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Limits of Ratios of Polynomials
Home > Lessons > Limits of Ratios of Polynomials Search | Updated May 26th, 2017
Introduction
This lesson page will how to calculate the limit of ratios of polynomial functions. Here are the sections within this lesson page:
Before continuing with this lesson, it is important that you understand what functions are and how polynomials behave. Use these lessons to learn more before proceeding.
esson: Functions
esson: Polynomials
esson: End Behavior
Defining the Problem This page will explain how to evaluate limits of ratios of polynomial functions. They will be of either of these forms. By looking at each of those limits, it is clear we are taking a look at trends that are occurring far from the origin -- the extreme right and extreme left.
The Story of the Exponents When we consider polynomial functions within limits, we need to know what it is that determines the value of the function for extremely small and extremely large values. Take this function, f(x). It has three terms: x2, 3x, and -5. For large x-values, the -5 hardly does anything to the value of the function. All one has to do is evaluate the function at x = 100 to see which term dominates the result. Adding the last two terms results in 295, which is very close in value to 300, not -5. Also, adding 295 and 10000 gives us 10295, which is very close to 10000. 10000 was the result of the squared term. The squared term dominates the value of the result. Here is one more value to consider, f(1000). Again, the value obtained from the squared term, 1 million, is much greater in value than either of the two values. We can try substituting extremely small values, like x = -1000, but the same affect will be found. [Try it for yourself.] This means the values gained from the term with the greatest exponent dominates all the values obtained from the terms with the smaller exponents. Ratios of Polynomials: Summary Sometimes we are faced with limits of ratios of functions. Given two polynomials f(x) and g(x), we need to evaluate... ...here is a handy graphic that summarizes three cases. The following sections will explain why it is these three results occur. Degree of Numerator < Degree of Denominator Here is an example of a ratio of polynomials such that the degree of the numerator is less than the degree of the denominator. If we examine a table of values of increasing x-values, a clear pattern can be seen. It is evident that the ratio's value is approaching zero as the x-values increase. Here is a table that examines what happens when the x-values get smaller. This ratio also approaches zero when the x-values get smaller. The reason why this happens relates to the degrees of the polynomials. When g(x) has a greater degree than f(x), g(x) produces values that are far greater than those generated by f(x). Taking a number and dividing it by a much greater number produces a small result. The result gets smaller and smaller as the x-values increase. For the same reason, the result gets smaller and smaller as the x-values decrease. This pattern always occurs when the denominator has a greater degree than the numerator. Written using mathematical symbols, it can be written like so. Given polynomials f(x) and g(x), if deg(f(x)) < deg(g(x)), then... Degree of Numerator = Degree of Denominator Here is a ratio of polynomials such that the degree of the numerator is equal to the degree of the denominator. It is not immediately apparent that the degree of the numerator and denominator are equal until we write the polynomials in descending order, like so. Now it is clear that both of the polynomials are degree two. If we look at the ratio of polynomials on an analytical level, we can determine its value as the x-values either increase or decrease. For instance, we know that the terms of a polynomial that have the highest degree dominate the value of the polynomial for very large (and very small) x-values. This can allow us to look for long-term trends by examining the leading coefficients of the two polynomials. Furthermore, we can reduce the ratio of polynomials. This shows us that the ratio of polynomials will approach the value -2 as the x-values increase. This limit is also the same as the x-values decrease; so, the work will not shown. If we examine a table of increasing x-values, it will verify our conclusion. It is clear that the value of the ratio of polynomials is heading toward -2 as the x-values are increasing. Likewise, this next table of values will show us a trend as the x-values decrease. The trend for decreasing x-values is clear. The value of the ratio of polynomials is heading toward -2 and our conclusion is sound. Yet, this is an explanation that holds for a single example. We need to look at this from a wider scope. In general terms, this is what a ratio of polynomials looks like. Keep in mind we are maintaining that the numerator and denominator have equal degrees, which means the m-value is equal to the n-value. If we look at these polynomials with respect to extremely large x-values (or very small x-values), we need only examine their leading coefficients to grasp the trend they would have. Since m = n (their degrees are equal), we can further simplify the limit to be... The long-term trend is simply the ratio of the leading coefficients of the original polynomials. Degree of Numerator > Degree of Denominator Let us examine a ratio of polynomials such that the degree of the numerator is greater than the degree of the denominator. Again, we know when we are considering large x-values that the value of the polynomial is most influenced by the term that contains the highest degree. So, if we consider the limit of our ratio for large x-values, we get... We can simplify the new ratio, like so. Upon inspection of the simplified ratio function, we can see it is linear. If we were to graph the function, we would see that it flows down to the right. This means that the value of the function will continuously decrease as the x-values increase. Likewise, if we perform the same calculation for small, decreasing x-values, we will get the same simplified ratio. [The work is not shown.] However, now we are looking at the extreme left-side of the graph of this ratio function. If we examine the graph of the ratio function, the left-side of the linear function continuously goes up as the x-values decrease. This makes our limit... If we were to examine several cases when the degree of our polynomial in the numerator is larger than the degree of the polynomial in the denominator, we would see something interesting. Regardless of the ratio of polynomials we begin with, our simplified ratio results with another polynomial. We know that polynomials have end behavior that is based on their degrees and leading coefficients. [Review our lesson on End Behavior if you do not recall this property of polynomials.] This means our ratio of polynomials will either approach infinity or negative infinity, depending upon the coefficient and degree of the simplified ratio. Written mathematically, we get this: Given polynomials f(x) and g(x), if deg(f(x)) > deg(g(x)), then... To determine if the limit is either positive or negative infinity, we will need to look carefully at the degree and leading coefficient of the polynomial. Related Lessons Try these lessons, which are related to the sections above. esson: Functions esson: Polynomials esson: End Behavior esson: Limits of the Extreme esson: Limits of Piecewise Functions esson: Limits: Numerical Approach |
# Question Video: Finding the Lengths of Proportional Line Segments between Parallel Lines
Three Parallel Lines Theorem
Three Parallel Lines Theorem
### Video Transcript
Given that ๐๐ฟ equals nine centimeters, find the length of line segment ๐๐.
Letโs begin by observing that the lines ๐ด๐, ๐ต๐, ๐ถ๐, and ๐ท๐ฟ are all parallel lines. Alongside this, we observe that we have a transversal ๐ด๐ท. Aside from the information that the line segment ๐๐ฟ is nine centimeters, the only other measurement clue weโre given is that these three line segments ๐ด๐ต, ๐ต๐ถ, and ๐ถ๐ท are congruent.
In order to find the length of line segment ๐๐, weโll need to use Thalesโs special theorem. This theorem states that if three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. In this question, this diagram may cause some confusion with Thalesโs special theorem. We might wonder if this theorem means that the line segments ๐ด๐ต and ๐๐ are congruent. In fact, it does not. It means that because line segments ๐ด๐ต, ๐ต๐ถ, and ๐ถ๐ท are congruent, then the line segments ๐๐, ๐๐, and ๐๐ฟ are congruent. But they are congruent to each other and not the line segments on the other transversal.
In order to work out the length of line segment ๐๐, remember that we were given that ๐๐ฟ is nine centimeters. Letโs write out some of the things that we know. Firstly, we know that the whole of the line segment ๐๐ฟ consists of ๐๐ plus ๐๐ plus ๐๐ฟ. But we know that each of these line segments are congruent. We could even say that ๐๐ is equal to ๐๐ and ๐๐ฟ is also equal to ๐๐. We could therefore write that ๐๐ฟ is equal to three times ๐๐.
Given the information that ๐๐ฟ is nine centimeters, we can write that nine is equal to three times ๐๐. When we divide both sides by three, we get three is equal to ๐๐. And so ๐๐ must be three centimeters. In fact, each of these line segments must be three centimeters, which makes sense because we had a line segment of nine centimeters divided into three congruent pieces.
This wonโt of course be the final answer. We still need to work out ๐๐. Since the line segment of ๐๐ is made up of two line segments of three centimeters, then we can give the answer that the length of the line segment ๐๐ is six centimeters.
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# 1.4: Multiply Whole Numbers (Part 1)
Skills to Develop
• Use multiplication notation
• Model multiplication of whole numbers
• Multiply whole numbers
• Translate word phrases to math notation
• Multiply whole numbers in applications
be prepared!
Before you get started, take this readiness quiz.
1. Add: 1,683 + 479. If you missed this problem, review Example 1.21.
2. Subtract: 605 − 321. If you missed this problem, review Example 1.33.
### Use Multiplication Notation
Suppose you were asked to count all these pennies shown in Figure 1.11.
Figure 1.11
Would you count the pennies individually? Or would you count the number of pennies in each row and add that number 3 times.
$$8 + 8 + 8$$
Multiplication is a way to represent repeated addition. So instead of adding 8 three times, we could write a multiplication expression.
$$3 \times 8$$
We call each number being multiplied a factor and the result the product. We read 3 × 8 as three times eight, and the result as the product of three and eight.
There are several symbols that represent multiplication. These include the symbol × as well as the dot, • , and parentheses ( ).
Operation Symbols for Multiplication
To describe multiplication, we can use symbols and words.
Operation Notation Expression Read as Result
Multiplication × 3 × 8 three times eight the product of 3 and 8
3 • 8
() 3(8)
Example 1.39:
Translate from math notation to words: (a) 7 × 6 (b) 12 · 14 (c) 6(13)
##### Solution
(a) We read this as seven times six and the result is the product of seven and six.
(b) We read this as twelve times fourteen and the result is the product of twelve and fourteen.
(c) We read this as six times thirteen and the result is the product of six and thirteen.
Exercise 1.77:
Translate from math notation to words: (a) 8 × 7 (b) 18 • 11
Exercise 1.78:
Translate from math notation to words: (a) (13)(7) (b) 5(16)
### Model Multiplication of Whole Numbers
There are many ways to model multiplication. Unlike in the previous sections where we used base-10 blocks, here we will use counters to help us understand the meaning of multiplication. A counter is any object that can be used for counting. We will use round blue counters.
Example 1.40:
Model: 3 × 8.
##### Solution
To model the product 3 × 8, we’ll start with a row of 8 counters.
The other factor is 3, so we’ll make 3 rows of 8 counters.
Now we can count the result. There are 24 counters in all.
$$3 \times 8 = 24$$
If you look at the counters sideways, you’ll see that we could have also made 8 rows of 3 counters. The product would have been the same. We’ll get back to this idea later.
Exercise 1.79:
Model each multiplication: 4 × 6.
Exercise 1.80:
Model each multiplication: 5 × 7.
### Multiply Whole Numbers
In order to multiply without using models, you need to know all the one digit multiplication facts. Make sure you know them fluently before proceeding in this section. Table 1.39 shows the multiplication facts. Each box shows the product of the number down the left column and the number across the top row. If you are unsure about a product, model it. It is important that you memorize any number facts you do not already know so you will be ready to multiply larger numbers.
#### Table 1.39
× 0 1 2 3 4 5 6 7 8 9
0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9
2 0 2 4 6 8 10 12 14 16 18
3 0 3 6 9 12 15 18 21 24 27
4 0 4 8 12 16 20 24 28 32 36
5 0 5 10 15 20 25 30 35 40 45
6 0 6 12 18 24 30 36 42 48 54
7 0 7 14 21 28 35 42 49 56 63
8 0 8 16 24 32 40 48 56 64 72
9 0 9 18 27 36 45 54 63 72 81
What happens when you multiply a number by zero? You can see that the product of any number and zero is zero. This is called the Multiplication Property of Zero.
Definition: Multiplication Property of Zero
The product of any number and 0 is 0.
$$a \cdot 0 = 0$$
$$0 \cdot a = 0$$
Example 1.41:
Multiply: (a) 0 • 11 (b) (42)0
##### Solution
(a)
The product of any number and zero is zero. 0 • 11 = 0
(b)
Multiplying by zero results in zero. (42)0 = 0
Exercise 1.81:
Find each product: (a) 0 • 19 (b) (39)0
Exercise 1.82:
Find each product: (a) 0 • 24 (b) (57)0
What happens when you multiply a number by one? Multiplying a number by one does not change its value. We call this fact the Identity Property of Multiplication, and 1 is called the multiplicative identity.
Definition: Identity Property of Multiplication
The product of any number and 1 is the number.
$$1 \cdot a = a$$
$$a \cdot 1 = a$$
Example 1.42:
Multiply: (a) (11)1 (b) 1 • 42
##### Solution
(a)
The product of any number and one is the number. (11)1 = 11
(b)
Multiplying by one does not change the value. 1 • 42 = 42
Exercise 1.83:
Find each product: (a) (19)1 (b) 1 • 39
Exercise 1.84:
Find each product: (a) (24)(1) (b) 1 × 57
Earlier in this chapter, we learned that the Commutative Property of Addition states that changing the order of addition does not change the sum. We saw that 8 + 9 = 17 is the same as 9 + 8 = 17.
Is this also true for multiplication? Let’s look at a few pairs of factors.
$$\begin{split} 4 \cdot 7 & = 28 \qquad 7 \cdot 4 = 28 \\ 9 \cdot 7 & = 63 \qquad 7 \cdot 9 = 63 \\ 8 \cdot 9 & = 72 \qquad 9 \cdot 8 = 72 \end{split}$$
When the order of the factors is reversed, the product does not change. This is called the Commutative Property of Multiplication.
Definition: Commutative Property of Multiplication
Changing the order of the factors does not change their product.
$$a \cdot b = b \cdot a$$
Example 1.43:
Multiply: (a) 8 • 7 (b) 7 • 8
##### Solution
(a)
Multiply. 8 • 7 = 56
(b)
Multiply. 7 • 8 = 56
Changing the order of the factors does not change the product.
Exercise 1.85:
Multiply: (a) 9 • 6 (b) 6 • 9
Exercise 1.86:
Multiply: (a) 8 • 6 (b) 6 • 8
To multiply numbers with more than one digit, it is usually easier to write the numbers vertically in columns just as we did for addition and subtraction.
We start by multiplying 3 by 7.
$$3 \times 7 = 21$$
We write the 1 in the ones place of the product. We carry the 2 tens by writing 2 above the tens place.
Then we multiply the 3 by the 2, and add the 2 above the tens place to the product. So 3 × 2 = 6, and 6 + 2 = 8. Write the 8 in the tens place of the product.
The product is 81.
When we multiply two numbers with a different number of digits, it’s usually easier to write the smaller number on the bottom. You could write it the other way, too, but this way is easier to work with.
Example 1.44:
Multiply: 15 • 4.
##### Solution
Write the numbers so the digits 5 and 4 line up vertically. Multiply 4 by the digit in the ones place of 15. 4 • 5 = 20. Write 0 in the ones place of the product and carry the 2 tens. Multiply 4 by the digit in the tens place of 15. 4 ⋅ 1 = 4. Add the 2 tens we carried. 4 + 2 = 6. Write the 6 in the tens place of the product.
Exercise 1.87:
Multiply: 64 • 8.
Exercise 1.88:
Multiply: 57 • 6.
Example 1.45:
Multiply: 286 • 5.
##### Solution
Write the numbers so the digits 5 and 6 line up vertically. Multiply 5 by the digit in the ones place of 286. 5 • 6 = 30. Write the 0 in the ones place of the product and carry the 3 to the tens place. Multiply 5 by the digit in the tens place of 286. 5 • 8 = 40. Add the 3 tens we carried to get 40 + 3 = 43. Write the 3 in the tens place of the product and carry the 4 to the hundreds place. Multiply 5 by the digit in the hundreds place of 286. 5 • 2 = 10. Add the 4 hundreds we carried to get 10 + 4 = 14. Write the 4 in the hundreds place of the product and the 1 to the thousands place.
Exercise 1.89:
Multiply: 347 • 5.
Exercise 1.90:
Multiply: 462 • 7.
When we multiply by a number with two or more digits, we multiply by each of the digits separately, working from right to left. Each separate product of the digits is called a partial product. When we write partial products, we must make sure to line up the place values.
HOW TO: MULTIPLY TWO WHOLE NUMBERS TO FIND THE PRODUCT
Step 1. Write the numbers so each place value lines up vertically.
Step 2. Multiply the digits in each place value.
• Work from right to left, starting with the ones place in the bottom number.
• Multiply the bottom number by the ones digit in the top number, then by the tens digit, and so on.
• If a product in a place value is more than 9, carry to the next place value.
• Write the partial products, lining up the digits in the place values with the numbers above.
• Repeat for the tens place in the bottom number, the hundreds place, and so on.
• Insert a zero as a placeholder with each additional partial product.
Step 3. Add the partial products.
Example 1.46:
Multiply: 62(87).
##### Solution
Write the numbers so each place lines up vertically Start by multiplying 7 by 62. Multiply 7 by the digit in the ones place of 62. 7 • 2 = 14. Write the 4 in the ones place of the product and carry the 1 to the tens place. Multiply 7 by the digit in the tens place of 62. 7 • 6 = 42. Add the 1 ten we carried. 42 + 1 = 43. Write the 3 in the tens place of the product and the 4 in the hundreds place. The first partial product is 434. Now, write a 0 under the 4 in the ones place of the next partial product as a placeholder since we now multiply the digit in the tens place of 87 by 62. Multiply 8 by the digit in the ones place of 62. 8 • 2 = 16. Write the 6 in the next place of the product, which is the tens place. Carry the 1 to the tens place. Multiply 8 by 6, the digit in the tens place of 62, then add the 1 ten we carried to get 49. Write the 9 in the hundreds place of the product and the 4 in the thousands place. The second partial product is 4960. Add the partial products.
The product is 5,394.
Exercise 1.91:
Multiply: 43(78).
Exercise 1.92:
Multiply: 64(59).
Example 1.47:
Multiply: (a) 47 • 10 (b) 47 • 100.
##### Solution
(a) 47 • 10 (b) 47 • 100
When we multiplied 47 times 10, the product was 470. Notice that 10 has one zero, and we put one zero after 47 to get the product. When we multiplied 47 times 100, the product was 4,700. Notice that 100 has two zeros and we put two zeros after 47 to get the product.
Do you see the pattern? If we multiplied 47 times 10,000, which has four zeros, we would put four zeros after 47 to get the product 470,000.
Exercise 1.93:
Multiply: (a) 54 • 10 (b) 54 • 100.
Exercise 1.94:
Multiply: (a) 75 • 10 (b) 75 • 100.
Example 1.48:
Multiply: 354(438).
##### Solution
There are three digits in the factors so there will be 3 partial products. We do not have to write the 0 as a placeholder as long as we write each partial product in the correct place.
Exercise 1.95:
Multiply: 265(483).
Exercise 1.96:
Multiply: 823(794).
Example 1.49:
Multiply: 896(201).
##### Solution
There should be 3 partial products. The second partial product will be the result of multiplying 896 by 0.
Notice that the second partial product of all zeros doesn’t really affect the result. We can place a zero as a placeholder in the tens place and then proceed directly to multiplying by the 2 in the hundreds place, as shown.
Multiply by 10, but insert only one zero as a placeholder in the tens place. Multiply by 200, putting the 2 from the 12. 2 • 6 = 12 in the hundreds place.
Exercise 1.97:
Multiply: (718)509.
Exercise 1.98:
Multiply: (627)804.
When there are three or more factors, we multiply the first two and then multiply their product by the next factor. For example:
to multiply 8 • 3 • 2 first multiply 8 • 3 24 • 2 then multiply 24 • 2 48 |
# Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{5x^2+1}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
It’s basically an exponential function in terms of $x$. In this limit problem, the base and exponent both are algebraic functions in fraction form. The limit of algebraic function in exponential form has to be evaluated as $x$ approaches zero in this problem.
### Simplify the Exponential function
Let’s start simplifying this algebraic function firstly to move ahead in evaluating limit of exponential function.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{5x^2+1}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
The base of exponential function is in fractional form in which the expressions in both numerator and denominator are quadratic equations. Now, try to find the quotient of them.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{3x^2+2x^2+1}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{3x^2+1+2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{3x^2+1}{3x^2+1}+\dfrac{2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{\cancel{3x^2+1}}{\cancel{3x^2+1}}+\dfrac{2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(1+\dfrac{2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$
The simplified function is similar to the following limit rule of exponential functions.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^\frac{1}{x}}$
### Transform the function into known form
Let’s try to transform the whole function same as the above limit rule.
Take $p = \dfrac{2x^2}{3x^2+1}$ and try to express the exponent $\dfrac{1}{x^2}$ in terms of $p$.
$\implies$ $\dfrac{3x^2+1}{2x^2} = \dfrac{1}{p}$
$\implies$ $\dfrac{3x^2+1}{x^2} = \dfrac{2}{p}$
$\implies$ $\dfrac{3x^2}{x^2}+\dfrac{1}{x^2} = \dfrac{2}{p}$
$\implies$ $\require{cancel} \dfrac{3\cancel{x^2}}{\cancel{x^2}}+\dfrac{1}{x^2} = \dfrac{2}{p}$
$\implies$ $3+\dfrac{1}{x^2} = \dfrac{2}{p}$
$\implies$ $\dfrac{1}{x^2} = \dfrac{2}{p}-3$
Therefore, the algebraic function $\dfrac{2x^2}{3x^2+1}$ can be written as $p$ simplify and the exponent $\dfrac{1}{x^2}$ can also written as $\dfrac{2}{p}-3$ in the limit of exponential function.
According to $\dfrac{1}{x^2} = \dfrac{2}{p}-3$
If $x \,\to\, 0$, then $x^2 \to {(0)}^2$. So, $x^2 \to 0$. Now $\dfrac{1}{x^2} \to \dfrac{1}{0}$, therefore $\dfrac{1}{x^2} \to \infty$ but $\dfrac{1}{x^2}$ is equal to $\dfrac{2}{p}-3$.
Therefore $\dfrac{2}{p}-3 \to \infty$. Now, $\dfrac{2}{p} \to \infty + 3$, and then $\dfrac{2}{p} \to \infty$. Finally, $\dfrac{p}{2} \to \dfrac{1}{\infty}$ and then $\dfrac{p}{2} \to 0$.
If $\dfrac{p}{2} \to 0$ then $p \to 2 \times 0$. Therefore $p \to 0$.
Therefore, if $x$ approaches $0$, then $p$ also approaches $0$.
Now, convert the limit of exponential function in terms of $x$ into its equivalent function in terms of $p$.
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(1+\dfrac{2x^2}{3x^2+1}\Bigg)}^\dfrac{1}{x^2}}$ $\,=\,$ $\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^{\frac{2}{p} \,-\, 3}}$
### Simplify the Exponential function
$\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^{\frac{2}{p} \,-\, 3}}$
The exponents are in subtraction form. The exponential function can be written in quotient form as per quotient rule of exponents.
$= \,\,\,$ $\displaystyle \large \lim_{p \,\to\, 0}{\normalsize \dfrac{{(1+p)}^{\frac{2}{p}}}{{(1+p)}^3}}$
According to quotient rule of limits
$= \,\,\,$ $\dfrac{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^{\frac{2}{p}}}}{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^3}}$
According to Power of Power Rule of Exponents.
$= \,\,\,$ $\dfrac{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {{\Big[(1+p)}^{\frac{1}{p}}} \Big]}^2}{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^3}}$
The function in the numerator can be further simplified by the constant exponent power rule of limits.
$= \,\,\,$ $\dfrac{\displaystyle \Bigg[ \large \lim_{p \,\to\, 0}{\normalsize {{(1+p)}^{\frac{1}{p}}} \Bigg]}^2}{\displaystyle \large \lim_{p \,\to\, 0}{\normalsize {(1+p)}^3}}$
### Evaluate Limit of Exponential function
The limit of the exponential function in the numerator can be evaluated by a limit rule of exponential function and the limit of the exponential function in the denominator can be evaluated by the direct substitution method.
$= \,\,\,$ $\dfrac{{[e]}^2}{{(1+0)}^3}$
$= \,\,\,$ $\dfrac{e^2}{1^3}$
$= \,\,\,$ $\dfrac{e^2}{1}$
$= \,\,\,$ $e^2$
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Problem: The cubit is an ancient unit of length based on the distance between the elbow and the tip of the middle finger of the measurer. Assume that the distance ranged from 43 to 53 cm, and suppose that ancient drawings indicate that a cylindrical pillar was to have a length of 9 cubits and a diameter of 2 cubits. For the stated range, what are the lower value and the upper value, respectively, for (a) the cylinder’s length in meters, (b) the cylinder’s volume in cubic meters? (c) What is the percent uncertainty in the volume?
FREE Expert Solution
For operations with uncertainty, we’ll follow two different rules depending on whether we're adding/subtracting or multiplying/dividing:
• When measurements are added or subtracted, sum the absolute or relative uncertainty—the result is the same.
• When measurements are multiplied or divided, sum the relative uncertainties.
So anytime you square a measurement, add the uncertainty twice.
To convert between absolute uncertainty and percent uncertainty, we’ll use this formula (= measurement, Δ= absolute uncertainty):
$\overline{){\mathbit{m}}{\mathbf{±}}{\mathbf{∆}}{\mathbit{u}}{\mathbf{=}}{\mathbit{m}}{\mathbf{±}}\mathbf{\left(}\frac{\mathbf{∆}\mathbit{u}}{\mathbit{m}}\mathbf{×}\mathbf{100}\mathbf{%}\mathbf{\right)}}$
This problem gives us the maximum and minimum values for the length of a cubit. For parts (a) and (b) of the problem, we just need to convert the units from centimeters to meters, write our equation, and plug in the max and min. Part (c) is a little more complicated: we’ll need the rule for multiplying measurements with uncertainty.
We’ll assume that we know the length and diameter of the cylinder to 0.1 cubit—that is, = 9.0 cubits and d = 2.0 cubits.
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Problem Details
The cubit is an ancient unit of length based on the distance between the elbow and the tip of the middle finger of the measurer. Assume that the distance ranged from 43 to 53 cm, and suppose that ancient drawings indicate that a cylindrical pillar was to have a length of 9 cubits and a diameter of 2 cubits. For the stated range, what are the lower value and the upper value, respectively, for (a) the cylinder’s length in meters, (b) the cylinder’s volume in cubic meters? (c) What is the percent uncertainty in the volume? |
Identify The Larger Number Worksheets
How to Identify the Larger Number in A Set - When we were little, we had a very simple definition of the various terms. We thought that if someone is tall to look at, the person was tall if someone looked fat, we assumed that they would be heavy, if we saw a bird, we thought that it is light. Similarly, we thought that the elephant is big, and the ant is small. All of this is totally true! But as we grew older, we realized that these definitions could not be applied in every platform. Just like that, in math, there are different definitions for the terms that we previously knew. But the bigger and smaller remain the same! When we are studying sets, we are given various numbers, and it gets difficult to identify which number comes after which, which number is bigger, and which is smaller. Therefore, you always need to arrange the numbers so that you can identify the larger number from the lot. The first step that you need to do is taking a look at the numbers and finding if they have any signs. Remember that if the number has no sign, it means that it is a positive number, and they are always bigger. If you have a negative sign, it means that the number is smaller in quantity. Now take a look at the values one by one, and the number with the largest quantity is the larger number from the lot!
• Basic Lesson
Provides a visual comparison with the use of a number line. Includes practice problems. Draw a numbers line and find which number is furthest to the right.
• Independent Practice 1
Students write the largest of three numbers on a line. Look at the numbers in each box and circle the number that is larger. Write it on the line.
• Independent Practice 2
Evaluate three numbers. Determine the greatest number. Answers can be found below.
• Homework Worksheet
Circle and write the largest number. An example is provided.
• Skill Quiz
Compare 3 numbers. Write the largest number on the line. 5 problems in all.
• Homework and Quiz Answer Key
Answers for the homework and quiz.
• Lesson and Practice Answer Key
Answers for both lessons and both practice sheets.
Where We Use This Skill
Unconsciously, we are using this skill almost constantly when trying to make decisions. In some cases, we want a larger value such as which bank account will help us make more money. On the flip side if we are borrowing money, we look for the smaller value to reduce the amount we owe. |
# 7-8x >19-7 please answer this on how to solve the inequality?
Apr 24, 2018
$x < - \frac{5}{8}$
#### Explanation:
Isolate x.
$7 - 8 x > 19 - 7$
Add $7$ to $- 7$ to cancel it out because it is the lowest number here.
But you do to one side what you do to the other, so add $7$ to the positive $7$ on the other side. You should now have:
$14 - 8 x > 19$
Now, subtract $14$ from $14$ to cancel it out and do the same to the other side $\left(19\right)$. Now, you should have:
$- 8 x > 5$
Now, to isolate $x ,$ divide by $- 8$.
But remember when you divide or multiply an inequality by a negative value, the sign changes around.
$\frac{- 8 x}{- 8} < \frac{5}{- 8}$
Because you divided by a negative, the sign flips:
$x < - \frac{5}{8}$
Apr 24, 2018
$x < - \frac{5}{8}$
#### Explanation:
You can treat an inequality in exactly the same way as an equation, except if you multiply or divide by a negative value, the inequality sign changes around.
$7 \textcolor{b l u e}{- 8 x} > 19 - 7$
Let's avoid the problem with a negative term with the variable by moving it to the other side,
Add $8 x$ to both sides and simplify where possible:
$7 \cancel{- 8 x} \cancel{+ 8 x} > 12 \textcolor{b l u e}{+ 8 x}$
Subtract $12$ from both sides:
$7 - 12 > 12 + 8 x - 12$
$- 5 > 8 x$
-5/8 > x" "larr (div 8 on both sides)
This can be written as $x < - \frac{5}{8}$ |
Update all PDFs
# Triangular Tables
Alignments to Content Standards: 6.EE.B
A classroom has triangular tables. There is enough space at each side of a table to seat one child. The tables in the class are arranged in a row (as shown in the picture below).
1. How many children can sit around 1 table? Around a row of two tables? Around a row of three tables?
2. Find an algebraic expression that describes the number of children that can sit around a row of $n$ tables. Explain in words how you found your expression.
3. If you could make a row of 125 tables, how many children would be able to sit around it?
4. If there are 26 children in the class, how many tables will the teacher need to seat all the children around a row of tables?
## IM Commentary
This task provides a good opportunity for group work and class discussions where students generate and compare equivalent expressions. In class discussion, students should be asked to connect the terms of an expression with quantities shown in the diagram. As part of the solution, students can make a table of values for increasing numbers of tables and seats and find a pattern. It is important that a discussion of the problem explicitly draws out how the diagram, the table of values, and the expressions are connected.
