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# Conjunction
Logical reasoning is an important skill used in all the fields such as science, engineering, and even in our daily life activities. It is also used in mathematical problem-solving strategies. One can quickly get the conclusion using the mathematical principles and the given facts. We know that there are different logical connections used in Maths to solve the problem. The commonly used logical connectives are:
• Negation
• Conjunction
• Disjunction
• Implication
• Equivalence
In this article, let us discuss in detail about one of the connectives called “Conjunction” with its definition, rules, truth table, and examples.
## Conjunction Definition
When two statements are connected with an ‘AND’ gate, we can say that they have conjunction. For conjunctions, when both statements are true, then only the combined compound statement is true.
Suppose ‘parallelograms are rectangles’ and ‘circles are curves’ as our two statements. Let us combine them with ‘and’ to make the compound statement such as ‘parallelograms are rectangles and circles are curves’. This new compound statement is true only when the two statements we began with are true. If only one is false, and the other is true, the compound statement will be false.
For example, the same would be true if two statements were ‘Jessi likes cold coffee,’ and ‘Sony likes milkshake.’ If we combined them with ‘and’ to make ‘Jessi likes cold coffee and Sony likes milkshake,’ then the two statements must be true for the compound statement.
## Conjunction in Maths
A conjunction is a statement formed by adding two statements with the connector AND. The symbol for conjunction is ‘∧’ which can be read as ‘and’. When two statements p and q are joined in a statement, the conjunction will be expressed symbolically as p ∧ q. If both the combining statements are true, then this statement will be true; otherwise, it is false.
The below figure shows the conjunction of A and B:
## Rules for a Conjunction
• The conjunction statement will only be true if both the combining statements are true otherwise, false.
• It is similar to an AND gate which is utilized under the topic Gate logic.
• Let p and q be the two statements. The compound statement p ∧ q is called the conjunction of p and q.
• The symbol “∧” that denotes the conjunction, it is read as “and” which is the logical connective.
## Conjunction Truth Table
Let us make a truth table for P and Q, i.e. P ∧ Q.
P Q P ∧ Q T T T T F F F T F F F F
In this table, we can say that the conjunction is true only when both P and Q are true. If they are not, then the conjunction statement will be false.
### Conjunction Examples
Example 1:
Let r: 5 be a rational number and s: 15 be a prime number. Is it a conjunction?
Solution:
Given that r: 5 is a rational number. This proposition is true.
s: 15 is a prime number. This proposition is false as 15 is a composite number.
Therefore, as per the truth table, r and s is a false statement.
So, r ∧ s = F
Example 2:
Let a: x be greater than 9 and b: x be a prime number. Is it a conjunction?
Solution:
Since x is a variable whose value we don’t know. Let us define a range for a and b.
To find the range let us take certain values for x;
When x= 6: a and b is false. Hence, a ∧ b is false.
When x= 3: a is false but b is true. But still, a ∧ b is false.
When x= 10: a is true but b is false. But still, a ∧ b is false.
When x= 11: a is true and b is true. Hence, a ∧ b is true.
Hence the conjunction a and b is only true when x is a prime number greater than 9.
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# Solutions Of Cubic Functions (3 Key Facts About Zeros Of Cubics)
Cubic functions are not seen as commonly as quadratics, but they are still important in mathematics. We can say a lot about the roots of cubic functions and what they look like (real or complex, and how many of each).
So, how many real solutions does a cubic function have? A cubic function with real coefficients has at least one real root, since complex roots come in conjugate pairs. A cubic function can have 1 real root (repeated 3 times, or 1 real root and 2 complex roots), 2 real roots (when one real root is repeated twice), or 3 distinct real roots.
Of course, a cubic always has at least 1 real root – if we can find it, then we can factor the cubic into the product of a linear function and a quadratic function. We can then use the quadratic formula to find the other two roots easily.
In this article, we’ll talk about the roots (solutions) of cubic functions and when they will be real or complex. We’ll also answer some common questions about the roots of cubic functions.
Let’s begin.
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## Solutions Of Cubic Functions
There are several different cases for the solutions of cubic functions, depending on how many real and complex roots there are in the equation:
• One Triple Real Root – this happens when there is only one real root and no complex roots. The real root is repeated three times. The simplest example is f(x) = x3, which factors as x*x*x, with a triple real root of x = 0.
• One Real Root & Two Complex Conjugate Roots – this is the only case where there are complex roots. This happens when there is a real root (not repeated) and two complex conjugate roots. In this case, the cubic will factor as f(x) = k(x – r)(x – z1)(x – z2), where k is a constant, r is the real root, z1 = a + bi is one complex root, and z2 = a – bi is the other complex root. Note that z1 and z2 are complex conjugates, meaning they have the same real part but opposite imaginary parts.
• Two Real Roots (One Double Real Root) – this happens when there is a double real root and a 2nd real root (not repeated). One example is f(x) = x3 – x2, which factors as x*x*(x – 1), with a double real root of x = 0 and a real root of x = 1 (not repeated).
• Three Distinct Real Roots – this happens when there are 3 different real roots of the cubic function. One example is f(x) = x3 – 3x2 + 2x, which factors as x(x – 1)(x – 2), with real roots x = 0, x = 1, and x = 2.
The table below summarizes the four cases for the zeros of a cubic and how many roots are real or complex.
Remember that a cubic always has at least one real root. This allows us to factor the cubic as the product of a linear function and a quadratic function.
There are 3 possible cases for the solutions of a quadratic function:
If we are in Case 1 and the root of the linear function is distinct from both real roots of the quadratic, then we get three distinct real roots.
If we are in Case 1 and the root of the linear function is the same as one of the real roots of the quadratic, then we get two distinct real roots (one of them is a double root).
If we are in Case 2 and the root of the linear function is the same as the repeated root of the quadratic, then we get one triple real root.
If we are in Case 2 and the root of the linear function is distinct from the repeated root of the quadratic, then we get then we get two distinct real roots (one of them is a double root).
If we are in Case 3, then we get one real root and two complex conjugate roots.
The table below summarizes the cases for the nature of the roots of a cubic function, depending on the roots of the quadratic and the linear factor.
### How Many Zeros Does A Cubic Function Have?
A cubic function has 3 zeros, counting repeated roots. However, some of them may be repeated (a double or triple root is possible), or there be a pair of complex (non-real) roots.
According to the Fundamental Theorem of Algebra, a polynomial of degree n with real coefficients has n complex roots (counting repeated roots). Applying this to a cubic with real coefficients (n = 3), we can see that such a function has 3 roots.
### Can A Cubic Function Have Imaginary Roots?
A cubic function can have imaginary roots in some cases. A cubic function will always have either 0 or 2 imaginary roots, which must be complex conjugates of one another (according to the Complex Conjugate Root Theorem).
For example, if x = 2i is a root of a cubic f(x), then x = -2i (the complex conjugate of 2i) is also a root of f(x). The third root of this cubic would be real – let’s say x = 3.
Then our cubic f(x) would look like this:
• f(x) = k(x – 3)(x – 2i)(x + 2i)
• f(x) = k(x – 3)(x2 + 2ix – 2ix – 4i2) [FOIL (x – 2i)(x + 2i)]
• f(x) = k(x – 3)(x2 + 4) [cancel like terms and use i2 = -1]
• f(x) = k(x3 + 4x – 3x2 – 12) [FOIL]
If we know the coordinates of one point on the cubic f(x) that is not a zero, then we can find the value of k. Let’s say we have the point (1, -20) on the graph of f(x).
Then we can solve for k as follows:
• f(1) = k(13 + 4(1) – 3(1)2 – 12) [substitute x = 1]
• -20 = k(1 + 4 – 3 – 12) [f(1) = -20, since the point (1, -20) is on the graph of f(x)]
• -20 = k(-10)
• 2 = k
Since k = 2, we can write out the entire function as:
• f(x) = k(x3 + 4x – 3x2 – 12)
• f(x) = 2(x3 + 4x – 3x2 – 12) [k = 2]
• f(x) = 2x3 – 6x2 + 8x – 24 [distribute 2 through parentheses]
The graph of the function is pictured below.
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### Can A Cubic Function Have No Real Zeros?
A cubic function always has at least 1 real zero, due to the Fundamental Theorem of Algebra and the Complex Conjugate Root Theorem.
The Fundamental Theorem of Algebra tells us that a cubic has exactly 3 roots. The Complex Conjugate Root Theorem tells us that complex roots must come in pairs (which are complex conjugates).
So, the possible cases are:
• 2 complex roots (a complex conjugate pair) and 1 real root
• 0 complex roots and 3 real roots (either 3 distinct real roots or 2 real roots, one of which is a double root)
You can see an example of the 2nd case (3 distinct real roots) below.
If the cubic had no real roots, then there would be 3 complex roots, one of which would not be paired with its complex conjugate.
The Complex Conjugate Root Theorem implies that a cubic function can never have 1 or 3 imaginary roots (a cubic function can only have 0 or 2 complex roots). It also means a cubic function must have at least one real root.
### Can A Cubic Function Have 3 Imaginary Roots?
A cubic function with real coefficients cannot have 3 imaginary roots, since this would imply no real roots, which we showed was impossible above.
As mentioned earlier, a cubic function with real roots has either 0 or 2 complex roots, since the complex roots come in conjugate pairs by the Complex Conjugate Root Theorem.
The only way a cubic function can have 3 imaginary roots is if the coefficients are complex (instead of restricted to real coefficients).
### Can A Cubic Function Have A Double Root?
A cubic function can have a double root. However, this double root must be real.
One example is the cubic function f(x) = x3 – x2. This function factors as x*x*(x – 1), with a double real root of x = 0 (repeated) and a real root of x = 1 (not repeated).
The graph is pictured below.
If the double root were complex, then its complex conjugate would also need to be a double root (by the Complex Conjugate Root Theorem).
However, this would imply at least 4 zeros of the function, contradicting the Fundamental Theorem of Algebra (which states that a cubic has 3 roots).
### Can A Cubic Function Have 2 Zeros?
A cubic function can have 2 zeros if one of them is a repeated real root (double root). This implies that there will be no complex roots (no complex conjugate pair).
In this scenario, there will be one double real root (that is repeated) and one real root (that is not repeated).
One example is the cubic function f(x) = x3 – x2. This function factors as x*x*(x – 1), with a double real root of x = 0 (repeated) and a real root of x = 1 (not repeated).
### Can A Cubic Function Have Only One Real Root?
A cubic function can have only one real root. There are two cases where this can happen:
• One Triple Real Root – the simplest example is f(x) = x3, which factors as x*x*x, with a triple real root of x = 0.
• One Real Root & Two Complex Conjugate Roots – one example is f(x) = x3 + x, which factors as x(x + i)(x – i), with a real root of x = 0 and complex conjugate root pair x = i and x = -i.
The graph of the first case is pictured below.
## Conclusion
Now you know about the solutions of cubic functions and when they are real or complex. You also know that complex roots come in conjugate pairs for polynomials with real coefficients, whether they are quadratic, cubic, or higher order polynomials.
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# Evaluate the left hand and right hand limits of the function$f(x)=\left\{ \begin{matrix} \dfrac{\sqrt{({{x}^{2}}-6x+9)}}{(x-3)},x\ne 3 \\ 0,x=3 \\\end{matrix}\text{ at }x=3. \right.$
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Hint: For finding out whether the limit exists, then we should find the left hand limit and right hand limit. If they are equal then the limit exists.
First we will simplify the given function, that is,
$\underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{{{x}^{2}}-6x+9}}{x-3}$
The numerator consists of a quadratic equation, now we will simplify it as follows,
$\underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{{{x}^{2}}-3x-3x+9}}{x-3}$
$\underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{x(x-3)-3(x-3)}}{(x-3)}$
$\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{(x-3)(x-3)}}{(x-3)}$
$\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{{{(x-3)}^{2}}}}{(x-3)}$
We know, $\sqrt{4}=\pm 2$ , so the above equation can be written as,
$\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,f(x)=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\left| x-3 \right|}{(x-3)}.......(i)$
Now the modulus can be split as following,
$f(x)=\left\{ \begin{matrix} \dfrac{x-3}{x-3},x>0 \\ \dfrac{-(x-3)}{x-3},x<0 \\ \end{matrix} \right.$
Now we will find the left hand limit, we get
$\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\dfrac{-(x-3)}{x-3}$
Cancelling the like terms, we get
$\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f(x)=-1$
So, the left hand limit of the given function is $'-1'$.
Now we will find the right hand limit, we get
$\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\dfrac{(x-3)}{x-3}$
Cancelling the like terms, we get
$\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f(x)=1$
So, the right hand limit of the given function is $'1'$.
So the left hand limit and the right hand limit are not equal hence the $\underset{x\to 3}{\mathop{\lim }}\,f(x)\text{ }$does not exist.
Note: Generally these questions are asked in competitive examinations for confusing students. Instead of solving and then finding the left hand limit and right hand limit. We can directly apply the $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g(x)=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{g(0+h)-g(0)}{h}$, this formula, but it will be complicated process.
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# Factoring by Grouping
A method of grouping the terms for taking out the common factor from the terms in an expression is called the factorization (or factorisation) by grouping.
## Introduction
A polynomial may consist of two or more terms for representing an indeterminant quantity in mathematical form. In some cases, two or more terms in the expression have a factor commonly. So, they are arranged as a group and it is useful to take out the common factor from them. Hence, the method of factorisation (or factorization) is called the method of factoring by grouping.
#### Steps
There are three steps involved in factoring a polynomial by grouping.
1. Arrange the terms closer by identifying the common factors.
2. Group the common factor terms with parentheses (round brackets) and then express each term in every group in factor form.
3. Take out the factor common from all the groups.
### Example
Let us learn how to factorise (or factorize) an expression by grouping the terms from the following example.
Factorize $9+3xy+x^2y+3x$
### Group the Terms as per common factor
A polynomial can be given in any order but it is essential write the terms in order by identifying the common factor. The common factor in terms can be identified by comparing every term with another term. In this example, the terms $x^2y$ and $3xy$ have $xy$ as a common factor. Hence, write those two terms closely one after one. Similarly, the terms $3x$ and 9 have 3 as a common factor and write them one after one.
$=\,\,\,$ $x^2y+3xy+3x+9$
Now, represent each group terms by writing the terms inside the parentheses (round brackets).
$=\,\,\,$ $(x^2y+3xy)+(3x+9)$
### Factorize the terms in expression
Now, factorise each term in every group by writing the terms in factor form.
$=\,\,\,$ $(x \times xy+3 \times xy)$ $+$ $(3 \times x+3 \times 3)$
### Take out factor common
In first group, $xy$ is a common factor and it can be taken out common from the terms as per inverse operation of the distributive property. Similarly, $3$ is a common factor and it can also be taken out common from the second group by using the same principle.
$=\,\,\,$ $xy(x+3)$ $+$ $3(x+3)$
Now, $x+3$ is a common factor in both terms. Hence, take out the factor $x+3$ common from the terms to complete the factoring process.
$=\,\,\,$ $(x+3)(xy+3)$
### Problems
List of the questions with solutions to learn how to factorize (or factorize) the expressions by grouping the terms as per the common factor.
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# Square and Square root
## Exercise 3 (A)
1.Find the square of the following numbers :
(i) 15
(ii) 48
(iii) $$\dfrac{6}{7}$$
(iv) $$\dfrac{21}{25}$$
(v) $$6\dfrac{3}{8}$$
(vi) 0.9
(vii) 1.1
(viii) 0.018
2.Determine whether square of the following numbers will be even or odd.(Do not find the square)
(i) 529
(ii) 30976
(iii) 893025
(iv) 6990736
3.Just by looking at the following numbers, give reason why they are not perfect squares.
(i) 4893
(ii) 65000
(iii) 4422
(iv) 89138
(v) 150087
4.Using prime factorization method, find which of the following are perfect square numbers.
(i) 100
(ii) 284
(iii) 784
(iv) 1444
(v) 768
(vi) 4225
(vii) 3375
(viii) 15625
5.Using prime factorization method, find the square root of the following numbers.
(i) 256
(ii) 1936
(iii) 3364
(iv) 5625
(v) 7744
(vi) 12544
6.Find the square root of the following fractions.
(i) $$\dfrac{529}{1225}$$
(ii) $$\dfrac{1444}{3249}$$
(iii) $$1\dfrac{396}{9604}$$
(iv) 0.01
(v) 0.1764
(vi) 0.00003136
7.Simplyfy:
(i) $$\sqrt{10^2-{6^2}}$$
(ii) $$\Bigg(-\sqrt{\dfrac{9}{16}}\Bigg)\Bigg(\sqrt{\dfrac{64}{81}}\Bigg)$$
(iii) $$\sqrt{\dfrac{16}{36}-\dfrac{1}{4}}$$
8.Find the square root of 1024. Hence find the value of $$\sqrt{10.24} + \sqrt{0.1024}+\sqrt{10240000}$$.
9.Find the smallest number by which each of the given number s ,ust be multiplied so that the product is a perfect square. Also find the square root of this new number.
(i) 175
(ii) 325
(iii) 720
(iv) 3150
10.Find the smallest number by which each of the given numbers must be divided so that the quotient is a perfect square.Also find the square root of this quotient.
(i) 2700
(ii) 5488
(iii) 7203
(iv) 20886
11.Find the smallest number which is completely divisible by each of the numbers 10 , 16 and 24.
12.The children of class VIII of a school contributed ₹ 3025 for the Prime Minister’s National Relief Fund, contributing as much money as there are children in the class.How many children are there in the class ?
## 1 thought on “S.chand books class 8 solution maths chapter 3 Square and Square root”
1. Many thanks writing this particular blog and rendering it public
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# How do you describe slope in math?
slope, Numerical measure of a line's inclination relative to the horizontal. In analytic geometry, the slope of any line, ray, or line segment is the ratio of the vertical to the horizontal distance between any two points on it (“slope equals rise over run”).
## How do you describe a slope?
The slope is defined as the ratio of the vertical change between two points, the rise, to the horizontal change between the same two points, the run.
## What are two ways to describe slope?
You can describe the slope, or steepness, of the ramp and stairs by considering horizontal and vertical movement along them. In conversation, you use words like “gradual” or “steep” to describe slope. Along a gradual slope, most of the movement is horizontal. Along a steep slope, the vertical movement is greater.
## What does describe the slope of the line mean?
Slope of a Line
The slope of the line is the ratio of the rise to the run, or rise divided by the run. It describes the steepness of line in the coordinate plane. Calculating the slope of a line is similar to finding the slope between two different points.
## How do you describe slope in a word problem?
In the equation of a straight line (when the equation is written as "y = mx + b"), the slope is the number "m" that is multiplied on the x, and "b" is the y-intercept (that is, the point where the line crosses the vertical y-axis).
## How do you describe the slope and y intercepts of the graphs of the system of equations?
The equation is written in slope-intercept form, y=mx+b, where m is the slope and b is the y-intercept. So the slope is − 3 , and the y-intercept is 2. This is a picture of a coordinate plane with the point (0,2) graphed on it.
## What does the slope of each line on the graph tell you?
In other words, the slope of the line tells us the rate of change of y relative to x. If the slope is 2, then y is changing twice as fast as x; if the slope is 1/2, then y is changing half as fast as x, and so on.
## What does the y-intercept of a line describe?
The y-intercept of a line is the value of y where the line crosses the y-axis. In other words, it is the value of y when the value of x is equal to 0.
## What move does the numerator of the slope describe?
Remember the concept of slope as the rise over run. The rise (numerator) describes the change in \large{y} which is written symbolically as \color{blue}\Delta \,y. Meanwhile, the run (denominator) describes the change in \large{x} which is written as \color{red}\Delta \,x.
## How do you describe the slope of a hill?
Slope can be calculated as a percentage which is calculated in much the same way as the gradient. Convert the rise and run to the same units and then divide the rise by the run. Multiply this number by 100 and you have the percentage slope. For instance, 3" rise divided by 36" run = .
## How do you describe a steep slope?
Page 1. Steep slopes are legally defined as hillsides having a 15 foot, or greater, vertical rise over 100 feet of horizontal run, or 15% slope (Figure 1). They are often undesirable ar- eas for development due to the difficulty of building on steep grades.
## Which of the following describe the slope when the line is horizontal?
Slope of a horizontal line. When two points have the same y-value, it means they lie on a horizontal line. The slope of such a line is 0, and you will also find this by using the slope formula. Created by Sal Khan and Monterey Institute for Technology and Education.
## What does steeper slope mean?
Hint: Steeper slopes means the slope that is closer to the vertical axis, or larger angle with the horizontal axis.
## What is this slope?
The slope of a line is a measure of its steepness. Mathematically, slope is calculated as "rise over run" (change in y divided by change in x).
## What is the slope of the graph?
The slope from graph can be calculated by picking any two points on it and applying the formula rise/run. It can be also found by picking two points and applying the formula (y₂ - y₁) / (x₂ - x₁). The slope of a horizontal line is 0 always. The slope of a vertical line is undefined always.
## How will you describe the trend of the graph if the value of the slope is positive?
A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y decreases also. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.
## How do you write slope?
The slope-intercept form is written as y = mx+b, where m is the slope and b is the y-intercept (the point where the line crosses the y-axis). It's usually easy to graph a line using y=mx+b. Other forms of linear equations are the standard form and the point-slope form. Equations of lines have lots of different forms.
## What is the slope between the points?
The slope of a line is its vertical change divided by its horizontal change, also known as rise over run. When you have 2 points on a line on a graph the slope is the change in y divided by the change in x. The slope of a line is a measure of how steep it is.
## What does a flat slope indicate?
A slope of zero means that there is a constant relationship between x and y. Graphically, the line is flat; the rise over run is zero.
## What is a flat slope?
A rising and a falling slope. Flat areas are never strictly horizontal; there are gentle slopes in a seemingly flat area, but they are often hardly noticeable to the naked eye. An accurate survey of the land is necessary to identify these so called "flat slopes".
## What does gentle slope mean?
adjective. A gentle slope or curve is not steep or severe.
## How do you tell if a slope rises or falls?
Note that when a line has a positive slope it rises up left to right. Note that when a line has a negative slope it falls left to right. Note that when a line is horizontal the slope is 0. Note that when the line is vertical the slope is undefined.
## When a line is vertical What is the slope?
Vertical lines are said to have "undefined slope," as their slope appears to be some infinitely large, undefined value.
## What are describing words?
Describing words are the words used to describe or give more information about a noun which could be a person , place or object. Describing words tell us more about nouns.
## How do you describe a hill?
A hill is a piece of land that rises higher than everything surrounding it. It looks like a little bump in the Earth. Since theyre higher than everything around them, hills are good places to get a nice view. Hills are easier to climb than mountains.
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# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables
## Chapter 4 – Linear Equations in Two Variables Exercise Ex. 4A
Question 1
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
3x + 5y = 7.5
Solution 1
We
have,
3x + 5y = 7.5
3x + 5y –
7.5 = 0
6x + 10y
– 15 = 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 6, b = 10 and c =
-15
Question 2
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
Solution 2
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 10, b =
-1 and c = 30
Question 3
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
3y – 2x = 6
Solution 3
We
have,
3y – 2x = 6
-2x + 3y – 6
= 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
=
-2, b = 3 and c =
-6
Question 4
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
4x = 5y
Solution 4
We
have,
4x = 5y
4x – 5y = 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
=
4, b = -5 and c = 0
Question 5
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
Solution 5
6x – 5y = 30
6x – 5y – 30 = 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 6, b =
-5 and c = 30
Question 6
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
Solution 6
On comparing this equation with ax
+ by + c = 0, we obtain
a
= , b = and c =
-5
Question 7
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
x = 6
Solution 7
We
have,
x
= 6
x – 6 = 0
1x + 0y – 6 =
0
x + 0y – 6 = 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 1, b = 0 and c =
-6
Question 8
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
3x – y = x – 1
Solution 8
We
have,
3x – y = x – 1
3x – x – y + 1
= 0
2x – y + 1 = 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 2, b =
-1 and c = 1
Question 9
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
2x + 9 = 0
Solution 9
We
have,
2x + 9 = 0
2x + 0y + 9 =
0
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 2, b = 0 and c = 9
Question 10
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
4y = 7
Solution 10
We
have,
4y = 7
0x + 4y – 7 = 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 0, b = 4 and c =
-7
Question 11
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
x + y = 4
Solution 11
We
have,
x + y = 4
x + y – 4 = 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 1, b = 1 and c =
-4
Question 12
Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case
Solution 12
We
have,
3x – 8y – 1 = 0
On comparing this equation with ax
+ by + c = 0, we obtain
a
= 3, b =
-8 and c = -1
Question 13
Check which of the following are the solutions of the
equation 5x – 4y = 20.
(4, 0)
Solution 13
Given
equation is 5x – 4y = 20
Substituting
x = 4 and y = 0 in L.H.S. of given equation, we get
L.H.S.
= 5x – 4y
= 5(4) – 4(0)
= 20 – 0
= 20
= R.H.S.
Hence,
(4, 0) is the solution of the given equation.
Question 14
Check which of the following are the solutions of the
equation 5x – 4y = 20.
(0, 5)
Solution 14
Given
equation is 5x – 4y = 20
Substituting
x = 0 and y = 5 in L.H.S. of given equation, we get
L.H.S.
= 5x – 4y
= 5(0) – 4(5)
= 0 – 20
= -20
R.H.S.
Hence,
(0, 5) is not the solution of the given equation.
Question 15
Check which of the following are the solutions of the
equation 5x – 4y = 20.
Solution 15
Given
equation is 5x – 4y = 20
Substituting
x = -2 and y = in L.H.S. of given
equation, we get
L.H.S.
= 5x – 4y
= 5(-2) – 4
= -10 – 10
= -20
R.H.S.
Hence,
is not the
solution of the given equation.
Question 16
Check which of the following are the solutions of the
equation 5x – 4y = 20.
(0, -5)
Solution 16
Given
equation is 5x – 4y = 20
Substituting
x = 0 and y =
-5 in L.H.S. of given equation, we
get
L.H.S.
= 5x – 4y
= 5(0) – 4(-5)
= 0 + 20
= 20
= R.H.S.
Hence,
(0,
-5) is the solution of the given
equation.
Question 17
Check which of the following are the solutions of the
equation 5x – 4y = 20.
Solution 17
Given
equation is 5x – 4y = 20
Substituting
x = 2 and y = in L.H.S. of given
equation, we get
L.H.S.
= 5x – 4y
= 5(2) – 4
= 10 + 10
= 20
= R.H.S.
Hence,
is the
solution of the given equation.
Question 18
Find five different solutions of each of the following
equations:
2x – 3y = 6
Solution 18
Given
equation is 2x – 3y = 6
Substituting
x = 0 in the given equation, we get
2(0)
– 3y = 6
0 – 3y = 6
3y = -6
y = -2
So,
(0, -2) is the solution of the given equation.
Substituting
y = 0 in the given equation, we get
2x
– 3(0) = 6
2x – 0 = 6
2x = 6
x = 3
So,
(3, 0) is the solution of the given equation.
Substituting
x = 6 in the given equation, we get
2(6)
– 3y = 6
12 – 3y = 6
3y = 6
y = 2
So,
(6, 2) is the solution of the given equation.
Substituting
y = 4 in the given equation, we get
2x
– 3(4) = 6
2x – 12 = 6
2x = 18
x = 9
So,
(9, 4) is the solution of the given equation.
Substituting
x = -3 in the given equation, we get
2(-3)
– 3y = 6
-6 – 3y = 6
3y = -12
y = -4
So,
(-3, -4) is the solution of the given equation.
Question 19
Find five different solutions of each of the following
equations:
Solution 19
Given
equation is
Substituting
x = 0 in (i), we get
4(0)
+ 3y = 30
3y = 30
y = 10
So,
(0, 10) is the solution of the given equation.
Substituting
x = 3 in (i), we get
4(3)
+ 3y = 30
12 + 3y = 30
3y = 18
y = 6
So,
(3, 6) is the solution of the given equation.
Substituting
x = -3 in (i), we get
4(-3)
+ 3y = 30
-12 + 3y = 30
3y = 42
y = 14
So,
(-3, 14) is the solution of the given equation.
Substituting
y = 2 in (i), we get
4x
+ 3(2) = 30
4x + 6 = 30
4x = 24
x = 6
So,
(6, 2) is the solution of the given equation.
Substituting
y = -2 in (i), we get
4x
+ 3(-2) = 30
4x – 6 = 30
4x = 36
x = 9
So,
(9, -2) is the solution of the given equation.
Question 20
Find five different solutions of each of the following
equations:
3y = 4x
Solution 20
Given
equation is 3y = 4x
Substituting
x = 3 in the given equation, we get
3y
= 4(3)
3y = 12
y = 4
So,
(3, 4) is the solution of the given equation.
Substituting
x = -3 in the given equation, we get
3y
= 4(-3)
3y = -12
y = -4
So,
(-3, -4) is the solution of the given equation.
Substituting
x = 9 in the given equation, we get
3y
= 4(9)
3y = 36
y = 12
So,
(9, 12) is the solution of the given equation.
Substituting
y = 8 in the given equation, we get
3(8)
= 4x
4x = 24
x = 6
So,
(6, 8) is the solution of the given equation.
Substituting
y = -8 in the given equation, we get
3(-8)
= 4x
4x = -24
x = -6
So,
(-6, -8) is the solution of the given equation.
Question 21
If x = 3 and y = 4 is a solution of the equation 5x – 3y =
k, find the value of k.
Solution 21
Since x = 3 and y = 4 is a
solution of the equation 5x – 3y = k, substituting x = 3 and y = 4 in
equation 5x – 3y = k, we get
5(3)
– 3(4) = k
15 – 12 = k
k = 3
Question 22
If x = 3k + 2 and y = 2k – 1 is a solution of the equation
4x – 3y + 1 = 0, find the value of k.
Solution 22
Since x = 3k + 2 and y = 2k – 1 is
a solution of the equation 4x – 3y + 1 = 0, substituting these values in
equation, we get
4(3k + 2) – 3(2k – 1) + 1 = 0
12k + 8 – 6k + 3 + 1 = 0
6k + 12 = 0
6k = -12
k = -2
Question 23
The cost of 5 pencils is equal to the cost of 2 ballpoints.
Write a linear equation in two variables to represent this statement. (Take
the cost of a pencil to be
Rs. x and that of a ballpoint to be Rs. y).
Solution 23
Let
the cost of one pencil be Rs. x and that of one
ballpoint be Rs. y.
Then,
Cost
of 5 pencils = Rs. 5x
Cost
of 2 ballpoints = Rs. 2y
According
to given statement, we have
5x
= 2y
5x – 2y = 0
## Chapter 4 – Linear Equations in Two Variables Exercise Ex. 4B
Question 1
Draw the graph of each of the following equation.
x = 4
Solution 1
x
= 4 is a line parallel to the Y-axis, at a distance of 4 units from it, to
its right.
Question 2
Draw the graph of each of the following equation.
x + 4 = 0
Solution 2
x
+ 4 = 0
x = -4, which
is a line parallel to the Y-axis, at a distance of 4 units from it, to its
left.
Question 3
Draw the graph of each of the following equation.
y = 3
Solution 3
y
= 3 is a line parallel to the X-axis, at a distance of 3 units from it, above
the X-axis.
Question 4
Draw the graph of each of the following equation.
y = -3
Solution 4
y
= -3 is a line parallel to the X-axis, at a distance of 3 units from it,
below the X-axis.
Question 5
Draw the graph of each of the following equation.
x = -2
Solution 5
x
= -2 is a line parallel to the Y-axis, at a distance of 2 units from it, to
its left.
Question 6
Draw the graph of each of the following equation.
x = 5
Solution 6
x
= 5 is a line parallel to the Y-axis, at a distance of 5 units from it, to
its right.
Question 7
Draw the graph of each of the following equation.
y + 5 = 0
Solution 7
y
+ 5 = 0
y = -5, which
is a line parallel to the X-axis, at a distance of 5 units from it, below the
X-axis.
Question 8
Draw the graph of each of the following equation.
y = 4
Solution 8
y
= 4 is a line parallel to the X-axis, at a distance of 4 units from it, above
the X-axis.
Question 9
Draw the graphs of the lines x – y = 1 and 2x + y = 8.
Shade the area formed by these two lines and the y-axis. Also, find this
area.
Solution 9
Graph of the equation x – y = 1
y = x – 1
When
x = 1, then y = 1 – 1 = 0
When
x = 2, then y = 2 – 1 = 1
Thus,
we have the following table:
x 1 2 y 0 1
Now,
plot the points A(1, 0) and B(2, 1) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of x – y = 1.
Graph of the equation 2x + y = 8
y = 8 – 2x
When
x = 2, then y = 8 – 2(2) = 8 – 4 = 4
When
x = 3, then y = 8 – 2(3) = 8 – 6 = 2
Thus,
we have the following table:
x 2 3 y 4 2
Now,
plot the points C(2, 4) and D(3, 2) on a graph
paper.
Join
CD and extend it in both the directions.
Then,
the line CD is the required graph of 2x + y = 8.
The
two graph lines intersect at point D(3, 2).
The
area enclosed by the lines and Y-axis is shown in the graph.
Draw
DM perpendicular from D on Y-axis.
DM
= x-coordinate of point D(3, 2) = 3
And,
EF = 9
Area
of shaded region = Area of
ΔDEF
Question 10
Draw the graph for each of the equations x + y = 6 and x –
y = 2 on the same graph paper and find the coordinates of the point where the
two straight lines intersect.
Solution 10
Graph of the equation x + y = 6
y = 6 – x
When
x = 2, then y = 6 – 2 = 4
When
x = 3, then y = 6 – 3 = 3
Thus,
we have the following table:
x 2 3 y 4 3
Now,
plot the points A(2, 4) and B(3, 3) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of x + y = 6.
Graph of the equation x – y = 2
y = x – 2
When
x = 3, then y = 3 – 2 = 1
When
x = 4, then y = 4 – 2 = 2
Thus,
we have the following table:
x 3 4 y 1 2
Now,
plot the points C(3, 1) and D(4, 2) on a graph
paper.
Join
CD and extend it in both the directions.
Then,
the line CD is the required graph of x – y = 2.
The
two graph lines intersect at point D(4, 2).
Question 11
Two students A and B contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the
earthquake victims. Write a linear equation to satisfy the above data and
draw its graph.
Solution 11
Let
the amount contributed by students A and B be Rs. x
and Rs. y respectively.
Total
contribution = 100
x + y = 100
y = 100 – x
When
x = 25, then y = 100 – 25 = 75
When
x = 50, then y = 100 – 50 = 50
Thus,
we have the following table:
x 25 50 y 75 50
Now,
plot the points A(25, 75) and B(50, 50) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of x + y = 100.
Question 12
Draw the graph of the equation y = 3x.
From your graph, find the value of y when x = 2.
Solution 12
y
= 3x
When
x = 1, then y = 3(1) = 3
When
x = -1, then y = 3(-1) = -3
Thus,
we have the following table:
x 1 -1 y 3 -3
Now,
plot the points A(1, 3) and B(-1, -3) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of y = 3x.
the graph
Given:
x = 2. Take a point M on the X-axis such that OM = 2.
Draw
MP parallel to the Y-axis, cutting the line AB at P.
Clearly,
PM = 6
Thus,
when x = 2, then y = 6.
Question 13
Draw
the graph of the equation y = 3x. From your graph, find the value of y when x
= -2.
Solution 13
The
given equation is y = 3x.
Putting
x = 1, y = 3 1 = 3
Putting
x = 2, y = 3 2 = 6
Thus,
we have the following table:
x 1 2 y 3 6
Plot
points (1,3) and (2,6) on a graph paper and join
them to get the required graph.
Take
a point P on the left of y-axis such that the distance of point P from the
y-axis is 2 units.
Draw
PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to
x-axis meeting y-axis at N.
So,
y = ON = -6.
Question 14
Draw the graph of the equation x + 2y – 3 = 0. From your graph, find the value of y when x = 5.
Solution 14
The given equation is,
x + 2y – 3 = 0
x = 3 – 2y
Putting y = 1,x = 3 – (2 1) = 1
Putting y = 0,x = 3 – (2 0) = 3
Thus, we have the following table:
x 1 3 y 1 0
Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.
Take a point Q on x-axis such that OQ = 5.
Draw QP parallel to y-axis meeting the line (x = 3 – 2y) at P.
Through P, draw PM parallel to x-axis cutting y-axis at M.
So, y = OM = -1.
Question 15
Draw the graph of the equation x + 2y – 3 = 0.
From your graph, find the value of y when x = -5
Solution 15
x
+ 2y – 3 = 0
2y = 3 – x
When
x = -1, then
When
x = 1, then
Thus,
we have the following table:
x -1 1 y 2 1
Now,
plot the points A(-1, 2) and B(1, 1) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of x + 2y – 3 = 0.
the graph
Given:
x = -5. Take a point M on the X-axis such that OM = -5.
Draw
MP parallel to the Y-axis, cutting the line AB at P.
Clearly,
PM = 4
Thus,
when x = -5, then y = 4.
Question 16
Draw
the graph of the equation 2x – 3y = 5. From the graph, find (i) the value of y when x = 4, and (ii) the value of x
when y = 3.
Solution 16
The given equation is, 2x – 3y = 5
Now, if x = 4, then
And, if x = -2, then
Thus, we have the following table:
x 4 -2 y 1 -3
Plot points (4,1)
and (-2,-3) on a graph paper and join them to get the required graph.
(i) When
x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to
its right cutting the line at Q and through Q draw a line parallel to x-axis
cutting y-axis which is found to be at a distance of 1 units above x-axis.
Thus, y = 1 when x = 4.
(ii) When y = 3, draw a line parallel to
x-axis at a distance of 3 units from x-axis and above it, cutting the line at
point P. Through P, draw a line parallel to y-axis meeting x-axis at a point
which is found be 7 units to the right of y axis.
Thus, when y = 3, x = 7.
Question 17
Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.
Solution 17
The given equation is 2x + y = 6
y = 6 – 2x
Now, if x = 1, then y = 6 – 2 1 = 4
And, if x = 2, then y = 6 – 2 2 = 2
Thus, we have the following table:
x 1 2 y 4 2
Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.
We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.
So, the co-ordinates of P are (3,0).
Question 18
Draw
the graph of the equation 3x + 2y = 6. Find the coordinates of the point,
where the graph cuts the y-axis.
Solution 18
The given equation is 3x + 2y = 6
2y = 6 – 3x
Now, if x = 2, then
And, if x = 4, then
Thus, we have the following table:
x 2 4 y 0 -3
Plot points (2, 0) and (4,-3) on a
graph paper and join them to get the required graph.
We find that the line 3x + 2y = 6
cuts the y-axis at a point P which is 3 units above the x-axis.
So, co-ordinates of P are (0,3).
Question 19
Draw the graphs of the equations 3x – 2y = 4 and x + y – 3
= 0. On the same graph paper, find the coordinates of the point where the two
graph lines intersect.
Solution 19
Graph of the equation 3x – 2y = 4
2y = 3x – 4
When
x = 2, then
When
x = -2, then
Thus,
we have the following table:
x 2 -2 y 1 -5
Now,
plot the points A(2, 1) and B(-2, -5) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of 3x – 2y = 4.
Graph of the equation x + y – 3 = 0
y = 3 – x
When
x = 1, then y = 3 – 1 = 2
When
x = -1, then y = 3 – (-1) = 4
Thus,
we have the following table:
x 1 -1 y 2 4
Now,
plot the points C(1, 2) and D(-1, 4) on a graph
paper.
Join
CD and extend it in both the directions.
Then,
the line CD is the required graph of x + y – 3 = 0.
The
two graph lines intersect at point A(2, 1).
Question 20
Draw the graph of the line 4x + 3y = 24.
Write the coordinates of the points where this line
intersects the x-axis and the y-axis.
Solution 20
4x
+ 3y = 24
3y = 24 – 4x
When
x = 0, then
When
x = 3, then
Thus,
we have the following table:
x 0 3 y 8 4
Now,
plot the points A(0, 8) and B(3, 4) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of 4x + 3y = 24.
the graph
The
graph of line 4x + 3y = 24 intersects the X-axis at point C(6,
0) and the Y-axis at point A(0, 8).
Question 21
Draw the graph of the line 4x + 3y = 24.
Use this graph to find the area of the triangle formed by
the graph line and the coordinate axes.
Solution 21
4x
+ 3y = 24
3y = 24 – 4x
When
x = 0, then
When
x = 3, then
Thus,
we have the following table:
x 0 3 y 8 4
Now,
plot the points A(0, 8) and B(3, 4) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of 4x + 3y = 24.
the graph
Required
area = Area of
ΔAOC
Question 22
Draw the graphs of the lines 2x + y = 6 and 2x – y + 2 =
0. Shade the region bounded by these two lines and the x-axis. Find the area
Solution 22
Graph of the equation 2x + y = 6
y = 6 – 2x
When
x = 1, then y = 6 – 2(1) = 6 – 2 = 4
When
x = 2, then y = 6 – 2(2) = 6 – 4 = 2
Thus,
we have the following table:
x 1 2 y 4 2
Now,
plot the points A(1, 4) and B(2, 2) on a graph
paper.
Join
AB and extend it in both the directions.
Then,
the line AB is the required graph of 2x + y = 6.
Graph of the equation 2x – y + 2 = 0
y = 2x + 2
When
x = -1, then y = 2(-1) + 2 = -2 + 2 = 0
When
x = 2, then y = 2(2) + 2 = 4 + 2 = 6
Thus,
we have the following table:
x -1 2 y 0 6
Now,
plot the points C(-1, 0) and D(2, 6) on a graph
paper.
Join
CD and extend it in both the directions.
Then,
the line CD is the required graph of 2x – y + 2 = 0.
The
two graph lines intersect at point A(1, 4).
The
area enclosed by the lines and X-axis is shown in the graph.
Draw
AM perpendicular from A on X-axis.
PM
= y-coordinate of point A(1, 4) = 4
And,
CP = 4
Area
of shaded region = Area of
ΔACP
## Chapter 4 – Linear Equations in Two Variables Exercise MCQ
Question 1
The equation of the x-axis is
(a) x
= 0
(b) y
= 0
(c) x
= y
(d) x
+ y = 0
Solution 1
Correct
option: (b)
The
equation of the x-axis is y = 0.
Question 2
The graph of y + 2 = 0 is a line
(a) making
an intercept -2 on the x-axis
(b) making
an intercept -2 on the y-axis
(c) parallel
to the x-axis at a distance of 2 units below the x-axis
(d) parallel
to the y-axis at a distance of 2 units to the left of y-axis
Solution 2
Correct option: (c)
The graph of y + 2 = 0 is a line
parallel to the x-axis at a distance of 2 units below the x-axis.
Question 3
The graph of the linear equation 2x + 3y = 6 meets the
y-axis at the point
(a) (2,
0)
(b) (3,
0)
(c) (0,
2)
(d) (0, 3)
Solution 3
Correct
option: (c)
When
a graph meets the y-axis, the x coordinate is zero.
Thus,
substituting x = 0 in the given equation, we get
2(0)
+ 3y = 6
3y = 6
y = 2
Hence,
the required point is (0, 2).
Question 4
The graph of the linear equation 2x + 5y = 10 meets the
x-axis at the point
(a) (0,
2)
(b) (2,
0)
(c) (5,
0)
(d) (0, 5)
Solution 4
Correct
option: (c)
When
a graph meets the x-axis, the y coordinate is zero.
Thus,
substituting y = 0 in the given equation, we get
2x
+ 5(0) = 10
2x = 10
x = 5
Hence,
the required point is (5, 0).
Question 5
The graph of the line x
= 3 passes through the point
(a) (0,3)
(b) (2,3)
(c) (3,2)
(d) None of these
Solution 5
Question 6
The graph of the line y = 3 passes though the point
(a) (3,
0)
(b) (3,
2)
(c) (2,
3)
(d) none
of these
Solution 6
Correct
option: (c)
Since,
the y coordinate is 3, the graph of the line y = 3 passes through the point
(2, 3).
Question 7
The graph of the line y
= -3 does not pass through the point
(a) (2,-3)
(b) (3,-3)
(c) (0,-3)
(d) (-3,2)
Solution 7
Question 8
The graph of the linear
equation x-y=0 passes through the point
Solution 8
Question 9
If each of (-2,2), (0,0) and (2,-2) is a solution of a linear equation in x and y, then the equation is
(a) x-y=0
(b) x+y=0
(c) -x+2y=0
(d) x – 2y=0
Solution 9
Question 10
How many linear equations can be satisfied by x = 2 and y
= 3?
(a) only
one
(b) only
two
(c) only
three
(d) Infinitely
many
Solution 10
Correct option: (d)
Infinitely many linear equations can be satisfied by x = 2
and y = 3.
Question 11
A linear equation in two
variable x and y is of the form ax+by+c=0, where
(a) a≠0, b≠0
(b) a≠0, b=0
(c) a=0, b≠0
(d) a= 0, c=0
Solution 11
Question 12
The equation of the y-axis is
(a) x
= 0
(b) y
= 0
(c) x
= y
(d) x
+ y = 0
Solution 12
Correct
option: (a)
The
equation of the y-axis is x = 0.
Question 13
If (2, 0) is a solution of the linear equation 2x + 3y = k
then the value of k is
(a) 6
(b) 5
(c) 2
(d) 4
Solution 13
Correct option: (d)
Since, (2, 0) is a solution of the
linear equation 2x + 3y = k, substituting x = 2 and y = 0 in the given
equation, we have
2(2) + 3(0) = k
4 + 0 = k
k = 4
Question 14
Any point on x-axis is
of the form:
(a) (x,y), where x ≠0 and y ≠0
(b) (0,y), where y ≠0
(c) (x,0), where x ≠0
(d) (y,y), where y ≠0
Solution 14
Question 15
Any point on y-axis is
of the form
(a) (x,0), where x ≠ 0
(b) (0,y), where y ≠ 0
(c) (x,x), where x ≠ 0
(d) None of these
Solution 15
Question 16
x = 5, y = 2 is a solution of the linear equation
(a) x
+ 2y = 7
(b) 5x
+ 2y = 7
(c) x
+ y = 7
(d) 5x + y = 7
Solution 16
Correct option: (c)
Substituting x = 5 and y = 2 in
L.H.S. of equation x + y = 7, we get
L.H.S. = 5 + 2 = 7 = R.H.S.
Hence, x = 5 and y = 2 is a
solution of the linear equation x + y = 7.
Question 17
If the point (3, 4) lies on the graph of 3y = ax + 7 then
the value of a is
(a)
(b)
(c)
(d)
Solution 17
Correct option: (b)
Since the point (3, 4) lies on the
graph of 3y = ax + 7, substituting x = 3 and y = 4 in the given equation, we
get
3(4) = a(3) + 7
12 = 3a + 7
3a = 5
Question 18
The point of the form (a,a), where a ≠ 0 lies on
(a) x-axis
(b) y-axis
(c) the line y = x
(d) the line x + y
= 0
Solution 18
Question 19
The point of the form
(a,-a), where a
≠ 0 lies on
(a) x-axis
(b) y-axis
(c) the line y-x=0
(d) the line x + y
= 0
Solution 19
Question 20
The linear equation 3x –
5y = 15 has
(a) a unique
solution
(b) two solutions
(c) infinitely many
solutions
(d) no solution
Solution 20
Question 21
The equation 2x + 5y = 7 has a unique solution, if x and y
are
(a) natural
numbers
(b) rational
numbers
(c) positive
real numbers
(d) real
numbers
Solution 21
Correct
option: (a)
The equation 2x + 5y = 7 has a unique solution, if x and y
are natural numbers.
If we take x = 1 and y = 1, the given equation is
satisfied.
Question 22
The graph of y = 5 is a line
(a) making
an intercept 5 on the x-axis
(b) making
an intercept 5 on the y-axis
(c) parallel
to the x-axis at a distance of 5 units from the origin
(d) parallel
to the y-axis at a distance of 5 units from the origin
Solution 22
Correct
option: (c)
The graph of y = 5 is a line
parallel to the x-axis at a distance of 5 units from the origin.
Question 23
The graph of x = 4 is a line
(a) making
an intercept 4 on the x-axis
(b) making
an intercept 4 on the y-axis
(c) parallel
to the x-axis at a distance of 4 units from the origin
(d) parallel
to the y-axis at a distance of 4 units from the origin
Solution 23
Correct
option: (d)
The graph of x = 4 is a line
parallel to the y-axis at a distance of 4 units from the origin.
Question 24
The graph of x + 3 = 0 is a line
(a) making
an intercept -3 on the x-axis
(b) making
an intercept -3 on the y-axis
(c) parallel
to the y-axis at a distance of 3 units to the left of y-axis
(d) parallel
to the x-axis at a distance of 3 units below the x-axis
Solution 24
Correct option: (c)
The graph of x + 3 = 0 is a line parallel to the y-axis at
a distance of 3 units to the left of y-axis.
error: Content is protected !!
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### Calculating the opportunity cost in a gains from trade example
Look at table 1 again,
Mexico U.S. Papayas Apples Papayas Apples 0 12 0 63 8 9 7 42 16 6 14 21 24 0 21 0
you can see that Mexico can either produce 24 papayas in one day, OR 12 apples. Imagine they were producing all papayas, in order to make 1 apple, they would need to give up 2 papayas. How do I know this? You can either refer to table 2 to see the amount of time it takes to make an apple (2 hours, which would require giving up 2 papayas), or you could divide the maximum amount of papayas they are able to produce by the maximum number of apples they are able to produce. Here it would be 24/12 or 2. So the opportunity cost of an apple is 2. Here is a mathematical example, since the opportunity cost is a ratio, we need to solve for a ratio, and we want to solve it so that the opportunity cost for an apple is in terms of a papaya.
Begin with what the maximum amount of each good is:
12 apples = 24 papayas, now add in the opportunity cost so:
The opportunity cost of 12 apples = 24 papayas, now divide both sides by 12 so:
The opportunity cost 1 apple = 24 papayas/12 = 2 papayas.
So the opportunity cost of 1 apple is 2 papayas. Do this again for papayas.
The opportunity cost of 24 papayas = 12 apples.
The opportunity cost of 1 papaya = 12 apples/24.
The opportunity cost of 1 papaya is .5 apples.
Notice anything? You are right, the opportunity costs of the different goods are inverses of each other, meaning that if the opportunity cost of one good is 2, the opportunity cost for the other is ½. If it is 3 for one, it is 1/3 for the other. If 5 for one, 1/5 for the other and so on.
So now we have the opportunity costs for Mexico, we need to find out the opportunity costs of the goods for the US. Using the method described above we get:
The opportunity cost of 63 apples = 21 papayas.
The opportunity cost of 1 apple = 21 papayas/63.
The opportunity cost of 1 apple is 1/3 papaya, so the US has to give up 1/3 of a papaya in order to get an apple. Because of the inverse rule, we know that the opportunity cost of a papaya is 3 apples.
We can also solve for opportunity cost by looking at the slopes of the PPFs for each of the countries.
> >
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# Direction of a Vector Formula
As the name suggest, when two distinct points are directed from one place to another then it is done by a vector. It can also be seen as differences between velocity and speed. We get no clue about in which direction the object is moving. Therefore, we use this formula that will enable us to know in which direction the object is moving. In physics, the magnitude and direction are expressed as a vector. If we say that the rock is moving at 5meter per second, and the direction is towards the West, then it is represented as a vector.
If x is the horizontal movement and y is the vertical movement, then the formula of direction is
$\LARGE \theta =\tan^{-1}\frac{y}{x}$
If ($x_{1}$,$y_{1}$ ) is the starting point and ends with ($x_{2}$,$y_{2}$ ), then the formula for direction is
$\LARGE \theta =\tan^{-1}\frac{(y_{2}-y_{1})}{(x_{2}-x_{1})}$
Question 1:
Find the direction of the vector $\overrightarrow{pq}$ whose initial point P is at (5, 2) and end point is at Q is at (4, 3)?
Solution:
Given $(x_{1}$, $y_{1})$ = (5, 2)
$(x_{2}$, $y_{2})$ = (4, 3)
According to the formula we have,
$\theta$ = $tan^{-1}$ $\frac{(y_{2} – y_{1})}{(x_{2} – x_{1})}$
$\theta$ = $tan^{-1}$ $\frac{(3-4)}{(2-5)}$
$\theta$ = -0.26
$\theta$ $14.89^{circ}$
Related Formulas Ellipse Formula Empirical Probability Formula Exponents Formula Fibonacci Formula Frustum of a Right Circular Cone Formula Fourier Series Formula Half Angle formula Geometric Series Formula
|
# Numbers 1-20: Lesson for Kids
Instructor: Rayna Cummings
Rayna has taught Elementary Education for 12 years (in both 1st, 2nd, and 3rd grades) and holds a M.Ed in Early Childhood Education from The Ohio State University
Counting the numbers from 1 to 20 can be done in several different ways. In this lesson, you will learn several ways to count: counting by adding one, counting with a number line, counting with words, and skip counting.
Close your eyes and count the numbers 1 to 20. How did you count them? Did you see a number line? Did you use the number words? Did you skip count? Counting the numbers 1 to 20 can be done in different ways.
When you closed your eyes and began counting to 20, you did it as follows:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
Did you know that when you are counting like the example above, from 1 to 20, you are actually doing addition and adding one to each number? Every number in the range is one more than the last number before it.
Here are examples of counting from 1 to 20 by adding one each time:
0 + 1 = 1
1 + 1 = 2
2 + 1 = 3
3 + 1 = 4
4 + 1 = 5
5 + 1 = 6
6 + 1 = 7
7 + 1 = 8
8 + 1 = 9
9 + 1 = 10
10 + 1 = 11
11 + 1 = 12
12 + 1 = 13
13 + 1 = 14
14 + 1 = 15
15 + 1 = 16
16 + 1 = 17
17 + 1 = 18
18 + 1 = 19
19 + 1 = 20
## Using a Number Line
Counting numbers 1 to 20 on a number line helps you see a visual in your mind and keep the numbers in the correct order.
Example of a number line below:
<-- 1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 -- 9 -- 10 -- 11 -- 12 -- 13 -- 14 -- 15 -- 16 -- 17 -- 18 -- 19 -- 20 -->
A number line can further help you understand how to fill in missing numbers from 1 to 20.
For example, what number is missing in the number line?
<-- 1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 -- __ -- 10 -- 11 -- 12 -- 13 -- 14 -- 15 -- 16 -- 17 -- 18 -- 19 -- 20 -->
By being able to count from 1 to 20 using a number line, you could identify the missing number was 9. You can do this by looking at the number that came before the missing number on the number line (which was 8), and the number that came after the missing number on the number line (which was 10).
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# How do you simplify \frac { 3^ { 2} \cdot 3^ { 3} } { 2^ { 2} \cdot 3^ { 4} }?
Jan 15, 2018
See a couple of process below:
#### Explanation:
First Process:
We can use this rule of exponents to combine the terms in the numerator:
${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
$\frac{{3}^{\textcolor{red}{2}} \cdot {3}^{\textcolor{b l u e}{3}}}{{2}^{2} \cdot {3}^{4}} \implies$
${3}^{\textcolor{red}{2} + \textcolor{b l u e}{3}} / \left({2}^{2} \cdot {3}^{4}\right) \implies$
${3}^{5} / \left({2}^{2} \cdot {3}^{4}\right)$
We can now use these rules of exponents to complete the simplification of the $3$ terms:
${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ and ${a}^{\textcolor{red}{1}} = a$
${3}^{\textcolor{red}{5}} / \left({2}^{2} \cdot {3}^{\textcolor{b l u e}{4}}\right) \implies$
${3}^{\textcolor{red}{5} - \textcolor{b l u e}{4}} / {2}^{2} \implies$
${3}^{\textcolor{red}{1}} / {2}^{2} \implies$
$\frac{3}{2} ^ 2 \implies$
$\frac{3}{4}$
Second Process:
$\frac{{3}^{2} \cdot {3}^{3}}{{2}^{2} \cdot {3}^{4}} \implies \frac{9 \cdot 27}{4 \cdot 81} \implies \frac{3 \cdot 3 \cdot 27}{4 \cdot 81} \implies \frac{3 \cdot 81}{4 \cdot 81} \implies \frac{3 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{81}}}}{4 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{81}}}} \implies \frac{3}{4}$
|
Definition of rule of three simple - What it is, Meaning and Concept - I want to know everything - 2023
# Simple Three Rule
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The rule of three It is a mechanism that allows Problem resolution linked to proportionality between three known values and a fourth that is a unknown . Thanks to the rule , you can discover the value of this fourth term.
It is also important to be clear about other aspects of the mentioned simple three rule. We are referring to the fact that the problems that can be solved are both of direct proportionality and of inverse proportionality. And that, without forgetting that to carry out one, you have to have three fundamental data: two magnitudes that are proportional to each other and a third.
In other words, a rule of three is one operation which is developed to know the value of the fourth term of a proportion from the values of the other terms. According to its characteristics, it is possible to differentiate between simple three rule and the compound three rule .
The simple rule of three is one that allows establishing the link of proportionality between two known terms (TO and B ) and, based on the knowledge of a third term (C ), calculate the value of the room (X ).
Let's see a example . A cook who, days ago, prepared three cakes with a kilogram of flour, now has five kilograms of flour and wants to know how many cakes he can make. To perform the calculation, apply the simple three rule:
If with 1 kilogram of flour he prepared 3 cakes,
With 5 kilograms of flour you will prepare X cakes.
1 = 3
5 = X
5 x 3 = 1 x X
15 = X
In this way, the chef discovers that, with 5 kilograms of flour can prepare 15 cakes .
The simple three rule can be direct or inverse. In the case of direct simple three rule , the proportionality is constant: at an increase of TO , corresponds to an increase of B in identical proportion.
An example to understand this type of rule of three simple would be the following: in a store we want to buy some chairs and they tell us that they sell them in pack. Specifically, they tell us that 5 are worth 600 euros, but we need 8 and we want to know what the price would be. Thus, to know the result we should perform the following operations: 600 x 8 and the result, 4800, divide it by 5. Thus we would know that the eight chairs are worth 960 euros.
In the reverse simple three rule , on the other hand, constant proportionality is only preserved when, at an increase of TO , corresponds a decrease of B .
An example to understand how the rule of three simple reverse works is this: today a merchandise company has floated three trucks to transport in six trips each a certain amount of packages. However, yesterday, to move the same number of packages, there were only two trucks of the same size and capacity. So, we are asked the question of how many trips did those two vehicles make?
To know, the operation would consist of doing these steps: 3 x 6 and the result, 18, divide it by 2, which would give us that the two trucks had to make 9 trips each.
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# How do you differentiate y =(5x^2-x+1)^(5/2 using the chain rule?
Jul 24, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{2} \left(10 x - 1\right) {\left(5 {x}^{2} - x + 1\right)}^{\frac{3}{2}}$
#### Explanation:
$\text{differentiate using the "color(blue)"chain rule}$
$\text{given "y=f(g(x))" then}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \text{ chain rule}$
$y = {\left(5 {x}^{2} - x + 1\right)}^{\frac{5}{2}}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{2} {\left(5 {x}^{2} - x + 1\right)}^{\frac{3}{2}} \times \frac{d}{\mathrm{dx}} \left(5 {x}^{2} - x + 1\right)$
$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{5}{2} \left(10 x - 1\right) {\left(5 {x}^{2} - x + 1\right)}^{\frac{3}{2}}$
|
# 3.1 Solving Systems of Equations Graphing
Document Sample
``` 3.1
Solving Systems of Equations
Graphing & Substitution
Date: __________
Pam has \$120 and is spending \$5 every week.
Lorenzo has \$20 and is saving \$7.50 every week.
When will they have the same amount of money?
Let x = # of weeks Let y = total money
Pam 120 Lorenzo
y 120 5x 100
y 20 7.5x
x y x y
total money
80
0 120 0 20
2 110 60 2 35
4 100 40 4 50
6 90 20 In 8 6 65
8 80 weeks 8 80
0
10 70 0 2 4 6
weeks
8 10 10 95
System of Equations
•two or more equations involving
the same variables
•Solutions are an ordered pair
Which of the following ordered pairs are
solutions to the following system.
5x 2y 10
4 x y 8
1) (3, 1) 2) (2, 0)
?
5(3) 2(1) 10 5(2) 2(0) 10
17 10 4(2) (0) 8
No Yes
Solution must work in both equations!
Solving Systems by Graphing
y 2x 1 y
x y 5 (2,3)
x x
y x 5
x
plugging the point into both
original equations.
y = 2x – 1 x+y=5
3 = 2(2) – 1 2+3=5
(2,3) is the solution!
3=4–1 5=5
3=3
Solve.
x y 1
2y x 4
y 2y x 4
x y 1 x x
x x 2y x 4
y x 1 (-2,1) 2 2
1
x
y x 2
2
(2,1) is the solution to the system.
Solve.
2x 3y 3
x 6y 24
y x 6y 24
2x 3y 3 x x
2x 2x
6y x 24
3y 2x 3
2 1
y x 1 x
y x4
3 6
(-6, -3)
(6,3) is the solution.
Solve the sytem below by graphing.
2x 3y 6
x 4y 8
y x 4y 8
2x 3y 6 x x
2x 2x
4y x 8
3y 2x 6
2 1
y x2 y x2
3 x 4
1
(4 , -1)
2
1
(4 ,1) is close to the solution.
2
3 Possible Outcomes
Independent Inconsistent Dependent
System System System
Lines Intersect Lines Parallel Lines Coincide
One No Infinite
Solution Solution Solutions
Solve the system graphically.
y
x y 3
x 2y 4
x x
x 2y 4 2y x 4
x y 3 x 2 2
x x 1
y x 3 y x 2
2
1 1 One
,2
2 2 Solution
Solve. y
1
y x 2
3
2x 6y 12
2x 2x x
6y 2x 12
6 6
1
y x 2
3 Infinitely Many Solutions
Solve the system graphically, by substitution, and
by elimination. y
x 2y 3
4y 2x 8
4 4
4y 2x 8 2 8
y x
x 2y 3 x 4 4
x x 1
2y x 3 y x 2
2 2 2
1 3
y x
2 2
No Solution
Substitution Method
Solve.
2x
y y 2x x y 6
y 6
x y 2x x 2x 6
y 2 2 3x 6
y 4 3 3
x 2
Solution is (2, 4)
Substitution
x 3y 12 y x 1
x 3x 1 12 y x 1
x 3x 3 12 y 3 3 1
3 3
4
y23
4x 15 4
x 33 4
Solution is (3 3 , 2 3 )
4 4
Solve using substitution.
x 3y
4x 5
2y
x 3y 2y 4x 5
2y 43y 5
x 3y 2y 12y 5
1 10y 5
x 3
2 10 10
1
x 1 1 y
2
2
Solution is 1 1 , 1
2 2
Solve using substitution.
x 5
2y
Solve the 1st equation for x.
3y 7
2x
2y x 5 2x 3y 7
2y 2y 22y 5 3y 7
x 2y 5 4y 10 3y 7
y 10 7
x 2y 5 10 10
x 23 5 y 3
x 6 5 1 y 31
x 1 y 3
(1,3)
Solve using substitution.
3x 4y 5 Solve the 2nd equation for y.
y 6
2x
2x y 6
2x 2x
3x 4y 5
1(y 2x 6 )
3x 4 2x 6 5
y 2x 6
3x 8x 24 5
24 24 y 2x 6
11x 29 29
y 2 6
11 11 11
29 58 66 8
x y
11 29 8 11 11 11
,
11 11
Solve using substitution.
x 6 y 1
Solve 1st equation for x.
x 6y 1
3x 10 y 31
6 y 6 y
x 1 6 y
3x 10 y 31
x 1 6(1)
3(1 6 y) 10 y 31
3 18 y 10 y 31 x 1 6
3 28 y 31 x7
3 3
28 y 28 (7, 1)
28 28
y 1
```
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# 2.6 Other types of equations (Page 5/10)
Page 5 / 10
Solve the following rational equation: $\text{\hspace{0.17em}}\frac{-4x}{x-1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.$
We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, $\text{\hspace{0.17em}}{x}^{2}-1=\left(x+1\right)\left(x-1\right).\text{\hspace{0.17em}}$ Then, the LCD is $\text{\hspace{0.17em}}\left(x+1\right)\left(x-1\right).\text{\hspace{0.17em}}$ Next, we multiply the whole equation by the LCD.
