, the two factors of -15 are
-5 and +3
Factor the given quadratic expression using -5 and +3.
x2 - 2x - 15 = (x - 5)(x + 3)
Therefore, the factors of the given quadratic expression are
(x - 5) and (x + 3)
Question 3 :
Factor :
x2 - 4x + 3
Solution :
In the quadratic expression above, the coefficient of x2 is 1.
Decompose the constant term +3 into two factors such that the product of the two factors is equal to +3 and the addition of two factors is equal to the coefficient of x, that is -4.
Then, the two factors of +3 are
-1 and -3
Factor the given quadratic expression using -1 and -3.
x2 - 4x + 3 = (x - 1)(x - 3)
Therefore, the factors of the given quadratic expression are
(x - 1) and (x - 3)
Question 4 :
Factor :
y2 - 21y - 72
Solution :
In the quadratic expression above, the coefficient of y2 is 1.
Decompose the constant term -72 into two factors such that the product of the two factors is equal to -72 and the addition of two factors is equal to the coefficient of y, that is -21.
Then, the two factors of -72 are
-24 and +3
Factor the given quadratic expression using -24 and +3.
y2 - 4y + 3 = (y - 24)(y + 3)
Therefore, the factors of the given quadratic expression are
(y - 24) and (y + 3)
Question 5 :
Factor :
3x2 – 5x – 12
Solution :
In the given quadratic expression, the coefficient of x2 is not 1.
So, multiply the coefficient of x2 and the constant term "-12".
⋅ (-12) = -36
Decompose -36 into two factors such that the product of two factors is equal to -36 and the addition of two factors is equal to the coefficient of x, that is -5.
Then, the two factors of -36 are
+4 and -9
Now we have to divide the two factors 4 and -3 by the coefficient of x2, that is 3.
3x2 – 5x – 12 = (3x + 4)(x - 3)
Therefore, the factors of the given quadratic expression are
(3x + 4) and (x - 3)
After having gone through the stuff given above, we hope that the students would have understood how to factor quadratic expressions.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here
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# How do you identity if the equation x^2+4y^2-11=2(4y-x) is a parabola, circle, ellipse, or hyperbola and how do you graph it?
Dec 4, 2016
The given equation represents ellipses.
#### Explanation:
Watch this general form of an equation of conic section -
$A {x}^{2} + B x + C {y}^{2} + D x + E y + F = 0$
If $A \times C = 1$; It is a circle.
If $A \times C > 0$; It is a ellips.
If $A \times C < 1$; It is a Hyperbola.
If $A$ or $B$ is equal to zero, it is a parabola.
Our equation is -
${x}^{2} + 4 {y}^{2} - 11 = 2 \left(4 y - x\right)$
Let us rewrite it in the know form to identify the equation.
${x}^{2} + 4 {y}^{2} - 11 = 8 y - 2 x$
${x}^{2} + 2 x + 4 {y}^{2} - 8 y - 11 = 0$
$A = 1$
$C = 4$
Since $A \times C$ i.e., $1 \times 4 > 0$
The given equation represents ellipses. |
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