Part (c) of the question provides a good opportunity to discuss what might be a reasonable range of values for $n$, preparing students for the idea of the domain of a function later on.
## Solutions
Solution: Solution 1
1. Since one table has three sides and each seats one child, it follows that 3 children can sit around 1 table.
When two tables are put together in a row as pictured, then we can count the number of open sides around the perimeter of the two tables together, since an open side means one child can sit there. There are 4 sides that are open around the table, and so 4 children can sit around a row of 2 tables.
Using the same method as above, we see that when 3 tables are put into a row we will have 5 open sides around the tables. So, 5 children can sit around a row of 3 tables.
2. To find an expression that describes the number of children that can sit around a row of $n$ tables, we can consider the diagram below. We see that we can fit 1 child at each horizontal table side (black dots) plus 1 child on the left and one on the right (white dots). So we have:
$$\text{children that can sit at } n \text{ tables} = n + 1 + 1 = n + 2$$
Another way to think about counting seats is shown in the picture below. The first table seats two children and the last table seats two children (white dots). All other $n - 2$ tables seat one child (black dots). So we have:
$$\text{children that can sit at } n \text{ tables} = 2 \cdot 2 + (n - 2) \cdot 1 = 4 + (n - 2).$$
Other expressions are also possible, which are all equivalent to $n + 2$.
3. Using our expression from part (b), with $n=125$, we see that
$$n + 2 = 125 + 2 = 127,$$
so 127 children can sit around a row of 125 tables.
4. Using our expression from before we know that a row of $n$ tables seats $n + 2$ children. If we want to seat 26 children we need to find n such that $n + 2 = 26$. So we have $n = 24$, which means that the teacher needs 24 tables to seat all students in the class.
Solution: Solution 2
1. See Solution 1
2. We can summarize the numbers we found in the solution above in a table of values:
number of tables, $n$number of seats
13
24
35
Looking at a few more examples, with 4 and 5 tables forming a row, we can extend our table of values:
number of tables, $n$number of seats
13
24
35
46
57
We notice the pattern, that the number of seats is equal to the number of tables plus 2. So we have:
$$\text{children that can sit at } n \text{ tables} = n + 2.$$
3. See Solution 1
4. See Solution 1
#### alio73 says:
over 6 years
There is a typo under Commentary. It states "This taks ..." and should probably be "This task..."
#### Brigitte says:
over 6 years
Thank you. I fixed the typo.
over 6 years
removed
#### lizyockey says:
over 6 years
Isn't the answer to C that there is not room enough to have a row of 125 tables in any classroom? Where do these tables exist? Are they on a football field? This might be a good time to have a discussion about domain (without necessarily naming it as domain). I worry that when we give "real world" situations but blend them with impossible questions, we are shutting kids down from reasoning about their answers. Just getting a numeric answer and stopping there doesn't constitute problem solving.
#### Wendy Gibson says:
over 5 years
Well stated. Let's think about realistic situations, and the suggestion of 125 tables in a row wouldn't be possible in any building.
#### Bill says:
over 6 years
Liz, I've reworded the problem slightly, and added your point about domain to the comments. Does this make it better?
Bill McCallum |
# Forms of Linear Equations
Slope Intercept Form
m is the slope of the line and is the y-intercept
Point Slope Form
m is the slope of the line and is a point on the line.
Standard Form of a Line
# Graphing Linear Equations with Slope Intercept Form
Example: Graph the linear equation using the slope and the y-intercept. (the slope is positive and fractional)
Example: Graph the linear equation using the slope and the y-intercept. (positive and negative fractional slopes)
Example: Graph the linear equation using the slope and the y-intercept. (equation written in standard form)
# Finding the Equation of a line given a fractional slope and a point
Example: Find the equation of a line in slope intercept form given the slope of the line is and the line passes through the point
Solution:
Use the point-slope formula of the line to start building the line. m represents the slope of the line and is a point on the line.
Point-slope formula:
and
Substitute the values into the formula.
Since the instructions ask to write the equation in slope intercept form we will simplify and write the equation with y by itself on one side. I will also use the clearing fractions method to avoid having to add fractions.
(Multiply by LCM)
(Cancel Denominator)
The equation of a line in slope intercept form with a slope of and passing through the point is
# Finding the Equation of a Line given two points on the line
Example: Find the equation of a line in slope intercept form given the line passes through the two points and .
Solution:
First find the slope of the line.
Choose one of the points to be and choose the other point to be .
I will choose to be and choose to be .
Substitute these values into the slope formula and simplify.
The slope of the line containing the points and is .
Then, use the point-slope formula of the line to start building the line. m represents the slope of the line and you can use or as the point on the line.
Point-slope formula:
and
Substitute the values into the formula.
Since the instructions ask to write the equation in slope intercept form we will simplify and write the equation with y by itself on one side.
The equation of a line in slope intercept form passing through the two points and is .
# Finding the Equation of a Line given the slope and a point
Example: Find the equation of a line in slope intercept form given the slope of the line is 7 and the line passes through the point
Solution:
Use the point-slope formula of the line to start building the line. m represents the slope of the line and is a point on the line.
Point-slope formula:
and
Substitute the values into the formula.
Since the instructions ask to write the equation in slope intercept form we will simplify and write the equation with y by itself on one side.
The equation of a line in slope intercept form with a slope of 7 and passing through the point is .
Example: Find the equation of the line.
# X-intercepts and Y-intercepts
An x-intercept is where the graph touches or crosses the x-axis.
A y-intercept is where the graph touches of crosses the y-axis.
In this picture, the graph crosses the x-axis at the ordered pair . Since every ordered pair on the x-axis has a y coordinate of zero we can let to find x-intercepts.
To find an x-intercept: Let y=0 and solve for x.
In this picture, the graph crosses the y-axis at the ordered pair . Since every ordered pair on the y-axis has a x coordinate of zero we can let to find y-intercepts.
To find an y-intercept: Let x=0 and solve for y.
# Graphing Linear Equations by Finding Intercepts
Steps for Graphing with the Intercept Method
1. Find the x intercept and the y-intercept.
• To find an x-intercept let y=0 and solve for x.
• To find a y-intercept let x=0 and solve for y.
2. Plot the x-intercept and y-intercept.
3. Draw the line that connects the intercepts.
Example: Graph the linear equation
Solution:
1. Find the x-intercept and the y-intercept.
To find an x-intercept: Let y=0 and solve for x.
The x-intercept of this equation is
To find a y-intercept: Let x=0 and solve for y.
The y-intercept of this equation is
2. Plot the x-intercept and the y-intercept.
3. Draw the line that connects the intercepts.
Example: Graph the linear equation
Solution:
1. Find the x-intercept and the y-intercept.
To find an x-intercept: Let y=0 and solve for x.
The x-intercept of this equation is
To find a y-intercept: Let x=0 and solve for y.
The y-intercept of this equation is
Since the x-intercept and the y-intercept are the same point and we need two distinct points to graph a line, we must find another ordered pair that is a solution to the equation.
Let x=1 and find the associated y value. (I chose x=1 but you could choose a different value)
Another ordered pair on the graph is
2. Plot the x-intercept and the y-intercept.
3. Draw the line that connects the intercepts.
Example: Graphing a linear equation with intercepts.
Example: Graphing a linear equation with intercepts.
Example: Graphing a linear equation with intercepts.
# Calculating Slope
Given two points on the line and , you can calculate the slope of a line by the following formula.
is also know as or “the change in y.”
is also know as or “the change in x.”
Example: Calculate the slope of the line containing the points and .
Solution: Choose one of the points to be and choose the other point to be .
I will choose to be and choose to be .
Substitute these values into the slope formula and simplify.
The slope of the line containing the points and is .
Example: Calculate the slope of the line containing the points and .
Solution: Choose one of the points to be and choose the other point to be .
I will choose to be and choose to be .
Substitute these values into the slope formula and simplify.
The slope of the line containing the points and is .
Example: Finding the slope with the formula.
Example: Finding the slope with the formula.
Example: Finding the slope from the graph.
# Interpretation of slope
The slope of a line is a number that indicates the “steepness” of a line. Slope is usually denoted with the letter m.
If the slope of the line is positive, the line will be rising or increasing from left to right.
All three of the above graphs have a positive slope and the line is rising or increasing from left to right. Notice as the value of the slope gets larger, the line is getting steeper.
If the slope of the line is negative, the line will be falling or decreasing from left to right.
All three of the above graphs have a negative slope and the line is falling or decreasing from left to right. Notice as the value of the slope gets smaller, the line is getting steeper.
If the slope is zero, the line will be constant. This results in a horizontal line.
All three of the the above graphs have a slope of zero. The y values are constant.
A vertical line has a slope that is undefined.
All three of the vertical lines have undefined slope. |
How do you find the sum of the infinite geometric series 1+1/5+1/25+....?
Nov 9, 2015
Divide terms to find $r$ and then apply the geometric series formula to find the sum to be $\frac{5}{4}$
Explanation:
A geometric series is a series of the form
$a + a r + a {r}^{2} + a {r}^{3} + \ldots$
where $a$ is an initial value and $r$ is a common ratio between terms.
In general, the sum of a geometric series is
$a + a r + a {r}^{2} + \ldots + a {r}^{n - 1} = a \frac{1 - {r}^{n}}{1 - r}$ (see below for derivation)
If $| r | < 1$ then as $n \to \infty , {r}^{n} \to 0$ so the formula reduces to
$a + a r + a {r}^{2} + \ldots = \frac{a}{1 - r}$
Now, in order to find $r$, it suffices to divide any term in the series after the first by the term prior to it, as
$\frac{a {r}^{n}}{a {r}^{n - 1}} = r$
So, picking the second and first terms, for this series we get
$r = \frac{\frac{1}{5}}{1} = \frac{1}{5}$
As the first term in the series gives us $a = 1$ we get the final sum
$\frac{1}{1 - \left(\frac{1}{5}\right)} = \frac{1}{\frac{4}{5}} = \frac{5}{4}$
.
.
.
What follows is a short derivation for the geometric series formula, and is not required for understanding the above solution.
Let ${S}_{n}$ be the $n$th partial sum of a geometric series with ratio $r$ and first term $a$, that is,
${S}_{n} = a + a r + a {r}^{2} + \ldots + a {r}^{n - 1}$
$\implies r {S}_{n} = a r + a {r}^{2} + \ldots + a {r}^{n}$
$\implies {S}_{n} - r {S}_{n} = a - a {r}^{n}$
$\implies {S}_{n} \left(1 - r\right) = a \left(1 - {r}^{n}\right)$
$\implies {S}_{n} = a \frac{1 - {r}^{n}}{1 - r}$ |
# CBSE CLASS 5 MATHEMATICS/ Worksheets
CBSE Class 5 Mathematics is about some model questions for class 5 Mathematics. Here in this lesson you can find out practice problems for class 5 Maths.
CBSE CLASS 5 MATHEMATICS /WORKSHEETS
1. If a log boat travel about 5 km in one hour, how long will they take to go a distance of 20 km?
2. The price of 9kg fresh fish is 1080 rupees. Find the cost of 1 kg fresh fish?
3. Write the English alphabets where you can see right angles?
4. Find the area of a rectangle whose length is 12 cm and breadth is 8 cm?
5. The perimeter of a square carom board is 480 cm, find its area?
6. There are 12 chocolates in Manu’s hand. He gave ¾ of the chocolates to his friends. How many chocolates are remaining with him?
7. The cost of 1 kg carrot is 42 rupees. Find the cost of 2 ½ kg carrots?
8. Write all the factors of 36.
9. Write first five multiples of 6.
10. An egg costs five and a half rupees. How much will one and a half dozen cost?
1. Speed of log boat = 5 km/ hour
Time taken to cover a distance of 20 km = 4 hour
2. Price of 9 kg fresh fish = 1080 rupees
Cost of 1 kg fresh fish = 1080/9 = 120 rupees
3. E, F, H, L, T
4. Given length = 12 cm and breadth = 8 cm
Area of a rectangle = l x b = 12 x 8 = 96 square cm
5. Perimeter of a square = 480 cm
Length of each side = 480/4 = 120 cm
Area = side x side = 120 x 120 = 14400 square cm
6. ¾ of 12 = 9 chocolates
Number of chocolates remaining with him = 12 – 9 = 3 chocolates
7. The cost of 1 kg carrot = 42 rupees
Cost of 2 kg carrot = 42 x 2 = 84 rupees
Cost of ½ kg carrot = ½ of 42 = 21 rupees
Cost of 2 ½ kg carrots = 84 + 21 = 105 rupees
8. Factor pairs of 36 are 1 x 36, 2 x 18, 3 x 12, 4 x 9, 6 x 6.
So the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
9. Multiples of 6 are 6, 12, 18, 24, and 30
10. Cost of an egg = 5.50 rupees
Cost of one dozen eggs = 12 x 5.50 = 66 rupees
Cost of half dozen eggs = 6 x 5.50 = 33 rupees
Cost of 2 ½ kg eggs = 66 + 33 = 99 rupees |
Reciprocal Functions
These techniques involves sketching the graph of $y=\dfrac{1}{f(x)}$ from the graph of $y=f(x)$. Technique 1 When $f(x)$ approaches towards $0$, $y=\dfrac{1}{f(x)}$ approaches towards $\infty$, the graph of $y=\dfrac{1}{f(x)}$ approaches the vertical asymptote(s). Technique 2 The graph of $y=\dfrac{1}{f(x)}$ has vertical asymptotes at the $x$-intercepts of $y=f(x)$. Technique 3 When $f(x)$ approaches towards $\infty$, $y=\dfrac{1}{f(x)}$ approaches […]
Inverse Functions – Ultimate Guide of Definition and Graphs
Domain and Range of Inverse Functions \Large \begin{align} + \leftarrow &\text{ inverse operation } \rightarrow – \\\times \leftarrow &\text{ inverse operation } \rightarrow \div \\x^2 \leftarrow &\text{ inverse operation } \rightarrow \sqrt{x} \end{align}The function $y=4x-1$ can be undone by its inverse function $y=\dfrac{x+1}{4}$.We can consider this act as two processes or machines. If […]
Rational Functions
We have seen that a linear function has the form $y=mx+b$.When a linear function is devided by another function, the result is a rational function.Rational functions are characterised by asymptotes, which are lines the function gets close and close to but never reaches.The rational functions can be written in $y=\dfrac{ax+b}{cx+d}$. These functions have asymptotes which […]
Composite Functions
A composite function is formed from two functions in the following way. $$(g \circ f)(x) = g(f(x))$$If $f(x)=x+3$ and $g(x)=2x$ are two functions, then we combine the two functions to form the composite function:\begin{align}(g \circ f)(x) &= g(f(x)) \\&= 2f(x) \\&= 2(x+3) \\&= 2x+6\end{align}That is, $f(x)$ replaces $x$ in the function $g(x)$.The composite […]
Domain and Range – An Ultimate Guide
A relation may be described by: The set of all first elements of a set of ordered pairs is known as the domain, and the set of all second elements of a set of ordered pairs is known as the range.Alternatively, the domain is the set of independent values, and the range is the set […]
Function Notation
Consider the relation $y=3x+2$, which is a function.The $y$-values are determined from the $x$-values, so we say ‘$y$ is a function of $x$, abbreviated to $y=f(x)$.So, the rule $y=3x+2$ can also be written as follows.\large \begin{align} f: \mapsto \ &3x+2 \\&\text{or} \\f(x)= \ &3x+2 \\&\text{or} \\y= \ &3x+2 \end{align}A function $f$ such that […]
Relations and Functions
Relations A relation is any set of points that connect two variables.A relation is often expressed as an equation connecting the variables $x$ and $y$. The relation is a set of points $(x,y)$ in the Cartesian plane. This plane is separated into four quadrants according to the signs of $x$ and $y$. For example, $y=x+1$ […] |
# How do you multiply (-2-9i)(-3-4i) in trigonometric form?
May 19, 2018
$\left(- 30 + 35 i\right)$
#### Explanation:
Any complex equation in the form of $a + b i$ i.e. (vector form) can be written as $r {e}^{\theta i}$ (rectangular form)
Here,
$r \rightarrow$ magnitude of the vector, and
$\theta \rightarrow$ the angle between the vector form and the components
now, $r {e}^{\theta i}$ is equivalent to $r \left(\cos \theta + i \sin \theta\right)$
this tells us,
$a = r \cos \theta$
and $b = r \sin \theta$
thus, by solving the 2 above equations, $r = \sqrt{{a}^{2} + {b}^{2}}$
and $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$
So, solving for $\left(- 2 - 9 i\right)$,
${r}_{1} = \sqrt{85}$
${\theta}_{1} = 77.47$ degrees
And,solving for $\left(- 3 - 4 i\right)$,
${r}_{2} = 5$
${\theta}_{2} = 53.13$ degrees
so, we get,
(−2−9i)(−3-4i)=sqrt85 e^(77.47i) x $5 {e}^{53.13 i} = 5 \sqrt{85} {e}^{130.6 i}$
:. (−2−9i)(−3-4i)=5sqrt85(cos130.6+isin130.6)
:. (−2−9i)(−3-4i)=5sqrt85(-0.651+0.759i)
:. (−2−9i)(−3-4i)=(-30+35i) |
# Stars and Bars
We discuss a combinatorial counting technique known as stars and bars. Consider the following problem.
### Problem: How many ordered sets of non-negative integers $(a, b, c, d)$ are there such that $a + b + c + d = 10$?
It may be tempting to list every possible combination of non-negative integers in an ordered fashion. However, this method of listing does not work easily for much larger numbers.
## Technique
Step 1: First, we will create a bijection between solutions to $a+b+c +d = 10$ with sequences of length 13 that consist of 10 $1$'s and 3 $0$'s. What this means is that we want to associate each solution with a unique sequence, and vice versa.
The construction is straightforward. Given a set of 4 integers $(a, b, c, d)$, we create the sequence that starts with $a$ $1$'s, then has a $0$, then has $b$ $1$'s, then has a $0$, then has $c$ $1$'s, then has a $0$, then has $d$ $1$'s. For example, if $(a, b, c, d) = (1, 4, 0, 2)$, then the associated sequence is $1 0 1 1 1 1 0 0 1 1$ . Now, if we add the restriction that $a + b + c + d = 10$, the associated sequence will consist of 10 $1$'s (from $a, b, c, d$) and 3 $0$'s (from our manual insert), hence have total length 13.
Conversely, given a sequence of length 13 that consists of 10 $1$'s and 3 $0$'s, set $a$ equal to the length of the initial string of $1$'s (before the first $0$), set $b$ equal to the length of the next string of 1's (between the first and second $0$), set $c$ equal to the length of the third string of $1$'s (between the second and third $0$), and set $d$ equal to the length of the last string of $1$'s (after the third $0$). Then, this yields a solution $a + b + c + d = 10$.
It is clear that the constructions associate a solution with a unique sequence, and vice versa. Hence we have a bijection.
Step 2: Now, it remains to count the number of solutions. Since we have a bijection with the sequences, it is equivalent to count the number of sequences of length 13 that consist of 10 $1$'s and 3 $0$'s.
First, let us assume that the $0$'s are distinct, say of the form $0_1, 0_2, 0_3$ . Then, in a sequence of length 13, there are 13 ways we could place $0_1$. After which, there are 13-1 ways we could place $0_2$. After which, there are 13-2 ways we could place $0_3$. Now, the rest of the sequence have to be $1$'s, and clearly there is only 1 way to do so. By the rule of product, this will yield $13 \times (13-1) \times (13-2) \times 1$ ways.
However, we have double counted many times, since the $0$'s are actually the same. There are 6 ways ( $0_1 0_2 0_3, 0_1 0_3 0_2, 0_2 0_1 0_3$, $0_2 0_3 0_1, 0_3 0_1 0_2, 0_3 0_2 0_1$ ) to arrange the 'distinct' $0$'s. Thus, each actual sequence would have been counted 6 times, so to get the actual number of ways, we will have to divide by 6.
Hence, there are $13 \times (13-1) \times (13 - 2 ) \times 1 \div 6 =286$ solutions.
Note: Students who are familiar with combinatorics should realize from the above argument that there are ${13 \choose 3} = 286$ solutions.
Note by Arron Kau
7 years, 4 months ago
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2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
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That's a really good note!!! We were taught the same thing, except we call it the "Beggar's Method" ....question states that in how many ways can 10 gold coins be distributed among 4 beggars....
- 7 years, 2 months ago
There is another way of tackling such a problem. a,b,c,d are non-negative integers. To ensure they are positive integers, adding 1 to each of them gives a+b+c+d=14. Now consider the 14 as a list of '1's. To get the number of ways of getting the ordered sets a,b,c,d we can partition the 14 by placing 3 lines in the gaps between the '1's. There are 13 gaps, so we have rephrased the problem - How many ways are there of choosing 3 gaps from 13 gaps. This yields 13 choose 3 = 286 (Apologies for the lack of LaTeX).
- 6 years, 7 months ago |
# Polynomial Operations Calculator
Instructions: This polynomial calculator will allow you to perform basic polynomial operations. Enter two polynomials and specify the operation you want to conduct among sum, subtraction or product, and the solver will show you step-by-step how to get the result. Type the polynomials like '3x^2 + 2x + 3'
Type the 1st polynomial $$p_1(x)$$ =
Type the 2nd polynomial $$p_2(x)$$ =
Operation ot perform =
Polynomials operations are operations that can be conducted among polynomials. Polynomials can be added, subtracted, multiplied and divided, regardless the order of the polynomial.
For example, we can add the polynomials $$p_1(x) = x + 3$$ and $$p_2(x) = 2x - 1$$ as follows
$p_1(x) + p_2(x)$ $= (x+3) + (2x - 1)$ $= x+3 + 2x - 1$ $= x + 2x + 3 - 1$ $= 3x + 2$
The procedure is simple: Just put the polynomials together and group by exponent and add the terms up. The same procedure is applied when adding polynomials of different order. For example, let us add $$p_1(x) = x^2+3$$ and $$p_2(x) = 2x - 1$$ as follows $p_1(x) + p_2(x)$ $= (x^2+3) + (2x - 1)$ $= x^2+3 + 2x - 1$ $= x^2 + 2x + 3 - 1$ $= x^2 + 2x + 2$
Almost exactly the same methodology is applied when we subtract polynomials, as indeed, subtracting $$p_2(x)$$ from $$p_1(x)$$ is the same as taking $$p_2(x)$$, multiplying each coefficient by $$-1$$ and then add this resulting polynomial to $$p_1(x)$$
For the multiplication of polynomials, things can get a bit messier because we need to cross multiply all the terms in one polynomials with the terms of all other polynomials. For example, let $$p_1(x) = x^2+3$$ and $$p_2(x) = 2x - 1$$, let's compute the multiplication
$p_1(x) \cdot p_2(x)$ $= (x^2+3) \cdot (2x - 1)$ $= (x^2)\cdot (2x)+ (x^2)\cdot (-1) + (3)\cdot (2x)+ (3)\cdot (-1)$ $= 2x^3 - x^2 + 6x - 13$
Aside from this polynomials calculator, you can choose among our selection of algebra calculators .
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Distributive Property: What will we be learning in this lesson? A while back, we talked about the commutative property of addition and we used this property to figure out how to add quickly.But the commutative property isn’t the only math property, which might lead you to wonder what great tricks the others have to offer. GCF and Distributive Property Problem Solving. 1. Tim and Moby know. Distributive property Let’s focus on the distributive property of multiplication The distributive property of multiplication states that when a number is multiplied by the sum of two numbers, the first number can be distributed to both of those numbers and multiplied by each of them separately, then adding the two products together for the same result as multiplying the first number by the sum. The distributive property is one of the most frequently used properties in math. The distributive property is a very deep math principle that helps make math work. But we can also apply the distributive property in the other direction, then calling out a common factor, and thus: Distributive property explained Normally when we see an expression like this …. The Distributive Property is easy to remember, if you recall that "multiplication distributes over addition". Practice the Distributive Property now. Algebra I teacher Carl Munn moves his students through a lesson on the distributive property. This is similar to how the distributive property works for multiplication. Print out these cards onto cardstock, ask a volunteer to cut out the cards and store them in zippered plastic bags, and you have a quick and easy game to help students practice identifying the distributive property. Distributive property explained Normally when we see an expression like this …. Here is another way to find $$5\cdot 79$$: $$\begin{array}{c}{5\cdot 79} \\ {5\cdot (80-1)} \\ {400-5} \\ {395}\end{array}$$ Glossary Entries. If we simplify the left side, we would add 5 + 3 first and get 4(8), which is 32. So check out the tutorial and let us know what you think! please help me by showing me how to do it step by step cuz i have already looked on the web for help and its too complicated for me so i would really appreciate it thanks =] if you could help me by showing me how to do either of these that would be great 1)9s(s+6) 2)(x+1)(x+4) Well, your wait is over. Do you remember how to multiply a fraction by a whole number? This can be done with subtraction as well, multiplying each number in the difference before subtracting. This math worksheet was created on 2015-02-22 and has been viewed 14 times this week and 236 times this month. If we simplify the right side, we would multiply 4(5) and get 20 and multiply 4(3) and get 12. a(b + c) = ab + ac. In general, this term refers to the distributive property of multiplication which states that the. i completely do not remember how to do it and now i have a quiz on it TOMORROW!!!!! Learn how to do distributive property to expand algebraic expressions. In mathematics, the distributive property of binary operations generalizes the distributive law from Boolean algebra and elementary algebra.In propositional logic, distribution refers to two valid rules of replacement.The rules allow one to reformulate conjunctions and disjunctions within logical proofs.. For example, in arithmetic: . Because today we’re talking about another one of those properties: the distributive property. Distributive Property; Well, the distributive property is that by which the multiplication of a number by a sum will give us the same as the sum of each of the sums multiplied by that number. 4 x 2 = 8 and 4 x 3 = 12. In math, distributive property says that the sum of two or more addends multiplied by a number gives you the same answer as distributing the multiplier, multiplying each addend separately, and adding the products together. Remember to put these in your notebook. Keep in mind that any letters used are variables that represent any real number. we just evaluate what’s in the parentheses first, then solve it:. So check out the tutorial and let us know what you think! You can get the worksheet used in this video for free by clicking on the link in the description below. The distributive property is a key mathematical property you’ll need to know to solve many algebra problems. In this example, 101 = 100 + 1, so: 17 101 = 17 (100 + 1) Split the problem into two easier problems. The distributive property says that when you multiply a factor by two addends, you can first multiply the factor with each addend, and then add the sum. It seems pretty easy to learn all of these skills in isolation, but using them together to solve one problem is … In this lesson you will learn how to use the distributive property and simplify expressions. This video is about how to do the distributive property. You can also multiply each addend first and then add the products. Then we use the distributive property to multiply the number 4 with both the 2 and the 3 inside the parentheses. $$6,000 \div 3$$ gives us the nice even quotient of 2,000. Solution: (3) (4 1/5) Rewrite 4 1/5 as 4 + 1/5 = (3) (4 + 1/5) apply distributive property . The Distributive Property, Examples and solutions, printable worksheets, use the distributive property to make calculating easier, how to use a diagram of a rectangle split into two smaller rectangles to write different expressions representing its area. All of the problems that we have done so far have not involved carrying remainders from one part to the next.. For instance, in the prior problem $$6837 \div 3$$, our divisor 3 evenly divided into the following parts. Welcome to The Multiply 3-Digit by 1-Digit Numbers Using the Distributive Property (A) Math Worksheet from the Long Multiplication Worksheets Page at Math-Drills.com. The distributive properties of addition and subtraction can be used to rewrite expressions for a variety of purposes. Using the Distributive Property With Fractions and Decimals. Distributive Property Definition. Vocabulary words are found in this magenta color throughout the lesson. Here's a picture of what that looks like: Tip: You can use the distributive property to solve tough multiplication problems where one of the factors is huge, even in your head! I need to buy snacks and soda for my party. The distributive property comes in all shapes and sizes, and can include fractions or decimals as well. Found in this magenta color throughout the lesson begins with note taking during which Munn... Addend first and get 4 ( 8 ), which is 32 the free examples Front... Be learning in this lesson distribution is needed i have a quiz on it TOMORROW!!. Sodas for $5 1 bag of chips for$ 5 the 3 inside the parentheses multiplying each addend and! Learn how to use the GCF in problem solving this … Mr. Munn 's focus is.. You will learn how to multiply the number 4 with both the 2 and the 3 inside the.... On 2015-02-22 and has been viewed 14 times this week and 236 this... Do it and now i have a quiz on it TOMORROW!!!..., one at a time how to do distributive property can be used to rewrite expressions a... See an expression like this … the lesson begins with note taking during which Mr. Munn 's focus assessment. Explained ( article ) | Khan Academy Save www.khanacademy.org this video is how! Of 2,000 Porch math to help teach this complex math topic s an example: 3 5... Property in problem solving 9: the distributive property in problem solving students only how to › to! Property lets you multiply a fraction by a sum by multiplying each separately! Problem in distributive form looks like: 4 x 2 = 8 4... To find the product and sizes, and multiply it by each number in the below! Of chips for $3 and 1 gallon sodas for$ 3 and 1 gallon sodas for $3 1! The link in the difference before subtracting not remember how to use distributive... Two examples any letters used are variables that represent any real number the left side we. All shapes and sizes, and can include fractions or decimals as well: how to divide using distribution magenta... Rather than teaching students only how to › how to do the distributive property explained article. Do not remember how to use in floral centerpieces for a party and needing food at the grocery.! Unit 6.6, lesson 9: the distributive property: the distributive property re talking about another one the... Party and needing food at the grocery store has 1 bag of chips for$ and... Make math work a sum by multiplying each number inside the parentheses, and multiply it by each number the! Principle that helps make math work link in the description below fractions or decimals as well now have! Math principle that helps make math work on Front Porch math to teach. Find: how to › how to use the distributive property: the distributive is. Addition or multiplication with subtraction probably know: multiply 17 101 using the distributive property or decimals well! Inside the parentheses, and can include fractions or decimals as well, multiplying each inside! And has been viewed 14 times this week and 236 times this week and 236 this. Property and simplify expressions of flowers and will contain only roses or only daisies have the same of.: the distributive property property lets you multiply a fraction by a sum by multiplying each number inside the,! To rewrite expressions for a party one of those properties: the distributive property is easy to remember if... Begins with note taking during which students use the distributive property \$ 6,000 3. All that helpful for most people: Madison has 56 roses and daisies. In problem solving remember, if you recall that multiplication distributes over addition '' the! Focuses on helping students understand why and when distribution is needed you remember how to the...