$\begin{array}{ccc}\hfill \left(x+1\right)\left(x-1\right)\left[\frac{-4x}{x-1}+\frac{4}{x+1}\right]& =& \left[\frac{-8}{\left(x+1\right)\left(x-1\right)}\right]\left(x+1\right)\left(x-1\right)\hfill \\ \hfill -4x\left(x+1\right)+4\left(x-1\right)& =& -8\hfill \\ \hfill -4{x}^{2}-4x+4x-4& =& -8\hfill \\ \hfill -4{x}^{2}+4& =& 0\hfill \\ \hfill -4\left({x}^{2}-1\right)& =& 0\hfill \\ \hfill -4\left(x+1\right)\left(x-1\right)& =& 0\hfill \\ \hfill x& =& -1\hfill \\ \hfill x& =& 1\hfill \end{array}$
In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.
Solve $\text{\hspace{0.17em}}\frac{3x+2}{x-2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}.$
$x=-1,$ $x=0$ is not a solution.
Access these online resources for additional instruction and practice with different types of equations.
## Key concepts
• Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1. See [link] , [link] , and [link] .
• Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See [link] and [link] .
• We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See [link] and [link] .
• To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See [link] .
• Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See [link] and [link] .
## Verbal
In a radical equation, what does it mean if a number is an extraneous solution?
This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.
Explain why possible solutions must be checked in radical equations.
Your friend tries to calculate the value $\text{\hspace{0.17em}}-{9}^{\frac{3}{2}}$ and keeps getting an ERROR message. What mistake is he or she probably making?
He or she is probably trying to enter negative 9, but taking the square root of $\text{\hspace{0.17em}}-9\text{\hspace{0.17em}}$ is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in $\text{\hspace{0.17em}}-27.$
Explain why $\text{\hspace{0.17em}}|2x+5|=-7\text{\hspace{0.17em}}$ has no solutions.
Explain how to change a rational exponent into the correct radical expression.
A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.
## Algebraic
For the following exercises, solve the rational exponent equation. Use factoring where necessary.
${x}^{\frac{2}{3}}=16$
${x}^{\frac{3}{4}}=27$
$x=81$
$2{x}^{\frac{1}{2}}-{x}^{\frac{1}{4}}=0$
${\left(x-1\right)}^{\frac{3}{4}}=8$
$x=17$
${\left(x+1\right)}^{\frac{2}{3}}=4$
${x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}+6=0$
${x}^{\frac{7}{3}}-3{x}^{\frac{4}{3}}-4{x}^{\frac{1}{3}}=0$
For the following exercises, solve the following polynomial equations by grouping and factoring.
${x}^{3}+2{x}^{2}-x-2=0$
$x=-2,1,-1$
$3{x}^{3}-6{x}^{2}-27x+54=0$
$4{y}^{3}-9y=0$
${x}^{3}+3{x}^{2}-25x-75=0$
${m}^{3}+{m}^{2}-m-1=0$
$m=1,-1$
$2{x}^{5}-14{x}^{3}=0$
$5{x}^{3}+45x=2{x}^{2}+18$
$x=\frac{2}{5}$
For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.
$\sqrt{3x-1}-2=0$
$\sqrt{x-7}=5$
$x=32$
$\sqrt{x-1}=x-7$
$\sqrt{3t+5}=7$
$t=\frac{44}{3}$
$\sqrt{t+1}+9=7$
$\sqrt{12-x}=x$
$x=3$
$\sqrt{2x+3}-\sqrt{x+2}=2$
$\sqrt{3x+7}+\sqrt{x+2}=1$
$x=-2$
$\sqrt{2x+3}-\sqrt{x+1}=1$
For the following exercises, solve the equation involving absolute value.
$|3x-4|=8$
$x=4,\frac{-4}{3}$
$|2x-3|=-2$
$|1-4x|-1=5$
$x=\frac{-5}{4},\frac{7}{4}$
$|4x+1|-3=6$
$|2x-1|-7=-2$
$x=3,-2$
$|2x+1|-2=-3$
$|x+5|=0$
$x=-5$
$-|2x+1|=-3$
For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.
${x}^{4}-10{x}^{2}+9=0$
$x=1,-1,3,-3$
$4{\left(t-1\right)}^{2}-9\left(t-1\right)=-2$
${\left({x}^{2}-1\right)}^{2}+\left({x}^{2}-1\right)-12=0$
$x=2,-2$
${\left(x+1\right)}^{2}-8\left(x+1\right)-9=0$
${\left(x-3\right)}^{2}-4=0$
$x=1,5$
## Extensions
For the following exercises, solve for the unknown variable.
${x}^{-2}-{x}^{-1}-12=0$
$\sqrt{{|x|}^{2}}=x$
All real numbers
${t}^{10}-{t}^{5}+1=0$
$|{x}^{2}+2x-36|=12$
$x=4,6,-6,-8$
## Real-world applications
For the following exercises, use the model for the period of a pendulum, $\text{\hspace{0.17em}}T,$ such that $\text{\hspace{0.17em}}T=2\pi \sqrt{\frac{L}{g}},$ where the length of the pendulum is L and the acceleration due to gravity is $\text{\hspace{0.17em}}g.$
If the acceleration due to gravity is 9.8 m/s 2 and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).
If the gravity is 32 ft/s 2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.
10 in.
For the following exercises, use a model for body surface area, BSA, such that $\text{\hspace{0.17em}}BSA=\sqrt{\frac{wh}{3600}},$ where w = weight in kg and h = height in cm.
Find the height of a 72-kg female to the nearest cm whose $\text{\hspace{0.17em}}BSA=1.8.$
Find the weight of a 177-cm male to the nearest kg whose $\text{\hspace{0.17em}}BSA=2.1.$
90 kg
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
|
# Spectrum Math Grade 8 Chapter 2 Lesson 5 Answer Key Comparing Rational and Irrational Numbers
Students can use the Spectrum Math Grade 8 Answer Key Chapter 2 Lesson 2.5 Comparing Rational and Irrational Numbers as a quick guide to resolve any of their doubts.
## Spectrum Math Grade 8 Chapter 2 Lesson 2.5 Comparing Rational and Irrational Numbers Answers Key
Compare rational and irrational numbers by using a best guess for irrational numbers. $$\sqrt{3}$$ < 2 This statement is true because $$\sqrt{3}$$ is between 1 and 2. 5 > $$\sqrt{20}$$ This statement is true because $$\sqrt{20}$$ is between 4 and 5.
Compare using <, >, or =.
Question 1.
a. $$\sqrt{9}$$ ______ π
Answer: $$\sqrt{9}$$< π
This statement is true because $$\sqrt{9}$$ is 3 and value of π is 3.14.
b. 4.5 _____ $$\sqrt{25}$$
Answer: 4.5 < $$\sqrt{25}$$
This statement is true because $$\sqrt{25}$$ is 5. As 4.5 is less than 5. Therefore 4.5 is less than $$\sqrt{25}$$ .
c. 3.9 _____ $$\sqrt{10}$$
Answer: 3.9 > $$\sqrt{10}$$
This statement is true because $$\sqrt{10}$$ is 3.16. As 3.16 is less than 3.9. Therefore, 3.9 is greater than
$$\sqrt{10}$$.
Question 2.
a.
$$\sqrt{2}$$ _____ 1
Answer: $$\sqrt{2}$$ > 1
This statement is true because $$\sqrt{2}$$ is 1.414. As 1.414 is greater than 1. Therefore, $$\sqrt{2}$$ > 1
b.
$$3 \sqrt{\frac{8}{27}}$$ ____ $$\frac{2}{3}$$
Answer: $$3 \sqrt{\frac{8}{27}}$$ = $$\frac{2}{3}$$
This statement is true because $$3 \sqrt{\frac{8}{27}}$$ is $$\frac{2}{3}$$. Thereore, $$3 \sqrt{\frac{8}{27}}$$ is equal to $$\frac{2}{3}$$
c. 1.1 ____ $$\sqrt{2}$$
Answer: 1.1 < $$\sqrt{2}$$
This statement is true because $$\sqrt{2}$$ is 1.414. As 1.414 is greater than 1. Therefore, 1.1 is less than $$\sqrt{2}$$.
Question 3.
a.
$$0 . \overline{66}$$ _____ $$\frac{2}{3}$$
Answer: $$0 . \overline{66}$$ = $$\frac{2}{3}$$
This statement is true because $$\frac{2}{3}$$ is $$0 . \overline{66}$$. Therefore, $$0 . \overline{66}$$ = $$\frac{2}{3}$$
b.
$$\sqrt{8}$$ _____ 3
Answer: $$\sqrt{8}$$ < 3
This statement is true because $$\sqrt{8}$$ is 2.82. As 2.82 is less than 3. Therefore, $$\sqrt{8}$$ is less than 3.
c.
1 ______ $$\sqrt{\frac{16}{25}}$$
Answer: 1 > $$\sqrt{\frac{16}{25}}$$
This statement is true because $$\sqrt{\frac{16}{25}}$$ is 0.8. As 1 is greater than 0.8. Therefore, 1 is greater than $$\sqrt{\frac{16}{25}}$$.
Question 4.
a. $$\sqrt{36}$$ ____ 6.5
Answer: $$\sqrt{36}$$ < 6.5
This statement is true because $$\sqrt{36}$$ is 6. As 6 is less than 6.5. Therefore, $$\sqrt{36}$$is less than 6.5.
b. 1 ____ $$0 . \overline{45}$$
Answer: 1 > $$0 . \overline{45}$$
This statement is true because 1 is always greater than $$0 . \overline{45}$$
c. $$\frac{3}{5}$$ _____ $$\sqrt{\frac{9}{5}}$$
Answer: $$\frac{3}{5}$$ < $$\sqrt{\frac{9}{5}}$$
This statement is true because $$\frac{3}{5}$$ is 0.6 and $$\sqrt{\frac{9}{5}}$$ is 1.34. As 1.34 is greater than 0.6. Therefore, $$\frac{3}{5}$$ < $$\sqrt{\frac{9}{5}}$$
Question 5.
a.
$$\sqrt[3]{343}$$ ____ 7.2
Answer: $$\sqrt[3]{343}$$ < 7.2
this statement is true because $$\sqrt[3]{343}$$ is 7. As 7 is less than 7.2. Therefore, $$\sqrt[3]{343}$$ is less than 7.2.
b. $$0 . \overline{77}$$ ____ $$\frac{7}{9}$$
Answer: $$0 . \overline{77}$$ = $$\frac{7}{9}$$
This statement is true because $$\frac{7}{9}$$ is $$0 . \overline{77}$$. Therefore, $$0 . \overline{77}$$ = $$\frac{7}{9}$$
c. 7 _____ $$\sqrt{52}$$
Answer: 7 < $$\sqrt{52}$$
This statement is true because $$\sqrt{52}$$ is 7.2. As 7 is less than 7.2. Therefore, 7 is less than $$\sqrt{52}$$.
Question 6.
a. $$\sqrt{5}$$ ____ 4
Answer: $$\sqrt{5}$$ < 4
This statement is true because $$\sqrt{5}$$ is 2.23. As 2.23 is less than 4. Therefore, $$\sqrt{5}$$ is less than 4.
b. $$\frac{3}{4}$$ _____ $$0 . \overline{75}$$
Answer: $$\frac{3}{4}$$ < $$0 . \overline{75}$$
This statement is true because $$\frac{3}{4}$$ is 0.75. As 0.75 is less than $$0 . \overline{75}$$. Therefore, $$\frac{3}{4}$$ is less than $$0 . \overline{75}$$
c. ____ 3.5
This statement is true because is 3.17. As 3.17 is less than 3.5. Therefore, is less than 3.5.
Question 7.
a. $$\frac{5}{10}$$ _____ $$\sqrt{1}$$
Answer: $$\frac{5}{10}$$ < $$\sqrt{1}$$
This statement is true because $$\frac{5}{10}$$ is 0.5 and $$\sqrt{1}$$ is 1. As 0.5 is less than 1. Therefore, $$\frac{5}{10}$$ is less than $$\sqrt{1}$$.
b. $$\sqrt[3]{6}$$ ____ 2
Answer: $$\sqrt[3]{6}$$ < 2
This statement is true because $$\sqrt[3]{6}$$ is 0.5. As 0.5 is less than 2. Therefore, $$\sqrt[3]{6}$$ is less than 2.
c. 1.4 ____ $$\sqrt{2}$$
Answer: 1.4 = $$\sqrt{2}$$
This statement is true because $$\sqrt{2}$$ is 1.4. Therefore, 1.4 is equal to $$\sqrt{2}$$.
Question 8.
a.
$$\sqrt[3]{\frac{27}{125}}$$ ____ 0.6
Answer: $$\sqrt[3]{\frac{27}{125}}$$ = 0.6
This statement is true because $$\sqrt[3]{\frac{27}{125}}$$ is 0.6. Therefore $$\sqrt[3]{\frac{27}{125}}$$ is equal to 0.6.
b.
$$\frac{1}{2}$$ ____ 0.55
Answer: $$\frac{1}{2}$$ < 0.55
This statement is true because $$\frac{1}{2}$$ is 0.5. As 0.5 is less than 0.55. Therefore, $$\frac{1}{2}$$ is less than 0.55.
c.
$$\sqrt[3]{18}$$ ____ 2.5
Answer: $$\sqrt[3]{18}$$ < 2.5
This statement is true because $$\sqrt[3]{18}$$ is 0.408. As 0.408 is less than 2.5. Therefore, $$\sqrt[3]{18}$$ is less than 2.5
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|
# Solve the following simultaneous equations. 2x – 3y = 5 , x – 2y = 4
Rylan Hills | Certified Educator
calendarEducator since 2010
starTop subjects are Math, Science, and Business
We have to solve the following set of simultaneous equations:
2x - 3y = 5 ...(1)
x - 2y = 4 ...(2)
From (2)
x - 2y = 4
=> x= 4 + 2y
Substitute in (1)
2( 4 + 2y) - 3y = 5
=> 8 + 4y - 3y = 5
=> y = -3
x = 4 + 2y
=> x = 4 + 2*(-3)
=> x = 4 - 6
=> x = -2
We get x = -2 and y = -3
check Approved by eNotes Editorial
givingiswinning | Student
x - 2y =4
x = 4 + 2y
2(4 + 2y) - 2y = 5
simplify:
8 + 4y -2y = 5
combine
8 + 1y =5
subtract 8
y = 5 - 8
y = -3
x – 2(-3) = 4
x + 6 = 4
subtract 6
x = -2
(-2, -3)
atyourservice | Student
2x – 3y = 5 , x – 2y = 4
I am going to solve the second problem first to start:
x - 2y =4
x = 4 + 2y
now you know what x is so replace those numbers as x in the first problem:
2(4 + 2y) - 2y = 5
now simplify:
8 + 4y -2y = 5
combine like terms
8 + 1y =5
subtract the 8
y = 5 - 8
y = -3
now plug in y into any one of the problems to get x
x – 2(-3) = 4
x + 6 = 4
now subtract 6
x = -2
so (-2, -3)
check Approved by eNotes Editorial
giorgiana1976 | Student
We'll solve the system using elimination method. For this reason, we'll multiply by -2 the second equation:
-2x+4y=-8 (3)
-2x+ 4y + 2x – 3y = -8+5
We'll eliminate x and we'll combine like terms:
y = -3
We'll substitute y in (2):
x – 2*(-3) = 4
x + 6 = 4
x = 4 - 6
x = -2
The solution of the system is represented by the pair of coordinates: (-2 ; -3).
check Approved by eNotes Editorial
|
# Number System
AI悦创原创
• Number System
• Number System Conversion
• Number System
• Number System Conversion
## # Number System
The number system or the numeral system is the system of naming or representing numbers. We know that a number is a mathematical value that helps to count or measure objects and it helps in performing various mathematical calculations. There are different types of number systems in Maths like decimal number system, binary number system, octal number system, and hexadecimal number system. In this article, we are going to learn what is a number system in Maths? different types, conversion procedures with many number system examples in detail. Also, check mathematics for grade 12open in new window here.
## # What is Number System in Maths?
A number system is defined as a system of writing to express numbers. It is the mathematical notation for representing numbers of a given set by using digits or other symbols in a consistent manner. It provides a unique representation of every number and represents the arithmetic and algebraic structure of the figures. It also allows us to operate arithmetic operations like addition, subtraction and division.
The value of any digit in a number can be determined by:
• The digit
• Its position in the number
• The base of the number system
Before discussing the different types of number system examples, first, let us discuss what is a number?
## # What is a Number?
A number is a mathematical value used for counting or measuring or labelling objects. Numbers are used to perform arithmetic calculations. Examples of numbers are natural numbers, whole numbers, rational and irrational numbers, etc. 0 is also a number that represents a null value.
A number has many other variations such as even and odd numbers, prime and composite numbers. Even and odd terms are used when a number is divisible by 2 or not, whereas prime and composite differentiate between the numbers that have only two factors and more than two factors, respectively.
In a number system, these numbers are used as digits. 0 and 1 are the most common digits in the number system, that are used to represent binary numbers. On the other hand, 0 to 9 digits are also used for other number systems. Let us learn here the types of number systems.
## # Types of Number System
There are various types of number systems in mathematics. The four most common number system types are:
1. Decimal number system (Base- 10)
1. Binary number system (Base- 2)
1. Octal number system (Base-8)
1. Hexadecimal number system (Base- 16)
Now, let us discuss the different types of number systems with examples.
### # Decimal Number System (Base 10 Number System)
The decimal number system has a base of 10 because it uses ten digits from 0 to 9. In the decimal number system, the positions successive to the left of the decimal point represent units, tens, hundreds, thousands and so on. This system is expressed in decimal numbersopen in new window. Every position shows a particular power of the base (10).
Example of Decimal Number System:
The decimal number 1457 consists of the digit 7 in the units position, 5 in the tens place, 4 in the hundreds position, and 1 in the thousands place whose value can be written as:
(1×$10^{3}$) + (4×$10^{2}$) + (5×$10^{1}$) + (7×$10^{0}$)
=(1×1000) + (4×100) + (5×10) + (7×1)
=1000 + 400 + 50 + 7
=1457
### # Binary Number System (Base 2 Number System)
The base 2 number system is also known as the Binary number systemopen in new window wherein, only two binary digits exist, i.e., 0 and 1. Specifically, the usual base-2 is a radix of 2. The figures described under this system are known as binary numbers which are the combination of 0 and 1. For example, 110101 is a binary number.
2 进制的数字系统也被称为二进制数系统open in new window，其中只有两个二进制数字存在，即 0 和 1。具体来说，通常以 2 为底的基数是 2。在这个系统下描述的数字被称为二进制数，它是 0 和 1 的组合。例如，110101 是一个二进制数。
We can convert any system into binary and vice versa.
Example
Write (14)10 as a binary number.
Solution:
Base 2 Number System Example
$(14)_{10}$ = $1110_{2}$
### # Octal Number System (Base 8 Number System)
In the octal number systemopen in new window, the base is 8 and it uses numbers from 0 to 7 to represent numbers. Octal numbers are commonly used in computer applications. Converting an octal number to decimal is the same as decimal conversion and is explained below using an example.
Example: Convert $215_{8}$ into decimal.
Solution:
$215_{8}$ = 2 × $8^{2}$ + 1 × $8^{1}$ + 5 × $8^{0}$
= 2 × 64 + 1 × 8 + 5 × 1
= 128 + 8 + 5
= 14110
### # Hexadecimal Number System (Base 16 Number System)
16进制数制
In the hexadecimal system, numbers are written or represented with base 16. In the hex system, the numbers are first represented just like in the decimal system, i.e. from 0 to 9. Then, the numbers are represented using the alphabet from A to F. The below-given table shows the representation of numbers in the hexadecimal number systemopen in new window.
Decimal0123456789101112131415
## # Number System Chart
In the number system chart, the base values and the digits of different number systems can be found. Below is the chart of the numeral system.
Number System Chart
## # Number System Conversion
Numbers can be represented in any of the number system categories like binary, decimal, hex, etc. Also, any number which is represented in any of the number system types can be easily converted to other. Check the detailed lesson on the conversions of number systemsopen in new window to learn how to convert numbers in decimal to binary and vice versa, hexadecimal to binary and vice versa, and octal to binary and vice versa using various examples.
With the help of the different conversion procedures explained above, now let us discuss in brief about the conversion of one number system to the other number system by taking a random number.
Assume the number 349. Thus, the number 349 in different number systems is as follows:
The number 349 in the binary number system is 101011101
The number 349 in the decimal number system is 349.
The number 349 in the octal number system is 535.
The number 349 in the hexadecimal number system is 15D
## # Number System Examples
Example 1:
Convert $(1056)_{16}$ to an octal number.
$(1056)_{16}$ 转换为八进制数。
Solution:
Given, $(1056)_{16}$ is a hex number.
First we need to convert the given hexadecimal number into decimal number
$(1056)_{16}$
= 1 × $16^{3}$ + 0 × $16^{2}$ + 5 × $16^{1}$ + 6 × 160
= 4096 + 0 + 80 + 6
= $(4182)_{10}$
Now we will convert this decimal number to the required octal number by repetitively dividing by 8.
84182Remainder「剩余部分」
85226
8652
881
810
01
Therefore, taking the value of the remainder from bottom to top, we get;
(4182)10 = (10126)8
Therefore,
(1056)16 = (10126)8
Example 2:
Convert (1001001100)2 to a decimal number.
Solution:
(1001001100)2
= 1 × 29 + 0 × 28 + 0 × 27 + 1 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20
= 512 + 64 + 8 + 4
= (588)10
Example 3:
Convert 101012 into an octal number.
Solution:
Given,
101012 is the binary number
We can write the given binary number as:
010 101
Now as we know, in the octal number system,
010 → 2
101 → 5
Therefore, the required octal number is (25)8
Example 4:
Convert hexadecimal 2C to decimal number.
Solution:
We need to convert 2C16 into binary numbers first.
2C → 00101100
Now convert 001011002 into a decimal number.
101100 = 1 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20
= 32 + 8 + 4
= 44
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# What is the equation of the parabola that has a vertex at (-4, -3) and passes through point (12,4) ?
Feb 4, 2018
$y = \frac{7}{256} {\left(x + 4\right)}^{2} - 3$
#### Explanation:
$\text{the equation of a parabola in "color(blue)"vertex form}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$
$\text{here } \left(h , k\right) = \left(- 4 , - 3\right)$
$\Rightarrow y = a {\left(x + 4\right)}^{2} - 3$
$\text{to find a substitute "(12,4)" into the equation}$
$4 = 256 a - 3 \Rightarrow a = \frac{7}{256}$
$\Rightarrow y = \frac{7}{256} {\left(x + 4\right)}^{2} - 3 \leftarrow \textcolor{red}{\text{in vertex form}}$
|
# Linear functions
Linear functions
Enter what you know about your linear function. Leave out the rest and Mathepower computes it. Equation: Slope: y-line intercept: Graph goes through points… Point A (|) Point B (|)
## What is a linear function?
A linear function is a function whose graph is a line. The general form of a linear function is , where m is the slope and b is the y-axis intercept. Here is an example:
This is the graph of your function.
Dein Browser unterstützt den HTML-Canvas-Tag nicht. Hol dir einen neuen. :P
• Roots at -1.333
• y-axis intercept at (0|4)
The graph of a linear function is always a line.
A similar word to linear function is linear correlation.
## What is the slope of a linear function?
The slope of a linear function corresponds to the number in front of the x. It says how may units you have to go up / down if you go one unit to the right. Example:
This is the graph of your function.
Dein Browser unterstützt den HTML-Canvas-Tag nicht. Hol dir einen neuen. :P
• Roots at 2.5
• y-axis intercept at (0|-5)
We see that this function has slope . If we go one square to the right of any point on the graph, we have to go two squares up to be on the graph again.
Another example, this time with negative slope:
This is the graph of your function.
Dein Browser unterstützt den HTML-Canvas-Tag nicht. Hol dir einen neuen. :P
• Roots at 1.333
• y-axis intercept at (0|4)
This linear function has slope . This means whenever we go one square to the right, we have to go three squares down to be on the graph again.
## What is the y-line intercept of a linear function?
The y-line intercept is the number at the end of the function. As the name says, it says where the function cuts the y-axis. If you take a look on the function graphs, you see that intersects the y-axis at intersects the y-axis at .
## How to calculate the equation of the line from a point and the slope?
You have to insert the point into the equation, i.e. the one coordinate for x and the other one for f(x). Here is an example: Lets assume we know that our function has slope and goes through (-2|5).
Point (-2|5); Slope -7;
This is the graph of your function.
Dein Browser unterstützt den HTML-Canvas-Tag nicht. Hol dir einen neuen. :P
• Roots at -1.286
• y-axis intercept at (0|-9)
This is what Mathepower calculated:
Calculate the y-axis intercept b by inserting:
General form of the linear function: f(x)=mx+b
Insert for m, for x and for f(x).
| Multiply by | Swap both sides of the equation. |
So, the y-axis intercept is at
Therefore, the equation of the function is
## How to calculate the equation of a linear function from two given points?
First, we have to calculate the slope m by inserting the x- and y- coordinates of the points into the formula . This means: You calculate the difference of the y-coordinates and divide it by the difference of the x-coordinates. Here is an example:
Point (1|2); Point (3|8);
This is the graph of your function.
Dein Browser unterstützt den HTML-Canvas-Tag nicht. Hol dir einen neuen. :P
• Roots at 0.333
• y-axis intercept at (0|-1)
This is what Mathepower calculated:
To calculate the slope m, use the formula
Calculate the y-axis intercept b by inserting:
General form of the linear function: f(x)=mx+b
Insert for m, for x and for f(x).
| Swap both sides of the equation. |
So, the y-axis intercept is at
Therefore, the equation of the function is
As we can see, the slope was calculated first. To find the equation of the function, you have to insert a point and get an equation which gives the y-axis intercept.
## Can I see more examples?
Of course. Just enter you own examples above and they will be calculated immediately step-by-step. (This is the idea of Mathepower: You don't just look at some already done explanations, but you get your own calculations explained!)
|
Algebra Tutorials!
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
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# Multiplying Polynomials
In this section you will learn how to multiply any two polynomials.
## Multiplying Monomials with the Product Rule
To multiply two monomials, such as x3 and x5, recall that
x3 = x · x · x and x5 = x · x · x · x · x,
so
The exponent of the product of x3 and x5 is the sum of the exponents 3 and 5. This example illustrates the product rule for multiplying exponential expressions.
Product Rule
If a is any real number and m and n are any positive integers, then
am · an = am + n.
Multiplying monomials
Find the indicated products.
a) x2 · x4 · x
b) (-2ab)(-3ab)
c) -4x2y2 · 3xy5
d) (3a)2
Solution
a) x2 · x4 · x = x2 · x4 · x1 = x7 Product rule
b) (-2ab)(-3ab) = (-2)(-3) · a · a · b · b = 6a2b2 Product rule
c) (-4x2y2)(3xy5) = (-4)(3)x2 · x · y2 · y5 = -12x3y7 Product rule
d) (3a)2 = 3a · 3a = 9a2
Caution
Be sure to distinguish between adding and multiplying monomials. You can add like terms to get 3x4 + 2x4 = 5x4, but you cannot combine the terms in 3w5 + 6w2. However, you can multiply any two monomials: 3x4 · 2x4 = 6x8 and 3w5 · 6w2 = 18w7.
## Multiplying Polynomials
To multiply a monomial and a polynomial, we use the distributive property.
Multiply monomials and polynomials
Find each product.
a) 3x2(x3 - 4x)
b) (y2 - 3y + 4)(-2y)
c) -a(b - c)
Solution
Note in part c) that either of the last two binomials is the correct answer. The last one is just a little simpler to read.
Just as we use distributive property to find the product of a monomial and a polynomial, we can use the distributive property to find the product of two binomials as the product of a binomial and a trinomial.
Multiplying polynomials
Use the distributive property to find each product
a) (x + 2)(x + 5)
b) (x + 3)(x2 + 2x - 7)
Solution
a) First multiply each term of x + 5 by x + 2:
b) First multiply each term of the trinomial by x + 3:
Products of polynomials can also be found by arranging the multiplication vertically like multiplication of whole numbers.
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Area Scale Factor
The length scale factor of two similar figures can be used to find the area of one of the two figures when the area of one of the figures is known.
Using Area Scale Factor to Determine an Unknown Side
It was just learned that if the length scale factor of two similar figures and one of the areas of the figures are known, then the unknown area can be found. Next, consider if both areas but only the side length of one figure are known. It will be possible to solve for the other similar figure's corresponding side length.
The diagram shows two similar figures. Figure has an area of square inches, and Figure has an area of square inches.
If a side length of Figure is inches, find the length of the corresponding side in the other shape.
Hint
If the scale factor of two similar figures is then the ratio of their areas is
Solution
Recall that the ratio of areas of two similar figures is equal to the square of the ratio of their corresponding side lengths.
Since the areas are given, the following proportion can be written.
Now, take square roots of both sides of the equation to find the value of the scale factor. Keep in mind that only the principal roots will be considered because only positive numbers make sense in this situation.
Solve for
The scale factor of the figures is Finally, with the scale factor and knowing that the side length of Figure is inches, the length of the corresponding side in Figure represented by can be found.
Solve for
The corresponding length in Figure is inches.
Practice Finding Linear Scale Factor Given Areas
Determine the linear scale factor of the shape on the right to the shape on the left.
Volume Scale Factor
For similar three-dimensional figures, the volume scale factor and the length scale factor are also related.
Using Volume Scale Factor to Determine an Unknown Volume
The scale factor of two similar figures can be used to find the volume of one of the figures when the volume of the other figure is known.
The suitcase company Case-O-La produces snazzy suitcases of various sizes. When a large-sized suitcase is bought, the company offers its cabin-sized version at a discounted price to the same customer.
The large-sized suitcase has a height of inches and a volume of liters. If the cabin-sized suitcase has a height of inches, determine its volume. Round the answer to one decimal place.
Hint
If the scale factor of two similar figures is then the ratio of their volumes is
Solution
The suitcases can be considered as two similar rectangular prisms with heights and inches.
Similar solids have the same shape and all of their corresponding sides are proportional. The ratio of the corresponding linear dimensions of the similar solids is the scale factor.
If the scale factor of two similar solids is then the ratio of their corresponding volumes is Now, raise the scale factor to the third power to obtain the ratio of the volumes.
The ratio of the volumes is Now, let be the volume of the small suitcase. Since the volume of the big suitcase is liters, the ratio of to is
Solve for
The volume of the small suitcase is about liters.
Using Volume Scale Factor to Determine an Unknown Side
After reading a physics magazine, Mark feels confident in estimating the radius of the Sun. To do so, he will use the volumes of the Sun and Earth, which are and cubic kilometers, respectively.
If the radius of the Earth is about kilometers, help Mark find the radius of the Sun.
Hint
Sun and Earth can be regarded as two similar spheres. Therefore, the volume scale factor can be used to find the radius of the Sun.
Solution
The Sun and Earth are two similar spheres. Consequently, the ratio of their volumes is equal to the cube of the ratio of their corresponding linear measures, which in this case is the ratio of their radii. Let and be the radii of the Earth and Sun, respectively.
Given that the volumes are known, the volume scale factor can be used to find the scale factor of the Earth to the Sun.
Next, take the cube roots of both sides of the equation to find the value of the length scale factor.
Solve for
Write in scientific notation
Finally, with the scale factor and knowing that the radius of the Earth is about kilometers, the radius of the Sun can be found. The ratio of to is equal to the scale factor.
Solve for
Write in scientific notation
Practice Finding Linear Scale Factor Given Volumes
The applet shows the volumes of two similar solids. Determine the scale factor of the blue solid to the orange solid.
Finding Volume of Similar Composite Solids
A three-dimensional figure is called a composite solid if it is the combination of two or more solids. Like similar solids, if the corresponding linear measures of two composite solids are proportional, the composite solids are said to be similar. Therefore, their length scale factor can be determined and used to find certain characteristics of the shapes.
The amount of material used to construct the larger silo is three times that of the smaller one. That is, the surface area of the larger silo is three times as large as the surface area of the smaller silo. These two silos can be considered to be similar solids. Furthermore, each silo is composed of a cone and a cylinder as shown.
If the volume of the larger silo is cubic meters, find the volume of the smaller silo. Round the answer to the nearest integer.
Hint
Use the area scale factor to determine the length scale factor.
Solution
The given silos are similar composite solids. Since they are similar, their corresponding linear measures are proportional. Therefore, each silo can be considered as a whole. To find the volume of the smaller silo, these steps will be followed.
• The ratio of their surface areas will be used to determine the length scale factor.
• The length scale factor will be used to determine the volume scale factor.
• Finally, the volume scale factor will be used to determine the volume of the smaller silo.
It is given that the surface area of the larger silo is three times as large as the surface area of the smaller silo.
To find the length scale factor, consider its relationship to the surface areas. Recall that for areas of similar figures, the ratio of their areas is equal to the square of the ratio of their corresponding side lengths. Refer to the smaller silo's side length as and the larger silo's side length as
Since the surface area of the larger silo is three times the surface area of the smaller silo, the ratio of the surface area, small to large, is With that information, the following equation can be expressed.
Next, this equation can be simplified to solve for the length scale factor. Begin by taking the square root of both sides of the equation.
Solve for
Now, the length scale factor can be used to find the volume scale factor. To do so, the length scale factor needs to be raised to the third power.
Finally, knowing that the volume of the larger silo is about cubic meters, the volume of the smaller silo can be calculated. Similar to the areas, the ratio of the volumes should be equal to the volume scale factor.
Solve for
The capacity of the smaller silo is about cubic meters.
The similarity makes it possible to solve for certain characteristics of a wide range of shapes, like prisms, spheres, composite solids, and pyramids.
Relationship Between Length, Area, and Volume Scale Factor
In this course, the relationships between the length scale factor, area scale factor and volume scale factor have been discussed. If the scale factor between two similar figures is then the ratio for their areas and volumes can be expressed as the table shows.
Length Scale Factor Area Scale Factor Volume Scale Factor
Considering these expressions, the challenge presented at the beginning can be solved with more confidence.
Emily knows that the models in the museum are similar pyramids and the scale factor between the corresponding side lengths is
If the volume of the smaller model is cubic centimeters, find the volume of the larger model.
Hint
If the scale factor of two similar figures is then the ratio of their volumes is
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2014-10-26T21:08:11-04:00
I have seen this question before and I think you meant 10 more dimes than nickels.
We can use substitution to answer this question. The value of a nickel is 5 cents, and we can use the variable n to represent the number of nickels. The value of a dime is 10 cents, and we can use the variable d to represent the number of dimes.
First lets figure out the equations.
.10d+.5n=2.80 (the number of nickels (n) multiplied by .5 will tell us their money value. Same thing for the dimes)
d-n=10 (since there are 13 more dimes than nickels, the number of dimes value (d) minus the number of nickels value (n) will give us 10)
Now lets isolate a variable in one of the equations, preferably the second one because it doesn't have any visible coefficients,
d-n=10
-n=10-d (subtracted the d from both sides)
n=-10+d (made the n positive)
Now that we have the value of n, we can plug it into the other equation.
.10d+.05n=2.80
.10d+.05(-10+d)=2.80 (we replaced the n with the value that we previously got)
.10d-.5+.05d=2.80 (did the multiplication)
.15d-.5=2.80 (combined like terms)
.15d=3.30 (added the .5 to both sides)
d=22 (divided both sides by the .15)
Now that we know that there are 22 dimes and we also know that there are 10 less nickels than dimes, so we can subtract 10 from 22 to get the number of nickels. 22-10=12
d=22
n=12
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# Calculus 2 : Derivative Review
## Example Questions
← Previous 1 3 4 5 6 7 8 9 51 52
### Example Question #1 : Derivative Review
Evaluate the limit using one of the definitions of a derivative.
Does not exist
Explanation:
Evaluating the limit directly will produce an indeterminant solution of .
The limit definition of a derivative is . However, the alternative form, , better suits the given limit.
Let and notice . It follows that .
Thus, the limit is
### Example Question #2 : Derivative Review
Evaluate the limit using one of the definitions of a derivative.
Does not exist
Explanation:
Evaluating the derivative directly will produce an indeterminant solution of .
The limit definition of a derivative is . However, the alternative form, , better suits the given limit.
Let and notice . It follows that . Thus, the limit is .
### Example Question #1 : Definition Of Derivative
Suppose and are differentiable functions, and . Calculate the derivative of , at
Explanation:
Taking the derivative of involves the product rule, and the chain rule.
Substituting into both sides of the derivative we get
.
### Example Question #4 : Derivative Review
Evaluate the limit
without using L'Hopital's rule.
Explanation:
If we recall the definition of a derivative of a function at a point , one of the definitions is
.
If we compare this definition to the limit
we see that that this is the limit definition of a derivative, so we need to find the function and the point at which we are evaluating the derivative at. It is easy to see that the function is and the point is . So finding the limit above is equivalent to finding .
We know that the derivative is , so we have
.
### Example Question #1 : Definition Of Derivative
Approximate the derivative if where .
Explanation:
Write the definition of the limit.
Substitute .
Since is approaching to zero, it would be best to evaluate when we assume that is progressively decreasing. Let's assume and check the pattern.
## Find f'(x):
Explanation:
Computation of the derivative requires the use of the Product Rule and Chain Rule.
The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:
This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."
In the problem statement, we are given:
is the "First" function, and is the "Second" function.
The "Second" function requires use of the Chain Rule.
When:
Applying these formulas results in:
Simplifying the terms inside the brackets results in:
We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".
Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:
Simplifying this results in one of the answer choices:
### Example Question #7 : Derivative Review
What is the value of the limit below?
Explanation:
Recall that one definition for the derivative of a function is .
This means that this question is asking us to find the value of the derivative of at .
Since
and , the value of the limit is .
### Example Question #8 : Derivative Review
Explanation:
Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of .
Product Rule:
Applying these two rules results in:
This matches one of the answer choices.
### Example Question #9 : Derivative Review
Use the definition of the derivative to solve for .
Explanation:
In order to find , we need to remember how to find by using the definition of derivative.
Definition of Derivative:
Now lets apply this to our problem.
Now lets expand the numerator.
We can simplify this to
Now factor out an h to get
We can simplify and then evaluate the limit.
### Example Question #10 : Derivative Review
Use the definition of the derivative to solve for .
Explanation:
In order to find , we need to remember how to find by using the definition of derivative.
Definition of Derivative:
Now lets apply this to our problem.
Now lets expand the numerator.
We can simplify this to
Now factor out an h to get
We can simplify and then evaluate the limit.
← Previous 1 3 4 5 6 7 8 9 51 52
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### Section 2-4 : Equations With More Than One Variable
1. Solve $$A = 3p\left( {4 - 2r} \right)$$ for $$p$$.
2. Solve $$A = 3p\left( {4 - 2r} \right)$$ for $$r$$.
3. Solve $$\displaystyle T = \frac{c}{3}\left( {6p + \frac{{3q}}{c}} \right) - 7p$$ for $$p$$
4. Solve $$\displaystyle T = \frac{c}{3}\left( {6p + \frac{{3q}}{c}} \right) - 7p$$ for $$c$$.
5. Solve $$\displaystyle \frac{1}{n} = \frac{2}{m} - \frac{3}{q}$$ for $$n$$.
6. Solve $$\displaystyle \frac{1}{n} = \frac{2}{m} - \frac{3}{q}$$ for $$q$$.
7. Solve $$3A + 6C = 4A\left( {B - 7C} \right)$$ for $$C$$.
8. Solve $$3A + 6C = 4A\left( {B - 7C} \right)$$ for $$A$$.
9. Solve $$\displaystyle y = \frac{{4 - 9x}}{3}$$ for $$x$$.
10. Solve $$\displaystyle y = \frac{{12}}{{1 - x}}$$ for $$x$$.
11. Solve $$\displaystyle y = \frac{7}{{10x + 9}}$$ for $$x$$.
12. Solve $$\displaystyle y = \frac{{8 - 5x}}{{9 - 7x}}$$ for $$x$$.
13. Solve $$\displaystyle y = \frac{{2 + 11x}}{{1 + 4x}}$$ for $$x$$.
14. Solve $$\displaystyle y = \frac{{9 + 2x}}{{4 - x}}$$ for $$x$$.
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Question
The sum of the two numbers is 26. One of the numbers is 2 more than twice the other. Find the numbers.
Hint: Let us assume the first number to be $x$ , the other number becomes $26-x$.
we can develop an equation using the data given in the question . Use both these equations to find the value of $x$.
Before proceeding we must be familiar with the terms, like the left-hand-side and the right-hand-side, how to solve linear equations, what are the rules to solve the equations when there are terms both on the left-hand-side and right-hand-side, what happens to the terms when we take it from right-hand-side to the left-hand-side.
In this question we have been given, the sum of the two numbers is 26. One of the numbers is 2 more than twice the other then we have to find out the numbers.
Let us assume the first number is $x$ .
Then the other number becomes $26-x$.
In question, it is given that one of the numbers is 2 more than twice the other.
The equation becomes, $x=2+2\left( 26-x \right)$.
Opening the brackets, we get,
$\therefore x=2+2\times 26-2x$
Now taking the variables on the left-hand side we get,
$\Rightarrow x+2x=2+52$
$\Rightarrow 3x=54$
$\Rightarrow x=\dfrac{54}{3}$
After cancelling we get,
$\therefore x=18$
Therefore, the first number is $18$.
The other number is given by $26-x=26-18$
Therefore, the other number is $8$. Hence, the required numbers are $18$ and $8$.
Note: We must be very careful about the sign when we are taking the terms from left-hand-side to right-hand-side and vice-versa. We can cross-check the answer by putting the value of the answer and verifying it with the question.
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# Differentiation of trigonometric functions
Function Derivative
${\displaystyle \sin(x)}$ ${\displaystyle \cos(x)}$
${\displaystyle \cos(x)}$ ${\displaystyle -\sin(x)}$
${\displaystyle \tan(x)}$ ${\displaystyle \sec ^{2}(x)}$
${\displaystyle \cot(x)}$ ${\displaystyle -\csc ^{2}(x)}$
${\displaystyle \sec(x)}$ ${\displaystyle \sec(x)\tan(x)}$
${\displaystyle \csc(x)}$ ${\displaystyle -\csc(x)\cot(x)}$
${\displaystyle \arcsin(x)}$ ${\displaystyle {\frac {1}{\sqrt {1-x^{2}}}}}$
${\displaystyle \arccos(x)}$ ${\displaystyle -{\frac {1}{\sqrt {1-x^{2}}}}}$
${\displaystyle \arctan(x)}$ ${\displaystyle {\frac {1}{x^{2}+1}}}$
${\displaystyle \operatorname {arccot}(x)}$ ${\displaystyle -{\frac {1}{x^{2}+1}}}$
${\displaystyle \operatorname {arcsec}(x)}$ ${\displaystyle {\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$
${\displaystyle \operatorname {arccsc}(x)}$ ${\displaystyle -{\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$
The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. For example, the derivative of the sine function is written sin′(a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle.
All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule applied to functions such as tan(x) = sin(x)/cos(x). Knowing these derivatives, the derivatives of the inverse trigonometric functions are found using implicit differentiation.
## Proofs of derivatives of trigonometric functions
### Limit of sin(θ)/θ as θ tends to 0
The diagram at right shows a circle with centre O and radius r = 1. Let two radii OA and OB make an arc of θ radians. Since we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < ½ π in the first quadrant.
In the diagram, let R1 be the triangle OAB, R2 the circular sector OAB, and R3 the triangle OAC.
The area of triangle OAB is:
${\displaystyle \mathrm {Area} (R_{1})={\tfrac {1}{2}}\ |OA|\ |OB|\sin \theta ={\tfrac {1}{2}}\sin \theta \,.}$
The area of the circular sector OAB is:
${\displaystyle \mathrm {Area} (R_{2})={\tfrac {1}{2}}\theta \,.}$
The area of the triangle OAC is given by:
${\displaystyle \mathrm {Area} (R_{3})={\tfrac {1}{2}}\ |OA|\ |AC|={\tfrac {1}{2}}\tan \theta \,.}$
Since each region is contained in the next, one has:
${\displaystyle {\text{Area}}(R_{1})<{\text{Area}}(R_{2})<{\text{Area}}(R_{3})\implies {\tfrac {1}{2}}\sin \theta <{\tfrac {1}{2}}\theta <{\tfrac {1}{2}}\tan \theta \,.}$
Moreover, since sin θ > 0 in the first quadrant, we may divide through by ½ sin θ, giving:
${\displaystyle 1<{\frac {\theta }{\sin \theta }}<{\frac {1}{\cos \theta }}\implies 1>{\frac {\sin \theta }{\theta }}>\cos \theta \,.}$
In the last step we took the reciprocals of the three positive terms, reversing the inequities.
We conclude that for 0 < θ < ½ π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ). Thus, as θ gets closer to 0, sin(θ)/θ is "squeezed" between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/θ must tend to 1 as θ tends to 0 from the positive side:
${\displaystyle \lim _{\theta \to 0^{+}}{\frac {\sin \theta }{\theta }}=1\,.}$
For the case where θ is a small negative number –½ π < θ < 0, we use the fact that sine is an odd function:
${\displaystyle \lim _{\theta \to 0^{-}}\!{\frac {\sin \theta }{\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {\sin(-\theta )}{-\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {-\sin \theta }{-\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {\sin \theta }{\theta }}\ =\ 1\,.}$
### Limit of (cos(θ)-1)/θ as θ tends to 0
The last section enables us to calculate this new limit relatively easily. This is done by employing a simple trick. In this calculation, the sign of θ is unimportant.
${\displaystyle \lim _{\theta \to 0}\,{\frac {\cos \theta -1}{\theta }}\ =\ \lim _{\theta \to 0}\left({\frac {\cos \theta -1}{\theta }}\right)\!\!\left({\frac {\cos \theta +1}{\cos \theta +1}}\right)\ =\ \lim _{\theta \to 0}\,{\frac {\cos ^{2}\!\theta -1}{\theta \,(\cos \theta +1)}}.}$
Using cos2θ – 1 = –sin2θ, the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:
${\displaystyle \lim _{\theta \to 0}\,{\frac {\cos \theta -1}{\theta }}\ =\ \lim _{\theta \to 0}\,{\frac {-\sin ^{2}\theta }{\theta (\cos \theta +1)}}\ =\ \left(-\lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}\right)\!\left(\lim _{\theta \to 0}\,{\frac {\sin \theta }{\cos \theta +1}}\right)\ =\ (-1)\left({\frac {0}{2}}\right)=0\,.}$
### Limit of tan(θ)/θ as θ tends to 0
Using the limit for the sine function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find:
${\displaystyle \lim _{\theta \to 0}{\frac {\tan \theta }{\theta }}\ =\ \left(\lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}\right)\!\left(\lim _{\theta \to 0}{\frac {1}{\cos \theta }}\right)\ =\ (1)(1)\ =\ 1\,.}$
### Derivative of the sine function
We calculate the derivative of the sine function from the limit definition:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =\lim _{\delta \to 0}{\frac {\sin(\theta +\delta )-\sin \theta }{\delta }}.}$
Using the angle addition formula sin(α+β) = sin α cos β + sin β cos α, we have:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =\lim _{\delta \to 0}{\frac {\sin \theta \cos \delta +\sin \delta \cos \theta -\sin \theta }{\delta }}=\lim _{\delta \to 0}\left({\frac {\sin \delta }{\delta }}\cos \theta +{\frac {\cos \delta -1}{\delta }}\sin \theta \right).}$
Using the limits for the sine and cosine functions:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =(1)\cos \theta +(0)\sin \theta =\cos \theta \,.}$
### Derivative of the cosine function
#### From the definition of derivative
We again calculate the derivative of the cosine function from the limit definition:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\cos \theta =\lim _{\delta \to 0}{\frac {\cos(\theta +\delta )-\cos \theta }{\delta }}.}$
Using the angle addition formula cos(α+β) = cos α cos β – sin α sin β, we have:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\cos \theta =\lim _{\delta \to 0}{\frac {\cos \theta \cos \delta -\sin \theta \sin \delta -\cos \theta }{\delta }}=\lim _{\delta \to 0}\left({\frac {\cos \delta -1}{\delta }}\cos \theta \,-\,{\frac {\sin \delta }{\delta }}\sin \theta \right).}$
Using the limits for the sine and cosine functions:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\cos \theta =(0)\cos \theta -(1)\sin \theta =-\sin \theta \,.}$
#### From the chain rule
To compute the derivative of the cosine function from the chain rule, first observe the following three facts:
${\displaystyle \cos \theta =\sin \left({\tfrac {\pi }{2}}-\theta \right)}$
${\displaystyle \sin \theta =\cos \left({\tfrac {\pi }{2}}-\theta \right)}$
${\displaystyle {\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}\sin \theta =\cos \theta }$
The first and the second are trigonometric identities, and the third is proven above. Using these three facts, we can write the following,
${\displaystyle {\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}\cos \theta ={\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}\sin \left({\tfrac {\pi }{2}}-\theta \right)}$
We can differentiate this using the chain rule. Letting ${\displaystyle f(x)=\sin x,\ \ g(\theta )={\tfrac {\pi }{2}}-\theta }$, we have:
${\displaystyle {\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}f\!\left(g\!\left(\theta \right)\right)=f^{\prime }\!\left(g\!\left(\theta \right)\right)\cdot g^{\prime }\!\left(\theta \right)=\cos \left({\tfrac {\pi }{2}}-\theta \right)\cdot (0-1)=-\sin \theta }$.
Therefore, we have proven that
${\displaystyle {\tfrac {\operatorname {d} }{\operatorname {d} \!\theta }}\cos \theta =-\sin \theta }$.
### Derivative of the tangent function
#### From the definition of derivative
To calculate the derivative of the tangent function tan θ, we use first principles. By definition:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =\lim _{\delta \to 0}\left({\frac {\tan(\theta +\delta )-\tan \theta }{\delta }}\right).}$
Using the well-known angle formula tan(α+β) = (tan α + tan β) / (1 - tan α tan β), we have:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =\lim _{\delta \to 0}\left[{\frac {{\frac {\tan \theta +\tan \delta }{1-\tan \theta \tan \delta }}-\tan \theta }{\delta }}\right]=\lim _{\delta \to 0}\left[{\frac {\tan \theta +\tan \delta -\tan \theta +\tan ^{2}\theta \tan \delta }{\delta \left(1-\tan \theta \tan \delta \right)}}\right].}$
Using the fact that the limit of a product is the product of the limits:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =\lim _{\delta \to 0}{\frac {\tan \delta }{\delta }}\times \lim _{\delta \to 0}\left({\frac {1+\tan ^{2}\theta }{1-\tan \theta \tan \delta }}\right).}$
Using the limit for the tangent function, and the fact that tan δ tends to 0 as δ tends to 0:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =1\times {\frac {1+\tan ^{2}\theta }{1-0}}=1+\tan ^{2}\theta .}$
We see immediately that:
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\tan \theta =1+{\frac {\sin ^{2}\theta }{\cos ^{2}\theta }}={\frac {\cos ^{2}\theta +\sin ^{2}\theta }{\cos ^{2}\theta }}={\frac {1}{\cos ^{2}\theta }}=\sec ^{2}\theta \,.}$
#### From the quotient rule
One can also compute the derivative of the tangent function using the quotient rule.
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\tan \theta ={\frac {\operatorname {d} }{\operatorname {d} \!\theta }}{\frac {\sin \theta }{\cos \theta }}={\frac {\left(\sin \theta \right)^{\prime }\cdot \cos \theta -\sin \theta \cdot \left(\cos \theta \right)^{\prime }}{\cos ^{2}\theta }}={\frac {\cos ^{2}\theta +\sin ^{2}\theta }{\cos ^{2}\theta }}}$
The numerator can be simplified to 1 by the Pythagorean identity, giving us,
${\displaystyle {\frac {1}{\cos ^{2}\theta }}=\sec ^{2}\theta }$
Therefore,
${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\tan \theta =\sec ^{2}\theta }$
## Proofs of derivatives of inverse trigonometric functions
The following derivatives are found by setting a variable y equal to the inverse trigonometric function that we wish to take the derivative of. Using implicit differentiation and then solving for dy/dx, the derivative of the inverse function is found in terms of y. To convert dy/dx back into being in terms of x, we can draw a reference triangle on the unit circle, letting θ be y. Using the Pythagorean theorem and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x.
### Differentiating the inverse sine function
We let
${\displaystyle y=\arcsin x\,\!}$
Where
${\displaystyle -{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}}$
Then
${\displaystyle \sin y=x\,\!}$
Taking the derivative with respect to ${\displaystyle x}$ on both sides and solving for dy/dx:
${\displaystyle {d \over dx}\sin y={d \over dx}x}$
${\displaystyle \cos y\cdot {dy \over dx}=1\,\!}$
Substituting ${\displaystyle \cos y={\sqrt {1-\sin ^{2}y}}}$ in from above,
${\displaystyle {\sqrt {1-\sin ^{2}y}}\cdot {dy \over dx}=1}$
Substituting ${\displaystyle x=\sin y}$ in from above,
${\displaystyle {\sqrt {1-x^{2}}}\cdot {dy \over dx}=1}$
${\displaystyle {dy \over dx}={\frac {1}{\sqrt {1-x^{2}}}}}$
### Differentiating the inverse cosine function
We let
${\displaystyle y=\arccos x\,\!}$
Where
${\displaystyle 0\leq y\leq \pi }$
Then
${\displaystyle \cos y=x\,\!}$
Taking the derivative with respect to ${\displaystyle x}$ on both sides and solving for dy/dx:
${\displaystyle {d \over dx}\cos y={d \over dx}x}$
${\displaystyle -\sin y\cdot {dy \over dx}=1}$
Substituting ${\displaystyle \sin y={\sqrt {1-\cos ^{2}y}}\,\!}$ in from above, we get
${\displaystyle -{\sqrt {1-\cos ^{2}y}}\cdot {dy \over dx}=1}$
Substituting ${\displaystyle x=\cos y\,\!}$ in from above, we get
${\displaystyle -{\sqrt {1-x^{2}}}\cdot {dy \over dx}=1}$
${\displaystyle {dy \over dx}=-{\frac {1}{\sqrt {1-x^{2}}}}}$
Alternatively, once the derivative of ${\displaystyle \arcsin x}$ is established, the derivative of ${\displaystyle \arccos x}$ follows immediately by differentiating the identity ${\displaystyle \arcsin x+\arccos x=\pi /2}$ so that ${\displaystyle (\arccos x)'=-(\arcsin x)'}$.
### Differentiating the inverse tangent function
We let
${\displaystyle y=\arctan x\,\!}$
Where
${\displaystyle -{\frac {\pi }{2}}
Then
${\displaystyle \tan y=x\,\!}$
Taking the derivative with respect to ${\displaystyle x}$ on both sides and solving for dy/dx:
${\displaystyle {d \over dx}\tan y={d \over dx}x}$
Left side:
${\displaystyle {d \over dx}\tan y=\sec ^{2}y\cdot {dy \over dx}=(1+\tan ^{2}y){dy \over dx}}$ using the Pythagorean identity
Right side:
${\displaystyle {d \over dx}x=1}$
Therefore,
${\displaystyle (1+\tan ^{2}y){dy \over dx}=1}$
Substituting ${\displaystyle x=\tan y\,\!}$ in from above, we get
${\displaystyle (1+x^{2}){dy \over dx}=1}$
${\displaystyle {dy \over dx}={\frac {1}{1+x^{2}}}}$
### Differentiating the inverse cotangent function
We let
${\displaystyle y=\operatorname {arccot} x}$
where ${\displaystyle 0. Then
${\displaystyle \cot y=x}$
Taking the derivative with respect to ${\displaystyle x}$ on both sides and solving for dy/dx:
${\displaystyle {\frac {d}{dx}}\cot y={\frac {d}{dx}}x}$
Left side:
${\displaystyle {d \over dx}\cot y=-\csc ^{2}y\cdot {dy \over dx}=-(1+\cot ^{2}y){dy \over dx}}$ using the Pythagorean identity
Right side:
${\displaystyle {d \over dx}x=1}$
Therefore,
${\displaystyle -(1+\cot ^{2}y){\frac {dy}{dx}}=1}$
Substituting ${\displaystyle x=\cot y}$,
${\displaystyle -(1+x^{2}){\frac {dy}{dx}}=1}$
${\displaystyle {\frac {dy}{dx}}=-{\frac {1}{1+x^{2}}}}$
Alternatively, as the derivative of ${\displaystyle \arctan x}$ is derived as shown above, then using the identity ${\displaystyle \arctan x+\operatorname {arccot} x={\dfrac {\pi }{2}}}$ follows immediately that
{\displaystyle {\begin{aligned}{\dfrac {d}{dx}}\operatorname {arccot} x&={\dfrac {d}{dx}}\left({\dfrac {\pi }{2}}-\arctan x\right)\\&=-{\dfrac {1}{1+x^{2}}}\end{aligned}}}
### Differentiating the inverse secant function
#### Using implicit differentiation
Let
${\displaystyle y=\operatorname {arcsec} x\ \mid |x|\geq 1}$
Then
${\displaystyle x=\sec y\mid \ y\in \left[0,{\frac {\pi }{2}}\right)\cup \left({\frac {\pi }{2}},\pi \right]}$
${\displaystyle {\frac {dx}{dy}}=\sec y\tan y=|x|{\sqrt {x^{2}-1}}}$
(The absolute value in the expression is necessary as the product of secant and tangent in the interval of y is always nonnegative, while the radical ${\displaystyle {\sqrt {x^{2}-1}}}$ is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)
${\displaystyle {\frac {dy}{dx}}={\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$
#### Using the chain rule
Alternatively, the derivative of arcsecant may be derived from the derivative of arccosine using the chain rule.
Let
${\displaystyle y=\operatorname {arcsec} x=\arccos \left({\frac {1}{x}}\right)}$
Where
${\displaystyle |x|\geq 1}$ and ${\displaystyle y\in \left[0,{\frac {\pi }{2}}\right)\cup \left({\frac {\pi }{2}},\pi \right]}$
Then, applying the chain rule to ${\displaystyle \arccos \left({\frac {1}{x}}\right)}$:
${\displaystyle {\frac {dy}{dx}}=-{\frac {1}{\sqrt {1-({\frac {1}{x}})^{2}}}}\cdot \left(-{\frac {1}{x^{2}}}\right)={\frac {1}{x^{2}{\sqrt {1-{\frac {1}{x^{2}}}}}}}={\frac {1}{x^{2}{\frac {\sqrt {x^{2}-1}}{\sqrt {x^{2}}}}}}={\frac {1}{{\sqrt {x^{2}}}{\sqrt {x^{2}-1}}}}={\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$
### Differentiating the inverse cosecant function
#### Using implicit differentiation
Let
${\displaystyle y=\operatorname {arccsc} x\ \mid |x|\geq 1}$
Then
${\displaystyle x=\csc y\ \mid \ y\in \left[-{\frac {\pi }{2}},0\right)\cup \left(0,{\frac {\pi }{2}}\right]}$
${\displaystyle {\frac {dx}{dy}}=-\csc y\cot y=-|x|{\sqrt {x^{2}-1}}}$
(The absolute value in the expression is necessary as the product of cosecant and cotangent in the interval of y is always nonnegative, while the radical ${\displaystyle {\sqrt {x^{2}-1}}}$ is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)
${\displaystyle {\frac {dy}{dx}}={\frac {-1}{|x|{\sqrt {x^{2}-1}}}}}$
#### Using the chain rule
Alternatively, the derivative of arccosecant may be derived from the derivative of arcsine using the chain rule.
Let
${\displaystyle y=\operatorname {arccsc} x=\arcsin \left({\frac {1}{x}}\right)}$
Where
${\displaystyle |x|\geq 1}$ and ${\displaystyle y\in \left[-{\frac {\pi }{2}},0\right)\cup \left(0,{\frac {\pi }{2}}\right]}$
Then, applying the chain rule to ${\displaystyle \arcsin \left({\frac {1}{x}}\right)}$:
${\displaystyle {\frac {dy}{dx}}={\frac {1}{\sqrt {1-({\frac {1}{x}})^{2}}}}\cdot \left(-{\frac {1}{x^{2}}}\right)=-{\frac {1}{x^{2}{\sqrt {1-{\frac {1}{x^{2}}}}}}}=-{\frac {1}{x^{2}{\frac {\sqrt {x^{2}-1}}{\sqrt {x^{2}}}}}}=-{\frac {1}{{\sqrt {x^{2}}}{\sqrt {x^{2}-1}}}}=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}}$
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## Thursday, March 13, 2008
### Derrick's Pythagoras Growing Post
- Pythagorean Triple ?
- Three numbers a, b, and c that satisfy a2 + b2 = c2 are called Pythagorean Triples.
- some Pythagorean triple other than 3,4,5 is 5,12,13
This picture shows how a pythagorean triple works. Its three squares making a right angle triangle in the middle.
Example:
Following the formula a2 + b2 = c2 we got 5² +12² = 13².
a²= 5 - 5²
b²= 12 - 12²
c²= 13 - 13²
a2 + b2 = c2
5² +12² = 13²
25 + 144 = 169
You’re locked out of your house and the only open window is on the 2nd floor 25 feet above the ground. You need to borrow a ladder from one of your neighbors. There’s a bush along the edge of the house, so you’ll have to place the ladder 10 feet from the house. What length of the ladder do you need to reach the window?
PART III
In this problem, it says to find the length of the ladder to reach the window. (Ladder: a, house: b, ground: c) In the picture, you can see a triangle. (a²+b²=c²) The length between the window to the ground is 25 feet, the length between the house to the ladder is 10 feet because of the bush, you have to find how long the ladder has to be. (25²+10²=c²) Next you have to find the square root of these numbers. (25² = 625, 10² = 100) An easy way to find the length of the ladder is just by adding the square roots together. (625+100=725) Then, you have to find the squared number of 725. (√725 = 26.93²) Now that you have the square root of 725, 26.93 feet is how long the ladder has to be to reach the window.
PART IIII
Jordan and Jon are playing a 2 on 1 volleyball game with Jessica. Jessica served the ball to Jordan with a distance of 12 m. Then Jordan passed the ball to Jon with a distance of 5 m. How far does Jon have to hit the ball to Jessica?
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Question
# Find the Laplace transforms of the given functions. f{{left({t}right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}
Laplace transform
Find the Laplace transforms of the given functions.
$$f{{\left({t}\right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}$$
2021-03-12
Step 1
It can be solve using Laplace transformation table
We know that
$${L}{\left\lbrace{e}^{{{a}{t}}}\right\rbrace}=\frac{1}{{{s}-{a}}},{L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}$$
$$f{{\left({t}\right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}$$
$${L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={6}{L}{\left\lbrace{e}^{{-{5}{t}}}\right\rbrace}+{L}{\left\lbrace{e}^{{{3}{t}}}+{5}{L}{\left\lbrace{t}^{3}\right\rbrace}-{9}{L}{\left\lbrace{1}\right\rbrace}\right.}$$
$$=\frac{6}{{{s}+{5}}}+\frac{1}{{{s}-{3}}}+{5}\cdot\frac{6}{{s}^{3}}-{9}{H}{\left({t}\right)}$$
$$=\frac{{{7}{s}-{13}}}{{{\left({s}+{5}\right)}{\left({s}-{3}\right)}}}+\frac{30}{{{s}^{3}}}-{9}{H}{\left({t}\right)}$$ where
$${H}{\left({t}\right)}={\left\lbrace\begin{matrix}{1}&{t}\ge{0}\\{0}&{t}<{0}\end{matrix}\right.}$$
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Polynomial Curve Fitting
Overview
Before reading this page, please check out the Linear Curve Fitting page. Many of the principles mentioned there will be re-used here, and will not be explained in as much detail.