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# Pre-Algebra Lesson 12 Objectives
1. Mapping out what’s been done to a variable using order of operations 2. Using flow charts to solve equations 1. Observing that we have to use reverse order of operations to find the value of a variable
Materials and Handouts 1. Copy the arrows and boxes onto thicker paper and cut them out. It would be great if I could put packing tape or laminate over the boxes so we could write on them with white board markers.
Homework
5-10 Go over homework minutes 5 Warm-up/Review minutes 1. Have them write in all the values in the flow chart by evaluating each expression 2. Then have them write what’s being done to the expression each time by putting the operation over the arrow. Instruction 1. Have them work through each of the three flow chart examples. a. I do the first one with them, b. We do the second together c. They should be able to do the third on their own. 2. Explain that in the last problem, a new word popped up: equation. What do they think it means? a. Equation means: An expression squirted out a specific number when something was plugged in originally, so we write it as “expression”=”the number squirted out” b. Solving an equation means: Taking the number that was squirted out and undoing everything the expression did to find the original number that was plugged in. 3. Play a game using the manipulatives on the next page. a. They make an expression using the arrows and boxes on the next page. b. I roll a die (don’t let them see it) an plug it into the expression and tell them the result c. They have to figure out what I rolled. If they figure it out, they get a point. If they can’t figure it out, I get a point. d. Repeat with a new expression. 4. Then go through the “real-world” Natalia example with them. Make sure they’re writing the expression, then drawing the flow chart, then writing the equation, then going backwards to solve for the variable.
Pre-Algebra Lesson 12
Name:_______________ Date:_________
Intro to Solving Equations Class Work
Exercise 1: Let’s review what we did in the last lesson by evaluating some expressions. a. Fill out the flow chart if a2-3 a2-3 2 a2 2a2 2a+1 3
2(a2-3)
a
2a
2a+1
3(a+2)-1
3(a+2)
a+2
3(2a+1)
b. Next to each arrow, show what’s being done from one box to get to the next box. For example, to go from the middle box, to the right, above the arrow. Exercise 2: These flow chart things are pretty cool right? They show what has been done to a number or variable and they show the order of operations as well. a. Draw the flow chart that created the expression below. is being multiplied by so put a “ ”
b. If you plug in
for
what does the expression produce? Fill in the chart:
c. Suppose that, for some reason we know that the expression produced the number 33 and we want to go backwards to find the value of that produced this result. So we know that . Put 33 in the box on the far right and try to go backwards to find .
Exercise 3: Let’s try it again a. Draw the flow chart that created the expression
b. If you plug in
for
what does the expression produce? Fill in the chart:
c. Suppose that, for some reason we know that the expression produced the number 5 and we want to go backwards to find the value of that produced this result. So we know that . Put 5 in the box on the far right and try to go backwards to find .
Exercise 4: And again a. Draw the flow chart that created the expression
b. If you plug in
what does the expression produce? Fill in the chart:
c. Suppose that, for some reason we know that the expression produced the number 162 and we want to go backwards to find the value of that produced this result. So we know that . Put 162 in the box on the far right and try to go backwards to find .
Exercise 5: Try drawing the whole flow chart for yourself. a. Draw the flow chart that created the expression
b. If you plug in
what does the expression produce? Fill in the chart:
c. Solve the equation
Definition of Equation _____________________________________________________________ _____________________________________________________________ What it means to “solve” an equation ___________________________________________________________________ ___________________________________________________________________ Exercise 6: Let’s see if we can use this new knowledge to help us analyze some real life situations. a. Natalia brings in a bunch of cupcakes for the students in homework café. Each student had two cupcakes except for Jess who only had one. There were no cupcakes left over. If there were students, write an expression for how many cupcakes were consumed.
b. Make a flow chart of the expression
c. If there had been 6 students in homework café, how many cupcakes were eaten?
d. If 23 cupcakes were eaten, write an equation that represents this situation
e. Use your flow chart to find how many students there were if 23 cupcakes were eaten.
Pre-Algebra Lesson 12
Name:_______________ Date:_________
Intro to Solving Equations Homework
1. In the flow chart, one value has been written in with red. Fill in the rest of the flow chart. a2-3 2 a2 -3 a2 2a2
61
2a+1 3
2(a2-3)
a
2a
2a+1
3(a+2)-1
3(a+2)
a+2
3(2a+1)
2. Consider each flow chart: a. What equation is the person trying to solve? Use the variable
b. What’s the solution to their equation?
c. What equation is the person trying to solve? Use the variable
d. What’s the solution to their equation?
e. What equation is the person trying to solve? Use the variable
f. What’s the solution to their equation?
3. Draw a flow chart for each equation, then go backwards to solve the equation. a. c.
b.
d.
4. In a game, someone asked Ari to: Think of a number Subtract 1 from your number Multiply the result by 2 Add 6
a. Draw a flow chart to represent this game.
b. If Ari 10, what equation could you solve to figure out what number Ari originally thought of? Find the number.
5. Lizzy decides to keep a pet T-rex. The T-rex, so that he doesn’t hurt anyone, eats every day. Each steak costs . Lizzy goes out shopping and spends in addition to the money she spends on steaks.
steaks
on non steak items
a. Write an expression to represent how much money Lizzy spent when she went shopping. b. Draw a flow chart to represent this expression.
c. If Lizzy spent rex eats.
total, use your flow chart to determine how many steaks a day the T- |
Use adaptive quiz-based learning to study this topic faster and more effectively.
# Expanding expressions
Algebraic expressions can sometimes be complicated!
$$(3\Tred{x} - 2)(2\Tred{x}^2 + \Tred{x} - 1) + 2$$
The expansion of an algebraic expression removes all the brackets and collects like terms. It is then quicker to evaluate the expression.
• Remove the brackets one by one by to find the products: \begin{align*} (\Tred{a}+\Tblue{b})(c+d) &= \Tred{a}(c+d) + \Tblue{b}(c+d) \\ &= \Tred{a}c + \Tred{a}d + \Tblue{b}c + \Tblue{b}d \end{align*}
• Get the signs right: \begin{align*} \Tred{\mathbf{+}}\times \Tred{\mathbf{+}} = \Tred{\mathbf{+}} &\qquad \Tred{\mathbf{+}}\times \Tblue{\mathbf{-}} = \Torange{\mathbf{-}}\\ \Tblue{\mathbf{-}}\times \Tblue{\mathbf{-}} = \Tred{\mathbf{+}} &\qquad \Tblue{\mathbf{-}}\times \Tred{\mathbf{+}} = \Tblue{\mathbf{-}} \end{align*}
• Rewrite products of variables using powers: $$a\times a = a^2,\quad a^2\times a = a\times a^2 = a^3$$
• Collect all the like terms.
$$\Tblue{2}(y-1) - 1= \Tblue{2}y + \Tblue{2}(-1) - 1 = 2y - 2 - 1 = 2y -3$$ \begin{align*} 2(\Tred{t}+1)(\Tred{t}-2) - 1 &= (2\Tred{t}+2)(\Tred{t}-2) - 1 \\ &= 2\Tred{t}\times \Tred{t} + 2\Tred{t} - 4\Tred{t} - 4 -1\\ &= 2\Tviolet{t^2} + (2-4)\Tred{t} + (-4-1) \\ &= 2\Tviolet{t^2} -2\Tred{t} -5 \\ \end{align*} |
# What is the probability of rolling either a 7 or 11 from a pair of dice?
Contents
What about 7 OR 11? There are 6 x 6 or 36 options, all are equally likely, 7 occurs 6 times, so the chances are 6/36 or 1/6. 11 occurs 2 times so chances are 2/36 or 1/18. 7 or 11 are 8 of the 36 options so 8/36 or 2/9.
## What is the probability of rolling a 7 or 11 with a pair of dice?
What is the probability that the sum will be a 7 or 11? There are 36 possible outcomes for the two dice. So, the probability is 8/36 = 2/9.
## What is the probability of getting a total of either 7 or 11 in a single throw with two dice?
2 Answers. The probability is 25% .
## What is the probability of rolling a 5 or 7?
Two (6-sided) dice roll probability table
Roll a… Probability
4 3/36 (8.333%)
5 4/36 (11.111%)
6 5/36 (13.889%)
7 6/36 (16.667%)
## What is the probability of rolling a sum of 7?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
IMPORTANT: Do you get money back on ante post bets?
## How many ways can you make 7 on a dice?
A Note on Probability
For Example: If you want to know the probability of rolling a 7, you just divide the number of ways you can get a 7 (there are six ways) by the total number of possibilities (36). Six divided by 36 is the same as 1/6, which is also the same at 16.67%.
## What is the probability of getting at most the difference of 3?
1/6 chance for each side, 1/36 to roll any one of those combinations. Multiply that chance by 3, for the 3 combinations we can roll to give us a difference of 3, and we get 3/36, or an 8.
## How do we calculate probabilities?
How to calculate probability
1. Determine a single event with a single outcome.
2. Identify the total number of outcomes that can occur.
3. Divide the number of events by the number of possible outcomes. |
# 2016 UMO Problems/Problem 2
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Four fair six-sided dice are rolled. What is the probability that they can be divided into two pairs which sum to the same value? For example, a roll of $(1,4,6,3)$ can be divided into $(1,6)$ and $(4,3)$, each of which sum to $7$, but a roll of $(1,1,5,2)$ cannot be divided into two pairs that sum to the same value.
## Solution 1
We split this into cases based on the the form of the unordered values shown on the dice. In the following, $1 <= a,b,c,d <= 6$ are distinct numbers. After determining the (unordered) numbers that can appear on the dice, we must determine how many ways we can order those dice rolls in sequence.
Case1: The unordered numbers shown are {a,a,a,a} In this case, it is clear that the pairing (a,a) and (a,a) will yield equal sums, so we have 6 possible choices for a, and this is the number of possible rolls that take this form.
Case2: The unordered numbers shown are {a,a,a,b} In this case, one of the pairs will contain b, hence the pairs will be (a,b) and (a,a). But these can never be equal as a 6= b.
Case3: The unordered numbers shown are {a,a,b,b} To obtain equal sums, we must split the numbers into the pairs (a,b) and (a,b). There are $\binom{6}{2}= 15$ ways to select a and b, and there are $\binom{4}{2}= 6$ ways to order the rolls (select two of the rolls to be a’s). Hence there are 15·6 = 90 total rolls that take this form.
Case4: The unordered numbers shown are {a,a,b,c} In this case, either b pairs with a or b pairs with c. If b pairs with a, then $a + b = c + a$ , which is impossible as $b \ne c$. Therefore, b must pair with c, so $2a = b + c$. Therefore, b + c is even. We may assume without loss of generality that $b <= c$, hence the possibilities for (b,c) are (b,c) = (1,3),(1,5),(2,4),(2,6),(3,5),(4,6). Therefore, there are six ways to choose (b,c), and a is automatically determined (it is the average of b and c). Then there are 4! 2! = 12 ways to order the rolls, hence there are $6\cdot 12 = 72$ total rolls that take this form.
Case5: The unordered numbers shown are {a,b,c,d} In this case, the sum of the pairs must be representable in at two ways with distinct integers. We find 5 = 1 + 4 = 2 + 3; 6 = 1 + 5 = 2 + 4; 7 = 1 + 6 = 2 + 5 = 3 + 4; 8 = 2 + 6 = 3 + 5; 9 = 3 + 6 = 4 + 5. These are the only numbers that can be represented as the sum of two numbers in at least two ways using distinct numbers. In particular, for the numbers $5,6,8,9$,we know immediately what the four numbers a,b,c,d are. For 7,we must choose two of the three pairs, and we can do this in three ways. Therefore, the number of ways to choose {a,b,c,d} is $1+1+1+1+3 = 7$. Then we can order the rolls in 4! = 24 ways. Thus there are $7\cdot 24 = 168$ total rolls that take this form. Adding these together, we find $6 + 90 + 72 + 168 = 336$ possibilities, so the answer is $\frac{336}{64} = \frac{7}{27}$. |
Maths-
General
Easy
Question
Gopi wants to rent a tent for an outdoor celebration. The Cost of tent is Rs.500 per hour, plus an additional Rs.100 set-up fee. a) Draw a graph to show the relationship between the number of hours the tent is rented, x, and The total cost of the tent Y.b) What is the equation of the line in slope- intercept form?
Hint:
The correct answer is: y = 500x + 100
Step by step solution:The cost of tent per hour = Rs.500Set up fee = Rs.100For the first hour, the total cost = Rs.(100 + 500) = Rs.600In the second hour, the set up fee is still Rs.100, and the cost of tent = Rs(500*2) = Rs.1000The total cost for 2 hours = Rs.(1000 + 100)= Rs.1100Continuing this way,The total cost for 3 hours = Rs.(500 × 3 + 100) = Rs.1600The total cost for 4 hours = Rs.(500 × 4 + 100) = Rs.2100The total cost for 4 hours = Rs.(500 × 5 + 100) = Rs.2600Now, taking time (in hours) to be in x-axis and the total cost (in Rs.) to be y- axis, we draw the following table.Now, we plot these points on the graphThe line intercept form of a line is y = mx + c, where is the slope of the line and is the y-intercept of the line.First, we find the slope of the equation.For two points and satisfying the equation, the slope is given byTaking any two points from the table and denoting them as, we havem = m = = 500Next, for finding the y-intercept, we input any value from the table in the equation y = mx + cWe choose the first point (1,600), and using it in the above equation, we have600 = m × 1 + cPutting the value of m in the above relation, we have600 = 500 × 1 + cSimplifying, we have600 = 500 + cThus, we getc = 600 - 500 = 100So, the y-intercept c = 100,Using the value of m and c, we get the required equation,y = 500x + 100
Instead of using these particular points to find the slope and the y-intercept, we can use any other points from the table; we will get the same equation at the end. We can also verify this equation by inserting the points from the table and checking if the equation is satisfied. Finally, we can find the line in any other forms of a straight line. |
## Pages
### Solving equations using algebra tiles: with pictures!
There are so many cool ways to use algebra tiles in math. Last week I wrote a post about using them to factor. In this post I wanted to show 3 examples for using algebra tiles to solve equations.
New to algebra tiles?
I'm using a free set of paper algebra tiles (linked below), mainly so that I could cut one in half for example 2 that involves a fraction. So let's get into it!
Example 1:
Solve 2x + 3 = 11 with algebra tiles.
>>A paper version of the solving mat can be found for free here or
The rectangular tiles are used to represent x and the small squares are each used to represent 1.
In this photo, the 3 squares on the left got recycled along with 3 squares on the right, leaving 2x = 8.
Here's the fun part! We can split the 2x into x and x and make equal groups of the small squares on the other side.
After recycling one group, we get x = 4.
Using algebra tiles to solve 2-step equations is an incredible way to introduce the topic.
Example 2:
Solve ½x + 4 = 7 with algebra tiles.
It's nice to use paper algebra tiles for this example so that an x can be cut in half. (We still keep the other half for the end.)
First I cut an x in half to show the ½x.
Here we can see our equation set up.
We recycle the 4 squares on the left and 4 on the right, leaving ½x = 3.
Now we bring in that other ½x from earlier. Because we just doubled our x, we have to double the other side. This gives us x = 6.
Example 3:
Solve 2x - 3 = 7 with algebra tiles.
To get negative tiles with paper, you can print on 2-sided paper, like astrobrights, or glue two pieces of paper together before cutting.
Here we have 2 x tiles and 3 negative tiles on the left and 7 tiles on the right.
To get rid of that -3 on the left, we +3 to make a zero pair. Then we have to +3 to the right, too.
The zero pair gets recycled since it's 0. Now we have 2x = 10.
And now back to the fun part: splitting into equal groups.
And recycling the extra stuff to get x = 5.
Video:
Here is a video showing all 3 examples in this post:
UPDATE: I made a set of digital algebra tiles in Google Slides that are set up for solving equations. They are linked below:
Resources:
Algebra tiles can be found here.
Paper solving mat can be found here.
NEW! Digital algebra tiles for solving equations can be found here.
Solving equations activities can be found here.
If you'd like to learn more about ways to use algebra tiles, I have put together an algebra tiles tutorial video that covers ways to use algebra tiles in middle school math.
### Using Algebra Tiles in Middle School Math:
watch algebra tiles tutorial video w/ free algebra tiles
1. AnonymousJune 15, 2022
Looks like your "Digital Google Slides solving mat" link has been hacked. Everytime I click on it, it brings me to corrupt site.
1. Thank you so much for letting me know. That was a link to a friend's blog. I will unlink that text. I really appreciate the help.
2. UPDATE: I made a Google Slides set of algebra tiles specific to solving that is now linked in the post.
2. Hi There. I found the template for the solving equations algebra tiles mat (that you can laminate and wipe clean). Do you have any similar resources for factorising, expanding, combining like terms etc?
1. Hi Jemma, I do have a pack of them linked in my algebra tiles video tutorial post here: https://www.scaffoldedmath.com/p/algebra-tiles-tutorial.html
Comment |
# Simplify (5 square root of S)/(2 square root of S+ square root of 4)
5S2S+4
Simplify the denominator.
Rewrite 4 as 22.
5S2S+22
Pull terms out from under the radical, assuming positive real numbers.
5S2S+2
Factor 2 out of 2S+2.
Factor 2 out of 2S.
5S2(S)+2
Factor 2 out of 2.
5S2(S)+2(1)
Factor 2 out of 2(S)+2(1).
5S2(S+1)
5S2(S+1)
5S2(S+1)
Multiply 5S2(S+1) by S-1S-1.
5S2(S+1)⋅S-1S-1
Multiply 5S2(S+1) and S-1S-1.
5S(S-1)2(S+1)(S-1)
Reorder.
Move S+1.
5S(S-1)2((S+1)(S-1))
Expand the denominator using the FOIL method.
5S(S-1)2(S2+S⋅-1+S-1)
Simplify.
Use axn=axn to rewrite S as S12.
5S(S-1)2((S12)2-1)
Apply the power rule and multiply exponents, (am)n=amn.
5S(S-1)2(S12⋅2-1)
Combine 12 and 2.
5S(S-1)2(S22-1)
Cancel the common factor of 2.
Cancel the common factor.
5S(S-1)2(S22-1)
Divide 1 by 1.
5S(S-1)2(S1-1)
5S(S-1)2(S1-1)
Simplify.
5S(S-1)2(S-1)
5S(S-1)2(S-1)
Group S-1 and S together.
5((S-1)S)2(S-1)
5((S-1)S)2(S-1)
Apply the distributive property.
5(SS-1S)2(S-1)
Multiply SS.
Raise S to the power of 1.
5(S1S-1S)2(S-1)
Raise S to the power of 1.
5(S1S1-1S)2(S-1)
Use the power rule aman=am+n to combine exponents.
5(S1+1-1S)2(S-1)
5(S2-1S)2(S-1)
5(S2-1S)2(S-1)
Rewrite -1S as -S.
5(S2-S)2(S-1)
Rewrite S2 as S.
Use axn=axn to rewrite S as S12.
5((S12)2-S)2(S-1)
Apply the power rule and multiply exponents, (am)n=amn.
5(S12⋅2-S)2(S-1)
Combine 12 and 2.
5(S22-S)2(S-1)
Cancel the common factor of 2.
Cancel the common factor.
5(S22-S)2(S-1)
Divide 1 by 1.
5(S1-S)2(S-1)
5(S1-S)2(S-1)
Simplify.
5(S-S)2(S-1)
5(S-S)2(S-1)
Simplify (5 square root of S)/(2 square root of S+ square root of 4)
### Solving MATH problems
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# Math Probability Coin Experiment
by TEN
1. In your own words, describe two main differences between classical and empirical probabilities.
The first difference between the two is that classical probability is a theoretical computation whereas empirical probability is computed based on experiment or observation.
The second difference is that classical probability assumes the occurrence of any possible event within the sample space is just a likely as any other, where empirical probability makes no assumptions regarding possible outcomes.
2. Coin Experiment
20 coins
Random Trials (coins were put in a small plastic bag and shaken around)
Results:
1st trial: 10 heads; 10 tails
2nd trial: 11 heads; 9 tails
3rd trial: 10 heads; 10 tails
4th trial: 13 heads; 7 tails
5th trial: 12 heads; 8 tails
6th trial: 11 heads; 9 tails
7th trial: 8 heads; 12 tails
8th trial: 12 heads; 8 tails
9th trial: 10 heads; 10 tails
10th trial: 10 heads; 10 tails
P(E) = (number of times a specific event is observed) ÷ (total number of events observed)
Number of times a specific event is observed = number of heads observed, or number of tails observed
Total number of events observed = the total of heads and tails observed (this would also be equal to the total number of coins times the number of trials: 20 coins*10 trials = 200.)
Consider just your first count of the tossed coins. What is the observed probability of tossing a head? Of tossing a tail? Show the formula you used and reduce the answer to lowest terms.
First Count of the tossed coins:
Observed probability of tossing a head? P(E) = 10 heads/20 coins = ½
Observed probability of tossing a tail? P(E) = 10 tails/20 coins = ½
Did any of your ten repetitions come out to have exactly the same number of heads and tails? How many times did this happen?
Exactly the same number of heads and tails occurred in 4 trials (out of a total of 10 trials)
How come the answers to the step above are not exactly ½ and ½?
The two outcomes of a typical coin flip are not equally likely because of a bias. The fact that the coins were put in a container (or bag) and mixed up by shaking the container probably did remove the human bias.
However, the coins are inherently biased because the weight is not evenly distributed within the coin.
The coins probably end up with the heavier side down more often than not. Apparently, this means that heads-up appears more frequently.
What kind of probability are you using in this “bag of coins” experiment?
Empirical probability
Compute the average number of heads from the ten trials (add up the number of heads and divide it by 10).
Average Number of Heads = (10 + 11 + 10 + 13 + 12 + 11 + 8 + 12 + 10 + 10)heads/10 = 107/10 = 10.7
Change this to the average probability of tossing heads by putting the average number of heads in a fraction over the number of coins you used in your tosses.
Average probability of tossing heads = 10.7/20 = 0.535 = 53.5%
Did anything surprising or unexpected happen in your results for this experiment?
Yes. There was a natural bias in the results that was quite obvious.
3. Write the sample space for the outcomes of tossing three coins using H for heads and T for tails.
There are 8 possible outcomes:
H,H,H
H,H,T
H,T,H
H,T,T
T,H,H
T,H,T
T,T,H
T,T,T
What is the probability for each of the outcomes?
probability for each of the outcomes = 1/8
Which kind of probability are we using here?
Classical probability (theoretical computation).
How come we do not need to have three actual coins to compute the probabilities for these outcomes?
We do not need to have three actual coins because we assume the probability of each possible outcome is the same. |
# Excercise 2.5 Fractions and Decimals - NCERT Solutions Class 7
## Chapter 2 Ex.2.5 Question 1
What is greater?
i) $$0.5 \rm \,or\,0.05$$
ii) $$0.7 \rm \,or\,0.5$$
iii) $$7 \rm \,or\,0.7$$
iv) $$1.37 \rm \,or\,1.49$$
v) $$2.03 \rm \,or\,2.30$$
vi) $$0.8 \rm \,or\,0.88$$
### Solution
What is known?
Decimal numbers
What is unknown?
Which decimal number is greater.
Reasoning:
First convert these decimals into fractions then convert them into like fraction, now we can simply find out which fraction/decimal is greater.
Steps:
(i) $$0.5$$ or $$0.05$$
\begin{align}0.5 \quad&\boxed{\;\;}\quad 0.05\\\\\frac{5}{10} \quad &\boxed{\;\;} \quad \frac{5}{100}\end{align}
Converting them into like fractions, we get
\begin{align}\frac{5\times 10}{10\times 10} &\quad\boxed{\;\;}\quad \frac{5\times 1}{100\times 1}\\\\\frac{50}{100} &\quad\boxed{\;\;}\quad \frac{50}{100}\\\\\frac{50}{100} &\quad\boxed{\gt}\quad \frac{5}{100}\end{align}
Therefore$$,0.5 > 0.05.$$
ii) $$0.7$$ or $$0.5$$
\begin{align}~0.7\, \quad\boxed{\;\;}\quad 0.5\\\\\frac{7}{10} \quad\boxed{\;\;}\quad \frac{5}{10} \\\\ \frac{7}{10} \quad\boxed{\gt}\quad \frac{5}{10}\end{align}
Therefore,$$0.7$$ $$>$$ $$0.5$$.
iii) $$7$$ or $$0.7$$
\begin{align}\text{7} \quad\boxed{\;\;}\quad \frac{7}{10}\\\\=\frac{7\times 10}{1\times 10} \quad\boxed{\;\;}\quad \frac{7}{10}\\\\\frac{70}{10} \quad\boxed{\gt}\quad \frac{7}{10}\end{align}
Therefore, $$7 > 0.7.$$
$$7$$ is greater.
iv) $$1.37$$ or $$1.49$$
\begin{align}&=1\text{.37} \quad\boxed{\;\;}\quad \text{1}\text{.49}\\\\&=\frac{137}{100} \quad\boxed{\;\;}\quad \frac{149}{100}\\\\&=\frac{137}{100} \quad\boxed{\lt}\quad \frac{149}{100}\end{align}
Therefore, $$1.37$$ $$<$$$$1.49$$
$$1.49$$ is greater.
v) $$2.03$$ or $$2.30$$
\begin{align}\text{2}\text{.03} \quad\boxed{\;\;}\quad \text{2}\text{.30} \\\\\frac{203}{100} \quad\boxed{\;\;}\quad \frac{230}{100}\\\\\frac{203}{100} \quad\boxed{\lt}\quad \frac{230}{100}\end{align}
Therefore, $$2.03$$ $$<$$ $$2.30$$
$$2.30$$ is greater.
vi) $$0.8$$ or $$0.88$$
\begin{align}\frac{08}{10} \quad\boxed{\;\;}\quad \frac{088}{100}\end{align}
Converting them into like fractions, we get
\begin{align}\frac{8\times 10}{10\times 10} \quad\boxed{\;\;}\quad \frac{88}{100}\\\\\frac{80}{100} \quad\boxed{\lt}\quad \frac{88}{100} \end{align}
Therefore, $$0.8 < 0.88.$$
$$0.88$$ is greater.
## Chapter 2 Ex.2.5 Question 2
Express as rupees using decimals:
i) $$7$$ rupees $$7$$ paise
ii) $$7$$ rupees $$7$$ paise
iii) $$7$$ rupees $$7$$ paise
iv) $$7$$ rupees $$7$$ paise
v) $$7$$ rupees $$7$$ paise
### Solution
What is known?
Amount in paise.
What is unknown?
Amount in rupees.
Reasoning:
$$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align}rupees
Steps:
i) $$7$$ paise
$$100$$ paise $$=$$ $$1$$ rupees
$$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align} rupees
$$7$$ paise $$=$$ $$7$$ $$×$$\begin{align}\frac{1}{{100}}\end{align}rupees $$= 0.07$$ rupees
ii) $$7$$ rupees $$7$$ paise
$$100$$ paise $$=$$ $$1$$ rupees
$$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align}rupees
$$7$$ rupees $$7$$ paise $$=$$ $$7$$ rupees $$+$$ $$7$$ paise
$$=$$ $$7$$ rupees $$+$$ $$7$$ $$×$$\begin{align}\frac{1}{{100}}\end{align}rupees
$$=$$ $$7$$ rupees $$+$$ $$0.07$$ rupees
$$=$$ $$7.07$$ rupees
iii) $$77$$ rupees $$77$$ paise
$$77$$ rupees $$77$$ paise $$=$$ $$77$$ rupees $$+$$ $$77$$ paise
$$=$$ $$77$$ rupees + $$77$$ $$×$$ \begin{align}\frac{1}{{100}}\end{align}rupees
$$\because$$$$100$$ paise $$=$$ $$1$$ rupees)
($$\therefore$$ $$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align}rupees)
$$=$$ $$77$$ rupees $$+$$ $$0.77$$ rupees
$$=$$ $$77.77$$ rupees
iv) $$50$$ paise
$$\therefore$$ $$100$$ paise $$=$$ $$1$$ rupees
$$\therefore$$ $$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align} rupees
$$50$$ paise $$=$$ $$50$$ $$×$$\begin{align}\frac{1}{{100}}\end{align}rupees
=> $$50$$ paise $$=$$ \begin{align}\frac{{50}}{{100}}\end{align}rupees
$$50$$ paise $$=$$ $$0.50$$ rupees
v) $$235$$ paise
$$\therefore$$ $$100$$ paise $$=$$ $$1$$ rupees
$$\therefore$$ $$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align} rupees
$$235$$ paise $$=$$ $$235$$ $$×$$ \begin{align}\frac{1}{{100}}\end{align} rupees
$$235$$ paise $$=$$ \begin{align}\frac{{235}}{{100}}\end{align}rupees
$$235$$paise $$=$$ $$2.35$$ rupees
## Chapter 2 Ex.2.5 Question 3
i) Express $$5\,\rm{cm}$$ in meter and kilometer
ii) Express $$35\,\rm{mm}$$ in $$\,\rm{cm}$$,$$\,\rm{m}$$ and $$\,\rm{km}$$.
### Solution
What is known?
Lengths in centimeter and millimeter
What is unknown?
Length in different units.
Reasoning:
\begin{align}1\, \rm{cm} =\frac{1}{{100}}\rm{m}\end{align}
\begin{align}1\rm{m} = \frac{1}{{1000}}\,\rm{km}\end{align}
\begin{align}1\rm{mm} = \frac{1}{{10}}\rm{cm}\end{align}
Steps:
(i) $$5\,\rm{cm}$$
$$\because100\,\rm{cm}$$ $$=$$$$1\,\rm{m}$$
$$\therefore1\,\rm{cm}$$ $$=$$\begin{align}\frac{1}{{100}}\end{align}m
$$5\,\rm{cm}$$ $$=$$ $$5$$ $$×$$ \begin{align}\frac{{1}}{100}\rm{m}\end{align}
$$5\,\rm{cm}$$ $$=$$ $$0.05\,\rm{m}$$
Also, $$\because1000\,\rm{m}$$ $$=$$ $$1\,\rm{km}$$
$$\because1\,\rm{m}$$ $$=$$ \begin{align}\frac{1}{{1000}}\end{align}$$\rm{km}$$
Thus, $$0.05\,\rm{m}$$ $$=$$ $$0.05$$ $$×$$ \begin{align}\frac{1}{{1000}}\end{align}$$\rm{km}$$
\begin{align} & = \frac{{0.05}}{{1000}}{\text{km}} \\ & = 0.00005{\text{km}}\end{align}
(ii) Express $$35\,\rm{mm}$$ in $$\rm{cm}$$,$$\rm{m}$$ and $$\rm{km}$$.
$$\because10\,\rm{mm}$$ $$=$$ $$1\,\rm{cm}$$
$$\because1\,\rm{mm}$$ $$=$$ \begin{align}\frac{1}{{10}}\end{align} $$\rm{cm}$$
\begin{align}35{\text{mm}} &= 35 \times \frac{1}{{10}}{\text{cm}} \\ 35{\text{mm}} &= \frac{{35}}{{10}}{\text{cm}} \\ 35{\text{mm}} &= 3.5{\text{cm}} \end{align}
Now, converting $$3.5\,\rm{cm}$$ in $$\rm{m}$$, we get
$$\because100\,\rm{cm} = 1 \,\rm{m}$$
\because 1\,\rm{cm} = \begin{align}\frac{1}{{100}}\end{align}\rm{m}
3.5\,\rm{cm} = 3.5 × \begin{align}\frac{1}{{100}}\end{align}\,\rm{m}
\begin{align} 3.5{\text{cm}} &= \frac{{3.5}}{{100}}{\text{m}} \\ 3.5{\text{cm}} &= 0.035{\text{m}} \end{align}
Again, converting $$0.035\,\rm{m}$$ into $$\rm{km}$$
$$\because1000\,\rm{m}$$ $$=$$ $$1\,\rm{km}$$
\because 1\,\rm{m} = \begin{align}\frac{1}{{1000}}\end{align}\,\rm{km}
0.035\,\rm{m} = 0.035 × \begin{align}\frac{1}{{1000}}\end{align}\,\rm{km}
0.035 \,\rm{m} = \begin{align}\frac{{0.035}}{{1000}}\end{align}\,\rm{km}
$$0.035\,\rm{m} = 0.000035\,\rm{km}$$
## Chapter 2 Ex.2.5 Question 4
Express in $$\rm{kg}$$:
i) $$200$$ kg
ii) $$3470$$ g
iii) $$4\,\rm{kg} \ 8\,\rm{g}$$
### Solution
What is known?
Weight in grams
What is unknown?
Weight in kilograms.
Reasoning:
\begin{align}1\rm{g} = \frac{1}{{1000}}\rm{kg}\end{align}
Steps:
(i) $$200\,\rm{g}$$
$$\because1000\,\rm{g} = 1\,\rm{kg}$$
\because 1\,\rm{g} = \begin{align}\frac{1}{{1000}}\end{align}\,\rm{kg}
So,
\begin{align}&200\,\rm{g} = 200 ×\frac{1}{{1000}}\,\rm{kg}\\&=\frac{{200}}{{1000}}\,\rm{kg}\\&= 0.200\,\rm{kg}\end{align}
or
\begin{align} &=\frac{{200}}{{1000}}\\\\&=\frac{2}{{10}}\\\\&= 0.2\,\rm{kg}\end{align}
(ii) $$3470\,\rm{g}$$
$$\because 1000\,\rm{g} =1\,\rm{kg}$$
\because1\,\rm{g} =\begin{align}\frac{1}{{1000}}\end{align}\,\rm{kg}
\begin{align}3470\,\rm{g}& = 3470 \times (\frac{1}{{1000}})\,\rm{kg}\\3470\,\rm{g} &= (\frac{{3470}}{{1000}})\,\rm{kg}\\3470{\,\rm{g}} &= \frac{{347}}{{100}} \\ 3470{\,\rm{g}} &= 3.47{\,\rm{kg}} \end{align}
(iii) $$4\,\rm{kg} \ 8\,\rm{g}$$
\begin{align}4\,\rm{kg} \ 8\,\rm{g} &= 4\,\rm{kg} + 8\,\rm{g}\\4\,\rm{kg} \ 8 \,\rm{g} &=4\,\rm{kg} + \frac{8}{{1000}}\,\rm{kg} \end{align}
$$\because 1000\,\rm{g}= 1\,\rm{kg}$$
\because 1\,\rm{g} = \begin{align}\frac{1}{{1000}}\end{align}\,\rm{kg}
\begin{align}4 \,\rm{kg} \ 8\,\rm{g} &= 4 \,\rm{kg} + 0.008 \,\rm{kg}\\4\,\rm{kg} \ 8\,\rm{kg}&=4.008\,\rm{kg}\end{align}
## Chapter 2 Ex.2.5 Question 5
Write the following decimal numbers in expanded form:-
i) $$20.03$$
ii) $$2.03$$
iii) $$200.03$$
iv) $$2.034$$
### Solution
What is known?
Decimal numbers
What is unknown?
Decimal numbers in expanded form.
Reasoning:
Hundreds Tens Ones Decimal Tenths Hundredths 100 10 1 --------- $\frac{1}{{10}}$ $\frac{1}{{100}}$
Steps:
i) $$20.03$$
\begin{align}20.03 = &2 \times 10 + 0 \times 1 +\\&0 \times\frac{1}{{10}} + 3 \times \frac{1}{{100}}\end{align}
ii) $$2.03$$
\begin{align}2.03 = &2 \times 1 + 0 \times\frac{1}{10} + \\&3 \times \frac{1}{100} \end{align}
iii) $$200.03$$
\begin{align} 200.03 = &2 \times 100 + 0 \times 10 + 0 \times 1 \\&+ 0 \times \frac{1}{10} + 3 \times \frac{1}{100} \end{align}
iv) $$2.034$$
\begin{align} 2.034 = &2 × 1 + 0 × \frac{1}{10} + \\& 3 \times \frac{1}{100} + 4 \times \frac{1}{1000} \end{align}
## Chapter 2 Ex.2.5 Question 6
Write the place value of $$2$$ in the following decimal numbers:-
i) $$2.56$$
ii) $$21.37$$
iii) $$10.25$$
iv) $$9.42$$
v) $$63.352$$
### Solution
What is known?
Decimal numbers
What is unknown?
Place value of $$2$$.
Reasoning:
Hundreds Tens Ones Decimal Tenths Hundredths 100 10 1 ---- $\frac{1}{{10}}$ $\frac{1}{{100}}$
Steps:
i) $$2.56$$
$$\underline{2}.56$$
$$2$$ is at ones place.
ii) $$21.37$$
$$\underline{2}1.37$$
$$2$$ is at ten's place.
iii) $$10.25$$
$$10. \underline{2} 5$$
$$2$$ is at tenths place.
iv) $$9 .4 2$$
$$9 .4 \underline{2}$$
$$2$$ is at hundredth place.
V) $$63.352$$
$$63 .3 5\underline{2}$$
$$2$$ is at thousandth place.
## Chapter 2 Ex.2.5 Question 7
Dinesh went from place $$A$$ to place $$B$$ and from there to place $$C$$. $$A$$ is $$7.5\,\rm{km}$$ from $$B$$ and $$B$$ is $$12.7\,\rm{km}$$ from $$C$$. Ayub went from place $$A$$ to place $$D$$ and from there to place $$C$$. $$D$$ is $$9.3\,\rm{km}$$ from $$A$$ and $$C$$ is $$11.8\,\rm{km}$$ from $$D$$.Who travelled more and how much ?
### Solution
What is known?
Distance between points. Dinesh went from place $$A$$ to $$B$$ and then $$B$$ to $$C$$. Ayub went from place $$A$$ to $$D$$ then $$D$$ to $$C$$.
What is unknown?
Who travelled more and how much.
Reasoning:
Calculate by all the distance travelled by Dinesh to how much he travelled then calculate distance travel by Ayub.
Steps:
Given
Distance travelled by Dinesh from $$A$$ to $$B = 7. 5\,\rm{km}$$
And from place $$B$$ to place $$C = 12.7 \,\rm{km}$$
$$\therefore$$ Total distance travelled by Dinesh
\begin{align} &= AB+ BC\\&= 7.5\,\rm{km} + 12.7\,\rm{km}\\&= 20.2 \,\rm{km}\end{align}
Distance travelled by Ayub from place $$A$$ to place $$D = 9.3 \,\rm{km}$$
And from place $$D$$ to place $$C = 11.8\,\rm{km}$$
$$\therefore$$ Total distance travelled by Ayub
\begin{align}&= 9.3\,\rm{km} + 11.8 \,\rm{km}\\&= 21.1 \,\rm{km}\end{align}
On comparing the total distance travelled by Dinesh and Ayub, we get
$$21.1\,\rm{km}$$$$>$$ $$20.2\,\rm{km}$$
Distance travelled by Ayub > Distance travelled by Dinesh
$$\therefore$$ Ayub covered more distance by Dinesh is
\begin{align}&= 21.1 -20.2\\&= 0.9 \,\rm{km}\\&= 0.9 × 1000 \,\rm{m}\\&= 900\,\rm{m}\end{align}
## Chapter 2 Ex.2.5 Question 8
Shyama bought $$5\,\rm{kg}\,300\,\rm{g}$$. apples and $$3\,\rm{kg}\ 250\,\rm{g}$$ mangoes. Sarala bought $$4\,\rm{kg}\ 800\,\rm{g}$$ oranges and $$4\,\rm{kg}\ 150\,\rm{g}$$ bananas. Who bought more fruits?
### Solution
What is known?
Fruits bought by Shyama and Sarala.
What is unknown?
Who bought more fruits.
Reasoning:
Find out total fruits purchased by both Shyama and Sarala then we easily find out who bought more.
Steps:
Weight of apples bought by Shyama $$=5\,\rm{kg}\ 300\,\rm{g}$$
Weight of mangoes bought by Shyama $$=3\,\rm{kg}\ 250\,\rm{g}$$
$$\therefore$$ Total weight of fruits bought by Shyama
\begin{align}&= 5 \,\rm{kg}\ 300 \,\rm{g} + 3 \,\rm{kg}\ 250 \,\rm{g} \\&= 8 \,\rm{kg}\ 550 \,\rm{g}\end{align}
Also,
Weight of oranges bought by Sarala $$=4\,\rm{kg}\ 800\,\rm{g}$$
Weight of oranges bought by Sarala $$=4\,\rm{kg}\ 150\,\rm{g}$$
$$\therefore$$ Total weight of fruits bought by Sarala
\begin{align}&= 4\,\rm{kg}\ 800 \,\rm{g}\ + 4 \,\rm{kg}\ 150 \,\rm{g}\ \\&= 8\,\rm{kg}\ 950\,\rm{kg}\end{align}
On comparing the quantity of fruits, we get
$$8\,\rm{kg}\ 950\,\rm{g} > 8 \,\rm{kg}\ 550 \,\rm{kg}$$
Thus, Sarala bought more fruits.
## Chapter 2 Ex.2.5 Question 9
How much less is $$28\,\rm{km}$$ than $$42.6\,\rm{km}$$?
### Solution
What is known?
Two numbers.
What is unknown?
Difference between these two numbers.
Reasoning:
We can simply calculate it by subtracting smaller number from bigger number.
Steps:
Here,
We have to find out the difference between $$28\,\rm{km}$$ and $$42.6\,\rm{km}$$
\begin{align}\therefore\text{Difference}&= 42.6\ – 28\\&= 14. 6 \,\rm{km}\end{align}
Thus, $$14.6\,\rm{km}$$ less is $$28\,\rm{km}$$ than $$42.6$$.
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# Mathnasium #MathTricks: Probability (Part 1)
Sep 28, 2022
Welcome to Mathnasium’s Math Tricks series. Today we are calculating basic probability.
Probability is the chance or likelihood that something will happen. To calculate the basic probability of something happening, we make a fraction of the number of ways an event can happen (the “favorable” outcomes) out of the total number of “possible” outcomes. Probability can also be represented as a decimal or percent.
The likelihood that something will happen falls between 0 or 0%, (impossible events), and 1 or 100%, (certain events). All other probabilities fall in between 0 and 1.
Follow the examples below to find the basic probabilities of the events.
##### Example 1: A bag contains 3 red, 4 white, and 6 blue marbles. If 1 marble is pulled from the bag at random, what is the probability that the marble is red OR white?
Step 1: Find the total number of possible outcomes.
There is a total of: 3 red marbles + 4 white marbles + 6 blue marbles = 13 marbles.
Step 2: Find the total number of favorable outcomes.
The favorable outcome is pulling a red OR white marble from the bag at random.
The total number of favorable outcomes is: 3 red + 4 white marbles = 7 marbles.
##### Example 2: If a blue marble was pulled from the same bag and not replaced, what is the probability of drawing another blue marble?
Step 1: Find the total number of possible outcomes.
Since a blue marble was pulled from the same bag, there are now: 13 - 1 = 12 marbles.
Step 2: Find the total number of favorable outcomes.
The favorable outcome is drawing another blue marble.
The total number of favorable outcomes is: 6 - 1 = 5 blue marbles.
If you missed this, or any of our other Math Tricks videos, check them out on our YouTube channel!
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# How do you simplify -3sqrt3(2+sqrt6)?
May 22, 2017
$- 6 \sqrt{3} - 9 \sqrt{2}$
#### Explanation:
First, distribute the $- 3 \sqrt{3}$ to both terms inside parentheses.
$- 3 \sqrt{3} \cdot 2 - 3 \sqrt{3} \cdot \sqrt{6}$
Now simplify.
$- 6 \sqrt{3} - 3 \sqrt{18}$
18 is $2 \times 3 \times 3$.
$- 6 \sqrt{3} - 3 \sqrt{2 \times 3 \times 3}$
You have two factors of 3 inside the radical, so you can pull them out and make one factor of 3 outside the radical.
-6sqrt3-3sqrt(2timescolor(limegreen)3timescolor(limegreen)3
$- 6 \sqrt{3} - 3 \cdot \textcolor{\lim e g r e e n}{3} \sqrt{2}$
Finally, simplify again.
$- 6 \sqrt{3} - 9 \sqrt{2}$
May 22, 2017
$- 6 \sqrt{3} - 9 \sqrt{2}$
#### Explanation:
Given -
$- 3 \sqrt{3} \left(2 + \sqrt{6}\right)$
$- 6 \sqrt{3} - 3 \sqrt{3} \sqrt{6}$
$- 6 \sqrt{3} - 3 \sqrt{3} \sqrt{3} \sqrt{2}$
$- 6 \sqrt{3} - 3.3 \sqrt{2}$
$- 6 \sqrt{3} - 9 \sqrt{2}$ |
# inequalities
• Jan 31st 2011, 09:16 PM
jam2011
inequalities
Please check my work.thank you so much
• Jan 31st 2011, 09:27 PM
Prove It
Quadratic Inequalities are always easiest to solve if you complete the square.
$\displaystyle \displaystyle x^2 - 2x - 3 \leq 0$
$\displaystyle \displaystyle x^2 - 2x + (-1)^2 - (-1)^2 - 3 \leq 0$
$\displaystyle \displaystyle (x - 1)^2 - 4 \leq 0$
$\displaystyle \displaystyle (x - 1)^2 \leq 4$
$\displaystyle \displaystyle \sqrt{(x-1)^2} \leq \sqrt{4}$
$\displaystyle \displaystyle |x-1| \leq 2$
$\displaystyle \displaystyle -2 \leq x - 1 \leq 2$
$\displaystyle \displaystyle -1 \leq x \leq 3$.
• Feb 1st 2011, 10:47 AM
HallsofIvy
Another good way to solve quadratic inequalities (and other complicated inequalities) is to solve the associated equation.
For example, to solve $\displaystyle \displaystyle x^2 - 2x - 3 \leq 0$, first solve $\displaystyle \displaystyle x^2 - 2x - 3= 0$, by factoring, $\displaystyle (x- 3)(x+ 1)= 0$ so x= 3 and x= -1. The point is this: a polynomial is a "continuous" function which means that it in order to go from positive to negative, it must pass through 0 and that can happen only at x= -1 and x= 3 which divide the real line into three intervals.
To decide which interval is ">0" and which "< 0" you can use either of two methods:
1) Choose a number in each interval and check the sign. x= -2 is less than -1 and $\displaystyle (-2)^2- 2(-2)- 3= 4+ 4- 3= 5> 0$ so the polynomial is greater than 0 for all x< -2. x= 0 is between -1 and 3 and $\displaystyle (0)^2- 2(0)- 3=-3< 0$ so the polynomial is less than 0 for all x between -1 and 3. Finally, x= 4 is larger than 3 and $\displaystyle (4)^2- 2(4)- 3= 5> 0$ so the polynomial is greater than 0 for all x greater than 3. Since the inequality say "$\displaystyle \le 0$", the solution set is $\displaystyle -1\le x\le 3$.
2) Notice that x+ 1 is negative for x< -1 and positive for x> -1 and that x- 3 is negative for x< 3 and positive for x> 3. If x< -1, it is also less than 3 so both factors are negative and their product is positive. If -1< x< 3, one factor, x+1, is positive but x- 3 is still negative. The product of one positive and one negative factor is negative. Finally, if x> 3, both factors are positive and their product is positive. That gives the same solution set as before and (fortunately!) the same solution as Prove It.
• Feb 1st 2011, 12:38 PM
Quote:
Originally Posted by jam2011
Please check my work.thank you so much
$\displaystyle x^2-2x-3\ \le\ 0$
$\displaystyle (x-3)(x+1)\le\ 0$
If $\displaystyle x-3 \ge\ 0$, then $\displaystyle x+1=(x-3)+4$ must be $\displaystyle \ge\ 0$ since it is 4 greater than (x-3)
No luck there...
Case 2:
If $\displaystyle x-3\ \le\ 0$, then $\displaystyle x\ \le\ 3$ and $\displaystyle x+1\ \ge\ 0$ only for $\displaystyle x\ \ge\ -1$
Hence $\displaystyle -1\ \le\ x\ \le\ 3$
so your answer is correct and the diagrams help illustrate the intersections.
Yet another way to proceed is....
$\displaystyle x^2-2x\ \le\ 3\Rightarrow\ x(x-2)\ \le\ 3$
The factors of 3 that differ by 2 are 1 and 3 or $\displaystyle -1$ and $\displaystyle -3$
$\displaystyle x(x-2)=3$ for $\displaystyle x=3$ and for $\displaystyle x=-1$
If $\displaystyle x<-1$, then $\displaystyle x(x-2)>3$
If $\displaystyle x>3$, then $\displaystyle x(x-2)>3$
The solution is $\displaystyle -1\ \le\ x\ \le\ 3$
• Feb 1st 2011, 02:03 PM
skeeter
$\displaystyle x^2 - 2x - 3 \le 0$
$\displaystyle (x + 1)(x - 3) \le 0$
think about the graph of the quadratic ... a parabola that opens upward with two zeros, $\displaystyle x = -1$ and $\displaystyle x = 3$
the section of the graph below the x-axis (the less than or equal to section) lies between and includes the two zeros.
• Feb 1st 2011, 02:08 PM
jam2011
nice job !!! thank you so much...i learn a lot.there are many ways to find and solve this problem. i owe you a lot... |
# How to Solve Parallel Circuits
Solving parallel circuits is an easy process once you know the basic formulas and principles. When two or more resistors are connected side by side the current can "choose" it's path (in much the same way as cars tend to change lanes and drive alongside one another when a one-lane road splits into two parallel lanes).[1] After reading these steps you should be able to find the voltage, current and resistance between two or more resistors in parallel.
## Cheat Sheet
• Total resistance RT for resistors in parallel: 1/RT = 1/R1 + 1/R2 + 1/R3 + ...
• Voltage is always the same across branches: VT = V1 = V2 = V3 = ...
• Total current IT = I1 + I2 + I3 + ...
• Ohm's Law: V = IR
Part 1
Part 1 of 3:
### Introduction to Parallel Circuits
1. 1
Identify parallel circuits. A parallel circuit has two or more branches that all lead from point A to point B. A single stream of electrons divides to flow through multiple branches, then merge back into one stream on the other side. Most problems involving parallel circuits will ask you to identify the total voltage, resistance, or current across the circuit (point A to point B).[2]
• Components "connected in parallel" are each located on a separate branch.
2. 2
Understand current and resistance in parallel circuits. Imagine a freeway with multiple lanes, and toll booths in each lane slowing down traffic. Building a new lane gives the cars another path to take, so it will always speed up traffic even though you're adding a new toll booth as well. Similarly, adding a new branch to a parallel circuit gives current an additional path to take. No matter how much resistance that new branch has, the total resistance of the circuit will decrease, and total current of the circuit will increase.[3]
3. 3
Sum currents in each branch to find total current. If you know the current in each branch, just add them together to find the total current. This is the amount of current flowing in the circuit after all the branches come together. In formulaic terms: IT = I1 + I2 + I3 + ...[4]
4. 4
Solve for total resistance. To find total resistance RT across the circuit, solve for it in the equation 1/RT = 1/R1 + 1/R2 + 1/R3 + ... where each R on the right-hand side represents the resistance on one branch of the circuit.[5]
• For example, a circuit has two resistors in parallel, each with 4Ω resistance. 1/RT = 1/4Ω + 1/4Ω → 1/RT = 1/2Ω → RT = 2Ω. In other words, two branches of equal resistance are exactly twice as easy to get through as one branch alone.
• If one branch has no resistance (0Ω), all the current goes through that branch. The total resistance is 0.[6]
5. 5
Remember what voltage describes. Voltage is the difference in electric potential between two points. Since you're comparing two points, not examining a path of movement, the voltage will remain the same no matter which branch you're looking at.[7] VT = V1 = V2 = V3 = ...
6. 6
Find missing values with Ohm's Law. Ohm's Law describes the relation between voltage V, current I, and resistance R: V = IR. If you know two of these values, use this formula to solve for the third.[8]
• Make sure every value refers to the same portion of the circuit. You may use Ohm's Law to examine the total circuit (V = ITRT) or a single branch (V = I1R1).
Part 2
Part 2 of 3:
### Example Circuit
1. 1
Make a chart to keep track of your work. If you have a parallel circuit with several unknown values, a chart will help you organize your information.[9] Here's an example chart for a circuit with three parallel branches. Note that branches are often referred to as R followed by a subscript number.
R1 R2 R3 Total Units
V
volts
I
amperes
R
ohms
2. 2
Fill in all the information given by the problem. For our example, we'll use a circuit powered by a 12 volt battery. The circuit has three parallel branches, with resistances 2Ω, 4Ω, and 9Ω. Add this information to your chart:
R1 R2 R3 Total Units
V
12
volts
I
amperes
R
2
4
9
ohms
3. 3
Copy the voltage value to each branch. Remember that the voltage across the whole circuit equals the voltage across each branch of a parallel circuit.
R1 R2 R3 Total Units
V
12
12
12
12
volts
I
amperes
R
2
4
9
ohms
4. 4
Use Ohm's Law to find the current in each branch. Each column in your chart includes voltage, current, and resistance. This means you can always solve for a missing value as long as you have the other two values in the same column. If you need a reminder, Ohm's Law is V = IR. The missing value is current in our example, so we can rearrange this as I = V/R.[10]
R1 R2 R3 Total Units
V
12
12
12
12
volts
I
12/2 = 6
12/4 = 3
12/9 = ~1.33
amperes
R
2
4
9
ohms
5. 5
Solve for total current. This is an easy value to find, since the total current equals the sum of the currents in each branch.[11]
R1 R2 R3 Total Units
V
12
12
12
12
volts
I
6
3
1.33
6 + 3 + 1.33 = 10.33
amperes
R
2
4
9
ohms
6. 6
Solve for total resistance. You can find this in two different ways. You can use the resistance row to calculate it using the formula 1/RT = 1/R1 + 1/R2 + 1/R3. However, it's often easier to solve for it using Ohm's Law and the total V and I values. When solving for resistance, rearrange Ohm's Law as R = V/I
R1 R2 R3 Total Units
V
12
12
12
12
volts
I
6
3
1.33
10.33
amperes
R
2
4
9
12 / 10.33 = ~1.17
ohms
Part 3
Part 3 of 3:
1. 1
Calculate power. As in any circuit, power P = IV. If you've solved for power along each of the branches, the total power PT equals the sum of all branch power values (P1 + P2 + P3 + ...).[12]
2. 2
Find total resistance for a two-branch circuit. If there are exactly two resistors in parallel, you can simplify the equation to the "product over sum" equation:
• RT = R1R2 / (R1 + R2)
3. 3
Find total resistance when all resistors are identical. If every resistor in parallel has the same resistance value, the equation becomes much simpler. RT = R1 / N, where N is the number of resistors.[13]
• For example, two identical resistors in parallel provides ½ the total resistance of one resistor alone. Eight identical resistors provide ⅛ of the total resistance.
4. 4
Calculate branch currents without voltage. This equation, called Kirchhoff's current divider rule, lets you solve for individual branch currents even if you don't know the circuit voltage.[14] You'll need to know the resistance of each branch, and the total current f the circuit:
• Two resistors in parallel: I1 = ITR2 / (R1 + R2)
• More than two resistors in parallel: To solve for I1, find the combined resistance of all resistors besides R1. Remember to use the formula for resistors in parallel. Now use the equation about, replacing R2 with your answer.
## Community Q&A
Search
• Question
Three resistors A, B and C are connected in parallel and take a total of 7.9 ampere. Resistor A takes 2.5 ampere and has a resistance of 48 ohms. The current through B is twice as much as that of C. How can I calculate the currents of resistors B and C?
Subtract 2.5 from 7.9, and divide the result by 3. The result is the current of C. Multiply this by 2 to find the current of B.
• Question
How can I find the total power in the current voltage?
Find the total power in any circuit: pt=p1+p2+p3+,,,+pn. In terms of voltage and current, by using this formula: p=vi (or) p=i*i*r (or)p=v*v*r.
• Question
What happens in a parallel circuit?
In a parallel circuit, current gets divided among the parallel branches in a manner so that the product of current and the resistance of each branch becomes the same. The sum of the current in each branch is equal to the total current of the circuit.
200 characters left
## Tips
• If solving Series-Parallel circuits, solve the Parallel parts first. Then you are left with a much easier Series circuit.
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• You may have been taught Ohm's Law as E = IR or V = AR. These are just different notations, meaning the same thing.
⧼thumbs_response⧽
• In a Parallel circuit the same voltage is applied across all the resistors.
⧼thumbs_response⧽
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wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 35 people, some anonymous, worked to edit and improve it over time. This article has been viewed 552,102 times.
Co-authors: 35
Updated: October 27, 2022
Views: 552,102
Categories: Electromagnetism
Article SummaryX
To solve parallel circuits, you'll need to know that parallel circuits have two or more branches that all lead from point A to point B. If you want to solve for total current, use the equation IT = I1 + I2 + I3 where IT is the total current, and I1 through I3 are the currents in each branch. If you want to solve for total resistance, simply use the equation 1/RT = 1/R1 + 1/R2 + 1/R3, where "R" equals the right side. The voltage of a circuit is the same on all sides and you can use Ohm's law, or V = IR, to find missing values. To see an examples of solving these equations, scroll down.
Thanks to all authors for creating a page that has been read 552,102 times. |
Introduction to matrices for better understanding of data science
What are Matrices?
At a very high level, you can think of a Matrix (Plural : matrices) as two dimensional array of numbers, symbols or expressions. For example – following image depicts a MxN matrix, where it has M-rows and N-Columns. A specific element is being denoted by Ai,j, which means the element value at the ith row and jth column.
Important Terms
• Size of matrix: The size of a matrix is the number of columns and the rows in a given matrix (M x N).
• Row Vector: matrices with the single row are called row vectors (1 x N)
• Column Vector: matrices with single column are called column vectors (M x 1)
• Square Matrix: a matrix with the same number of rows and columns is called square matrix (N x N)
• Infinite Matrix: a matrix with infinite number of rows or columns is called infinite matrix
• Empty Matrix: a matrix with zero number of rows and columns is called an empty matrix
• Identity Matrix: the identity matrix is a N x N matrix with 1 at the diagonal and 0 else where
• Inverse Matrix: A matrix is an inverse matrix of a given matrix, if their multiplications results into identity matrix
Key Operations
Read the reference links to know the details about matrix operations. Here is the summary of the commonly used operations:
• two matrices of exactly same dimensions can be added or subtracted element by element
• Scalar Multiplication
• Multiplying a matrix by constant number (negative or positive) leads to multiplication of all the element by that specific number.
• Matrix Multiplications
• two matrices can be multiplied only when the number of columns in the first equals the number of rows in the second
• You can calculate square of a matrix only when they have same number of rows and columns (i.e. they are square matrix)
• Transposition
• The transpose of a matrix results into a MxN matrix becoming a NxM matrix. Here the rows become columns and the columns becomes rows
Usage of matrices
As you know that matrices are the tabular representation of data, essentially anything which requires two dimensional numerical data representation, can make use of the matrices.
Following is the list of more obvious usage of the matrices
• Demographic representation of data for a specific topic
• Projection of 3-dimensional images into two dimensional screens
• Website’s page ranking
• Plotting graphs
• In Seismic Surveys, to produce detailed images of local geology to determine the location and size of possible oil and gas reservoirs
• In robotics & automation in terms of base elements for the robot movements, including rotations and translations through planes
• It is used in cryptography to encrypt / decrypt messages
• Matrices allow complex dutching/betting combinations without separate formulae such as multiple complex simultaneous equations
• In the gaming, the uses of matrices allows effective algorithms around collision and ray tracing
• Markov Chain makes use of matrices extensively, which in turn is used to solve various real life problems including calculation of possibilities (as in trades or prediction of some events). |
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# Recent problems solved by 'mathick'
Linear-systems/69431: i have to solve the problems using system of equations.. 1.x+y=6 2. 2x-3y=-13 y=2x y=2x+7 1 solutions Answer 49527 by mathick(4) on 2007-02-06 00:01:16 (Show Source): You can put this solution on YOUR website!Let's start with #1: x + y = 6 y = 2x. There are a few ways to solve these, but the substitution method is often easiest. 1. Substitute 2x in for y in the 1st equation: x + (2x) = 6. (This can be done because the 2nd equation says y and 2x are interchangeable). 2. Simplify: 3x = 6. 3. Solve for x: x = 2. 4. Plug 2 in for x in the 1st equation: 2 + y = 6. 5. Solve for y: y = 4. 6. Final Answer: (2, 4). (x = 2, y = 4). That's the substitution method - the next example has a few more steps: 2x-3y=-13 y=2x+7 1. Substitute 2x + 7 in for y in the 1st equation: 2x - 3(2x + 7) = -13. (This can be done because the 2nd equation says y and 2x + 7 are interchangeable). 2. Simplify: -4x - 21 = -13. 3. Solve for x: x = -2. 4. Plug -2 in for x in the 1st equation: 2(-2) - 3y = -13. 5. Solve for y: y = 3. 6. Final Answer: (-2, 3). (x = -2, y = 3).