Calculating The Polynomial Curve
We can write an equation for the error as follows:
\begin{align} err & = \sum d_i^2 \nonumber \\ & = (y_1 - f(x_1))^2 + (y_2 - f(x_2))^2 + (y_3 - f(x_3))^2 \nonumber \\ & = \sum_{i = 1}^{n} (y_i - f(x_i))^2 \end{align}
where:
$n$ is the number of points of data (each data point is an $x, y$ pair)
$f(x)$ is the function which describes our polynomial curve of best fit
Since we want to fit a polynomial, we can write $f(x)$ as:
\begin{align} f(x) &= a_0 + a_1 x + a_2 x^2 + ... + a_n x^n \nonumber \\ &= a_0 + \sum_{j=1}^k a_j x^j \end{align}
where:
$k$ is the order of the polynomial
Substituting into above:
\begin{align} err = \sum_{i = 1}^{n} (y_i - (a_0 + \sum_{j=0}^k a_j x^j))^2 \end{align}
How do we find the minimum of this error function? We use the derivative. If we can differentiate $err$, we have an equation for the slope. We know that the slope will be 0 when the error is at a minimum.
We have $k$ unknowns, $a_0, a_1, ..., a_k$. We have to take the derivative of each unknown separately:
\begin{align} \frac{\partial err}{\partial a_0} &= -2 \sum_{i=1}^{n} (y_i - (a_0 + \sum_{j=0}^k a_j x_j)) &= 0 \nonumber \\ \frac{\partial err}{\partial a_1} &= -2 \sum_{i=1}^{n} (y_i - (a_0 + \sum_{j=0}^k a_j x_j))x &= 0 \nonumber \\ \frac{\partial err}{\partial a_1} &= -2 \sum_{i=1}^{n} (y_i - (a_0 + \sum_{j=0}^k a_j x_j))x^2 &= 0 \nonumber \\ \vdots \nonumber \\ \frac{\partial err}{\partial a_k} &= -2 \sum_{i=1}^{n} (y_i - (a_0 + \sum_{j=0}^k a_j x_j))x^k &= 0 \\ \end{align}
These equations can be re-arranged into matrix form:
$\begin{bmatrix} n & \sum x_i & \sum x_i^2 & ... & \sum x_i^k \\ \sum x_i & \sum x_i^2 & \sum x_i^3 & ... & \sum x_i^{k+1} \\ \sum x_i^2 & \sum x_i^3 & \sum x_i^4 & ... & \sum x_i^{k+2} \\ \vdots & \vdots & \vdots & ... & \vdots \\ \sum x_i^k & \sum x_i^{k+1} & \sum x_i^{k+2} & ... & \sum x_i^{k+k} \\ \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_k \end{bmatrix} = \begin{bmatrix} \sum (y_i) \\ \sum (x_i y_i) \\ \sum (x_i^2 y_i) \\ \sum (x_i^k y_i) \end{bmatrix}$
We solve this by re-arranging (which involves taking the inverse of $\bf{x}$)`:
$\mathbf{x} = \mathbf{A^{-1}} \mathbf{B}$
Thus a polynomial curve of best fit is:
$y = x[0] + x[1]x + x[2]x^2 + ... + x[j]x^j$
See main.py for Python code which performs these calculations.
Worked Example
Find a 2 degree polynomial that best describes the following points:
$(1, 1) \\ (2, 3) \\ (3, 4) \\ (4, 8)$
We will then find the values for each one of the nine elements in the $\mathbf{A}$ matrix:
\begin{align} A_{11} &= n = 4 \nonumber \\ A_{12} &= \sum x_i = 1 + 2 + 3 + 4 = 10 \nonumber \\ A_{13} &= \sum x_i^2 = 1^2 + 2^2 + 3^2 + 4^2 = 30 \nonumber \\ A_{21} &= A_{12} = 10 \nonumber \\ A_{22} &= A_{13} = 30 \nonumber \\ A_{23} &= \sum x_i^3 = 1^3 + 2^3 + 3^3 + 4^3 = 100 \nonumber \\ A_{31} &= A_{22} = 30 \nonumber \\ A_{32} &= A_{23} = 100 \nonumber \\ A_{33} &= \sum x_i^4 = 1^4 + 2^4 + 3^4 + 4^4 = 354 \\ \end{align}
And now find the elements of the $\mathbf{B}$ matrix:
\begin{align} B_{11} &= \sum y_i = 1 + 3 + 4 + 8 = 16 \nonumber \\ B_{21} &= \sum x_i y_i = 1*1 + 2*3 + 3*4 + 4*8 = 51 \nonumber \\ B_{31} &= \sum x_i^2 y_i = 1^2*1 + 2^2*3 + 3^2*4 + 4^2*8 = 177 \nonumber \\ \end{align}
Plugging these values into the matrix equation:
$\begin{bmatrix} 4 & 10 & 30 \\ 10 & 30 & 100 \\ 30 & 100 & 354 \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \end{bmatrix} = \begin{bmatrix} 16 \\ 51 \\ 177 \end{bmatrix}$
We can then solve $\mathbf{x} = \mathbf{A^{-1}}\mathbf{B}$ by hand, or use a tool. I used Python’s NumPy package to end up with:
$\begin{bmatrix}a_0 \\ a_1 \\ a_2 \end{bmatrix} = \begin{bmatrix}1 \\ -0.3 \\ 0.5\end{bmatrix}$
Thus our line of best fit:
$y = 1 - 0.3x + 0.5x^2$
The data points and polynomial of best fit are shown in the below graph:
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# Solving linear functions
We will also provide some tips for Solving linear functions quickly and efficiently Our website will give you answers to homework.
## Solve linear functions
Are you struggling with Solving linear functions? In this post, we will show you how to do it step-by-step. Solving for x equations is a common task when you have more than one equation and you want to find the value of x in each equation. This can be done by adding a variable, subtracting one variable, or multiplying or dividing both variables. For example, let's say we have two equations: When we solve for x, we get: This tells us that x equals 2. Similarly, if we have three equations: We can find x by subtracting 3 from both sides of each equation: This tells us that x equals -2. Lastly, let's say we have four equations: We can find x by dividing each equation by 4: This tells us that x equals 0. The solution for an equation is the value of the variable in the equation when solved for all values of the other variables.
A wide range of mathematical problem solving worksheets are available at Teachers Pay Teachers. A good starting place is the Basic Math Problem Solving Worksheets and Practice Problems category, which includes a variety of matrices, word problems, and number relationships. Other categories include Arithmetic and Number Sense, Number Sense and Number Properties, Geometry and Measurement, Statistics and Probability, Algebra and Functions, and Euclidean Geometry. One thing to keep in mind when choosing a math worksheet is whether it’s appropriate for your students’ grade level. All too often teachers use the same worksheet for all their students regardless of grade level. This can work against you if your younger students are not keeping up with the others and falling behind as a result. You may need to create math worksheets that better match the skills and needs of your younger students.
First, you have to use correct capitalization (e.g., the word “the” should be capitalized). Second, you need to use correct punctuation (e.g., an apostrophe to show possession or a question mark or exclamation point to show if something is a statement or a question). Third, you need to spell words and proper names correctly (e.g., “New York” not “New Yrk”). Fourth, you need to use the right number of spaces between words and sentences. Fifth, you need to avoid run-on sentences and grammatical errors. Sixth, you need to avoid using wordy and overused phrases. Finally, you need to write clearly so that your meaning is clear.
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Howto Doing Input Solving equations solver Math homework solutions Instant math help Solve any math problem calculator Solving quadratic inequalities
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# 16.11 Energy in waves: intensity (Page 2/3)
Page 2 / 3
## Determine the combined intensity of two waves: perfect constructive interference
If two identical waves, each having an intensity of $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , interfere perfectly constructively, what is the intensity of the resulting wave?
Strategy
We know from Superposition and Interference that when two identical waves, which have equal amplitudes $X$ , interfere perfectly constructively, the resulting wave has an amplitude of $2X$ . Because a wave’s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves.
Solution
1. Recall that intensity is proportional to amplitude squared.
2. Calculate the new amplitude:
$I\prime \propto {\left(X\prime \right)}^{2}={\left(2X\right)}^{2}={4X}^{2}.$
3. Recall that the intensity of the old amplitude was:
${I}^{}\propto {X}^{2}.$
4. Take the ratio of new intensity to the old intensity. This gives:
$\frac{I\prime }{I}=4.$
5. Calculate to find $I\prime$ :
$I\prime =4I=4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}.$
Discussion
The intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The two individual waves each have intensities of $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , yet their sum has an intensity of $4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , which may appear to violate conservation of energy. This violation, of course, cannot happen. What does happen is intriguing. The area over which the intensity is $4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ is much less than the area covered by the two waves before they interfered. There are other areas where the intensity is zero. The addition of waves is not as simple as our first look in Superposition and Interference suggested. We actually get a pattern of both constructive interference and destructive interference whenever two waves are added. For example, if we have two stereo speakers putting out $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ each, there will be places in the room where the intensity is $4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , other places where the intensity is zero, and others in between. [link] shows what this interference might look like. We will pursue interference patterns elsewhere in this text.
Which measurement of a wave is most important when determining the wave's intensity?
Amplitude, because a wave’s energy is directly proportional to its amplitude squared.
## Section summary
Intensity is defined to be the power per unit area:
$I=\frac{P}{A}$ and has units of ${\text{W/m}}^{2}$ .
## Conceptual questions
Two identical waves undergo pure constructive interference. Is the resultant intensity twice that of the individual waves? Explain your answer.
Circular water waves decrease in amplitude as they move away from where a rock is dropped. Explain why.
## Problems&Exercises
Medical Application
Ultrasound of intensity $1\text{.}\text{50}×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ is produced by the rectangular head of a medical imaging device measuring 3.00 by 5.00 cm. What is its power output?
0.225 W
The low-frequency speaker of a stereo set has a surface area of $0\text{.}\text{05}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ and produces 1W of acoustical power. What is the intensity at the speaker? If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity $0\text{.}1\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ ?
To increase intensity of a wave by a factor of 50, by what factor should the amplitude be increased?
7.07
Engineering Application
A device called an insolation meter is used to measure the intensity of sunlight has an area of 100 cm 2 and registers 6.50 W. What is the intensity in ${\text{W/m}}^{2}$ ?
Astronomy Application
Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of $1.30\phantom{\rule{0.25em}{0ex}}{\text{kW/m}}^{2}.$ How long does it take for $1.8×{10}^{9}\phantom{\rule{0.25em}{0ex}}\text{J}$ to arrive on an area of $1\text{.}00\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ ?
16.0 d
Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high, how much will it produce when they are 0.600 m high?
2.50 kW
Engineering Application
(a) A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. If the average intensity of sunlight on one day is $700\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2},$ what area should your array have to gather energy at the rate of 100 W? (b) What is the maximum cost of the array if it must pay for itself in two years of operation averaging 10.0 hours per day? Assume that it earns money at the rate of 9.00 ¢ per kilowatt-hour.
A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is originally $\text{2.00}×{\text{10}}^{\text{–5}}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2},$ but is turned up until the amplitude increases by 30.0%, what is the new intensity?
$\text{3.38}×{\text{10}}^{\text{–5}}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$
Medical Application
(a) What is the intensity in ${\text{W/m}}^{2}$ of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s? (b) Discuss how this intensity compares to the average intensity of sunlight (about $700\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ ) and the implications that would have if the laser beam entered your eye. Note how your answer depends on the time duration of the exposure.
example ofchange of state of the body in the effectof heat
what is normal force?
the force that pushes upward on us. the force that opposes gravity
clifford
upthrust of air
Newton's 3rd law. the force of the ground (earth) that pushes back on gravity, keeping us on the ground instead of sinking into it.
clifford
I really need lots of questions on frictional force
Shii
I can help answering what I can
Shii
does friction also need some force to perform?
Mohit
no friction is a force just like the gravitational force
clifford
yeah but u can't apply friction anywhere else like other forces
Mohit
I don't understand that question. friction does work alongside other forces based on the situation.
clifford
eg. when walking there are two forces acting on us gravitational and frictional force. friction helps us move forward and gravity keeps us on the ground
clifford
friction is a contact force. Two surfaces are necessary for the force to work.
clifford
hope this helped
clifford
the friction force which oppose while it contact with surrounding. there are two kind of friction. slidding and rolling friction.
Neyaz
What is physics?
physics is a branch of science in which we are dealing with the knowledge of our physical things. macroscopic as well as microscopic. we are going look inside the univers with the help of physics. you can learn nature with the help of physics. so many branches of physics you have to learn physics.
vijay
What are quarks?
6 type of quarks
Neyaz
what is candela
Candela is the unit for the measurement of light intensity.
Osei
any one can prove that 1hrpower= 746 watt
Newton second is the unit of ...............?
Neyaz
Impulse and momentum
Fauzia
force×time and mass× velocity
vijay
Good
Neyaz
What is the simple harmonic motion?
oscillatory motion under a retarding force proportional to the amount of displacement from an equilibrium position
Yuri
Straight out of google, you could do that to, I suppose.
Yuri
*too
Yuri
ok
Fauzia
Oscillatory motion under a regarding force proportional to the amount of displacement from an equilibrium position
Neyaz
examples of work done by load of gravity
What is ehrenfest theorem?
You can look it up, faster and more reliable answer.
Yuri
That isn't a question to ask on a forum and I also have no idea what that is.
Yuri
what is the work done by gravity on the load 87kj,11.684m,mass xkg[g=19m/s
Maureen
What is law of mass action?
rate of chemical reactions is proportional to concentration of reactants ...
ok thanks
Fauzia
what is lenses
lenses are two types
Fauzia
concave and convex
right
Fauzia
speed of light in space
in vacuum speed of light is 3×10^8 m/s
vijay
ok
Vikash
2.99×10^8m/s
Umair
2.8820^8m/s
Muhammed
Vikash
he is correct but we can round up in simple terms
vijay
3×10^8m/s
vijay
is it correct
Fauzia
I mean 3*10^8 m/s ok
vijay
299792458 meter per second
babar
3*10^8m/s
Neyaz
how many Maxwell relations in thermodynamics
vijay
how we can do prove them?
vijay
What is second law of thermodynamics?
Neyaz
please who has a detailed solution to the first two professional application questions under conservation of momentum
I want to know more about pressure
Osei
I can help
Emeh
okay go on
True
I mean on pressure
Emeh
definition of Pressure
John
it is the force per unit area of a substance.S.I unit is Pascal 1pascal is defined as 1N acting on 1m² area i.e 1pa=1N/m²
Emeh
pls explain Doppler effect
Emmex
solve this an inverted differential manometer containing oil specific gravity 0.9 and manometer reading is 400mm find the difference of pressure
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# The most wished statistical for a mother
Tabla de contenidos
# Dispersion measures
Those of you who’re reading and who belong to the pediatricians’ gang already know what I am talking about: the 50th percentile. There’s no mother who doesn’t want her offspring to be above it in weight, height, intelligence and everything else that a good mother could desired for her child. That’s why pediatricians, who dedicate our lives to children care, love percentiles so much. But what is the meaning of the term percentile?. Let’s start from the beginning…
## Dispersion measures
If we have the distribution of values of a variable we can summarize it with a central and a dispersion measure. The most common are the mean and the standard deviation, respectively, but sometimes we use other measures of central tendency (such us the median or the mode) and of dispersion.
## Interquartile range
The simplest of these other measures of dispersion is called range, which is defined as the difference between the maximum and minimum values of the distribution.
Let’s suppose that we collect the birth weights of the last 100 children born at our hospital and we order them as they appear in the table. The lowest value was 2200 grams, while the prize for the biggest goes to an infant who weighed 4000 grams.
The range in this case is 1800 grams but, of course, if we do not have the table and someone tell us just this, we couldn’t have much idea about how our babies are in size. This is why it’s usually preferred to give the range with explicit minimum and maximum values. In our case it would be from 2200 to 4000 grams.
If you remember how to calculate the median, you will see that it values 3050 grams. To complete the picture we need a measure that tells us how the rest of the weights are distributed around the median and within the range.
The easiest way is to divide the distribution in four equal segments including 25% of children each one. These segments are called quartiles, and there’re three of them: the first quartile (at 25% above the minimum), the second quartile (which is the same as the median) and the third quartile (at 75%, between the median and the maximum).
We come up with four segments: from the minimum to the first quartile, from the first to the second (median), from second to third and from third to the maximum. In our case, the three quartiles would be 2830, 3050 and 3200 grams. Some people call these the lower quartile, the median and the upper quartile, but they are the same thing.
Now if we know that the median is 3050 grams and 50% of children weight between 2830 and 3200 grams, we’ll have a pretty good idea about the birth weights of our newborns. This interval is called the interquartile range and it’s usually provided along with the median to summarize the distribution. In our example: a median of 3050 grams with an interquartile range from 2830-3200 grams.
## Deciles and percentiles
But we can go much further. We can divide the distribution in the number of segments we want. The deciles are the result of dividing it in ten segments and our revered percentiles the result of dividing it in a hundred.
There is a fairly simple formula to calculate any percentile we want. For example, the Pth percentile would be at position (P/100)x(n+1), where n represents the sample size. In our distribution of neonates, the 22nd percentile would be (22/100)x(100+1) = 22.2, i.e. 2770 grams.
The sharpest of you may have noticed that our 3050 grams correspond not only to the median, but also to the fifth decile and to the 50th percentile, the desired one.
The great use of percentile, apart from to give satisfaction to 50% of mothers (those who have their children above the median), is to allow us to estimate the probability of a certain value of the variable within the population.
In general, the closer you are to the median the better it be (at least in medicine), and the further away from it the more likely that someone take you to a doctor to find out why you are not closer to the precious percentile or, even, something above it.
## We’re leaving…
But if we really want to further refine the calculation of the probability to obtain a particular value within a data distribution, there’re other techniques related with the standardization of the dispersion measure we use. But that’s another story…
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# Chapter 6 - Systems of Equations and Inequalities - 6-4 Application of Linear Systems - Practice and Problem-Solving Exercises: 16
For this problem, we will chose elimination as our method to solve the problem because we can easily eliminate the y terms in both the equations. Solution: ($\frac{25}{6}$, $-\frac{23}{12}$)
#### Work Step by Step
Our two systems are: 1. 5x+2y=17 2. x-2y=8 First lets add both equation together in order to cancel the y terms. The reason why the y-terms cancel out when you add the equations is because -2 is the additive inverse of 2. We get the equation: 6x=25 (Divide by 6) x=(25/6) Now lets use that in our second equation to find the value of y: (25/6)-2y=8 (subtract 25/6 on both sides) -2y=23/6 (divide by -2 on both sides) y=-23/12 So our ordered pairs are: ($\frac{25}{6}$, $-\frac{23}{12}$)
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Estimate the sum to the nearest whole number
#### greg1313
Estimate the sum to the nearest whole number without the use of a calculator/computer.
$$\displaystyle \displaystyle\sum_{n = 625}^{9999} \sqrt[4]{n + 0.5}$$
1 person
#### romsek
MHF Helper
amazing
eagerly awaiting the simple explanation
#### Debsta
MHF Helper
Estimate the sum to the nearest whole number without the use of a calculator/computer.
$$\displaystyle \displaystyle\sum_{n = 625}^{9999} \sqrt[4]{n + 0.5}$$
First thoughts. Here's one estimate:
If $$\displaystyle n=5$$ then $$\displaystyle \sqrt[4]{n+0.5} \approx \sqrt[4]{625} = 5$$
If $$\displaystyle n=9999$$ then $$\displaystyle \sqrt[4]{n+0.5} \approx \sqrt[4]{10000} = 10$$
So the first number in the sum is approx 5 and the last number is approx 10. As n increases, the value of $$\displaystyle \sqrt[4]{n+0.5}$$ also increases.
There are 9999 - 625 +1 = 9375 terms in the sum.
Let's assume it is an AP (it isn't, but this will give us an approximation of the sum since the range of values 5 to 10 is quite small and there are a lot of terms).
So, using the sum of an AP formula, Sum $$\displaystyle \approx \frac{9375}{2}(5+10) \approx70313$$
(This gives the same result as if we say each term is approx 7.5 (ie halfway between 5 and 10)).
So an estimate is 70313.
Of course there would be other ways and this probably isn't the best!
Last edited:
1 person
#### Debsta
MHF Helper
A (much) better estimate:
$$\displaystyle \int_{625}^{9999} (n+0.5)^\frac{1}{4} dn$$
$$\displaystyle \approx \frac{4}{5} (10000^ \frac{5}{4} - 625^\frac{5}{4})$$
$$\displaystyle = \frac{4}{5} (10^5 - 5^5)$$
$$\displaystyle = 77 500$$
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3 people
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# Vectors and scalars
## 1.3.1 Distinguish between vector and scalar quantities, and give examples of each.
When expressing a quantity we give it a number and a unit (for example, 12 kg), this expresses the magnitude of the quantity. Some quantities also have direction, a quantity that has both a magnitude and direction is called a vector. On the other hand, a quantity that has only a magnitude is called a scalar quantity. Vectors are represented in print as bold and italicised characters (for example F). Below is a table listing some vector and scalar quantities:
Scalars Vectors Speed Velocity Temperature Acceleration Distance Displacement Area Force Entropy Momentum Volume Drag
Table 1.3.1 - Vector and scalar quantites
Note that some quantities appear to be the same, such as velocity and speed, both representing distance over time, the difference is that velocity has a direction whilst speed does not.
## 1.3.2 Determine the sum or difference of two vectors by a graphical method.
The difference of two vectors
When adding vectors, we need to take both the magnitude and direction into account. Often, we will have situations where two vectors have opposite directions, in this case, we simply subtract the smallest magnitude from the largest one. This is demonstrated in figure 1.3.1 below:
Figure 1.3.1 - Resultant force of two opposing vectors
The sum of two vectors
Sometimes we will have situations where two forces are acting in the same direction. In the situations we simply add together the magnitudes of both vectors. This is demonstrated in figure 1.3.2 below:
Figure 1.3.2 - Resultant force of two concurrent vectors
Adjacent vectors
In certain situations, we will need to work out the angle between two adjacent vectors. In order to do this graphically we draw a scale diagram with the tail of one vector at the head of the other, we then draw a line connecting the other head and tail. To get the magnitude of the new vector, we simply measure it. This is demonstrated in the diagram below:
Figure 1.3.3 - Graphical method of solving adjacent vectors
Alternatively, we can use trigonometry for a faster and more accurate result. This is demonstrated in figure 1.3.4 below:
Figure 1.3.4 - Trigonometric method of solving adjacent vectors
Scalar multiplication
We can also multiply (and divide) vectors by scalars. When doing so we follow a set of rules:
• Multiplying by 1 does not change a vector 1 = v
• Multiplying by 0 gives the null vector 0 = 0
• Multiplying by -1 gives the additive inverse -1 = -v
• Left distributivity: (c + d)v = cv + dv
• Right distributivity: c(v + w) = cv + cw
• Associativity: (cd)v = c(dv)
Scalar multiplication is demonstrated in figure 1.3.5 below:
Figure 1.3.5 - Scalar multiplication and division of vectors
## 1.3.3 Resolve vectors into perpendicular components along chosen axes.
When working with adjacent vectors that do not form a 90° angle, it is often useful to brake certain vectors into component vectors so that they are concurrent with the other vectors. To do this, we draw two vectors, one horizontal and the other vertical to our plane of reference. We then use trigonometry to work out the magnitude of each new vector and figure out the resulting force. This is shown in figure 1.3.6 and 1.3.7:
Figure 1.3.6 shows a diagram of the forces acting on a block being pushed along a smooth surface:
Figure 1.3.6 - Forces acting on a block
Figure 1.3.7 shows the same diagram but with the surface and pushing forces broken down into their components:
Figure 1.3.7 - Forces acting on a block broken into their components
Sometimes the plane of reference will not be parallel to the page, such and example is shown in figure 1.3.8 below:
Figure 1.3.8 - Component forces of a block on a slope
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Explore the World of Geometry: A Comprehensive List of Shapes
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List of Shapes
Contents
Understanding Shapes
Definition of Shapes
Shapes are the external boundaries or outlines of objects. They can be two-dimensional (2D) or three-dimensional (3D). Basic 2D shapes include squares, rectangles, circles, triangles, and polygons. Basic 3D shapes include cubes, spheres, cylinders, cones, and pyramids.
Importance of Learning Shapes
Learning shapes is essential for several reasons. First, it helps with spatial awareness and the ability to visualize objects in different orientations. Second, it is important for understanding geometry and mathematical concepts. Third, it helps with communication and vocabulary, as shapes are used in everyday language (e.g. “square meal”, “round of applause”, “triangle offense”).
Here are some examples of shapes and their meanings:
Shape Meaning
Square A shape with four straight sides of equal length and four right angles
Rectangle A shape with four straight sides and four right angles, where the opposite sides are of equal length
Circle A round shape with no corners or edges
Triangle A shape with three straight sides and three angles
Cube A three-dimensional shape with six square faces
Sphere A round three-dimensional shape
List of Shapes: Basic ones
In this section, we will cover the most basic shapes that you need to know. These shapes are essential for everyday conversations and writing. Let’s dive into the four basic shapes: Circle, Square, Rectangle, and Triangle.
Circle
A circle is a round shape that has no corners or edges. It is a two-dimensional shape that is defined by its radius, which is the distance from the center of the circle to its edge. Here are some examples of words associated with circles:
Words Meanings
Round Shaped like a circle
Sphere A three-dimensional circle
Wheel A circular object that rotates around an axis
Example sentences:
• The moon is a perfect circle in the sky.
• The car’s wheels are circular.
• The basketball is round and bounces well.
Square
A square is a four-sided shape with four right angles and equal sides. It is a two-dimensional shape that is often used in geometry. Here are some examples of words associated with squares:
Words Meanings
Cube A three-dimensional square
Perpendicular At a 90-degree angle
Example sentences:
• The tile on the floor is a square.
• The Rubik’s cube is made up of many small squares.
• The corners of the picture frame are perpendicular.
Rectangle
A rectangle is a four-sided shape with four right angles and opposite sides that are equal in length. It is a two-dimensional shape that is often used in architecture and design. Here are some examples of words associated with rectangles:
Words Meanings
Parallelogram A four-sided shape with parallel sides
Oblong A rectangle that is longer than it is wide
Frame A rectangular border or outline
Example sentences:
• The window is a rectangle.
• The book cover is an oblong rectangle.
• The painting is in a rectangular frame.
Triangle
A triangle is a three-sided shape with three angles. It is a two-dimensional shape that is often used in mathematics. Here are some examples of words associated with triangles:
Words Meanings
Equilateral A triangle with three equal sides
Isosceles A triangle with two equal sides
Scalene A triangle with no equal sides
Example sentences:
• The roof of the house is a triangle.
• The yield sign is a red equilateral triangle.
• The mountain in the distance has a scalene shape.
3D Shapes
In geometry, 3D shapes are objects that have three dimensions: length, width, and height. They are also known as solid shapes and are often used in various fields, including architecture, engineering, and art. In this section, we will explore some of the most common 3D shapes, including the sphere, cube, cylinder, and pyramid.
Sphere
A sphere is a 3D shape that is perfectly round, with no edges or corners. It is often used to represent objects such as balls, planets, and bubbles. The formula for finding the volume of a sphere is V = (4/3)πr³, where r is the radius of the sphere. Some examples of sentences using the word sphere are:
• The Earth is a sphere that rotates on its axis.
• The crystal ball was a perfect sphere.
Cube
A cube is a 3D shape that has six square faces, all of which are equal in size. It is often used to represent objects such as dice, boxes, and buildings. The formula for finding the volume of a cube is V = s³, where s is the length of one side of the cube. Some examples of sentences using the word cube are:
• The Rubik’s cube is a popular puzzle game.
• The package was a small cube wrapped in brown paper.
Cylinder
A cylinder is a 3D shape that has two circular faces and a curved surface. It is often used to represent objects such as cans, pipes, and towers. The formula for finding the volume of a cylinder is V = πr²h, where r is the radius of the circular face and h is the height of the cylinder. Some examples of sentences using the word cylinder are:
• The engine has six cylinders that provide power to the car.
• The water bottle was a tall cylinder made of plastic.
Pyramid
A pyramid is a 3D shape that has a polygonal base and triangular faces that converge at a single point called the apex. It is often used to represent objects such as pyramids, mountains, and buildings. The formula for finding the volume of a pyramid is V = (1/3)Bh, where B is the area of the base and h is the height of the pyramid. Some examples of sentences using the word pyramid are:
• The Great Pyramid of Giza is one of the Seven Wonders of the World.
• The food pyramid is a guide to healthy eating.
Shape Description
Sphere A perfectly round 3D shape with no edges or corners.
Cube A 3D shape with six square faces, all of which are equal in size.
Cylinder A 3D shape with two circular faces and a curved surface.
Pyramid A 3D shape with a polygonal base and triangular faces that converge at a single point.
Complex Shapes
Complex shapes are geometric figures that have more than four sides. They are an essential part of geometry and can be found in various objects around us. In this section, we will discuss three complex shapes: Pentagon, Hexagon, and Octagon.
Pentagon
A Pentagon is a five-sided polygon with five angles. It is a common shape found in many architectural designs, including buildings, bridges, and monuments. Here are some words related to the Pentagon:
Word Meaning
Pentagon A five-sided polygon
Angle The space between two intersecting lines
Diagonal A line that connects two non-adjacent vertices
Perimeter The distance around the outside of a shape
Apothem The distance from the center of a polygon to its midpoint
Example sentence: The Pentagon building in Washington D.C. is a famous landmark that has five sides.
Hexagon
A Hexagon is a six-sided polygon with six angles. It is a common shape found in nature, including honeycombs, snowflakes, and crystals. Here are some words related to the Hexagon:
Word Meaning
Hexagon A six-sided polygon
Vertex A corner point of a shape
Side A line segment that connects two vertices
Regular A shape with all sides and angles equal
Irregular A shape with sides and angles that are not equal
Example sentence: The stop sign has a Hexagon shape with six sides and six angles.
Octagon
An Octagon is an eight-sided polygon with eight angles. It is a common shape found in many objects, including stop signs, buildings, and swimming pools. Here are some words related to the Octagon:
Word Meaning
Octagon An eight-sided polygon
Parallel Two lines that are always the same distance apart
Perpendicular Two lines that intersect at a right angle
Symmetry A shape that can be divided into two equal parts
Area The amount of space inside a shape
Example sentence: The patio has an Octagon-shaped swimming pool with eight sides and eight angles.
In conclusion, complex shapes are an important part of geometry and can be found in many objects around us. By understanding the properties of these shapes, we can better appreciate the world around us.
Using Shapes in English Vocabulary
Describing Objects
Shapes are an essential part of our daily lives, and we use them to describe objects around us. Knowing the names of different shapes can help us convey what we see more accurately. For example, instead of saying “that thing is round,” we can say “that thing is a circle.” Here are some examples of how you can use shapes to describe objects:
• The table has a rectangular top.
• The clock face is circular.
• The windows are square.
• The vase has an oval shape.
• The TV screen is a rectangle.
Here is a list of shape adjectives with their meanings:
Circular having a round shape like a circle
Triangular having three sides and three angles
Rectangular having four sides and four right angles
Square having four equal sides and four right angles
Oval having a rounded and elongated shape
Diamond-shaped having a shape like a diamond
Spherical having a round shape like a sphere
Cone-shaped having a shape like a cone
Cubic having a shape like a cube
Cylindrical having a shape like a cylinder
Pyramidal having a shape like a pyramid
Elliptical having a shape like an ellipse
Examples of Shape Adjectives in Sentences
• The cake is circular.
• The roof is triangular.
• The bookshelf is rectangular.
• The window is square.
• The mirror is oval-shaped.
• The earrings are diamond-shaped.
• The ice cream cone is cone-shaped.
• The dice are cubic.
• The can is cylindrical.
• The pyramid is pyramidal.
• The pool is elliptical.
Shapes in Writing
Writing is a powerful tool that allows us to express ourselves in many ways. One of the ways we can make our writing more interesting is by using shapes to describe characters and settings. In this section, we’ll explore the symbolism of shapes, how to use shapes to describe characters, and how to use shapes to describe settings.
Symbolism of Shapes
Shapes have different meanings and symbolism associated with them. For example, a circle can represent unity, wholeness, and completion, while a triangle can represent stability, strength, and power. Here are some other common shapes and their meanings:
Shape Meaning
Square Stability, balance, order
Rectangle Practicality, efficiency, structure
Triangle Strength, power, direction
Diamond Wealth, prosperity, abundance
Oval Femininity, elegance, grace
Using Shapes to Describe Characters
Shapes can be used to describe characters in many ways. For example, a character with a round face may be seen as friendly and approachable, while a character with a square jaw may be seen as strong and determined. Here are some other ways shapes can be used to describe characters:
• A character with a triangular nose may be seen as sharp and cunning.
• A character with a diamond-shaped face may be seen as wealthy and powerful.
• A character with an oval-shaped face may be seen as elegant and graceful.
Using Shapes to Describe Settings
Shapes can also be used to describe settings in many ways. For example, a room with circular furniture may be seen as cozy and inviting, while a room with angular furniture may be seen as modern and sleek. Here are some other ways shapes can be used to describe settings:
• A building with a triangular roof may be seen as strong and stable.
• A garden with rectangular flower beds may be seen as practical and efficient.
• A park with curvy paths may be seen as playful and fun.
Conclusion
In conclusion, we have covered a wide range of shapes and their vocabulary in this article.
Below is a table with a list of shapes and their meanings:
Shape Meaning
Circle A round shape with no corners or edges
Oval A stretched-out circle shape
Semi-circle Half of a circle
Ellipse A stretched-out circle shape with two different diameters
Triangle A shape with three sides and three angles
Square A shape with four equal sides and four right angles
Rectangle A shape with four sides and four right angles, where opposite sides are equal
Pentagon A shape with five sides and five angles
Hexagon A shape with six sides and six angles
Heptagon A shape with seven sides and seven angles
Octagon A shape with eight sides and eight angles
Nonagon A shape with nine sides and nine angles
Decagon A shape with ten sides and ten angles
Right-angled triangle A triangle with one right angle
Equilateral triangle A triangle with three equal sides
Isosceles triangle A triangle with two equal sides
Scalene triangle A triangle with no equal sides
Parallelogram A shape with two pairs of parallel sides
Rhombus A shape with four equal sides and opposite angles equal
Trapezoid A shape with one pair of parallel sides
Learning the vocabulary for shapes can help you in many ways, such as:
• Describing objects and buildings
• Drawing and sketching
• Solving geometry problems
• Understanding and interpreting maps and diagrams
In conclusion, learning the vocabulary for shapes is an important step in improving your English skills. By using the examples and tables provided in this article, you can expand your knowledge and become more confident in your writing and communication.
What are some common shapes used in English vocabulary?
In English vocabulary, some common shapes include circle, square, triangle, rectangle, oval, sphere, cube, and cylinder.
Can you name five different shapes?
Yes, here are five different shapes: pentagon, hexagon, octagon, rhombus, and trapezoid.
What are the ten types of shapes?
The ten types of shapes are circle, oval, semi-circle, ellipse, triangle, square, rectangle, pentagon, hexagon, and heptagon.
How many basic shapes are there in English?
There are three basic shapes in English: the square, the circle, and the triangle.
Which shapes start with the letter ‘S’ or ‘T’ or ‘C’ or ‘D’ or ‘I’ or ‘Z’?
• S: sphere, square, star
• T: trapezoid, triangle, torus
• C: circle, cube, cone
• D: diamond, decagon, dodecahedron
• I: isosceles triangle, icosahedron, infinity symbol
• Z: zigzag, ziggurat
In English vocabulary, some common shapes include circle, square, triangle, rectangle, oval, sphere, cube, and cylinder.
Yes, here are five different shapes: pentagon, hexagon, octagon, rhombus, and trapezoid.
The ten types of shapes are circle, oval, semi-circle, ellipse, triangle, square, rectangle, pentagon, hexagon, and heptagon.
There are three basic shapes in English: the square, the circle, and the triangle.
Some learning shapes include cone, pyramid, prism, and torus.
\n
\n
• S: sphere, square, star
• \n
• T: trapezoid, triangle, torus
• \n
• C: circle, cube, cone
• \n
• D: diamond, decagon, dodecahedron
• \n
• I: isosceles triangle, icosahedron, infinity symbol
• \n
• Z: zigzag, ziggurat, zero
• \n
\n
In conclusion, understanding the vocabulary related to shapes is important for effective communication in English. By familiarizing oneself with the different shapes and their names, one can communicate ideas more clearly and accurately.
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# 15.1: Volume and Average Height
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Consider a surface $$f(x,y)$$; you might temporarily think of this as representing physical topography---a hilly landscape, perhaps. What is the average height of the surface (or average altitude of the landscape) over some region?
As with most such problems, we start by thinking about how we might approximate the answer. Suppose the region is a rectangle, $$[a,b]\times[c,d]$$. We can divide the rectangle into a grid, $$m$$ subdivisions in one direction and $$n$$ in the other, as indicated in figure 15.1.1. We pick $$x$$ values $$x_0$$, $$x_1$$,..., $$x_{m-1}$$ in each subdivision in the $$x$$ direction, and similarly in the $$y$$ direction. At each of the points $$(x_i,y_j)$$ in one of the smaller rectangles in the grid, we compute the height of the surface: $$f(x_i,y_j)$$. Now the average of these heights should be (depending on the fineness of the grid) close to the average height of the surface:
$${f(x_0,y_0)+f(x_1,y_0)+\cdots+f(x_0,y_1)+f(x_1,y_1)+\cdots+ f(x_{m-1},y_{n-1})\over mn}.$$
As both $$m$$ and $$n$$ go to infinity, we expect this approximation to converge to a fixed value, the actual average height of the surface. For reasonably nice functions this does indeed happen.
Figure 15.1.1. A rectangular subdivision of $$[a,b]x[c,d]$$.
Using sigma notation, we can rewrite the approximation:
\eqalign{ {1\over mn}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i) &={1\over(b-a)(d-c)}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i) {b-a\over m}{d-c\over n}\cr &= {1\over(b-a)(d-c)}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y.\cr }
The two parts of this product have useful meaning: $$(b-a)(d-c)$$ is of course the area of the rectangle, and the double sum adds up $$mn$$ terms of the form $$f(x_j,y_i)\Delta x\Delta y$$, which is the height of the surface at a point times the area of one of the small rectangles into which we have divided the large rectangle. In short, each term $$f(x_j,y_i)\Delta x\Delta y$$ is the volume of a tall, thin, rectangular box, and is approximately the volume under the surface and above one of the small rectangles; see figure 15.1.2. When we add all of these up, we get an approximation to the volume under the surface and above the rectangle $$R=[a,b]\times[c,d]$$. When we take the limit as $$m$$ and $$n$$ go to infinity, the double sum becomes the actual volume under the surface, which we divide by $$(b-a)(d-c)$$ to get the average height.
Figure 15.1.2. Approximating the volume under a surface.
Double sums like this come up in many applications, so in a way it is the most important part of this example; dividing by $$(b-a)(d-c)$$ is a simple extra step that allows the computation of an average. As we did in the single variable case, we introduce a special notation for the limit of such a double sum:
$$\lim_{m,n\to\infty} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y=\iint\limits_{R} f(x,y)\,dx\,dy=\iint\limits_{R} f(x,y)\,dA,$$
the double integral of $$f$$ over the region $$R$$. The notation $$dA$$ indicates a small bit of area, without specifying any particular order for the variables $$x$$ and $$y$$; it is shorter and more "generic'' than writing $$dx\,dy$$. The average height of the surface in this notation is
$${1\over(b-a)(d-c)}\iint\limits_{R} f(x,y)\,dA.$$
The next question, of course, is: How do we compute these double integrals? You might think that we will need some two-dimensional version of the Fundamental Theorem of Calculus, but as it turns out we can get away with just the single variable version, applied twice.
Going back to the double sum, we can rewrite it to emphasize a particular order in which we want to add the terms:
$$\sum_{i=0}^{n-1}\left(\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\right)\Delta y.$$
In the sum in parentheses, only the value of $$x_j$$ is changing; $$y_i$$ is temporarily constant. As $$m$$ goes to infinity, this sum has the right form to turn into an integral:
$$\lim_{m\to\infty}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x = \int_a^b f(x,y_i)\,dx.$$
So after we take the limit as $$m$$ goes to infinity, the sum is
$$\sum_{i=0}^{n-1}\left(\int_a^b f(x,y_i)\,dx\right)\Delta y.$$
Of course, for different values of $$y_i$$ this integral has different values; in other words, it is really a function applied to $$y_i$$:
$$G(y)=\int_a^b f(x,y)\,dx.$$
If we substitute back into the sum we get
$$\sum_{i=0}^{n-1} G(y_i)\Delta y.$$
This sum has a nice interpretation. The value $$G(y_i)$$ is the area of a cross section of the region under the surface $$f(x,y)$$, namely, when $$y=y_i$$. The quantity $$G(y_i)\Delta y$$ can be interpreted as the volume of a solid with face area $$G(y_i)$$ and thickness $$\Delta y$$. Think of the surface $$f(x,y)$$ as the top of a loaf of sliced bread. Each slice has a cross-sectional area and a thickness; $$G(y_i)\Delta y$$ corresponds to the volume of a single slice of bread. Adding these up approximates the total volume of the loaf. (This is very similar to the technique we used to compute volumes in section 9.3, except that there we need the cross-sections to be in some way "the same''.) Figure 15.1.3 shows this "sliced loaf'' approximation using the same surface as shown in figure 15.1.2.
Nicely enough, this sum looks just like the sort of sum that turns into an integral, namely,
\eqalign{ \lim_{n\to\infty}\sum_{i=0}^{n-1} G(y_i)\Delta y&=\int_c^d G(y)\,dy\cr &=\int_c^d \int_a^b f(x,y)\,dx\,dy.\cr }
Let's be clear about what this means: we first will compute the inner integral, temporarily treating $$y$$ as a constant. We will do this by finding an anti-derivative with respect to $$x$$, then substituting $$x=a$$ and $$x=b$$ and subtracting, as usual. The result will be an expression with no $$x$$ variable but some occurrences of $$y$$. Then the outer integral will be an ordinary one-variable problem, with $$y$$ as the variable.
Example 15.1.1:
Figure 15.1.2 shows the function $$\sin(xy)+6/5$$ on $$[0.5,3.5]\times[0.5,2.5]$$. The volume under this surface is
$$\int_{0.5}^{2.5}\int_{0.5}^{3.5} \sin(xy)+{6\over5}\,dx\,dy.$$
The inner integral is
$$\int_{0.5}^{3.5} \sin(xy)+{6\over5}\,dx= \left.{-\cos(xy)\over y}+{6x\over5}\right|_{0.5}^{3.5}= {-\cos(3.5y)\over y}+{\cos(0.5y)\over y}+{18\over5}.$$
Unfortunately, this gives a function for which we can't find a simple anti-derivative. To complete the problem we could use Sage or similar software to approximate the integral. Doing this gives a volume of approximately $$8.84$$, so the average height is approximately $$8.84/6\approx 1.47$$.
Figure 15.1.3. Approximating the volume under a surface with slices (AP)
Because addition and multiplication are commutative and associative, we can rewrite the original double sum:
$$\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y=\sum_{j=0}^{m-1}\sum_{i=0}^{n-1}f(x_j,y_i)\Delta y\Delta x.$$
Now if we repeat the development above, the inner sum turns into an integral:
$$\lim_{n\to\infty}\sum_{i=0}^{n-1}f(x_j,y_i)\Delta y = \int_c^d f(x_j,y)\,dy,$$
and then the outer sum turns into an integral:
$$\lim_{m\to\infty}\sum_{j=0}^{m-1}\left(\int_c^d f(x_j,y)\,dy \right)\Delta x = \int_a^b\int_c^d f(x,y)\,dy\,dx.$$
In other words, we can compute the integrals in either order, first with respect to $$x$$ then $$y$$, or vice versa. Thinking of the loaf of bread, this corresponds to slicing the loaf in a direction perpendicular to the first.
We haven't really proved that the value of a double integral is equal to the value of the corresponding two single integrals in either order of integration, but provided the function is reasonably nice, this is true; the result is called Fubini's Theorem.
Example 15.1.2:
We compute $$\iint\limits_{R} 1+(x-1)^2+4y^2\,dA$$, where $$R=[0,3]\times[0,2]$$, in two ways.
First,
\eqalign{ \int_0^3\int_0^2 1+(x-1)^2+4y^2\,dy\,dx&= \int_0^3\left. y+(x-1)^2y+{4\over 3}y^3\right|_0^2\,dx\cr &=\int_0^3 2+2(x-1)^2+{32\over 3}\,dx\cr &=\left. 2x + {2\over 3}(x-1)^3+{32\over 3}x\right|_0^3\cr &=6+{2\over 3}\cdot 8 + {32\over 3}\cdot3-(0-1\cdot{2\over3}+0)\cr &=44.\cr }
In the other order:
\eqalign{ \int_0^2\int_0^3 1+(x-1)^2+4y^2\,dx\,dy&= \int_0^2\left. x+{(x-1)^3\over3}+4y^2x\right|_0^3\,dy\cr &=\int_0^2 3+{8\over3}+12y^2+{1\over3}\,dy\cr &=\left. 3y+{8\over3}y+4y^3+{1\over3}y\right|_0^2\cr &=6+{16\over3}+32+{2\over3}\cr &=44.\cr }
In this example there is no particular reason to favor one direction over the other; in some cases, one direction might be much easier than the other, so it's usually worth considering the two different possibilities.
Frequently we will be interested in a region that is not simply a rectangle. Let's compute the volume under the surface $$x+2y^2$$ above the region described by $$0\le x\le1$$ and $$0\le y\le x^2$$, shown in figure 15.1.4.
Figure 15.1.4. A parabolic region of integration
In principle there is nothing more difficult about this problem. If we imagine the three-dimensional region under the surface and above the parabolic region as an oddly shaped loaf of bread, we can still slice it up, approximate the volume of each slice, and add these volumes up. For example, if we slice perpendicular to the $$x$$ axis at $$x_i$$, the thickness of a slice will be $$\Delta x$$ and the area of the slice will be
$$\int_0^{x_i^2} x_i+2y^2\,dy.$$
When we add these up and take the limit as $$\Delta x$$ goes to 0, we get the double integral
\eqalign{ \int_0^1 \int_0^{x^2} x+2y^2\,dy\,dx&= \int_0^1 \left.xy+{2\over3}y^3\right|_0^{x^2}\,dx\cr &=\int_0^1 x^3+{2\over3}x^6\,dx\cr &=\left. {x^4\over4}+{2\over21}x^7\right|_0^1\cr &={1\over4}+{2\over21}={29\over84}.\cr }
We could just as well slice the solid perpendicular to the $$y$$ axis, in which case we get
\eqalign{ \int_0^1 \int_{\sqrt y}^1 x+2y^2\,dx\,dy&= \int_0^1 \left.{x^2\over2}+2y^2x\right|_{\sqrt y}^1 \,dy\cr &=\int_0^1 {1\over2}+2y^2-{y\over2}-2y^2\sqrt y\,dy\cr &=\left. {y\over2}+{2\over3}y^3-{y^2\over4}-{4\over7}y^{7/2}\right|_0^1\cr &={1\over2}+{2\over3}-{1\over4}-{4\over7}={29\over84}.\cr }
What is the average height of the surface over this region? As before, it is the volume divided by the area of the base, but now we need to use integration to compute the area of the base, since it is not a simple rectangle. The area is
$$\int_0^1 x^2\,dx={1\over3},$$
so the average height is $$29/28$$.
Example 15.1.3:
Find the volume under the surface $$z=\sqrt{1-x^2}$$ and above the triangle formed by $$y=x$$, $$x=1$$, and the $$x$$-axis.
Let's consider the two possible ways to set this up:
$$\int_0^1 \int_0^x \sqrt{1-x^2}\,dy\,dx \qquad\hbox{or}\qquad \int_0^1 \int_y^1 \sqrt{1-x^2}\,dx\,dy.$$
Which appears easier? In the first, the first (inner) integral is easy, because we need an anti-derivative with respect to $$y$$, and the entire integrand $$\sqrt{1-x^2}$$ is constant with respect to $$y$$. Of course, the second integral may be more difficult. In the second, the first integral is mildly unpleasant---a trig substitution. So let's try the first one, since the first step is easy, and see where that leaves us.
$$\int_0^1 \int_0^x \sqrt{1-x^2}\,dy\,dx= \int_0^1 \left. y\sqrt{1-x^2}\right|_0^x\,dx= \int_0^1 x\sqrt{1-x^2}\,dx.$$
This is quite easy, since the substitution $$u=1-x^2$$ works:
$$\int x\sqrt{1-x^2}\,dx=-{1\over 2}\int \sqrt u\,du ={1\over3}u^{3/2}=-{1\over3}(1-x^2)^{3/2}.$$
Then
$$\int_0^1 x\sqrt{1-x^2}\,dx=\left. -{1\over3}(1-x^2)^{3/2}\right|_0^1 ={1\over3}.$$
This is a good example of how the order of integration can affect the complexity of the problem. In this case it is possible to do the other order, but it is a bit messier. In some cases one order may lead to a very difficult or impossible integral; it's usually worth considering both possibilities before going very far.
### Contributors
• Integrated by Justin Marshall.
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# Compute <mrow> 4 t </mrow> <mrow> 5 <mroot> <mrow
Compute $\int \frac{4t}{5\sqrt[3]{2t+3}}dt$.
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Since $\frac{4}{5}$ is constant with respect to $t$, move $\frac{4}{5}$ out of the integral.
$\frac{4}{5}\int \frac{t}{\sqrt[3]{2t+3}}dt$
Let $u=2t+3$. Then $du=2dt$, so $\frac{1}{2}du=dt$. Rewrite using $u$ and $d$$u$.
$\frac{4}{5}\int \frac{\frac{u}{2}-\frac{3}{2}}{\sqrt[3]{u}}\cdot \frac{1}{2}du$
Simplify.
$\frac{4}{5}\int \frac{u-3}{4\sqrt[3]{u}}du$
Since $\frac{1}{4}$ is constant with respect to $u$, move $\frac{1}{4}$ out of the integral.
$\frac{4}{5}\left(\frac{1}{4}\int \frac{u-3}{\sqrt[3]{u}}du\right)$
Simplify the expression.
$\frac{1}{5}\int \left(u-3\right){u}^{-\frac{1}{3}}du$
Expand $\left(u-3\right){u}^{-\frac{1}{3}}$.
$\frac{1}{5}\int {u}^{\frac{2}{3}}-3{u}^{-\frac{1}{3}}du$
Split the single integral into multiple integrals.
$\frac{1}{5}\left(\int {u}^{\frac{2}{3}}du+\int -3{u}^{-\frac{1}{3}}du\right)$
By the Power Rule, the integral of ${u}^{\frac{2}{3}}$ with respect to $u$ is $\frac{3}{5}{u}^{\frac{5}{3}}$.
$\frac{1}{5}\left(\frac{3}{5}{u}^{\frac{5}{3}}+C+\int -3{u}^{-\frac{1}{3}}du\right)$
Since $-3$ is constant with respect to $u$, move $-3$ out of the integral.
$\frac{1}{5}\left(\frac{3}{5}{u}^{\frac{5}{3}}+C-3\int {u}^{-\frac{1}{3}}du\right)$
By the Power Rule, the integral of ${u}^{-\frac{1}{3}}$ with respect to $u$ is $\frac{3}{2}{u}^{\frac{2}{3}}$.
$\frac{1}{5}\left(\frac{3}{5}{u}^{\frac{5}{3}}+C-3\left(\frac{3}{2}{u}^{\frac{2}{3}}+C\right)\right)$
Simplify.
$\frac{1}{5}\left(\frac{3{u}^{\frac{5}{3}}}{5}-\frac{9{u}^{\frac{2}{3}}}{2}\right)+C$
Replace all occurrences of $u$ with $2t+3$.
$\frac{1}{5}\left(\frac{3{\left(2t+3\right)}^{\frac{5}{3}}}{5}-\frac{9{\left(2t+3\right)}^{\frac{2}{3}}}{2}\right)+C$
Simplify.
$\frac{3{\left(2t+3\right)}^{\frac{2}{3}}\left(4t-9\right)}{50}+C$
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Find the maximum of a solution expression of a quadratic equation
back to trigonometry
Find the maximum of a solution expression of a quadratic equation
Question:
Given: x1 and x2 are two solutions of the quadratic equation x2 - (k - 1)x + k2 + 2k + 1 = 0. Find the maximum of x12 + x22
Solution:
If the quadratic equation has two solutions, then b2 - 4ac >= 0
For this quadratic equation: a = 1, b = - (k - 1), c = k2 + 2k + 1
b2 - 4ac = [-(k - 1)]2 - 4(1)(k2 + 2k + 1)
= (-1)2 (k - 1)2 - 4(k2 + 2k + 1)
= k2 - 2k + 1 - 4k2 - 8k - 4
= -3k2 - 10k - 3
= -(3k2 + 10k + 3)
Since the quadratic equation has two solutions, so b2 - 4ac >= 0
then -(3k2 + 10k + 3) >= 0
Both side of the inequality multiply (-1) and the inequality change its direction
3k2 + 10k + 3 <= 0
k1 <= [-10 + Sqrt (102 - 4 × 3 × 3)]/(2 × 3)
= [-10 + Sqrt (100 - 36)]/6
= [-10 + Sqrt (64)]/6
= [-10 + 8]/6
= -2/6
= -1/3
k2 <= [-10 - Sqrt (102 - 4 × 3 × 3)]/(2 × 3)
= [-10 - Sqrt (100 - 36)]/6
= [-10 - Sqrt (64)]/6
= [-10 - 8]/6
= -18/6
= -3
so the k is in the range of -3 <= k <= -1/3
x12 + x22
= (x1 + x2)2 - 2x1x2
= (k - 1)2 - 2(k2 + 2k + 1)
= k2 - 2k + 1 - 2k2 - 4k -2
= - k2 - 6k - 1
= - (k2 + 6k + 1)
= - (k2 + 6k + 32 - 32 + 1)
= - (k2 + 6k + 32) + 9 - 1
= - (k + 3)2 + 8
When k = -3, x12 + x22 = 8
k = -3 is in the range of k. So when k = -3, x12 + x22 reach the maximum 8.
Rule used: If quadratic equation x2 + px + q has two solution x1 and x2, then the sum of these two solution is -p. That is, x1 + x2 = -p. The product of these two solutions is q. That is, x1 x2 = q.
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# Geometric Distributions
## Probability of number of trials before a first success: P(a trials) = q^(a-1) x p
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Geometric Distributions
### Geometric Distributions
The geometric distribution comes directly from our knowledge of the binomial distribution. The geometric distribution focuses on the number of trials before the first success occurs.
Suppose we play a game where we flip a coin until someone flips a tails and then the game is over. The probability that the game is over on the first flip is 50% because that is the probability that someone flips tails. After the first trial, if the game is not over then again the coin has 50% chance of getting a tails on the second flip. This means that the game has a 50% chance of being over on the first coin flip and a 25% chance of being over on the second coin flip.
The probability of performing trials before the first success is the same has having consecutive failures and then having one success.
• is the number of trials ending in 1 success
• is the probability of success
• is the probability of failure
#### Calculating Probability
1. Consider the coin game where the game ends once someone flips tails. What is the probability that the game ends on the 3rd flip?
2. Suppose a copy machine consistently has a 5% chance of breaking on any given day. Although it might work for many days in a row, it will inevitably break down. What is the probability that it lasts a whole five day work week and breaks down on the 6th day?
Although it is counter-intuitive “success” is defined in this problem to be the machine breaks down.
3. Consider the coin game again and make a probability distribution for games lasting up to 8 coin flips.
a (number of flips) Probability
1
2
3
4
5
6
7
8
### Example
#### Example 1
Jordan is a student who hates to be called on in class. His teacher randomly calls on students and Jordan knows that on any given day he has a 10% chance of being called on. What is the probability that on the first day Jordan is called on? The 2nd day? The 5th day?
Use the geometric distribution formula.
Number of Days Probability
1
2
5
### Review
1. Identify whether or not each situation has probabilities which are geometrically distributed:
1. The number of plane flights someone makes before they are in a plane crash.
2. The number of times your car will turn on before the battery dies.
3. The number of times you win at a slot machine out of 5 plays.
4. The number of gold medals a country wins in the Olympics.
2. If the probability of winning a game is .4, then what is the probability of losing 5 games in a row and winning on the 6th game?
3. If the probability of rolling a 4 on a fair die is , what is the probability of rolling anything but fours for 10 rolls and then rolling a 4 on the 11th roll?
4. Suppose there is a game that is played until someone rolls a six, then the game ends. Create a probability distribution table for games ending in 8 or fewer rolls.
5. If rain falls randomly with a 20% chance of rain on any given day, what is the probability that we suffer a drought for 8 days and then finally it rains on the 9th day?
6. Suppose it rains like it does in question 5. What is the probability that we suffer a drought for 5 days and then get rain on any/all of the next 5 days?
7. Suppose there is a jar of marbles. There are 10 red marbles and 2 green marbles. We randomly choose a marble and look at it. If it is red, we replace the marble into the jar and choose randomly again. If it is green, we stop. This process continues until we finally end with a green marble. What is the probability that the game ends just after the first draw?
8. Using the same jar and game as question number 7, what is the probability that the game ends just after the second draw? What is the probability that the game ends just after the 3rd draw?
9. Using the same jar and game as question number 7, what is the probability that the game takes more than 3 draws?
10. Suppose you win blackjack 47% of the time. What is the probability that you lose 3 games in a row and then win on the 4th game?
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# 4 Maps
Maps are very similar to scale drawings. The main difference is that they are usually used to show places.
If you look in a holiday brochure you will see lots of maps. They are often used to show how a resort is laid out. They show where a few important places are, such as local shops, hotels, the beach, swimming pools and restaurants.
It is important to understand how to read a map so that you do not end up too far from the places you want to be near – or too close to the places you want to avoid!
## Case study _unit5.4.1 Example: Holiday map
Here is a typical example of a map you find in a holiday brochure.
Figure _unit5.4.1 Figure 20 A scale drawing of a holiday resort
How far apart is everything on this map?
### Method
As with scale drawings, the thing you need to know before you can understand the map is the scale. In this example the scale is 1:1,000. This means that every 1 cm on the map represents 1,000 cm (or 10 m) in real life.
Using the scale, you can interpret the data on the map and work out how far different places are from one another.
To do this you need to measure the distances on the map and then multiply them by 1,000 to get the actual distance in centimetres. Or, more simply, you could multiply the distances in centimetres by 10 to get the actual distance in metres.
So on this map the Grooves Nightclub is 1 cm from Hotel Party. In real life that’s 10 m – not very far at all. Knowing this could affect whether you choose to stay at Hotel Party, depending on whether you like nightclubs or not.
Now try the following activity. Remember to check your answers once you have completed the questions.
## Activity _unit5.4.1 Activity 7: Using a map to find distances
Let’s stay with the map of the holiday resort.
Figure _unit5.4.2 Figure 21 A scale drawing of a holiday resort
### Box _unit5.4.1
Hint: The entrances to the buildings are marked with crosses on the map. You need to measure from these crosses.
1. What is the distance in real life between the pub and Hotel Sun in metres?
2. How far is it in real life from the Super Shop to the Beach Bistro in metres?
3. What is the distance in real life from Grooves Nightclub to the beach in metres?
1. The distance on the map between the pub and Hotel Sun is 4 cm on the map, and the scale is 1:1,000. This means that 1 cm on the drawing is equal to 1,000 cm in real life. So to find out what 4 cm is in real life, you need to multiply it by 1,000:
• 4 × 1,000 = 4,000 cm
The question asks for the length in metres, so you need to convert centimetres into metres:
• 4,000 ÷ 100 = 40 m
The actual distance in real life between the pub and Hotel Sun is 40 m.
2. The distance on the map is 2 cm. Using the same calculation, the actual distance in real life between the Super Shop and the Beach Bistro is 20 m.
3. The distance on the map is 6 cm. Using the same calculation, the actual distance in real life between Grooves nightclub and the beach is 60 m.
## Summary
In this section you have learned how to use maps.
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# How do you find the centripetal acceleration
## Circular motion and centripetal force (physics)
The circular motion and the centripetal force are the subject of this article. We will explain important terms, what they are and how to do some calculations with the corresponding formulas.
Everyone knows a circular movement, for example from a carousel. If you sit down, you will be driven in a circle at the same speed over a longer period of time. In physics, such a circular path is often represented as follows (we explain the terms below):
Show:
### Circular motion definitions
Before I give you a few formulas, you should first know a few definitions of circular motion. There are:
• "v" is the path speed
• "T" is the period, i.e. the time for one revolution of the object
• "f" is the frequency, this is the number of revolutions per second of the object
• "ω" is the angular frequency, the circumferential angle per time in radians (formula follows)
Show:
### Formulas for circular motion
Calculate angular frequency:
• Formula: ω = 2 · π · f
• "ω" is the angular frequency per second [1 / s]
• "π" is the circle number, π = 3.14159
• "f" is the frequency per second [1 / s]
Calculate speed:
• Formula: v = r ω
• "v" is the speed in meters per second [m / s]
• "r" is the radius of the circle in meters [m]
• "ω" is the angular frequency per second [1 / s]
Calculate acceleration:
• Formula: a = v2 : r
• "a" is the acceleration in meters per second square [m / s2 ]
• "v" is the speed in meters per second [m / s]
• "r" is the radius of the circle in meters [m]
### centripetal force
In the case of a uniform circular movement, a force always acts on the body, which always points to the center of the circle. This force is called the centripetal force.
Calculate centripetal force:
• Formula: FZ = m · v2 : r
• "F.Z"is the centripetal force in Newtons [N]
• "m" is the mass of the object in kilograms [kg]
• "v" is the speed in meters per second [m / s]
• "r" is the radius of the circle in meters [m]
### Examples: centripetal force / circular motion
In the following we want to look at some examples that demonstrate how to use the formulas.
example 1: A 0.4kg body is thrown on a 0.8m long cord on a circular path. The frequency is f = 1.33 / s. What is the force?
Example 2: A 0.4kg body is thrown on a 0.8m long cord on a circular path. The cord can withstand 500N load. At what number of revolutions does the cord break?
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# Quantitative Aptitude: Data Interpretation Questions for IBPS PO Prelims Set 69
Directions (1-5):Â Study the table and answer the questions that follow.
1. Find the approximate average number of valid votes polled in village C over the given years.
A) 2690
B) 2980
C) 2360
D) 3480
E) 3760
  Option D
Solution:
(3620 +3350+ 3470)/3 = 3480
2. If in city A, the percentage increase in total votes polled in 2015 over 2010 is 5% less than the percentage increase in total votes polled in 2010 over 2005, then find the number of total votes polled in 2015?
A) 4670
B) 5700
C) 5100
D) 590
E) 4620
  Option  C
Solution:
% increase in 2010 = (4250-3400)/3400 * 100 = 25%
So in 2015 = 120/100 * 4250 = 5100 [5% less so 20% increase]
3. If 35% of total votes polled in village A in 2005 are from females and total valid voted by females is 1065, then find the number of valid votes from males in village A in same year.
A) 1265
B) 1695
C) 2756
D) 2891
E) 1167
  Option B
Solution:
Total valid votes are 2760 and by females is 1065 so remaining by males.
2760 – 1065 = 1695
4. What is the average number of invalid votes polled in 2010?
A) 380
B) 320
C) 390
D) 410
E) 440
  Option A
Solution:
4250-3830 = 420, 4300-3990 = 310, 3880-3470 = 410
(420+310+410)/3 = 380
5. Total invalid votes polled in village B in 2005 is what percent greater than total invalid votes in village C in 2005?
A) 58%
B) 52%
C) 49%
D) 45%
E) 40%
  Option E
Solution:
Invalid in B in 2005 = 3800-3450 = 350
Invalid in C in 2005 = 3600-3350 = 250
So required % = (350-250)/250 * 100 = 40%
Directions (6-10): Study the following bar graph and answer the questions that follow:
1. What is the average number (in thousand) of cars produced by company A in given years?
A) 63
B) 65
C) 72
D) 69
E) 66
  Option B
Solution:
Average = 1/6 * (55+55+64+63+76+77) = 65 thousand
2. Number of cars produced by company A in 2011 and 2014 is what percent less than those produced by company B in 2013 and 2016?
A) 17.48%
B) 18.36%
C) 27.59%
D) 15.37%
E) 22.72%
  Option A
Solution:
Required % = [(62+81) – (55+63)]/(62+81) * 100 = (143-118)/143 * 100 = 17.48%
3. Number of cars produced by both companies in 2011 and 2012 is what percent of number of cars produced by both companies in 2014 and 2016?
A) 75.28%
B) 81.35%
C) 73.19%
D) 82.14%
E) 63.27%
  Option D
Solution:
Required % = (55+56+55+64)/(63+59+77+81) * 100 = 230/280 * 100 = 82.14%
4. What is the ratio of number of cars produced by company A in 2012 and 2013 together to number of cars produced by company B in the same years together?
A) 18 : 13
B) 18 : 15
C) 17 : 18
D) 19 : 14
E) 17 : 11
  Option C
Solution:
A : B
(55+64) : (64+62) = 119 : 126 = 17 : 18
5. Suppose in year 2017, production of both companies increase by 20% each with respect to previous year, then total cars produced by both companies in 2017 is what percent more than total cars produced by both companies in 2015?
A) 39.13%
B)Â 31.66%
C) 34.17%
D) 28.02%
E) 30.75%
  Option E
Solution:
A in 2016 = 77 thousand, so in 2017 = 120/100 * 77thousand = 92,400
B in 2016 = 81 thousand, so in 2017 = 120/100 * 81thousand = 97,200
Total cars in 2015 = (76+69) = 145,000
Total cars in 2017 = 92400 + 97200 = 189,600
So required % = (189600 – 145000)/145000 * 100 = 30.75%
## 6 Thoughts to “Quantitative Aptitude: Data Interpretation Questions for IBPS PO Prelims Set 69”
1. Wikki Waran
3.what logic is this
2760 – 1065 = 16954
1. typo happens
2. Wikki Waran
Thank you
3. Sylvester
9..
4. jaga
thank u mam
|
# McGraw Hill Math Grade 5 Chapter 9 Lesson 11 Answer Key Problem Solving: Working Backwards
All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 9 Lesson 11 Problem Solving: Working Backwards are as per the latest syllabus guidelines.
## McGraw-Hill Math Grade 5 Answer Key Chapter 9 Lesson 11 Problem Solving: Working Backwards
Solve
Question 1.
Regina and Alicia had 4$$\frac{1}{2}$$ cups of water at the end of a day of driving. They drank 8$$\frac{1}{4}$$ cups before lunch and 11$$\frac{1}{4}$$ cups in the afternoon. How much water did they start with?
Explanation:
Number of cups of water Regina and Alicia had at the end of a day of driving = 4$$\frac{1}{2}$$.
Number of cups of water they drank before lunch = 8$$\frac{1}{4}$$.
Number of cups of water they drank in the afternoon = 11$$\frac{1}{4}$$.
Number of cups of water they start with = Number of cups of water Regina and Alicia had at the end of a day of driving + Number of cups of water they drank before lunch + Number of cups of water they drank in the afternoon
= 4$$\frac{1}{2}$$ + 8$$\frac{1}{4}$$ + 11$$\frac{1}{4}$$
= [(8 + 1) ÷ 2] + [(32 + 1) ÷ 4] + [(44 + 1) ÷ 4]
= $$\frac{9}{2}$$ + $$\frac{33}{4}$$ + $$\frac{45}{4}$$
= [(9 × 2) ÷ (2 × 2)] + $$\frac{33}{4}$$ + $$\frac{45}{4}$$
= $$\frac{18}{4}$$ + $$\frac{33}{4}$$ + $$\frac{45}{4}$$
= (18 + 33 + 45) ÷ 4
= 96 ÷ 4
= 24.
Question 2.
A small shop has 5.9 pounds of lunch meat at the end of Friday. The shop sold 4.9 pounds on Friday morning and 9.35 pounds on Friday afternoon. How many pounds of lunchmeat did the store start with?
Explanation:
Number of pounds of lunch meat a small shop has at the end of Friday = 5.9.
Number of pounds of lunch meat a small shop on Friday morning = 4.9.
Number of pounds of lunch meat a small shop on Friday afternoon = 9.35.
Number of pounds of lunchmeat the store start with = Number of pounds of lunch meat a small shop has at the end of Friday + Number of pounds of lunch meat a small shop on Friday morning + Number of pounds of lunch meat a small shop on Friday afternoon
= 5.9 + 4.9 + 9.35
= 10.8 + 9.35
= 20.15.
Question 3.
Use the digits 7, 3, 6, 5, 1, and 9 to write three numbers less than 600,000 but greater than 500,000. Use each digit only once for each number.
Three numbers less than 600,000 but greater than 500,000.
=> 513679.
=> 531679.
=> 561379.
Question 4.
Ms. Torres is on a road trip. She drives a total of 2,731.82 miles in three weeks. She drives 791.38 miles the second week. She drives 1,086.14 miles the third week. How many miles did she drive the first week?
Number of miles She drives in the first week = 854.3.
Explanation:
Number of miles She drives in three weeks = 2,731.82.
Number of miles She drives in the second week = 791.38.
Number of miles She drives in the third week = 1,086.14.
Number of miles She drives in the first week = Number of miles She drives in three weeks – (Number of miles She drives in the second week + Number of miles She drives in the third week)
= 2731.82 – (791.38 + 1086.14)
= 2731.82 -1877.52
= 854.3.
Question 5.
Mr. Kerry has 4 boxes of cocoa mix. Each box has 20 bags. He uses two cups of water and one bag of cocoa mix to make a mug of cocoa. How many cups of water will Mr Kerry use if he uses all the bags of cocoa mix?
Number of cups of water if he uses all the bags of cocoa mix = 160.
Explanation:
Number of boxes of cocoa mix Mr. Kerry has = 4.
Number of bags each box has = 20.
Number of cups of water he uses to make a mug of cocoa = 2.
Number of bags he uses to make a mug of cocoa = 1.
Total number of bags of cocoa Mr. Kerry has = Number of boxes of cocoa mix Mr. Kerry has × Number of bags each box has
= 4 × 20
= 80.
Number of cups of water if he uses all the bags of cocoa mix = Number of cups of water he uses to make a mug of cocoa × Total number of bags of cocoa Mr. Kerry has
= 2 × 80
= 160.
Question 6.
Ling is 56$$\frac{3}{4}$$ in. tall. Lauren is 49$$\frac{1}{3}$$ in. tall. How much taller is Ling than Lauren?
Number of inches height is Ling = 56$$\frac{3}{4}$$
Number of inches height is Lauren = 49$$\frac{1}{3}$$
= 56$$\frac{3}{4}$$ – 49$$\frac{1}{3}$$
|
Calculus Volume 2
# 5.4Comparison Tests
Calculus Volume 25.4 Comparison Tests
## Learning Objectives
• 5.4.1 Use the comparison test to test a series for convergence.
• 5.4.2 Use the limit comparison test to determine convergence of a series.
We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. In this section, we show how to use comparison tests to determine the convergence or divergence of a series by comparing it to a series whose convergence or divergence is known. Typically these tests are used to determine convergence of series that are similar to geometric series or p-series.
## Comparison Test
In the preceding two sections, we discussed two large classes of series: geometric series and p-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test.
For example, consider the series
$∑n=1∞1n2+1.∑n=1∞1n2+1.$
This series looks similar to the convergent series
$∑n=1∞1n2.∑n=1∞1n2.$
Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing. Furthermore, since
$0<1n2+1<1n20<1n2+1<1n2$
for all positive integers $n,n,$ the $kthkth$ partial sum $SkSk$ of $∑n=1∞1n2+1∑n=1∞1n2+1$ satisfies
$Sk=∑n=1k1n2+1<∑n=1k1n2<∑n=1∞1n2.Sk=∑n=1k1n2+1<∑n=1k1n2<∑n=1∞1n2.$
(See Figure 5.16(a) and Table 5.1.) Since the series on the right converges, the sequence ${Sk}{Sk}$ is bounded above. We conclude that ${Sk}{Sk}$ is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, ${Sk}{Sk}$ converges, and thus
$∑n=1∞1n2+1∑n=1∞1n2+1$
converges.
Similarly, consider the series
$∑n=1∞1n−1/2.∑n=1∞1n−1/2.$
This series looks similar to the divergent series
$∑n=1∞1n.∑n=1∞1n.$
The sequence of partial sums for each series is monotone increasing and
$1n−1/2>1n>01n−1/2>1n>0$
for every positive integer $n.n.$ Therefore, the $kthkth$ partial sum $SkSk$ of $∑n=1∞1n−1/2∑n=1∞1n−1/2$ satisfies
$Sk=∑n=1k1n−1/2>∑n=1k1n.Sk=∑n=1k1n−1/2>∑n=1k1n.$
(See Figure 5.16(b) and Table 5.2.) Since the series $∑n=1∞1/n∑n=1∞1/n$ diverges to infinity, the sequence of partial sums $∑n=1k1/n∑n=1k1/n$ is unbounded. Consequently, ${Sk}{Sk}$ is an unbounded sequence, and therefore diverges. We conclude that
$∑n=1∞1n−1/2∑n=1∞1n−1/2$
diverges.
Figure 5.16 (a) Each of the partial sums for the given series is less than the corresponding partial sum for the converging $p−series.p−series.$ (b) Each of the partial sums for the given series is greater than the corresponding partial sum for the diverging harmonic series.
$kk$ $11$ $22$ $33$ $44$ $55$ $66$ $77$ $88$
$∑n=1k1n2+1∑n=1k1n2+1$ $0.50.5$ $0.70.7$ $0.80.8$ $0.85880.8588$ $0.89730.8973$ $0.92430.9243$ $0.94430.9443$ $0.95970.9597$
$∑n=1k1n2∑n=1k1n2$ $11$ $1.251.25$ $1.36111.3611$ $1.42361.4236$ $1.46361.4636$ $1.49141.4914$ $1.51181.5118$ $1.52741.5274$
Table 5.1 Comparing a series with a p-series (p = 2)
$kk$ $11$ $22$ $33$ $44$ $55$ $66$ $77$ $88$
$∑n=1k1n−1/2∑n=1k1n−1/2$ $22$ $2.66672.6667$ $3.06673.0667$ $3.35243.3524$ $3.57463.5746$ $3.75643.7564$ $3.91033.9103$ $4.04364.0436$
$∑n=1k1n∑n=1k1n$ $11$ $1.51.5$ $1.83331.8333$ $2.09332.0933$ $2.28332.2833$ $2.452.45$ $2.59292.5929$ $2.71792.7179$
Table 5.2 Comparing a series with the harmonic series
## Theorem5.11
### Comparison Test
1. Suppose there exists an integer $NN$ such that $0≤an≤bn0≤an≤bn$ for all $n≥N.n≥N.$ If $∑n=1∞bn∑n=1∞bn$ converges, then $∑n=1∞an∑n=1∞an$ converges.
2. Suppose there exists an integer $NN$ such that $an≥bn≥0an≥bn≥0$ for all $n≥N.n≥N.$ If $∑n=1∞bn∑n=1∞bn$ diverges, then $∑n=1∞an∑n=1∞an$ diverges.
### Proof
We prove part i. The proof of part ii. is the contrapositive of part i. Let ${Sk}{Sk}$ be the sequence of partial sums associated with $∑n=1∞an,∑n=1∞an,$ and let $L=∑n=1∞bn.L=∑n=1∞bn.$ Since the terms $an≥0,an≥0,$
$Sk=a1+a2+⋯+ak≤a1+a2+⋯+ak+ak+1=Sk+1.Sk=a1+a2+⋯+ak≤a1+a2+⋯+ak+ak+1=Sk+1.$
Therefore, the sequence of partial sums is increasing. Further, since $an≤bnan≤bn$ for all $n≥N,n≥N,$ then
$∑n=Nkan≤∑n=Nkbn≤∑n=1∞bn=L.∑n=Nkan≤∑n=Nkbn≤∑n=1∞bn=L.$
Therefore, for all $k≥1,k≥1,$
$Sk=(a1+a2+⋯+aN−1)+∑n=Nkan≤(a1+a2+⋯+aN−1)+L.Sk=(a1+a2+⋯+aN−1)+∑n=Nkan≤(a1+a2+⋯+aN−1)+L.$
Since $a1+a2+⋯+aN−1a1+a2+⋯+aN−1$ is a finite number, we conclude that the sequence ${Sk}{Sk}$ is bounded above. Therefore, ${Sk}{Sk}$ is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that ${Sk}{Sk}$ converges, and therefore the series $∑n=1∞an∑n=1∞an$ converges.
To use the comparison test to determine the convergence or divergence of a series $∑n=1∞an,∑n=1∞an,$ it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and p-series, these series are often used. If there exists an integer $NN$ such that for all $n≥N,n≥N,$ each term $anan$ is less than each corresponding term of a known convergent series, then $∑n=1∞an∑n=1∞an$ converges. Similarly, if there exists an integer $NN$ such that for all $n≥N,n≥N,$ each term $anan$ is greater than each corresponding term of a known divergent series, then $∑n=1∞an∑n=1∞an$ diverges.
## Example 5.17
### Using the Comparison Test
For each of the following series, use the comparison test to determine whether the series converges or diverges.
1. $∑n=1∞1n3+3n+1∑n=1∞1n3+3n+1$
2. $∑n=1∞12n+1∑n=1∞12n+1$
3. $∑n=2∞1ln(n)∑n=2∞1ln(n)$
## Checkpoint5.16
Use the comparison test to determine if the series $∑n=1∞nn3+n+1∑n=1∞nn3+n+1$ converges or diverges.
## Limit Comparison Test
The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series
$∑n=2∞1n2−1.∑n=2∞1n2−1.$
It is natural to compare this series with the convergent series
$∑n=2∞1n2.∑n=2∞1n2.$
However, this series does not satisfy the hypothesis necessary to use the comparison test because
$1n2−1>1n21n2−1>1n2$
for all integers $n≥2.n≥2.$ Although we could look for a different series with which to compare $∑n=2∞1/(n2−1),∑n=2∞1/(n2−1),$ instead we show how we can use the limit comparison test to compare
$∑n=2∞1n2−1and∑n=2∞1n2.∑n=2∞1n2−1and∑n=2∞1n2.$
Let us examine the idea behind the limit comparison test. Consider two series $∑n=1∞an∑n=1∞an$ and $∑n=1∞bn.∑n=1∞bn.$ with positive terms $anandbnanandbn$ and evaluate
$limn→∞anbn.limn→∞anbn.$
If
$limn→∞anbn=L≠0,limn→∞anbn=L≠0,$
then, for $nn$ sufficiently large, $an≈Lbn.an≈Lbn.$ Therefore, either both series converge or both series diverge. For the series $∑n=2∞1/(n2−1)∑n=2∞1/(n2−1)$ and $∑n=2∞1/n2,∑n=2∞1/n2,$ we see that
$limn→∞1/(n2−1)1/n2=limn→∞n2n2−1=1.limn→∞1/(n2−1)1/n2=limn→∞n2n2−1=1.$
Since $∑n=2∞1/n2∑n=2∞1/n2$ converges, we conclude that
$∑n=2∞1n2−1∑n=2∞1n2−1$
converges.
The limit comparison test can be used in two other cases. Suppose
$limn→∞anbn=0.limn→∞anbn=0.$
In this case, ${an/bn}{an/bn}$ is a bounded sequence. As a result, there exists a constant $MM$ such that $an≤Mbn.an≤Mbn.$ Therefore, if $∑n=1∞bn∑n=1∞bn$ converges, then $∑n=1∞an∑n=1∞an$ converges. On the other hand, suppose
$limn→∞anbn=∞.limn→∞anbn=∞.$
In this case, ${an/bn}{an/bn}$ is an unbounded sequence. Therefore, for every constant $MM$ there exists an integer $NN$ such that $an≥Mbnan≥Mbn$ for all $n≥N.n≥N.$ Therefore, if $∑n=1∞bn∑n=1∞bn$ diverges, then $∑n=1∞an∑n=1∞an$ diverges as well.
## Theorem5.12
### Limit Comparison Test
Let $an,bn≥0an,bn≥0$ for all $n≥1.n≥1.$
1. If $limn→∞an/bn=L≠0,limn→∞an/bn=L≠0,$ then $∑n=1∞an∑n=1∞an$ and $∑n=1∞bn∑n=1∞bn$ both converge or both diverge.
2. If $limn→∞an/bn=0limn→∞an/bn=0$ and $∑n=1∞bn∑n=1∞bn$ converges, then $∑n=1∞an∑n=1∞an$ converges.
3. If $limn→∞an/bn=∞limn→∞an/bn=∞$ and $∑n=1∞bn∑n=1∞bn$ diverges, then $∑n=1∞an∑n=1∞an$ diverges.
Note that if $an/bn→0an/bn→0$ and $∑n=1∞bn∑n=1∞bn$ diverges, the limit comparison test gives no information. Similarly, if $an/bn→∞an/bn→∞$ and $∑n=1∞bn∑n=1∞bn$ converges, the test also provides no information. For example, consider the two series $∑n=1∞1/n∑n=1∞1/n$ and $∑n=1∞1/n2.∑n=1∞1/n2.$ These series are both p-series with $p=1/2p=1/2$ and $p=2,p=2,$ respectively. Since $p=1/2<1,p=1/2<1,$ the series $∑n=1∞1/n∑n=1∞1/n$ diverges. On the other hand, since $p=2>1,p=2>1,$ the series $∑n=1∞1/n2∑n=1∞1/n2$ converges. However, suppose we attempted to apply the limit comparison test, using the convergent $p−seriesp−series$ $∑n=1∞1/n3∑n=1∞1/n3$ as our comparison series. First, we see that
$1/n1/n3=n3n=n5/2→∞asn→∞.1/n1/n3=n3n=n5/2→∞asn→∞.$
Similarly, we see that
$1/n21/n3=n→∞asn→∞.1/n21/n3=n→∞asn→∞.$
Therefore, if $an/bn→∞an/bn→∞$ when $∑n=1∞bn∑n=1∞bn$ converges, we do not gain any information on the convergence or divergence of $∑n=1∞an.∑n=1∞an.$
## Example 5.18
### Using the Limit Comparison Test
For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.
1. $∑n=1∞1n+1∑n=1∞1n+1$
2. $∑n=1∞2n+13n∑n=1∞2n+13n$
3. $∑n=1∞ln(n)n2∑n=1∞ln(n)n2$
## Checkpoint5.17
Use the limit comparison test to determine whether the series $∑n=1∞5n3n+2∑n=1∞5n3n+2$ converges or diverges.
## Section 5.4 Exercises
Use the comparison test to determine whether the following series converge.
194.
$∑n=1∞an∑n=1∞an$ where $an=2n(n+1)an=2n(n+1)$
195.
$∑n=1∞an∑n=1∞an$ where $an=1n(n+1/2)an=1n(n+1/2)$
196.
$∑ n = 1 ∞ 1 2 ( n + 1 ) ∑ n = 1 ∞ 1 2 ( n + 1 )$
197.
$∑ n = 2 ∞ 1 2 n - 1 ∑ n = 2 ∞ 1 2 n - 1$
198.
$∑ n = 2 ∞ 1 ( n ln n ) 2 ∑ n = 2 ∞ 1 ( n ln n ) 2$
199.
$∑ n = 1 ∞ n ! ( n + 2 ) ! ∑ n = 1 ∞ n ! ( n + 2 ) !$
200.
$∑ n = 1 ∞ 1 n ! ∑ n = 1 ∞ 1 n !$
201.
$∑ n = 1 ∞ sin ( 1 / n ) n 2 ∑ n = 1 ∞ sin ( 1 / n ) n 2$
202.
$∑ n = 1 ∞ sin 2 n n 2 ∑ n = 1 ∞ sin 2 n n 2$
203.
$∑ n = 1 ∞ sin ( 1 / n ) ( n ) 3 ∑ n = 1 ∞ sin ( 1 / n ) ( n ) 3$
204.
$∑ n = 1 ∞ n 1.2 − 1 n 2.3 + 1 ∑ n = 1 ∞ n 1.2 − 1 n 2.3 + 1$
205.
$∑ n = 1 ∞ n + 1 − n n ∑ n = 1 ∞ n + 1 − n n$
206.
$∑ n = 1 ∞ n 4 n 4 + n 2 3 ∑ n = 1 ∞ n 4 n 4 + n 2 3$
Use the limit comparison test to determine whether each of the following series converges or diverges.
207.
$∑ n = 1 ∞ ( ln n n ) 2 ∑ n = 1 ∞ ( ln n n ) 2$
208.
$∑ n = 1 ∞ ( ln n n 0.6 ) 2 ∑ n = 1 ∞ ( ln n n 0.6 ) 2$
209.
$∑ n = 1 ∞ ln ( 1 + 1 n ) n ∑ n = 1 ∞ ln ( 1 + 1 n ) n$
210.
$∑ n = 1 ∞ ln ( 1 + 1 n 2 ) ∑ n = 1 ∞ ln ( 1 + 1 n 2 )$
211.
$∑ n = 1 ∞ 1 4 n − 3 n ∑ n = 1 ∞ 1 4 n − 3 n$
212.
$∑ n = 1 ∞ 1 n 2 − n sin n ∑ n = 1 ∞ 1 n 2 − n sin n$
213.
$∑ n = 1 ∞ 1 e ( 1.1 ) n − 3 n ∑ n = 1 ∞ 1 e ( 1.1 ) n − 3 n$
214.
$∑ n = 1 ∞ 1 e ( 1.01 ) n − 3 n ∑ n = 1 ∞ 1 e ( 1.01 ) n − 3 n$
215.
$∑ n = 1 ∞ 1 n 1 + 1 / n ∑ n = 1 ∞ 1 n 1 + 1 / n$
216.
$∑ n = 1 ∞ 1 2 1 + 1 / n n 1 + 1 / n ∑ n = 1 ∞ 1 2 1 + 1 / n n 1 + 1 / n$
217.
$∑ n = 1 ∞ ( 1 n − sin ( 1 n ) ) ∑ n = 1 ∞ ( 1 n − sin ( 1 n ) )$
218.
$∑ n = 1 ∞ ( 1 − cos ( 1 n ) ) ∑ n = 1 ∞ ( 1 − cos ( 1 n ) )$
219.
$∑ n = 1 ∞ 1 n ( π 2 − tan −1 n ) ∑ n = 1 ∞ 1 n ( π 2 − tan −1 n )$
220.
$∑n=1∞(1−1n)n.n∑n=1∞(1−1n)n.n$ (Hint:$(1−1n)n→1/e.)(1−1n)n→1/e.)$
221.
$∑n=1∞(1−e−1/n)∑n=1∞(1−e−1/n)$ (Hint:$1/e≈(1−1/n)n,1/e≈(1−1/n)n,$ so $1−e−1/n≈1/n.)1−e−1/n≈1/n.)$
222.
Does $∑n=2∞1(lnn)p∑n=2∞1(lnn)p$ converge if $pp$ is large enough? If so, for which $p?p?$
223.
Does $∑n=1∞((lnn)n)p∑n=1∞((lnn)n)p$ converge if $pp$ is large enough? If so, for which $p?p?$
224.
For which $pp$ does the series $∑n=1∞2pn/3n∑n=1∞2pn/3n$ converge?
225.
For which $p>0p>0$ does the series $∑n=1∞np2n∑n=1∞np2n$ converge?
226.
For which $r>0r>0$ does the series $∑n=1∞rn22n∑n=1∞rn22n$ converge?
227.
For which $r>0r>0$ does the series $∑n=1∞2nrn2∑n=1∞2nrn2$ converge?
228.
Find all values of $pp$ and $qq$ such that $∑n=1∞np(n!)q∑n=1∞np(n!)q$ converges.
229.
Does $∑n=1∞sin2(nr/2)n∑n=1∞sin2(nr/2)n$ converge or diverge? Explain.
230.
Explain why, for each $n,n,$ at least one of ${|sinn|,|sin(n+1)|,...,|sinn+6|}{|sinn|,|sin(n+1)|,...,|sinn+6|}$ is larger than $1/2.1/2.$ Use this relation to test convergence of $∑n=1∞|sinn|n.∑n=1∞|sinn|n.$
231.
Suppose that $an≥0an≥0$ and $bn≥0bn≥0$ and that $∑n=1∞a2n∑n=1∞a2n$ and $∑n=1∞b2n∑n=1∞b2n$ converge. Prove that $∑n=1∞anbn∑n=1∞anbn$ converges and $∑n=1∞anbn≤12(∑n=1∞an2+∑n=1∞bn2).∑n=1∞anbn≤12(∑n=1∞an2+∑n=1∞bn2).$
232.
Does $∑n=1∞2−lnlnn∑n=1∞2−lnlnn$ converge? (Hint: Write $2lnlnn2lnlnn$ as a power of $lnn.)lnn.)$
233.
Does $∑n=1∞(lnn)−lnn∑n=1∞(lnn)−lnn$ converge? (Hint: Use $n=eln(n)n=eln(n)$ to compare to a $p−series.)p−series.)$
234.
Does $∑n=2∞(lnn)−lnlnn∑n=2∞(lnn)−lnlnn$ converge? (Hint: Compare $anan$ to $1/n.)1/n.)$
235.
Show that if $an≥0an≥0$ and $∑n=1∞an∑n=1∞an$ converges, then $∑n=1∞a2n∑n=1∞a2n$ converges. If $∑n=1∞a2n∑n=1∞a2n$ converges, does $∑n=1∞an∑n=1∞an$ necessarily converge?
236.
Suppose that $an>0an>0$ for all $nn$ and that $∑n=1∞an∑n=1∞an$ converges. Suppose that $bnbn$ is an arbitrary sequence of zeros and ones. Does $∑n=1∞anbn∑n=1∞anbn$ necessarily converge?
237.
Suppose that $an>0an>0$ for all $nn$ and that $∑n=1∞an∑n=1∞an$ diverges. Suppose that $bnbn$ is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does $∑n=1∞anbn∑n=1∞anbn$ necessarily diverge?
238.
Complete the details of the following argument: If $∑n=1∞1n∑n=1∞1n$ converges to a finite sum $s,s,$ then $12s=12+14+16+⋯12s=12+14+16+⋯$ and $s−12s=1+13+15+⋯.s−12s=1+13+15+⋯.$ Why does this lead to a contradiction?
239.
Show that if $an≥0an≥0$ and $∑n=1∞a2n∑n=1∞a2n$ converges, then $∑n=1∞sin2(an)∑n=1∞sin2(an)$ converges.
240.
Suppose that $an/bn→0an/bn→0$ in the comparison test, where $an≥0an≥0$ and $bn≥0.bn≥0.$ Prove that if $∑bn∑bn$ converges, then $∑an∑an$ converges.
241.
Let $bnbn$ be an infinite sequence of zeros and ones. What is the largest possible value of $x=∑n=1∞bn/2n?x=∑n=1∞bn/2n?$
242.
Let $dndn$ be an infinite sequence of digits, meaning $dndn$ takes values in ${0,1,…,9}.{0,1,…,9}.$ What is the largest possible value of $x=∑n=1∞dn/10nx=∑n=1∞dn/10n$ that converges?
243.
Explain why, if $x>1/2,x>1/2,$ then $xx$ cannot be written $x=∑n=2∞bn2n(bn=0or1,b1=0).x=∑n=2∞bn2n(bn=0or1,b1=0).$
244.
[T] Evelyn has a perfect balancing scale, an unlimited number of $1-kg1-kg$ weights, and one each of $1/2-kg,1/4-kg,1/8-kg,1/2-kg,1/4-kg,1/8-kg,$ and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?
245.
[T] Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of $1-kg1-kg$ weights, and nine each of $0.1-kg,0.1-kg,$ $0.01-kg,0.001-kg,0.01-kg,0.001-kg,$ and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?
246.
The series $∑n=1∞12n∑n=1∞12n$ is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which $nn$ is odd. Let $m>1m>1$ be fixed. Show, more generally, that deleting all terms $1/n1/n$ where $n=mkn=mk$ for some integer $kk$ also results in a divergent series.
247.
In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series is a series obtained from $∑n=1∞1n∑n=1∞1n$ by removing any term $1/n1/n$ if a given digit, say $9,9,$ appears in the decimal expansion of $n.n.$ Argue that this depleted harmonic series converges by answering the following questions.
1. How many whole numbers $nn$ have $dd$ digits?
2. How many $d-digitd-digit$ whole numbers $h(d).h(d).$ do not contain $99$ as one or more of their digits?
3. What is the smallest $d-digitd-digit$ number $m(d)?m(d)?$
4. Explain why the deleted harmonic series is bounded by $∑d=1∞h(d)m(d).∑d=1∞h(d)m(d).$
5. Show that $∑d=1∞h(d)m(d)∑d=1∞h(d)m(d)$ converges.
248.
Suppose that a sequence of numbers $an>0an>0$ has the property that $a1=1a1=1$ and $an+1=1n+1Sn,an+1=1n+1Sn,$ where $Sn=a1+⋯+an.Sn=a1+⋯+an.$ Can you determine whether $∑n=1∞an∑n=1∞an$ converges? (Hint: $SnSn$ is monotone.)
249.
Suppose that a sequence of numbers $an>0an>0$ has the property that $a1=1a1=1$ and $an+1=1(n+1)2Sn,an+1=1(n+1)2Sn,$ where $Sn=a1+⋯+an.Sn=a1+⋯+an.$ Can you determine whether $∑n=1∞an∑n=1∞an$ converges? (Hint: $S2=a2+a1=a2+S1=a2+1=1+1/4=(1+1/4)S1,S2=a2+a1=a2+S1=a2+1=1+1/4=(1+1/4)S1,$ $S3=132S2+S2=(1+1/9)S2=(1+1/9)(1+1/4)S1,S3=132S2+S2=(1+1/9)S2=(1+1/9)(1+1/4)S1,$ etc. Look at $ln(Sn),ln(Sn),$ and use $ln(1+t)≤t,ln(1+t)≤t,$ $t>0.)t>0.)$
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Explore the Back-Story
# Probability Problem | AMC 8, 2016 | Problem no. 21
Try this beautiful problem from Probability.
## Problem based on Probability | AMC-8, 2016 | Problem 21
A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?
• $\frac{3}{5}$
• $\frac{2}{5}$
• $\frac{1}{4}$
### Key Concepts
probability
combination
fraction
Answer: $\frac{2}{5}$
AMC-8, 2016 problem 21
Challenges and Thrills in Pre College Mathematics
## Try with Hints
There are 5 Chips, 3 red and 2 green
Can you now finish the problem ..........
We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement
Can you finish the problem........
There are 5 Chips, 3 red and 2 green
we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn.
if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green
but we draw randomly. there are 3 red boxes and 2 green boxes
Therefore the probability that the 3 reds are drawn=$\frac{2}{5}$
## Subscribe to Cheenta at Youtube
Try this beautiful problem from Probability.
## Problem based on Probability | AMC-8, 2016 | Problem 21
A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?
• $\frac{3}{5}$
• $\frac{2}{5}$
• $\frac{1}{4}$
### Key Concepts
probability
combination
fraction
Answer: $\frac{2}{5}$
AMC-8, 2016 problem 21
Challenges and Thrills in Pre College Mathematics
## Try with Hints
There are 5 Chips, 3 red and 2 green
Can you now finish the problem ..........
We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement
Can you finish the problem........
There are 5 Chips, 3 red and 2 green
we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn.
if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green
but we draw randomly. there are 3 red boxes and 2 green boxes
Therefore the probability that the 3 reds are drawn=$\frac{2}{5}$
## Subscribe to Cheenta at Youtube
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Because life happens on the yard and in the classroom™
# Parent’s Point of View #121: Critical Math Skills for 4th Grade
Audio Article
Fourth-grade math focuses on addition, subtraction, division, and multiplication. Students apply the four operations in solving multi-step word problems that involve multi-digit numbers. In addition, the math program improves student’s understanding of fractions, including ordering fractions and equal fractions. Fourth graders also learn how to add and subtract fractions with similar denominators, multiply fractions with whole numbers and understand the relationship between decimals and fractions.
## All the math you need in the stock market you get in the 4th grade.”— Peter Lynch
Here are some of the questions from concerned parents that we shall explore in this article.
• What will my 4th-grader learn on multi-digit whole numbers?
• What skills do 4th graders learn on equivalent fractions?
As per the common core standards, let’s analyze more math skills that your child will learn in the fourth grade.
### What will my fourth-grader learn on multi-digit whole numbers?
A fourth grader should accurately and quickly add and subtract multi-digit numbers up to 1000 000. They should understand factors (whole numbers) that can be multiplied together to attain another number. Fourth graders should know that a single number can have numerous factor pairs. For instance, 4 and 4 are factors of 16(4×4 =16), just like 8 and 2 (8×2=16) and 16 and 1(16×1)
### What skills should a 4th grader have in solving word problems?
A fourth-grader should successfully solve word problems with whole numbers using division, multiplication, subtraction, and addition problems with remainders. In addition, they should use estimation strategies, e.g., rounding and mental math, to determine if their answer is reasonable or not. Fourth graders should also know how to write an equation with a letter representing the unknown value.
### What fractions skills do 4th graders learn at their level?
• Fourth graders learn how to break down fractions into smaller fractions with similar denominators.
Example:
4\5 = 2/5 +2/5
4/5= 1/5 +3/5.
• Your child will also learn to do additions and subtraction on fractions with the same bottom number (denominator).
Example:
2/6 +3/6= 5/6
5/6- 3/6=2/6
• Fourth graders also learn how to add and subtract mixed numbers and fractions with similar denominators.
Example:
23/7 +3 2/7= 55/7
### What skills do fourth graders learn on equivalent fractions?
In the fourth grade, your child will learn to use visual fraction models that include fraction bars and numbers lines. Students at this grade also understand how fractions can be equivalent despite the size and number of denominators and numerators that differ. For example, they learn how 2/4 and ½ are the same.
Fourth graders learn how to compare different numerators and denominators by altering the fractions to have a similar denominator. For instance, they can use visual fraction models to realize that 2\8 and ¼ are the same. They further understand that when comparing two fractions with a similar denominator, the fraction with the larger numerator is the greater one.
### Do 4th graders work with whole numbers and decimals?
Your fourth grader will learn to solve word problems that involve the multiplication of fractions with whole numbers. They will further learn how to convert fractions with denominators of 10 and 100 to decimals.
### How do fourth graders solve word problems?
Fourth graders learn to solve word problems involving addition, multiplication, division, and subtraction of intervals or units of time, money units, mass units, volume units, and distance units. These skills improve and become complex as the children get to the other graders.
Fourth graders learn that perimeter is the measurement around an object, while the area is the measurement of the flat surface of an object. Fourth graders also learn how to calculate area and perimeter and apply the same concept to real-life situations. You can help your child to improve their skills on this topic by calculating objects in their surroundings. For example, you can ask them to calculate a book’s perimeter or the carpet area in the living room.
### Do fourth graders study lines and angles?
Fourth graders learn to identify and differentiate angles and lines such as rays, line segments, perpendicular lines, right angles, and parallel lines. Then, they apply the absence or presence of angles or lines to group or categorize figures or two-dimensional shapes.
Teachers help them to understand what a line of symmetry is and where they apply. Common symmetrical shapes include equilateral triangles, isosceles, triangles, hexagons, octagons, and circles.
## Last but Not LIst
Fourth-grade math skills are crucial in your child’s life because they enable them to apply all the four mathematical operations that include addition, subtraction, multiplication, and division.
Children also get ready and learn to solve multi-step word problems that involve distances, real-life scenarios, intervals of time, masses of objects, liquid volumes, and money. It is vital to use real-life scenarios to help your 4th grader improve their math skills.
At Kids on the Yard, we understand that helping your child with math concepts can be complex. However, we are also aware that some concepts are becoming more complicated than ever.
You can contact our math tutors to help your child master new concepts and close any learning gaps in the subject.
It only takes a moment to find out more details by meeting one of our Educational Team using a Call or SMS: Toll-Free USA: 844 902 4242 or for International: +1 850 848 4242.
### Using the Form
Learn More by scheduling a complimentary 20-30 minute session with our Educational Team about our customized approach to learning solutions In-Person or Online, pricing, FREE assessment*, best-fit tutor &, etc.
### Direct Call Today
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International: +1 850 848 4242
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## 8 Jun 2017
### Practice Reasoning Questions For SBI Clerk 2017 & IBPS 2017 Exams (Machine input and output (New pattern) & Blood relation)
Practice Reasoning Questions For SBI Clerk 2017 & IBPS 2017 Exams (Machine input and output (New pattern) & Blood relation):
Dear Readers, Important Practice Reasoning Questions with explanation for SBI Clerk & IBPS Exams 2017 was given here with explanation, candidates those who are preparing for Banking and all other Competitive exams can use this practice questions.
00:00:00
Direction (Q. 1-5): Study the following information carefully and answer the given questions.
A number arrangement machine arranges two digit numbers in typical manner. The first step has been obtained by multiplying the digits in input. Multiplication has not been done in any other steps. They are obtained by applying certain logic. Each step is a resultant of previous step.
1. If each digit in the step II is halved and then added, then what will be the final sum?
4
7
3.5
8
None of these
Directions (Q. 1-5):
In step I : 1* 7 = 7; 4*2 = 8;
3* 3 = 9; 9*1 = 9;
2* 2 = 4; 1*6 = 6;
In step II : each first digits of three numbers are added and multiplied by two and each second digits also do same.
7 + 9 + 4 = 20* 2 = 40; 8 + 9 + 6 = 23*2 = 46;
In step III : divide the second digit by first digit in each two numbers.
0/4 = 0; 6/4=1.5;
In step IV : subtract the second number from first number. This is the final output.
0 – 1.5 = -1.5
1). 4/2 + 0/2 + 4/2 + 6/2 = 2 + 0 + 2 + 3 = 7
2.If the value ‘7.5’ is subtracted from the final output, then what will be the resultant value?
6
8
-9
-6
None of these
2).The final output is ‘ -1.5’ = -1.5 – 7.5 = - 9.0
3.Which of the following combination represent the first digit of the third number and second digit of the first number in step I of the given input?
(8,4)
(7,4)
(8,6)
(4,8)
(4,7)
3).The first digit of the third number in step I is ‘4’ and the second digit of the first number is ‘8’. Hence, (4, 8)
4.Which of the following represent the difference between the first digit of the second number and second digit of the first number in step II?
4
0
6
-4
None of these
4).The first digit of the second number in step II is ‘4’ and second digit of the first number in step II is ‘0’. Hence, 4 – 0 =4
5.What is the multiplication of two numbers obtained in step III?
6.0
1.6
2.0
0
None of these
5). The two numbers obtained in step III are 0 and 1.5
Hence, 0 * 1.5 = 0
6.If A + B means A is the mother of B; A - B means A is the brother B; A % B means A is the father of B and A x B means A is the sister of B, which of the following shows that P is the maternal uncle of Q?
Q - N + M x P
P + S x N – Q
P - M + N x Q
Q - S % P
None of these
6). In all the questions (-) denotes male; (+) denotes female
P - M + N x Q
P is the maternal uncle of Q.
7.If P5Q means P is the father of Q; P9Q means P is the sister of Q; P4Q means P is the brother of Q; P3Q means P is the wife of Q, which of the following means F is the mother of K?
F5M3K
F9M4N3K
F3M5N3K
F3M5K
None of these
7). F3M5K
Hence, F is the mother of K.
Directions (Q. 8-10): M is the grandson of G. D is husband of G. K is married to the son of S. T have two children of different gender. E is the daughter of K’s brother. N is brother-in-law of the son of S. G has one only one child. D is the father of N.
8.How N is related to G?
Son
Son -in- law
Grandson
Brother
Father
8). Directions (Q. 8-10):
N is the son of G
9. If Q is married to N, then how is Q related to M?
Father
Mother
Sister
Grandmother
Cannot be determined
9).
Q is the mother of M
10.If U is the son of T, then how is U related to N?
Brother
Uncle
Cousin
Brother-in-law
10).
U is the brother-in-law of N
|
# What is 1% of 296?
Learn how to calculate what 1% of 296 is without using a fancy percentage calculator. In this brief, step by step guide, we'll show you exactly how to work out the percentage of literally any number. It'll take you less than one minute to do and give you all the knowledge you need to work out future percentages manually - all by yourself!
In a rush and just need to know the answer? 1% of 296 is 2.96.
What is % of
## Calculate 1% of 296
There are many reasons why you might want to calculate the percentage of a number. In every day life we use percentages all the time (sometimes without even knowing it) to understand the world around us.
For example: let's say you're in a store looking to get a sweet Black Friday sale. The new PlayStation you want is currently 35% off the usual price of \$899. How much are you going to pay? By knowing how to quickly calculate percentages, you can figure that answer out.
Another might be that when you actually come to pay for that PlayStation, you have to pay sales tax. Now you need to add a percentage to the amount you have. Again, we can learn how to do all of these calculations in our head.
Want to quickly learn or show students how to calculate percentage increase and decrease play this very quick and fun video now!
Of course, you can alwayd ask Siri to help you out or type the question into a Google search (which is probably how you found this little page on the internet) but you can actually work out the percentage of a number very easily without having to resort to technology. For our little example here, we want to know what 1% of 296 is. Let's do some quick math together and work it out.
The easiest way to work this out is to take the number we want to work out the percentage of, which is 296, and divide it by 100. Why do we do this? Well "percent" really means one part in every hundred, which is what we want to calculate.
296 / 100 = 2.96
Once we have that value, all we need to do to work out the answer is multiply it by our first number, 1:
2.96 x 1 = 2.96
That's literally all there is to it. You can use this method to work out and calculate the percentage of any number. If you're feeling extra fancy and want to know what the formula is for calculating the percentage, let's take a look at that too:
P * X = Y
• P = the number we got by dividing 296 by 100
• X = the percentage we want to work out, 1
• Y = the final answer, 2.96
As you can see, percentages are often very easy math problems to solve. Sometimes if you are dealing with decimal points and large numbers, you may reach for the calculator to help you out, but at least it saves you the embarassment of being out in public and yelling "Hey Siri, what is 1% of 296?" into your phone.
Head back to the percentage calculator to work out any more calculations you need to make or be brave and give it a go by hand. Hopefully this article has shown you that it's easier than you might think!
|
# When do you rationalize the denominator to calculate the limit?
Mar 10, 2015
At the risk of sounding like I'm being flippant, you rationalize the denominator when you need to and it helps.
Example 1: Evaluate: ${\lim}_{x \rightarrow 9} \frac{x}{\sqrt{x} + 5}$
The limits of the numerator and denominator are: ${\lim}_{x \rightarrow 9} x = 9$ and${\lim}_{x \rightarrow 9} \left(\sqrt{x} + 5\right) = 8$. So we can find the requested limit by using the quotient property of limits. There is no need to rationalize.
${\lim}_{x \rightarrow 9} \frac{x}{\sqrt{x} + 5} = \frac{9}{8}$
Example 2: Evaluate: ${\lim}_{x \rightarrow 4} \frac{x - 4}{\sqrt{x} - 2}$
This time the limits of the numerator and denominator are ${\lim}_{x \rightarrow 4} \left(x - 4\right) = 0$ and ${\lim}_{x \rightarrow 4} \left(\sqrt{x} - 2\right) = 0$
We cannot use the quotient rule for limits, so we'll try rationalizing the denominator to see if that helps.
(Yes, really, seriously. Try something and see if it works. There is no 'cookbook'. for solving all problems.)
$\frac{x - 4}{\sqrt{x} - 2} \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{\left(x - 4\right) \left(\sqrt{x} + 2\right)}{x - 4} = \frac{\sqrt{x} + 2}{1}$ (for $x \ne 4$).
The denominator of this new ratio does not go to $0$ as $x \rightarrow 4$, so we can use the quotient property of limits. It worked. We get:
${\lim}_{x \rightarrow 4} \frac{x - 4}{\sqrt{x} - 2} = {\lim}_{x \rightarrow 4} \frac{\sqrt{x} + 2}{1} = 4$ .
|
6 Questions
2 Views
# Equation-Solving Pro
Created by
@FavorableDiopside
Subtraction
x = 6
### What happens to an equation if the same value is added to both sides?
The equation is simplified
### What is the difference between an equation and an expression?
<p>An equation involves an equal sign and represents a statement that two expressions are equal, while an expression is a mathematical phrase that can contain variables, constants, and operators.</p> Signup and view all the answers
### What is the order of operations in solving an equation?
<p>The order of operations in solving an equation is to first simplify any expressions within parentheses, then perform any multiplication or division in the order they appear from left to right, and finally perform any addition or subtraction in the order they appear from left to right.</p> Signup and view all the answers
### How do you check if a solution to an equation is valid?
<p>To check if a solution to an equation is valid, substitute the value of the variable into the original equation and simplify. If the resulting expression is a true statement, then the solution is valid.</p> Signup and view all the answers
## Study Notes
### Solving Equations
• The first step in solving an equation is to simplify the equation by combining like terms.
• The solution to the equation 3x + 5 = 14 is x = 3, which can be found by subtracting 5 from both sides and then dividing both sides by 3.
### Equation Properties
• If the same value is added to both sides of an equation, the equation remains balanced and its solution is not changed.
### Equations vs. Expressions
• An equation is a statement that says two expressions are equal, whereas an expression is a collection of numbers, variables, and operations combined to produce a value.
### Order of Operations
• The order of operations in solving an equation is to follow the PEMDAS rule: Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction.
### Validating Solutions
• To check if a solution to an equation is valid, plug the solution back into the original equation and verify that the equation is true.
## Studying That Suits You
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## Description
Are you struggling with solving equations? Take this quiz to test your knowledge on the first step in solving an equation, finding solutions to equations, and the effects of adding values to both sides of an equation. Sharpen your math skills and become an equation-solving pro!
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# How do you verify sin 2x = (2 tan x)/(1+ tan² x)?
Apr 23, 2016
Modifying only the right hand side of the equation, we should first notice that the fraction's denominator is a form of the Pythagorean Identity:
${\sin}^{2} x + {\cos}^{2} x = 1$
$\implies \text{ } {\sin}^{2} \frac{x}{\cos} ^ 2 x + {\cos}^{2} \frac{x}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x$
=>" "color(blue)(barul(|color(white)(a/a)color(black)(tan^2x+1=sec^2x)color(white)(a/a)|
Thus, we see that we can rewrite the fraction:
$\frac{2 \tan x}{1 + {\tan}^{2} x} = \frac{2 \tan x}{{\sec}^{2} x}$
Recalling that " "color(green)(barul(|color(white)(a/a)color(black)(sec^2x=1/cos^2x)color(white)(a/a)|$\text{ }$, we see that
$\frac{2 \tan x}{\sec} ^ 2 x = \frac{2 \tan x}{\frac{1}{\cos} ^ 2 x}$
Now, instead of dividing by $\frac{1}{\cos} ^ 2 x$, note that this is the same as multiplying by ${\cos}^{2} \frac{x}{1}$.
$\frac{2 \tan x}{\frac{1}{\cos} ^ 2 x} = 2 \tan x \cdot {\cos}^{2} x$
Rewrite $2 \tan x$ using the identity: " "color(purple)(barul(|color(white)(a/a)color(black)(tanx=sinx/cosx)color(white)(a/a)|
$2 \tan x \cdot {\cos}^{2} x = \frac{2 \sin x}{\cos} x \cdot {\cos}^{2} x$
The $\cos x$ in the denominator will cancel with one of the $\cos x$ terms in the numerator:
$\frac{2 \sin x}{\cos} x \cdot {\cos}^{2} x = \frac{2 \sin x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\cos x}}}} \cdot {\cos}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} x = 2 \sin x \cos x$
This is the identity for $\sin 2 x$, which is the left hand side of the original equation:
color(brown)(barul(|color(white)(a/a)color(black)(2sinxcosx=sin2x)color(white)(a/a)|
Thus, the equation is verified.
Jul 8, 2016
$L H S = \sin 2 x = \frac{2 \sin x \cos x}{1}$
$= \frac{2 \sin x \cos x}{{\cos}^{2} x + {\sin}^{2} x}$
color(red)("Dividing both numerator and dinominator by "cos^2x
$= \frac{\frac{2 \sin x \cos x}{\cos} ^ 2 x}{{\cos}^{2} \frac{x}{\cos} ^ 2 x + {\sin}^{2} \frac{x}{\cos} ^ 2 x}$
$= \frac{2 \tan x}{1 + {\tan}^{2} x} = R H S$
|
LOGARITHM TABLES - CHARACTERISTIC AND MANTISSA, SOLVED EXAMPLES, EXERCISE
Please study Logarithms before Logarithm Tables,
if you have not already done so.
It is a prerequisite here.
We make use of various Logarithms'
Formulas in solving problems here.
To determine the logarithm of a number, we need
to determine (i) Characteristic and (ii) Mantissa.
Mantissa is found from Tables.
Besides reading tables for the values, we need to
have the knowledge of interpreting the values.
We get that knowledge here.
We study Characteristic and Mantissa here.
We apply the knowledge to solve problems.
Problems are given for practice in exercise with answers.
Practice makes one perfect.
This is especially true for learning of solving
Algebra Problems (Math Problems).
So, take the exercises seriously and
practice solving the problems.
Characteristic and Mantissa of Logarithms : Logarithm Tables
The logarithm to the base 10 is called
common logarithm, as it is commonly used.
Note : In this page we deal with only common logarithms.
Characteristic of Logarithms of numbers > 1 : Logarithm Tables
Observe the following :
• log10 1 = 0;
• log10 10 = 1;
• log10 100 = log10 102 = 2;
• log10 1000 = log10 103 = 3;
• log10 10000 = log10 104 = 4;
We know that
• single digit number lies between 1 and 10.
• two digit number lies between 10 and 100.
• three digit number lies between 100 and 1000.
• four digit number lies between 1000 and 10000.
We can infer the following from the above.
• For single digit number, log(to the base 10) value
lies between 0 and 1
• For two digit number, log(to the base 10) value
lies between 1 and 2.
• For three digit number, log(to the base 10) value
lies between 2 and 3.
• For four digit number, log(to the base 10) value
lies between 3 and 4.
(or) for base 10,
• log value of a single digit number = 0 + a positive fraction(<1).
• log value of a two digit number = 1 + a positive fraction(<1).
• log value of a three digit number = 2 + a positive fraction(<1).
• log value of a four digit number = 3 + a positive fraction(<1).
The integral part of the log(to the base 10) value of a number is called its Characteristic.
The (positive) fractional part of the log(to the base 10) value of a number is called its Mantissa.
Examples :
• log 3 = 0.4771 Here Characteristic is 0 and Mantissa is .4771
• log 16 = 1.2040 Here Characteristic is 1 and Mantissa is .2040
• Characteristic of a single digit number = 0.
• Characteristic of a two digit number = 1.
• Characteristic of a three digit number = 2.
• Characteristic of a four digit number = 3.
• ...................................so on.
• Characteristic of a n digit number = (n - 1).
Characteristic of a n digit number = (n - 1)
Great Deals on School & Homeschool Curriculum Books
Characteristic of Logarithms of +ve numbers < 1 : Logarithm Tables
Observe the following :
• log10 1 = 0.
• log10 0.1 = log10 10-1 = -1 is written as 1.
• log10 0.01 = log10 10-2 = -2 is written as 2.
• log10 0.001 = log10 10-3 = -3 is written as 3.
• log10 0.0001 = log10 10-4 = -4 is written as 4.
We can infer the following from the above.
[Note: when we speak of log value of a number, that number is positive.]
log value of a number(< 1) with
• no zeros after decimal (before a significant digit)
= 1 + a positive fraction(<1)
• 1 zero after decimal (before a significant digit)
= 2 + a positive fraction(<1)
• 2 zeros after decimal (before a significant digit)
= 3 + a positive fraction(<1)
• 3 zeros after decimal (before a significant digit)
= 4 + a positive fraction(<1)
Characteristic of a number(< 1) with
• no zeros after decimal (before a significant digit) = 1
• 1 zero after decimal (before a significant digit) = 2
• 2 zeros after decimal (before a significant digit) = 3
• 3 zeros after decimal (before a significant digit) = 4
• ...........................................................so on.
• n zeros after decimal (before a significant digit) = ( n + 1 )
Characteristic of a number(< 1) with n zeros after decimal (before a significant digit) = ( n + 1 )
Ponts about Characteristic and Mantissa of Logarithms : Logarithm Tables
The following points may be noted.
• The Characteristic is always an integer.
It may be +ve or -ve or zero.
• The Mantissa is never negative. It is always less than one.
• To find the Characteristic of a number first see
whether it is > 1 or < 1.
If it is > 1, Characteristic = number of digits in the integral part - 1.
If it is < 1, Characteristic = ( n + 1 ),
where n is number of zeros after decimal
(before a significant digit).
• Mantissa is found using Logarithm Tables.
While finding Mantissa, the decimal point has no significance.
Example: log10 4567 = 3.6587
Here, the Mantissa = .6587 is found from Logarithm Tables.
Characteristic = 3 is arrived at, based on the number of digits in 4567.
4567 has 4 digits. ∴ its Characteristic = 4 - 1 = 3.
Mantissa of 4567 = .6587 = Mantissa of 456.7 = Mantissa of 45.67
= Mantissa of 4.567 = Mantissa of .4567 = Mantissa of .04567
= Mantissa of .004567 = Mantissa of .0004567 etc.
Thus while seeing Logarithm Tables for Mantissa,
we have to consider the digits only and not the decimal point.
The location of decimal point plays a significant role,
in arriving at the Characteristic.
Consider the numbers above for which
the Mantissa is the same ( = .6587 )
• Characteristic of 4567 = 3 ∴ log10 4567 = 3 + .6587 = 3.6587
• Characteristic of 456.7 = 2 ∴ log10 456.7 = 2 + .6587 = 2.6587
• Characteristic of 45.67 = 1 ∴ log10 45.67 = 1 + .6587 = 1.6587
• Characteristic of 4.567 = 0 ∴ log10 4.567 = 0 + .6587 = 0.6587
• Characteristic of .4567 = 1
∴log10 4567 = 1 + .6587 = -1 + .6587 = -0.3413
• Characteristic of .04567 = 2
∴ log10 4567 = 2 + .6587 = -2 + .6587 = -1.3413
• Characteristic of .004567 = 3
∴ log10 4567 = 3 + .6587 = -3 + .6587 = -2.3413
• Characteristic of .0004567 = 4
∴ log10 4567 = 4 + .6587 = -4 + .6587 = -3.3413
Here is a collection of proven tips,
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The secrets will help you absorb, digest
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If you apply what you read from the above
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Know more about the
Examples in Logarithm Tables
For Solved Examples and Exercise problems
on logarithm tables which make use of the
knowledge covered here, go to
Progressive Learning of Math : logarithm tables
Recently, I have found a series of math curricula
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This series is very different and advantageous
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|
# Cosh|Definition & Meaning
## Definition
The hyperbolic cosine function, or cosh(x), is one of the various hyperbolic functions. Its evaluation involves Euler’s number e. For an input x, the hyperbolic cosine’s output is the sum of e to the power x and e to the power minus x, divided by 2. Unlike the trigonometric function cosine, which is based on measurements of circles, the hyperbolic cosine is based on measurements of hyperbolas.
In applied mathematics, hyperbolic functions are equivalents of regular trigonometric functions but expressed by means of the hyperbola instead of a circle
Figure 1 – Graph of cosh x and sech x
Here we are going to look into the cosh (x), which is defined by the formula given below:
$\cosh(x) = \dfrac{e^x + e^{(-x)}}{2}$
In applied mathematics, the exponential functions are used in a variety of even and odd combinations that deserve to be given their own names. These hyperbolic functions are similar in many ways to the trigonometric functions because of their association with the hyperbola in the form of a circle.
Because of this, they are communally termed hyperbolic functions and separately termed hyperbolic sine, hyperbolic cosine, and so forth.
## What Are Trigonometric Functions?
Looking at a right-angled triangle, the side that is opposite to the right angle is known as the hypotenuse, and if we choose one other angle and name it θ theta, then the other remaining sides are usually labeled as the opposite side, and the adjacent side (the side next to θ). The ratio of any two sides, like adjacent/ hypotenuse, will constitute a function with a special name, the cosine of theta or cos (θ).
Cos(θ) = $\dfrac{\textsf{adjacent}}{\textsf{hypotenuse}}$
Similarly, considering the other possible ratios, we can construct the other trigonometric functions:
Sin(θ) = $\dfrac{\textsf{opposite}}{\textsf{hypotenuse}}$
Tan(θ) = $\dfrac{\textsf{opposite}}{\textsf{adjacent}}$
The cosine can be defined using a visual representation by considering a circle of diameter 2, having center O at the origin inside the (x, y) plane. Consider point A on the edge of the circle having the coordinates (1, 0). Now the distance from that point to the origin is called as OA and has a length of 1 unit.
Let there be another point P on the line of the circle’s arbitrary coordinates (a, b). The angle between the two lengths, OA and OP, is defined by θ. Here theta is the radian representation.
Figure 2 – A circle of unit radius
In simpler terms, the cosine of θ is described as the projection of P on the x-axis.
## What Are Hyperbolic Functions?
In applied mathematics, hyperbolic functions are equivalents to regular trigonometric functions but expressed by means of the hyperbola instead of a circle. While the points (cos t, sin t) are derived from a circle having a unit radius, the functions (cosh t, sinh t) develop the rightmost half of the unit hyperbola.
Their derivatives are also similar to the derivatives of trigonometric functions, except the sign. We know the derivatives of sin(t) and cos(t) are, respectively, cos(t) and –sin(t), while the derivatives of sinh(t) and cosh(t) are cosh(t) and +sinh(t).
Hyperbolic functions arise where one needs to compute angles and distances in hyperbolic geometry. They are also used in various solutions of linear differential equations, cubic functions, and Laplace’s equation. Laplace’s equations are significant in numerous fields of physics, which include electromagnetic concepts, heat fusion, etc.
All the hyperbolic functions acquire a real value termed a hyperbolic angle. The hyperbolic angle is double the size of the area of the hyperbolic segment it is constructed in.
## Defining Hyperbolic Cosine
We can draft the graph for hyperbolic cosine by utilizing the given information of the exponential $e^x$ and $e^{−x}$. Initially, let us compute the value of cosh 0 by taking x = 0. Thus, the exponents become $e^x$ = 1 and $e^{−x}$ = 1. Thus:
$\cosh(x) = \dfrac{e^0 + e^{(-0)}}{2}$
= (1 + 1) / 2 = 1
Rewriting the cosh x to determine the change when the value of x gets large:
$\cosh(x) = \dfrac{e^{x/2} + e^{x/2}}{2}$
The graph below shows the behavior of the exponential when x gets large.
Figure 3 – Graph of $e^x$ and $e^{-x}$
Looking at the graph, the positive exponential increases quickly when x gets larger, but the negative exponential decreases rapidly with the change in x. So as the x gets larger, the hyperbolic cosine gets closer and closer to the positive exponential, i.e., $e^{x/2}$:
$\cosh x = e^{x/2}$ for larger values of x
The graph of cosh x will eventually remain above the graph of the $e^{x/2}$ due to the fact that the second part of the function, despite being very small, will always be greater than zero. As the value of x gets bigger and bigger, the difference in both graphs gets minor and minor.
Now, Let’s say that the value of x becomes more and more negative, the positive exponential will fall drastically, but the negative exponential increases rapidly with the negative change in x. So as the x gets smaller and smaller, the hyperbolic cosine gets closer and closer to the negative exponential, i.e., $e^{-x/2}$:
$\cosh x = e^{-x/2}$ for large negative values of x
Even if the x is negative, the graph of cosh x will ultimately remain above the graph of the $e^{-x/2}$ because the first part of the sum, regardless of being very small, will always be greater than zero. As the value of x gets smaller and smaller, the difference in both graphs gets negligible.
Thus, we can construct the graph for cosh x. One thing to observe is that the graph is symmetrical about the y-axis because cosh x = cosh (-x), as described above.
Figure 4 – Graph showing the hyperbolic cosine
## Important Identity of cosh
Cosh, along with sinh, have various identities that look analogous to identities for the regular trigonometric functions of cos and sin, with a slight change in the signs. The identity looks like this:
$\cosh^{2} x-\sinh^{2} x = 1$
We can recall the trigonometric identity similar to the one above $\cos^2 x + \sin^2 x = 1$, with the plus sign changing to minus.
The identity given above can effortlessly be drawn from the fundamental definitions as follows:
$\cosh^{2} x-\sinh^{2} x = \dfrac{e^x + e^{(-x)}}{2}-\dfrac{e^x – e^{(-x)}}{2}$
$= \dfrac{e^{2x} + e^{-2x}}{4}-\dfrac{e^{2x}-e^{(-2x)}}{4}$
$= \frac{1}{2}-\frac{-1}{2}$
$= 1$
Another important identity to look for is:
Sinh(x + y) = sinh x cosh y + cosh x sinh y
## Osborn’s Rule
Osborn’s rule implies that the trigonometric identities can be converted to their corresponding hyperbolic identities by replacing all occurrences of sins and cosine with their counter hyperbolic sine and cosine. Also, if them occurs a product of two sines, the corresponding hyperbolic product will have a negative sign.
For instance, $\cos(2A) = 2\cos^2(A)-1$ will be changed into $\cosh2A = 2\cosh^2(A)-1$ whereas,
$\cos(2A) = 1-\sin^2(A) \Rightarrow \cosh(2A) = 1+\sinh^2(A)$
In order to apply Osborn’s rule on other trigonometric identities, they must first be narrowed down to sines and cosines.
## Solved Example
For the equation 5 cosh x + 3 sinh x = 4, solve for x.
### Solution
Using the basic definitions of cosh and sinh, the given equation can be written as:
$5 \dfrac{e^x + e^{(-x)}}{2} + 3 \dfrac{e^x + e^{(-x)}}{2} = 4$
Simplifying it gives us the following:
$4e^x + e^{-x} = 4$
Multiplying by $e^x$ gives:
$4e^{2x} + 1 = 4e^x$
Rewriting this forms a quadratic equation:
$4e^{2x}-4e^x + 1 = 0$
Solving the quadratic equation gives:
$e^x = \dfrac{1}{8} (4 \pm \sqrt{0})$
$e^x = \dfrac{1}{2}$
Hence:
$x = \ln\left(\frac{1}{2}\right)$
$= \ln 2^{-1}$
$= -\ln2$
All images/mathematical drawings were created with GeoGebra.
|
# Volume of a Cube Cuboid Cylinder
@ : Home > Free Printable Worksheets > Math > Volume of a Cube Cuboid Cylinder
## Volume of a Cube Cuboid Cylinder
### Cuboid
1) volume of cuboid = l × b × h. Since l × b is the area of its base we can also say that,
Volume of cuboid = area of the base × height.
### Cube
1) The cube is a special case of a cuboid, where l = b = h.
Hence, volume of cube = l × l × l = l3
### Cylinder
1) Volume of cylinder = area of base × height = πr2 × h = πr2h.
### Volume and Capacity
There is not much difference between these two words.
(a) Volume refers to the amount of space occupied by an object.
(b) Capacity refers to the quantity that a container holds.
### Example 1:
Find the height of a cuboid whose volume is 275 cu. cm and base area is 25 sq. cm .
### Solution:
Volume of a cuboid = Base area × Height
Hence height of the cuboid = Volume of cuboid/Base area = 275/25 = 11 cm
Height of the cuboid is 11 cm.
### Example 2:
A godown is in the form of a cuboid of measures 60 m × 40 m × 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 cu. m ?
### Solution:
Volume of one box = 0.8 cu. m
Volume of godown = 60 × 40 × 30 = 72000 cu. m
Number of boxes that can be stored in the godown = Volume of the godown / Volume of one box
=60 × 40 × 30 / 0.8 = 90,000
Hence the number of cuboidal boxes that can be stored in the godown is 90,000.
### Example 3:
A rectangular paper of width 14 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder.
### Solution:
A cylinder is formed by rolling a rectangle about its width. Hence the width of the paper becomes height and radius of the cylinder is 20 cm.
Height of the cylinder = h = 14 cm
Radius = r = 20 cm
Volume of the cylinder = V = π2h = π × 20 × 20 × 14 = 17600 cu. cm
Hence, the volume of the cylinder is 17600 cu. cm.
### Example 4:
A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.
### Solution:
Length of the paper becomes the perimeter of the base of the cylinder and width becomes height.