Equations/61781: When solving a rational equation, why it is OK to remove the denominator by multiplying both sides by the LCD and why can you not do the same operation when simplifying a rational expression?1 solutions Answer 42569 by mathick(4) on 2006-11-17 00:45:35 (Show Source): You can put this solution on YOUR website! Good question. The difference between expressions and equations is key to the answer. A simple example of an expression is: . And a simple example of an equation is: . If the equation were a scale, the left side and right side would balance each other perfectly. Now if the same weight (say 3) is added to both sides of the equation: , you get an equation that is equivalent to (basically the same as) the original equation . is basically the same equation as (they have the same answer). In an equation, the left and right side are balanced. The main idea in solving equations is: if you start with a balanced scale and then do the same thing to both sides of the scale, the scale will still end up balanced. With an expression, however there's no scale. An expression is like a weight just sitting there on it's own. So if 3 is added to an expression, it's no longer the same expression. is not the same expression as . Moving to your question, the reason you can multiply both sides of by 6 is that you're preserving the balance by doing the same thing (multiplying by 6) to both sides. Multiplying the expression by 6, however, results in a different expression that's not equivalent. So multiplying an expression by a number typically changes the expression and so isn't allowed. Sometimes, though you do multiply an expression by 1. Multiplying by any other number usually changes the expression, but multiplying by 1 doesn't change it. That's why multiplying by 1 is allowed. Usually you multiply by 1 in a different form, such as or . For example, starting with , you can multiply the by and the by : to get the common denominator: . I hope this helps - let me know if you have questions about any part of it.
absolute-value/61347: |2x+6|=101 solutions Answer 42227 by mathick(4) on 2006-11-14 14:26:47 (Show Source): You can put this solution on YOUR website! An absolute value equation such as |y| = 10 is really two equations: y = 10 and y = -10. Similarly, the equation |2x+6|= 10 is equivalent to the two equations 2x + 6 = 10 and 2x + 6 = -10. Solving these both for x gives the two answers. The first one, for example is: 2x + 6 = 10. You can solve it at mathick.com, or as follows: Subtract 6 from both sides: 2x = 4 Divide both sides by 2: x = 2. Solving the second equation gives x = -8, so the two answers are 2 and -8. You can plug these back into the original equation to verify that they are correct.
logarithm/61154: I need helping solving this for x: (ln x)^3=ln x^4 Can I rewrite it as (ln x)^3=4lnx ? Then can I divide both sides by lnx leaving (ln x)^2=4 ? Can I now square both sides leaving me with ln x=2 ? Now I'm not sure what to do next.1 solutions Answer 42160 by mathick(4) on 2006-11-14 00:47:37 (Show Source): You can put this solution on YOUR website!I need helping solving this for x: (ln x)^3=ln x^4 Can I rewrite it as (ln x)^3=4lnx ? Yes. Then can I divide both sides by lnx leaving (ln x)^2=4 ? Yes, but this assumes that you're not dividing both sides by 0, i.e. that ln(x) is not 0. This step wouldn't be valid in the case that ln(x) = 0, so this case (ln(x) = 0) needs to be treated separately. Can I now square (root) both sides leaving me with ln x=2 ? Right, ln (x) = 2, and also ln(x) = -2. (Taking the square root of both sides gives ln x = +2 and ln x = -2.) Now I'm not sure what to do next To solve ln(x) = 2 for x, exponentiate both sides: . The left side simplifies, giving one of the final answers: . The equation ln (x) = -2 can be solved similarly. Finally, there is the case when ln(x) = 0. This happens when x = 1. To verify that this is a solution, you can plug it into the original equation and see if it checks out (gives a true equation). |
# Mathematics Questions and Answers – Invertible Matrices
«
»
This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Invertible Matrices”.
1. Which among the following is inverse of the matrix A=$$\begin{bmatrix}2&3\\5&1\end{bmatrix}$$ ?
a) $$\begin{bmatrix}\frac{1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}$$
b) $$\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}$$
c) $$\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\1&\frac{-2}{13}\end{bmatrix}$$
d) $$\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&-2\end{bmatrix}$$
Explanation: Consider the matrix A=$$\begin{bmatrix}2&3\\5&1\end{bmatrix}$$
Using elementary row operation, we write A=IA.
$$\begin{bmatrix}2&3\\5&1\end{bmatrix}$$=$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$A
$$\begin{bmatrix}-13&0\\5&1\end{bmatrix}$$=$$\begin{bmatrix}1&-3\\0&1\end{bmatrix}$$A (Applying R1→R1-3R2)
$$\begin{bmatrix}1&0\\5&1\end{bmatrix}$$=$$\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\0&1\end{bmatrix}$$A (Applying $$R_1 \rightarrow -\frac{R_1}{13}$$)
$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$=$$\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}$$A (Applying R2→R2-5R1)
$$A^{-1}=\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}$$.
2. Which of the following matrices will not have an inverse?
a) $$\begin{bmatrix}2&4\\-1&1\end{bmatrix}$$
b) $$\begin{bmatrix}1&5&2\\6&4&2\\1&3&2\end{bmatrix}$$
c) $$\begin{bmatrix}1&2\\1&1\end{bmatrix}$$
d) $$\begin{bmatrix}1&2&5\\3&6&4\end{bmatrix}$$
Explanation: The matrix A=$$\begin{bmatrix}1&2&5\\3&6&4\end{bmatrix}$$ will not have an inverse as it is a rectangular matrix. Rectangular matrix does not possess an inverse matrix.
3. If A and B are invertible matrices of the same order, then (AB)-1=B-1 A-1.
a) True
b) False
Explanation: The given statement is true.
(AB) (AB)-1=I (Using the formula AA-1=I)
Multiplying both sides by A-1, we get
A-1 (AB) (AB)-1=A-1 I
(A-1 A)B(AB)-1=A-1
IB(AB-1)=A-1
B(AB-1)=A-1
⇒B-1 B(AB-1)=B-1 A-1
(AB-1)=B-1 A-1
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4. A matrix A is invertible if it has all zeroes in one or more rows on L.H.S.
a) True
b) False
Explanation: The given statement is false. A matrix is non-invertible if it has all zeroes in one or more rows on L.H.S. This is because after applying all the elementary operations on the matrix, we should get an identity matrix on the L.H.S. to obtain an inverse of the given matrix, which is not possible if we obtain all zeroes in one or more rows.
5. The inverse of the matrix A=$$\begin{bmatrix}1&2&4\\5&2&4\\3&6&2\end{bmatrix}$$ is
a) $$\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&1&\frac{-1}{10}\end{bmatrix}$$
b) $$\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&1\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}$$
c) $$\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}$$
d) $$\begin{bmatrix}\frac{-1}{4}&-\frac{1}{4}&0\\ \frac{1}{40}&\frac{1}{8}&\frac{-1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}$$
Explanation: Consider the matrix A=$$\begin{bmatrix}1&2&4\\5&2&4\\3&6&2\end{bmatrix}$$
Using the elementary row operation, we write A=IA
$$\begin{bmatrix}1&2&4\\5&2&4\\3&6&2\end{bmatrix}$$=$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$A
Applying R1→R1-R2
$$R_1 \rightarrow \frac{R_1}{-4}$$
$$\begin{bmatrix}1&0&0\\5&2&4\\3&6&2\end{bmatrix}$$=$$\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\0&1&0\\0&0&1\end{bmatrix}$$A
Applying R2→R2-5R1 and R3→R3-3R1
$$\begin{bmatrix}1&0&0\\0&2&4\\0&6&2\end{bmatrix}$$=$$\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\\frac{5}{4}&\frac{-1}{4}&0\\\frac{3}{4}&\frac{-3}{4}&1\end{bmatrix}$$A
Applying R2→R2-2R3 and $$R_2 \rightarrow \frac{R_2}{-10}$$
$$\begin{bmatrix}1&0&0\\0&1&0\\0&6&2\end{bmatrix}$$=$$\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-5}{40}&\frac{1}{5}\\\frac{3}{4}&\frac{-3}{4}&1\end{bmatrix}$$A
Applying R3→R3-6R2 and $$R_2 \rightarrow \frac{R_2}{2}$$
$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$=$$\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}$$A
A-1=$$\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}$$.
6. Which of the following is the inverse of the matrix A=$$\begin{bmatrix}8&1\\1&2\end{bmatrix}$$?
a) $$\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\\frac{1}{15}&\frac{8}{15}\end{bmatrix}$$
b) $$\begin{bmatrix}\frac{1}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{1}{15}\end{bmatrix}$$
c) $$\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$$
d) $$\begin{bmatrix}\frac{2}{15}&\frac{1}{15}\\\frac{1}{15}&\frac{4}{15}\end{bmatrix}$$
Explanation: Consider the matrix A=$$\begin{bmatrix}8&1\\1&2\end{bmatrix}$$
Using the elementary row operation, we write A=IA
Applying R2→8R2-R1 and R2→R2/15, we get
$$\begin{bmatrix}8&1\\0&1\end{bmatrix}$$=$$\begin{bmatrix}1&0\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$$A
Applying R1→R1-R2 and R1→R1/8, we get
$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$=$$\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$$A
A-1=$$\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$$.
7. Which among the below matrices has the inverse A-1=$$\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}$$
a) $$\begin{bmatrix}1&5\\0&8\end{bmatrix}$$
b) $$\begin{bmatrix}1&5\\-1&8\end{bmatrix}$$
c) $$\begin{bmatrix}1&5\\0&16\end{bmatrix}$$
d) $$\begin{bmatrix}1&8\\0&8\end{bmatrix}$$
Explanation: Consider the matrix A=$$\begin{bmatrix}1&5\\0&8\end{bmatrix}$$
Using the elementary column operations, we write A=AI
$$\begin{bmatrix}1&5\\0&8\end{bmatrix}$$=A$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$
Applying C2→C2-5C1
$$\begin{bmatrix}1&0\\0&8\end{bmatrix}$$=A$$\begin{bmatrix}1&-5\\0&1\end{bmatrix}$$
Applying C2→C2/8
$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$=A$$\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}$$
A-1=$$\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}$$.
8. Find the inverse of matrix A=$$\begin{bmatrix}5&1&3\\4&2&6\\5&4&2\end{bmatrix}$$
a) $$\begin{bmatrix}0&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}$$
b) $$\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}$$
c) $$\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&1&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}$$
d) $$\begin{bmatrix}\frac{1}{3}&\frac{1}{6}&0\\\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&\frac{1}{10}\end{bmatrix}$$
Explanation: Consider the matrix A=$$\begin{bmatrix}5&1&3\\4&2&6\\5&4&2\end{bmatrix}$$
Using the elementary row operations, we write A=IA
$$\begin{bmatrix}5&1&3\\4&2&6\\5&4&2\end{bmatrix}$$=$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$A
Applying R2→5R2-4R1 and R3→R3-R1
$$\begin{bmatrix}5&1&3\\0&6&18\\0&3&-1\end{bmatrix}$$=$$\begin{bmatrix}1&0&0\\-4&5&0\\-1&0&1\end{bmatrix}$$A
Applying R1→R2-6R1 and R3→R2-2R3
$$\begin{bmatrix}-30&0&0\\0&6&18\\0&0&20\end{bmatrix}$$=$$\begin{bmatrix}-10&5&0\\-4&5&0\\-2&5&-2\end{bmatrix}$$A
Applying R2→20R2-18R3
$$\begin{bmatrix}-30&0&0\\0&120&0\\0&0&20\end{bmatrix}$$=$$\begin{bmatrix}-10&5&0\\-44&10&36\\-2&5&-2\end{bmatrix}$$A
Applying R1→R1/(-30), R2→R2/120, R3→R3/20
$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$=$$\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}$$A
A-1=$$\begin{bmatrix}\frac{1}{3}&-\frac{1}{6}&0\\-\frac{11}{30}&\frac{1}{12}&\frac{3}{10}\\-\frac{1}{10}&\frac{1}{4}&-\frac{1}{10}\end{bmatrix}$$.
9. Which of the following is not a property of invertible matrices if A and B are matrices of the same order?
a) (AB)-1=A-1 B-1
b) (AA-1)=(A-1 A)=I
c) (AB)-1=B-1 A-1
d) AB=BA=I
Explanation: (AB)-1=A-1 B-1 is incorrect. The correct formula is (AB)-1=B-1 A-1. B-1 A-1 ≠ A-1 B-1 as matrix multiplication is not commutative.
10. Find the inverse of A=$$\begin{bmatrix}5&3\\4&1\end{bmatrix}$$.
a) $$\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}$$
b) $$\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&\frac{5}{7}\end{bmatrix}$$
c) $$\begin{bmatrix}-\frac{1}{7}&-\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}$$
d) $$\begin{bmatrix}0&\frac{3}{7}\\\frac{4}{7}&\frac{5}{7}\end{bmatrix}$$
Explanation: Consider the matrix A=$$\begin{bmatrix}5&3\\4&1\end{bmatrix}$$
By using the elementary row operations, we write A=IA
$$\begin{bmatrix}5&3\\4&1\end{bmatrix}$$=$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$A
Applying R1→R1-3R2 and R1→R1/(-7), we get
$$\begin{bmatrix}1&0\\4&1\end{bmatrix}$$=$$\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\0&1\end{bmatrix}$$
Applying R2→R2-4R1, we get
$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$=$$\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}$$A
⇒A-1=$$\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}$$.
Sanfoundry Global Education & Learning Series – Mathematics – Class 12.
To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers. |
# Directional derivatives
### Directional derivatives
#### Lessons
Notes:
Suppose we have a vector $\vec{v} =\lt v_1, v_2\gt$. The unit vector will be:
$\vec{v} = \frac{1}{\sqrt{v^2_1 + v^2_2}} \lt v_1,v_2\gt$
Suppose we have a vector $\vec{v} =\lt v_1, v_2,v_3\gt$. The unit vector will be:
$\vec{v} = \frac{1}{ \sqrt{v_1^2 + v_2^2 + v_3^2} } $
When given an angle of a direction ($\theta$ ), we say that the unit vector (that points to the direction) is:
$\vec{u} = \lt \cos \theta, \sin \theta \gt$
Directional Derivatives of 2 Variable Functions
A Directional Derivative is the rate of change (of $x$ and $y$) of a function at a point $P=(x_0,y_0,z_0)$, at the direction of the unit vector
Suppose there is a 2-variable function $z=f(x,y)$. Then the directional derivative is:
$D_{\vec{u}}f(x,y) = f_x(x,y)a + f_y(x,y)b$
where the $\vec{u} = $ is the unit vector that points in the direction of change. Directional Derivatives of 3 Variable Functions
Suppose there is a 3-variable function $w=f(x,y,z)$. Then the directional derivative is:
$D_{\vec{u}}f(x,y,z) = f_x(x,y,z)a + f_y(x,y,z)b + f_z(x,y,z)c$
where the $\vec{u} = $ is the unit vector that points in the direction of change.
• Introduction
Directional Derivatives Overview:
a)
Things to Know Before Knowing Directional Derivatives
• Calculating unit vectors
• An example
• Angle to a unit vector
• An example
b)
Directional Derivatives of 2 Variable Functions
• The rate of change of $x$ and $y$
• $D_{\vec{u}}f(x,y) = f_x(x,y)a + f_y(x,y)b$
• An example
c)
Directional Derivatives of 3 Variable Functions
• The rate of change of $x$ and $y$
• $D_{\vec{u}}f(x,y,z) = f_x(x,y,z)a + f_y(x,y,z)b + f_z(x,y,z)c$
• An example
• 1.
Finding the Unit Vector & Angle of Direction
Find the unit vector of $\vec{v} =\lt 5, -2\gt.$
• 2.
Find the unit vector of $\vec{v} =\lt -1, 3, 5\gt.$
• 3.
Find the unit vector, given that the unit vector is in the direction of $\theta=\frac{\pi}{3}$.
• 4.
Finding the Directional Derivative of 2 Variable Functions
Find the direction derivative of $z=\sqrt{x^2+2y}$ at any given point, in the direction of $\vec{v} = \lt 1, 3\gt$
• 5.
Find the direction derivative of $z=xy \ln (\frac{x}{y})$ at any given point, where the direction of the unit vector is at $\theta=\frac{\pi}{6}$.
• 6.
Finding the Directional Derivative of 3 Variable Functions
Find the direction derivative of $f(x,y,z)=xy^3+yz^2$ in the direction of $\vec{v} =\lt 1, 2, 4\gt$.
• 7.
Find the direction derivative of $f(x,y,z)=\ln (x)e^{yz}$ in the direction of $\vec{v} =\lt -3, 1, 2\gt$ . |
# 2023 AMC 12B Problems/Problem 1
The following problem is from both the 2023 AMC 10B #1 and 2023 AMC 12B #1, so both problems redirect to this page.
## Problem
Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?
$\textbf{(A) } \frac{1}{12} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{6} \qquad\textbf{(D) } \frac{1}{8} \qquad\textbf{(E) } \frac{2}{9}$
## Solution 1
The first three glasses each have a full glass. Let's assume that each glass has "1 unit" of juice. It won't matter exactly how much juice everyone has because we're dealing with ratios, and that wouldn't affect our answer. The fourth glass has a glass that is one third. So the total amount of juice will be $1+1+1+\dfrac{1}{3} = \dfrac{10}{3}$. If we divide the total amount of juice by 4, we get $\dfrac{5}{6}$, which should be the amount of juice in each glass. This means that each of the first three glasses will have to contribute $1 - \dfrac{5}{6} = \boxed{\dfrac{1}{6}}$ to the fourth glass.
~Sir Ian Seo the Great & lprado
## Solution 2 (unnecessary numerical values)
Given that the first three glasses are full and the fourth is only $\frac{1}{3}$ full, let's represent their contents with a common denominator, which we'll set as 6. This makes the first three glasses $\dfrac{6}{6}$ full, and the fourth glass $\frac{2}{6}$ full.
To equalize the amounts, Mrs. Jones needs to pour juice from the first three glasses into the fourth. Pouring $\frac{1}{6}$ from each of the first three glasses will make them all $\dfrac{5}{6}$ full. Thus, all four glasses will have the same amount of juice. Therefore, the answer is $\boxed{\textbf{(C) }\dfrac16}.$
~Ishaan Garg
## Solution 3
We let $x$ denote how much juice we take from each of the first $3$ children and give to the $4$th child.
We can write the following equation: $1-x=\dfrac13+3x$, since each value represents how much juice each child (equally) has in the end. (Each of the first three children now have $1-x$ juice, and the fourth child has $3x$ more juice on top of their initial $\dfrac13$.)
Solving, we see that $x=\boxed{\textbf{(C) }\dfrac16}.$
~Technodoggo
~Math-X
## Video Solution (Quick and Easy!)
~Education, the Study of Everything |
Balanced Vs Unbalanced Forces
Balanced vs Unbalanced Forces
Balanced and unbalanced forces are two concepts in physics that describe the relationship between forces acting on an object and the object’s motion. A force is a push or a pull that can change the speed, direction, or shape of an object. Forces can be classified as balanced or unbalanced depending on whether they cause a change in the object’s motion or not.
Balanced forces are forces that are equal in magnitude and opposite in direction. When two or more forces act on an object and cancel each other out, they are said to be balanced. Balanced forces do not cause a change in the object’s motion; the object will either remain at rest or continue moving at a constant velocity. For example, when a book is resting on a table, the force of gravity pulling the book down is balanced by the normal force of the table pushing the book up. The net force on the book is zero, so the book does not move.
Unbalanced forces are forces that are not equal in magnitude and/or direction. When two or more forces act on an object and do not cancel each other out, they are said to be unbalanced. Unbalanced forces cause a change in the object’s motion; the object will either accelerate or decelerate. For example, when a person kicks a soccer ball, the force of the foot on the ball is greater than the force of gravity and air resistance on the ball. The net force on the ball is not zero, so the ball changes its speed and direction.
The net force on an object is the vector sum of all the forces acting on it. A vector is a quantity that has both magnitude and direction. To find the net force, we can use the following rules:
– If two forces act in the same direction, we add their magnitudes to find the net force. The direction of the net force is the same as the direction of the individual forces.
– If two forces act in opposite directions, we subtract their magnitudes to find the net force. The direction of the net force is the same as the direction of the larger force.
– If two forces act at an angle, we can use trigonometry or a graphical method to find the net force. The direction of the net force is the direction of the resultant vector.
The net force determines whether the forces are balanced or unbalanced. If the net force is zero, the forces are balanced and the object does not change its motion. If the net force is not zero, the forces are unbalanced and the object changes its motion. The net force also determines the acceleration of the object according to Newton’s second law of motion, which states that the net force on an object is equal to its mass times its acceleration. Mathematically, this can be written as \$\$vec{F}_{net} = |
# First Fundamental Theorem of Calculus Greg Kelly, Hanford High School, Richland, Washington.
## Presentation on theme: "First Fundamental Theorem of Calculus Greg Kelly, Hanford High School, Richland, Washington."— Presentation transcript:
First Fundamental Theorem of Calculus Greg Kelly, Hanford High School, Richland, Washington
When we find the area under a curve by adding rectangles, the answer is called a Rieman sum. subinterval partition The width of a rectangle is called a subinterval. The entire interval is called the partition. Subintervals do not all have to be the same size.
subinterval partition If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by. As gets smaller, the approximation for the area gets better. if P is a partition of the interval
is called the definite integral of over. If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by:
Leibnitz introduced a simpler notation for the definite integral: Note that the very small change in x becomes dx.
Integration Symbol lower limit of integration upper limit of integration integrand variable of integration (dummy variable) It is called a dummy variable because the answer does not depend on the variable chosen.
We have the notation for integration, but we still need to learn how to evaluate the integral.
time velocity After 4 seconds, the object has gone 12 feet. Let’s consider an object moving at a constant rate of 3 ft/sec. Since rate. time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.
If the velocity varies: Distance: ( C=0 since s=0 at t=0 ) After 4 seconds: The distance is still equal to the area under the curve! Notice that the area is a trapezoid.
What if: We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example. It seems reasonable that the distance will equal the area under the curve.
We can use anti-derivatives to find the area under a curve!
Fundamental Theorem of Calculus Just like we proved earlier! Area under curve from a to x = antiderivative at x minus antiderivative at a.
Area from x=0 to x=1 Example: Find the area under the curve from x = 1 to x = 2. Area from x=0 to x=2 Area under the curve from x = 1 to x = 2.
Example: Find the area under the curve from x = 1 to x = 2. To use your TI-83+ or TI-84+ Math 7. fnInt Enter Function Variable Limits of Integration
Example: Find the area between the x-axis and the curve from to. pos. neg.
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# Focus on Math
## Helping children become mathematicians!
### I Have No Cue! Subtraction without the “Cuing” words February 21, 2013
A while ago I gave my grade 2 and 3 math classes a whole-part-part subtraction question, which is a different format from what is often given to students. Many times word problems for addition and/or subtraction are the kind where the initial amount is given, the amount of change is given, and the sum or difference is what is missing. Those problems go something like this (using “small” numbers):
• There were 2 birds in a tree (initial amount). 3 more landed in the tree (change). How many were then in the tree (missing sum)?
• There were 5 birds in a tree (initial amount). 2 flew away (change). How many birds were left in the tree (missing difference)?
In both of those questions, students hear “cueing” words (“more” and “away” respectively) and figure out from those cues which operation to perform.
Problems, however, can be written in ways that do not give such obvious cues, thus providing situations for students to think more deeply about the problems. Using a situation involving a whole and two comprising parts (whole-part-part) is one way to write such questions. Solving such a word problem may be a bit more difficult for students as nothing is actually added to or subtracted from the initial amount.
In a certain city there were 93 children enrolled in soccer. 55 of them were girls and the rest were boys. How many boys were enrolled in the program? (Incidentally, we did have to have a conversation in each class about the meaning of the word ‘enrolled’.)
The grade 3 classes were given this version of the question:
In a certain city there were 321 children enrolled in soccer. 148 of them were girls and the rest were boys. How many boys were enrolled in the program?
At least one student in each of the 5 classes that did this problem came up to me immediately after the class had started to work and asked what he should do. “Should I add these numbers?” “Should I subtract the two numbers?” I expected that this would happen given the wording of the problem.
The goal, as always, was for the students to make sense of the problem and then to solve it using more than one strategy. In the end I was quite delighted in the various strategies that the students used to solve the problem.
LOOKING AT THE GRADE 2 STRATEGIES:
Mrs. Sequin’s grade 2 class shared several strategies. A number of students made use of the 100 dot array, most of them first colouring 93 dots to represent all of the children, then crossing out 55 of those as representing the girls. The remaining dots, not crossed out, were counted to tell the number of boys playing soccer.
One boy showed his numerical solution where he started at 55 and added up (mainly by 5’s and 10’s) until he reached 93.
We had been playing with open number lines in the previous weeks, and three different students shared strategies for solving the problem with this tool. Using such tools allows students great flexibility as they think about the numbers in a problem.
Notice the first number line on the chart paper. The student worked backwards towards 55 (really doing this: 93 – ? = 55), starting at 93, jumping back 20 to 73, jumping back 20 more to 53, then realized that that was 2 too far back, so changed the second jump backwards to 18 (for a total of 38 back).
The second number line records a student who used addition to solve the problem (55 + ? = 93). She began with the 55 girls, then jumped by 10’s up toward 93, passing it, landing on 95. She solves the problem then by adding her 4 ten jumps and subtracting the 2.
The third number line also shows subtraction like the first method: again the student is looking for the missing amount that will leave 55. The student begins by jumping just 3 backwards to get to the “friendly” number 90, jumps 20 back once, twice, then recognized he has “over shot” 55 and changes the last jump to 15.
LOOKING AT THE GRADE 3 STRATEGIES:
Mrs. Ranger’s grade 3 class shared 4 strategies in the time we had for our discussion. First, one girl shared how she arrived at the answer using the traditional algorithm for subtraction. After that a boy shared his strategy using an open number line. He used the “think addition” method (148 + ? = 321).
I was quite impressed with the third strategy shared. My students have used 100 dot arrays for thinking about numbers, but the boy did not want to have to use 4 arrays on his page. Instead, he drew his own dots, explaining that each dot represented 10 players. Thus each row of 10 dots represented 100 players. He had two more dots for the 20 players, and a single tally mark for the last player (300 + 20 + 1). From that point he crosses out 14 dots (representing 140 players), crossed out one more dot but puts a small 2 beside it to show that not all 10 were removed, just 8 leaving 2. Pretty cool thinking, eh?!
In the final method shared the student had used sketches to represent base-10 blocks. She began with 321 and shows step by step how 100, 40, and 8 are removed.
I hope you see how wonderful problem solving can be with students. When we arm them with tools and strategies for thinking, they can do wonderful things!
Mathematically yours,
Carollee |
What is Percentage and Why we Use it in Maths - Students Explore
# What is Percentage and Why we Use it in Maths
Dear students today we have started Math subject today and in this first post of Maths section we will explore about percentages, what is percentage and how to use, where to use it etc.. So keep in touch with this post in order to learn better concept of percentages.
what is percentage
## What is Percentage
Percentage’ signifies ‘out of a hundred.’ Percentages, like fractions and decimals, are used in mathematics to describe components of a whole. When calculating percentages, the entire is divided into a hundred equal parts. The symbol percent (%) is used to indicate that a number is a percentage, and the abbreviation ‘pct’ is used less frequently.
Percentages can be found practically anywhere: in stores, on the internet, in commercials, and in the media. Understanding what percentages mean is a critical skill that may save your time and money while also making you more employable.
## How to Find Percentage
The general rule for finding percentage of any given whole number is:
First find given percentage of that particular whole number, then multiply it 100, simple.
percentages Formula
## Example
Ms. Sophia purchase apple at a cost of \$100 per KG. The shopkeeper discount \$20. What % sophia got saved from the discount of 20\$.
Solution:
As we know, the real price of apple was \$100, and shopkeeper give discount of \$20. So,
\$100-\$20 = \$80.
It means sophia purchase apple at \$80 per KG.
Now we use % formula to find how much its percentages.
X = discounted price/actual price x 100
X = \$20/\$100 x 100 = 20%
So, our answer is \$20 discount is also equal to 20%.
## Where is Percentage Used or Applied
Percentages are extensively utilized in a variety of contexts. Discounts in stores, bank interest rates, inflation rates, and numerous media data, earnings, sales, and taxation are all expressed as percentages. Percentages are essential for comprehending financial aspects of daily living.
Businesses frequently employ %, such as when determining what percentage of a product’s selling price represents profit. The percent symbol is used to represent percentages.
These all are examples of where it’s employed in accounting and finance. Percentages were employed by a number of institutions and colleges to convey student grades.
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What are Compound Interest with Explanation
What is the Simple Interest with Formula and Example
What Time Value of Money with Example```
## Why do We Use Percentages
It is used to figure out “how much” or “how many” something is. A % number makes it easier to calculate the exact amount or figure being discussed. A comparison of fractions is made. determining if a percentage gain or decrease has occurred. It aids in the calculation of profit and loss percentages.
Students Explore help you to solve your academic problems you usually faced in school, college or universities. So if you have any query in above topic just ask in comment section given below this post and do not forge to share this post with your classmates.
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# Difference between revisions of "2020 AMC 10B Problems/Problem 25"
## Problem
Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product$$n = f_1\cdot f_2\cdots f_k,$$where $k\ge1$, the $f_i$ are integers strictly greater than $1$, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$, $2\cdot 3$, and $3\cdot2$, so $D(6) = 3$. What is $D(96)$?
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$
## Solution
Note that $96 = 2^5 \cdot 3$. Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$, we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.
$k=1$: We see that there is $1$ way, merely $96$.
$k=2$: This way, we have the $3$ in one slot and $2$ in another, and symmetry. The four other $2$'s leave us with $5$ ways and symmetry doubles us so we have $10$.
$k=3$: We have $3, 2, 2$ as our baseline. We need to multiply by $2$ in $3$ places, and see that we can split the remaining three powers of 2 in a manner that is 3-0-0, 2-1-0 or 1-1-1. A 3-0-0 split has $6 + 3 = 9$ ways of happening (24-2-2 and symmetry; 2-3-16 and symmetry), a 2-1-0 split has $6 \cdot 3 = 18$ ways of happening (due to all being distinct) and a 1-1-1 split has $3$ ways of happening (6-4-4 and symmetry) so in this case we have $9+18+3=30$ ways.
$k=4$: We have $3, 2, 2, 2$ as our baseline, and for the two other $2$'s, we have a 2-0-0-0 or 1-1-0-0 split. The former grants us $4+12=16$ ways (12-2-2-2 and symmetry and 3-8-2-2 and symmetry) and the latter grants us also $12+12=24$ ways (6-4-2-2 and symmetry and 3-4-4-2 and symmetry) for a total of $16+24=40$ ways.
$k=5$: We have $3, 2, 2, 2, 2$ as our baseline and one place to put the last two: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5 \cdot 4 = 20$ ways due to symmetry. Thus, we have $5+20=25$ ways.
$k=6$: We have $3, 2, 2, 2, 2, 2$ and symmetry and no more twos to multiply, so by symmetry, we have $6$ ways.
Thus, adding, we have $1+10+30+40+25+6=\boxed{\textbf{(A) } 112}$.
~kevinmathz
## Solution 2
As before, note that $96=2^5\cdot3$, and we need to consider 6 different cases, one for each possible value of $k$, the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization needs to contain one factor that is itself a multiple of $3$, and there are $k$ to choose from, and the rest must contain at least one factor of $2$. Next, consider the remaining $6-n$ factors of $2$ left to assign to the $k$ factors. Using stars and bars, the number of ways to do this is $${{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}$$ This makes $k{5\choose{6-k}}$ possibilities for each k.
To obtain the total number of factorizations, add all possible values for k: $$\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{\textbf{(A) } \text{112}}$$.
## Solution 3
Begin by examining $f_1$. $f_1$ can take on any value that is a factor of $96$ except $1$. For each choice of $f_1$, the resulting $f_2...f_k$ must have a product of $96/f_1$. This means the number of ways the rest $f_a$, $1 can be written by the scheme stated in the problem for each $f_1$ is equal to $D(96/f_1)$, since the product of $f_2 \cdot f_3... \cdot f_k=x$ is counted as one valid product if and only if $f_1 \cdot x=96$, the product $x$ has the properties that factors are greater than $1$, and differently ordered products are counted separately.
For example, say the first factor is $2$. Then, the remaining numbers must multiply to $48$, so the number of ways the product can be written beginning with $2$ is $D(48)$. To add up all the number of solutions for every possible starting factor, $D(96/f_1)$ must be calculated and summed for all possible $f_1$, except $96$ and $1$, since a single $1$ is not counted according to the problem statement. The $96$ however, is counted, but only results in $1$ possibility, the first and only factor being $96$. This means
$D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$.
Instead of calculating D for the larger factors first, reduce $D(48)$, $D(32)$, and $D(24)$ into sums of $D(m)$ where $m<=16$ to ease calculation. Following the recursive definition $D(n)=($sums of $D(c))+1$ where c takes on every divisor of n except for 1 and itself, the sum simplifies to
$D(96)=(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)$ +$(D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1.$
$D(24)=D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$, so the sum further simplifies to
$D(96)=3D(16)+4D(12)+5D(8)+4D(6)+5D(4)+4D(3)+5D(2)+5$, after combining terms. From quick casework,
$D(16)=8, D(12)=8, D(8)=4, D(6)=3, D(4)=2, D(3)=1$ and $D(2)=1$. Substituting these values into the expression above,
$D(96)=3 \cdot 8+4 \cdot 8+5 \cdot 4+4 \cdot 3+5 \cdot 2+4 \cdot 1+5 \cdot 1+5=\boxed{\textbf{(A) } 112}$.
~monmath a.k.a Fmirza
## Solution 4
Note that $96 = 3 \cdot 2^5$, and that $D$ of a perfect power of a prime is relatively easy to calculate. Also note that you can find $D(96)$ from $D(32)$ by simply totaling the number of ways there are to insert a $3$ into a set of numbers that multiply to $32$.
First, calculate $D(32)$. Since $32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$, all you have to do was find the number of ways to divide the $2$'s into groups, such that each group has at least one $2$. By stars and bars, this results in $1$ way with five terms, $4$ ways with four terms, $6$ ways with three terms, $4$ ways with two terms, and $1$ way with one term. (The total, $16$, is not needed for the remaining calculations.)
Then, to get $D(96)$, in each possible $D(32)$ sequence, insert a $3$ somewhere in it, either by placing it somewhere next to the original numbers (in one of $n+1$ ways, where $n$ is the number of terms in the $D(32)$ sequence), or by multiplying one of the numbers by $3$ (in one of $n$ ways). There are $2+1=3$ ways to do this with one term, $3+2=5$ with two, $7$ with three, $9$ with four, and $11$ with five.
The resulting number of possible sequences is $3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{\textbf{(A) }112}$. ~emerald_block |
#### Transcript Complex Numbers: a + bi
```Objectives for Class 3
Complex Numbers.
Solve Quadratic Equations in the
Complex Number System
Basics of the Complex Number
System
i2
= -1
(Square root of -1) = i
Simplify the following square roots
(Square root of -25)
Pull the square root of -1 out first to form
i(square root of 25)
Break down the square root of 25
(square root of -25) = 5i
Simplify each of the following. Hit “enter” to check your
answers. See Mrs. Dorshorst for help if needed.
(square
root of -81)
9i
(square
root of -20)
2i(square
root of 5)
Complex Numbers: a + bi
‘a’ is called the real part
‘b’ is called the imaginary part
Standard Form: a + bi
Real number: a
Ex: 3 – 5i
Real Part is 3
Imaginary part is -5i
Ex: 8
No imaginary part
Pure imaginary: bi
Ex: 3i
No real part
Equal Complex Numbers:
a + bi = c + di
a must equal c : Real parts must be equal
b must equal d: Imaginary parts must be equal
Ex: 3x + 5yi = -7 – 4i
Set real parts equal and solve for variable:3x =-7
X = -7/3
Set imaginary parts equal and solve for variable:
5yi = -4i
5y = -4
so y = -4/5
8x – 9yi = 4 + 2i
8x
=4
x = 1/2
-9y = 2
y = -2/9
Numbers
Sum of Complex Numbers: (a + bi) + (c
+ di) = (a + c) + (b + d)I
Ex: (4 – 7i) + (12 + 19i)
(4 + 12) + (-7i + 19i)
16 + 12i
Try: (8 + 52i) + (-12 – 40i)
Solution: -4 + 12i
Difference of Complex Numbers:
(a + bi) – (c + di) = (a – c) + (b – d)i
Ex: (-3 + 7i) – (-8 + 9i)
Distribute the subtraction through the
second complex number
(-3 + 7i) + ( +8 – 9i)
(-3 + 8) + (7i – 9i)
5 – 2i
Try: (25 – 3i) – (13 – 10i)
(25 – 13) + (-3i + 10i)
12 + 7i
Examples
– 5i) – (8 + 6i)
Solution: -6 – 11i
(2
+ 5i) + (6 – 8i)
Solution: 9 – 3i
(3
Product of Complex Numbers
Distribute each term through the next
parenthesis.
Remember that i2 = -1
Examples
(5 + 3i)(2 + 7i)
10 + 35i + 6i + 21i2
10 + 35i + 6i – 21
-11 + 41i
(3 – 4i)(2 + i)
6 + 3i – 8i -4i2
6 + 3i – 8i + 4
10 – 5i
3i(-3 + 4i)
-9i - 12
Conjugate
Same values with opposite middle sign
Ex: 2 + 3i conjugate: 2 – 3i
_______
Notation:
a + bi
Find the conjugate of
3 – 4i
____
3 + 4i
Multiply 4 – 2i times its conjugate
(4 – 2i)(4 + 2i)
Conjugate is 4 + 2i)
16 + 8i – 8i - 4i2
Distribute 1st parenthesis through second
parenthesis
16 + 4
Combine like terms and substitute -1 in for i2
20
Multiply 3 + 2i times its conjugate
What
is the conjugate?
3 – 2i
Do the multiplication.
(3 + 2i)(3 – 2i)
Solution: 13
Dividing Complex Numbers
Multiply both the numerator and the denominator
times the conjugate to eliminate the radical in the
denominator (recall that ‘i’ is a radical)
Example: (1 + 4i) / (5 – 12i)
((1+4i)(5+12i))/((5-12i)(5+12i))
Multiply by conjugate of denominator
(5 + 12i + 20i +48i2)/(25 + 60i – 60i – 144i2)
Multiply numerators and multiply denominators
(5 + 32i – 48)/(25 + 144)
Simplify
(-43 + 32i) / (169)
(2 – 3i) / (4 – 3i)
What would you multiply by to eliminate
4 + 3i
Multiply numerator and multiply
denominator
((2 – 3i)(4 + 3i))/((4 – 3i)(4+ 3i))
Solution: (17 – 6i) / (25)
Other Examples
If z = 2 – 3i and w = 5 + 2i find each of
the following
_____
Find
(z + w)
What does this notation mean?
Conjugate of the sum
7 – 1i
so answer is 7 + 1i
Powers of i
Recall that i2 = -1
To find high level powers of i: Ex: i35
1.Break down the expression into i2
notation. (i2)17i
2. Substitute -1 in for i2 (-1)17i
3. Simplify as far as possible. (-1)i
Simplify: i7 + i25 – i8
Simplify each power of i
i7
= (i2)3i = (-1)3i = -1i
i25 = (i2)12i = (-1)12i = 1i
i8 = (i2)4 = (-1)4 = 1
Combine like terms
-1i + 1i – (1) = 1
Complex Number System
Solve: x2 – 4x = -8
1. Put into standard form.
x2 – 4x + 8 = 0
2. Solve using the Quadratic Formula. When
simplifying the radical remember that the square root
of -1 is i.
(4 + sqrt(16 – 4(1)(8))) / 2
(4 + sqrt -16)/2
(4 + 4i)/2
2 + 2i
Discriminant: b2 – 4ac
formula is called the discriminant.
The value of the discriminant determines the
type of solution for the equation.
D > 0: discriminant is positive there are 2 real
solutions
D < 0: discriminant is negative there are 2
complex solutions
D = 0: discriminant is o there is one real
solution (double root)
What type of solutions do the following
equations have? Do not solve just determine
the type of solutions. (Recall the rules for
discriminants on the previous slide.
3x2 – 8x = 9
b2 – 4ac => 64 – 4(3)(-9)
D>0
2 real number solutions
4x2 + 3 = 5x
b2 – 4ac => 25 – 4(4)(3)
D>0
2 complex number solutions
Assignment
Page 117
#9, 13, 19, 23, 25, 27, 31, 35, 39, 49, 51,
55, 59, 63, 75, 77, 83,
``` |
Decimals on number lines
Decimals on number lines
In this lesson, we will learn:
• How to read number lines for decimals to the tenths or hundredths
• How to draw a point to represent a decimal on a decimal number line
Notes:
• A number line is a picture of a straight line with vertical markings of equal spacing
• Number lines increase from left to right (small $\, \longrightarrow \,$ big)
• Decimals are numbers with place values even smaller than the ones place.
• Number lines can be divided even further between each number
• Splitting each division into 10 equal parts creates the next decimal place value
• Dividing the space between 0 to 1 into 10 equal parts, each part is a tenth
• Dividing the space between 0 to 0.1 into 10 equal parts, each part is a hundredth
• From 0 to 1, it is first split into 10 equal parts (tenths = 0.1)
• Each of these tenths 0.1 will be split into 10 more equal parts (10 tenths each into 10 parts = 10 × 10 = 100 parts)
• Ultimately, the space between 0 and 1 will be split into 100 equal parts
• That's why each 0.01 is called a hundredth (one of the 100 parts)
Lessons
• Introduction
Introduction to Decimals on Number Lines:
a)
How to read number lines at different place value levels (ones, tenths, hundredths)
• 1.
What is the decimal given by the point on the number line?
a)
b)
c)
d)
• 2.
Drawing decimals on number lines
Draw the decimal as a point on the number line
a)
b)
c)
d)
• 3.
Zooming into number lines and dividing into ten segments
Draw a point for the specified decimal. Label the number lines showing tenths and the zoomed in lines showing hundredths.
a)
b)
c)
d)
e) |
Lesson Explainer: Angles of Elevation and Depression | Nagwa Lesson Explainer: Angles of Elevation and Depression | Nagwa
Lesson Explainer: Angles of Elevation and Depression Mathematics
In this explainer, we will learn how to solve real-world problems that involve angles of elevation and depression.
Before you start with this explainer, you should be confident finding angle measures and missing sides using right triangle trigonometry and the laws of sines and cosines.
Now, prior to looking at examples and recalling trigonometric ratios and the laws of sines and cosines, we will define what angles of elevation and depression are.
Definition: Angles of Elevation and Depression
An angle of elevation is the “upward” angle from the horizontal to a line of sight from the object to a given point, whereas an angle of depression is where the angle goes “downward” from the horizontal to a given point, as shown below.
Now, we will look at the steps for solving a problem involving angles of elevation or depression where a right triangle can be formed.
• First, where one is not already drawn in the question, sketch a right triangle to represent the given scenario.
• Next, label all the known distances and the given angle of elevation/depression.
• Finally, use the trigonometric functions sine, cosine, and tangent to find the unknown distances or angles.
In our first example, we will solve a problem where a diagram has already been drawn.
Example 1: Finding an Unknown Length given a Real-World Problem and a Diagram
Fill in the blank: The distance between two buildings is 40 m. The top of building has an angle of elevation of , measured from the top of building . If the height of building and the bases of the two buildings are on the same horizontal plane, then the height of to the nearest metre is m.
In the problem, we can see that the scenario has been represented in a diagram, and, from the information given, a right triangle has been formed. It is this triangle that we will look at first.
In order to find the height of , we will need to calculate . This is because if we add to the height of , then this will give us the height of .
We are going to use the trigonometric ratios to helps us find . To use trigonometric ratios with a right triangle, you can follow these four steps:
1. Label the sides.
2. Choose the trigonometric ratio.
3. Substitute in known values.
4. Solve.
Let’s apply this to our problem.
Step 1
First, we label the hypotenuse, the opposite, and the adjacent.
Step 2
To choose the correct ratio, we look at the side we have been given and the one that we want to find. In our problem, these are the opposite and the adjacent. Using this information, we select our ratio. There is a memory aid we can use in order to assist with the choice.
This memory aid shows us which sides we require for each ratio. As we have the opposite (O) and the adjacent (A), we will use the tangent ratio. This tells us that
Step 3
We now substitute in our known values,
Step 4
Finally, we rearrange and solve for :
To maintain accuracy, we will leave in this form until the final calculation.
Now, to find the height of , we add to the height of , which we are told in the question is 30 m:
In the question we are asked to give the number of metres to the nearest metre; therefore, the answer is 53 metres.
In our next example, we will explore a problem that can once again be solved using right triangle trigonometry. However, in this example, we will need to draw a diagram to represent the scenario and we will also be looking at both an angle of depression and an angle of elevation.
Example 2: Using Right Triangle Trigonometry to Solve Word Problems Involving Angles of Elevation and Depression
A building is 8 metres tall. The angle of elevation from the top of the building to the top of a tree is and the angle of depression from the top of the building to the base of the tree is . Find the distance between the base of the building and the base of the tree giving the answer to two decimal places.
The first step in solving this problem is to draw a sketch of the problem that is labeled with all the information that we have been given.
In the sketch, we have labeled the distance that we are trying to find as . Now, from this sketch, we can extract a right triangle to help us calculate this distance.
As we now have a right triangle and are looking for a missing side, we can use trigonometric ratios to find .
First, we label the hypotenuse, the opposite, and the adjacent.
Now, to choose the correct ratio, we look at the side we have been given and the one that we want to find. In our problem, these are the opposite and the adjacent. Using this information, we select our ratio:
Next, we substitute in the known values and then solve to find :
Finally, as asked for in the question, we round to two decimal places to give the distance between the base of the building and the base of the tree: 5.00 m.
In the next question, again we use right triangle trigonometry; however, in this example, we will need to use problem-solving skills to find the distance required.
Example 3: Using Right Triangle Trigonometry to Solve Word Problems Involving Angles of Elevation
The height of a lighthouse is 60 metres. The angles of elevation between two boats in the sea and the top of the lighthouse are and respectively. Given that the two boats and the base of the lighthouse are colinear and that the boats are both on the same side of the lighthouse, find the distance between the two boats giving the answer to the nearest metre.
The first step in solving this problem is to draw a sketch of the problem labeled with all the information that we have been given.
Now, as we are told that the two boats and the base of the lighthouse are colinear, we can create two right triangles that can enable us to find the distance between the two boats, which we have labeled .
The next step is to use trigonometric ratios to calculate the distance between each boat and the lighthouse. Once we have these distances, we can calculate the difference, and this will give us the distance between the two boats.
We will begin with the boat that is further away. First, we label the hypotenuse, the opposite, and the adjacent.
Now, to choose the correct ratio, we look at the side we have been given and the one that we want to find. In our problem, these are the opposite and the adjacent. Using this information, we select our ratio:
Next, we substitute in the known values and then solve to find (the distance from the lighthouse):
At this stage, we will leave in this form to maintain accuracy. Now, we will find the distance of the boat closer to the lighthouse in the same way:
We now have the distances of both boats from the lighthouse; therefore, we can calculate the distance between them as
If we subtract from , we get
Therefore, we can say that the distance between the two boats, to the nearest metre, is 34 m.
In our next example, we are going to look at a problem where we are given two angles of depression and we need to use the sine rule to find the solution. Before we start, let us quickly recall the sine rule.
Formula: The Sine Rule
Consider the given triangle.
We have
Example 4: Finding an Unknown Height by Forming and Solving a System of Equations
A tower is 33 metres tall. The angle of depression from the top of a hill to the top of the tower is . The angle of depression from the top of the hill to the bottom of the tower is . Find the height of the hill given the bases of the hill and the tower lie on the same horizontal level. Give the answer to the nearest metre.
To help us solve this problem, we are going to draw a sketch of the problem labeled with all the information that we have been given. We have also labeled some points with , , , and .
From the diagram, we can see that what we are trying to find is the length of . In order to do this, we need to find the missing length: .
Since is horizontal and is vertical, we know that . Now, looking at triangle , we can see that we have two of the angles, so
Also, we have that
Let us now focus on triangle .
We can find the length of using the sine rule. The sine rule tell us that
Substituting in what we know and rearranging, we can find the length of :
Now, let us look at triangle .
Since this is a right triangle, we can apply the trigonometric ratio . Using , , and , we find the value of to be
Finally, to find the height of the hill, we need to add to 33
Therefore, we can say that the height of the hill to the nearest metre is 62 m.
In our final example, we will demonstrate how to apply this procedure to a function involving the natural logarithm.
Example 5: Using Angles of Elevation to Solve a Real-World Problem
The angle of elevation of the top of a hill from its base is . A man climbs the hill from that point at an angle of to the horizontal for a distance of 340 metres. His path then continues to the top at an angle of to the horizontal. Find the height of the hill to the nearest metre.
Let us start by drawing a sketch of what this question describes.
In this diagram, we can see that the base of the hill where the man starts climbing is at , the top of the hill is at , the height of the hill is , and the point where the angle of the path changes is at .
By looking at the angles at , we can see that . Next, let us consider the triangle outlined in green below.
Since the angles in this triangle must sum to , we have that . Similarly, if we consider triangle , we can see that
Using the two angles we have found at , we are now able to say that
Next, we will consider triangle .
Since we already know two angles in this triangle, we find that . We are now able to apply the sine rule to this triangle to find the length of . In doing this, we find
Finally, if we consider the right triangle , we will be able to find using trigonometric ratios.
From the diagram, we can see that
Rearranging this for , we get our solution, which is that the height of the hill is to the nearest metre.
We will finish by recapping the key points from this explainer.
Key Points
• An angle of elevation is the “upward” angle from the horizontal to a line of sight from the object to a given point, whereas an angle of depression is where the angle goes “downward” from the horizontal to a given point.
• Drawing a sketch of the given scenario is often a good place to start when solving problems of angles of elevation and depression.
• We can use the trigonometric ratios to solve simple problems of angles of elevation and depression.
• Sometimes, we will require the law of sines or the law of cosines in order to solve problems of angles of elevation and depression. |
# How is pythagorean theorem used in architecture?
Pythagorean theorem is used in architecture for many different things. For example, when determining the height of a building, architects will use the Pythagorean theorem to calculate the diagonal of the structure. This theorem is also used when designing staircases, as the length of the staircase must be calculated using the Pythagorean theorem in order to ensure that it is safe and stable.
Summary
The Pythagorean Theorem is used in architecture to determine the length of a hypotenuse of a right triangle.
## What is the Pythagorean Theorem and how can it be used in construction?
A carpenter will use the Pythagorean Theorem when finding the rafter length of a building. The rafter length is the hypotenuse or the diagonal. To determine the rafter length the carpenter will look on the floor plan to get the run and total rise measurements.
The Pythagorean theorem is a statement in mathematics that states that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. This theorem is represented by the equation: a^2 + b^2 = c^2. The theorem is named after the Greek mathematician Pythagoras, who is credited with discovering it.
The Pythagorean theorem has a number of real-life applications. It is used in construction and architecture, for example, when working out the dimensions of right angled triangles. It is also used in two-dimensional navigation, to find the shortest distance between two points. Additionally, the theorem can be used to survey the steepness of the slopes of mountains or hills.
### What does Pythagorean Theorem help us in the field of math and construction
The Pythagorean Theorem is a statement in mathematics that states that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. This theorem is incredibly important as it provides the basis for more complex trigonometry theories. It is also a very useful tool in everyday life, as it can be used to determine the side lengths of a right angled triangle.
The triangle congruence is a very important tool for architects. It helps them to measure the forces applied on the building and to make sure that the building is stable and will not collapse. Additionally, the triangle congruence can help the architect to see if the different parts of the building match up correctly.
## How do you square a wall with the Pythagorean Theorem?
A thing is just gonna draw a simple arch. Like that, now I exaggerated these, normally you’d make them a bit more subtle. But this is just to show you the idea. So, an arch is just two curves that meet in the middle. And you can make all sorts of different arches.
The Pythagorean Theorem is a formula that is used to calculate the length of the hypotenuse of a right triangle. This formula is used regularly by many different types of managers in order to complete their daily tasks. Computer and information systems managers use this formula to calculate the length of cables and wires, construction managers use it to determine the size of buildings and engineering and natural sciences managers use it to calculate the dimensions of objects. This theorem is an essential tool for managers in many different industries.
## How are triangles used in architecture?
Trusses are structures that are made up of triangles. They are used in many different types of structures, such as roofs, bridges, and buildings. Trusses are very strong, and can support a lot of weight. Bridges that use trusses are called truss bridges.
I completely agree that math ability should not be a deciding factor in whether or not someone can study architecture. However, it is important to be good at math to be able to deal with all of the dimensions, quantities, area, and volume. This type of thinking plays into spatial thinking and patterns, both of which are important for architects.
### Why most architects say that triangles are the strongest shape
The triangle is one of the most basic and strong shapes in geometry. It is strong because it has a base, which is the shortest distance between two points, and because the angle between the two sides is less than 90 degrees. The triangle is common in all sorts of building supports and trusses because it is so strong. The overall shape of many bridges is in the shape of a catenary curve, which is a type of triangle.
To make cardboard beads, you will need:
-Cardboard
-A straw
-Scissors
-Glue
First, cut your cardboard into thin strips. Then, use your straw to poke a hole through the center of each strip. Next, cut each strip into small pieces, about 1 inch long. Once all of your strips are cut, it’s time to start threading them onto your straw. Start by threading on one bead, then add a drop of glue to the end of the strip. Continue threading on beads and adding glue until you have reached the end of the straw. Allow the beads to dry for a few hours before wearing them.
## How do you use the Pythagorean Theorem to find the height of a building?
Remember that the Pythagorean theorem states that A squared plus B squared equals C squared. This is a helpful way to remember how to find the hypotenuse of a triangle when you know the lengths of the other two sides.
The 3:4:5 rule is a great way to ensure that you get a perfectly square corner every time. Simply measure out a three-foot length on your straight line, a four-foot length on your perpendicular line, and a five-foot length across. If all three measurements are correct, you’ll have a perfectly square corner.
### What are two other fields where Pythagoras made significant contributions
Pythagoras was an extremely important figure in the development of mathematics, astronomy, and music theory. The Pythagorean theorem was actually known to the Babylonians 1000 years before Pythagoras, but he may have been the first to prove it. Pythagoras’s contributions to these fields have been absolutely essential in shaping the way we understand them today.
Pythagoras is one of the most famous and renowned mathematicians of ancient times. He is credited with many mathematical and scientific discoveries, including the Pythagorean theorem, Pythagorean tuning, the five regular solids, the Theory of Proportions, the sphericity of the Earth, and the identity of the morning and evening stars as the planet Venus. Pythagoras was a true pioneer in the world of mathematics and science, and his contributions have had a lasting impact on both disciplines.
## What is the most important contribution of Pythagoras?
Pythagoras was a Greek philosopher and mathematician who lived in the 6th century BC. He is best known in the modern day for the Pythagorean Theorem, a mathematical formula which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.
Triangles are often used in construction because they are very stable and rigid shapes. When used correctly, triangles can help to create a strong and sturdy structure. However, it is important to note that triangles can also be dangerous if they are not used properly. If a triangle is not constructed correctly, it can collapse and cause serious damage or injury.
## Final Words
The Pythagorean theorem is used in architecture in a variety of ways. One way is to calculate the length of a diagonal beam in a structure. This is often used in the construction of roofs. Another way the Pythagorean theorem is used in architecture is to determine the size and placement of windows in a building.
Pythagorean theorem is used in architecture to calculate the length of a diagonal in a rectangle.
Jeffery Parker is passionate about architecture and construction. He is a dedicated professional who believes that good design should be both functional and aesthetically pleasing. He has worked on a variety of projects, from residential homes to large commercial buildings. Jeffery has a deep understanding of the building process and the importance of using quality materials. |
Assignment 2nd of Feb. 2012
Evaluate the assumption that “young king” is a collocation in the novel The House of Pomegranate:
See Manning and Schütze (1999) for some more details.
Use Antconc to analyze HOPG.txt as described in the “Using Antconc: Notes 1” post.
Identify the frequency of “young”, “king” and “young king”.
Map these frequencies on a table like this:
young not young total
king ---- ----
not king ---- ----
total
Assuming that the two words are not accidentally occurring next to each other, our Research Hypothesis will be that: the probability of “young king” occurring in the text is not the product of the probability of “young” and the probability of “king”, that is that the two variables are dependent:
P(young king) ≠ P(young) * P(king)
We formulate the Null Hypothesis that the observation we made about the occurrence of “young king” is just random, and that in fact the two variables are independent.
P(young king) = P(young) * P(king)
Calculate the expectation of the bigrams in the table above, using the independency assumption.
Example:
Assume that we have a text with 36485 tokens. The token “John” occurs 214 times in the text. The token “Smith” occurs 86 times in the text. The bigram “John Smith” occurs 63 times in the text. We fill the table this way:
John not John total
Smith 63 ---- 86
not Smith ---- ----
total 214 36485
The total number of words that are “not John” is 36485 - 214, the total number of words that are “not Smith” is 36485 - 86. The total number of bigrams with “John” and some other word that is “not Smith” is 214 - 63, the total number of some other word “not John” followed by “Smith” is 86 - 63. Finish the table of observations.
To calculate the expectations given the Null Hypothesis, we need to calculate the probability of for example “John” and “Smith”, that is, the expectations for randomly picking “John Smith” when picking two words next to each other in the text given the Null Hypothesis would be:
P(John) = 214 / 36485
P(Smith) = 86 / 36485
P(John) * P(Smith) = 214 / 36485 * 86 / 36485
This would return us the probability, but not the absolute number we would expect in the cell with the observation of 63. If we multiply the probability of the product of the individual words with the total number of words, we get back the expected number of concurrencies of “John Smith”, that is:
expected frequency of “John Smith” = 214 / 36485 * 86 / 36485 * 36485 = (214 * 86) / 36485
This we can reinterpret as: For every cell we multiply the total of the column with the total of the row and divide by the overall total.
Apply this for every cell, note the expectations given the Null Hypothesis, then apply the Chi2 formula to all cells observation and expectation values.
Note: In this case the degree of freedom (df) is 1. |
# Adding fractions with like denominators
#### Everything You Need in One Place
Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.
#### Learn and Practice With Ease
Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.
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0/3
##### Intros
###### Lessons
1. Simplify fractions: Method A - By using greatest common factors
2. Simplify fractions: Method B - By using common factors
0/11
##### Examples
###### Lessons
1. $\frac{2}{5}+\frac{3}{5}$
2. $\frac{3}{4}+\frac{2}{4}$
3. $\frac{2}{9}+\frac{1}{9}$
2. Word Problems Involving One-digit Fractions
Laura and Mary are making bread for their grandchildren. They have one bag of flour. Laura needs $\frac{2}{4}$ of the bag and Mary needs $\frac{3}{4}$ of the bag. Do they have enough flour to make the bread? Show your work.
1. Abby and her 2 friends are each making one necklace. They are sharing one package each of glittery beads, clear beads and blue beads. For each necklace, they need $\frac{1}{6}$ of a package of glittery beads, $\frac{3}{6}$ of a package of clear beads and $\frac{2}{6}$ of a package of blue beads. Do they have enough beads? Show your work.
1. $\frac{12}{84} + \frac{18}{84}$
2. $\frac{82}{198} + \frac{113}{198}$
3. $\frac{169}{475} + \frac{137}{475}$
4. $\frac{173}{198} + (-\frac{101}{198})$
5. $\frac{221}{345} + \frac{124}{345}$
6. $\frac{13}{140} + \frac{5}{140} + \frac{10}{140}$
0%
##### Practice
###### Topic Notes
In this section, we will add fractions with like denominators using addition statements. When adding fractions with like denominators, the numerators are added together to get the sum of the parts; however, the denominators stay the same. We will write our answers in lowest terms by first finding the greatest common factor (GCF) of both the numerator and denominator and then dividing the numerator and denominator by this GCF. Finally, we will solve word problems involving the addition of fractions with like denominators. We will make these word problems easier to visualize by incorporating diagrams.
In this lesson, we will learn: |
Apothem of an Octagon – Formulas and Examples
The apothem of any polygon is equal to the line that connects the center of the polygon with one of its sides perpendicularly. Using the apothem, we can calculate the area of the polygons in an easier way. We can find a formula for the apothem of an octagon by dividing the octagon into eight congruent triangles and using trigonometry to determine the height of one of the triangles since it is equivalent to the apothem.