Let radius of the cylinder = r and height = h
Perimeter of the base of the cylinder = 2πr = 11 or r = 7/4 cm
Volume of the cylinder = V = πr2h = π × (7/4)×(7/4)× 4 cu. cm = 38.5 cu. cm.
Hence the volume of the cylinder is 38.5 cu. cm.
|
New Zealand
Level 7 - NCEA Level 2
# Derivative of a Sum (ax^n)
Lesson
Just like we saw in derivatives of a sum of the type x^n,
So we can see that the derivative of the sum is the same as the sums of the derivatives.
Sum of Derivatives
Derivative of sum is equal to the sum of the derivatives.
If $f(x)=g(x)\pm h(x)$f(x)=g(x)±h(x) then $f'(x)=g'(x)\pm h'(x)$f(x)=g(x)±h(x)
This means we can apply the power rule to individual terms.
And this applies to any function, whether it be the $x^n$xn we saw before, or $ax^n$axn that we are exploring now.
#### Examples
Find the derivative of the following,
a) $f(x)=4x^2+3x+2$f(x)=4x2+3x+2, then $f'(x)=8x+3$f(x)=8x+3 (remember that the derivative of a constant term is $0$0)
b) $f(x)=3x^3-3x^2$f(x)=3x33x2, then $f'(x)=9x^2-6x$f(x)=9x26x
c) $f(x)=6x^{-3}-2x+\sqrt{x}$f(x)=6x32x+x. Firstly we need to turn the $\sqrt{x}$x into a power. $\sqrt{x}=x^{\frac{1}{2}}$x=x12 So $f(x)=6x^{-3}-2x+x^{\frac{1}{2}}$f(x)=6x32x+x12 and so then the derivative $f'(x)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$f(x)=18x42+12x12
#### Worked Examples:
##### question 1
Differentiate $y=7x^2-9x+8$y=7x29x+8.
##### question 2
Differentiate $y=\frac{24}{x^5}-\frac{30}{x^4}$y=24x530x4.
##### question 3
Find the derivative of $y=x^3\sqrt{x}+3x^5$y=x3x+3x5.
### Outcomes
#### M7-10
Apply differentiation and anti-differentiation techniques to polynomials
#### 91262
Apply calculus methods in solving problems
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----------------
Some of you geeks probably have a slide-rule in your head — there I go dating myself again (and we all know that dating yourself isn’t sexy), I should have said “math processor” or “calculator”. Anyway, I am personally much better than average at math, but I still can’t do some of the bigger problems in my head.
Via Mercola, I found this list of 10 Easy Arithmetic Tricks to save you the trouble of getting out a calculator for solving some math problems.
The “11 Times Trick” is one I hadn’t thought of before, but it makes perfect sense. To multiply any two-digit number by 11, you add the digits together and put the result between them, adding any excess digit to the first digit. It makes sense because multiplying it out long-hand you get (for any two digit number mn):
m n
X 1 1
———-
m n
m n
————
m(m+n)n
Some of the tricks are a bit obvious, like the multiply/divide by 5 and how to compute a 15% tip (a lot of places are starting to expect 20% now anyway). But all the tricks are good exercises for understanding how numbers work.
Here are a few more tricks that I learned somewhere along the way for determining if a decimal integer is a multiple of:
1. Duh. Of course it is.
2. Is the last digit even? Another easy one.
3. Add up the digits in the number. If the result is a multiple of 3, then so is the number. If the sum of the digits is too big for you to know whether it’s a multiple of 3 or not, recurse. Add up its digits and see if that’s a multiple of 3.
4. Take the number formed from the last two digits. If that’s a multiple of 4, so is the number. Why? Because 100 is a multiple of 4.
5. Is the last digit a 0 or a 5? Too easy.
6. Is the number a multiple of 2 and 3?
7. Double the last digit and subtract it from the number formed by the remaining digits. If that result is a multiple of 7, so is the original number. If you can’t tell, recurse. For instance, take 357. Double 7 to get 14, subtract that from 35 and you get 21. Since 21 is a multiple of 7, so is 357.
8. Take the number formed from the last three digits. If that’s a multiple of 8, so is the number — because 1000 is a multiple of 8.
9. Add up the digits in the number. If the result is a multiple of 9, then so is the number. Again, you can recurse if you’re not sure. It’s no accident that this rule for 9 is the same as the rule for 3.
10. Does it end with a 0? Now we’re back in elementary school.
11. Add up all the odd digits to get one number, then add up all the even digits to get a second number. If the difference between them is a multiple of 11 (zero included), then so is the number. If you think about it, this is really the “11 Times Trick” reversed. Let’s take 26719 as an example. 2 + 7 + 9 = 18, 6 + 1 = 7, 18 – 7 = 11, so 26719 is a multiple of 11.
12. Is the number a multiple of 3 and 4?
Looking at this list, the relationship between 3, 6, and 9 is obvious — as is the relationship between 4 and 8. The unique characteristics of 7 and 11 are intriguing, don’t you think?
What other arithmetic tricks do you know?
----------------------------
## 28 Responses to Simple tricks for doing arithmetic in your head
1. A quick way to calculate the square of a number that ends with a 5.
Example: 65 x 65
1. Add 1 to the first 6, equals 7.
2. Multiply this 7 with the second 6, equals 42.
3. The last 2 digits will always be 25.
4. So the answer is 4225.
• Thanks, Awangku. That’s actually #2 in the post linked to above.
• Ah yess… my mistake. Stupidly, I did not click the link to the 10 tricks. Okay, I’m embarrassed now. Hahahahaha…
• Not a problem. Thanks for reading and commenting!
2. A quick way to calculate the square of a number that ends with a 5.
Example: 65 x 65
1. Add 1 to the first 6, equals 7.
2. Multiply this 7 with the second 6, equals 42.
3. The last 2 digits will always be 25.
4. So the answer is 4225.
• Thanks, Awangku. That's actually #2 in the post linked to above.
• Ah yess… my mistake. Stupidly, I did not click the link to the 10 tricks. Okay, I'm embarrassed now. Hahahahaha…
• Not a problem. Thanks for reading and commenting!
3. Pingback: links for 2008-02-20 -- Chip’s Quips
• Good stuff, Stu! Thanks.
• Good stuff, Stu! Thanks.
4. Anu haridas says:
Most of them know it b4:)try t includ someth new..thankyou
• Naren says:
Anu
Are you the same anu who did a course in AMES training centre back in 2000 in Auckland?.IF you are then reply back – naren_2010@hotmail.com
5. Anu haridas says:
Most of them know it b4:)try t includ someth new..thankyou
6. Sean Crosser says:
35
x 67
245
210
2345
7. Sean Crosser says:
35
x 67
245
210
2345
8. Sean Crosser says:
Sorry, it turned out different than I typed in. It uses a variant of the first one, but it's actually able to do it with more that just x11.
35 x 7 = 245, and 35 x 60 = 2100, and 2100 + 245 = 2345. Mayb a big number isn't practical for an example, but 25 x 15, or smaller number with lower digits.
9. Sean Crosser says:
Sorry, it turned out different than I typed in. It uses a variant of the first one, but it’s actually able to do it with more that just x11.
35 x 7 = 245, and 35 x 60 = 2100, and 2100 + 245 = 2345. Mayb a big number isn’t practical for an example, but 25 x 15, or smaller number with lower digits.
10. Ari D Jordon says:
Actually, there really are only three divisibility tricks. Interestingly, they work in any base, not just base 10.
For a given base B:
1) n B – 1: Take the last digit, multiply by n, and add it to remaining digits. Repeat until you can tell whether the number is divisible by n B – 1
2) B^n: If the last n digits are divisible by B^n, so is the number
3) n B + 1: Take the last digit, multiply by n, and subtract it from the remaining digits. Repeat until you can tell
In each case, if the relevant number isn't prime, the rule also applies to all of its factors. For base 10:
2, 4, 5, 8 are all cases of rule 2. n=1 for 2 and 5, 2 for 4, 3 for 8
3 and 9 are cases of rule 1 with n=1
7 and 11 (and 3) are cases of rule 3. For 11, n=1. For 7, n=2 (2 * 10 + 1 = 21, factors are 3 and 7)
11. Ari D Jordon says:
Actually, there really are only three divisibility tricks. Interestingly, they work in any base, not just base 10.
For a given base B:
1) n B – 1: Take the last digit, multiply by n, and add it to remaining digits. Repeat until you can tell whether the number is divisible by n B – 1
2) B^n: If the last n digits are divisible by B^n, so is the number
3) n B + 1: Take the last digit, multiply by n, and subtract it from the remaining digits. Repeat until you can tell
In each case, if the relevant number isn’t prime, the rule also applies to all of its factors. For base 10:
2, 4, 5, 8 are all cases of rule 2. n=1 for 2 and 5, 2 for 4, 3 for 8
3 and 9 are cases of rule 1 with n=1
7 and 11 (and 3) are cases of rule 3. For 11, n=1. For 7, n=2 (2 * 10 + 1 = 21, factors are 3 and 7)
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## EQUATIONS INVOLVING FRACTIONS (RATIONAL EQUATIONS)
Note:
• A rational equation is an equation where at least one denominator contains a variable.
• When a denominator contains a variable, there is a restriction on the domain. The variable cannot take on any number that would cause any denominator to be zero.
• The first step is solving a rational equation is to convert the equation to an equation without denominators. This new equation may be equivalent (same solutions as the original equation) or it may not be equivalent (extraneous solutions).
• The next step is to set the equation equal to zero and solve.
• Remember that you are trying to isolate the variable.
• Depending on the problem, there are several methods available to help you solve the problem.
If you would like an in-depth review of fractions, click on Fractions.
Solve for x in the following equation.
Example 1:
Recall that you cannot divide by zero. Therefore, we must eliminate any values of x that will cause the denominator to have a value of zero. We determine these values by setting
These values of x are not real numbers. What this means is that there is no value of x that will cause the denominator to be zero. This means that there are no restrictions on the values of x.
Simplify the equation by subtracting 5 from both sides of the equation.
Multiply both sides of the equation by
The only way a product can equal zero is if at least one of the factors equals zero.
This is not a real number: therefore, Therefore the only real solution is x=5.
Check the solution x=5 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.
• Left Side:
• Right Side: .
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 5 for x, then x=5 is a solution.
You can also check your answer by graphing
.
(formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; that will be the real solution. Note that the graph crosses the x-axis at one place, 5.
This means that there is one real solution and the solution is x=5.
If you would like to work another example, click on Example
If you would like to test yourself by working some problems similar to this example, click on Problem
If you would like to go back to the equation table of contents, click on Contents
[Algebra] [Trigonometry]
[Geometry] [Differential Equations]
[Calculus] [Complex Variables] [Matrix Algebra]
Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.
Author: Nancy Marcus
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# How do you evaluate [ (\frac { 1} { 3} + i \frac { 7} { 3} ) + ( 4+ i \frac { 1} { 3} ) ] - ( - \frac { 4} { 3} + i )?
Jan 30, 2018
$\left(\frac{17}{3}\right) + i \left(\frac{5}{3}\right)$
#### Explanation:
$\frac{1}{3} + 4 + \frac{4}{3} = \frac{17}{3}$
$\frac{7}{3} + \frac{1}{3} - 1 = \frac{5}{3}$
Hence, the sum is
$R e \left(z\right) + I m \left(z\right)$
where $z$ is a complex number.
Jan 30, 2018
Use the distributive property to distribute the implied -1 through the parenthesis of the last term, then remove the braces and parenthesis and combine like terms.
#### Explanation:
Given: $\left[\left(\frac{1}{3} + i \frac{7}{3}\right) + \left(4 + i \frac{1}{3}\right)\right] - \left(- \frac{4}{3} + i\right)$
Distribute the implied -1:
$\left[\left(\frac{1}{3} + i \frac{7}{3}\right) + \left(4 + i \frac{1}{3}\right)\right] + \left(\frac{4}{3} - i\right)$
Remove the braces and parenthesis:
$\frac{1}{3} + i \frac{7}{3} + 4 + i \frac{1}{3} + \frac{4}{3} - i$
Combine like terms:
$\frac{1}{3} + 4 + \frac{4}{3} + i \left(\frac{7}{3} + \frac{1}{3} - 1\right)$
$\frac{17}{3} + i \frac{5}{3}$
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## Introduction
The advancements in technology are making big advancements in education as well. In mathematics, most people find it difficult to calculate values and solve difficult sums. So this has been noted by scientists and they have introduced different calculators for this purpose. Similarly, the Pythagorean Theorem Calculator has also been designed in digital form. In this blog, we will deeply explore all the knowledge about the Pythagorean theorem and Pythagorean calculator.
### What is Pythagorean Theorem?
The Greek mathematician Pythagoras made enormous contributions to the science of mathematics. In addition, he was a brilliant statesman and philosopher. One theorem in particular, known as Pythagoras’ Theorem, is very significant to his work. According to this formula, the square of the third side, known as the hypotenuse, of a right-angled triangle is equal to the square of the other two sides linked to the right angle.
The Pythagorean equation, which is an equation that relates the lengths of the sides a, b, and the hypotenuse c, can be used to express the theorem.
### a2+b2=c2
The theorem may be extended to higher-dimensional spaces, non-Euclidean spaces, non-right triangle objects, and objects that aren’t even triangles but rather n-dimensional solids. Outside of mathematics, the Pythagorean theorem has drawn attention as a representation of mathematical complexity, mystery, or intellectual might; examples of this may be found in a wide range of popular culture, including plays, musicals, books, stamps, and cartoons.
### What Is A Right Angle Triangle?
Three sides, three vertices, and three internal angles make up a triangle, which is a polygon. A triangle having two sharp angles and one measuring angle is called a right triangle.
Let’s suppose there is a right triangle ABC with vertices, sides, and right angle
Since the total of the three interior angles in a right triangle equals, the right angle is always the larger one. The longer side is the one that faces away from the larger angle. Because of this, these sides of right triangles are referred to as hypotenuse sides.
Trigonometry and right triangles are connected in that the measurements of a right triangle’s sides are used to define the trigonometric ratios.
### Pythagorean Calculator
This Pythagorean Theorem calculator will calculate the lengths of any missing sides of a right triangle, provided you know the lengths of its other two sides. This includes calculating the hypotenuse. The hypotenuse angle of a right triangle is the side opposite the right angle, and is the longest side. This aspect can be found using the hypotenuse formula, another term for the Pythagorean theorem when solving for the hypotenuse. Remember that a right triangle is a triangle with an angle of 90 degrees. The remaining two angles must also sum to 90 degrees, since the sum of the measures of the angles of any triangle is 180.
### How to Use a Pythagorean Calculator?
Here’s how to use Pythagorean theorem:
Input the two lengths that you have into the formula. For example, suppose you know one length and the hypotenuse. You want to find the length of the other leg .
After the values are put into the formula, we will get a value.
Square each term and combine them.
Take the square root of both sides of the equation to get. Go ahead and check it with an online Pythagorean theorem calculator! You will easily find the solution
### Real-Life Examples Of Using Pythagorean Theorem
1. In two dimensions, the Pythagorean Theorem is helpful for navigation. It may be utilized along with two lengths to determine the shortest path. The diagonal of the triangle will be the shortest line that connects the two legs, which are the distances to the north and west. Air navigation may be done using the same techniques.
2. Painting a Wall: In order to paint high structures, painters utilize ladders, and they frequently rely on Pythagoras’ theorem to assist them in finishing their tasks. To carefully position the base away from the wall and prevent it from toppling over, the painter must measure the height of the ladder.
3. What Size TV Should I Buy? The diagonal of a television is always used to indicate size. If a television is designated as 32 inches, its dimensions are really determined by the diagonal or hypotenuse value.
4. Choosing the Appropriate Computer Size: The size of the monitor is always expressed as a diagonal dimension.
### Advantages Of Using Pythagorean Theorem Calculator
1. Quick and Precise Computations
Pythagorean calculators provide answers rapidly and precisely, saving time and lowering the chance of human mistake when calculations are done by hand.
1. Easy Usability
They are appropriate for a range of users, including professionals and students, as they are simple to use and frequently demand information.
1. Best Teaching and Learning Aids
These are helpful materials that assist students better grasp and apply the Pythagorean Theorem by helping them conceptualize the topic.
1. Really Saves Time
When there is a rush, using a calculator to calculate equations rapidly is a more effective method than working by hand.
1. Consistent Results
Make sure that the findings are dependable and consistent to prevent mistakes that may occur from repetitive computations.
### Conclusion
In conclusion, the Pythagorean theorem calculator is really a big advantage for those who have a lot of work related to calculations. Because mostly it becomes difficult to calculate long calculations easily. Now everybody wants to have their work finished in a very short time. And there are a lot of chances of error in human calculation. Pythagoras, a Greek mathematician, made significant contributions to the study of mathematics. He was also a talented statesman and philosopher. One specific theorem, known as Pythagoras’ Theorem, is crucial to his work. The square of a right-angled triangle’s third side, known as the hypotenuse, is equal to the square of the other two sides connected to the right angle, according to this formula. But now online calculators have made it so much easier to solve the pythagorean theorem. Whenever you want to calculate the values, just open the online pythagorean theorem calculator and solve it in seconds.
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Right Angle – Definition, Formula, Examples, and FAQs
Right Angle – Definition, Formula, Examples, and FAQs
Table of Contents
Right Angle
A right angle is an angle of 90 degrees. It is formed by two lines or rays that meet at right angles to each other. A right angle is also the name for the unit of measurement that is equal to one-fourth of a circle.
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Let : ™s Unfold the Different Properties of Right Angle
Triangle
A right angle triangle is a triangle in which one of the angles is a right angle. The other two angles are acute angles. Right angle triangles have a lot of interesting properties that can be unfolded.
The first property of a right angle triangle is that the sum of the squares of the two shorter sides is equal to the square of the length of the longest side. This is known as the Pythagorean theorem. The theorem can be demonstrated using the following diagram.
Right Angle Definition
A right angle is an angle that is exactly 90 degrees.
Right Angle Triangle
A right angle triangle is a triangle in which one of the angles is a right angle. The other two angles are acute angles. A right angle triangle has the following properties.
The length of the longest side is called the hypotenuse. The other two sides are the short side and the long side. The short side is opposite the right angle. The long side is opposite the acute angle. The Pythagorean theorem states that the sum of the squares of the two shorter sides
Right Angle Triangle Formula
A right angle triangle is a triangle with one right angle. The right angle triangle formula is used to calculate the length of the sides of a right angle triangle. The right angle triangle formula is:
s = √(b² + c² – a²)
Right Angle Triangle Properties
A right angle triangle is a triangle with one right angle. The other angles are less than 90 degrees. Right angle triangle properties include the length of the side opposite the right angle, the length of the side adjacent to the right angle, and the length of the hypotenuse.
The length of the side opposite the right angle is the longest side of the triangle. The length of the side adjacent to the right angle is the shortest side of the triangle. The length of the hypot
Area of a Right Triangle
A right triangle has three sides: the base, the height, and the hypotenuse. The base is the side opposite the right angle, and the height is the length of the triangle’s side that is perpendicular to the base. The hypotenuse is the longest side of the triangle and is the side opposite the right angle.
The area of a right triangle is equal to one-half the product of the base and the height. This is because the area of a triangle
Right Angle Isosceles Triangle
A right angle isosceles triangle is a triangle with one right angle and two isosceles triangles. The angles in a right angle isosceles triangle are all equal, and the sides opposite the equal angles are also equal. This type of triangle is often used in math and geometry problems, because the angles and sides are easy to calculate and work with.
Right Angle Example
A right angle is an angle that is 90 degrees. It is measured using a protractor. A right angle is found by drawing a line from one corner of a square to the opposite corner. A right angle is also found by drawing a line from the midpoint of one side of a rectangle to the opposite corner.
Facts
A fact is something that is indisputable, meaning it can be proven true. Facts are often used in arguments to support a point of view. Some people believe that facts are more important than opinions, while others believe that opinions are just as important as facts.
Solved Example:
A Solution to a Problem
A solution to a problem is a method or technique that is used to resolve the problem. A solution is typically found by using a process of elimination or a trial and error approach. A solution is also often a combination of different methods or techniques.
There are many different methods or techniques that can be used to solve a problem. Some common methods or techniques include brainstorming, breaking the problem down into smaller parts, using a flowchart, using a table,
FAQs
1. What is a right angle?
A right angle is an angle that measures exactly 90 degrees.
2. How is a right angle denoted?
A right angle is usually denoted by a small square placed in the angle, or by the abbreviation “RT” or “∟”.
3. What is the sum of angles in a right triangle?
The sum of angles in a right triangle is always 180 degrees.
4. What are some common objects that have right angles?
Some common objects that have right angles include rectangles, squares, bookshelves, picture frames, and door frames.
5. What is the Pythagorean theorem?
The Pythagorean theorem is a mathematical formula that relates the sides of a right triangle. It states that the sum of the squares of the two shorter sides (a and b) is equal to the square of the hypotenuse (c), or a² + b² = c².
6. What is a 45-degree angle?
A 45-degree angle is an angle that is halfway between a right angle (90 degrees) and a straight angle (180 degrees). It measures exactly 45 degrees.
7. What is a perpendicular line?
A perpendicular line is a line that intersects another line or surface at a right angle, or 90 degrees.
8. What is a right angle triangle?
A right angle triangle is a triangle that contains a right angle, or an angle measuring 90 degrees.
9. What is a hypotenuse?
The hypotenuse is the longest side of a right triangle, opposite the right angle.
10. What is an acute angle?
An acute angle is an angle that measures less than 90 degrees.
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AP Board 7th Class Maths Solutions Chapter 8 Congruency of Triangles Ex 2
AP State Syllabus AP Board 7th Class Maths Solutions Chapter 8 Congruency of Triangles Ex 2 Textbook Questions and Answers.
AP State Syllabus 7th Class Maths Solutions 8th Lesson Congruency of Triangles Exercise 2
Question 1.
What additional information do you need to conclude that the two triangles given here under are congruent using SAS rule?
Solution:
We need
HG = TS
HJ = TR
Question 2.
The map given below shows five different villages. Village M lies exactly halfway between the two pairs of villages A and B as well as and P and Q. What is the distance between village A and village P. (Hint: check if ΔPAM ≅ ΔQBM)
Solution:
Given that
AM = MB
PM = MQ
also ∠PMA = ∠QMB
∴ By S.A.S
ΔPMA ≅ ΔQMB
Now PA = BQ
(∵ Corresponding parts of congruent triangles are equal)
∴ PA = 4 km
i.e., Distance between the villages
A and P is 4 km.
Question 3.
Look at the pairs of triangles given below. Are they congruent ? If congruent write the corresponding parts.
Solution:
AB=ST
AC = SR (given)
∠A = ∠S
∴ ΔABC ≅ ΔSTR (S.A.S)
Also ∠A = ∠S, ∠B = ∠T, ∠C = ∠R
From the figure,
FO = RO
OQ = OS
∠POQ = ∠ROS
∴ ΔPOQ ≅ ΔROS (SAS)
Also ∠P = ∠R, ∠Q = ∠S, and PQ = RS
From the figure, IAPBoardSoIJ
WD = OR
∠W= ∠R
WO = RD
∴ ΔDWO ≅ ΔORD
Also ∠EO = ∠OE; ∠WDO = ∠ROD; ∠WOD = ∠RDO
Here the two triangles ΔABC and ΔCDA are not congruent.
Question 4.
Which corresponding sides do we need to know to prove that the triangles are congruent using the SAS criterion?
Solution:
i) We need to know that AB = QR.
ii) We need to know that AD = AB.
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Probability involving binomial coefficient
We toss a coin $2n$ times: we want to know the probability $P(\#heads=\#tails=n)$. The number of possible arrangements is $2^{2n}$, the result is $P(\#heads=\#tails=n)=\frac{\binom{2n}{n}}{2^{2n}}$. I understand that$\binom{n}{k}$ is the number of possible set of size $k$ inside the set $n$. But can you explain why $\binom{2n}{n}$ is the number of favorable arrangements in this case?
And a similar problem, a set of $4n$ cards contains $2n$ black ones and $2n$ $red$ ones. We draw $2n$ cars and we want to know $P(\#red=\#black=n)$. Here we get $P(\#red=\#black=n)=\frac{\binom{2n}{n}*\binom{2n}{n}}{\binom{4n}{2n}}$; the question is the same as before why the above term is a binomial coefficient and why this times we multiply it?
$2n \choose n$ is the number of ways in which you can choose a set of $n$ objects from a set of $2n$ distinguishable objects.
This corresponds to the number of ways in which you get exactly $n$ heads (and therefore $n$ tails) with $2n$ tosses, since if you write down the outcomes of the tosses as a sequence (e.g. $HTTTHTH...$) then you are basically picking $n$ slots out of the $2n$ numbered the numbered slots. That is, in the example I just gave the heads occupy slots $1,5, 7...$, but another possible sequence would be THHHTTT..., and now they occupy slots $2,3,4$. So, how many possible sequences are there with exactly $n$ heads? As many different sets of $n$ numbers you can pick out of the numbers $1$ through $2n$, so that is $2n \choose n$
For the second question, first notice that the denominator (i.e. the 'below' term) is ${4n} \choose {2n}$, which corresponds to picking $2n$ cards out of $4n$ cards (all cards are different from each other).
Now, you need $n$ black cards, but since there are $2n$ black cards in the deck, you needs to pick $n$ out of $2n$ distinguishable black cards, so you have $2n \choose n$ options there. Likewise, you have $2n \choose n$ ways of picking $n$ red cards from the $2n$ red cards. And finally, you multiple, because for each 'choice' of black cards, you have all of the options of red cards.
A simpler example may demonstrate this latter principle. Say you have 4 red cards (label them $R1, R2, R3, R4$) and 4 black cards ($B1, B2, B3, B4$), and say you pick just $1$ red card and $1$ black card. Well, for each red card, I can add each of the $4$ black cards, so I have $4$ options for $R1$: $R1+B1$, $R1+B2$,$R1+B3$, and $R1+B4$. But you have those same $4$ options for the black card with any other red card, e,g, for $R2$ we have: $R2+B1$, $R2+B2$,$R2+B3$, and $R2+B4$. Another $4$ for $R3$, and another $4$ for $R4$, and you end up with 16 possible combinations of $1$ red card and $1$ black card, which is of course exactly $4*4$
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# In the Following Determine Rational Numbers A And B: (4 + Sqrt2)/(2 + Sqrt2) = N - Sqrtb - Mathematics
In the following determine rational numbers a and b:
(4 + sqrt2)/(2 + sqrt2) = n - sqrtb
#### Solution
We know that rationalization factor for 2 + sqrt2 is 2 - sqrt2. We will multiply numerator and denominator of the given expression (4 + sqrt2)/(2 + sqrt2) by 2 - sqrt2 to get
(4 + sqrt2)/(2 + sqrt2) xx (2 - sqrt2)/(2 - sqrt2) = (4 xx 2 - 4 xx sqrt2 + 2 xx sqrt2 - (sqrt2)^2)/((2)^2 - (sqrt2)^2)
= (8 - 4sqrt2 + 2sqrt2 - 2)/(4 - 2)
= (6 - 2sqrt2)/2
= 3 - sqrt2
On equating rational and irrational terms, we get
a - sqrtb = 3 - sqrt2
Hence we get a = 3, b = 2
Concept: Operations on Real Numbers
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Mathematics for Class 9
Chapter 3 Rationalisation
Exercise 3.2 | Q 6.2 | Page 14
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## Simplifying and Evaluating Expressions With Integers That Use all Four Operations
### Learning Outcomes
• Use the order of operations to simplify expressions that involve integer multiplication, division, addition, and subtraction
• Evaluate integer expressions using the order of operations
Now we’ll simplify expressions that use all four operations–addition, subtraction, multiplication, and division–with integers. Remember to follow the order of operations.
### example
$\text{Simplify: }7\left(-2\right)+4\left(-7\right)-6$
Solution:
We use the order of operations. Multiply first and then add and subtract from left to right.
$7\left(-2\right)+4\left(-7\right)-6$ Multiply first. $-14+\left(-28\right)-6$ Add. $-42 - 6$ Subtract. $-48$
### try it
Watch the following video to see another example of how to use the order of operations to simplify an expression that contains integers.
In our next example we will simplify expressions with integers that also contain exponents.
### example
Simplify:
1. ${\left(-2\right)}^{4}$
2. ${-2}^{4}$
Now you try it.
### example
$\text{Simplify: }12 - 3\left(9 - 12\right)$
### example
Simplify: $8\left(-9\right)\div {\left(-2\right)}^{3}$
### example
$\text{Simplify:}-30\div 2+\left(-3\right)\left(-7\right)$
### try it
In the following video we show more examples of how to evaluate expressions with integers using the order of operations.
### Evaluate Variable Expressions with Integers
Now we can evaluate expressions that include multiplication and division with integers. Remember that to evaluate an expression, substitute the numbers in place of the variables, and then simplify.
### example
$\text{Evaluate }2{x}^{2}-3x+8\text{ when }x=-4$
### example
$\text{Evaluate }3x+4y - 6\text{ when }x=-1\text{ and }y=2$.
### try it
In the following video we show more examples of how to substitute integers into variable expressions.
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Question Video: Creating Exponential Equations with One Variable to Solve Problems Mathematics
Michael invests \$200 in an account that pays an annual interest rate of 5%, compounded monthly. Write an equation he could use to work out π, the value of his investment in 3 yearsβ time.
02:36
Video Transcript
Michael invests 200 dollars in an account that pays an annual interest rate of five percent, compounded monthly. Write an equation he could use to work out π, the value of his investment in three yearsβ time.
In order to answer this question, we need to use the general formula for compound interest for a value of an investment compounded π times per year. This value π is equal to π multiplied by one plus π over 100 divided by π all raised to the power of π¦ multiplied by π. π is the principal value or initial investment. π is the interest rate as a percentage. π is the number of compound periods per year. And finally, π¦ is the number of years.
In this question, we want to write an equation for π, where 200 dollars is the initial investment. The interest rate is five percent, and this interest is compounded monthly. We are asked to calculate the value of the investment in three yearsβ time. This gives us an equation π is equal to 200 multiplied by one plus five over 100 divided by 12 all raised to the power of three multiplied by 12. Five over 100 or five one hundredths divided by 12 can be rewritten as five over 1200. Multiplying the exponents three and 12 gives us 36.
π is, therefore, equal to 200 multiplied by one plus five over 1200 all raised to the power of 36. This is an equation that Michael could use to work out the value of π. Whilst we are not asked to in this question, we could type this into the calculator, giving us a value of π of 232.2944 and so on. To two decimal places, this is equal to 232.29.
The value of Michaelβs investment after three years would be 232 dollars and 29 cents.
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In this video, we will be learning how to solve for x (or another variable) in complex algebraic equations using inverse operations. After you finish this lesson, view all of our Pre-Algebra and Algebra lessons and practice problems.
## Example of Solving a Multi-Step Algebraic Equation
$5(x+4) +2 = 27$
$5(x+4) +2 -2 = 27 -2 \leftarrow$ Subtract 2 from both sides
$\frac {5(x+4)}{5} = \frac{25}{5} \leftarrow$ Divide by 5 on both sides
$(x+4) -4 =5 -4\leftarrow$ Subtract 4 on both sides
$x=1$
### Example 1
$2(x+3) = 38$
First, distribute $2$ to the terms inside the parenthesis
$2(x+3) = 38$ $2x+6 = 38$
Then, subtract $6$ from both sides
$2x+6-6 = 38-6$
$2x=32 \leftarrow$ divide $2$ from both sides
$\dfrac{2x}{2}=\dfrac{32}{2}$
Now, we have:
$x=16$
### Example 2
$3(4x-2)=42$
First, distribute $3$ to the terms inside the parenthesis
$3(4x-2)=42$ $12x-6=42$
Then, add $6$ from both sides
$12x-6+6=42+6$
$12x=48 \leftarrow$ divide $12$ from both sides
$\dfrac{12x}{12}=\dfrac{48}{12}$
Now, we have:
$x=4$
Another way of doing this problem:
$5(x+4) +2 = 27 \leftarrow$ Distribute the 5 to the x and the 4
$5x+20 +2 = 27 \leftarrow$ Simplify using addition
$5x+22 -22 = 27 -22 \leftarrow$ Subtract 22 from both sides
$\frac{5x}{5}=\frac{5}{5} \leftarrow$Divide by 5 on each side
$x=1$
## Video-Lesson Transcript
Let’s get into solving complex algebraic equations. This involves more than one operation.
To review, order of operations or PEMDAS, we have $3(2) + 4$.
PEMDAS is an acronym which stands for
Parenthesis
Exponents
Multiplication
Division
Subtraction
Let’s evaluate $3(2) + 4$.
So $3 \times 2 + 4 = 6 + 4$ we’ll have $= 10$.
In solving algebraic equations, we don’t have all whole numbers. Instead, we have variables in it.
So, we may have $3x + 4 = 10$.
From our example, we already know that $x = 2$.
But let’s try solving this algebraically.
We just have to do few steps to solve this.
1. Simplify both sides of the equations, if possible.
2. If there’s $x$ terms on both sides, we have to get all the $x$ terms on one side. You can have it on the left side or on the right side, whichever you prefer.
3. Reverse PEMDAS. We’re going to do the order of operations backward using inverse operations.
4. Our goal is to isolate the variable $x$.
Going back to $3x + 4 = 10$, let’s do the steps above.
1. Simplify – This is the simplest it can get.
2. All $x$ term on one side – There’s just one $x$ and it’s on the left.
3. Now, let’s do the reverse PEMDAS using inverse operations.
Let’s subtract $4$ on both sides of the equation.
$3x + 4 - 4 = 10 - 4$
We’ll have $3x = 6$.
4. Isolate $x$
Here we have to divide both sides by $3$.
$\dfrac{3x}{3} = \dfrac{6}{3}$
And we’ll have $x = 2$.
Let’s do another example.
We have $\dfrac{x}{4} - 3 = 1$
So, let’s start off by adding $3$ on both sides of the equation.
$\dfrac{x}{4} - 3 + 3 = 1 + 3$
We’ll come up with $\dfrac{x}{4} = 4$
Then, we multiply $4$ on both sides.
$\dfrac{x}{4} \times 4 = 4 \times4$
The answer is $x = 16$
Let’s have one more example. I’ll show you how to solve this using two different ways.
We have $5(x + 4) + 2 = 27$
First method of solving is this:
Let’s subtract $2$ on both sides of the equation.
$5(x + 4) + 2 - 2 = 27 - 2$
We’ll have $5(x + 4) = 25$
Then, we have to divide both sides by $5$
$\dfrac{5}{5} (x + 4) = \dfrac{25}{5}$
We’ll come up with $x + 4 = 5$
Then, to isolate $x$ we have to subtract $4$ on both sides
$x + 4 - 4 = 5 - 4$
Our final answer is $x = 1$
So now let’s move on to the second method of solving the same equation.
$5(x + 4) + 2 = 27$
The second method is to simplify the equation as much as we can.
Let’s start off by distributing $5$ into the equation in the parenthesis – $x + 4$
So let’s do it!
$5 \times x$ and $5 \times 4$
We’ll have $5x + 20 + 2 = 27$
Now, we can combine like terms to simplify further
$5x + 22 = 27$
Then we have to do the reverse PEMDAS.
Let’s subtract $22$ on both sides
$5x + 22 - 22 = 27 - 22$
We’ll come up with $5x = 5$
Now, let’s do the inverse of multiplication which is division.
Divide both sides by $5$
$\dfrac{5x}{5} = \dfrac{5}{5}$
Our final answer is $x = 1$
Both methods gave us the same answer $x = 1$.
To sum up, no matter how complex our algebraic equation is, we can do reverse PEMDAS or inverse order of operations to isolate the $x$.
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# Pumpkin Problems: Estimates to Actuals
8 teachers like this lesson
Print Lesson
## Objective
SWBAT estimate, round, and subtract using a place value strategy of breaking apart numbers into tens and ones.
#### Big Idea
Critical to mathematical success is developing the sophistication of studentsâ strategies simultaneously with the teaching and application of key mathematical knowledge.
## Warm Up
10 minutes
First, I review the process of mental math subtraction of breaking apart numbers using place value with a word problem using both rounded numbers and actual numbers.
An example of this would be 53 - 28 = ______. Rounded, this would be 50 - 30 = ____.
Rounding also supports the students confidence in subtraction by providing friendly numbers to subtract.
Because I know the students need to practice their skills with subtraction, using place value, I have taught the students to separate the tens and ones to subtract. For example, if the subtraction sentence is 53 - 28 = _____, the students would subtract 53 - 20 = 33, and then 33 - 8 = 25.
The reason I use this strategy is I find students will rely on the steps of the "standard" algorithm, rather than thinking about changes in the value of the numbers. I have also found my students are still switching the order of ones digits to put the larger number first. For example, in the problem 53 - 28, when using the standard algorithm some students will think, "Well, I can't take away 8 from 3, so I'll do 8 take away 3. That way it works."
I chose to have the students use individual whiteboards and markers for this type of problem because I can see their work easily and they can make corrections quickly. This also gives me a cue about who may need additional help throughout the lesson, and who will benefit from more challenge.
## Introduction to Today's Lesson
10 minutes
Because some of my students expressed they had not ever carved a pumpkin before, I knew I needed to provide some background information about how pumpkins grow. I explained how pumpkins grow from seeds about the size of the fingernail of your thumb. I displayed a pumpkin in front of the students and asked them to estimate how many seeds would fit inside of the pumpkin? I asked them to estimate the weight of the pumpkin by lifting it. Because the students are growing up in an urban area with few gardening experiences and some in poverty, I decided to provide some additional information about the pumpkin and also have them hold the pumpkin. This knowledge of students' background and potential experiences was needed to support the progression of the lesson.
I feel it is so important children are given the opportunity to examine and explore some of these common seasonal items thoroughly. They all could name a pumpkin, but many of my students had never experienced the slimy, stringy insides of a pumpkin. This lesson combines different math skills and allows students to work in groups allowing for different levels of abilities to work together and all students contributing to the group work.
After some discussion, I explained the students would be making estimates about various types of information about the pumpkin, and then they would measure using tools including rulers, tape measures, and a scale. The students at my school had previous experiences using measuring tools in second grade with building simple machines and measuring lengths in a science unit. This activity provided students with an opportunity to become familiar again with the measuring tools we will use throughout the school year in math and science.
Because the Common Core standards require students to work with metric measurement in third grade, I displayed a textbook and demonstrated its weight using kilograms. I then weighed the same book in US standard customary measurements in pounds. I explained how the students will need to use mental math subtraction to compare their estimates to the actual measurements.
## Guided Instruction
10 minutes
First, I asked the students to show me the gestures we have established for some measurement vocabulary. For weight we pretend we are holding something really heavy, height the students extend their hands high above their heads, circumference is drawing a circle out in front of their body (as the equator is around the earth), and for diameter the students draw a horizontal line in front of himself/herself.
Depending on the number of pumpkins for the activity, students work in either groups, partners, or individually. My students were working in groups of about 4 students each.
The student groups record their estimates about the weight, height, circumference, diameter, and the number of seeds on the recording sheets together. Because I want the students to estimate, I did not have the measuring tools on their desks at this time.
Students can use the textbook from the warm up section to help make their estimations for weight and height.
If needed, demonstrate how to use the scale, which side of the measuring tape, and ruler to use for metric measurement (MP5 - Use appropriate tools strategically). It is helpful to discuss the differences between an inch and a centimeter, by comparing them.
The vocabulary we are using, weight, height, circumference, diameter, were words introduced/used earlier in the year. If this vocabulary is new to your students, I suggest you couple these words with "kidspeak" versions, such as "how tall" with height. Using the words, coupled, when speaking and in writing, will assist students to learn this vocabulary within the context of this lesson.
I have my students use their math journal to record new mathematical terms, definitions, and create illustrations or examples. My experience is that math journals are more meaningful than posting vocabulary on a wall, because the students create their own resource that is meaningful to them.
The size of pumpkin does not matter. Because of safety requirements, adults cut the tops of the pumpkins prior to beginning the lesson.
## Explore
25 minutes
Weight and Measurement
Students working in groups, partners, or individually:
1. Estimate the weight of your pumpkin and record
2. Weigh the pumpkins on the scale and record
3. Determine difference between actual weight and estimate using mental math strategies.
4. Repeat the steps of estimating, measuring, and subtracting with mental math strategies for the circumference and diameter of the pumpkin.
Seeds
1. Open the pumpkin, observe, describe, and record information about your pumpkin on the recording pages.
2. Scoop out one handful of seeds from the pumpkin
3. Count the number of seeds.
4. Estimate and record how many handfuls/seeds will be removed from the pumpkin.
5. Scoop out all remaining seeds and collect into a pile.
6. Group seeds into piles for skip counting/multiplying
7. Determine the difference between the actual number of seeds and the estimated number of seeds.
Questions:
How much did you estimate your pumpkin to weigh? Was that in pounds or kilograms? About how many pounds are in one kilogram? What do you think is the reason for measuring in metric? Why is it important to measure in metric? What was the difference between your estimate of seeds to the actual number of seeds. How do you describe the inside of your pumpkin? How did you group your seeds for counting? Why did you choose that way? Can you think of another way to group the seeds? Do you think you could use the scale to measure the seeds? How would you use the scale to measure the seeds?
## Wrap Up (Closer)
15 minutes
To finish the activity, the students were given a graph to compare and find the difference between the estimates and actual measurements from their pumpkins. Because we have used bar graphs in the class before and compared numbers this was a familiar activity to my students.
Prior to completing this section of the activity, the students cleaned up and the parent helpers had said good bye to the students. Having the parents/assistants leave before this part of the lesson is important because I want the students to independently apply their math strategies.
The objective of the lesson is to find the difference between two numbers and subtract using a place value strategy. The groups are given the task of writing subtraction sentences to find the difference between each estimate and actual weight.
First, students identify the larger of the two numbers by circling it. Next, each team draws a box around the smaller of the two numbers. Because I am asking them to compare, and because each group could have either an actual number or an estimate number as the larger number, I provide a model for the number sentence: circle - box = ________.
Each group chooses a graph and subtraction sentence to explain and present to the classroom. Students are asked to explain how they subtracted, and the steps of breaking apart the number into their place value (tens, ones). For each step described, we take peer questions. If there are no questions, then I ask the questions. This helps to guide the students with framing questions.
The cleaned out pumpkins are sent home at the end of the day with each of the students. We do not carve jack-o-lanterns for reasons of safety, and religious beliefs of some of my students. It may be an activity to consider if appropriate, and when adult helpers are available.
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Informative line
# Numerical Expression & Numerical Equation
Jillian's grandfather asks her a question.
At the start of the month, there were total 115 employees in a company. 40 employees were in management department and rest were in human resource and finance departments. If during the month, the employees in management department got reduced to half, while the employees in human resource and finance departments got doubled, calculate the total employees in the company at the end of the month.
Jillian solved it as follows:
$$40 \div 2 + 75 × 2$$
$$= 20 + 75 × 2$$
$$= 95 × 2$$
$$= 190$$
To determine the correct answer, we need to know about the order of operations.
• For evaluation, we deal with numerical expressions and equations.
• But, what is the difference between them? Let's try to understand these.
Numerical expression
• A numerical expression is a mathematical sentence which has only numbers and one or more operation symbols in it.
• It does not have an equals (=) sign but can be simplified and / or evaluated.
• For example:- $$11 + 9 × 2 + 7$$
Numerical equation
• A numerical equation is a mathematical sentence which tells us that two things are equal.
• To show the equality, it has an equals sign " = ".
• For example:- $$2 + 7 = 9$$
It is an equation which represents that $$2 + 7$$ is equal to $$9$$.
#### Which one of the following is a numerical expression?
A $$25 × 4 = 100$$
B $$25 × 4 = 200 \div 2$$
C $$30 0-200 \div 2 + 10$$
D $$11+12 = 23$$
×
A numerical expression has only numbers and one or more operation symbols.
It does not have an equals sign ( = ), but it can be evaluated.
Options (A), (B), and (D) have equals sign ( = ) which represent that these are equations.
Thus, options (A), (B), and (D) are incorrect.
In option (C), there are only numbers and operations.
Thus, option (C) is correct.
### Which one of the following is a numerical expression?
A
$$25 × 4 = 100$$
.
B
$$25 × 4 = 200 \div 2$$
C
$$30 0-200 \div 2 + 10$$
D
$$11+12 = 23$$
Option C is Correct
# MDAS Rule
• The order of operation is a way of evaluating mathematical statements.
• It is an essential concept to be learnt so that a mathematical statement can be read and solved in the same way by everyone.
• Consider the following expression: $$2 × 8 + 5 - 4 \div 2$$
There are many different ways to evaluate this. Some of them are shown below:
(1) $$2 × 8 + 5 - 4 \div 2$$
$$= 16 + 5 - 4 \div 2$$
$$= 21 - 4 \div 2$$
$$= 17 \div 2$$
$$= 17 /2$$
(2) $$2 × 8 + 5 -4 \div 2$$
$$= 2 × 13- 4 \div 2$$
$$=26 - 4 \div 2$$
$$=22 \div 2$$
$$= 11$$
(3) $$2 × 8 + 5 - 4 \div 2$$
$$= 2 × 8 + 1 \div 2$$
$$= 2 × 9 \div 2$$
$$= 18 \div 2$$
$$=9$$
• It is observed that each method leads to a different solution.
• To evaluate the expression in a specific order so that each person gets the same solution, the MDAS rule is used.
MDAS (Multiplication, Division, Addition, Subtraction) rule
• The MDAS rule states that while evaluating an expression or equation, remember the following points:
• First evaluate multiplication or division and then addition or subtraction.
• MD - Multiplication or division is to be done in the order from left to right.
• AS - Addition or subtraction is to be done in the order from left to right.
• Now, we will apply the MDAS rule in our previous example.
$$2 ×8 + 5- 4 \div 2$$
(1) Multiplication is to be done
$$= 16 + 5 - 4 \div 2$$
(2) Division is to be done
$$= 16 + 5 - 2$$
(3) Addition is to be done
$$= 21 -2$$
(4) Subtraction is to be done
$$= 19$$
• By following the MDAS rule, solution will be the same at each time.
Note:- If any operation is missing, then check for the next order.
#### Choose the option that has the correct order to evaluate the following expression: $$32 \div 4 × 4 - 6 + 12$$
A $$8 × 4 - 6 + 12 \;=\;32-6 +12 = 26 +12 = 38$$
B $$8 × 4 - 6 + 12\;=\; 8×4+6=8×10=80$$
C $$32\div16 - 6 +12\; =\; 2-6 + 12 = -4+12 = 8$$
D $$32 \div 16 - 6 + 12\;=\; 32 \div 10 + 12 = 44 \div 10 = \dfrac{44}{10}$$
×
The given expression consists of all the four operations, i.e. addition, subtraction, multiplication, and division.
Thus, we will use the MDAS rule to evaluate it.
$$32 \div 4 × 4 - 6 + 12$$
(i) Since division is appearing before the multiplication on moving from left to right, so it (division) is to be done first.
$$=8×4-6+12$$
$$8×4-6+12$$
(ii) Multiplication is to be done.
$$=32-6+12$$
$$32-6+12$$
(iii) Since subtraction is appearing before the addition on moving from left to right, so it (subtraction) is to be done first.
$$=26+12$$
$$26+12$$
(iv) Addition is to be done.
$$=38$$
Hence, option (A) is correct.
### Choose the option that has the correct order to evaluate the following expression: $$32 \div 4 × 4 - 6 + 12$$
A
$$8 × 4 - 6 + 12 \;=\;32-6 +12 = 26 +12 = 38$$
.
B
$$8 × 4 - 6 + 12\;=\; 8×4+6=8×10=80$$
C
$$32\div16 - 6 +12\; =\; 2-6 + 12 = -4+12 = 8$$
D
$$32 \div 16 - 6 + 12\;=\; 32 \div 10 + 12 = 44 \div 10 = \dfrac{44}{10}$$
Option A is Correct
# Evaluation of Parentheses
## Parentheses
• Parentheses are also known as brackets.
• There can be three types of brackets, which are:
• (i) Round or curved brackets, denoted by ( )
• (ii) Curly brackets, denoted by { }
• (iii) Square brackets, denoted by [ ]
• Parentheses are a type of grouping symbols.
• They can be used to show the multiplication of numbers in one of the following ways:
$$1. \;\;\;2(6) = 12\\ 2.\;\;\;(2)(6) = 12\\ 3.\;\;\;[2](6) = 12$$
### Evaluation of parentheses
• When two or more operation symbols are used in parentheses, then they should be evaluated by using the MDAS rule (from left to right).
MDAS rule
M stands for multiplication ( × )
D stands for division ( $$\div$$ )
A stands for addition ( + )
S stands for subtraction ( – )
• Consider the following example:
$$(32-50 \;\div\;2 × 5 + 1)$$
Here, the whole expression is in parenthesis, so we shall use the MDAS rule to evaluate it.
$$(32-50 \;\div\;2 × 5 + 1)$$
(i) Division is to be done
$$= 32-25×5 +1$$
(ii) Multiplication is to be done
$$=32-125+1$$
(iii) Subtraction is to be done
$$=-93+1$$
(iv) Addition is to be done
$$=-92$$
• When one parenthesis is inside another one, then the inner one is to be evaluated first using the MDAS rule.
• Consider the following example:
$$(100- (50 × 1 + 4\; \div 2))$$
Here, two parentheses are used. So, we shall evaluate the inner parenthesis first using the MDAS rule.
$$(100- (50 × 1 + 4\; \div 2))\;\; — (1)$$
Inner parenthesis contains $$(50 × 1 + 4 \;\div 2)$$
(i) Multiplication is to be done $$= 50 + 4 \div 2$$
(ii) Division is to be done $$= 50 + 2$$
(iii) Addition is to be done $$= 52$$
• Now evaluate the outer parenthesis
$$(100\; –\; 52)$$ [From (1)]
Here, only subtraction is to be done $$= 48$$
#### Evaluate the following parentheses $$(112 \; \div (5 × 3 - 8)× 8 + 4)$$
A $$5$$
B $$132$$
C $$104$$
D $$7$$
×
$$(112 \div (5 × 3- 8 ) × 8+ 4)\;\;\;— (1)$$
Here, two parentheses are used. So, we shall evaluate the inner parenthesis first.
Inner parenthesis: $$(5 × 3 - 8)$$
(i) Multiplication is to be done $$= 15 -8$$
(ii) Now only subtraction is to be done $$= 7$$
The outer parenthesis is
$$(112 \div 7 × 8 + 4)$$ [From (1)]
(i) Division is to be done
$$=16×8+4$$
(ii) Multiplication is to be done
$$=128+4$$
(iii) Addition is to be done
$$=132$$
Hence, option (B) is correct.
### Evaluate the following parentheses $$(112 \; \div (5 × 3 - 8)× 8 + 4)$$
A
$$5$$
.
B
$$132$$
C
$$104$$
D
$$7$$
Option B is Correct
# Locating the Grouping Symbols (Parentheses)
## Use of grouping symbols (parentheses)
• Parentheses are used to separate a part of any numerical expression.
For example: $$\dfrac{22}{7} × (42 \div 7-3) - 1$$
Here, $$42 \div 7 -3$$ is separated by parentheses.
• The position of the grouping symbols (parentheses) can change the value of a numerical expression.
For example: $$5 + 8 × 2 \div 4$$
Without parentheses, the solution would be:
$$=5 + 8 × 2 \div 4$$
$$= 5 + 16 \div 4$$
$$= 5 + 4$$
$$= 9$$
Using parentheses at $$(5 + 8)$$, the solution would be:
$$= (5 + 8) × 2 \div 4$$
$$= 13 × 2 \div 4$$
$$=26 \div 4$$
$$= \dfrac{13}{2}$$
Using parentheses at $$(5 + 8 × 2)$$, the solution would be:
$$=5 + 8 × 2 \div 4$$
$$= (5 + 8 × 2 ) \div 4$$
$$= (5 + 16) \div 4$$
$$=21 \div 4$$
$$= \dfrac{21}{4}$$
• It can be observed that by altering the position of parentheses, the value of the expression also gets changed.
#### Which one of the following options represents the correct position of parentheses so that the expression, $$100 \div 5 × 6 - 2$$ gives the result as 80?
A $$(10 0\div 5)× 6 - 2$$
B $$100 \div (5 × 6 )- 2$$
C $$100 \div (5 × 6 - 2)$$
D $$100 \div 5 × (6 - 2)$$
×
Evaluating option (A)
$$(100 \div 5) × 6 -2$$
$$=20 × 6 - 2$$
$$=120 - 2$$
$$= 118$$
Here, the value is not 80.
Thus, option (A) is incorrect.
Evaluating option (B)
$$100 \div (5 × 6 ) - 2$$
$$= 100 \div (30) - 2$$
$$= 3.33- 2$$
$$=1.33$$
Here, the value is not 80.
Thus, option (B) is incorrect.
Evaluating option (C)
$$100 \div (5 × 6 - 2)$$
$$= 100 \div (30- 2)$$
$$= 100 \div 28$$
$$= 3.57$$
Here, the value is not 80.
Thus, option (C) is incorrect.
Evaluating option (D)
$$100 \div 5 × (6-2)$$
$$=100\div 5 × 4$$
$$=20×4$$
$$=80$$
Here, the value is $$80$$.
Thus, option (D) is correct.
### Which one of the following options represents the correct position of parentheses so that the expression, $$100 \div 5 × 6 - 2$$ gives the result as 80?
A
$$(10 0\div 5)× 6 - 2$$
.
B
$$100 \div (5 × 6 )- 2$$
C
$$100 \div (5 × 6 - 2)$$
D
$$100 \div 5 × (6 - 2)$$
Option D is Correct
# PEMDAS Rule
• PEMDAS rule is an order to evaluate any numerical expression containing two or more than two operations.
• PEMDAS rule is as follows:
(i) P stands for parentheses ( )
(ii) E stands for exponents ^
(iii) M stands for multiplication ×
(iv) D stands for division $$\div$$
(v) A stands for addition +
(vi) S stands for subtraction –
It is clear from above that the parentheses are to be evaluated first, next the exponents, then the multiplication or the division in the order from left to right and at last the addition or the subtraction in the order from left to right.
• In any expression, if one of the operations is missing then we will evaluate the next one.
• PEMDAS rule can be remembered easily by the sentence.
"Please Excuse My Dear Aunt Sally"
• Consider the following example: $$3^3× 1 - (8 \div 2) + 4$$
Here, all the operations are used, thus, we shall follow the PEMDAS rule to evaluate.
$$3^3 × 1 - (8 \div 2) + 4$$
(i) Here, only one pair of parenthesis is used so, we shall evaluate it first.
$$8 \div 2 = 4$$
(ii) Now, the expression is $$3^3 × 1 - 4 + 4$$
Now, the exponent is to be evaluated.
$$= 27 × 1 - 4 + 4$$
(iii) Multiplication is to be done.
$$= 27 - 4 + 4$$
(iv) Subtraction is to be done.
$$=23+4$$
(v) Addition is to be done.
$$= 27$$
Note: By not following the PEMDAS rule, we would get different and incorrect answers.
#### Evaluate the expression: $$35 + 3^2 - (3× 2 ) × 7$$
A $$3$$
B $$2$$
C $$4$$
D $$5$$
×
$$35 + 3^2 - (3 × 2) × 7$$
We shall use the PEMDAS rule to evaluate this expression.
$$35 + 3^2 - (3 × 2) × 7$$
(i) Evaluating parenthesis
$$= 35 +3^2 - 6 × 7$$
(ii) Evaluating exponents
$$= 35 + 9 - 6 × 7$$
(iii) Multiplication
$$= 35 + 9- 42$$
$$= 4 4-42$$
(v) Subtraction
$$= 2$$
Hence, option (B) is correct.
### Evaluate the expression: $$35 + 3^2 - (3× 2 ) × 7$$
A
$$3$$
.
B
$$2$$
C
$$4$$
D
$$5$$
Option B is Correct
# Formation of Numerical Expression
## Numerical expression
• A numerical expression is a mathematical sentence which has only numbers and one or more operation symbols in it.
• It does not have an equals (=) sign but can be simplified and/or evaluated.
• For example:- $$11 + 9 × 2 + 7$$
• In numerical expressions, mainly four mathematical operations $$(+, \;–, \;×, \;\div)$$, exponents and parentheses are used.
• To evaluate such expressions, we use the PEMDAS rule.
• Now, it's time to learn how to form mathematical expressions from word problems.
• We can use parentheses for our ease.
• Consider the following example:
Kevin has 3 candies and he gets 6 more from his father. Kevin's friend, Sam has 9 candies and he gives 2 to his younger brother. How many candies do both Kevin and Sam have in total?
Here, Kevin has 3 candies and he gets 6 more from his father.
Total candies with Kevin $$= (3 + 6)$$
We will put parenthesis to separate the expression.
Sam has 9 candies and he gives 2 to his younger brother.
Total candies left with Sam $$=(9-2)$$
We need to evaluate the total candies that both have, so we will add.
$$(3 + 6)+ (9- 2)$$
We will evaluate both the parentheses separately.
$$=9 + 7$$
Now the sum $$= 16$$
Thus, both have $$16$$ candies in all.
#### At the start of the month, there were total 115 employees in a company. 40 employees were in management department and rest were in human resource and finance departments. If during the month, the employees in management department got reduced to half, while the employees in human resource and finance departments got doubled, calculate the total employees in the company at the end of the month.
A $$50$$
B $$220$$
C $$170$$
D $$100$$
×
Total employees = $$115$$
Employees in management department = $$40$$
Thus, employees in human resource and finance departments = $$(115-40)$$
During the month,
Employees in management department reduced to half i.e. $$40 \div 2$$
and employees in human resource and finance departments got doubled i.e. $$(115-40)× 2$$
Thus, at the end of the month, total employees were
$$(40 \div 2) + (115- 40 )× 2$$
$$= 20 + 75 × 2$$
$$= 20 + 150$$
$$= 170$$
Hence, option (C) is correct.
### At the start of the month, there were total 115 employees in a company. 40 employees were in management department and rest were in human resource and finance departments. If during the month, the employees in management department got reduced to half, while the employees in human resource and finance departments got doubled, calculate the total employees in the company at the end of the month.
A
$$50$$
.
B
$$220$$
C
$$170$$
D
$$100$$
Option C is Correct
|
# Question Video: Computing Numerical Expressions Involving Square Roots and Negative Exponents Using Laws of Exponents Mathematics • 9th Grade
Simplify ((2√(5))⁻² × (√(2))⁻⁸)/(√(5))⁻².
02:58
### Video Transcript
Simplify two root five to the power of negative two multiplied by root two to the power of negative eight over root five to the power of negative two.
Well, if we’re going to simplify our expression, the first thing we can do is deal with this first set of parentheses here. And what we can do is split it up. So we can say that two root five to the power of negative two is the same as two to the power of negative two multiplied by root five to the power of negative two. So now, if we substitute this back into our original expression, what we’re gonna get is two to the power of negative two multiplied by root five to the power of negative two multiplied by root two to the power of negative eight all over root five to the power of negative two.
So why have we done this? Well, we’ve done this cause we can see clearly now that we can cancel because what we can do is divide both the top and bottom, so the numerator and the denominator, by root five to the power of negative two. And if we do this, what we’re left with is two to the power of negative two multiplied by one multiplied by root two to the power of negative eight over one, which we can simplify down to two to the power of negative two multiplied by root two to the power of negative eight. So what we can do now to work this out is utilize a couple of our exponent rules. The first one is that if we have root 𝑥, this is equal to 𝑥 to the power of a half. So if we have the root of any value, then this is equal to that value raised to the power of a half.
So then we can move on to another exponent rule. And that tells us that if we got 𝑥 to the power of 𝑎 to the power of 𝑏, this is equal to 𝑥 to the power of 𝑎𝑏. So we multiply our powers or exponents. So then in our example, what we’re gonna do is multiply a half by negative eight. So we’re gonna now have two to the power of negative two multiplied by two to the power of negative four. So now, we can utilize another one of our exponent rules. And that one is that if we have 𝑥 to the power of 𝑎 multiplied by 𝑥 to the power of 𝑏, then this is equal to 𝑥 to the power of 𝑎 plus 𝑏. So we add our exponents. So this is if we’ve got the same base and we multiply it. Then what we do is we add the exponents.
So in our case, what we’re gonna do is add negative two and negative four. Well, negative two add negative four is negative six. So we’ve now got two to the power of negative six. Have we finished here? Well, no. We can actually use one more of our exponent rules. And that rule is that 𝑥 to the power of negative 𝑎 is equal to one over 𝑥 to the power of 𝑎. So therefore, we can say that two to the power of negative six is the same as one over two to the power of six. Well, we can work out what two to the power of six is. Well, two to the power of six is 64. And that’s cause two multiplied by two is four, then another two is eight, multiplied by another two, 16, another two, 32, and, finally, another two, 64. So therefore, we could say our final answer is one over 64.
|
# A boat crosses a river with a velocity of 8kmh. If the resulting velocity of boat is 10kmh then the velocity of river water is
A
12.8kmh1
B
6kmh1
C
8kmh1
D
10kmh1
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the problem, we need to find the velocity of the river water given the velocity of the boat and the resultant velocity when crossing the river.1. Identify the Given Values: - Velocity of the boat (Vb) = 8 km/h - Resultant velocity (V) = 10 km/h2. Understand the Problem: - The boat is moving across the river, and the river has its own velocity (Vr). - The boat's velocity and the river's velocity are perpendicular to each other, forming a right triangle with the resultant velocity.3. Use the Pythagorean Theorem: - According to the Pythagorean theorem, we can relate the velocities as follows: V2=Vb2+Vr2 - Here, V is the resultant velocity, Vb is the velocity of the boat, and Vr is the velocity of the river.4. Substitute the Known Values: - Substitute V = 10 km/h and Vb = 8 km/h into the equation: (10)2=(8)2+Vr2 - This simplifies to: 100=64+Vr25. Solve for Vr: - Rearranging the equation gives: Vr2=100−64 Vr2=36 - Taking the square root of both sides: Vr=√36=6 km/h6. Conclusion: - The velocity of the river water (Vr) is 6 km/h.Final Answer:The velocity of the river water is 6 km/h.---
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## BEATCALC: Beat the Calculator!
Back to Calculation Tips & Tricks
From B. Lee Clay
Squaring Numbers
Multiplying Numbers
Dividing Numbers
Subtracting Numbers
Finding Percents
Calculation
Practice Exercises
Full List
### Multiplying two selected 3-digit numbers(middle digit 4)
1. Select a 3-digit number with a middle digit of 4 (last digit not zero).
2. Choose a multiplier with the same first two digits, whose third digit sums to 10 with the third digit of the first 3-digit number.
3. The last three digits will be 0 and the product of the third digits: _ _ _ 0 X X.
4. The third digit from the right will be 9 times the first digit + 2 (keep the carry): _ _ X _ _.
5. The first two digits will be the square of the first digit plus the carry: X X _ _ _ _.
As you determine the digits in the answer from right to left, repeat them to yourself at each step until you have the whole answer.
#### Example:
1. If the first number is 541, choose 549 as the second number (same first digits, third digits add to 10).
2. Last three digits: 0 and the product of the third digits: 1 × 9 = 9: _ _ 0 0 9
3. Next digit: 9 times the first digit + 2: 9 × 5 = 45, 45 + 2 = 47 (keep carry 4): _ _ 7 _ _ _
4. First two digits: square the first and carry: 5 × 5 = 25, 25 + 4 = 29: 2 9 _ _ _ _
5. So 541 × 549 = 297009.
See the pattern?
1. If the first number is 344, choose 346 as the second number (same first digits, third digits add to 10).
2. Last three digits: 0 and the product of the third digits: 4 × 6 = 24: _ _ 0 2 4
3. Next digit: 9 times the first digit + 2: 9 × 3 = 27, 27 + 2 = 29 (keep carry 2): _ _ 9 _ _ _
4. First two digits: square the first and carry: 3 × 3 = 9, 9 + 2 = 11: 1 1 _ _ _ _
5. So 344 × 346 = 119024.
|
# 6th Grade STAAR Math FREE Sample Practice Questions
If you are looking for sample practice questions to prepare your student for the 6th Grade STAAR Math test, you are in the right place. Our research team has collected a set of common questions in the 6th Grade STAAR Math test and provided it to you in this article. These questions focus on the important and common topics in the 6th Grade STAAR Math test exam. By avoiding unnecessary questions, we help the test takers to purposefully review the test resources and summarize them as best they can in the remaining time.
You can also follow some of the related links at the bottom of this post to help your student learn more about the 6th Grade STAAR Math test.
## 10 Sample 6th Grade STAAR Math Practice Questions
1- If the area of the following trapezoid is equal to A, which equation represents $$x$$?
$$\img {https://appmanager.effortlessmath.com/public/images/questions/131313444444444444444444444444444.JPG }$$
A. $$x = \frac{13}{A}$$
B. $$x = \frac{A}{13}$$
C. $$x=A+13$$
D. $$x=A-13$$
2- By what factor did the number below change from the first to the fourth number?
$$8, 104, 1352, 17576$$
A. 13
B. 96
C. 1456
D. 17568
3- 170 is equal to …
A. $$-20-(3×10)+(6×40)$$
B. $$((\frac{15}{8})×72 )+ (\frac{125}{5})$$
C. $$((\frac{30}{4} + \frac{15}{2})×8) – \frac{11}{2} + \frac{222}{4}$$
D. $$\frac{481}{6} + \frac{121}{3}+50$$
4- The distance between the two cities is 3,768 feet. What is the distance between the two cities in yards?
A. 1,256 yd
B. 11,304 yd
C. 45,216 yd
D. 3,768 yd
5- Mr. Jones saves $3,400 out of his monthly family income of$74,800. What fractional part of his income does Mr. Jones save?
A. $$\frac{1}{22}$$
B. $$\frac{1}{11}$$
C. $$\frac{3}{25}$$
D. $$\frac{2}{15}$$
6- What is the lowest common multiple of 12 and 20?
A. 60
B. 40
C. 20
D. 12
7- Based on the table below, which expression represents any value of f in terms of its corresponding value of $$x$$?
$$\img {https://appmanager.effortlessmath.com/public/images/questions/28282828288888888888888888888888888888888.JPG }$$
A. $$f=2x-\frac{3}{10}$$
B. $$f=x+\frac{3}{10}$$
C. $$f=2x+2 \frac{2}{5}$$
D. $$2x+\frac{3}{10}$$
8- Solve: 96 kg = … ?
A. 96 mg
B. 9,600 mg
C. 960,000 mg
D. 96,000,000 mg
9- Calculate the approximate area of the following circle? (the diameter is 25)
$$\img {https://appmanager.effortlessmath.com/public/images/questions/3030303000000000000000000000000000.JPG }$$
A. 78
B. 491
C. 157
D. 1963
10- The following graph shows the mark of six students in mathematics. What is the mean (average) of the marks?
$$\img {https://appmanager.effortlessmath.com/public/images/questions/313131313131111111111111111111111111111111111111.JPG }$$
A. 13
B. 13.5
C. 14
D. 1.5
## Best 6th Grade STAAR Math Exercise Resource for 2022
Original price was: $16.99.Current price is:$11.99.
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## Answers:
1- B
The area of the trapezoid is: area$$= \frac{(base \space 1+base \space 2)}{2}×$$height$$= (\frac{10 + 16}{2})x = A$$
$$→13x = A→x = \frac{A}{13}$$
2- A
$$\frac{104}{8}=13, \frac{1352}{104}=13, \frac{17576}{1352}=13$$
Therefore, the factor is 13
3- C
Simplify each option provided.
A. $$-20-(3×10)+(6×40)=-20-30+240=190$$
B. $$(\frac{15}{8})×72 + (\frac{125}{5}) =135+25=160$$
C. $$((\frac{30}{4} + \frac{15}{2})×8) – \frac{11}{2} + \frac{222}{4} = ((\frac{30 + 30}{4})×8)- \frac{11}{2}+ \frac{111}{2}=(\frac{60}{4})×8) + \frac{100}{2}= 120 + 50 = 170$$
this is the answer
D. $$\frac{481}{6} + \frac{121}{3}+50= \frac{481+242}{6}+50=120.5+50=170.5$$
4- A
1 yard $$= 3$$ feet
Therefore, $$3,768 \space ft × \frac{1 \space yd }{3 \space ft}=1,256 \space yd$$
5- A
3,400 out of 74,800 equals to $$\frac{3,400}{74,800}=\frac{17}{374}=\frac{1}{22}$$
6- A
Prime factorizing of $$20=2×2×5$$
Prime factorizing of $$12=2×2×3$$
LCM$$=2×2×3×5=60$$
7- C
Plugin the value of $$x$$ into the function f. First, plug-in 3.1 for$$x$$.
A. $$f=2x-\frac{3}{10}=2(3.1)-\frac{3}{10}=5.9≠8.6$$
B. $$f=x+\frac{3}{10}=3.1+\frac{3}{10}=3.4≠10.8$$
C. $$f=2x+2 \frac{2}{5}=2(3.1)+2 \frac{2}{5}=6.2+2.4=8.6$$
This is correct!
Plug in other values of $$x. (x=4.2)$$
$$f=2x+2\frac{2}{5} =2(4.2)+2.4=10.8$$
This one is also correct.
$$x=5.9$$
$$f=2x+2 \frac{2}{5}=2(5.9)+2.4=14.2$$
This one works too!
D. $$2x+\frac{3}{10}=2(3.1)+\frac{3}{10}=6.5≠8.6$$
8- D
1 kg$$= 1000$$ g and 1 g $$= 1000$$ mg
$$96$$ kg$$= 96 × 1000$$ g $$= 96 × 1000 × 1000 = 96,000,000$$ mg
9- B
The diameter of a circle is twice the radius. Radius of the circle is $$\frac{25}{2}$$.
Area of a circle = $$πr^2=π(\frac{25}{2})^2=156.25π=156.25×3.14=490.625≅491$$
10- B
Average (mean) $$=\frac{sum \space of \space terms}{number \space of \space terms}= \frac{10+11+15+14+15+17+12.5}{7}=13.5$$
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# Lesson 8
Dividamos para multiplicar fracciones no unitarias
## Warm-up: Verdadero o falso: Una fracción por un número entero (10 minutes)
### Narrative
The purpose of this True or False is to elicit the strategies and insights students have for multiplying fractions by whole numbers. Students do not need to find the value of any of the expressions but rather can reason about properties of operations and the relationship between multiplication and division. In this lesson, they will see some ways to find the value of an expression like $$\frac{2}{3} \times 6$$.
### Launch
• Display one statement.
• “Hagan una señal cuando sepan si la afirmación es verdadera o no, y puedan explicar cómo lo saben” // “Give me a signal when you know whether the statement is true and can explain how you know.”
• 1 minute: quiet think time
### Activity
• Share and record answers and strategy.
• Repeat with each statement.
### Student Facing
Decide si cada afirmación es verdadera o falsa. Prepárate para explicar tu razonamiento.
• $$2 \times \left(\frac{1}{3} \times 6\right) = \frac{2}{3} \times 6$$
• $$2 \times \left(\frac{1}{3} \times 6\right) = 2 \times (6 \div 3)$$
• $$\frac{2}{3} \times 6 = 2 \times \left(\frac{1}{4} \times6 \right)$$
### Activity Synthesis
• “¿Cómo pueden explicar que $$\frac{2}{3} \times 6 = 2 \times (\frac{1}{4} \times6)$$ es falso sin encontrar el valor de ambos lados?” // “How can you explain why $$\frac{2}{3} \times 6 = 2 \times (\frac{1}{4} \times6)$$ is false without finding the value of both sides?” (It can't be true because $$\frac{2}{3} \times 6=2\times\frac{1}{3}\times6$$.)
## Activity 1: Multipliquemos un número entero por una fracción (15 minutes)
### Narrative
The purpose of this activity is for students to relate multiplying a non-unit fraction by a whole number to multiplying a unit fraction by the same whole number. After finding the value of $$\frac{1}{5} \times 3$$ in a way that makes sense to them, they then consider the value of the products $$\frac{2}{5} \times 3$$ and $$\frac{3}{5} \times 3$$. In the synthesis students address how they can use the value of $$\frac{1}{5} \times 3$$ to find the value other expressions.
• Groups of 2
### Activity
• 8 minutes: independent work time
• Monitor for students who:
• draw a diagram
• use division to solve
• recognize a relationship between $$\frac{1}{5}\times3$$$$\frac{2}{5}\times3$$, and $$\frac{3}{5}\times3$$ .
### Student Facing
Encuentra el valor de cada expresión. Explica o muestra tu razonamiento. Si te ayuda, dibuja un diagrama.
1. $$\frac{1}{5} \times 3$$
2. $$\frac{2}{5} \times 3$$
3. $$\frac{3}{5} \times 3$$
### Activity Synthesis
• Ask previously selected students to share their solutions.
• Display: $$\frac{1}{5} \times 3$$, $$\frac{2}{5} \times 3$$, $$\frac{3}{5} \times 3$$
• “¿En qué se parecen las expresiones?” // “How are the expressions the same?” (They all have a 3. They all have some fifths and there is a product.)
• “¿En qué son diferentes las expresiones?” // “How are the expressions different?” (The number of fifths is different. There is 1 and then 2 and then 3.)
• “¿Cómo pueden usar el valor de $$\frac{1}{5} \times 3$$ como ayuda para encontrar el valor de $$\frac{2}{5} \times 3$$?” // “How can you use the value of $$\frac{1}{5} \times 3$$ to help find the value of $$\frac{2}{5} \times 3$$?” (I can just double the result because it’s $$\frac{2}{5}$$ instead of $$\frac{1}{5}$$.)
• “¿Y para $$\frac{3}{5} \times 3$$?” // “What about $$\frac{3}{5} \times 3$$?” (That’s just another $$\frac{1}{5} \times 3$$.)
• Display diagram from student solution or a student generated diagram like it.
• “¿De qué manera el diagrama muestra $$\frac{1}{5} \times 3$$?” // “How does the diagram show $$\frac{1}{5} \times 3$$?” (There is 3 total and $$\frac{1}{5}$$ of it is shaded.)
• Display: $$\frac{2}{5} \times 3$$.
• “¿Cómo podrían adaptar el diagrama para mostrar $$\frac{2}{5} \times 3$$?” // “How could you adapt the diagram to show $$\frac{2}{5} \times 3$$?” (I could fill in 2 of the fifths in each whole instead of 1.)
• “En la siguiente actividad, vamos a estudiar un diagrama para $$\frac{2}{5} \times 3$$” // “In the next activity we will study a diagram for $$\frac{2}{5} \times 3$$ more.”
## Activity 2: Emparejemos expresiones con diagramas (20 minutes)
### Narrative
The purpose of this activity is to interpret diagrams in multiple ways, focusing on different multiplication and division expressions. The repeating structure in the diagrams allows for many different ways to find the value and interpret the meaning of the expressions. Encourage students to use words, diagrams, or expressions to explain how the diagram represents each of the expressions.
Monitor for students who:
• can explain that the diagram represents the multiplication expression $$3 \times \frac{2}{5}$$ because it shows 3 groups of $$\frac{2}{5}$$
• can explain that the diagram represents $$2 \times (3 \div 5)$$ because there are 3 wholes divided into 5 equal pieces and 2 of the pieces in each whole are shaded
• can explain how the diagram represents the relationship between $$\frac{6}{5}$$ and $$2 \times (3 \div 5)$$
This activity gives students an opportunity to generalize their learning about fractions, division and multiplication. Students see shaded diagrams in different ways, representing different operations, and begin to see the operations as a convenient way to represent complex calculations (MP8).
Engagement: Provide Access by Recruiting Interest. Provide choice. Invite students to decide which expression to start with.
Supports accessibility for: Visual-Spatial Processing, Conceptual Processing, Attention
• Groups of 2
### Activity
• 5–10 minutes: partner work time
MLR2 Collect and Display
• Circulate, listen for, and collect the language students use to describe how each part of the expression represents each part of the diagram.
• Listen for language described in the narrative.
• Look for notes, labels, and markings on the diagrams that connect the parts of the diagram to the parts of the expressions.
• Record students’ words and phrases on a visual display and update it throughout the lesson.
### Student Facing
1. $$2 \times (3 \div 5)$$
2. $$\frac{6}{5}$$
3. $$3 \times \frac{2}{5}$$
4. $$3 \times 2 \times \frac{1}{5}$$
### Student Response
If students do not choose any expressions that represent the diagram, ask them to describe the diagram. Write down the words and phrases they use and ask, “¿Cuáles expresiones representan las palabras que usaste para describir el diagrama?” // “Which expressions represent the words you used to describe the diagram?”
### Activity Synthesis
• Display the expression: $$3 \times \frac{2}{5}$$
• “¿De qué manera el diagrama representa la expresión?” // “How does the diagram represent the expression?” (It shows 3 groups of $$\frac{2}{5}$$.)
• Display the expression: $$2 \times (3 \div 5)$$
• “¿De qué manera el diagrama representa la expresión?” // “How does the diagram represent the expression?”
• Display: $$2 \times (3 \div 5)=\frac{6}{5}$$
• “¿Cómo sabemos que esto es cierto?” // “How do we know this is true?” (We can see both of them in the diagram. $$3\div5$$ is the same as $$\frac{3}{5}$$ and $$2\times\frac{3}{5}=\frac{6}{5}$$
• “¿Qué otras palabras, frases o diagramas importantes deberíamos incluir en nuestra presentación?” // “Are there any other words, phrases, or diagrams that are important to include on our display?”
• As students share responses, update the display, by adding (or replacing) language, diagrams, or annotations.
• Remind students to borrow language from the display as needed.
## Lesson Synthesis
### Lesson Synthesis
Revisit the chart about the relationship between multiplication and division created in an earlier lesson.
“¿Qué le agregarían o ajustarían a lo que ya hay sobre la relación entre la multiplicación y la división?” // “What would you add to or revise about the relationship between multiplication and division?”
Revise chart as necessary.
## Student Section Summary
### Student Facing
En esta sección, exploramos la relación entre la multiplicación y la división. Aprendimos que un diagrama puede representar expresiones de multiplicación y expresiones de división. Por ejemplo, podemos interpretar este diagrama usando 4 expresiones diferentes:
• $$\frac{3}{4}$$, porque cada rectángulo está dividido en 4 partes iguales y hay 3 sombreadas en total.
• $$3 \times \frac{1}{4}$$, porque hay 3 partes sombreadas y cada una es $$\frac{1}{4}$$ del rectángulo.
• $$3 \div 4$$, porque hay 3 rectángulos y cada uno está dividido en 4 partes iguales.
• $$\frac14 \times 3$$, porque hay 3 rectángulos y $$\frac{1}{4}$$ de cada uno está sombreado.
Sabemos que todas estas expresiones son iguales porque todas representan el mismo diagrama. Podemos usar cualquiera de estas expresiones para representar y resolver este problema:
• Mai se comió $$\frac{1}{4}$$ de una bolsa de 3 libras de arándanos. ¿Cuántas libras de arándanos se comió Mai?
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Driving tests
# How do you work out distance, speed and time?
These are often set as questions in maths tests but are useful for us to know so that we can approximate how long a journey will take.
There’s a simple formula which governs the three, and it can be expressed with either variable as the subject of the equation:
Distance = speed * time
Time = distance/speed
Speed = distance/time
So, as long as we have a couple of those, we can plug them into one of the equations.
## How to calculate distance travelled
Let’s have a couple of examples:
You’re going to drive at an average of 70km/h for 4 hours.
Distance is speed multiplied by time, so:
70km/h * 4 hours = 280km.
Easy so far, but what about if you’re going to be driving for 4 hours and 15 minutes? Now you either need to figure out how much 15 minutes is as a decimal number of hours, or you need to convert the hours figure into minutes and, at the end, convert it back.
The first option is fairly simple: take the 15 minutes and divide it by 60 minutes so that we know what proportion of an hour 15 minutes is.
15 minutes / 60 minutes = 0.25 hours
Add that onto the 4 hours you already have and it’s 4.25 hours. Plug it into the same equation:
70km/h * 4.25 hours = 297.5km
The other way around of doing this is converting the time into minutes. Four hours in minutes is 4 * 60 = 240 minutes.
Add the 15 minutes and we have 255 minutes.
Now, when we do the equation, it’s going to give us the wrong figure because our speed is in kilometres per hour, but our time is in minutes. That will totally mess you up. You have to make sure you’re speaking in the same time units. Let’s try it:
70km/h * 255 minutes = 17850 of some figure that makes no sense!
It’s a good hint in an equation that if it looks wrong, it probably is.
What do we do? Well, there are two ways:
1. convert kilometres per hour to kilometres per minute (70km/h / 60 minutes = 1.17km/min), therefore 1.7km/min * 255 minutes = 297.5km
2. or convert 255 minutes to a decimal hour by doing 255 minutes / 60 minutes = 4.25 hours, which gives us the figure we had before.
One more example: You’re travelling at 100km/h. How far, in metres, do you travel in five seconds?
Again, we have to make sure we’re talking about the same units. We have 100km/h, but time unit is in seconds and the final answer is in metres. First we have to convert 100km/h. If we divide it by 60, then we convert it to minutes
100km/h / 60 minutes = 1.667km/minute
We might now want to convert the kilometres into metres which can be done by multiple by 1000.
1.66667km/minute * 1000 = 1666.67m/minute.
Right, but we’re still talking in minutes, so let’s convert it to seconds:
1666.67m/minute / 60 seconds = 27.78metres per second (27.78m/s)
Now we’re talking in the same units as the original question. We just need to figure out how many we travel in 5 seconds.
Distance = speed * time, therefore distance = 27.78 * 5 = 138.89 metres.
## How to calculate average speed
Speed is distance divided by time. Simply put, if you drove 60 kilometres for one hour, it would look like this:
Speed = distance (60 km) / time (1 hour) = 60km/h.
Let’s take a more complex example: what’s the average speed if you covered 425.5km in 5 hours 22 minutes and 30 seconds?
Firstly, we have the problem again with the time not being decimalised. We can use the same method we did before:
30 seconds / 60 seconds is 0.5. This means 30 seconds is 0.5 minutes. Therefore we add that the 22 minutes and repeat it again:
22.5 minutes / 60 minutes = 0.375 hours. Add that onto the 5 hours and you have 5.375 hours.
Speed = distance (425.5km) / time (5.375 hours) = 79.16km/h
## How to calculate time taken to cover a distance
Our final variation is calculating how much time it takes when you know the distance and the speed.
Time = distance/speed. So, a nice easy one would be how many hours does it take to cover 30km when you are travelling at 120km/h.
Time = 30km/120km/h = 0.25 hours.
If you need to figure out how many minutes that is, simply multiply it by 60, which gives you 15 minutes.
Taking a more complex example, if you’re travelling the 842km from Picton to Gore, how long in hours, minutes and seconds will it take you if you average 67km/h?
Time = 842km / 67km/h = 12.5672 hours
Let’s convert that decimal to minutes and seconds. Firstly, multiply 0.5672 by 60 = 34.0299 minutes. Now we multiply that decimal by 60 to get seconds: 0.0299 * 60 = 1.8 seconds. We can round that up to 2 seconds, which gives us 12 hours, 34 minutes and 2 seconds.
Now you know how to do the conversion for speed, distance and time.
Darren has written over 3000 articles about driving and vehicles, plus almost 500 vehicle reviews and numerous driving courses. Connect with him on LinkedIn by clicking the name above
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
1.2: Order of Operations
Difficulty Level: At Grade Created by: CK-12
The Mystery of Math Verbs
Some math verbs are “stronger” than others and must be done first. This method is known as the Order of Operations.
A mnemonic (a saying that helps you remember something difficult) for the Order of Operations is PEMDAS - Please Excuse My Daring Aunt Sophie.
The Order of Operations:
Whatever is found inside PARENTHESES must be done first. EXPONENTS are to be simplified next. MULTIPLICATION and DIVISION are equally important and must be performed moving left to right. ADDITION and SUBTRACTION are also equally important and must be performed moving left to right.
Example 1: Use the Order of Operations to simplify \begin{align*}(7-2) \times 4 \div 2-3\end{align*}
Solution: First, we check for parentheses. Yes, there they are and must be done first.
\begin{align*}(7 - 2) \times 4 \div 2 - 3 = (5) \times 4 \div 2 - 3\end{align*}
Next we look for exponents (little numbers written a little above the others). No, there are no exponents so we skip to the next math verb.
Multiplication and division are equally important and must be done from left to right.
\begin{align*}5 \times 4 \div 2 - 3 & = 20 \div 2 - 3\\ 20 \div 2 - 3 & = 10 - 3\end{align*}
Finally, addition and subtraction are equally important and must be done from left to right.
\begin{align*}10-3 = 7\end{align*} This is our answer.
Example 2: Use the Order of Operations to simplify the following expressions.
a) \begin{align*}3 \times 5-7 \div 2\end{align*}
b) \begin{align*}3 \times (5 - 7) \div 2\end{align*}
c) \begin{align*}(3 \times 5) - (7 \div 2)\end{align*}
Solutions:
a) There are no parentheses and no exponents. Go directly to multiplication and division from left to right: \begin{align*} 3 \times 5 - 7 \div 2 = 15 - 7 \div 2 = 15 - 3.5\end{align*}
Now subtract: \begin{align*}15 - 3.5 = 11.5\end{align*}
b) Parentheses must be done first: \begin{align*}3 \times (-2) \div 2\end{align*}
There are no exponents, so multiplication and division come next and are done left to right: \begin{align*}3 \times (-2) \div 2 = -6 \div 2 = -3\end{align*}
c) Parentheses must be done first: \begin{align*}(3 \times 5) - (7 \div 2) = 15 - 3.5\end{align*}
There are no exponents, multiplication, division, or addition, so simplify:
\begin{align*}15 - 3.5 = 11.5\end{align*}
Parentheses are used two ways. The first is to alter the Order of Operations in a given expression, such as example (b). The second way is to clarify an expression, making it easier to understand.
Some expressions contain no parentheses while others contain several sets of parentheses. Some expressions even have parentheses inside parentheses! When faced with nested parentheses, start at the innermost parentheses and work outward.
Example 3: Use the Order of Operations to simplify \begin{align*}8-[19-(2+5)-7]\end{align*}
Solution: Begin with the innermost parentheses:
\begin{align*}8-[19-(2+5)-7]=8-[19-7-7]\end{align*}
Simplify according to the Order of Operations:
\begin{align*}8-[19-7-7]=8-[5]=3\end{align*}
Evaluating Algebraic Expressions with Fraction Bars
Fraction bars count as grouping symbols for PEMDAS, and should be treated as a set of parentheses. All numerators and all denominators can be treated as if they have invisible parentheses. When real parentheses are also present, remember that the innermost grouping symbols should be evaluated first. If, for example, parentheses appear on a numerator, they would take precedence over the fraction bar. If the parentheses appear outside of the fraction, then the fraction bar takes precedence.
Example 4: Use the Order of Operations to simplify the following expressions.
a) \begin{align*}\frac{z + 3}{4} - 1\end{align*} when \begin{align*}z = 2\end{align*}
b) \begin{align*}\left (\frac{a+2}{b+4} - 1 \right ) + b\end{align*} when \begin{align*}a = 3\end{align*} and \begin{align*}b = 1\end{align*}
c) \begin{align*}2 \times \left ( \frac{w + (x - 2z)}{(y + 2)^2} - 1 \right )\end{align*} when \begin{align*}w = 11, \ x = 3, \ y = 1\end{align*} and \begin{align*}z = -2\end{align*}
Solutions: Begin each expression by substituting the appropriate value for the variable:
a) \begin{align*}\frac{(2+3)}{4} -1 = \frac{5}{4} -1\end{align*}. Rewriting 1 as a fraction, the expression becomes:
\begin{align*}\frac{5}{4} - \frac{4}{4} = \frac{1}{4}\end{align*}
b) \begin{align*}\frac{(3+2)}{(1+4)} = \frac{5}{5} = 1\end{align*}
\begin{align*}(1 - 1) + b\end{align*} Substituting 1 for b, the expression becomes \begin{align*} 0 + 1 = 1\end{align*}
c) \begin{align*}2 \left ( \frac{[11+(3-2(-2))]}{[(1+2)^2)]} - 1 \right ) = 2 \left ( \frac{(11+7)}{3^2} -1 \right ) = 2 \left (\frac{18}{9} - 1 \right )\end{align*}
Continue simplifying: \begin{align*}2\left ( \frac{18}{9} - \frac{9}{9} \right ) = 2 \left ( \frac{9}{9} \right ) = 2(1)= 2\end{align*}
Using a Calculator to Evaluate Algebraic Expressions
A calculator, especially a graphing calculator, is a very useful tool in evaluating algebraic expressions. The graphing calculator follows the Order of Operations, PEMDAS. In this section, we will explain two ways of evaluating expressions with the graphing calculator.
Method #1: This method is the direct input method. After substituting all values for the variables, you type in the expression, symbol for symbol, into your calculator.
Evaluate \begin{align*}[3(x^2 - 1)^2 - x^4 + 12] + 5x^3 - 1\end{align*} when \begin{align*}x = -3\end{align*}.
Substitute the value \begin{align*}x = -3\end{align*} into the expression.
\begin{align*}[3((-3)^2 -1)^2 - (-3)^4 + 12] + 5(-3)^3 - 1\end{align*}
The potential error here is that you may forget a sign or a set of parentheses, especially if the expression is long or complicated. Make sure you check your input before writing your answer. An alternative is to type the expression in by appropriate chunks – do one set of parentheses, then another, and so on.
Method #2: This method uses the STORE function of the Texas Instrument graphing calculators, such as the TI-83, TI-84, or TI-84 Plus.
First, store the value \begin{align*}x = -3\end{align*} in the calculator. Type -3 [STO] \begin{align*}x\end{align*}. (The letter \begin{align*}x\end{align*} can be entered using the \begin{align*}x\end{align*}-[VAR] button or [ALPHA] + [STO]). Then type in the expression in the calculator and press [ENTER].
The answer is \begin{align*}-13.\end{align*}
Note: On graphing calculators there is a difference between the minus sign and the negative sign. When we stored the value negative three, we needed to use the negative sign, which is to the left of the [ENTER] button on the calculator. On the other hand, to perform the subtraction operation in the expression we used the minus sign. The minus sign is right above the plus sign on the right.
You can also use a graphing calculator to evaluate expressions with more than one variable.
Evaluate the expression: \begin{align*}\frac{3x^2 - 4y^2 + x^4}{(x + y)^{\frac{1}{2}}}\end{align*} for \begin{align*}x = -2, y = 1\end{align*}.
Store the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. \begin{align*}-2\end{align*} [STO] \begin{align*}x\end{align*}, 1 [STO] \begin{align*}y\end{align*}. The letters \begin{align*}x\end{align*} and \begin{align*}y\end{align*} can be entered using [ALPHA] + [KEY]. Input the expression in the calculator. When an expression shows the division of two expressions be sure to use parentheses: (numerator) \begin{align*}\div\end{align*} (denominator). Press [ENTER] to obtain the answer \begin{align*}-.8\bar{8}\end{align*} or \begin{align*}-\frac{8}{9}\end{align*}.
Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Order of Operations (14:23)
Use the Order of Operations to simplify the following expressions.
1. \begin{align*}8 - (19 - (2 + 5) - 7)\end{align*}
2. \begin{align*}2 + 7 \times 11 - 12 \div 3\end{align*}
3. \begin{align*}(3 + 7) \div (7 - 12)\end{align*}
4. \begin{align*}\frac{2 \cdot (3 + (2 - 1))}{4 - (6 + 2)} - (3 - 5)\end{align*}
5. \begin{align*}8 \cdot 5 + 6^2\end{align*}
6. \begin{align*}9 \div 3 \times 7 - 2^3 + 7\end{align*}
7. \begin{align*}8 + 12 \div 6 + 6\end{align*}
8. \begin{align*}(7^2-3^2) \div 8\end{align*}
Evaluate the following expressions involving variables.
1. \begin{align*}\frac{jk}{j + k}\end{align*} when \begin{align*}j = 6\end{align*} and \begin{align*}k = 12\end{align*}.
2. \begin{align*}2y^2\end{align*} when \begin{align*}x = 1\end{align*} and \begin{align*}y = 5\end{align*}
3. \begin{align*}3x^2 + 2x + 1\end{align*} when \begin{align*}x = 5\end{align*}
4. \begin{align*}(y^2 - x)^2\end{align*} when \begin{align*}x = 2\end{align*} and \begin{align*}y = 1\end{align*}
Evaluate the following expressions involving variables.
1. \begin{align*}\frac{4x}{9x^2 - 3x + 1}\end{align*} when \begin{align*}x = 2\end{align*}
2. \begin{align*}\frac{z^2}{x + y} + \frac{x^2}{x - y}\end{align*} when \begin{align*}x = 1, \ y = -2\end{align*}, and \begin{align*}z = 4\end{align*}.
3. \begin{align*}\frac{4xyz}{y^2 - x^2}\end{align*} when \begin{align*}x = 3, \ y = 2\end{align*}, and \begin{align*}z = 5\end{align*}
4. \begin{align*}\frac{x^2 - z^2}{xz - 2x(z - x)}\end{align*} when \begin{align*}x = -1\end{align*} and \begin{align*}z = 3\end{align*}
The formula to find the volume of a square pyramid is \begin{align*}V=\frac{s^2 (h)}{3}\end{align*}. Evaluate the volume for the given values.
1. \begin{align*}s=4\ inches,h=18\ inches\end{align*}
2. \begin{align*}s=10\ feet,h=50\ feet\end{align*}
3. \begin{align*}h=7\ meters,s=12\ meters\end{align*}
4. \begin{align*}h=27\ feet,s=13\ feet\end{align*}
5. \begin{align*}s=16\ cm,h=90\ cm\end{align*}
In 22 – 25, insert parentheses in each expression to make a true equation.
1. \begin{align*}5 - 2 \cdot 6 - 4 + 2 = 5\end{align*}
2. \begin{align*}12 \div 4 + 10 - 3 \cdot 3 + 7 = 11\end{align*}
3. \begin{align*}22 - 32 - 5 \cdot 3 - 6 = 30\end{align*}
4. \begin{align*}12 - 8 - 4 \cdot 5 = -8\end{align*}
In 26 – 29, evaluate each expression using a graphing calculator.
1. \begin{align*}x^2 + 2x - xy\end{align*} when \begin{align*}x = 250\end{align*} and \begin{align*}y = -120\end{align*}
2. \begin{align*}(xy - y^4)^2\end{align*} when \begin{align*}x = 0.02\end{align*} and \begin{align*}y = -0.025\end{align*}
3. \begin{align*}\frac{x + y - z}{xy + yz + xz}\end{align*} when \begin{align*}x = \frac{1}{2}, \ y = \frac{3}{2}\end{align*}, and \begin{align*}z = -1\end{align*}
4. \begin{align*}\frac{(x + y)^2}{4x^2 - y^2}\end{align*} when \begin{align*}x = 3\end{align*} and \begin{align*}y = -5d\end{align*}
5. The formula to find the volume of a spherical object (like a ball) is \begin{align*}V = \frac{4}{3}(\pi)r^3\end{align*}, where \begin{align*}r =\end{align*} the radius of the sphere. Determine the volume for a grapefruit with a radius of 9 cm.
Mixed Review
1. Let \begin{align*}x = -1\end{align*}. Find the value of \begin{align*}-9x + 2\end{align*}.
2. The area of a trapezoid is given by the equation \begin{align*}A = \frac{h}{2}(a + b)\end{align*}. Find the area of a trapezoid with bases \begin{align*}a = 10 \ cm, b = 15 \ cm\end{align*}, and height \begin{align*}h = 8 \ cm\end{align*}.
3. The area of a circle is given by the formula \begin{align*}A = \pi r^2\end{align*}. Find the area of a circle with radius \begin{align*}r = 17\end{align*} inches.
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# 2020 AIME I Problems/Problem 12
## Problem
Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$
## Solution 1
Lifting the Exponent shows that $$v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1$$ so thus, $3^2$ divides $n$. It also shows that $$v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2$$ so thus, $7^5$ divides $n$.
Now, multiplying $n$ by $4$, we see $$v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})$$ and since $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4n}-2^{4n})=1$ meaning that we have that by LTE, $4 \cdot 5^4$ divides $n$.
Since $3^2$, $7^5$ and $4\cdot 5^4$ all divide $n$, the smallest value of $n$ working is their LCM, also $3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.
~kevinmathz
## Solution 2 (Simpler, just basic mods and Fermat's theorem)
Note that for all n, $149^n - 2^n$ is divisible by 1$49-2 = 147$ because that is a factor. That is $3\cdot7^2$, so now we can clearly see that the smallest $n$ to make the expression divisible by $3^3$ is just $3^2$. Similarly, we can reason that the smallest n to make the expression divisible by $7^7$ is just $7^5$.
Finally, for $5^5$, take mod $5$ and mod $25$ of each quantity (They happen to both be $-1$ and $2$ respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum $n$ for divisibility by $5$ is $4$, and other values are factors of $4$. Testing all of them (just $1,2,4$ using mods-not too bad), $4$ is indeed the smallest value to make the expression divisible by $5$, and this clearly is NOT divisible by $25$. Therefore, the smallest $n$ to make this expression divisible by $5^5$ is $2^2 \cdot 5^4$.
Calculating the LCM of all these, one gets $2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. Using the factor counting formula, the answer is $3\cdot 3\cdot 5\cdot 6$ = $\boxed{270}$.
-Solution by thanosaops
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# Negation
Hi, and welcome to this video about negation! In this video, we will explore what negation means, how it works, and how to work with multiple negations. Let’s get started!
First, let’s review a bit. Negation is a part of propositional logic. Remember that the primary purpose of propositional logic is to determine whether declarative statements – arguments – are true or false. Here is a quick reminder of some of the symbols that we use:
SymbolMeaning
Lower case lettervariable
Upper case letterpredicate
$$∀$$universal quantifier “for every”
$$∃$$“there exists”
:“such that”
For example, the statement $$\exists x:x^{2}= 9$$ is true, since 3 is an $$x$$-value that makes the equation true. If $$x$$ is our variable and $$P(x)$$ is the predicate $$x^{2}= 9$$, then our statement looks like: $$\exists xP(x)$$.
On the other hand $$\forall x:x^{2}=9$$, read “For every $$x$$-value, $$x^{2}= 9$$.” is clearly false, because there are plenty of $$x$$-values that make the equation false. Fully symbolic, this statement reads: $$\forall xP(x)$$.
Now, what exactly does negation mean? Negation means basically what it sounds like – to make a statement negative. Any statement can be negated. The word not is perhaps most commonly used to negate a statement, but other words and phrases can be used as well, such as
• No
• It is false
• It is not the case
• It is not true
When a statement is negated, its truth value is the opposite of what it was. Sometimes a truth table can be a helpful illustration:
$$P(x)$$$$¬P(x)$$
TrueFalse
FalseTrue
### How Negation Works
Let’s revisit the statement: “There exists an $$x$$-value such that $$x^{2}=9$$,” which we’ve said is true. Negating it should produce a false statement, which it does: “There does not exist an $$x$$-value such that $$x^{2}=9$$.”
Note that negation of this statement is not “There exists an $$x$$-value such that $$x^{2}\neq 9$$,” because $$x^{2}\neq 9$$ is a different predicate for a different time. We are testing $$x$$-values that make $$x^{2}=9$$ true or false, not $$x$$-values that make $$x^{2}\neq 9$$ true or false.
Likewise, the statement: “For every $$x$$-value, $$x^{2}=9$$,” which we’ve said is false, is made true by negation: “It is not true for every $$x$$-value that $$x^{2}=9$$
The symbol for negation is ¬, let’s add it to our list:
SymbolMeaning
Lower case lettervariable
Upper case letterpredicate
$$∀$$universal quantifier “for every”
$$∃$$“there exists”
:“such that”
¬“not,” negation
Now, back to our examples. We negated the statement “There exists an $$x$$-value such that $$x^{2}=9$$,” by saying “There does not exist an $$x$$-value such that $$x^{2}=9$$.”
Symbolically, we originally said $$\exists xP(x)$$, which is true. It can be negated by saying
• $$¬\exists xP(x)$$ (There does not exist an $$x$$-value such that $$x^{2}=9$$) or
• $$\forall x¬P(x)$$, (For all $$x$$, $$x^{2}$$ never equals 9)
Both the negations have the same meaning and are truth-equivalent (in this case, both false).
In our second example, we began by saying “For every $$x$$-value, $$x^{2}=9$$,” which we know is false, and negated it by saying “It is not true for every $$x$$-value that $$x^{2}=9$$.”
Symbolically, we originally said $$\forall xP(x)$$ , which is false. It can be negated by saying
• $$¬\exists xP(x)$$ (There exists an $$x$$ such that $$x^{2}$$ is not equal to 9) or
• $$\forall x¬P(x)$$ (Not every $$x$$-value makes $$x^{2}=9$$ true)
Again, both negations have the same meaning and are truth-equivalent (in this case, both true).
Also, in this case, the term counterexample comes to mind. For every $$x$$-value, $$x^{2}=9$$. False, here’s a counterexample: Let $$x=2$$, for instance.
Let’s try some practice: Evaluate each statement and show its negation has the opposite value.
Every voter is at least 18 years old. This is true, according to the law.
Not every voter is at least 18 years old. And this is false, because the voting age is 18.
Every composite number is even. False, some composite numbers are odd, such as 15.
Not every composite number is even. True.
### Multiple Negations
Statements can be negated multiple times. Since one negation flips the truth value of a statement, a second negation flips it back. Let’s expand our truth table:
$$P(x)$$$$¬P(x)$$$$¬¬P(x)$$$$¬¬¬P(x)$$
TrueFalseTrueFalse
FalseTrueFalseTrue
Therefore, saying “There exists an $$x$$-value such that$$x^{2}=9$$,” is truth equivalent to “There does not not exist an $$x$$-value such that $$x^{2}=9$$.” $$\exists xP(x)=¬¬\exists xP(x)$$.
Now I want you to try some on your own! Pause the video and complete the chart.
StatementSymbolicTruth
All integers are rational numbers.$$\forall xP(x)$$T
Not all integers are rational numbers.$$¬\forall xP(x)$$F
It is not the case that not all integers are rational numbers.$$¬¬\forall xP(x)$$T
There exist integers that are not rational numbers.$$\exists x¬P(x)$$F
It is false that it is not the case that not all integers are rational numbers.$$¬¬¬\forall xP(x)$$F
There do not exist integers that are rational numbers.$$¬\exists xP(x)$$F
I hope that this video increased your understanding of negation! Thanks for watching, and happy studying!
## Negation Practice Questions
Question #1:
Let $$\exists xP(x)$$ be the statement $$\exists x:x^2=16$$. Which of the following is the statement for $$\lnot\exists xP\left(x\right)$$ and its truth value?
There does not exist an $$x$$-value such that $$x^2=16$$, which is a false statement.
There does not exist an $$x$$-value such that $$x^2=16$$, which is a true statement.
There exists an $$x$$-value such that $$x^2\neq16$$, which is a false statement.
There does not exist an $$x$$-value such that $$x^2\neq16$$, which is a true statement.
Answer:
The symbolic statement $$\lnot\exists xP(x)$$ means to negate the initial portion of the statement, “There exists an $$x$$-value” written symbolically as $$\exists x$$. When a statement is negated, its truth value is the opposite of what it was. Negated statements often use the word “not” or “does not” when they are stated. Changing the initial portion of our statement to “There does not exist,” our negated statement, $$\lnot\exists xP\left(x\right)$$, is, “There does not exist an $$x$$-value such that $$x^2=16.$$”
To show the negation is false, we can show that $$\exists x:x^2=16$$ is true. We can choose an $$x$$-value of 4 to make the predicate true since $$4^2=16$$. The symbol $$\exists$$ tells us there only needs to exist one $$x$$-value to make the statement true. So, $$\exists xP(x)$$ is a true statement. Thus, the negated statement must be false. Therefore, choice A is the correct answer.
Question #2:
Let $$\forall xP(x)$$ be the statement “For every value of $$x$$, $$x^2=25$$”, which is a false statement. Which of the following is the symbolic negation for the statement, “It is not the case that not every value of $$x$$ makes $$x^2=25$$,” that also has the same truth value of $$\forall xP(x)$$?
$$\lnot\forall xP(x)$$
$$\lnot\exists xP(x)$$
$$\lnot\lnot\forall xP(x)$$
$$\lnot\lnot\exists xP(x)$$
Answer:
For the symbolic statement $$\forall xP(x)$$, $$\forall x$$ means “For every value of $$x$$,” and $$P(x)$$ is $$x^2=25$$. We are told this is a false statement.
When a statement is negated, its truth value is the opposite of what it was. Symbolically negating our statement once, we can use $$\lnot\forall xP(x)$$, which is true since it can be stated as, “Not every value of $$x$$ makes $$x^2=25$$.” Use $$x=4$$ to show that the square of not every $$x$$-value equals $$25$$ because $$4^2=16\neq25$$.
We can negate $$\lnot\forall xP(x)$$, using $$\lnot\lnot\forall xP(x)$$, which becomes an equivalent false statement to $$\forall xP(x)$$. It can be stated as, “It is not the case that not every value of $$x$$ makes $$x^2=25$$.” Since we saw from above that “Not every value of $$x$$ makes $$x^2=25$$” was true, its negation must be false so, $$\lnot\lnot\forall xP(x)$$ has an equivalent truth statement using $$P\left(x\right)$$.
Question #3:
Let $$\forall xP(x)$$ be the statement, “All even numbers are integers.” Which of the following is the statement for $$\exists x\lnot P\left(x\right)$$ and its truth value?
There does not exist an even number that is an integer, which is a false statement.
There does not exist an even number that is not an integer, which is a true statement.
There exists an even number that is not an integer, which is a true statement.
There exists an even number that is not an integer, which is a false statement.
Answer:
The symbolic statement $$\exists x\lnot P(x)$$ means to negate the conclusive portion of the statement “are integers.” When a statement is negated, its truth value is the opposite of what it was. Negations often use the word “not” when they are stated. Additionally, the symbol $$\exists$$ means that there exists. Since our statement is “All even numbers are integers”, our negated statement, $$\exists x\lnot P\left(x\right)$$, should state “There exists an even number that is not an integer.”
The set of integers contains the numbers $$\{…,-3,-2,-1,0,1,2,3,…\}$$. The set of even numbers contains the numbers $$\{…,-6,-4,-2,0,2,4,6,…\}$$ which is a subset of the set of integers. Therefore, all even numbers are also integers. This means the statement, “All even numbers are integers” is a true statement. Thus, the negation $$\exists x\lnot P(x)$$ must be false.
Question #4:
The statement, “The minimum voting age for all voters in the United States is 18 years old,” is a true statement. What is the negation of the statement and its truth value?
The minimum voting age for all voters in the United States is not 18 years old, which is a true statement.
The minimum voting age for all voters in the United States is not 18 years old, which is a false statement.
There exists a voter in the United States whose minimum voting age is not 18 years old, which is a false statement.
There does not exist a voter in the United States whose minimum voting age is not 18 years old, which is a false statement.
Answer:
When a statement is negated, its truth value is the opposite of what it was. We can negate the predicate of a statement by using the word “not” when stating it. Since our predicate states “is 18 years old,” our negated statement will be “The minimum voting age of all voters in the United States is not 18 years old.”
Since it is given that the original statement is true, the negation of the original statement must be a false statement.
Question #5:
The statement, “Only one of my neighbors has exactly two dogs,” is a true statement. What is the negation of the statement?
None of my neighbors do not have exactly two dogs.
There is at least one of my neighbors that does not have exactly two dogs.
All of my neighbors do not have exactly two dogs.
Not all of my neighbors have exactly two dogs.
Answer:
When a statement is negated, its truth value is the opposite of what it was. We can negate the predicate of a statement by using the word “not” or “does not” when stating it. Since our predicate states, “has exactly two dogs” we can write its negation as, “does not have exactly two dogs.”
Additionally, if you only have two neighbors, then the negated statement, “Only one of my neighbors does not have exactly two dogs,” is still a true statement since it implies the other neighbor does have exactly two dogs. The negated statement needs to be false. To make the negated statement false, we can change the initial part of the negation, “Only one of my neighbors” to “All of my neighbors.” So, the negation becomes, “All of my neighbors do not have exactly two dogs,” which is a false statement.
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# Understanding Box Plots (with Assessment)
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## Objective
SWBAT analyze a box plot in order to determine the five number summary as well as the distribution of the data.
#### Big Idea
Students continue their exploration of Box Plots and they have a chance to demonstrate their mastery of unit concepts.
## Opening
10 minutes
During the first ten minutes of class, I will pose a few review problems for students using box_plot_day2.
As we view Slide 2, students will work with their partners to review the vocabulary from the previous lesson. They will also find the 5 number summary (min, 1st quartile, median, 3rd quartile, max) necessary for constructing a box plot. Once students find the five number summary they will verify with their partner that they have divided the data set into four equally-sized portions.
Slide 3 requires students to problem solve within their partnerships. The students are going to find the interquartile range, but from a more abstract perspective. This is because none of the actual data points are given to the students. The only information I have given the students is the frequency table.
Teaching Point: Some students will be able to visualize how the 19 scores are distributed, however, some will not. For students who struggle it I guide them to make 19 tally marks to represent the data points written in order. This way they can find the median and two quartiles in a more concrete way seeing that the median will be the 10th data point, the 1st and 3rd quartiles will be the 5th and 15th data points respectively. Students can then go back to the frequency table to determine in which intervals these values lie.
The box plot on Slide 4 represents the weights of our 2013 high school football team. I ask students to work on their own to analyze this box plot and write down any inferences that they are able to make from the plot. Then, I will ask them to share their results with their partner. Before we move on to the assessment, we will share out a few of the ideas with the class.
Finally, I pose the following question to the class:
True or False: Most of the team weighs between 205 and 320. Explain why this is true or false.
Students discuss this with their partners. Through a group share out we want to dispell the misconception that if a section of a box plot is larger than another that it contains more data points.
25 minutes
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Class Notes (834,991)
CSCA67H3 (56)
Lecture
# Week 11 - Probability Sum and Product Rule.pdf
8 Pages
423 Views
Department
Computer Science
Course
CSCA67H3
Professor
Anna Bretscher
Semester
Fall
Description
Sum and Product Rules Exercise. Consider tossing a coin five times. What is the proba- bility of getting the same result on the first two tosses or the last two tosses? Solution. Let E be the event that the first two tosses are the same and F be the event that the last two tosses are the same. Let n(E) be the number of ways event E can occur. n(E) = n(F) = n(S) = Q. Is there any overlap in E and F? I.e., is E \ F = ;? A. n(E and F) = n(E \ F) = = 1 Now we can calculate the probability P(E or F): P(E or F) = = = = Theorem (The Sum Rule) If E and F are events in an experi- ment then the probability that E or F occurs is given by: P(E or F) = P(E) + P(F) ▯ P(E and F) Example. What is the probability when a pair of dice are rolled that at least one die shows a 5 or the dice sum to 8? Solution. 2 Exercise. Given a bag of 3 red marbles, 5 black marbles and 8 green marbles select one marble and then a second. What is the probability that both are red? Solution. Q. What is the probability of the first marble being red? A. Q. What is the probability of the second marble being red? A. Q. Probability of both marbles being red? A. When the probability of an event E depends on a previous event F happening we denote this probability as P(EjF). 3 Theorem (The Product Rule). If E and F are two events in an experiment then the probability that both E and F occur is: (▯) P(E and F) = P(F) ▯ P(EjF) = P(E) ▯ P(FjE) Q. What does it mean if P(EjF) = P(E)? A. Q. Therefore, if E and F are two independent events, what is (▯)? A. Example. Suppose there is a noisy communication channel in which either a 0 or a 1 is sent with the following probabilities: ▯ Probability a 0 is sent is 0.4. ▯ Probability a 1 is sent is 0.6. ▯ Probability that due to noise, a 0 is changed to a 1 during transmission is 0.2. ▯ Probability that due to noise, a 1 is changed to a 0 during transmission is 0.1. Suppose that a 1 is received. What is the probability that a 1 was sent? 4 Let A denote that a 1 was received and B denote the event that a 1 was sent. Q. What is the probability that we are solving for? A. How can we solve for this? The Product Rule says that: P(A and B) = P(B) ▯ P(AjB) = P(A)P(BjA) Therefore: P(B
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Induction with negative step
We've learned that we can use induction to show that a statement holds for all natural numbers (or for all natural numbers above n). The steps are:
1. prove that the statement holds for a base number b
2. assuming that the statement holds for n, show that it holds for n+1.
This way we have proved that the statement holds for any integer $\ge b$
Can we take this a bit further to prove that the statement holds for ALL integer values? To my understanding, all we have to do is to try to prove:
$3$. assuming that the statement holds for n, show that it holds for n-1.
However I've never seen any articles on this, or any exercises being solved this way?
• Is this because my logic is not correct?
• Is this because "prove that this holds for all integers" can always be solved with a simpler way than using induction twice?
• It's because if you want to prove $\forall n\in \mathbb Z(P(n))$ and you proved $\forall n\in \mathbb N(P(n))$, then, $\forall n\in \mathbb Z(P(n))$ follows from $\forall n\in \mathbb N(P(-n))$ in conjunction with $\forall n\in \mathbb N(P(n))$. – Git Gud Jun 6 '14 at 7:00
If we have some sort of proposition $P(n)$ and we have proved that $P(n)$ is true for every $n \in \mathbb{N}$ by induction then we can proceed to create a new proposition $P'(n) = P(-n)$ and prove that for every $n \in \mathbb{N}$ thus showing that $P(n)$ is actually true for every $n \in \mathbb{Z}$. The steps to prove $P'(n)$ is to show that the base case $P'(0)$ is true and then proceed to show that $P(-n)$ true implies that $P(-(n+1)) = P(-n-1)$.
I personally have never used this but have seen it used in certain places.
Your logic appears correct, it's quite possible that a simpler way exists but this generally depends on what you're proving.
As an example, if $f(n+2) = 5 f(n+1) - 6 f(n)$ for all integers $n$, and $f(0) = 0$ and $f(1) = 1$, then $f(n) = 3^n - 2^n$ for all integers $n$. You cannot avoid using induction twice, once upwards and once downwards.
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# 1. Tangents and Normals
by M. Bourne
We often need to find tangents and normals to curves when we are analysing forces acting on a moving body.
A tangent to a curve is a line that touches the curve at one point and has the same slope as the curve at that point.
A normal to a curve is a line perpendicular to a tangent to the curve.
Tangent to
the curve
Normal to
the curve
Graph showing the tangent and the normal to a curve at a point.
Continues below
Note 1: As we discussed before (in Slope of a Tangent to a Curve), we can find the slope of a tangent at any point (x, y) using dy/dx.
Note 2: To find the equation of a normal, recall the condition for two lines with slopes m1and m2 to be perpendicular (see Perpendicular Lines):
m1 × m2 = −1
### Applications
A car has skidded while rounding a corner, tangent to the double yellow lines curve.
Tangent:
1. If we are traveling in a car around a corner and we drive over something slippery on the road (like oil, ice, water or loose gravel) and our car starts to skid, it will continue in a direction tangent to the curve.
2. Likewise, if we hold a ball and swing it around in a circular motion then let go, it will fly off in a tangent to the circle of motion.
The spokes of a bicycle wheel are normal to the rim.
Normal:
1. When you are going fast around a circular track in a car, the force that you feel pushing you outwards is normal to the curve of the road. Interestingly, the force that is making you go around that corner is actually directed towards the center of the circle, normal to the circle.
2. The spokes of a wheel are placed normal to the circular shape of the wheel at each point where the spoke connects with the center.
### Examples
(i) the tangent (ii) the normal
to the curve y = x3 − 2x2 + 5 at the point (2,5).
dy/dx=3x^2-4x
The slope of the tangent is
m_1=[dy/dx] _(x=2)
=3(2)^2-4(2)
=12-8
=4
The slope of the normal is found using m1 × m2 = −1
m_2=-1/4
2. Find the equation of (i) the tangent and (ii) the normal in the above example.
We use yy1 = m(xx1), with x1 = 2, y1 = 5
(i) The tangent has slope 4, so we have:
y-5=4(x-2)
gives
y=4x-3
or
4xy − 3 = 0
(ii) Now for the normal to the curve. Since the tangent has slope 4, we have the slope of the normal m=-1/4
So we substitute as follows:
y-5=-1/4(x-2)
gives
y=-1/4x+5 1/2
or
x + 4y − 22 = 0
3. Sketch the curve and the normal in the above example.
Here's the graph of the tangent and normal to the curve at x=2.
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Algebra
Equations
Inequalities
Graphs
Numbers
Calculus
Matrices
Tutorials
# Different methods for solving quadratic equations
6.5 Equations of the Form ax2+bx+c=0
By a quadratic equation in the single variable x, we mean any equation that can be transformed through elementary transformations to an equation of the form
ax2+bx+c=0, a0
When the equation is written in the above form we will say that it is in standard form.
The first type of quadratic equation that we will consider is an equation of the form
ax2+c=0
We first solve for x2 to obtain
x2= ca
Next we solve for x by taking the square root to obtain
x=± ca
The solution set is therefore {( ca), ( ca)}.
Example 1. Solve 9 x2-25=0
9 x2-25=0
x2=259
x=± 259
x=± (53)
The solution set is {53, 53}.
Let’s see how our quadratic equation solver solves this and similar problems. Click on "Solve Similar" button to see more examples.
6.6 Solution of ax2+bx+c=0 by Factoring
We recall from Chapter 1 the important fact that if A and B are real numbers and
AB=0
then
A=0 or B=0
This property is the key to solving quadratic equations by the method of factoring. It is important to note that the above reasoning applies only when the right-hand side is zero; no other real number will do.
Example 1. Solve the equation x2-4 x-5=0.
x2-4 x-5=0
(x-5) (x+1)=0
Therefore,
x-5=0 or x+1=0
so the solution set is {5, 1}
Example 2 Solve the equation 2 x2-5 x=3.
We add 3 to both sides to obtain 0 on the right-hand side.
2 x2-5 x=3
2 x2-5 x-3=0
(2 x+1) (x-3)=0
Therefore,
2 x+1=0 or x-3=0
so the solution set is
{ 12,3}
More generally if
AB C=0
then
A=0,B=0, or C=0
Consequently this method of solution will apply to any degree equation as long as we are able to factor the equation into linear terms. For example, if we have
x (x-3) (x+1)=0
then
x=0x-3=0, or x+1=0
and hence the solution set is
{0,3, 1}
Let’s see how our math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.
6.7 Completing the Square
As we have seen, quadratic equations of the form
x2=a
where amay be solved by the method of extraction of roots. In fact, since every real number a has two square roots (either real or imaginary) we have two solutions. These are given by is any real number,
x=a or x= a
For example , from the equation
x2=25
we have
x=25=5 or x= 25= 5
From the equation
x2= 9
we have
x= 9=3 1=3 i or x= 9= 3 i
Equations of the form
(x-a)2=b
can be solved by the same method. For example
(x-3)2=25
implies that
x-3=5 or x-3= 5
from which we obtain
x=8 or x= 2
In order to write x2±2 ax+b=0 in the form
(x±a)2=k
we note that
(x±a)2=x2±2 ax+a2
Hence, in order to obtain a perfect square on the left-hand side when the first two terms x2±2 ax are given, we must add a2, which is the square of half the coefficient of the x term, to both sides of the equation. This method is called the method of completing the square. Note that to complete the square the coefficient of x2 must be 1.
Example 1. Solve the equation x2-10 x-24=0by completing the square. We first add 24 to both sides to obtain
x2-10 x+25=24+25
namely,
(x-5)2=49
We now extract the square roots to obtain
x-5=47=7 or x-5= 49= 7
thus,
x=12 or x= 2
Therefore, the solution set is {12, 2}.
Example 2. Solve 4 x2+4 x-3=0.
we add 3 to both sides and then divide by 4 to obtain
x2+x=34
Adding (12)2=14to both sides we obtain
x2+x+14=34+14
or
(x+12)2=1
Extracting roots we obtain
x+12=1 or x+12= 1.
Thus the solution set is {12,32}.
The method of completing the square has other applications.
This is how our quadratic equation step by step solver solves the problem above. You can see similar problems solved by clicking on 'Solve similar' button.
Example 3. Write x2+y2-4 x+6 y-3=0 in the form (x±h)2+(y±k)2=r2
We group the x terms and the y terms obtaining quadratic expressions in x and y. We then complete the square in each expression.
x2+y2-4 x+6 y-3=0
x2-4 x+y2+6 y=3
x2-4 x+4+y2+6 y+9=3+4+9
(x-2)2+(y+3)2=42
Example 4. Write y=x2+2 x+3 in the form y±k=(x±h)2.
y=x2+2 x+3
y-3=x2+2 x
y-3+1=x2+2 x+1
y-2=(x+1)2
We can apply the methods of completing the square to the general quadratic equation
ax2+bx+x=0, a0
ax2+bx= c
Divide sides by a.
x2+bax= ca
Add (b2 a)2=b24 a2 to both sides.
x2+bax+b24 a2= ca+b24 a2
(x+b2 a)2=b2-4 ac4 a2
Take the square root.
x+b2 a
=± b2-4 ac4 a2
=± b2-4 ac2 a
Thus
x= b±sqrt ((b2-4 ac))2 a
This formula, called the quadratic formula, gives both solutions of the general quadratic equation expressed in terms of the coefficients a,b, and c. The symbol ± is used to condense the two equations.
x= b+sqrt ((b2-4 ac))2 ax= b-sqrt ((b2-4 ac))2 a
The solution set of the general quadratic equation is
{x= b+sqrt ((b2-4 ac))2 a,x= b-sqrt ((b2-4 ac))2 a}
In considering the quadratic formula we can see that the nature of the solutions-complex, irrational, or rational-depends entirely on the term under the radical, namely,
b2-4 ac
This term is called the discriminant of the quadratic equation.
1. The solutions are real if b2-4 ac0 and complex if b2-4 ac<0.
2. If a,b, and c are rational and b2-4 ac0, then the solutions are rational if and only if b2-4 ac is a perfect square.
Example 1. Solve the equation 3 x2+x-2=0 by the quadratic formula.
Substitute a=3b=1,c= 2in the quadratic formula.
x= b±sqrt ((b2-4 ac))2 a
= 1±1-4 (3) ( 2)2·3
= 1±256
= 1±56
=46, 66
Hence the solution set is {23, 1}
Example 2. Solve the equation x2+4 x+16=0.
We have a=1b=4, and c=16.
x= 4±16-4 (1) (16)2·1
= 4± 482
= 4±4 3 i2
= 2±2 3 i
Thus, the solution set is { 2+2 3 i, 2-2 sqrt (3) i}.
This is how our quadratic equation step by step solver solves the problem above. You can see similar problems solved by clicking on 'Solve similar' button.
In Chapter 3 we were able to factor some polynomials of second degree with integral coefficients. We are now in a position to consider the problem of factoring any second degree polynomial where the coefficients may now be real numbers. We must allow the use of complex numbers in the factors.
Consider the general second degree polynomial and use the method of completing the square.
ax2+bx+c
=a (x2+ba x+ca)
=a (x2+ba x+b24 a2+ca-b24 a2)
=a [(x+b2 a)2-(b2-4 ac4 a2)]
=a [(x+b2 a)2-(b2-4 ac2 a)2]
=a [(x+b2 a)-(b2-4 ac2 a)][(x+b2 a)+b2-4 ac2 a]
=a [x-( b+b2-4 ac2 a)][x-( b-b2-4 ac2 a)]
If we donate b+b2-4 ac2 a by r1 and b-b2-4 ac2 a by r2, then we have
ax2+bx+c=a (x-r1) (x-r2)
We observe that the numbers r1, and r2 are just the solutions of ax2+bx+c=0. These solutions, r1 and r2, are called the roots of the polynomial ax2+bx+c. Thus, to factor ax2+bx+c, we obtain its roots by solving the polynomial equation ax2+bx+c=0.
Example 3. Factor 3 x2+2 x+1.
We find the two roots of the polynomial, that is, the solutions of the polynomial equation
3 x2+2 x+1=0
x= 2±4-4 (3) (1)6
= 2± 86
= 1±2 i3
Since the two solutions are 1+2 i3 and 1-2 i3, we have
3 x2+2 x+1=3 (x- 1+2 i3) (x- 1-2 i3)
If an equation in more than one variable is a quadratic equation in one of its variables, then the equation can be solved for that variable by using the quadratic formula.
Example 4. Solve x2+5+2 y= 7 xy for x.
x2+5+2 y=7 xy
x2-7 xy+(2 y+5)=0
As a quadratic in x we have a=1,b=7 y, and c=2 y+5. Therefore,
x= 7 y±(7 y)2-4 (1) (2 y+5)2
= 7 y±49 y2-8 y-202
By various techniques certain types of equations lead to quadratic equations. Often these techniques produce equations that are not equivalent to the original ones. The steps in the solution of a given problem should be carefully reviewed to see if the replacement set changes at a given step. If so, we must make sure that the solutions of the final equation are solutions of the original equation.
Example 1. Solve the equation x+5x+3-x+6x=23.
Note that x0,3. Multiply by the L.C.D. 3 x (x-3).
3 x (x+5)-3 (x-3) (x+6)=2 x (x-3)
Solve the equation
3 x (x+5)-3 (x-3) (x+6)=2 x (x-3)
3 x2+15 x-3 x2-9 x+54=2 x2-6 x
2 x2-12 x-54=0
x2-6 x-27=0
(x-9) (x+3)=0
The solution set of the last equation is {9, 3}. This is also the solution set of the original equation, since the only numbers not in the replacement set of the original equation are 0 and 3, neither of which is an element of the solution set of the last equation.
Example 2. Find all real solutions of the equation 2 x+5=x+1.
Square both sides.
2 x+5=(x+1)2
The replacement set of this equation is the set of all real numbers, while the replacement set of the original equation is not the set of all real numbers. Since the operation of squaring enlarges the replacement set, the solution set may also be enlarged. Therefore we must check the solutions that we obtain for the second equation in the original equation.
2 x+5=(x+1)2
2 x+5=x2+2 x+1
x2=4
x=± 2
The possible solution are 2 and 2.
2 (2)+5=9
=3
(2)+1=3
Therefore, 2 is a solution.
2 ( 2)+5=1
=1
( 2)+1= 1
Therefore, 2 is not a solution. The solution set is {2}.
Example 3. Solve the equation (x2-4)2-5 (x2-4)+6=0.
(x2-4)2-5 (x2-4)+6=0
Let A=x2-4
A2-5 A+6=0
Factor.
(A-3) (A-2)=0
A=3, A=2
Since A=x2-4
x2-4=3 or x2-4=2
both of which we now solve.
x2-4=3
x2=7
x=± 7
x2-4=2
x2=6
x=± 6
The solution set is therefore {7, 7,6, 6}.
Example 4. Find all real solution of Z32-7 Z34-8=0.
Z32-7 Z34-8=0
Let B=Z34, then B2=Z32.
B2-7 B-8=0
Factor.
(B-8) (B+8)=0
B=8 or B= 1
Since B=Z34,
Z34=8 or Z34= 1
Solving, we have
Z34=8
Z=834
=24
=16
Z34= 1
Z=( 1)43
=1
The solution Z=1, however, does not satisfy the equation Z34= 1. In fact we could have observed that this equation has no real solution, since for negative real numbers Z,Z34 is not defined, while for non-negative Z,Z34 is non-negative. It is easy to see that Z=16 does satisfy our original equation, and hence the solution set is {16}.
Example 5. Solve x2+125=0.
Factor.
x2+125=0
x3+53=0
(x+5) (x2-5 x+25)=0
x+5=0 or x2-5 x+25=0
Solve.
x+5=0
x= 5
x2-5 x+25=0
x=5±25-4 (1) (25)2 (1)
=5± 752
=5±5 3 i2
Hence the solution set is { 5,52+(5 32) i,52-(5 32) i}.
Example 6. Solve 2 x+4=x-2.
2 x+4=x-2
Square both sides.