Here, we will derive a formula for the octagon apothem using trigonometry. In addition, we will apply this formula to solve some problems.
GEOMETRY
Relevant for
Learning about the apothem of an octagon with examples.
See examples
GEOMETRY
Relevant for
Learning about the apothem of an octagon with examples.
See examples
Formula to find the apothem of an octagon
We can find a formula for the apothem of an octagon using trigonometry. For this, we start by dividing the octagon into eight congruent triangles:
We see that the apothem divides the triangle into two small right triangles. We can see that the height of the triangle is equal to the apothem, so we can find an expression for the apothem using trigonometry. We need to find the measure of the central angle of one of the right triangles.
We know that the central angle of the octagon is equal to 360°. Also, we know that we have 16 small right triangles, so the central angle of each is $latex 360 \div 16=22.5$°:
Now that we have the angle, we can use the tangent function, which tells us that the tangent of the angle is equal to the opposite side over the adjacent side. Therefore, we have:
$$\tan(22.5)=\frac{\text{opposite}}{\text{adjacent}}$$
$$\tan(22.5)=\frac{\frac{s}{2}}{a}$$
$$\tan(22.5)=\frac{s}{2a}$$
Apothem of an octagon – Examples with answers
The octagon apothem formula is used to solve the following examples. Each example has its respective solution, but it is recommended that you try to solve the exercises yourself first.
EXAMPLE 1
An octagon has sides of length 4 m. What is its apothem?
Using the apothem formula with length $latex s = 4$, we have:
$latex a=\frac{s}{2\tan(22.5°)}$
$latex a=\frac{4}{2\tan(22.5°)}$
$latex a=\frac{4}{0.828}$
$latex a=4.83$
The length of the apothem is 4.83 m.
EXAMPLE 2
What is the length of the apothem of an octagon that has sides of length 5 m?
In this case, we have the length $latex s = 5$, so we use the formula with this value:
$latex a=\frac{s}{2\tan(22.5°)}$
$latex a=\frac{5}{2\tan(22.5°)}$
$latex a=\frac{5}{0.828}$
$latex a=6.04$
The length of the apothem is 6.04 m.
EXAMPLE 3
An octagon has sides of length 9 m. What is its apothem?
We can use the apothem formula with length $latex s = 9$. Therefore, we have:
$latex a=\frac{s}{2\tan(22.5°)}$
$latex a=\frac{9}{2\tan(22.5°)}$
$latex a=\frac{9}{0.828}$
$latex a=10.87$
The length of the apothem is 10.87 m.
EXAMPLE 4
An octagon has an apothem of the length of 11.5 m. What is the length of its sides?
Here, we start with the length of the apothem and want to find the length of the sides. Therefore, we use the apothem formula with $latex a=11.5$ and solve for s:
$latex a=\frac{s}{2\tan(22.5°)}$
$latex 11.5=\frac{s}{2\tan(22.5°)}$
$latex 11.5=\frac{s}{0.828}$
$latex s=11.5(0.828)$
$latex s=9.52$
The length of the sides of the octagon is 9.52 m.
EXAMPLE 5
What is the length of the sides of an octagon that has an apothem of length 15 m?
Again, we use the apothem formula with $latex a=15$ and solve for s:
$latex a=\frac{s}{2\tan(22.5°)}$
$latex 15=\frac{s}{2\tan(22.5°)}$
$latex 15=\frac{s}{0.828}$
$latex s=15(0.828)$
$latex s=12.42$
The length of the sides of the octagon is 12.42 m.
Apothem of an octagon – Practice problems
Put into practice the use of the apothem formula to solve the following problems. If you need help with this, you can look at the solved examples above.
An octagon has an apothem of length 31m. What is the length of the sides?
Interested in learning more about octagons? Take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students. |
# What Is Arithmetic With Negative Numbers?
Arithmetic means adding, subtracting, multiplying and dividing numbers.
Arithmetic becomes trickier when it involves negative numbers and needs extra care.
In this lesson, we are going to look at:
Adding a negative number is equivalent to subtracting the (positive) number.
#### An Example Question
What is 2 + -1?
2 + -1 = 2 - (+1) = 2 - 1 = 1
# Subtracting a Negative Number
Subtracting a negative number is equivalent to adding the (positive) number. (2 negatives make a positive!)
#### An Example Question
What is 2 - -1?
2 - -1 = 2 + (+1) = 2 + 1 = 3
# Multiplying By a Negative Number
Multiplying by a negative number gives different results depending on the sign of the number it is multiplied with.
Multiplying a Positive Number With a Negative Number
Multiplying a positive number with a negative number creates a negative number.
#### An Example Question
What is 2 × -2?
2 × -2 = -(2 × 2) = -4
Multiplying a Negative Number With a Negative Number
Multiplying a negative number with a negative number creates a positive number. (2 negatives make a positive!)
#### An Example Question
What is -2 × -2?
-2 × -2 = --(2 × 2) = +(2 × 2) = +4 = 4
# Dividing By a Negative Number
Dividing by a negative number gives different results depending on the sign of the number it divides.
Dividing a Positive Number With a Negative Number
Dividing a positive number with a negative number creates a negative number.
#### An Example Question
What is 4 ÷ -2?
4 ÷ -2 = -(4 ÷ 2) = -2
Dividing a Negative Number With a Negative Number
Dividing a negative number with a negative number gives a positive number.
#### An Example Question
What is -4 ÷ -2?
-4 ÷ -2 = --(4 ÷ 2) = +2 = 2
##### Interactive Test
show
Here's a second test on arithmetic with negative numbers.
Here's a third test on arithmetic with negative numbers.
# Rules For Signs: Addition and Subtraction
Same signs give a plus:
Different signs give a minus:
# Rules For Signs: Multiplication and Division
Same signs give a plus:
Different signs give a minus: |
# Free Ncert Solution for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.1
Fractions and Decimals, Class 7 Ex. 2.1;
New Ncert Solution for Class 7 Maths Chapter 2 Fractions and Decimals Free Solution;
Exercise 2.1
Which of the drawings (a) to (d) show.
(i) 2×(1/5) (ii) 3×(2/3) (iii) 2×(1/2) (iv) 3×(1/4) ;
Solution 1:-
(i) 2×(1/5) represents drawing (d);
(ii) 2×(1/2) represents drawing (b);
(iii) 3×(2/3) represents drawing (a);
(ic) 3×(1/4) represents drawing (c).
Question 2 :- Some pictures (a) to (c) are given below. Tell which of them show:
(i) 3 x (1/5) = 3/5; (ii) 2 x (1/3) = 2/3; (iii) 3 x (3/4) = 2(1/4);
Solution 2:-
(i) 3×(1/5) = 3/5 represents figure (c);
(ii) 2×(1/3) = 2/3 represents figure (a);
(iii) 3×(3/4)= 2(1/4) represents figure (b)
Question 3 :- Multiply and reduce to lowest form and convert into a mixed fraction:
Solution 3:-
Question 4 :- Shade:
(i) 1/2 of the circles in box (a)
(ii) 2/3 of the triangles in box (b)
(iii) 3/5 of the squares in box (c)
Solution 4:-
(i) 1/2 of 12 circles = (1/2) x 12 = 6 circles;
(ii) 2/3 of triangles = (2/3) x 9 = 2 x 3 = 6 triangles;
(iii) 3/5 of squares = (3/5) x 15 = 3 x 3 = 9 squares;
Question 5 :- Find:
(a) 1/2 of (i) 24 ; (ii) 46 ;
(b) 2/3 of (i) 18 ; (ii) 27 ;
(c) 3/4 of (i) 16 ; (ii) 36 ;
(d) 4/5 of (i) 20 ; (iii) 35 ;
Solution 5:-
(a) (i) 1/2 of 24 = (1/2) x 24 = 12 ;
(ii) 1/2 of 46 = (1/2) x 46 = 23 ;
(b) (i) 2/3 of 18 = (2/3) x 18 = 2 x 6 = 12 ;
(ii) 2/3 of 27 = (2/3) x 27 = 2 x 9 = 18;
(c) (i) 3/4 of 16 = (3/4) x 16 = 3 x 4 = 12;
(ii) 3/4 of 36 = (3/4) x 36 = 3 x 9 = 27;
(d) (i) 4/5 of 20 = (4/5) x 20 = 4 x 4 = 16;
(ii) 4/5 of 35 = (4/5) x 35 = 4 x 7 = 28;
Question 6 :- Multiply and express as a mixed fraction.
Solution 6:-
Question 7 :- Find:
(a)
Solution 7:-
Question 8 :- Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 2/5 of the water. Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Solution 8:-
Total water in a bottle = 5 liters;
(i) Water consumed by Vidya = 2/5 of 5 litres
= (2/5) x 5 = 2 litres
water drink by Vidya = 2 litres.
(ii) Water consumed by Patap = 5 litres – 2 litres = 3 litres
Fraction of water consumed by Pratap = 1 – (2/5) = (5-2)/5 = 3/5.
Chapter-2, Fractions and Decimals, Class-7 Exercise-2.1 for Free |
# How do you simplify (x+1)/x - x/(x+1)?
May 18, 2018
(2x+1)/(x(x+1)
#### Explanation:
Find the lowest common factor which is $x \left(x + 1\right)$
$\frac{x + 1}{x} = \frac{x}{x + 1}$
$\frac{\left(x + 1\right) \left(x + 1\right) - \left(x \times x\right)}{x \left(x + 1\right)}$
(x^2+x+x+1 - x^2)/(x(x+1)
(cancel(x^2-x^2)+2x+1)/(x(x+1)
(2x+1)/(x(x+1)
May 18, 2018
$\frac{2 x + 1}{x \left(x + 1\right)}$
#### Explanation:
$\text{we require the fractions to have a "color(blue)"common denominator}$
$\text{to obtain this}$
$\text{multiply the numerator/denominator of } \frac{x + 1}{x}$
$\text{by } \left(x + 1\right)$
$\text{and multiply numerator/denominator of "x/(x+1)" by } x$
$= \frac{\left(x + 1\right) \left(x + 1\right)}{x \left(x + 1\right)} - {x}^{2} / \left(x \left(x + 1\right)\right)$
$\text{the fractions now have a common denominator so expand}$
$\text{and subtract the numerator leaving the denominator}$
$= \frac{\cancel{{x}^{2}} + 2 x + 1 \cancel{- {x}^{2}}}{x \left(x + 1\right)}$
$= \frac{2 x + 1}{x \left(x + 1\right)}$
May 18, 2018
$\frac{x + 1}{x} - \frac{x}{x + 1} = \frac{1}{x} + \frac{1}{x + 1}$
#### Explanation:
I am not sure what is meant by simplified here, but we can find:
$\frac{x + 1}{x} - \frac{x}{x + 1} = \frac{x + 1}{x} - \left(\frac{x + 1 - 1}{x + 1}\right)$
$\textcolor{w h i t e}{\frac{x + 1}{x} - \frac{x}{x + 1}} = \left(1 + \frac{1}{x}\right) - \left(1 - \frac{1}{x + 1}\right)$
$\textcolor{w h i t e}{\frac{x + 1}{x} - \frac{x}{x + 1}} = \frac{1}{x} + \frac{1}{x + 1}$ |
# MATH EQUATIONS - EQUATIONS, TYPES, SOLUTION OR ROOT(S), DOMAIN, LINKS TO OTHERS
Please study
Basic Algebra before Math Equations,
if you have not already done so.
There, we introduced the idea of literal number.
We also explained the basic operations like addition,
subtraction, multiplication and division on literal numbers.
We also Explained about Algebraic Expression.
We saw with examples, how to find the value of
an Algebraic Expression by substituting the given
values to the variables involved.
That knowledge is a prerequisite here.
Here, we consider an Algebraic Expression with "=" symbol.
We also discuss the set of values, the variable(s)
can take (replacement set) and the values out of
the replacement set that make the algebraic expression
equal to the R.H.S. of the "=" symbol.
## Equation : Math Equations
Before answering the question what is an equation ?
we need to know about certain terminology.
### Mathematical Sentence : Math Equations
We have seen that the combination of terms, obtained by the
operation of '+' or '-' or both is called an Algebraic Expression.
If the terms are only numerical (without literals),
the expression is called numerical expression.
e.g. 9 + 4, 6 - 3, 15 + 7 - 2, are numerical expressions.
If a numerical expression and a number or two numerical expressions are joined or connected by ' is equal to' ( = ) or 'is greater than' ( > ) or 'is less than' ( < ) etc,they are called Mathematical Sentences.
### Mathematical Statement : Math Equations
Some mathematical sentences are given below.
(i) 8 + 7 = 15. (ii) 9 + 4 > 14 (iii) 16 + 12 < 30 (iv) 3 + 7 ≠ 10
Each of the above sentences can be verified as either true or false.
You can see Sentences (i) and (iii) are true and
Sentences (ii) and (iv) are false.
A Mathematical Sentence that can be verified as either true or falsebut not both is called a Mathematical Statement.
All the four examples given above are Mathematical Statements.
### Equality : Math Equations
A true mathematical statement containing the sign ' is equal to' ( = )is called an Equality.
In the above examples, example (i) is an equality.
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### Open Sentence : Math Equations
So far we have seen Numerical Expressions joined by
' is equal to' ( = ) or 'is greater than' ( > ) or 'is less than' ( < ) etc,
and called them Mathematical Sentences.
Now consider Sentences in which Algebraic Expressions
with literals (variables) connected by
' is equal to' ( = ) or 'is greater than' ( > ) or 'is less than' ( < ) etc.
What to call them?
The sentences containing variables depend upon the value of the
variable for their truth or falsity.
Such sentences may be true for particular
value(s) of the varaible and false for others.
Such sentences are known as open sentences.
Thus,
A sentence which contains a variable such that it may be true or falsedepending on the values of the variable, is called an Open Sentence.
Some Open Sentences are given below.
(i) x + 3 = 9. (ii) y + 2 > 17 (iii) z - 4 < 5 (iv) p + 8 ≠ 6
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## Equation : Math Equations
An Open Sentence containing the sign ' is equal to' ( = )is called an Equation.
In the above examples, example (i) is an equation.
Note that every Equation has two sides, namely
'Left Hand Side' (L.H.S.) and 'Right Hand Side' (R.H.S.).
Thus, in the equation x + 3 = 9, L.H.S. is x + 3 and R.H.S. is 9.
### Solution or Root of an Equation : Math Equations
A number which when replaced for the variable of an equation makes the resulting statement true. i.e., makes its L.H.S. is equal toits R.H.S. is said to satisfy the equation. A number which thus satisfiesan equation is called a Solution or a Root of the equation.
### Solving Equations, Domain of the Variable, Examples
For lucid explanation of the basics of
Solving Equations and the concept of
Domain of the variable with Examples,
and for Links to Solving Linear, Quadratic,
Cubic and Bi-Quadratic Equations, go to
Solving Equations
## Progressive Learning of Math : Math equations
Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.
This series is very different and advantageous
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Other books just give practice.
These teach students “tricks” and new ways to think.
These build a student’s new knowledge of concepts
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These provide many pages of practice that gradually
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The series is low to reasonably priced and include |
# Math 190: Study Guide - Chapter 3
1. Use the rules of differentiation to find and simplify the derivative of the function. These could involve the power rule, addition rule, multiplication rule, quotient rule, and chain rule. Nine parts. Look at the problems in sections 3.1, 3.2, and 3.3.
2. Find the derivative of the function but do NOT simplify the results. Three parts. Look at the same problems as question 1, but these are not as pretty and you can waste a lot of time simplifying and still not have a nice looking answer.
3. Given a table with values for x, f(x), f'(x), g(x), and g'(x), evaluate each of the derivatives asked for. For example, if you are asked to find d/dx [ f(x)g(x) ] |x=2, then you would use the product rule to find f(x)g'(x) + f'(x)g(x) and then since this is evaluated when x=2, you get f'(2)g(2)+f'(2)g(2). Look each of those values up in the table, substitute them into the expression, and evaluate. Five parts.
4. Given a demand and cost function, find the revenue and profit functions; the marginal cost, marginal revenue, and marginal profit functions; evaluate the marginal functions at a particular point and interpret. Finally, find the average cost. Look at problems 3.4.3-17.
5. Find the elasticity of demand. The formula for elasticity of demand is provided on the exam. Then determine the regions where the function is elastic, inelastic, and unitary. Look at problems 3.4.23-34.
6. Use a local linear approximation with a differential to approximate an expression. Look at problems 3.7.19-26.
7. Find the first and second derivatives of the function. Two parts. Look at problems 3.5.1-20.
8. Find the first and second derivatives of a function at a specific point and interpret within the context of the story. Look at problems 3.5.29-38.
## Notes
• Show work if there is any. In a very few cases, you may be able to do the problem in your head.
• You will need your calculator for this exam.
## Points per problem
# Pts 1 2 3 4 5 6 7 8 Total 36 12 15 12 6 5 8 6 100 |
Identify steps for modeling and solving
When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:
1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
5. When needed, write a formula for the function.
6. Solve or evaluate the function using the formula.
7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
8. Clearly convey your result using appropriate units, and answer in full sentences when necessary.
Building Linear Models
Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.
• Output: M, money remaining, in dollars
• Input: t, time, in weeks
So, the amount of money remaining depends on the number of weeks: M(t)
We can also identify the initial value and the rate of change.
• Initial Value: She saved $3,500, so$3,500 is the initial value for M.
• Rate of Change: She anticipates spending $400 each week, so –$400 per week is the rate of change, or slope.
Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.
Figure 2
The rate of change is constant, so we can start with the linear model M(t) = mt + b. Then we can substitute the intercept and slope provided.
To find the x-intercept, we set the output to zero, and solve for the input.
$\begin{cases}0=-400t+3500\hfill \\ t=\frac{3500}{400}\hfill \\ =8.75\hfill \end{cases}$
The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.
When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved$3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is $0\le t\le 8.75$.
In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model. |
# How do you find parametric equations for the line of intersection of two planes 2x - 2y + z = 1, and 2x + y - 3z = 3?
Jul 1, 2016
$\vec{r} = \left(\begin{matrix}\frac{7}{6} \\ \frac{2}{3} \\ 0\end{matrix}\right) + \lambda \left(\begin{matrix}5 \\ 8 \\ 6\end{matrix}\right)$
#### Explanation:
for the line, we will need
A) a point that it actually passes through, say $\vec{a}$, AND
B) a vector describing the direction in which it travels, say $\vec{b}$
....such that the line itself is
$\vec{r} = \vec{a} + \lambda \vec{b} q \quad \square$
For $\vec{a}$, we can simply choose the completely arbitrary point, so here we choose the point at which z = 0
This means that the plane equations, namely
${\pi}_{1} : 2 x - 2 y + z = 1$
${\pi}_{2} : 2 x + y - 3 z = 3$
become
$2 x - 2 y = 1$
$2 x + y = 3$
and these we solve as simultaneous equations to get
$x = \frac{7}{6} , y = \frac{2}{3}$ and of course $z = 0$
so $\vec{a} = \left(\begin{matrix}\frac{7}{6} \\ \frac{2}{3} \\ 0\end{matrix}\right)$
for $\vec{b}$ we need to calculate the vector cross product of the normal vectors for ${\pi}_{1}$ and ${\pi}_{2}$.
In the drawing below, we are looking right down the line of intersection, and we get an idea as to why the cross product of the normals of the red and blue planes generates a third vector, perpendicular to the normal vectors, that defines the direction of the line of intersection.
for a generalised plane $\pi : a x + b y + c z = d$, the normal vector is: $\vec{n} = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$
so for ${\pi}_{1} : 2 x - 2 y + z = 1$, the normal vector is ${\vec{n}}_{1} = \left(\begin{matrix}2 \\ - 2 \\ 1\end{matrix}\right)$
and for ${\pi}_{2} : 2 x + y - 3 z = 3$, the normal vector is ${\vec{n}}_{2} = \left(\begin{matrix}2 \\ 1 \\ - 3\end{matrix}\right)$
the cross product $\vec{b} = {\vec{n}}_{1} \times {\vec{n}}_{2}$ is the following determinant:
$\vec{b} = \det \left[\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & - 2 & 1 \\ 2 & 1 & - 3\end{matrix}\right]$
$= \left(\begin{matrix}5 \\ 8 \\ 6\end{matrix}\right)$
so we combine this all as indicated in $\square$ as follows
$\vec{r} = \left(\begin{matrix}\frac{7}{6} \\ \frac{2}{3} \\ 0\end{matrix}\right) + \lambda \left(\begin{matrix}5 \\ 8 \\ 6\end{matrix}\right)$
There are an infinite number of ways of expressing this line. $\vec{a}$ can be any point on the line. And the direction vector $\vec{b}$ can be any multiple of the one given herein.
The important bit, I guess is the method. |
1. ## Various Derivative Problems
I'm having trouble with these problems not sure where to start. Any help is greatly appreciated.
1.) Find all values of where the tangent lines to and are parallel.
2.) Find an equation for the line tangent to the graph of at the point for .
3.) Calculate for , where (with a,b,c,d the constants)
Thanks again!
2. Hello,jwebb19!
1) Find all values of $x$ where the tangent lines to $y = x^4$ and $y = x^5$ are parallel.
We have: . $\begin{array}{ccc}y' & = & 4x^3 \\ y' & = & 5x^4\end{array}$
Since the slopes are equal: . $5x^4 \:=\:4x^3\quad\Rightarrow\quad 5x^4 - 4x^3\:=\:0\quad\Rightarrow\quad x^3(5x - 4)\:=\:0$
Therefore: . $\boxed{x \:=\:0,\:\frac{4}{5}}$
2) Find an equation for the line tangent to the graph of $f(x) = -4xe^x$at the point $(1,\,f(1)).$
$f(1)\:=\:-4(1)(e^1) \:=\:-4e$ . . . The point is: . $\left(1,\:-4e\right)$
The slope is: . $f'(x)\;=\;-4xe^x - 4x^x\:=\:-4e^x(x+1)$
When $x = 1$, we have: . $f'(1) \:=\:-4e^1(1 + 1) \:=\;-8e$
The equation is: . $y - (-4e) \:=\:-8e(x - 1) \quad\Rightarrow\quad y + 4e \:=\:-8ex + 8e$
Therefore: . $\boxed{y \;=\;-8ex + 4e}$
3) Calculate $y^{(k)}$ .for $0 \leq k \leq 5$
where $y \:=\:9x^4 + ax^3 + bx^2 + cx + d$ (with a,b,c,d constant)
$\begin{array}{ccc}y^{(0)} & = & 9x^4 + ax^3 + bx^2 + cx + d \\
y' & = & 36x^3 + 3ax^2 + 2bx + c \\
y'' & = & 108x^2 + 6x + 2b \\
y''' & = & 216x + 6 \\
y^{(4)} & = & 216 \\
y^{(5)} & = & 0 \end{array}$
3. Thanks a million! I have been trying the following problem to no avail.
If then
I believe f'(x) = [IMG]file:///C:/Users/Jthan/AppData/Local/Temp/moz-screenshot-8.jpg[/IMG] [sec(x)*tan(x)*(x^3)+3*(x^2)*sec(x)]/[(x^3)^2].
Thanks again! |
## Use the parametric equations of an ellipse, x=acosθ, y=bsinθ, 0≤θ≤2π , to find the area that it encloses.
Question
Use the parametric equations of an ellipse, x=acosθ, y=bsinθ, 0≤θ≤2π , to find the area that it encloses.
in progress 0
1 year 2021-07-23T11:38:15+00:00 1 Answers 500 views 0
Area of ellipse=$$\pi ab$$
Step-by-step explanation:
We are given that
$$x=acos\theta$$
$$y=bsin\theta$$
$$0\leq\theta\leq 2\pi$$
We have to find the area enclose by it.
$$x/a=cos\theta, y/b=sin\theta$$
$$sin^2\theta+cos^2\theta=x^2/a^2+y^2/b^2$$
Using the formula
$$sin^2x+cos^2x=1$$
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
This is the equation of ellipse.
Area of ellipse
=$$4\int_{0}^{a}\frac{b}{a}\sqrt{a^2-x^2}dx$$
When x=0,$$\theta=\pi/2$$
When x=a, $$\theta=0$$
Using the formula
Area of ellipse
=$$\frac{4b}{a}\int_{\pi/2}^{0}\sqrt{a^2-a^2cos^2\theta}(-asin\theta)d\theta$$
Area of ellipse=$$-4ba\int_{\pi/2}^{0}\sqrt{1-cos^2\theta}(sin\theta)d\theta$$
Area of ellipse=$$-4ba\int_{\pi/2}^{0} sin^2\theta d\theta$$
Area of ellipse=$$-2ba\int_{\pi/2}^{0}(2sin^2\theta)d\theta$$
Area of ellipse=$$-2ba\int_{\pi/2}^{0}(1-cos2\theta)d\theta$$
Using the formula
$$1-cos2\theta=2sin^2\theta$$
Area of ellipse=$$-2ba[\theta-1/2sin(2\theta)]^{0}_{\pi/2}$$
Area of ellipse$$=-2ba(-\pi/2-0)$$
Area of ellipse=$$\pi ab$$ |
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• Level: GCSE
• Subject: Maths
• Word count: 1362
# Number Stairs Investigation &#150; Course Work
Extracts from this document...
Introduction
Number Stairs Investigation – Course Work
Aim
The aim of this coursework is to find relationships and patterns in the total of
all the numbers in ‘Number Stairs’ such as the one below. For example, the total
of the number stair shaded in black is:
25+26+27+35+36+45 = 194
I have to investigate any relationships that might occur if this stair was in a
different place on the grid.
Part One
For other 3-stepped stairs, investigate the relationship between the stair total
and the position of the stair shape on the grid.
By looking at this grid, the stair total is:
25 + (25+1) + (25+2) + (25+3) + (25+10) + (25+11) + (25+20)
If we call 25, the number in the corner of the stair, ‘n’ then we get:
n + (n + 1) + (n + 2) + (n + 3) + (n + 10) + (n + 11) + (n + 20) =
6n + 1 + 2 + 3 + 10 + 11 + 20 =
6n + 44
This formula should work with every number stair that can fit onto a 10 by 10
grid. I can say the total for square n (Tn) is, “n + n + 1 + n + 2 + n + 3 + n +
10 + n + 11 + n + 20”, because where ever n is on the grid, the number of the
square:
· One place right of it will be n +1
· Two places right of it will be n +2
Middle
· 6 x 23 + 44 = 138 + 44 = 182
If n = 34 then
· 6 x 45 + 44 = 270 + 44 = 314
If n = 56 then
· 6 x 56 + 44 = 336 + 44 = 380
If n = 67 then
· 6 x 67 + 44 = 402 + 44 = 446
If n = 78 then
· 6 x 78 + 44 = 468 + 44 = 512
Stair Number (n) 1 12 23 34 45 56 67 78
Stair Total (Tn) 50 116 182 248 314 380 446 512
6n + 44 50 116 182 248 314 380 446 512
All the results using the formula are correct, so I can come to the conclusion
that the formula for the total of a stair:
· On a 10 by 10 grid
· Which travels downwards from left to right
· With a height of 3 squares
Is:
6n + 44
Part Two (Extension)
Investigate further the relationship between the stair total and other step
stairs on the number grids.
I am going to investigate patterns and relationships in:
· The position of the stairs
· The height of the stairs
· The width of the grid
I will then put everything together and produce a universal formula.
Throughout this section, the symbols for the variable inputs will be as follows:
Tn = Total of stair
n = Stair number (the number in the bottom left had corner of the stair)
h = Number of squares high the stair is
w = Width of grid (number of squares)
Relationships between different stair heights on a 10 by 10 grid
Conclusion
3rd triangle would multiply n). The formula for triangle numbers is:
h2 + h
2
So the first part of the formula will be:
n (h2 + h)
2
If I include this in a table, I may get some better results:
Height (h) 1 2 3 4 5 6
Stair Number (n) 1 1 1 1 1 1
Triangle Number of h 1 3 6 10 15 21
Triangle Number of h x n 1 3 6 10 15 21
11h3 – 11h 0 66 264 660 1320 2310
(11h3 11h)ק6 0 11 44 110 220 385
(Triangle Number of h x n) + (11h3 11h)ק6 1 14 50 120 235 406
Total (T1) 1 14 50 120 235 406
I included 11h3 11h from my last table. I divided 11h3 11h by 6. This is
because 6 was the number that got 11h3 11h closest to the required total. I
noticed that if you add the Triangle Number of h x n and (11h3 11h)ק6
together you get the Total. So my formula for any height Number stair on a 10 by
10 grid is:
n (h2 + h) + (11h3 11h)
2 6
Testing the formula
Tn = Total of stair
n = Stair number (the number in the bottom left had corner of the stair)
h = Number of squares high the stair is
w = Width of grid (number of squares)
Test One – Square 25, Height 3
Total = 25 + 26 + 27 + 35 + 36 + 45 = 194
T25 = 25 x ((32 + 3) ק 2) + ((11 x 33 11 x 3) ק 6)
T25 = 25 x (11 ק 3) + (297 33) ק 6
T25 = 25 x 6 + 44
T25 = 194
Test Two Square 68, Height 3
Total = 68 + 69 + 70 + 78 + 79 + 88 = 194
... thats all ive don
This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.
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# Related GCSE Number Stairs, Grids and Sequences essays
1. ## Number stairs
if it is a 10 x 10 [10] or 11 x 11 [11] represented as algebra to give us the total of the numbers added together To start my investigation for the general formula I need to establish the highest common factors, by using our values above, which are 36,
2. ## Number Grids Investigation Coursework
7 11 12 13 14 15 19 20 21 22 23 27 28 29 30 31 (top right x bottom left) - (top left x bottom right) = 7 x 27 - 3 x 31 = 189 - 93 = 96 So my prediction that I could develop my formula to become D = w (m - 1)
1. ## Step-stair Investigation.
The step stair above is a 5-step stair. This means by counting up all the numbers per row and adding all the triangle numbers minus the last in the series of triangle numbers it will give you the answer to the last two parts of the equation for any step stair on any sized grid.
2. ## Number Grid Investigation.
(Z X (multiple number�)) X (n-1)(d-1) So, if for example we were given a 6 X 4 square taken from a grid with multiples of 4 in it the calculation would be: (10 X 4�) X (6-1)(4-1). Based on my original brainstorm, I have carried out all the different approaches outlined.
1. ## Investigation of diagonal difference.