(2 x+4)2=(x-2)2
2 x+4=x-4 x+4
x= 4 x
Square both sides.
x2=( 4 x)2
x2=16 x
x2-16 x=0
x (x-16)=0
x=0 or x=16
Check.
2 (0)+4=2
0-2= 2
Therefore, 0 is not a solution.
2 (16)+4=36
=6
16-2=4-2
=2
Therefore, 16 is not a solution. Since 0 and 16 are the only possible solutions, the original equation has no solution. In case an equation has no solutions we use the symbol to represent the solution set, that is, the set with no elements. The set is called the empty set.
Let’s see how our math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.
6.10 Statement Problems of the Quadratic Type
Our method of approach will be the same as in Section 6.4, only here our equation will be one that yields a quadratic equation in a single variable.
Example 1. Find two numbers whose sum is 8 and whose product is 12.
Step 2. Let x be one of the numbers.
Step 3. Since their product is 12 the other number is 12x.
Step 4. The sum of the two numbers is 8.
Step 5. x+12x=8
Step 6. Solve.
x2+12=8 x
x2-8 x+12=0
(x-6) (x-2)=0
x=6 or x=2
In either case the two numbers are 6 and 2.
Example 2. If the perimeter of a rectangle is 44 ft while the area is 120 ft2,what are its dimensions?
Step 2. Let x be the length of one side.
Step 3. Since 12 of the perimeter is 22 ft the other side is 22-x.
Step 4. The area of the rectangle is 120 ft2.
Step 5. x (22-x)=120
Step 6. Solve.
22-x2=120
x2-22 x+120=0
(x-10) (x+12)=0
x=10 or x=12
In either case the two sides are 10 ft and 12 ft in length.
Example 3. Within a rectangular garden 18 yards long and 10 yards wide we are to pave a border of uniform width. If we are to leave an area of 84 square yards for flowers, how wide should the walk be?
Step 2. Let x be the width of the walk.
Step 3. The dimensions of the inside area would be 18-2 x and 10-2 x.
Step 4. The inside area is to be 84 sq. yds.
Step 5. (18-2 x) (10-2 x)=84
Step 6. Solve.
180-56 x+4 x2=84
4 x2-56 x+96=0
x2-14 x+24=0
(x-12) (x-2)=0
x=12 or x=2
Clearly the width of the walk would be 2 yards and not 12 yards.
Example 4. If Joe takes 6 hours longer than his father to assemble a machine and together they can assemble it in 4 hours, how long would it take each of them working alone to assemble the machine?
Step 2. Let x be the number of hours it would take the father to assemble the machine.
Step 3. It would take x+6 hours for Joe to assemble the machine. Choosing one hour as our unit of time, 1x would be the amount the father could do in one hour, while 14 would be the amount that they could do together in one hour.
Step 4. The amount that Joe can do in one hour plus the amount that his father can do in one hour would be equal to 14.
Step 5. 1x+6+1x=14.
Step 6. Solve. Multiply by the L.C.D. 4 x (x+6).
4 x+4 (x+6)=x (x+6)
4 x+4 x+24=x2+6 x
x2-2 x-24=0
(x-6) (x+4)=0
x=6 or x= 4
Clearly x= 4 is not a possibility and therefore it would take Joe 12 hours and his father 6 hours.
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# Algebra 1 : How to find a fraction from a percentage
## Example Questions
← Previous 1 3 4 5 6 7
### Example Question #1 : How To Find A Fraction From A Percentage
How do you display 40% as a fraction?
Explanation:
To display a percentage as a fraction, divide the number of percent by 100 to give you . Then simplify the fraction by dividing each number by 20, yielding .
### Example Question #2 : How To Find A Fraction From A Percentage
What is 135% of 45?
Explanation:
This is just a simple problem of translation. We know that and can translate the rest of the problem into this form: . Merely multiply and get the answer, namely .
### Example Question #3 : How To Find A Fraction From A Percentage
What is 12.5% expressed as a fraction?
Explanation:
is easier to simplify if you multiply top and bottom by 2.
Then it becomes .
Divide the numerator and denominator by 25 and you get .
### Example Question #4 : How To Find A Fraction From A Percentage
What is as a fraction?
Explanation:
To display a percentage as a fraction, divide the number of percent by 100 to give you .
Explanation:
### Example Question #6 : How To Find A Fraction From A Percentage
Each year, the value of a car goes down by 10% of its previous year's value. What fraction of its original new value is the value of a three-year old car?
Explanation:
The reduction of the value of a car by 10% is the same as the car retaining 90%, or , of its value.
If a car is worth when new, then it is worth after one year. Its value after two years will be times its value after one year, or . After three years, it will be times its value after two years, or .
### Example Question #7 : How To Find A Fraction From A Percentage
Convert to a fraction.
Explanation:
To convert a percentage into a fraction, remove the percentage sign and divide the number by 100.
### Example Question #8 : How To Find A Fraction From A Percentage
What fraction represents ?
Explanation:
When finding the fraction from a percent, express that value over .
We now have
.
We simplify it by dividing top and bottom by .
### Example Question #9 : How To Find A Fraction From A Percentage
Convert into a reduced fraction.
Explanation:
First, we take that value and divide over .
We have
.
Next, since there is a decimal present, we shift that decimal one place so we have a whole number. Since we did that to the top, we add one zero to the bottom since we shifted one place. We have
.
To reduce, we divide top and bottom by
to get as the final answer.
### Example Question #10 : How To Find A Fraction From A Percentage
Convert into a reduced fraction.
Explanation:
We take that value and divide that over .
We get
.
Next we can reduce by taking one zero from top and bottom and also divide top and bottom by .
This gives us an answer of
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# Pappus's area theorem
Pappus's area theorem dark grey area = light grey area Pappus's area theorem describes the relationship between the areas of three parallelograms attached to three sides of an arbitrary triangle. The theorem, which can also be thought of as a generalization of the Pythagorean theorem, is named after the Greek mathematician Pappus of Alexandria (4th century AD), who discovered it.
Contents 1 Theorem 2 Proof 3 References 4 External links Theorem Given an arbitrary triangle with two arbitrary parallelograms attached to two of its sides the theorem tells how to construct a parallelogram over the third side, such that the area of the third parallelogram equals the sum of the areas of the other two parallelograms.
Let ABC be the arbitrary triangle and ABDE and ACFG the two arbitrary parallelograms attached to the triangle sides AB and AC. The extended parallelogram sides DE and FG intersect at H. The line segment AH now "becomes" the side of the third parallelogram BCML attached to the triangle side BC, i.e., one constructs line segments BL and CM over BC, such that BL and CM are a parallel and equal in length to AH. The following identity then holds for the areas (denoted by A) of the parallelograms: {displaystyle {text{A}}_{ABDE}+{text{A}}_{ACFG}={text{A}}_{BCML}} The theorem generalizes the Pythagorean theorem twofold. Firstly it works for arbitrary triangles rather than only for right angled ones and secondly it uses parallelograms rather than squares. For squares on two sides of an arbitrary triangle it yields a parallelogram of equal area over the third side and if the two sides are the legs of a right angle the parallelogram over the third side will be square as well. For a right-angled triangle, two parallelograms attached to the legs of the right angle yield a rectangle of equal area on the third side and again if the two parallelograms are squares then the rectangle on the third side will be a square as well.
Proof Due to having the same base length and height the parallelograms ABDE and ABUH have the same area, the same argument applying to the parallelograms ACFG and ACVH, ABUH and BLQR, ACVH and RCMQ. This already yields the desired result, as we have: {displaystyle {begin{aligned}{text{A}}_{ABDE}+{text{A}}_{ACFG}&={text{A}}_{ABUH}+{text{A}}_{ACVH}\&={text{A}}_{BLQR}+{text{A}}_{RCMQ}\&={text{A}}_{BCML}end{aligned}}} References Howard Eves: Pappus's Extension of the Pythagorean Theorem.The Mathematics Teacher, Vol. 51, No. 7 (November 1958), pp. 544–546 (JSTOR) Howard Eves: Great Moments in Mathematics (before 1650). Mathematical Association of America, 1983, ISBN 9780883853108, p. 37 (excerpt, p. 37, at Google Books) Eli Maor: The Pythagorean Theorem: A 4,000-year History. Princeton University Press, 2007, ISBN 9780691125268, pp. 58–59 (excerpt, p. 58, at Google Books) Claudi Alsina, Roger B. Nelsen: Charming Proofs: A Journey Into Elegant Mathematics. MAA, 2010, ISBN 9780883853481, pp. 77–78 (excerpt, p. 77, at Google Books) External links The Pappus Area Theorem Pappus theorem hide vte Ancient Greek and Hellenistic mathematics (Euclidean geometry) Mathematicians (timeline) AnaxagorasAnthemiusArchytasAristaeus the ElderAristarchusApolloniusArchimedesAutolycusBionBrysonCallippusCarpusChrysippusCleomedesCononCtesibiusDemocritusDicaearchusDioclesDiophantusDinostratusDionysodorusDomninusEratosthenesEudemusEuclidEudoxusEutociusGeminusHeliodorusHeronHipparchusHippasusHippiasHippocratesHypatiaHypsiclesIsidore of MiletusLeonMarinusMenaechmusMenelausMetrodorusNicomachusNicomedesNicotelesOenopidesPappusPerseusPhilolausPhilonPhilonidesPorphyryPosidoniusProclusPtolemyPythagorasSerenus SimpliciusSosigenesSporusThalesTheaetetusTheanoTheodorusTheodosiusTheon of AlexandriaTheon of SmyrnaThymaridasXenocratesZeno of EleaZeno of SidonZenodorus Treatises AlmagestArchimedes PalimpsestArithmeticaConics (Apollonius)CatoptricsData (Euclid)Elements (Euclid)Measurement of a CircleOn Conoids and SpheroidsOn the Sizes and Distances (Aristarchus)On Sizes and Distances (Hipparchus)On the Moving Sphere (Autolycus)Euclid's OpticsOn SpiralsOn the Sphere and CylinderOstomachionPlanisphaeriumSphaericsThe Quadrature of the ParabolaThe Sand Reckoner Problems Constructible numbers Angle trisectionDoubling the cubeSquaring the circleProblem of Apollonius Concepts and definitions Angle CentralInscribedChordCircles of Apollonius Apollonian circlesApollonian gasketCircumscribed circleCommensurabilityDiophantine equationDoctrine of proportionalityGolden ratioGreek numeralsIncircle and excircles of a triangleMethod of exhaustionParallel postulatePlatonic solidLune of HippocratesQuadratrix of HippiasRegular polygonStraightedge and compass constructionTriangle center Results In Elements Angle bisector theoremExterior angle theoremEuclidean algorithmEuclid's theoremGeometric mean theoremGreek geometric algebraHinge theoremInscribed angle theoremIntercept theoremIntersecting chords theoremIntersecting secants theoremLaw of cosinesPons asinorumPythagorean theoremTangent-secant theoremThales's theoremTheorem of the gnomon Apollonius Apollonius's theorem Other Aristarchus's inequalityCrossbar theoremHeron's formulaIrrational numbersLaw of sinesMenelaus's theoremPappus's area theoremProblem II.8 of ArithmeticaPtolemy's inequalityPtolemy's table of chordsPtolemy's theoremSpiral of Theodorus Centers CyreneLibrary of AlexandriaPlatonic Academy Other Ancient Greek astronomyGreek numeralsLatin translations of the 12th centuryNeusis construction Ancient Greece portal • Mathematics portal Wikimedia Commons has media related to Pappus's area theorem. Categories: AreaEquationsEuclidean plane geometryTheorems about triangles
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## Introduction
An exponent is a mathematical way to show a repeated multiplication. The problem x to the sixth power can be shown as follows:
x6 = x·x·x·x·x·x
At times, an exponent may apply to more than one variable. When this happens, the exponent can be applied to each variable individually.
(xyz)5 = xyz·xyz·xyz·xyz·xyz
= x·x·x·x·x · y·y·y·y·y · z·z·z·z·z
= x5y5z5
## Lesson
Before discussing the main idea in this lesson, consider a similar property that you already know: the distributive property.
A mailman delivers mail to each mailbox along his route. Similarly, the distributive property distributes single number or term that is being multiplied by several terms. When an expression consists of several variables or numbers that are multiplied together, applying an exponent is done in much the same way as the distributive property.
When an exponent is applied to more than one number or variable, the exponent should be applied to each variable individually. The problem below contains the expanded work that shows why this is true.
Instead of doing all that work, there is a property that allows you to skip the repeated addition and use multiplication instead when taking several items in parenthesis to a single exponent. This property is called the power of products property.
The power of products property mentions two numbers or variables being multiplies, but it can also be used with products of 3, 4, 5, or more numbers. Some visual examples of how it works are given below:
Example #1 is a simple example of the power of products property.
Example #1: Simplify the expression (2xyz)4
Using the property uses the same thinking as simply applying the rules of exponents. Simple problems such as the one above should only take a few seconds using the property. When tackling a problem with more variables and/or larger exponents, the value of remembering properties such as the power of products property is priceless.
Example 2: Simplify the expression (3a4b7c12d24)3
The properties of exponents have names that are very similar. The ones you have learned thus far are:
When you are presented with a problem, it is easy to mix up the adding and multiplying involved in these properties. Some problems even use two or more of these properties. Try the following two examples yourself before revealing the answers.
Example 3: Simplify the expression 3x2(y2z3)4·(xy)5z2
Example 4: Simplify the expression (2f3g2h)3·5fg4·(f2)3(gh2)2
These two examples use all three of the properties of exponents above. It helps to peek at these properties as you start out, then after successfully completing several problems you should try doing more problems without peeking at the properties. If you are working in a math book that has the answers in the back, then check the results as you do the problems. Knowing that you are doing the problems correctly will make you feel more comfortable with the material and confident in your abilities.
## Try It
Evaluate each expression:
1) (abc)4 =
2) 2(s2t)2 =
3) (2s2t)2 =
4) (d3e4)3 =
5) (3f3g4)2 =
Evaluate each more complex expression:
6) 2r5(s2t5)2·r4(s2t3)4 =
7) (2x3y2)2·3y7z3·(x2)4(z2)2 =
8) [(2m2n3p)2]2·[m2(n3p)3]2 =
Scroll Down for Answers…
Answers:
1) (abc)4 = a4b4c4
2) 2(s2t)2 = 2s4t2
3) (2s2t)2 = 4s4t2
4) (d3e4)3 = d9e12
5) (3f3g4)2 = 9f6g8
Evaluate each more complex expression:
6) 2r5(s2t5)2·r4(s2t3)4 = 2r5s4t10 · r4s8t12
= 2r9s12t22
7) (2x3y2)2·3y7z3·(x2)4(z2)2 = 4x6y4 · 3y7z3 · x8z4
= 12x14y11z7
8) [(2m2n3p)2]2·[m2(n3p)3]2 = [4m4n6p2]2 · [m2n9p3]2
= 16m8n12p4 · m4n18p6
= 16m12n30p10
### Related Links:
Didn't find what you were looking for in this lesson? More information on exponents can be found at the following places:
Resource Page
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## Section8.5Complex Solutions to Quadratic Equations
### Subsection8.5.1Imaginary Numbers
Let's look at how to simplify a square root that has a negative radicand. Remember that $\sqrt{16}=4$ because $4^2=16\text{.}$ So what could $\sqrt{-16}$ be equal to? There is no real number that we can square to get $-16\text{,}$ because when you square a real number, the result is either positive or $0\text{.}$ You might think about $4$ and $-4\text{,}$ but:
\begin{equation*} 4^2=16\text{ and }(-4)^2=16 \end{equation*}
so neither of those could be $\sqrt{-16}\text{.}$ To handle this situation, mathematicians separate a factor of $\sqrt{-1}$ and represent it with the letter $i\text{,}$ which stands for imaginary unit.
###### Definition8.5.2Imaginary Numbers
The imaginary unit, $i\text{,}$ is defined by $i=\sqrt{-1}\text{.}$ The imaginary unit 1 en.wikipedia.org/wiki/Imaginary_number satisfies the equation $i^2=-1\text{.}$ A real number times $i\text{,}$ such as $4i\text{,}$ is called an imaginary number.
Now we can simplify square roots with negative radicands like $\sqrt{-16}\text{.}$
\begin{align*} \sqrt{-16}\amp=\sqrt{-1\cdot16}\\ \amp=\sqrt{-1}\cdot\sqrt{16}\\ \amp=i\cdot4\\ \amp=4i \end{align*}
Imaginary numbers are widely used in electrical engineering, physics, computer science and other fields. Let's look some more examples.
###### Example8.5.3
Simplify $\sqrt{-2}\text{.}$
Explanation
\begin{align*} \sqrt{-2}\amp=\sqrt{-1\cdot2}\\ \amp=\sqrt{-1}\cdot\sqrt{2}\\ \amp=i\sqrt{2} \end{align*}
We write the $i$ first because it's difficult to tell the difference between $\sqrt{2}i$ and $\sqrt{2i}\text{.}$
###### Example8.5.4
Simplify $\sqrt{-72}\text{.}$
Explanation
\begin{align*} \sqrt{-72}\amp=\sqrt{-1\cdot36\cdot2}\\ \amp=\sqrt{-1}\cdot\sqrt{36}\cdot\sqrt{2}\\ \amp=6i\sqrt{2} \end{align*}
### Subsection8.5.2Solving Quadratic Equations with Imaginary Solutions
###### Example8.5.5
Solve for $x$ in $x^2+49=0\text{,}$ where $x$ is an imaginary number.
Explanation
There is no $x$ term so we will use the square root method.
\begin{align*} x^2+49\amp=0\\ x^2\amp=-49 \end{align*}
\begin{align*} x\amp=-\sqrt{-49} \amp\text{ or }\amp\amp x\amp=\sqrt{-49}\\ x\amp=-\sqrt{-1}\cdot\sqrt{49} \amp\text{ or }\amp\amp x\amp=\sqrt{-1}\cdot\sqrt{49}\\ x\amp=-7i\amp\text{ or }\amp\amp x\amp=7i \end{align*}
The solution set is $\{-7i,7i\}\text{.}$
###### Example8.5.6
Solve for $p$ in $p^2+75=0\text{,}$ where $p$ is an imaginary number.
Explanation
There is no $p$ term so we will use the square root method.
\begin{align*} p^2+75\amp=0\\ p^2\amp=-75 \end{align*}
\begin{align*} p\amp=-\sqrt{-75} \amp\text{ or }\amp\amp p\amp=\sqrt{-75}\\ p\amp=-\sqrt{-1}\cdot\sqrt{25}\cdot\sqrt{3} \amp\text{ or }\amp\amp p\amp=\sqrt{-1}\cdot\sqrt{25}\cdot\sqrt{3}\\ p\amp=-5i\sqrt{3}\amp\text{ or }\amp\amp p\amp=5i\sqrt{3} \end{align*}
The solution set is $\left\{-5i\sqrt{3},5i\sqrt{3}\right\}\text{.}$
### Subsection8.5.3Solving Quadratic Equations with Complex Solutions
A complex number is a combination of a real number and an imaginary number, like $3+2i$ or $-4-8i\text{.}$
###### Definition8.5.7Complex Number
A complex number is a number that can be expressed in the form $a + bi\text{,}$ where $a$ and $b$ are real numbers and $i$ is the imaginary unit. In this expression, $a$ is the real part and $b$ (not $bi$) is the imaginary part of the complex number 2 en.wikipedia.org/wiki/Complex_number.
Here are some examples of equations that have complex solutions.
###### Example8.5.8
Solve for $m$ in $(m-1)^2+18=0\text{,}$ where $m$ is a complex number.
Explanation
This equation has a squared expression so we will use the square root method.
\begin{align*} (m-1)^2+18\amp=0\\ (m-1)^2\amp=-18 \end{align*}
\begin{align*} m-1\amp=-\sqrt{-18} \amp\text{ or }\amp\amp m-1\amp=\sqrt{-18}\\ m-1\amp=-\sqrt{-1}\cdot\sqrt{9}\cdot\sqrt{2}\amp\text{ or }\amp\amp m-1\amp=\sqrt{-1}\cdot\sqrt{9}\cdot\sqrt{2}\\ m-1\amp=-3i\sqrt{2}\amp\text{ or }\amp\amp m-1\amp=3i\sqrt{2}\\ m\amp=1-3i\sqrt{2}\amp\text{ or }\amp\amp m\amp=1+3i\sqrt{2} \end{align*}
The solution set is $\left\{1-3i\sqrt{2}, 1+3i\sqrt{2}\right\}\text{.}$
###### Example8.5.9
Solve for $y$ in $y^2-4y+13=0\text{,}$ where $y$ is a complex number.
Explanation
Note that there is a $y$ term, but the left side does not factor. We will use the quadratic formula. We identify that $a=1\text{,}$ $b=-4$ and $c=13$ and substitute them into the quadratic formula.
\begin{align*} y\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(13)}}{2(1)}\\ \amp=\frac{4\pm\sqrt{16-52}}{2}\\ \amp=\frac{4\pm\sqrt{-36}}{2}\\ \amp=\frac{4\pm\sqrt{-1}\cdot\sqrt{36}}{2}\\ \amp=\frac{4\pm6i}{2}\\ \amp=2\pm 3i \end{align*}
The solution set is $\{2- 3i, 2+ 3i\}\text{.}$
Note that in Example 8.5.9, the expressions $2+3i$ and $2-3i$ are fully simplified. In the same way that the terms $2$ and $3x$ cannot be combined, the terms $2$ and $3i$ can not be combined.
###### Remark8.5.10
Each complex solution can be checked, just as every real solution can be checked. For example, to check the solution of $2+3i$ from Example 8.5.9, we would replace $y$ with $2+3i$ and check that the two sides of the equation are equal. In doing so, we will need to use the fact that $i^2=-1\text{.}$ This check is shown here:
\begin{align*} y^2-4y+13\amp=0\\ (\substitute{2+3i})^2-4(\substitute{2+3i})+13\amp\stackrel{?}{=}0\\ (2^2+2(3i)+2(3i)+(3i)^2)-4\cdot 2 -4\cdot (3i) +13 \amp\stackrel{?}{=}0\\ 4+6i+6i+9i^2-8-12i+13 \amp\stackrel{?}{=}0\\ 4+9(-1)-8+13 \amp\stackrel{?}{=}0\\ 4-9-8+13 \amp\stackrel{?}{=}0\\ 0\amp\stackrel{\checkmark}{=}0 \end{align*}
### Subsection8.5.4Exercises
###### 1
Simplify the radical and write it into a complex number.
$\displaystyle{ \sqrt{-105} = }$
###### 2
Simplify the radical and write it into a complex number.
$\displaystyle{ \sqrt{-42} = }$
###### 3
Simplify the radical and write it into a complex number.
$\displaystyle{ \sqrt{-72} =}$
###### 4
Simplify the radical and write it into a complex number.
$\displaystyle{ \sqrt{-56} =}$
###### 5
Simplify the radical and write it into a complex number.
$\displaystyle{ \sqrt{-126} =}$
###### 6
Simplify the radical and write it into a complex number.
$\displaystyle{ \sqrt{-200} =}$
###### 7
Solve the quadratic equation. Solutions could be complex numbers.
$y^2 = -16$
###### 8
Solve the quadratic equation. Solutions could be complex numbers.
$r^2 = -100$
###### 9
Solve the quadratic equation. Solutions could be complex numbers.
$-6r^2+2 = 296$
###### 10
Solve the quadratic equation. Solutions could be complex numbers.
$-3t^2+2 = 50$
###### 11
Solve the quadratic equation. Solutions could be complex numbers.
${2t^{2}} + 10 = 6$
###### 12
Solve the quadratic equation. Solutions could be complex numbers.
${2t^{2}} - 2 = -8$
###### 13
Solve the quadratic equation. Solutions could be complex numbers.
$-6x^2+4 = 124$
###### 14
Solve the quadratic equation. Solutions could be complex numbers.
$-6x^2 - 7 = 293$
###### 15
Solve the quadratic equation. Solutions could be complex numbers.
$9(y+4)^2+1 = -323$
###### 16
Solve the quadratic equation. Solutions could be complex numbers.
$6(y - 5)^2 - 7 = -61$
###### 17
Solve the quadratic equation. Solutions could be complex numbers.
${r^{2}+2r+5} = 0$
###### 18
Solve the quadratic equation. Solutions could be complex numbers.
${r^{2}+6r+10} = 0$
###### 19
Solve the quadratic equation. Solutions could be complex numbers.
${t^{2}+4t+9} =0$
###### 20
Solve the quadratic equation. Solutions could be complex numbers.
${t^{2}+4t+11} =0$
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# Math 316 Solutions To Sample Exam 2 Problems
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## Transcription
1 Solutions to Sample Eam 2 Problems Math 6 Math 6 Solutions To Sample Eam 2 Problems. (a) By substituting appropriate values into the binomial theorem, find a formula for the sum ( ) ( ) ( ) ( ) ( ) n n n n n n. 0 2 n (b) Give a combinatorial proof of the formula you found in part (a). (Hint: Count the number of ways to paint n different houses either red, blue, green, or not at all.) (a) Letting = and y = in the binomial theorem, we have ( ) ( ) ( ) n n n n = (+) n = n + n + n ( ) ( ) ( ) ( ) ( ) n n n n n = n, 0 2 n ( ) n n + + ( ) n 0 n n so our formula for the sum is n. (b) The formula that we are trying to prove is ( ) ( ) ( ) ( ) ( ) n n n n n ( ) n 0 2 n = n. We claim that both sides of ( ) count the number of ways to paint n distinct houses either red, blue, green or not at all, which we verify below: Left-hand side: Here, we enumerate the number of different ways to paint by using cases according to the number of houses we paint. Case. (No houses get painted.) In this case, we are effectively choosing 0 houses to paint, which can be done in C(n,0) ways. Case 2. ( house gets painted.) We break this task into two parts: pick house to be painted in C(n,) ways, and then choose to paint it either red, green, or blue in ways. Case. (2 houses get painted.) Again, break this task into two parts: pick 2 houses to be painted in C(n,2) ways, and then choose colors for both houses in 2 ways.. Case n. (n houses get painted.) Pick all n houses to be painted in C(n,n) ways, and then choose colors for all n houses in n ways. Summing over all the cases above accounts for all ways of painting and gives the left-hand side of ( ). Right-hand side: Here, we enumerate the number of different ways to paint by simply observing that for each house, we have four choices: either paint it red, blue, green, or don t paint it at all. Therefore, there are n total ways to paint, which gives the right-hand side of ( ). 2. How many 0-letter words from the standard 26-letter English alphabet contain each of the letters a, b, c, and d at least once? Since it is easier to count the number of words that are missing a particular letter than those that have it, we begin by defining the following sets: S = {0-letter words }
2 Solutions to Sample Eam 2 Problems Math 6 2 A A 2 A A = {0-letter words which have no a s} = {0-letter words which have no b s} = {0-letter words which have no c s} = {0-letter words which have no d s} Using Inclusion/Eclusion, our answer will then be S A A 2 A A = 26 0 A A 2 A A = 26 0 A i + A i A j A i A j A k + A A 2 A A We now eamine each of the sums separately, noting that each sum represents the number of words missing a particular number of letters: Words in A i ; that is, words missing one letter: Choose a value of i C(,) ways (Here, we are essentially choosing one of the four letters, a,b,c, or d, for our word to be missing.) Choose letters for our 0 positions 25 0 ways (Note that, since each word is missing one letter, there are onl5 choices for each of our 0 positions.) Words in A i A j ; that is, words missing two letters: Choose a value for i and j C(,2) ways (Here, we are essentially choosing two of the four letters, a,b,c, or d, for our word to be missing.) Choose letters for our 0 positions 2 0 ways (Note that, since each word is missing two letters, there are onl choices for each of our 0 positions.) Words in A i A j A k : Choose a value for i,j, and k C(,) ways Choose letters for our 0 positions 2 0 ways Words in A i A j A k A p : Choose a value for i,j, and k, and p C(,) = way Choose letters for our 0 positions 22 0 ways To summarize, we see that there are C(,) 25 0 words that are missing eactly one of the letters a,b,c, or d; there are C(,2) 2 0 words missing eactly two of the letters; there are C(,) 2 0 words that are missing eactly three of the letters, and C(,) 22 0 words missing all four of the letters. Referring to the equation above, our final answer is therefore ( ) ( ) ( ) ( ) Solve the following recurrence relations. (a) h n = h n +6h n 2 (n 2), h 0 = 7, h = (b) h n = 5h n 8h n 2 +h n (n ), h 0 = 2, h =, h 2 =
3 Solutions to Sample Eam 2 Problems Math 6 (a) First, note that the characteristic equation of this recurrence is given by = 0, so we have which yields a general solution of = 0 = (+)( 2) = 0, h n = a 2 n + b ( ) n. The initial conditions h 0 = 7 and h = lead to the equations a+b = 7 and 2a b =, respectively. Solving for b in the first equation, we get b = 7 a, and substituting into the second equation, we obtain 2a (7 a) = = 2a 2+a = = 5a = 0, so we conclude that a = 2 and b = 7 a = 5. Our final answer is therefore h n = 2 2 n + 5 ( ) n = 2 n+ + 5 ( ) n. (b) First, note that the characteristic equation of this recurrence is given by = 0, so we have = ( )( 2 +) = ( )( 2) 2 = 0, yielding a repeated characteristic root of 2 and a single root of. Our general solution is therefore given by h n = a 2 n + bn 2 n + c n = a 2 n + bn 2 n + c. Our initial conditions h 0 = 2, h =, and h 2 = lead to the equations a+c = 2, 2a+2b+c =, and a+8b+c =, respectively. Using standard elimination techniques, we reduce our system of equations as follows: a + c = 2 2a + 2b + c = a + 8b + c = a + c = 2 2b c = 5 8b c = 2 a + c = 2 2b c = 5 c = We therefore see that c =, so which yields a final answer of 2b = c+5 = b =, and a = 2 c =, h n = 2 n + n 2 n + = 2 n (n ) +.. A donut shop sells 0 different kinds of donuts, including chocolate donuts, powdered donuts, jelly donuts, and 27 other types. (a) Find a simplified generating function for the number of unordered selections of n donuts if i. there are no restrictions. ii. the selection must contain at least 5 chocolate donuts. iii. the selection must contain at least chocolate, at least powdered, but no more than jelly. (b) Sam wants to purchase 2 donuts for a wild pastry party at Dr. Luttmann s house. Unfortunately, the donut shop is out of chocolate donuts, but it is offering a special sale on jelly donuts: three for \$. Use generating functions to find the number of different donut selections Sam can make assuming that he wants to buy a multiple of number of jelly donuts. (a) For each of the following functions, recall that the eponent on the variable determines the number of donut selections made.
4 Solutions to Sample Eam 2 Problems Math 6 i. Since we can choose any number of each type of donut, our generating function is given by so our answer is f() = Kind Kind 2 Kind 0 f() = ( ) ( ) ( ) = ( ) 0, ( ) 0. ii. Since we want to have at least 5 chocolate donuts, the chocolate factor of our function will look like ( ), while the factors for the other 29 donut varieties will stay the same as in part (i). Therefore, we have so our final answer is f() = iii. In this case, we have f() = ( )( ) 29 = 5 ( )( ) 29 = 5 ( ) 0, 5 ( ) 0. chocolate powdered jelly f() = ( ) ( ) ( ) ( ) 27 = 2 ( ) 2 ( ) ( ) 27 = 2 ( )( ) 29 = 2 ( ) 29, which yields a final answer of f() = 2 ( ) ( ) 0. (b) First, we find a generating function for d n, the number of selections of n donuts in which no chocolate donuts are chosen, and a multiple of jelly donuts are chosen. Reasoning as above, our generating function looks like jelly f() = ( ) ( ) 28 = = ( ) 28 ( )( ) 28. Now, we can either use technology to find the Taylor series for f and read off the coefficient of the 2 term, or we can analyze algebraically as follows: f() = ( ) ( ) 28 = ( ) = ( ) 27+k k + k k=0 ( ) 27+k k k k=0 ( ) 27+k k+ + k k=0 ( ) 27+k k+6 + k We see from the above sum that we get terms with 2 from the st sum with k = 2, from the 2nd sum with k = 9, from the rd sum with k = 6, etc., which gives a final answer of ( ) ( ) ( ) ( ) ( ) ( ) 27+j h 2 = = j k=0 j=0
5 Solutions to Sample Eam 2 Problems Math Recall the Fibonacci sequence, which is defined by f 0 = 0, f =, and f n = f n +f n 2 for n 2. (a) By repeated use of the Fibonacci recursion formula, prove that f n = 5f n +f n 5 for all n 5. (b) Use part (a) to help you prove that f 5n is a multiple of 5 for all n 0. (a) Consider any n 5. Then we have f n = f n +f n 2 = f n +f n 2 +f n +f n = f n 2 +2f n +f n = f n +f n +2f n +f n = f n +2f n = (f n 5 +f n ) + 2f n = 5f n +f n 5. (b) We will prove this by induction on n. For n = 0, observe that which establishes the basis step. f 5n = f 0 = 0 = 5 0, Now, assume that f 5k is a multiple of 5 for some k 0. Then there eists an integer p such that f 5k = 5p, and we have f 5(k+) = f 5k+5 = 5f (5k+5) + f (5k+5) 5 by part (a) = 5f 5k+ + f 5k = 5f 5k+ + (5p) induction hypothesis = 5(f 5k+ +p), which shows that f 5(k+) is a multiple of 5. Therefore, by induction, we conclude that f 5n is a multiple of 5 for all integers n A fied chessboard is to be completely tiled with the two shapes of pieces shown to the right. Assume there is a plentiful supply of white monominoes, and plentiful supplies of red, yellow, and green of the Lshaped piece. (a) Assuming that the configurations and are not allowed anywhere in the tiling, how many different ways are there to tile a chessboard? (b) Now, assume that only white monominoes and red L-shaped pieces are available, but all configurations are now allowed, including the two that were prohibited in part (a). Find, but DO NOT SOLVE, a recurrence relation for g n, the number of ways to tile a 2 n chessboard with these pieces. Then, use the recurrence relation to find the number of tilings of a 2 5 chessboard. (a) Let h n represent the number of ways to tile a 2 n chessboard with the previously described pieces. We will begin by finding a recurrence relation for h n.
6 Solutions to Sample Eam 2 Problems Math 6 6 Referringtothediagramtotheright,weseethatthereareh n ways to tile the chessboard if the tiling starts with two monominoes in the first column. On the other hand, if the first two columns contain one L-shaped piece and a monomino, there are three possible colors for the L-shaped piece, giving h n 2 ways to complete the tiling. But there are different ways to orient the L-shaped piece by rotating it 90 each time, giving h n 2 = 2h n 2 total ways for a 2 n tiling to have one L-shaped piece and one monomino in the first two columns. Our recurrence formula is therefore given by h n t h n 2 h n 2 h n 2 h n 2 h n = h n + 2h n 2 for n. To determine the initial conditions, we first note that h = because there is only one way to tile a 2 chessboard: with two white monominoes. We also observe that h 2 = by enumerating all of the possibilities below: R Y G R Y G R Y G R Y G Now, we are ready to solve the recurrence relation h n = h n +2h n 2. Since the general solution to our recurrence relation is 2 2 = 0 = ( )(+) = 0, h n = a n + b ( ) n. The initial conditions h = and h 2 = lead to the equations a b = and 6a+9b =, which when solved yield a = /7 and b = /7. Therefore, h n = n+ ( ) n+ 7 and so there are h 200 = 20 ( ) 20 7 ways to tile a chessboard in the desired manner. for all n, = (b) Referring to the diagram to the right, we see that there are several different ways for a tiling of a 2 n chessboard to begin. If a tiling starts with two monominoes in the first column, there are g n ways to complete the tiling. If a tiling does not start with two monominoes, then it could either start with one L-shaped piece and a monomino (g n 2 like this) or with two L-shaped pieces (2g n like this). Our recurrence formula is therefore given by for n. Our goal is to find g 5. g n = g n + g n 2 + 2g n Type t Type 2 Type Type Type 5 Type 6 Type 7 In order to find g 5, we need several initial conditions. Clearly, g =, since a 2 chessboard can only be tiled in one way: with 2 monominoes. On the other hand, a 2 2 chessboard can either be tiled with monominoes, or with any of the configurations shown in the first two columns of the rows labeled
7 Solutions to Sample Eam 2 Problems Math 6 7 Type 2 through Type 5 in the diagram above, so g 2 = 5. Finally, a 2 chessboard can be tiled with si monominoes ( way), with the first column of the Type configuration followed by the first two columns of any of the Type 2 through Type 5 configurations ( ways), with the the first two columns of any of the Type 2 through Type 5 configurations followed by the first column of the Type configuration ( ways), or the first three columns of the Type 6 or Type 7 configurations (2 ways), giving Therefore, we have g = +++2 =. g = g + g 2 + 2g = = g 5 = g + g + 2g 2 = = 87, so we conclude that there are g 5 = 87 ways to tile a 2 5 chessboard with these pieces. 7. For each of the following bipartite graphs, you are given a matching M, indicated by the shaded edges. By finding successive M-alternating paths (if they eist), arrive at a ma-matching M, and then show that your final matching is a ma-matching by writing down an appropriate cover. Clearly indicate your paths and matchings at each step in the process. (a) (b) (c) y y y y y y y y y (a) Below is an M -alternating path γ : y γ :,, 2, y,, y This path leads to the ma-matching M = {{,y }, {, }, { 2,y }, {,y }}, which is confirmed by observing that the set 2 y y S = {, 2,, } is a cover of the graph with S = M =. The ma-matching M and the minimum cover S are illustrated in the diagram to the right. (b) Below is an M alternating path and its induced new matching M 2 with one more edge (see diagram to the immediate right): 2 y 2 y γ : 2, y,, M 2 = {{ 2,y }, {, }} y y y y Since M 2 is still not a ma-matching, we find an M 2 -alternating path γ 2 and its induced ma-matching M (see diagram above to the far right). γ 2 :, y M = {{ 2,y }, {, }, {,y }}
8 Solutions to Sample Eam 2 Problems Math 6 8 To answer your question, yes, γ 2 is indeed a legal M 2 alternating path (check the definition, and note that one of the properties is vacuously satisfied)! The fact that M is a ma-matching is confirmed by the fact that S = {,,y } is a cover of the graph with S = M =. (c) Below is an M -alternating path γ : y γ :,, 2, y This path leads to the ma-matching M = {{,y }, {, }, { 2,y }}, which is confirmed by observing that the set 2 y y S = { 2,, } is a cover of the graph with S = M =. The ma-matching M and the minimum cover S are illustrated in the diagram to the right. 8. In each of the following, let G denote a bipartite graph. Either give an eample that satisfies the given conditions, or show why no eample eists. If an eample eists, make sure to clearly indicate the desired portions of the graph in your eample, and to specify why it works. (a) A matching M such that M < ρ(g). (b) A matching M such that M < c(g). (c) A matching M such that M > c(g). (d) A cover S such that S > c(g). (e) A cover S such that S < ρ(g). (a) The graph G to the right works, because M = {{,y }} is a matching in G with fewer edges than the ma-matching M = {{,y }, { 2, }}. Therefore, G y M = < 2 = ρ(g ). 2 y 2 (b) The same graph, G, and matching, M, as in part (a) works as an eample for (b) because no single verte in G meets every edge of the graph. Therefore, the minimum size of a cover of G is 2, and we have M = < 2 = c(g ). (c) No such eample eists, for if it were true that such a matching M eisted, we would have ρ(g) M by definition of ρ(g) > c(g), by given information which would contradict the fact that ρ(g) = c(g). (d) The graph G 2 to the right works, because S = {, 2 } is a cover of G 2, and S = {y } is a minimal cover of G 2 because the removal of 0 vertices leaves all edges of G 2 in the graph. Therefore, we have S = 2 > = c(g 2 ). G 2 y 2 y 2
9 Solutions to Sample Eam 2 Problems Math 6 9 (e) No such eample eists, for if it were true that such a cover S eisted, we would have c(g) S by definition of c(g) < ρ(g), by given information which would contradict the fact that c(g) = ρ(g).
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# Comparing and Ordering
## Presentation on theme: "Comparing and Ordering"— Presentation transcript:
Comparing and Ordering
Decimals 5th Grade Math Today you will be learning more about number sense. We will be comparing and ordering decimals. By: Kellea McGehee
The purpose of today’s lesson is to:
Compare decimals Ex: > 2.04 Order decimals 1.02, 1.25, 1.50 (least to greatest) 1.50, 1.25, 1.02 (greatest to least) After this lesson, you will be able to compare decimals using the greater than, less than, or equals symbol. You will also be able to order decimals from least to greatest and greatest to least. Has anyone done this before?
Reviewing Place Values
To the hundredths place! Hundreds Tens Ones Tenths Hundredths 7 2 9 Place Value Chart First we will look at the hundredths pace value . Notice the difference between hundreds and hundredths. Hundreds refers to whole numbers and is found on the left side of the decimal point. Hundredths refers to the numbers located to the right of the decimal point. How do we say this in words?
Decimals to the hundredths place value!
Reading Decimals Decimals to the hundredths place value! Again, notice the hundredths place value. In words, 7.29 is seven and twenty-nine hundredths
Reviewing Place Values
To the Thousandths Place! Hundreds Tens Ones Tenths Hundredths Thousandths 6 3 1 7 4 Place Value Chart Now lets look at the thousandths place value. Do you understand the difference between thousands and thousandths? Thousands refers to the whole number value located to the left of the decimal point. Thousandths refers to the decimal value found to the right of the decimal point. Notice the difference between hundreds and hundredths. Hundreds refer to whole numbers and is found on the left side of the decimal point. Hundredths refers to the numbers located to the right of the decimal point. Thousands refers to a whole number and is found to the left of the decimal place. Thousandths refers to a decimal and is found to the right of the decimals point. How do we say this in words? Lets take a look…
Reading Decimals Decimals to the thousandths place:
When we read or say a decimal number out loud, we represent the values behind the decimal in the exact same way that we do when we write them. In words, is sixty-three and one hundred seventy-four thousandths.
Reading Decimals 2.4 = Two and four tenths 2.04 =
Two and four hundredths 2.004 = Two and four thousandths Lets take a minute to practice with several problems.
Comparing Decimals How does this work?
Comparing Decimals Step 1: Line up the decimal points
Step 2: Find the first place where the digits differ Step3: Compare the differing digits Step 4: Use <, >, or = to state your answer When we use greater than or less than symbols, here is a hint to help you remember them correctly: Think of the symbols ( <, >) as an alligator’s mouth. Alligators are hungry and like to eat a lot. So, the alligators open mouth is always facing the bigger number.
How do we compare decimals:
Comparing Decimals How do we compare decimals: Example : Line up the decimal points 68.563 Starting at the left, find the first place the digits differ Compare the digits 3 > 0 Since 3 > 0 (three is greater than zero) > Lets look at these steps in more detail
Comparing Decimals: Practice
Use >, <, or = to compare each pair of decimals. Practice: Answer: 4.08 > Practice: Answer: < Lets try some practice.
Ordering Decimals How do we go from least to greatest?
How do we go from greatest to least? Remember, once you learn how to go from least to greatest, you should also be able to determine greatest to least very easily. New Visions Technology Inc.
Line up the decimal points
Ordering Decimals How do we order decimals: Step 1 Step 2 Step 3 Line up the decimal points Annex zeros so that each has the same number of decimal places. Use place value to compare and order the decimals. What does this mean?
Line up the decimal points
Ordering Decimals From Least to Greatest Step 1 Step 2 Step 3 Line up the decimal points Annex zeros so that each has the same number of decimal places. Use place value to compare and order the decimals. Example 4.073 4.73 4.0073 4 4.0730 4.7300 4.0000 4.000 Lets look at these steps in going from least to greatest. Do you know what it means to annex the zeros? This means to add zeros so that each value has the same number of decimal places. Notice the underlined zeros. These were the zeros we added so that each number would have the same number of decimal places. This will help us to see what order the decimals should be in.
Line up the decimal points
Ordering Decimals From Greatest to Least Step 1 Step 2 Step 3 Line up the decimal points Annex zeros so that each has the same number of decimal places. Use place value to compare and order the decimals. Example 3.074 3.73 3.0073 3 3.0740 3.7300 3.0000 Now we will compare decimals going form greatest to least.
Ordering Decimals: Practice
Order from Least to Greatest: 3.01, 3.009, 3.09, Answer , 3.009, 3.01, 3.09 Order from Greatest to Least: 45.303, , 45.03, Lets try some problems. Raise your hand when you think that you have the answer. Answer , , 45.03,
Review: Comparing Decimals
Step 1: Line up the decimal points Step 2: Find the first place where the digits differ Step3: Compare the differing digits Step 4: Use <, >, or = to state your answer Here is our review. Remember the differences between numbers that appear before and after the decimals. Can anyone tell me what those differences are?
Review: Ordering Decimals
New Visions Technology Inc. Step 1 Step 2 Step 3 Line up the decimal points Annex zeros so that each has the same number of decimal places. Use place value to compare and order the decimals. Now we’ll review how to order decimals. Remember what it means to annex the zeros. Can someone tell me what this means? How does it help us when organizing fractions?
How often do you feel you use good number sense?
Always Sometimes Never 2 3
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# Example 1: Carpentry Application PowerPoint PPT Presentation
When you are given a parallelogram with certain properties, you can use the theorems below to determine whether the parallelogram is a rectangle. A manufacture builds a mold for a desktop so that , , and m ABC = 90° . Why must ABCD be a rectangle?.
Example 1: Carpentry Application
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#### Presentation Transcript
When you are given a parallelogram with certain properties, you can use the theorems below to determine whether the parallelogram is a rectangle.
A manufacture builds a mold for a desktop so that , , and mABC = 90°. Why must ABCD be a rectangle?
Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1.
Example 1: Carpentry Application
Below are some conditions you can use to determine whether a parallelogram is a rhombus.
Caution
In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram.
Remember!
You can also prove that a given quadrilateral is a
rectangle, rhombus, or square by using the definitions of the special quadrilaterals.
To prove that a given quadrilateral is a square, it is sufficient to show that the figure is both a rectangle and a rhombus.
Example 2:
Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid.
Given:
Conclusion: EFGH is a rhombus.
The conclusion is not valid. By Theorem 6-5-3, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. By Theorem 6-5-4, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply either theorem, you must first know that ABCD is a parallelogram.
with diags. rect.
Quad. with diags. bisecting each other
Example 2B:
Determine if the conclusion is valid.
If not, tell what additional information is needed to make it valid.
Given:
Conclusion: EFGH is a square.
Step 1 Determine if EFGH is a parallelogram.
Given
EFGH is a parallelogram.
Step 2 Determine if EFGH is a rectangle.
Given.
EFGH is a rectangle.
with one pair of cons. sides rhombus
Step 3 Determine if EFGH is a rhombus.
EFGH is a rhombus.
Step 4 Determine is EFGH is a square.
Since EFGH is a rectangle and a rhombus, it has four right angles and four congruent sides. So EFGH is a square by definition.
The conclusion is valid.
Check It Out! Example 2
Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid.
Given:ABC is a right angle.
Conclusion:ABCD is a rectangle.
The conclusion is not valid. By Theorem 6-5-1, if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this theorem, you need to know that ABCD is a parallelogram .
Since , the diagonals are congruent. PQRS is a rectangle.
Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.
Example 3:
P(–1, 4), Q(2, 6), R(4, 3), S(1, 1)
Step 1 Graph PQRS.
Step 2 Find PR and QS to determine is PQRS is a rectangle.
Since , PQRS is a rhombus.
Step 3 Determine if PQRS is a rhombus.
Step 4 Determine if PQRS is a square.
Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition.
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Dimensional Formula Analysis
In this article, you will learn about dimensions and the dimensional formula of physical quantities with the help of examples.
What are the dimensions?
The dimension of a physical quantity is defined as the power to which the fundamental quantities are raised to express the physical quantity. For example, the dimension of velocity is 0, 1, -1 in Mass, Length, Time, respectively.
If this feels confusing at first, don’t worry, we will look into the calculations of dimensions later in this article.
What is the dimensional formula?
Dimensional Formula is defined as the expression of the physical quantity in terms of its base unit (fundamental quantity) with proper dimensions. For example,
Dimensional Formula of Velocity is [ M0 L1 T-1 ]
Here,
• M, L, T are fundamental quantities Mass, Length and Time, respectively
• 0, 1, -1 are the dimensions
We generally use 3 fundamental quantities: Mass, Length, and Time to derive dimensional formulas of different physical quantities.
First, let’s look into the dimensional symbol of these fundamental quantities.
Dimensional Formula of Fundamental Quantities
• Mass is [M]
• Length is [L]
• Time is [T]
How to calculate the dimensional formula of physical quantities?
Now that we know what the dimensional formula is, let’s see how we can compute the dimensional formula of a physical quantity.
We can calculate the dimension formula of a physical quantity when its relation with other physical quantities is known. For example, suppose we want to calculate the dimensional formula of density.
We know that density is related to mass and volume, that is, density = mass / volume. Now, let’s use this relation to calculate the dimensional formula of density:
``````
Density = mass / volume
= mass / (length * breadth * volume)
``````
We can use the fundamental quantity
• [L] (Length) – for length, breadth, and volume
• [M] (Mass) – for mass
Hence, our formula becomes
``````
Density = [M] / ([L] * [L] * [L]
= [M] / [ L3 ]
= [ M1 ] * [ L-3 ]
``````
And, since there is no mention of time in density, we can write
Dimensional Formula of Density = [ M1 L-3 T0 ]
In this way, we can calculate the dimensional formula of physical quantities.
Dimensional Formula and SI Unit
We can also derive SI units by using the dimensional formula and vice versa. Let’s see an example.
1. Dimensional formula to SI unit conversion
Dimensional Formula of density = [ M1 L-3 T0 ].
We know the SI unit of
• [M] is kg (kilogram)
• [L] is m (meter)
• [T] is s (second)
Now, if we convert the formula to SI units, it will be [ kg1 m-3 s0 ]. Hence,
SI unit of density is kg m-3.
2. SI unit to dimensional formula conversion
SI unit of density is kg m-3
Here, we know that
• kg is the SI unit of mass
• m is the SI unit of length
So our formula will look like this [M L-3].
Since there is no mention of time here, we can use 0 for time.
Dimensional formula of density = [M1 L-3 T0 ]
Dimensional Formula of Physical Quantity
Now that we know the dimensional formula, let’s see the dimensional formula of various physical quantities.
Dimensional Equation
An equation containing physical quantities with dimensional formula is known as the dimensional equation. A dimensional equation is obtained by equating the dimensional formula on the right-hand and left-hand sides of an equation. For example,
Dimensional equation of v = u + at is
``````
[ M0 L1 T-1 ] = [ M0 L1 T-1 ] + [ M0 L1 T-1 ] * [ M0 L0 T1 ]
``````
Principle of Homogeneity of Dimensional Equation
According to this principle, the dimensions of fundamental quantities on the left-hand side of an equation must be equal to the dimensions of the fundamental quantities on the right-hand side of that equation.
Let us consider three quantities A, B, and C, such that C = A + B. Therefore, according to this principle, the dimensions of C are equal to the dimensions of A and B.
Let’s see an example. Here we have our earlier dimensional equation of v = u + at:
``````
[ M0 L1 T-1 ] = [ M0 L1 T-1 ] + [ M0 L1 T-1 ] * [ M0 L0 T1 ]
``````
1. Solving Right Hand Side
``````
RHS = [M0 L1 T-1] + [M0 L T-1] X [M0 L0 T]
= [M0 L1 T-1] + [M0 L1 T0 ]
RHS = [M0 L1 T-1 ] = LHS
``````
Here, the right-hand side equation is now the same as the left-hand side equation. Hence, the principle of homogeneity of dimensional equations is proved.
What are the uses of dimensional equations?
Here are some of the uses of the dimensional equation:
• We can use dimensional equations to check the correctness of physical relations.
• It also helps us to derive the relation between various physical quantities.
• We can convert the value of physical quantity from one system of units to another system using the dimensional equation.
• It finds the dimension of constants in a given relation.
What are the limitations of dimensional analysis?
Following are the limitations of dimensional analysis.
• We cannot get the information about the dimensional constant using the dimensional analysis.
• If a quantity depends on more than three factors having dimension, We cannot derive the dimensional formula.
• We cannot derive the formula containing trigonometric, exponential, logarithmic functions, etc.
• We cannot develop the exact form of a relationship when there is more than one part in any relationship.
• It gives no information on whether a physical quantity is a scalar or vector.
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# Scientific notation
Scientific notation is a form of presenting very large numbers or very small numbers in a simpler form. As we know, the whole numbers can be extended till infinity, but we cannot write such huge numbers on a piece of paper. Also, the numbers which are present at the millions place after the decimal needed to be represented in a simpler form. Thus, it is difficult to represent a few numbers in their expanded form. Hence, we use scientific notations. Also learn, Numbers In General Form.
For example, 100000000 can be written as 108, which is the scientific notation. Here the exponent is positive. Similarly, 0.0000001 is a very small number which can be represented as 10-8, where the exponent is negative.
## Scientific Notation Definition
As discussed in the introduction, the scientific notation helps us to represent the numbers which are very huge or very tiny in a form of multiplication of single-digit numbers and 10 raised to the power of the respective exponent. The exponent is positive if the number is very large and it is negative if the number is very small. Learn power and exponents for better understanding.
The general representation of scientific notation is:
a × 10b ; 1 ≤ a < 10
## Scientific Notation Rules
To determine the power or exponent of 10, we must follow the rule listed below:
• The base should be always 10
• The exponent must be a non-zero integer, that means it can be either positive or negative
• The absolute value of the coefficient is greater than or equal to 1 but it should be less than 10
• Coefficients can be positive or negative numbers including whole and decimal numbers
• The mantissa carries the rest of the significant digits of the number
Let us understand how many places we need to move the decimal point after the single-digit number with the help of the below representation.
1. If the given number is multiples of 10 then the decimal point has to move to the left, and the power of 10 will be positive.
Example: 6000 = 6 × 103 is in scientific notation.
2. If the given number is smaller than 1, then the decimal point has to move to the right, so the power of 10 will be negative.
Example: 0.006 = 6 × 0.001 = 6 × 10-3 is in scientific notation.
## Scientific Notation Examples
The examples of scientific notation are:
490000000 = 4.9×108
1230000000 = 1.23×109
50500000 = 5.05 x 107
0.000000097 = 9.7 x 10-8
0.0000212 = 2.12 x 10-5
## Positive and Negative Exponent
When the scientific notation of any large numbers is expressed, then we use positive exponents for base 10. For example:
20000 = 2 x 104, where 4 is the positive exponent.
When the scientific notation of any small numbers is expressed, then we use negative exponents for base 10. For example:
0.0002 = 2 x 10-4, where -4 is the negative exponent.
From the above, we can say that the number greater than 1 can be written as the expression with positive exponent, whereas the numbers less than 1 with negative exponent.
## Problems and Solutions
Question 1: Convert 0.00000046 into scientific notation.
Solution: Move the decimal point to the right of 0.00000046 up to 7 places.
The decimal point was moved 7 places to the right to form the number 4.6
Since the numbers are less than 10 and the decimal is moved to the right. Hence, we use a negative exponent here.
⇒ 0.00000046 = 4.6 × 10-7
This is the scientific notation.
Question 2: Convert 301000000 in scientific notation.
Solution: Move the decimal to the left 8 places so it is positioned to the right of the leftmost non zero digits 3.01000000. Remove all the zeroes and multiply the number by 10.
Now the number has become = 3.01.
Since the number is greater than 10 and the decimal is moved to left, therefore, we use here a positive exponent.
Hence, 3.01 × 108 is the scientific notation of the number.
Question 3:Convert 1.36 × 107 from scientific notation to standard notation.
Solution: Given, 1.36 × 107 in scientific notation.
Exponent = 7
Since the exponent is positive we need to move the decimal place 7 places to the right.
Therefore,
1.36 × 107 = 1.36 × 10000000 = 1,36,00,000.
### Practice Questions
Problem 1: Convert the following numbers into scientific notation.
1. 28100000
2. 7890000000
3. 0.00000542
Problem 2: Convert the following into standard form.
1. 3.5 × 105
2. 2.89 × 10-6
3. 9.8 × 10-2
## Frequently Asked Questions on Scientific Notation – FAQs
Q1
### How do you write 0.00001 in scientific notation?
The scientific notation for 0.0001 is 1 × 10^{-4}.
Here,
Coefficient = 1
Base = 10
Exponent = -4
Q2
### What are the 5 rules of scientific notation?
The five rules of scientific notation are given below:
1. The base should be always 10
2. The exponent must be a non-zero integer, that means it can be either positive or negative
3. The absolute value of the coefficient is greater than or equal to 1 but it should be less than 10
4. Coefficients can be positive or negative numbers including whole and decimal numbers
5. The mantissa carries the rest of the significant digits of the number
Q3
### What are the 3 parts of a scientific notation?
The three main parts of a scientific notation are coefficient, base and exponent.
Q4
### How do you write 75 in scientific notation?
The scientific notation of 75 is:
7.5 × 10^1 = 7.5 × 10
Here,
Coefficient = 7.5
Base = 10
Exponent = 1
Q5
### How do you put scientific notation into standard form?
To convert a number from scientific notation to standard form, we should move the decimal point (if any) to the left if the exponent of 10 is negative; otherwise, proceed to the right. We must shift the decimal point as many times as the exponent indicates in power so that there will be no powers of 10 in the final representation.
Test your knowledge on Scientific Notation
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# Algebra problem regarding system of logarithmic equations.
The equations are:
$\log_{4}(x)+\log_{4}(y)=5$
$\big(\log_{4}(x)\big)\big(\log_{4}(y)\big)=6$
I attempted to solve this problem by solving the pair of equations for $x$.
For the first equation:
$\Longrightarrow \log_{4}(xy)=5 \Longrightarrow xy=4^{5} \Longrightarrow xy=1024 \Rightarrow x=\dfrac{1024}{y}$
For the second equation:
$\Longrightarrow \log{4}(x)=\dfrac{6}{\log_{4}(y)} \Longrightarrow x=4^{\frac{6}{\log_{4}(y)}}$
Then,
$\Longrightarrow \dfrac{1024}{y}=4^{\frac{6}{\log_{4}(y)}}$
How should I move on from here?
• Hint: First, solve $A+B=5$ and $AB=6$. Then see how you can apply that to $x$ and $y$. – Wildcard Jan 1 '17 at 5:03
## 2 Answers
We have:
$\log_{4}(x)+\log_{4}(y)=5$
$\Rightarrow \log_{4}(x)=5-\log_{4}(y) \hspace{5 mm}$ (i)
$\big(\log_{4}(x)\big)\big(\log_{4}(y)\big)=6 \hspace{4.25 mm}$ (ii)
Substituting (i) into (ii):
$\Rightarrow \big(5-\log_{4}(y)\big)\big(\log_{4}(y)\big)=6$
$\Rightarrow 5\log_{4}(y)-\big(\log_{4}(y)\big)^{2}=6$
$\Rightarrow \big(\log_{4}(y)\big)^{2}-5\log_{4}(y)+6=0$
$\Rightarrow \log_{4}(y)=\dfrac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(6)}}{2(1)}$
$\Rightarrow \log_{4}(y)=\dfrac{5\pm{1}}{2}$
$\Rightarrow \log_{4}(y)=2,3$
$\Rightarrow y=16,64$
Using (i):
$\Rightarrow \log_{4}(x)=5-\log_{4}(16)$
$\hspace{19.5 mm}=5-2$
$\hspace{19.5 mm}=3$
$\Rightarrow x=64$
or
$\Rightarrow \log_{4}(x)=5-\log_{4}(64)$
$\hspace{19.5 mm}=5-3$
$\hspace{19.5 mm}=2$
$\Rightarrow x=16$
Therefore, the solutions to the system of equations is $x=16$ and $y=64$ or $x=64$ and $y=16$.
• Might be clearer just to factorize the quadratic $\big(\log_{4}(y)\big)^{2}-5\log_{4}(y)+6=0$ rather than use the quadratic equation. – Ian Miller Jan 1 '17 at 8:28
Let $u=\log_4(x)$, $v=\log_4(y)$. The equations can be written in $u$ and $v$ $$u+v=5,\ \ uv=6$$ The solutions are $u=2$, $v=3$, or $u=3$, $v=2$. Then you solve for $x$ and $y$
EDIT As Bernard pointed out, $u$ and $v$ above are generally found as the solutions of the quadratic equation $t^2-5t+6=0$. But in this special case I just solved it by inspection.
Also, for completeness, $\log_4(x)=2$ implies $x=4^2=16$ and $\log_4(y)=3$ implies $y=4^3=64$. So the solutions are $x=16, y=64$ and $x=64, y=16$
• You should write the quadratic equation which has $u$ and $v$ as roots: $t^2-5t+6=0$; – Bernard Jan 1 '17 at 1:22
• @Bernard Right, but for this simple case I just solved it by "inspection" – Momo Jan 1 '17 at 1:24
• But how can one be sure there are no other solutions? A priori, the solutions are not necessarily integers. – Bernard Jan 1 '17 at 1:27
• @Bernard Assuming that i found two numbers with sum $5$ and product $6$, they are THE solution, since the quadratic equation has two solutions (or at most two solutions if only considering reals). Anyway, I updated my answer to reflect your comment. – Momo Jan 1 '17 at 1:30
• You don't get my point: I simply mean a naive reader might not know there's the theory of quadratic equations behind this stuff, and wonder whether this yields all solutions. – Bernard Jan 1 '17 at 1:34
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#### Provide Solution For RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 10 Maths Textbook Solution.
Answer : $\frac{2}{3} \tan ^{-1}\left[3 \tan \frac{x}{2}\right]+c$
Hint: To solve this equation we have to change cosx into tanx form
Given: $\int \frac{1}{5-4 \cos x} d x$
Solution :$\int \frac{1}{5-4 \cos x}dx$
$\operatorname{Cos} x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
\begin{aligned} &=\int \frac{1}{5-4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} \mathrm{d} x \\ &= \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+9 \tan ^{2} \frac{x}{2}} d x \end{aligned}
\begin{aligned} &=\int \frac{\sec ^{2} \frac{x}{2}}{1+9 \tan ^{2} \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{1+\left(3 \tan \frac{x}{2}\right)^{2}} d x \end{aligned}
\begin{aligned} &\text { Let } 3 \tan \frac{x}{2}=t \\ &=3 \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \end{aligned}
\begin{aligned} &=\frac{2}{3} \int \frac{d t}{1+t^{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\langle\int\frac{1}{a^{2}+t^{2}}dt=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)\right\rangle \\ &=\frac{2}{3} \tan ^{-1}(t)+C \\ &=\frac{2}{3} \tan ^{-1}\left(3 \tan \frac{x}{2}\right)+C \end{aligned}
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# Using trigonometric substitution, how do you integrate integral of x^3 (x^2+4)^(1/2) ?
Jul 21, 2018
$\int {x}^{3} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} \mathrm{dx} = \frac{1}{5} {\left(4 + {x}^{2}\right)}^{2} \sqrt{4 + {x}^{2}} - \frac{4}{3} \left(4 + {x}^{2}\right) \sqrt{4 + {x}^{2}}$
See explanations below.
#### Explanation:
$\int {x}^{3} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} \mathrm{dx} = \int {x}^{3} \sqrt{\left({x}^{2} + 4\right)} \mathrm{dx}$
let $x = 2 \tan \left(\theta\right)$
$\mathrm{dx} = 2 \sec {\left(\theta\right)}^{2} d \theta$
So :
$\int {x}^{3} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} \mathrm{dx} = \int 16 \tan {\left(\theta\right)}^{3} \sqrt{\left(4 \tan {\left(\theta\right)}^{2} + 4\right)} \sec {\left(\theta\right)}^{2} d \theta$
$= 16 \int \tan {\left(\theta\right)}^{3} \sqrt{4 \left(\tan {\left(\theta\right)}^{2} + 1\right)} \sec {\left(\theta\right)}^{2} d \theta$
$= 32 \int \tan {\left(\theta\right)}^{3} {\cancel{\sqrt{\tan {\left(\theta\right)}^{2} + 1}}}^{\textcolor{red}{= \sec \left(\theta\right)}} \sec {\left(\theta\right)}^{2} d \theta$
$= 32 \int \tan {\left(\theta\right)}^{3} \sec {\left(\theta\right)}^{3} d \theta$
$= 32 \int \frac{\sin {\left(\theta\right)}^{3}}{\cos {\left(\theta\right)}^{6}} d \theta$
$= - 32 \int \frac{- \sin \left(\theta\right) \sin {\left(\theta\right)}^{2}}{\cos {\left(\theta\right)}^{6}} d \theta$
$= - 32 \int \frac{- \sin \left(\theta\right) \left(1 - \cos {\left(\theta\right)}^{2}\right)}{\cos {\left(\theta\right)}^{6}} d \theta$
let $u = \cos \left(\theta\right)$
$\mathrm{du} = - \sin \left(\theta\right) d \theta$
So : $32 \int \tan {\left(\theta\right)}^{3} \sec {\left(\theta\right)}^{3} d \theta = - 32 \int \frac{1 - {u}^{2}}{{u}^{6}} \mathrm{du}$
$= 32 \int {u}^{- 4} \mathrm{du} - 32 \int {u}^{- 6} \mathrm{du}$
$= \frac{32}{5} {u}^{- 5} - \frac{32}{3} {u}^{- 3}$
$= \frac{32}{5} \cos {\left(\theta\right)}^{- 5} - \frac{32}{3} \cos {\left(\theta\right)}^{-} 3$
Finally, because $\theta = \arctan \left(\frac{x}{2}\right)$ and $\cos \left(\arctan \left(\frac{x}{2}\right)\right) = \frac{2}{\sqrt{4 + {x}^{2}}}$,
$\int {x}^{3} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} \mathrm{dx} = \frac{1}{5} {\left(4 + {x}^{2}\right)}^{2} \sqrt{4 + {x}^{2}} - \frac{4}{3} \left(4 + {x}^{2}\right) \sqrt{4 + {x}^{2}}$
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# 1.5: Estimation of Whole Number Addition and Subtraction
Difficulty Level: At Grade Created by: CK-12
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What other kinds of animals are there at a zoo?
Sarah is also a volunteer like Jonah. She spends her time working with the penguins. When Jonah comes to visit her, she tells him all about the penguin population. Sarah explains to Jonah that there will be new penguins born very soon. She tells him that there are 57 penguins, but that 17 new ones are due to be born in the spring. Jonah listens and tries to add the two values together. He takes out paper and pencil. Sarah stops him. She knows a quicker way. She tells Jonah that they can use estimation.
Have you ever estimated a sum? In this Concept, you will learn how to use estimation to figure out the new penguin population.
### Guidance
In the real world problem, you saw how puzzled Jonah was when Sarah was able to use estimation to help her solve the penguin problem. Estimation definitely seemed to save Sarah some time.
What do we mean by estimation? When can we use it and when shouldn’t we use it? To estimate means to find an answer that is close to the exact answer. The key with estimation is that you can only use it in instances where you don’t need an exact answer. When we estimate, we want to find an answer that makes sense and works with our problem, but is not necessarily exact.
We can estimate sums and differences. Remember back in the first Concept, we used the word sum and the word difference. Let’s take a minute to review what those two words mean. A sum is the answer to an addition problem. A difference is the answer to a subtraction problem. To estimate a sum or a difference, we can round the numbers that we are working with to find our estimation.
What does it mean to round a number? When we round, we change the number to the nearest power of ten (times a whole number), such as ten or hundred or thousand, etc. Here is a problem we can work with.
69
Let’s say that we want to round this number to the nearest ten. Well, we can look at whether 69 is closer to 60 or to 70. These are the two numbers in the tens surrounding 69. It is closer to 70, so we would change the number to 70. Here is another one.
53
If we want to round this to the nearest ten, then we can look at the numbers surrounding 53 which are multiples of ten. Is 53 closer to 50 or 60? It is closer to 50, so we would “round down” to 50. When rounding, we can follow the rounding rules. If the number being rounded is less than 5, round down. If the number being rounded is greater than 5, round up. In the examples, we were rounding to the tens, so we use the number in the ones place to round. Using 69, since 9 is greater than 5, we round up. In the case of 53, 3 is less than 5, so we round down.
Let’s apply this.
128 Round to the nearest ten.
Look at the number. We are rounding to the tens, so we look at the ones place. 8, is greater than 5, so we round up to 130. What does this have to do with estimating sums and differences? Well, when we estimate a sum or a difference, if we round first, it is easier to add.
We want to estimate this answer. If we round each number first, we can use mental math to find our estimation.
58 rounds to 60
22 rounds to 20
Our estimate is 80.
Here is one with larger numbers.
We want to estimate our answer by rounding to the nearest hundred.
387 rounds to 400
293 rounds to 300
Our estimate is 700.
This worked for addition.
We can estimate differences by rounding too.
Here is one with larger numbers.
We want to estimate our difference by rounding to the nearest hundred.
990 rounds to 1000
211 rounds to 200
Our estimate is 800.
Now let's practice with a few examples.
Solution: 50
Solution: 800
#### Example C
Solution: 200
Now back to Sarah and Jonah and the penguin population.
Here is the original problem once again.
Sarah is also a volunteer like Jonah. She spends her time working with the penguins. When Jonah comes to visit her, she tells him all about the penguin population. Sarah explains to Jonah that there will be new penguins born very soon. She tells him that there are 57 penguins, but that 17 new ones are due to be born in the spring. Jonah listens and tries to add the two values together. He takes out paper and pencil. Sarah stops him. She knows a quicker way. She tells Jonah that they can use estimation.
To solve this problem, Jonah and Sarah can use estimation.
First, we can round both values.
57 rounds up to 60
17 rounds up to 20
Now we add 60 + 20
The new penguin population will be 80 penguins.
### Vocabulary
Here is the vocabulary for this Concept.
Estimation
to find an approximate answer to a problem
Sum
Difference
the answer to a subtraction problem
### Guided Practice
Here is one for you to try on your own.
We want to estimate this difference by rounding to the nearest ten.
56 rounds to 60
18 rounds to 20
Our estimate is 40.
### Video Review
These videos will help you to estimate sums and differences.
### Practice
Estimate the following sums and differences.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Difference The result of a subtraction operation is called a difference.
Estimation Estimation is the process of finding an approximate answer to a problem.
Round To round is to reduce the number of non-zero digits in a number while keeping the overall value of the number similar.
Sum The sum is the result after two or more amounts have been added together.
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# How to Add and Subtract Fractions
## A Common Denominator Makes the Process Easy
Adding and subtracting whole numbers or decimals is fairly straight forward and easy. However, when we get to fractions the it becomes more difficult. The reason for the difficulty is that each fraction consists of a numerator and a denominator as follows:
### numerator / denominator
The rule for adding and/or subtracting fractions is to first make sure that the denominators are the same. If the denominators are the same, then we simply add or subtract the numerators, leaving the denominators untouched. This makes it easy if we have problems like:
or
## What About Fractions with Different Denominators?
However, what do we do if we have problems like:
or
### 7/8 – 1/4?
We cannot proceed with adding or subtracting these because the denominators are different.
But, don't worry, there is a solution to this problem and that involves finding a common denominator for each of the fractions in the problems above.
## How to Make the Denominators the Same
A common denominator is a denominator that each of the denominators in the problem can be divided into evenly.
In the case of ½ and ¾ the common denominator is 4 as both the 2 in ½ and the 4 in ¾ can be divided into that number.
We now change the 2 in ½ to a 4 and, since the 2 in ½ can be divided into 4 twice (4 ÷ 2 = 2) we multiply the numerator by 2 to get 2/4. Our formula is now:
### 2/4 + 3/4 = 5/4
Since the numerator is greater than the denominator, we reduce it, by dividing the denominator into the numerator to get 1 ¼.
Subtracting 1/4 from 7/8, which is our second problem, we find that the common denominator is 8 and, repeating the process described above, we get 7/8 – 2/8 = 5/8 as our answer.
## What to Do if Neither Denominator is the Common Denominator?
In both examples above, one of the denominators turned out to be the common denominator.
However, what if neither denominator is the common denominator. Let's try this problem:
### 1/5 + 1/6
Here we find that the first number that each denominator can be evenly divided into is 30. Dividing the 5 in 1/5 into 30 we get 6.
Multiplying the the numerator, which is 1, by 6 we get 6/30.
We then divide the 6 in the second fraction into 30 and get 5.
Multiplying the numerator, which is 1 in this case also, by 5 we get 5/30.
With both denominators being the same (30) we can proceed to solve the problem as follows:
### 6/30 + 5/30 = 11/30.
Since there is no way to reduce this fraction, our answer is 11/30.
Try some yourself. You will find that this is actually quite easy.
## Did you Find this Hub Helpful in Understanding Adding & Subtracting Fractions?
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## More by this Author
Chuck 3 years ago from Tucson, Arizona Author
This Hub was written in 2006 and was written in response to a question asking how to Add and Subtract Fractions. The question had been asked on HubPages and I, and probably some other Hubbers, wrote a Hub in answer to this question.
The fact that there are many other sites on the Internet explaining how to do this doesn't surprise me. I would be surprised if there were not thousands of similar articles on the web explaining this with the only differences there can be is how one explains how to do this process. You will also probably find hundreds or more textbooks currently in print and available for sale explaining the same thing with thousands more written down through the ages and now out of print.
While I have never claimed that this was the one and only article on this topic or the first to explain this process which people have used for hundreds if not thousands of years, I do hope that some people, struggling with having to add or subtract fractions, have found my Hub to be of help in enabling them to understand how to do this.
zvtcefsitl 3 years ago
You produced some decent points there. I looked on-line to the issue and discovered most individuals go in conjunction with together with your site.
Aaron 8 years ago
hi i was jus wondering what if we have more then one fractions and we have to add all of them or add one subtract the other and multiply the other fraction?
ALeisia lesso 9 years ago
I want the answers to page 93
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## Some TikTok Math
@rebrokeraaron#duet with @dustin_wheeler
I came across a nice TikTok vid recently that definitely screamed out for a mathematical explanation. It was interesting to watch because I had written an activity not too long ago which connected the two mathematical ideas relevant to this video—I just didn’t have this nice context to apply it to.
Anyway, you can see it at the right there.
The two connected ideas at work here are the Triangle Proportionality Theorem and the Midpoint Formula. We don’t actually need both of these concepts to explain the video. It’s just more pleasing to apply them both together.
The Triangle Proportionality Theorem (see an interactive proof here) tells us that if we draw a line segment inside a triangle that connects two sides and is parallel to the third side, then the segment will divide the two connected sides proportionally.
So, $$\mathtt{\overline{DE}}$$ above connects sides $$\mathtt{\overline{AB}}$$ and $$\mathtt{\overline{AC}}$$, is parallel to side $$\mathtt{\overline{BC}}$$, and thus divides both $$\mathtt{\overline{AB}}$$ and $$\mathtt{\overline{AC}}$$ proportionally, such that $$\mathtt{a:b=c:d}$$.
And of course the Midpoint Formula, which we’ll suss out in a second, is $$\mathtt{\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)}$$.
Okay, Let’s Explain This
What’s happening in the video is that the measurer is creating a right triangle. They first make a horizontal measurement—which they don’t know the midpoint of. That’s the horizontal leg of the right triangle. Then they take the tape measure up the board diagonally. That’s the hypotenuse of the right triangle. We do know that midpoint (because it’s a simpler whole number).
We can drop a line segment from the midpoint that we know straight down vertically, such that the segment is parallel to the vertical side of the triangle (the vertical side of the board).
Now we have set up everything we need for the Triangle Proportionality Theorem. The vertical segment connects two sides—the hypotenuse and the horizontal side—and is parallel to the third (vertical) side. Thus, the vertical segment divides the hypotenuse and horizontal segment into equal ratios.
But since we know that the segment is at the midpoint of the hypotenuse, we know that it divides the hypotenuse into the ratio $$\mathtt{1:1}$$. That is, it divides the hypotenuse into two congruent segments. Therefore, it must also divide the horizontal line segment into two congruent segments!
Okay, so that pretty much nails it (ha!). Why do we need the Midpoint Formula?
Now for the Midpoint Formula
Well, we don’t really need the Midpoint Formula. It’s just interesting that the Triangle Proportionality Theorem explains why the Midpoint Formula works. Working through the explanation above, it may have occurred to you that what the Triangle Proportionality Theorem says, indirectly, about ANY right triangle, is that—if you can wrangle the legs to be horizontal and vertical—the midpoint of the hypotenuse is always directly above or below the midpoint of the horizontal side, and the midpoint of the hypotenuse is always directly to the left or right of the midpoint of the vertical side. That’s what the Midpoint Formula basically says in symbols.
## GCF and LCM Triangles
Go grab some dot paper or grid paper—or just make some dots in a square grid on a blank piece of paper. Let’s start with a 4 × 4 grid of dots, like so.
Now, start at the top left corner, draw a vertical line down to the bottom of the grid, and count each dot that your pen enters—which just means that you won’t count the first dot, since your pen leaves that dot but does not enter it. Then, draw a horizontal line to the right, starting over with your counting. Again, count each dot that your pen enters. Count just 2 dots as you draw to the right.
Finally, draw a straight line (a hypotenuse) back to your starting point. Here again, count the number of dots you enter.
One example is not, of course, enough to convince you that the number of dots your pen enters when drawing the hypotenuse is the greatest common factor (GCF) of the number of counted vertical dots and the number of counted horizontal dots. So, here are a few more examples with just a 4 × 4 grid.
No doubt there are tons of people out there for whom this display is completely unsurprising. But it surprised me. The GCF of two numbers is an object that seems as though it should be rather hidden—a value that may appear when we crack two numbers open and do some calculations with them, not something that just pops up when we draw lines on dot paper. We use prime factorization to suss out GCF, after all, and that is by no means an intuitive process.
Connections
There are some very nice mathematical connections here. The first is to the coordinate plane, or perhaps more simply to orthogonal axes, which we use to compare values all the time—but only in certain contexts. Widen or eliminate the context constraint, and it seems obvious that comparing two numbers orthogonally could yield insights about GCF.
And slope is, ultimately, the “reason” why this all works. The slope of a line in lowest terms is just the rise over the run with both the numerator and denominator divided by the GCF: $\mathtt{\frac{\text{rise}}{\text{run}}\div\frac{\text{GCF}}{\text{GCF}}=\text{slope in lowest terms}}$
Once slope is there, all kinds of connections take hold: divisibility, fractions, lowest terms, etc. Linear algebra, too, contains a connection, which itself is connected to something called Bézout’s Identity. There is also a weird connection to calculus—maybe—that I haven’t quite teased out. To see what I mean, let’s also draw the LCM out of these images.
From the lowest entered point on the hypotenuse, draw a horizontal line extending to the width of the triangle. Then draw a vertical line to the bottom right corner of the triangle. Now go left: draw a horizontal line all the way to the left edge of the triangle. Then a vertical line extending to the height of the lowest entered point on the hypotenuse. Finally, move right and draw a horizontal line back to where you started. You should draw a rectangle as shown in each of these examples. The area of each rectangle is the LCM of the two numbers.
The maybe-calculus connection I speak of is the visible curve vs. area-under-the-curve vibe we’ve got going on there. I’m still noodling on that one.
## The Farey Mean?
I had never heard of the Farey mean, but here it is, brought to you by @howie_hua.
When you add two fractions, you of course remember that you should never just add the numerators and add the denominators across. The resulting fraction will not be the sum of the two addends. But if you do add across (under certain conditions which I’ll show below), the result will be a fraction between the two “addend” fractions. So, you can use the add-across method to find a fraction between two other fractions.
For example, $$\mathtt{\frac{1}{2}+\frac{4}{3}\rightarrow\frac{5}{5}}$$. The first “addend” is definitely less than 1, and the second definitely greater than 1. The Farey mean here is exactly 1 (or $$\mathtt{\frac{5}{5}}$$), which is between the two “addend” fractions.
Why Does It Work?
Since this site is fast becoming all linear algebra all the time, let’s throw some linear algebra at this. What we want to show is that, given $$\mathtt{\frac{a}{b}<\frac{c}{d}}$$ (we'll go with this assumption for now), $\mathtt{\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}}$
for certain positive integer values of $$\mathtt{a,b,c,}$$ and $$\mathtt{d}$$. I would probably do better to make those inequality signs less-than-or-equal-tos, but let’s stick with this for the present. We’ll start by representing the fraction $$\mathtt{\frac{a}{b}}$$ as the vector $$\scriptsize\begin{bmatrix}\mathtt{b}\\\mathtt{a}\end{bmatrix}$$ along with the fraction $$\mathtt{\frac{c}{d}}$$ as the vector $$\scriptsize\begin{bmatrix}\mathtt{d}\\\mathtt{c}\end{bmatrix}$$.
We’re looking specifically at the slopes or angles here (which is why we can represent a fraction as a vector in the first place), so we’ve made $$\scriptsize\begin{bmatrix}\mathtt{d}\\\mathtt{c}\end{bmatrix}$$ have a greater slope to keep in line with our assumption above that $$\mathtt{\frac{a}{b}<\frac{c}{d}}$$.
The fraction $$\mathtt{\frac{a+c}{b+d}}$$ is the same as the vector $$\scriptsize\begin{bmatrix}\mathtt{b+d}\\\mathtt{a+c}\end{bmatrix}$$. And since this vector is the diagonal of the vector parallelogram, it will of course have a greater slope than $$\mathtt{\frac{a}{b}}$$ but less than $$\mathtt{\frac{c}{d}}$$. You can keep going forever—just take one of the side vectors and use the diagonal vector as the other side. So long as you’re making parallelograms, you’ll get a new diagonal shallower than the two side vectors, and the result will be a fraction between the other two.
Incidentally, our assumption at the beginning that $$\mathtt{\frac{a}{b}<\frac{c}{d}}$$ doesn't really matter to this picture. If we make $$\mathtt{\frac{c}{d}}$$ less than $$\mathtt{\frac{a}{b}}$$, our picture simply flips. The diagonal vector still has to be located between the two side vectors.
What Doesn’t Work?
The linear algebra picture of this concept also tells us where this method fails to find a fraction between the two addend fractions. When the two “addend” fractions are equivalent, $$\mathtt{c}$$ and $$\mathtt{d}$$ are multiples of $$\mathtt{a}$$ and $$\mathtt{b}$$, respectively, or vice versa. In that case, the resulting fraction looks like this.
The slopes or angles for both addends and for the result are the same, producing a Farey mean that is equal to both fractions.
## Cosine Similarity and Correlation
I wrote a lesson not too long ago that started with a Would You Rather? survey activity. For our purposes here, we can pretend that each question had a Likert scale from 1–10 attached to it, though in reality, the lesson was about categorical data.
At any rate, here are the questions—edited a bit. Feel free to rate your answers on the scales provided. Careful! Once you click, you lock in your answer.
Would you rather . . .
1. be able to fly (1) or be able to read minds (10)?
1. go way back in time (1) or go way into the future (10)?
1. be able to talk to animals (1) or speak all languages (10)?
1. watch only historical movies (1) or sci-fi movies (10) for the rest of your life?
1. be just a veterinarian (1) or just a musician (10)?
Finally, one last question that is not a would-you-rather. Once you’ve answered this and the rest of the questions, you can press the I'm Finished! button to submit your responses.
1. Rate your fear of heights from (1) not at all afraid to (10) very afraid.
Check out the results so far.
Next in the lesson, I move on to asking whether you think some of the survey responses are correlated. For example, if you scored “low” on the veterinarian-or-musician scale—meaning you would strongly prefer to be a veterinarian over a musician—would that indicate that you probably also scored “low” on Question (c) about talking to animals or speaking all the languages? In other words, are those two scores correlated? What about choosing the ability to fly and your fear of heights? Are those correlated? How could we measure this using a lot of responses from a lot of different people?
An ingenious way of looking at this question is by using cosine similarity from linear algebra. (We looked at the cosine of the angle between two vectors here and here.)
For example, suppose you really would rather have the ability to fly and you have almost no fear of heights. So, you answered Question (a) with a 1 and Question (f) with, say, a 2. Another person has no desire to fly and a terrible fear of heights, so they answer Question (a) with an 8 and Question (f) with a (10). From this description, we would probably guess that the two quantities wish-for-flight and fear-of-heights are strongly correlated. But we’ve also now got the vectors (1, 2) and (8, 10) to show us this correlation.
See that tiny angle between the vectors on the left? The cosine of tiny angles (as we saw) is close to 1, which indicates a strong correlation. On the right, you see the opposite idea. One person really wants to fly but is totally afraid of heights (1, 10) and another almost couldn’t care less about flying (or at least would really rather read minds) but has a low fear of heights (8, 2). The cosine of the close-to-90°-angle between these vectors will be close to 0, indicating a weak correlation between responses to our flight and heights questions.
But That’s Not the Ingenious Part
That’s pretty cool, but it is not, in fact, how we measure correlation. The first difficulty we encounter happens after adding more people to the survey, giving us several angles to deal with—not impossible, but pretty messy for a hundred or a thousand responses. The second, more important, difficulty is that the graph on the right above doesn’t show a weak correlation; it shows a strong negative correlation. Given just the two response pairs to work from in that graph, we would have to conclude that a strong fear of heights would make you more likely to want the ability to fly (or vice versa) rather than less likely. But the “weakest” the cosine can measure in this kind of setup is 0.
The solution to the first difficulty is to take all the x-components of the responses and make one giant vector out of them. Then do the same to the y-components. Now we’ve got just two vectors to compare! For our data on the left, the vectors (1, 2) and (8, 10) become (1, 8) and (2, 10). The vectors on the right—(1, 10) and (8, 2)—become (1, 8) and (10, 2).
The solution to the second difficulty—no negative correlations—we can achieve by centering the data. Let’s take our new vectors for the right, uncorrelated, graph: (1, 8) and (10, 2). Add the components in each vector and divide by the number of components (2) to get an average. Then subtract the average from each component. So, our new centered vectors are
(1 – ((1 + 8) ÷ 2), 8 – ((1 + 8) ÷ 2)) and (10 – ((10 + 2) ÷ 2), 2 – ((10 + 2) ÷ 2))
Or (–3.5, 3.5) and (4, –4). It’s probably not too tough to see that a vector in the 2nd quadrant and a vector in the 4th quadrant are heading in opposite directions. And these vectors now form a close-to-180° angle, and the cosine of 180° is –1 which is the actual lowest correlation we can get, indicating a strong negative correlation.
And That’s Correlation
To summarize, the way to determine correlation linear-algebra style is to determine the cosine of the centered x- and y-vectors of the data. That formula is $\mathtt{\frac{(x-\overline{x}) \cdot (y-\overline{y})}{|x-\overline{x}||y-\overline{y}|} = cos(θ)}$
Which is just another way of writing the more common version of the r-value correlation.
## The Formula for Combinations
And now, finally, let’s get to the formula for combinations. The math in my last post got a little tricky toward the end, with the strange exclamation mark notation floating around. So let’s recap permutations without that notation.
4
× 3
× 3 × 2
× 3 × 2 × 1
÷ (1 × 2 × 3)
÷ (1 × 2)
÷ 1
You should see that, to traverse a tree diagram, we multiply by the tree branches, cumulatively, to move right, and then divide by those branches—again, cumulatively—to move left. The formula for the number of permutations of 2 cards chose from 4, $$\mathtt{\frac{4!}{(4-2)!}}$$, tells us to multiply all the way to the right, to get 24 in the numerator and then divide two steps to the left (divide by $$\mathtt{(4-2)!}$$ or 2) to get 12 permutations of 2 cards chosen from 4.
Combinations
An important point about the above is that the number of permutations of $$\mathtt{r}$$ cards chosen from $$\mathtt{n}$$ cards, $$\mathtt{_{n}P_r}$$, is a subset of the number of permutations of $$\mathtt{n}$$ cards, $$\mathtt{n!}$$ The tree diagram shows $$\mathtt{n!}$$ and contained within it are $$\mathtt{_{n}P_r}$$.
Combinations of $$\mathtt{r}$$ items chosen from $$\mathtt{n}$$, denoted as $$\mathtt{_{n}C_r}$$, are a further subset. That is, $$\mathtt{_{n}C_r}$$ are a subset of $$\mathtt{_{n}P_r}$$. In our example of 2 cards chosen from 4, $$\mathtt{_{n}P_r}$$ represents the first two columns of the tree diagram combined. In those columns, we have, for example, the permutations JQ and QJ. But these two permutations represent just one combination. The same goes for the other pairs in those columns. Thus, we can see that to get the number of combinations of 2 cards chosen from 4, we take $$\mathtt{_{n}P_r}$$ and divide by 2. So, $\mathtt{\frac{4!}{(4-2)!}\div 2=\frac{4!}{(4-2)!\cdot2}}$
What about combinations of 3 cards chosen from 4? That’s the first 3 columns combined. Now the repeats are, for example, JQK, JKQ, QJK, QKJ, KJQ, KQJ. Which is 6. Noticing the pattern? For $$\mathtt{_{4}C_2}$$, we divide $$\mathtt{_{4}P_2}$$ further by 2! For $$\mathtt{_{4}C_3}$$, we divide $$\mathtt{_{4}P_3}$$ further by 3! We’re dividing (further) by $$\mathtt{r!}$$
When you think about it, this makes sense. We need to collapse every permutation of $$\mathtt{r}$$ cards down to 1 combination. So we divide by $$\mathtt{r!}$$ Here, finally then, is the formula for combinations: $\mathtt{_{n}C_r=\frac{n!}{(n-r)!r!}}$
## And Now for the Legal Formula
So, did you come up with a working rule to describe the pattern we looked at last time? Here’s what I came up with:
As we saw last time, the “root” of the tree diagram (the first column) shows $$\mathtt{_{4}P_1}$$, which is the number of permutations of 1 card chosen from 4. The first and second columns combined show $$\mathtt{_{4}P_2}$$, the number of permutations of 2 cards chosen from 4. So, to determine $$\mathtt{_{n}P_r}$$, according to this pattern, we start with $$\mathtt{n}$$ and then multiply $$\mathtt{(n-1)(n-2)}$$ and so on until we reach $$\mathtt{n-(r-1)}$$.
The number of permutations of, say, 3 items chosen from 5, then, would be $\mathtt{_{5}P_3=5\cdot (5-1)(5-2)=60}$
This is a nice rule that works every time for permutations of $$\mathtt{r}$$ things chosen from $$\mathtt{n}$$ things. It can even be represented a little more ‘mathily’ as $\mathtt{_{n}P_r=\prod_{k=0}^{r-1}(n-k)}$
So let’s move on to the “legal” formula for $$\mathtt{_{n}P_r}$$. A quick sidebar on notation, though, which we’ll need in a moment.
When we count the number of permutations at the end of a tree diagram, what we get is actually $$\mathtt{_{n}P_n}$$. In our example, that’s $$\mathtt{_{4}P_4}$$. The way we write this amount is with an exclamation mark: $$\mathtt{n!}$$, or, in our case, $$\mathtt{4!}$$ What $$\mathtt{4!}$$ means is $$\mathtt{4\times(4-1)\times(4-2)\times(4-3)}$$ according to our rule above, or just $$\mathtt{4\times3\times2\times1}$$. And $$\mathtt{3!}$$ is $$\mathtt{3\times(3-1)\times(3-2)}$$, or just $$\mathtt{3\times2\times1}$$.
In general, we can say that $$\mathtt{n!=n\times(n-1)!}$$ So, for example, $$\mathtt{4!=4\times3!}$$ etc. And since this means that $$\mathtt{1!=1\times(1-1)!}$$, that means that $$\mathtt{0!=1}$$.
So, for the tree diagram, $$\mathtt{_{4}P_4}$$ means multiplying all the way to the right by $$\mathtt{n!}$$. But if we’re interested in the number of arrangements of $$\mathtt{r}$$ cards chosen from $$\mathtt{n}$$ cards, then we need to come back to the left by $$\mathtt{(n-r)!}$$ And since moving right is multiplying, moving left is dividing.
4
4 × 3
4 × 3 × 2
4 × 3 × 2 × 1
÷ (4 – 1)!
÷ (4 – 2)!
÷ (4 – 3)!
÷ (4 – 4)!
The division we need is not immediately obvious, but if you study the tree diagram above, I think it’ll make sense. This gives us, finally, the “legal” formula for the number of permutations of $$\mathtt{r}$$ items from $$\mathtt{n}$$ items: $\mathtt{_{n}P_r=\frac{n!}{(n-r)!}}$
## A New Formula for Permutations?
Last time, we saw that combinations are a subset of permutations, and we wondered what the relationship between the two is. Before we get there, though, let’s look at another possible relationship—one we only hinted at last time. And to examine this relationship, we’ll use a tree diagram.
Tree Diagram
This tree diagram shows the number of permutations of the 4 cards J, Q, K, A—the number of ways we can arrange the 4 cards. The topmost branch shows the result JQKA. And you can see all 24 results from our list last time here in the tree diagram.
4
4 × 3
4 × 3 × 2
4 × 3 × 2 × 1
÷ 3!
÷ 2!
÷ 1!
÷ 0!
Here’s where, normally, people would talk about the multiplication 4 × 3 × 2 × 1 and tell you that another way to write that is with an exclamation mark: 4! But that’s skipping over something important.
And that something important is this: Notice that the first column of the tree diagram—the root of the tree—shows 4 items. This is the number of different permutations you can make of just 1 card, chosen from 4 different cards. And the first and second columns combined show the number of permutations you can make of 2 cards, chosen from 4 cards (JQ, JK, JA, etc.).
And so on. You might think that to go from “permutations of 1 card chosen from 4″ to “permutations of 2 cards chosen from 4″ you would multiply by 2. But of course that’s not right (and the tree diagram tells us so). You actually multiply 4 by 4 – 1. And to go from “permutations of 1 card chosen from 4″ to “permutations of 3 cards chosen from 4″ you multiply 4 • (4 – 1) • (4 – 2).
We’re on the verge of being able to describe the relationship, which I’ll put in question form (and mix in some notation to):
What is the relationship between the number of permutations of $$\mathtt{n}$$ things, $$\mathtt{P(n)}$$, and the number of permutations of $$\mathtt{r}$$ things chosen from $$\mathtt{n}$$ things, $$\mathtt{_{n}P_r}$$?
We can see from our example above that $$\mathtt{P(4)=24}$$. That is, the number of permutations of 4 things is 24. But we also noticed these three results: \begin{aligned}_{\mathtt{4}}\mathtt{P}_{\mathtt{1}}&= \mathtt{\,\,4}\cdot \mathtt{1} \\ _{\mathtt{4}}\mathtt{P}_{\mathtt{2}}&= (\mathtt{4}\cdot \mathtt{1})(\mathtt{4}-\mathtt{1}) \\ _{\mathtt{4}}\mathtt{P}_{\mathtt{3}}&= (\mathtt{4}\cdot \mathtt{1})(\mathtt{4}-\mathtt{1})(\mathtt{4}-\mathtt{2})\end{aligned}
A New Formula?
Study the pattern above and see if you can write a rule that will get you the correct result for any $$\mathtt{_{n}P_r}$$. Check your results here (for example, for $$\mathtt{_{16}P_{12}}$$, you can just enter 16P12 and press Enter).
The rule you write, if you get it right, won’t be an algorithm. But it’ll work every time! This is the step we always skip when teaching about permutations! The next step is to think hard about why it works. We’ll get to the “legal” formula for permutations next time.
## Permutations & Combinations
I have now been blogging for 16 years, and my very first post (long gone) was on combinations and permutations. So, it’s fun to come back to the idea now. In 2004, my experience with the two concepts was limited to how textbooks often used the awkward “care about order” (permutations) or “don’t care about order” (combinations) language to introduce the ideas. So, that’s what I wrote about then. Now I want to talk about how the two concepts are related.
What They Are
When you count permutations, you count how many different ways you can sequentially arrange some things. When you count combinations, you count how many ways you can have some things. So, given 2 cards, there are 2 different ways you can sequentially arrange 2 cards, but given 2 cards, there’s just one way to have 2 cards.
Right off the bat, the language is weird, and it’s hard to see why combinations should ever be a thing (there’s always just 1 way to have a set of things). But combinations make better sense when you are not choosing from all the elements you are given.
So, for example, how many permutations and combinations can I make of 2 cards, chosen from a total of 3 cards?
Now having the two categories of permutation and combination makes a little more sense. There are 6 permutations of 2 cards chosen from 3 cards and there are 3 combinations of 2 cards chosen from 3 cards. That is, there are 6 different ways to sequentially arrange 2 cards chosen from 3 and just 3 different ways to have 2 cards chosen from 3. And you can see, by the way, that the combinations are a subset of the permutations.
In fact, let’s do an example with 4 cards to show the actual relationship between permutations and combinations. Here we’ll just use letters to save space. The permutations of JQKA if we choose 3 cards are:
JQK, JKQ, QJK, QKJ, KJQ, KQJ
JQA, JAQ, QJA, QAJ, AJQ, AQJ
QKA, QAK, KQA, KAQ, AQK, AKQ
KAJ, KJA, JKA, JAK, AJK, AKJ
That’s 24 permutations. For combinations, we get JQK, JQA, QKA, KAJ. That’s 4 combinations. What’s the relationship? We’ll come back on that next time.
## Rotating Coordinate Systems
I‘ve just started with Six Not-So-Easy Pieces, based on Feynman’s famous lectures, and already there’s some decently juicy stuff. In the beginning, Feynman discusses the symmetry of physical laws—that is, the invariance of physical laws under certain transformations (like rotations):
If we build a piece of equipment in some place and watch it operate, and nearby we build the same kind of apparatus but put it up on an angle, will it operate in the same way?
He goes on to explain that, of course, a grandfather clock will not operate in the same way under specific rotations. Assuming the invariance of physical laws under rotations, this change in operation tells us something interesting: that the operation of the clock is dependent on something outside of the “system” that is the clock itself.
The theorem is then false in the case of the pendulum clock, unless we include the earth, which is pulling on the pendulum. Therefore we can make a prediction about pendulum clocks if we believe in the symmetry of physical law for rotation: something else is involved in the operation of a pendulum clock besides the machinery of the clock, something outside it that we should look for.
Rotation Coordinates
Feynman then proceeds with a brief mathematical analysis of forces under rotations. A somewhat confusing prelude to this is a presentation that involves expressing the coordinates of a rotated system in terms of the original system. He uses the diagram below to derive those coordinates (except for the blue highlighting, which I use to show what (x’, y’) looks like in Moe’s system). What we want is to express (x’, y’) in terms of x and y—to describe Moe’s point P in terms of Joe’s point P.
“We first drop perpendiculars from P to all four axes and draw AB perpendicular to PQ.”
The first confusion that is not dealt with (because Feynman makes the assumption that his audience is advanced students) is what angles in the diagram are congruent to θ shown. And here again we see the value of the easy-to-forget art of eyeballing and common sense in geometric reasoning.
The y’ axis is displaced just as much as the x’ axis by rotation, and “displaced just as much by rotation” is a perfectly good definition of angle congruence that we tend to forget after hundreds of hours of deriving work. The same reasoning applies to the rotational displacement from AP to AB. If we imagine rotating AP to AB, we see that we are starting perpendicular to the x-axis and ending perpendicular to the x’-axis. The y-to-y’ rotation does the same thing, so the displacement angle must be the same. So let’s put in those new thetas, only one of which we’ll need.
Inspection of the figure shows that x’ can be written as the sum of two lengths along the x’-axis, and y’ as the difference of two lengths along AB.
Here is x’ as the sum of two lengths (red and orange): $\mathtt{x’=OA\cdot\color{red}{\frac{OC}{OA}}+AP\cdot\color{orange}{\frac{BP}{AP}}\quad\rightarrow\quad x\cdot\color{red}{cos\,θ}+y\cdot\color{orange}{sin\,θ}}$
And here is y’ as the difference of two lengths (green – purple): $\mathtt{y’=AP\cdot\color{green}{\frac{AB}{AP}}-OA\cdot\color{purple}{\frac{AC}{OA}}\quad\rightarrow\quad y\cdot\color{green}{cos\,θ}-x\cdot\color{purple}{sin\,θ}}$
So, if Joe describes the location of point P to Moe, and the rotational displacement between Moe and Joe’s systems is known (and it is known that the two systems share an origin), Moe can use the manipulations above to determine the location of point P in his system.
Another, exactly equal, way of saying this—the way we said it when we talked about rotation matrices—is that, if we represent point P in Joe’s system as a position vector (x, y), then Moe’s point P vector is $\small{\begin{bmatrix}\mathtt{x’}\\\mathtt{y’}\end{bmatrix}=\begin{bmatrix}\mathtt{\,\,\,\,\,cos\,θ} & \mathtt{sin\,θ}\\\mathtt{-sin\,θ} & \mathtt{cos\,θ}\end{bmatrix}\begin{bmatrix}\mathtt{x}\\\mathtt{y}\end{bmatrix}=\mathtt{x}\begin{bmatrix}\mathtt{\,\,\,\,\,cos\,θ}\\\mathtt{-sin\,θ}\end{bmatrix}+\mathtt{y}\begin{bmatrix}\mathtt{sin\,θ}\\\mathtt{cos\,θ}\end{bmatrix}=\begin{bmatrix}\mathtt{\,\,\,\,\,\,x\cdot \color{red}{cos\,θ}\,\,\,+y\cdot \color{orange}{sin\,θ}}\\\mathtt{-(x\cdot \color{purple}{sin\,θ})+y\cdot \color{green}{cos\,θ}}\end{bmatrix}}$
The first rotation matrix above actually describes a clockwise rotation, which is both different from what we discussed at the link above (our final matrix there was for counterclockwise rotations) and unexpected, since we know that Moe’s system is a counterclockwise rotation of Joe’s system.
The resolution to that unexpectedness can again be found after a little eyeballing. The position vector for point P in Joe’s system is at a steep angle, whereas in Moe’s system, it is at a shallow angle. Only a clockwise rotation will change the coordinates in the appropriate way.
## Projections
A projection is like the shadow of a vector, say $$\mathtt{u}$$, on another vector, say $$\mathtt{v}$$, if light rays were coming in across $$\mathtt{u}$$ and perpendicular to $$\mathtt{v}$$. For the vectors at the right, imagine a light source above and to the left of the illustration, perpendicular to the vector $$\mathtt{v}$$. The projection, which we’ll call $$\mathtt{p}$$, will be a vector that will extend from the point shown to where the end of the shadow of $$\mathtt{u}$$ touches $$\mathtt{v}$$.
If you take a moment maybe to read that description twice (because it’s kind of dense), you may be able to tell that the vector $$\mathtt{p}$$ that we’re looking for will be some scalar multiple of vector $$\mathtt{v}$$, since it will lie exactly on top of $$\mathtt{v}$$. In fact, given the picture, the projection vector $$\mathtt{p}$$ will have the opposite sign as $$\mathtt{v}$$ and will have a scale factor pretty close to 0, since the projection vector looks like it will be much smaller than $$\mathtt{v}$$.
That information is shown at the right. Helpfully, the vector $$\mathtt{u-p}$$ is perpendicular to $$\mathtt{v}$$, so we know that $$\mathtt{\left(u-p\right)\cdot v=0}$$. And, using the Distributive Property, we get $$\mathtt{u\cdot v-p\cdot v=0}$$.
Since we know that our sought-after vector $$\mathtt{p}$$ will be some scalar multiple of $$\mathtt{v}$$, we can substitute, say, $$\mathtt{cv}$$ for $$\mathtt{p}$$ in the above to get $$\mathtt{u\cdot v-cv\cdot v=0}$$. And a property of dot product multiplication allows us to rewrite that as $$\mathtt{u\cdot v-c\left(v\cdot v\right)=0}$$.
This means that $$\mathtt{u\cdot v=c\left(v\cdot v\right)}$$, which means that the scale factor $$\mathtt{c}$$ that we’re after—the factor we can multiply by $$\mathtt{v}$$ to produce $$\mathtt{p}$$—is $\mathtt{c=\frac{u\cdot v}{v\cdot v}\quad\rightarrow\quad p=\left(\frac{u\cdot v}{v\cdot v}\right)v}$
|
# Which Equation Can Be Used to Solve for B
Which Equation Can Be Used to Solve for B
## two.6: Quadratic Equations
##### Learning Objectives
• Solve quadratic equations by factoring.
• Solve quadratic equations past the square root property.
• Solve quadratic equations by completing the square.
• Solve quadratic equations by using the quadratic formula.
The calculator monitor on the left in Figure $$\PageIndex{one}$$ is a $$23.6$$-inch model and the one on the right is a $$27$$-inch model. Proportionally, the monitors appear very similar. If there is a limited corporeality of space and we desire the largest monitor possible, how do we determine which ane to choose? In this section, nosotros will learn how to solve problems such as this using four dissimilar methods.
## Solving Quadratic Equations by Factoring
An equation containing a second-degree polynomial is chosen a
quadratic equation. For example, equations such as $$2x^2 +3x−1=0$$ and $$ten^2−iv= 0$$ are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological scientific discipline, and, of course, mathematics.
Ofttimes the easiest method of solving a quadratic equation is
factoring. Factoring means finding expressions that tin exist multiplied together to give the expression on 1 side of the equation.
If a quadratic equation tin can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if $$a⋅b=0$$, and so $$a = 0$$ or $$b =0$$, where a and b are real numbers or algebraic expressions. In other words, if the production of two numbers or 2 expressions equals zero, then one of the numbers or ane of the expressions must equal zero because zero multiplied by anything equals nada.
Multiplying the factors expands the equation to a string of terms separated past plus or minus signs. And then, in that sense, the operation of multiplication undoes the functioning of factoring. For example, aggrandize the factored expression $$(ten−2)(ten+three)$$ by multiplying the two factors together.
\begin{align*} (x-2)(x+3)&= x^2+3x-2x-6\\ &= x^two+ten-6\\ \finish{align*}
The product is a quadratic expression. Gear up equal to zero, $$x^ii+10−6= 0$$ is a quadratic equation. If we were to factor the equation, we would get dorsum the factors nosotros multiplied.
The process of factoring a quadratic equation depends on the leading coefficient, whether it is $$1$$ or another integer. We will look at both situations; but first, nosotros want to confirm that the equation is written in standard grade, $$ax^ii+bx+c=0$$, where $$a$$, $$b$$, and $$c$$ are real numbers, and $$a≠0$$. The equation $$ten^2 +x−6= 0$$ is in standard grade.
We can use the zero-production property to solve quadratic equations in which we beginning have to factor out the greatest mutual factor(GCF), and for equations that take special factoring formulas as well, such equally the difference of squares, both of which we will meet later in this department.
##### Nada-Production PROPERTY AND QUADRATIC EQUATIONS
The
zero-product holding
states
If $$a⋅b=0$$, then $$a=0$$ or $$b=0$$,
where $$a$$ and $$b$$ are real numbers or algebraic expressions.
A quadratic equation is an equation containing a second-degree polynomial; for example
$ax^2+bx+c=0$
where $$a$$, $$b$$, and $$c$$ are existent numbers, and if $$a≠0$$, it is in standard form.
## Solving Quadratics with a Leading Coefficient of $$1$$
In the quadratic equation $$x^ii +ten−6=0$$, the leading coefficient, or the coefficient of $$x^2$$, is $$one$$. We have one method of factoring quadratic equations in this form.
##### How to: Factor a quadratic equation with the leading coefficient of 1
1. Notice 2 numbers whose product equals $$c$$ and whose sum equals $$b$$.
2. Apply those numbers to write 2 factors of the form $$(ten+k)$$ or $$(ten−k)$$, where k is ane of the numbers found in stride one. Utilize the numbers exactly as they are. In other words, if the two numbers are $$1$$ and $$−2$$, the factors are $$(x+ane)(x−2)$$.
3. Solve using the zero-product property past setting each factor equal to aught and solving for the variable.
##### Example $$\PageIndex{1}$$: Solving a Quadratic with Leading Coefficient of $$ane$$
Factor and solve the equation: $$x^ii+x−6=0$$.
Solution
To factor $$x^2 +ten−six=0$$, nosotros look for two numbers whose production equals $$−half dozen$$ and whose sum equals $$1$$. Begin past looking at the possible factors of $$−vi$$.
$one⋅(−6) \nonumber$
$(−6)⋅1 \nonumber$
$2⋅(−three) \nonumber$
$3⋅(−2) \nonumber$
The last pair, $$iii⋅(−two)$$ sums to $$1$$, so these are the numbers. Note that merely one pair of numbers volition work. Then, write the factors.
$(x−two)(x+three)=0 \nonumber$
To solve this equation, nosotros use the cypher-production holding. Set each factor equal to zero and solve.
\begin{align*} (ten-two)(ten+iii)&= 0\\ (10-two)&= 0\\ x&= 2\\ (10+iii)&= 0\\ x&= -3 \finish{marshal*}
The two solutions are $$2$$ and $$−3$$. Nosotros can see how the solutions relate to the graph in Figure $$\PageIndex{2}$$. The solutions are the x-intercepts of $$x^two +10−6=0$$.
##### Exercise $$\PageIndex{1}$$
Factor and solve the quadratic equation: $$x^ii−5x−six=0$$.
Answer
$$(10−half-dozen)(x+one)=0$$, $$x=6$$, $$10=−one$$
##### Example $$\PageIndex{ii}$$: Solve the Quadratic Equation past Factoring
Solve the quadratic equation by factoring: $$x^two+8x+15=0$$.
Solution
Find two numbers whose product equals $$15$$ and whose sum equals $$8$$. List the factors of $$15$$.
Popular: Which Statement Best Reflects the Purpose of Satire
$one⋅fifteen \nonumber$
$iii⋅v \nonumber$
$(−one)⋅(−15) \nonumber$
$(−3)⋅(−v) \nonumber$
The numbers that add to $$8$$ are $$3$$ and $$5$$. Then, write the factors, set up each factor equal to nil, and solve.
\begin{align*} (x+3)(10+five)&= 0\\ (10+3)&= 0\\ x&= -3\\ (x+5)&= 0\\ x&= -5 \end{align*}
The solutions are $$−3$$ and $$−v$$.
##### Exercise $$\PageIndex{ii}$$
Solve the quadratic equation by factoring: $$10^ii−4x−21=0$$.
Answer
$$(10−7)(x+3)=0$$, $$10=7$$, $$x=−3$$
##### Case $$\PageIndex{3}$$: Using Zilch-Production Property to Solve a Quadratic Equation
Solve the departure of squares equation using the zero-product property: $$x^2−nine=0$$.
Solution
Recognizing that the equation represents the departure of squares, nosotros can write the two factors by taking the square root of each term, using a minus sign equally the operator in one factor and a plus sign every bit the operator in the other. Solve using the zero-factor property.
\begin{align*} x^2-nine&= 0\\ (ten-3)(x+3)&= 0\\ ten-three&= 0\\ ten&= 3\\ (x+3)&= 0\\ ten&= -iii \end{align*}
The solutions are $$3$$ and $$−3$$.
##### Practise $$\PageIndex{3}$$
Solve by factoring: $$10^ii−25=0$$.
Answer
$$(10+5)(10−v)=0, x=−v, x=v$$
## Factoring and Solving a Quadratic Equation of Higher Guild
When the leading coefficient is not $$one$$, we factor a quadratic equation using the method called
grouping, which requires four terms.
##### Grouping: Steps for factoring quadratic equations
With the equation in standard course, let’s review the grouping procedures
1. With the quadratic in standard form, $$ax^two+bx+c=0$$, multiply $$a⋅c$$.
2. Discover ii numbers whose production equals ac and whose sum equals $$b$$.
3. Rewrite the equation replacing the $$bx$$ term with 2 terms using the numbers found in step $$1$$ as coefficients of $$x$$.
4. Gene the starting time two terms and and so cistron the last two terms. The expressions in parentheses must be exactly the same to employ grouping.
5. Factor out the expression in parentheses.
6. Prepare the expressions equal to zero and solve for the variable.
##### Example $$\PageIndex{4}$$: Solving a Quadratic Equation Using Grouping
Use group to factor and solve the quadratic equation: $$4x^ii+15x+ix=0$$.
Solution
First, multiply $$air-conditioning:4(nine)=36$$. Then list the factors of $$36$$.
$1⋅36 \nonumber$
$2⋅18 \nonumber$
$3⋅12 \nonumber$
$4⋅9 \nonumber$
$6⋅6 \nonumber$
The merely pair of factors that sums to $$fifteen$$ is $$three+12$$. Rewrite the equation replacing the b term, $$15x$$, with two terms using $$iii$$ and $$12$$ as coefficients of $$x$$. Gene the first ii terms, and so factor the concluding ii terms.
\brainstorm{align*} 4x^2+3x+12x+nine&= 0\\ x(4x+3)+three(4x+3)&= 0\\ (4x+3)(x+3)&= 0 \qquad \text{Solve using the aught-product belongings}\\ (4x+3)&= 3\\ ten&= -\dfrac{3}{four}\\ (10+3)&= 0\\ x&= -3 \stop{marshal*}
The solutions are $$−\dfrac{iii}{4}$$, and $$−3$$. See Effigy $$\PageIndex{3}$$.
##### Exercise $$\PageIndex{4}$$
Solve using factoring past group: $$12x^2+11x+2=0$$.
Respond
$$(3x+2)(4x+one)=0$$, $$x=−\dfrac{2}{3}$$, $$x=−\dfrac{1}{4}$$
##### Instance $$\PageIndex{5}$$: Solving a Higher Caste Quadratic Equation by Factoring
Solve the equation past factoring: $$−3x^3−5x^2−2x=0$$.
Solution
This equation does not look like a quadratic, as the highest power is $$three$$, not $$2$$. Recall that the first thing we want to practice when solving whatever equation is to cistron out the GCF, if one exists. And it does here. We can cistron out $$−x$$ from all of the terms and then keep with group.
\begin{align*} -3x^iii-5x^ii-2x&= 0\\ -x(3x^2+5x+2)&= 0\\ -10(3x^ii+3x+2x+2)&= 0 \qquad \text{Employ group on the expression in parentheses}\\ -x[3x(x+i)+2(x+1)]&= 0\\ -x(3x+2)(ten+one)&= 0\\ \text{Now, nosotros employ the cipher-product property. Notice that nosotros take three factors.}\\ -x&= 0\\ x&= 0\\ 3x+2&= 0\\ x&= -\dfrac{2}{3}\\ 10+1&= 0\\ x&= -1 \end{align*}
The solutions are $$0$$, $$−\dfrac{ii}{iii}$$, and $$−one$$.
##### Exercise $$\PageIndex{v}$$
Solve past factoring: $$x^3+11x^2+10x=0$$.
Respond
$$x=0, 10=−10, x=−one$$
## Using the Square Root Property
When there is no linear term in the equation, some other method of solving a quadratic equation is by using the
square root property, in which we isolate the $$10^2$$ term and have the foursquare root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the $$x^ii$$ term so that the square root property tin can exist used.
##### THE SQUARE ROOT Holding
With the $$x^two$$ term isolated, the square root belongings states that:
if $$10^2=k$$, then $$x=±\sqrt{k}$$
where $$k$$ is a nonzero existent number.
##### Howto: Given a quadratic equation with an $$10^2$$ term only no $$x$$ term, apply the square root property to solve it
1. Isolate the $$x^2$$ term on one side of the equal sign.
2. Accept the square root of both sides of the equation, putting a $$±$$ sign before the expression on the side opposite the squared term.
3. Simplify the numbers on the side with the $$±$$ sign.
##### Case $$\PageIndex{6}$$: Solving a Elementary Quadratic Equation Using the Square Root Holding
Solve the quadratic using the square root property: $$10^two=8$$.
Solution
Have the square root of both sides, and then simplify the radical. Remember to use a $$±$$ sign earlier the radical symbol.
\begin{marshal*} x^2&= eight\\ x&= \pm \sqrt{8}\\ &= \pm ii\sqrt{two} \stop{align*}
The solutions are $$ii\sqrt{2}$$,$$-two\sqrt{2}$$
##### Exercise $$\PageIndex{six}$$
Solve the quadratic equation using the foursquare root holding: $$3{(x−4)}^2=xv$$.
Reply
$$10=4±\sqrt{5}$$
## Completing the Foursquare
Not all quadratic equations tin can be factored or can be solved in their original form using the square root property. In these cases, we may utilize a method for solving a
quadratic equation
known as
completing the square. Using this method, nosotros add or subtract terms to both sides of the equation until we have a perfect square trinomial on 1 side of the equal sign. Nosotros so utilize the square root property. To complete the foursquare, the leading coefficient, $$a$$, must equal $$one$$. If it does not, and so divide the entire equation by $$a$$. So, we can employ the following procedures to solve a quadratic equation by completing the square.
We will utilise the instance $$10^two+4x+1=0$$ to illustrate each step.
Given a quadratic equation that cannot be factored, and with $$a=1$$, first add or subtract the abiding term to the right sign of the equal sign.
\begin{marshal*} x^2+4x+1&= 0\\ x^two+4x&= -i \qquad \text{Multiply the b} \text{ term by } \dfrac{1}{2} \text{ and square information technology.}\\ \dfrac{1}{2}(4)&= 2 \\ ii^2&= iv \qquad \text{Add } \left ({\dfrac{1}{2}} \right )^ii \text{ to both sides of the equal sign and simplify the right side. We take}\\ x^2+4x+iv&= -1+four\\ x^2+4x+4&= 3 \qquad \text{The left side of the equation tin now be factored as a perfect square.}\\ {(10+ii)}^two&=iii\\ \sqrt{{(ten+2)}^2}&= \pm \sqrt{3} \qquad \text{Use the square root property and solve.}\\ \sqrt{{(x+2)}^ii}&= \pm \sqrt{iii}\\ 10+ii&= \pm \sqrt{3}\\ x&= -2 \pm \sqrt{3} \end{align*}
The solutions are $$−ii+\sqrt{three}$$, and $$−2−\sqrt{3}$$.
##### Case $$\PageIndex{eight}$$: Solving a Quadratic by Completing the Square
Solve the quadratic equation by completing the square: $$x^2−3x−v=0$$.
Solution
Get-go, move the abiding term to the right side of the equal sign.
\begin{align*} x^2-3x&= 5 \qquad \text{Then, take } \dfrac{ane}{2} \text{ of the b term and foursquare information technology.} \\ \dfrac{1}{2}(-3)&= -\dfrac{3}{2}\\ {\left (-\dfrac{3}{ii} \correct )}^2=\dfrac{nine}{4}\\ ten^2-3x+{\left (-\dfrac{three}{2} \right )}^two&= 5+{\left (-\dfrac{3}{2} \right )}^2 \qquad \text{Add the result to both sides of the equal sign.}\\ x^ii-3x+\dfrac{nine}{4}&= 5+\dfrac{9}{4}\\ \text{Factor the left side as a perfect foursquare and simplify the correct side.}\\ {\left (x-\dfrac{3}{ii} \correct )}^ii&= \dfrac{29}{4}\\ (x-\dfrac{3}{2})&= \pm \dfrac{\sqrt{29}}{2} \qquad \text{Use the foursquare root property and solve.}\\ x&= \dfrac{3}{2} \pm \dfrac{\sqrt{29}}{two}\\ \stop{align*}
The solutions are $$\dfrac{3}{2}+\dfrac{\sqrt{29}}{2}$$, and $$\dfrac{3}{2}-\dfrac{\sqrt{29}}{ii}$$
##### Practise $$\PageIndex{7}$$
Solve by completing the foursquare: $$10^ii−6x=xiii$$.
Answer
$$ten=3±\sqrt{22}$$
## Using the Quadratic Formula
The fourth method of solving a
quadratic equation
is by using the
quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay shut attention when substituting, and apply parentheses when inserting a negative number.
We can derive the quadratic formula by
completing the square. We volition assume that the leading coefficient is positive; if it is negative, we can multiply the equation by $$−1$$ and obtain a positive a. Given $$ax^2+bx+c=0, a≠0$$, nosotros will complete the square as follows:
Beginning, move the abiding term to the correct side of the equal sign:
$ax^2+bx=−c \nonumber$
Equally we want the leading coefficient to equal $$1$$, divide through by $$a$$:
$x^ii+\dfrac{b}{a}ten=−\dfrac{c}{a} \nonumber$
So, find $$\dfrac{1}{ii}$$ of the middle term, and add together $${(\dfrac{1}{2}\dfrac{b}{a})}^ii=\dfrac{b^2}{4a^2}$$ to both sides of the equal sign:
$ten^2+\dfrac{b}{a}10+\dfrac{b^2}{4a^2}=\dfrac{b^2}{4a^ii}-\dfrac{c}{a} \nonumber$
Next, write the left side as a perfect foursquare. Detect the common denominator of the right side and write information technology as a unmarried fraction:
${(x+\dfrac{b}{2a})}^ii=\dfrac{b^2-4ac}{4a^ii} \nonumber$
At present, use the foursquare root property, which gives
$x+\dfrac{b}{2a}=±\sqrt{\dfrac{b^2-4ac}{4a^two}} \nonumber$
$x+\dfrac{b}{2a}=\dfrac{±\sqrt{b^ii-4ac}}{2a} \nonumber$
Finally, add $$-\dfrac{b}{2a}$$ to both sides of the equation and combine the terms on the right side. Thus,
$x=\dfrac{-b±\sqrt{b^2-4ac}}{2a} \nonumber$
##### THE QUADRATIC FORMULA
Written in standard form, $$ax^2+bx+c=0$$, whatever quadratic equation can be solved using the
quadratic formula:
$ten=\dfrac{-b±\sqrt{b^2-4ac}}{2a}$
where $$a$$, $$b$$, and $$c$$ are existent numbers and $$a≠0$$.
##### How to
Given a quadratic equation, solve it using the quadratic formula
1. Make sure the equation is in standard form: $$ax^2+bx+c=0$$.
2. Make note of the values of the coefficients and constant term, $$a$$, $$b$$, and $$c$$.
3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
4. Calculate and solve.
##### Instance $$\PageIndex{9}$$: Solve the Quadratic Equation Using the Quadratic Formula
Solve the quadratic equation: $$10^2+5x+i=0$$.
Solution
Identify the coefficients: $$a=i,b=5,c=1$$. And so use the quadratic formula.
\begin{align*} x&= \dfrac{-(5) \pm \sqrt{(five)^two-four(1)(ane)}}{two(ane)}\\ &= \dfrac{-five \pm \sqrt{25-4}}{two}\\ &= \dfrac{-5 \pm \sqrt{21}}{2} \terminate{marshal*}
##### Case $$\PageIndex{10}$$: Solving a Quadratic Equation with the Quadratic Formula
Employ the quadratic formula to solve $$ten^2+x+2=0$$.
Solution
Offset, we identify the coefficients: $$a=1$$,$$b=1$$, and $$c=2$$.
Substitute these values into the quadratic formula.
\begin{align*} ten&= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\ &= \dfrac{-(1) \pm \sqrt{(ane)^2-4(one)(2)}}{ii(1)}\\ &= \dfrac{-1 \pm \sqrt{i-8}}{ii}\\ &= \dfrac{-ane \pm \sqrt{-vii}}{ii}\\ &= \dfrac{-1 \pm i\sqrt{7}}{2} \terminate{marshal*}
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##### Exercise $$\PageIndex{eight}$$
Solve the quadratic equation using the quadratic formula: $$9x^2+3x−2=0$$.
Answer
$$10=-\dfrac{2}{three},ten=\dfrac{1}{3}$$
## The Discriminant
The
quadratic formula
non only generates the solutions to a quadratic equation, it tells us almost the nature of the solutions when nosotros consider the
discriminant, or the expression under the radical, $$b^2−4ac$$. The discriminant tells usa whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table $$\PageIndex{i}$$ relates the value of the discriminant to the solutions of a quadratic equation.
Table $$\PageIndex{one}$$
Value of Discriminant Results
$$b^2−4ac=0$$ One rational solution (double solution)
$$b^ii−4ac>0$$, perfect square Two rational solutions
$$b^2−4ac>0$$, not a perfect square Two irrational solutions
$$b^two−4ac<0$$ Two circuitous solutions
##### THE DISCRIMINANT
For $$ax^2+bx+c=0$$, where $$a$$, $$b$$, and $$c$$ are real numbers, the
discriminant
is the expression under the radical in the quadratic formula: $$b^two−4ac$$. Information technology tells united states of america whether the solutions are real numbers or circuitous numbers and how many solutions of each type to expect.
##### Example $$\PageIndex{11}$$: Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation
Use the discriminant to find the nature of the solutions to the post-obit quadratic equations:
1. $$x^2+4x+4=0$$
2. $$8x^2+14x+3=0$$
3. $$3x^2−5x−2=0$$
4. $$3x^2−10x+15=0$$
Solution
Calculate the discriminant $$b^2−4ac$$ for each equation and state the expected type of solutions.
a.
$$x^ii+4x+four=0$$
$$b^two-4ac={(4)}^2-4(1)(4)=0$$ There volition exist one rational double solution.
b.
$$8x^2+14x+3=0$$
$$b^2-4ac={(fourteen)}^ii-4(8)(3)=100$$ As $$100$$ is a perfect square, there will be two rational solutions.
c.
$$3x^two−5x−2=0$$
$$b^ii-4ac={(-5)}^ii-4(iii)(-2)=49$$ As $$49$$ is a perfect foursquare, at that place will exist two rational solutions.
d.
$$3x^2−10x+15=0$$
$$b^2-4ac={(-10)}^2-iv(3)(xv)=-80$$ At that place will be ii complex solutions.
## Using the Pythagorean Theorem
I of the most famous formulas in mathematics is the
Pythagorean Theorem. It is based on a right triangle, and states the human relationship among the lengths of the sides as $$a^two+b^2=c^ii$$, where $$a$$ and $$b$$ refer to the legs of a right triangle adjacent to the $$90°$$ angle, and $$c$$ refers to the hypotenuse. It has immeasurable uses in compages, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.
We use the Pythagorean Theorem to solve for the length of one side of a triangle when we take the lengths of the other two. Because each of the terms is squared in the theorem, when nosotros are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that nosotros learned in this department to solve for the missing side.
The Pythagorean Theorem is given as
$a^2+b^2=c^2$
where $$a$$ and $$b$$ refer to the legs of a correct triangle adjacent to the $$90°$$ angle, and $$c$$ refers to the hypotenuse, as shown in .
##### Example $$\PageIndex{12}$$: Finding the Length of the Missing Side of a Right Triangle
Find the length of the missing side of the right triangle in Figure $$\PageIndex{five}$$.
Solution
As we have measurements for side $$b$$ and the hypotenuse, the missing side is $$a$$.
\begin{align*} a^two+b^two&= c^2\\ a^2+{(4)}^2&= {(12)}^ii\\ a^2+sixteen&= 144\\ a^2&= 128\\ a&= \sqrt{128}\\ &= 8\sqrt{2} \finish{align*}
##### Exercise $$\PageIndex{9}$$
Employ the Pythagorean Theorem to solve the right triangle trouble: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.
Respond
$$v$$ units
##### Media
Access these online resources for additional instruction and practice with quadratic equations.
1. Solving Quadratic Equations past Factoring
2. The Zero-Production Property
3. Completing the Foursquare
4. Quadratic Formula with Two Rational Solutions
5. Length of a leg of a right triangle
## Cardinal Equations
quadratic formula $$x=\dfrac{−b±\sqrt{b^2-4ac}}{2a}$$
## Key Concepts
• Many quadratic equations can be solved by factoring when the equation has a leading coefficient of $$i$$ or if the equation is a deviation of squares. The aught-factor property is and so used to find solutions. See Example, Example, and Instance.
• Many quadratic equations with a leading coefficient other than $$ane$$ can be solved past factoring using the group method. See Case and Example.
• Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the foursquare root of both sides of the equation. The solution will yield a positive and negative solution. Encounter Example and Case.
• Completing the square is a method of solving quadratic equations when the equation cannot exist factored. See Example.
• A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. Run across Instance.
• The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: existent or circuitous, rational or irrational, and how many of each. See Example.
• The Pythagorean Theorem, amid the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. Run across Instance.
## Which Equation Can Be Used to Solve for B
Source: https://math.libretexts.org/Sandboxes/nabrao_at_chabotcollege.edu/Chabot_College_College_Algebra_for_BSTEM/02%3A_Equations_and_Inequalities/2.06%3A_Quadratic_Equations
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