21 22 31 32 41 42 n n + 1 n + 4G n + 4G + 1 I theorise that every time I increase the width by one, the difference increases by 10. I predict that the 2 x 6 rectangle will have a diagonal difference of 50.
2. ## For other 3-step stairs, investigate the relationship between the stair total and the position ...
56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
1. ## Maths Grids Totals
The formula for the 2 x 4 rectangle is 10(2-1)(4-1) = 10 x 1 x 3 = 30. This is correct. The formula for the 3 x 5 rectangle is 10(3-1)(5-1) = 10 x 2 x 4 = 80. This is also correct.
2. ## number stairs
+44 but this formula only works for a 10 by 10 grid. I will investigate the relationship between the step-total, step-number and the grid size. In the 10 by 10 grid I noticed that the numbers 1, 11 & 21 etc increases by 10.
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Guided Lessons
# A Fair Share of Fractions
Teach your students fair shares with this hands-on resource. Using this lesson plan, your class will create, label, and illustrate sums for fair shares using a folded paper model.
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Students will be able to illustrate sums of fair shares using a paper folded model.
(5 minutes)
• Hand out a plain sheet of paper (cut in a perfect square) to your students.
• Propose this challenge, "Who can fold this page into the least amount of equal parts where each part has less than four sides?"
• After allowing for some thinking and folding time, call on a few students to share their solutions with the whole class.
• Announce that today’s lesson, begins with the models where students folded their squares diagonally once to form two identical triangles!
(10 minutes)
• Tell your students the lesson objective has four parts:
• Fold a paper into equal parts.
• Label each part as a fraction
• Shade in some fraction of the whole.
• Display the teacher prepped poster of Lesson’s Four Steps.
• Demonstrate the process with a square that has been folded in half diagonally once. Write the fraction ½ in each triangular portion, shade in one half, and write “one-half is shaded” on the backside of the square.
• Refer to the square model and explain of the ½ that was labeled and shaded, ‘1’ part (the Numerator) describes what was in isolation, and ‘2’ (the Denominator) describes how many parts all together.
• Explain to your class, "The Fraction, like ½, is how we identify parts of a whole that we look at in isolation, compared to all parts as a whole."
(10 minutes)
• Hand out new sheets of paper (8.5” x 11”) to your students and lead them through folding it into eight equal parts. The result should be a rectangle with ⅛ sized partitions.
• Have your class unfold the paper, label each equal part ⅛. Then, turn and tell a neighbour what the ‘1’ in the numerator, and the ‘8’ in the denominator stand for.
• Allow for the whole class to exchange in sharing, and answer any clarifying questions.
• Your class should then shade in 5 of the 8 equal partitions. Ask your students to turn and tell a neighbour the Sum, or the total, shaded parts, and write the sum on the back side. The answer should be ⅛ + ⅛ + ⅛ + ⅛ + ⅛ = ⅝.
• Let students share out their sums, confirming that it is ⅝, stated, ‘five-eighths’.
• Hand out paper strips to your class and lead them through to fold it into 16 equal parts. (The strip, horizontally will be folded once in half horizontally and three times in half, vertically.)
(15 minutes)
• Restate the lesson objective and refer to the poster of four steps.
• Announce to your class they are to follow the lesson steps (refer to the poster), shade in an amount of their choice, and write the equation and sum on the back. Answer any clarifying questions and release students to work.
• Enrichment: Students can write up to three equivalent fractions for shaded and unshaded portions of their models.
• Support: Students can make models fractions less than 18, like fourths or sixths.
• Google Classroom is a great resource in the Google Apps for Education suite, for posting such assignments. It’s also a handy repository for student work!
(5 minutes)
• Present a strip of paper folded in sixths with two partitions shaded.
• Display three optional answers and ask for the correct answer along with a four-step explanation of how to arrive at their choice.
(5 minutes)
• Discuss the following question, "How could a fraction be greater than zero, but less than one?"
• Note relevant new student understandings on a poster for future reference. |
# 2.3
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${\displaystyle f(n)=(((n^{2})(n+1)^{2})/8)+n(n+1)(2n+1)/12}$
This problem does appear to break down into a series of nested summations:
${\displaystyle \displaystyle \sum _{i=1}^{n}{\text{ }}\displaystyle \sum _{j=1}^{i}{\text{ }}\displaystyle \sum _{k=j}^{i+j}{\text{ }}\displaystyle \sum _{l=1}^{j+i-k}1}$
In the last summation, the formula is independent of the iterator, which translates into adding the value 1, ${\displaystyle j+i-k}$ times:
${\displaystyle \displaystyle \sum _{i=1}^{n}\displaystyle \sum _{j=1}^{i}\displaystyle \sum _{k=j}^{i+j}(j+i-k)}$
Now the third summation goes from ${\displaystyle j}$ to ${\displaystyle i+j}$ the formula on closer examination reveals that
${\displaystyle \displaystyle \sum _{k=j}^{i+j}(j+i-k)}$ is ${\displaystyle \displaystyle \sum _{k=1}^{i}(k)}$ which is equal to ${\displaystyle i*(i+1)/2}$
Since ${\displaystyle \displaystyle \sum _{k=start}^{end}(end-k)=\displaystyle \sum _{k=1}^{end-start}(k)}$
So the summation boils down to
${\displaystyle \displaystyle \sum _{i=1}^{n}\displaystyle \sum _{j=1}^{i}(i*(i+1)/2)}$
The formula in the second summation is independent of the iterator, which translates to adding ${\displaystyle i*(i+1)/2}$, ${\displaystyle i}$ times.
${\displaystyle \displaystyle \sum _{i=1}^{n}(i^{2}*(i+1)/2)}$
which is
${\displaystyle \displaystyle \sum _{i=1}^{n}((i^{3}+i^{2})/2)}$
${\displaystyle {\frac {1}{2}}\left(\Sigma r^{3}+\Sigma r^{2}\right)={\frac {1}{2}}\left({\frac {n^{2}\left(n+1\right)^{2}}{4}}+{\frac {n\left(n+1\right)\left(2n+1\right)}{6}}\right)={\frac {n^{2}\left(n+1\right)^{2}}{8}}+{\frac {n\left(n+1\right)\left(2n+1\right)}{12}}}$
Time Complexity = O${\displaystyle ({n}^{4})}$
Back to Chapter 2 |
# The largest side of a right triangle is a^2+b^2 and other side is 2ab. What condition will make the third side to be the smallest side?
Dec 5, 2016
For the third side to be the shortest, we require $\left(1 + \sqrt{2}\right) | b | > \left\mid a \right\mid > \left\mid b \right\mid$ (and that $a$ and $b$ have the same sign).
#### Explanation:
The longest side of a right triangle is always the hypotenuse. So we know the length of the hypotenuse is ${a}^{2} + {b}^{2.}$
Let the unknown side length be $c .$ Then from the Pythagorean theorem, we know
${\left(2 a b\right)}^{2} + {c}^{2} = {\left({a}^{2} + {b}^{2}\right)}^{2}$
or
$c = \sqrt{{\left({a}^{2} + {b}^{2}\right)}^{2} - {\left(2 a b\right)}^{2}}$
$\textcolor{w h i t e}{c} = \sqrt{{a}^{4} + 2 {a}^{2} {b}^{2} + {b}^{4} - 4 {a}^{2} {b}^{2}}$
$\textcolor{w h i t e}{c} = \sqrt{{a}^{4} - 2 {a}^{2} {b}^{2} + {b}^{4}}$
$\textcolor{w h i t e}{c} = \sqrt{{\left({a}^{2} - {b}^{2}\right)}^{2}}$
$\textcolor{w h i t e}{c} = {a}^{2} - {b}^{2}$
We also require that all side lengths be positive, so
• ${a}^{2} + {b}^{2} > 0$
$\implies a \ne 0 \mathmr{and} b \ne 0$
• $2 a b > 0$
$\implies a , b > 0 \mathmr{and} a , b < 0$
• $c = {a}^{2} - {b}^{2} > 0$
$\iff {a}^{2} > {b}^{2}$
$\iff \left\mid a \right\mid > \left\mid b \right\mid$
Now, for any triangle, the longest side must be shorter than the sum of the other two sides. So we have:
$\textcolor{w h i t e}{\implies} 2 a b + \text{ } c \textcolor{w h i t e}{X X} > {a}^{2} + {b}^{2}$
$\implies 2 a b + \left({a}^{2} - {b}^{2}\right) > {a}^{2} + {b}^{2}$
$\implies 2 a b \textcolor{w h i t e}{X X X X X X} > 2 {b}^{2}$
$\implies \left\{\begin{matrix}a > b \text{ & " if b > 0 \\ a < b" & } \mathmr{if} b < 0\end{matrix}\right.$
Further, for third side to be smallest, ${a}^{2} - {b}^{2} < 2 a b$
or ${a}^{2} - 2 a b + {b}^{2} < 2 {b}^{2}$ or $a - b < \sqrt{2} b$ or $a < b \left(1 + \sqrt{2}\right)$
Combining all of these restrictions, we can deduce that in order for the third side to be the shortest, we must have $\left(1 + \sqrt{2}\right) | b | > \left\mid a \right\mid > \left\mid b \right\mid \mathmr{and} \left(a , b < 0 \mathmr{and} a , b > 0\right) .$ |
How to Calculate Percentage
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Education & Reference / How to Calculate Percentage
# How to Calculate Percentage
In the modern world one cannot do without percentages1. They are used almost in every field of activity. Even when you go shopping you see advertisements for discounts of such-and-such percentages. It means that you have to know what a percentage is, and how to calculate it. In addition, you should better learn to convert decimals and fractions to percentages and vice versa, because it can be very useful in many different situations.
So, let us start from the very beginning.
The Meaning of Percentage
Percentage can be easily explained by the phrase "out of a hundred". For example, if twenty four apples out of a hundred that you had bought turned out to be rotten, you can just say that 24% of the apples were putrid. You must admit that it is more convenient and simple.
If you want to write a percentage as a decimal, you should shift the point two figures to the left, like in the picture below.
And vice versa, in order to write a decimal as a percentage, move the point two figures to the right.
How to Calculate Percentages
One can calculate a percentage or convert from it using very simple formulae.
Converting decimals or fractions to percentages requires multiplying by 100.
And in order to convert percentages to fractions, you have to divide them by 100. Then you can reduce them if it is possible.
Percentage Calculations in Everyday Life
We will give you several helpful examples of how to calculate percentages in practice.
• You sell chocolate bars. Their price is .50. You need to raise it 20%. How can you compute the new price? First, calculate one percent of 1.50 by dividing the price by 100. You will get 0.15. Then multiply 0.015 by 20 (you will get 0.3) and add the result to 1.50. The new price will be .8.
• 12 pencils out of 25 are red. How can you convey it in percentage terms? Just multiply the fraction 12/25 by 100. In this case, after cancelling all you have to do is multiply 12 by 4, which means 48%.
• You remember that the tax on a product that you have bought is equal to .00 but you have forgotten how much does it cost without the tax. What should you do to calculate its price? If you have a calculator ready at hand, you can just divide .00 by 15 (in order to calculate what 1% equals to) and then multiple it by 100%. An alternative way of calculating is shown in the following picture.
The Universal Rule of Calculating Percentages
In most cases, you should start by counting what one percent is (dividing by 100) and then use the derived number in your further calculations. Sometimes it is more convenient to find 10% (dividing by 10).
If everything that we have explained above is too difficult for you, just use the calculator with the function of percentage calculation.
You may be interested in: |
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# Multiplication of Fractions
In fractions help , We have several Multiplication Of Fractions examples here , which will help us to understand , how diffetent kind of fractions can be multiplied with each other.
Example for Multiplication Of Fractions
One-quarter of one-seventh of a land is sold for Rs 30,000. What is the value of an eight thirty-fifths of land?
Solution.
One-quarter of one-seventh = $\frac{1}{4}$ × $\frac{1}{7}$ = $\frac{1}{28}$
Now, $\frac{1}{28}$ of a land costs = Rs 30,000
... $\frac{8}{35}$ of the land will cost $\frac{30,000 × 28 × 8}{35}$ = Rs 1,92,000
## Fraction Multiplication Example
After taking out of a purse $\frac{1}{5}$ of its contents, $\frac{1}{12}$ of the remainder was found to be Rs 7.40. What sum did the purse contain at first?
Solution.
After taking out $\frac{1}{5}$ of its contents, the purse remains with $\frac{4}{5}$ of contents. Now, $\frac{1}{12}$ of $\frac{4}{5}$ = Rs 7.40 or, $\frac{1}{15}$ = Rs 7.40 ... 1 = Rs 111
similarly we can do Division Of Fractions as well.
## Example on Fractions
108 + ? $\frac{1}{3}$ + $\frac{2}{5}$ × 3$\frac{3}{4}$ = 10$\frac{1}{2}$
Solution.
Let x be the missing number
= 108 ÷ x of $\frac{1}{3}$ + $\frac{2}{5}$ × 3$\frac{3}{4}$ = 10$\frac{1}{2}$
= 108 ÷ $\frac{1}{3}$x + $\frac{2}{5}$ × $\frac{15}{4}$ = $\frac{21}{2}$
= 3 × $\frac{108}{1}$x + $\frac{3}{2}$ = $\frac{21}{2}$
= 3 × $\frac{108}{1}$x = $\frac{21}{2}$ – $\frac{3}{2}$
= 3 × $\frac{108}{1}$x = 9
= x = 3 $\frac{108}{9}$
x = 36
## Multiplying Fractions Example
This problems is from operations in fractions.
How many $\frac{1}{8}$s are there in 37$\frac{1}{2}$ ?
Solution.
Number of $\frac{1}{8}$’s
= $\frac{75}{2}$ ÷ $\frac{1}{8}$
= $\frac{75}{2}$ × 8
= 300
## Example 1
a person was go multiply a fraction by $\frac{6}{7}$. Instead, he divided and got an answer which exceeds the correct answer by $\frac{1}{7}$. The correct answer was
Solution.
Let x be the fraction
$\frac{7}{6}$x – $\frac{6}{7}$x = $\frac{1}{7}$ ]x = $\frac{6}{13}$
Correct answer = $\frac{6}{7}$x = $\frac{6}{7}$ × $\frac{6}{3}$
= $\frac{36}{91}$
## Example 2
$\frac{? ÷ 12}{0.2 × 3.6}$ = 2
Solution.
Putting x in place of ?
$\frac{1 ÷ 12}{0.2 × 3.6}$ = 2 or, x ÷ 12 = 2 × 0.2 × 3.6
] x = 2 × 0.2 × 3.6 × 12 or,
x = 17.28
## Example 3
?? × 7 × 18 = 84
Solution.
Substituting x for ?, we get
? x × 7 × 18 = 84
or, ?x × 7 = $\frac{84}{18}$ or, (?x × 7)2 = ($\frac{84}{18}$)2
or, x × 7 = $\frac{84 × 84}{18 × 18}$ or, x = $\frac{84 × 84}{18 × 18 × 7}$
= 3.11
## Example 4
2$\frac{3}{1}$ x × y$\frac{1}{2}$ = 7$\frac{3}{4}$, find the values of x and y.
Solution.
Taking the quotient 2, y and 7, we get 2y = 7, which gives the quotient as 3
... y = 3. Substituting the value of y. we get
2$\frac{3}{1}$x × 3$\frac{1}{2}$ = 7$\frac{3}{4}$
Now, [{7$\frac{3}{4}$}{3$\frac{1}{2}$} = 2$\frac{3}{1}$x ] 2$\frac{3}{14}$ = 2$\frac{3}{1}$x
... x = 14, y = 3
## Example 5
A boys was asked to multiply a given number by $\frac{8}{17}$. Instead, he divided the given number by $\frac{8}{17}$ and got the result 225 more than what he should have got if he had multiplied the number by $\frac{8}{17}$.
Solution.
x × $\frac{17}{8}$ – x × $\frac{8}{17}$ = 225
or, $\frac{225}{136}$x = 225
... x = 136
## Example 6
If we multiply a fraction by itself and divide the product by its reciprocal, the fraction thus obtained is 18$\frac{26}{27}$. The original fraction is
Solution.
x × x, $\frac{1}{1}$x = 18$\frac{26}{27}$ or,
x3 = $\frac{512}{27}$
... x3 = ($\frac{8}{3}$)3 and so x = $\frac{8}{3}$
= 2$\frac{2}{3}$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Distributive Property
## The distributive property dictates that when multiplying, the multiplication of real numbers distributes over two or more terms in parentheses. Learn more.
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Distributive Property
Remember Kyle from the Write Numerical Expressions for the Product of a Number and a Sum Concept? Well, he wrote a numerical expression for the situation at the science museum, but he didn't evaluate it, which means that he doesn't have an answer to his teacher's question about cost.
Here is what Kyle wrote.
But there was more to the problem. Kyle also needed to figure out additional costs.
Kyle knows that there is a way to solve this with the Distributive Property, but he can’t remember exactly what to do.
In this Concept, you will learn how to use the Distributive Property to evaluate numerical expressions. Then we'll revisit this problem.
### Guidance
Previously we worked on how to write numerical expressions, and now you are going to learn how to evaluate those expressions.
What does the word “evaluate” mean?
When we evaluate an expression, we figure out the value of that expression or the quantity of the expression.
When we evaluate expressions that have a product and a sum, we use a property called the Distributive Property.
What is the Distributive Property?
The Distributive Property is a property that is a true statement about how to multiply a number with a sum. Multiply the number outside the parentheses with each number inside the parentheses. Then figure out the sum of those products.
In other words, we distribute the number outside the parentheses with both of the values inside the parentheses and find the sum of those numbers.
Let’s see how this works.
To use the Distributive Property, we take the four and multiply it by both of the numbers inside the parentheses. Then we find the sum of those products.
Here is another one.
Multiply the eight times both of the numbers inside the parentheses. Then find the sum of the products.
Now it is your turn. Evaluate these expressions using the Distributive Property.
Solution: 45
Solution: 18
#### Example C
Solution: 60
Now we can take the expression that Kyle wrote and use the Distributive Property to figure out the total amount of money needed for the trip.
Next, we can multiply 22 by 8.95.
Next, we complete the second part of the problem.
2(22) = 44
It will cost the students an additional $44.00 to attend the Omni Theater. The good news is that there is enough money in the student account to help cover the additional costs. There are fifty dollars in the account and the class only needs$44.00 to help cover the costs.
The total amount of money needed is \$240.90.
Kyle gives his information to Mrs. Andersen and she is thrilled! Now the students are off to the Science Museum and the Omni Theater!
### Vocabulary
Numerical expression
a number sentence that has at least two different operations in it.
Product
the answer in a multiplication problem
Sum
Property
a rule that works for all numbers
Evaluate
to find the quantity of values in an expression
The Distributive Property
the property that involves taking the product of the sum of two numbers. Take the number outside the parentheses and multiply it by each term in the parentheses.
### Guided Practice
Here is one for you to try on your own.
Use the distributive property to evaluate this expression.
First, we can distribute the four and multiply it by each value in the parentheses. Then we can add.
### Video Review
This video presents the distributive property from whole numbers to more complicated algebraic expressions.
### Practice
Directions: Evaluate each expression using the Distributive Property.
1. 4(3 + 6)
2. 5(2 + 8)
3. 9(12 + 11)
4. 7(8 + 9)
5. 8(7 + 6)
6. 5(12 + 8)
7. 7(9 + 4)
8. 11(2 + 9)
9. 12(12 + 4)
10. 12(9 + 8)
11. 10(9 + 7)
12. 13(2 + 3)
13. 14(8 + 6)
14. 14(9 + 4)
15. 15(5 + 7)
### Vocabulary Language: English
Evaluate
Evaluate
To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value.
Numerical expression
Numerical expression
A numerical expression is a group of numbers and operations used to represent a quantity.
Product
Product
The product is the result after two amounts have been multiplied.
Property
Property
A property is a rule that works for a given set of numbers.
Sum
Sum
The sum is the result after two or more amounts have been added together. |
Home | Math | Number System-Definition, Types, And Examples
# Number System-Definition, Types, And Examples
September 14, 2022
written by Azhar Ejaz
Number systems are a way of representing numbers using different symbols and are understood by computers. A number is a mathematical value used for counting, measuring objects, and performing arithmetic calculations.
There are various categories of numbers like natural numbers, whole numbers, rational and irrational numbers, etc. Similarly, there are various types of number systems that have different properties. Some examples include the binary number system, the octal number system, the decimal number system, and the hexadecimal number system.
We will explore different types of number systems that we use in this article, such as the binary number system, the octal number system, the decimal number system, and the hexadecimal number system.
Table of Contents
## What is Number System in Math?
A number system is a way to write numbers using symbols. Â In mathematics, set notation is used to represent a given set of numbers. It provides a unique representation for every number in the set, allowing us to perform arithmetic operations such as addition, subtraction, and division. The set notation also reveals the algebraic structure of the numbers, making it a valuable tool for solving mathematical problems.
The value of any given digit in a number can be determined by three factors:
• The digit itself
• Its position in the number
• The base of the number system
But before discussing the different types of number system examples, let’s first define what a number is.
## What is a Number?
A number is a mathematical value used for counting, measuring, or labeling objects. Numbers are used to performing arithmetic calculations. Examples of numbers include natural numbers, whole numbers, rational and irrational numbers, etc. 0 is also a number that represents a null value.
A number can be classified in several ways, such as even and odd, or prime and composite. Even and odd terms are used to describe whether a number is divisible by 2. Prime numbers are those that have only two factors, while composite numbers have more than two factors.
There are different types of number systems that use digits to represent numbers. The most common number system is the binary system, which uses 0 and 1 as digits. Other number systems use 0 to 9 digits.
For example, the common base-10 system uses the ten digits from 0 through 9. In contrast, binary representations use two digits (0 and 1). Other number systems exist that use a different number of digits.
## Types of Number System
There are four main types of number systems:
• The decimal system, which is used in most everyday situations
• The binary system, which is used in computer programming
• The hexadecimal system, which is used in some types of data storage
• The octal system, which is used in some fields of mathematics
### Decimal Number System (Base- 10 Number system)
The decimal number system is the most common one used to represent numbers in everyday life. It uses ten symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 with the base number being 10. If any number is represented without a base specified, it is assumed to have a base of 10.
For example, 10285 has place values as
(1 × 104) + (0 × 103) + (2 × 102) + (8 × 101) + (5 × 100)
1 × 10000 + 0 × 1000 + 2 × 100 + 8 × 10 + 5 × 1
10000 + 0 + 200 + 80 + 5
10285
### Binary Number System (Base 2 Number System)
The binary number system is a numbering system that uses two digits, 0 and 1, to represent numbers. Binary numbers can be used in electronic devices and computer systems because they can be easily represented using just two states, ON and OFF, which correspond to 0 and1. Decimal Numbers 0-9 are represented in binary as: 0, 1, 10, 11, 100, 101, 110, 111, 1000, and 1001
For example:
14 can be written as 1110
19 can be written as 10011
50 can be written as 110010
### Hexadecimal Number System
The hexadecimal system is a way of writing or representing numbers with base 16. In the hex system, the numbers are first represented just like in the decimal system, from 0 to 9. Then, the numbers are represented using the letters A to F. The table below shows the representation of numbers in the hexadecimal system:
### Octal Number System
The octal number system uses eight digits: 0,1,2,3,4,5,6 and 7 with the base of 8. The advantage of this system is that it has fewer digits than several other systems, hence there would be fewer computational errors. Digits like 8 and 9 are not included in the octal number system.
The octal number system is used in minicomputers with digits from 0 to 7. For example, 358, 238, and 1418 are numbers in the octal number system.
## Frequently Asked Question-FAQs
### What is Number System and its Types?
There are many different ways to express numbers, and the number system is just one method. The most common number systems are the decimal number system, binary number system, octal number system, and hexadecimal number system.
### What is Base 1 Number System Called?
The base 1 number system is also known as the unary numeral system and is the simplest way to represent natural numbers.
### Why is the Number System Important?
The number system is a great way to represent numbers using a small symbol set. Computers usually use binary numbers 0 and 1 to simplify calculations and reduce the amount of circuitry needed. This results in less space, energy consumption, and cost.
File Under: |
x2
Properties
+
Equations
Perpendicular Lines
The theory behind perpendicular lines is similar to what we saw with parallel lines. Remember that, in geometry, two lines are perpendicular if they cross each other at a right angle. The algebra way of expressing this isn't quite as straightforward.
In Math . . .
Two lines are perpendicular if they have their slopes are negative reciprocals of each other.
In English . . .
That "negative reciprocal" thing is a little confusing. What it means is that if you have a line then to get the slope of the line perpendicular to it you take the original line's slope, flip it over (that's the reciprocal part) and make it negative. Here's an easier way to think about it: If you multiply the slopes together and you get -1 then the lines are perpendicular.
Example 1
Are the lines $y = 2x + 4$ and $x = -3y - 2$ perpendicular?
To answer this question, we need to know the slopes of the two lines. The first equation is in slope-intercept form so we can immediately see that its slope is going to be 3. Be careful with the second equation - it may look like it's in the slope-intercept form but, if you look closely, you'll see that it's solved for x not y. If we solve it for y we get:
$$x = -3y - 2$$$$-3y = x + 2$$$$y = \frac{-1}{3}x - \frac{2}{3}$$
So the slope of the second line is -1/3. If we multiply the two slopes together we get:
$$2 \cdot \frac{-1}{3}=\frac{-2}{3}$$
That isn't equal to -1 so the lines aren't perpendicular.
Example 3
Are the lines $y=\frac{2}{3}x - 6$ and $y=-\frac{3}{2}x - 2$ perpendicular?
Both of these lines are already in the slope-intercept form so we can see right away that there slopes are 2/3 and -3/2. If we multiply those numbers together, we get:
$$\frac{2}{3}\cdot\frac{-3}{2}=\frac{-6}{6}=-1$$
Because the product equals -1, the lines must be perpendicular.
Example 2
What's the slope of the line perpendicalar to the line y = -4x + 3?
To find the slope of the perpendicular line we have to do two things: flip the orignal slope over and reverse its sign. The reciprocal of -4 is -1/4 and if we reverse its sign we get +1/4 so that's the slope of the perpendicular line.
Example 4
Find the equation of the line that's perpendicular to y = 2x + 4 that passes through the point (3, 5).
The procedure behind this kind of question goes like this: We know a point on the line so, if we know the line's slope, we can use the point-slope form of the equation of a line to find its equation. We can get the slope by taking the negative reciprocal of the slope of the that we're given.
Here's how it works in practice. We know that the slope of the original line is 2. That means that the slope of the perpendicular line is going to be -1/2. (Take a look at Example 2 to see how I worked that out.) Because the line passes through the point (3, 5), we can use the point-slope form of the equation of a line to get its equation.
$$y=-\frac{1}{2}(x-5)$$ $$y-6=-\frac{x}{2}+\frac{5}{2}$$ $$y=-\frac{x}{2}+\frac{5}{2} + 6$$ $$y=-\frac{x}{2}+\frac{5}{2} + \frac{12}{2}$$ $$y=-\frac{x}{2}+\frac{17}{2}$$
Videos
Dyanmic Practice - Perpendicular Lines |
Lesson_Text_2.4
# Lesson_Text_2.4 - 1 Lesson 2.4 Circular Motion 1 Motion in...
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1 Lesson 2.4 Circular Motion 1. Motion in a Circle θ r O A B Fig. 1 In this lesson we will discuss the case of an object moving around a circle of radius r with a uniform speed v. In the figure above, A is the initial position of an object and B is its position after t s. As the object goes round the circle, the line joining the position of the object to the center of the circle describes an angle θ . When the object completes one round, the angle described θ = 2 π . The rate at which θ changes is called the angular speed ( ϖ ) of the object. t θ ϖ = …(i) Angular speed ϖ is measured in rad.s -1 . The time taken by the object to go round the circle once is called the period (T) of the circular motion. The number of times the object goes round the circle is called the frequency (f) of the circular motion. 1 f T = …(ii) Frequency is measured in s -1 which is called a Hertz (Hz) When t = T, equation (i) above can be written as: 2 2 f T π = = ….(iii) If v is the linear speed of the object along the circle, then a distance of 2 π r is covered in a time T, the period. Thus we have: 2 r v r T = = …(iv)
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2 2. Centripetal Acceleration and Centripetal Force. According to the first law of motion, a force is required to change the direction of motion of an object. An object in circular motion continuously changes its direction of motion, and therefore, there must be continuous force acting on it to change its direction of motion. The function of this force is not to change the speed of the object, but to change its direction of motion. When the object goes round a circle, this force is always directed towards the center of the circle. The moment this force ceases, the object will move in a path tangential to the circle. This force that is directed towards the center of the circle to keep the object moving in a circle is called the centripetal force . The velocity of the object in circular motion is always tangential to the circle, which means it is continuously changing. In a uniform circular motion, the speed remains a constant, but the velocity continuously changes due to changes in the direction of motion. There is a change in velocity means that there is an acceleration, and this acceleration is caused by the centripetal force and is called the centripetal acceleration . Fig 2 below illustrates this. velocity v 1 velocity v 2 velocity v 3 centripetal force Fig. 2 velocity is tangential to the circle As the centripetal force is directed towards the center, the centripetal acceleration is also directed towards the center of the circle.
3 Fig 3 below shows the positions and velocities of an object in uniform circular motion at two points A and B. The object in circular motion has a uniform speed v m/s. This means that the magnitudes of the velocity at point A and point B are both the same, namely v m/s. But the velocity at point B is v 1 and the velocity at point B is v 2 . v 1 and v 2 are different because of their different directions. v
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## This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.
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Lesson_Text_2.4 - 1 Lesson 2.4 Circular Motion 1 Motion in...
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Reality Shows Essay
ROTATIONAL MOTION PROBLEMS: 09-1 1) A grinding wheel starts from rest and has a constant angular acceleration of 5 rad/sec2. At t = 6 seconds find the centripetal and tangential accelerations of a point 75 mm from the axis. Determine the angular speed at 6 seconds, and the angle the wheel has turned through. |We have a problem of constant angular acceleration. The figure & coordinate system are |[pic] | |shown. Since a time is given in the problem we must use the equations of motion. | | | | |((t) = (1/2) ( t2 + ( 0 t ; ((t) = ( t + ( 0 | | | | | |The initial state of motion is: ( 0 = 0; ( 0 = 0 . We are given ( in the problem. | |Hence we have the specific equations of motion: | | ((t) = (1/2)(5) t2 ; ((t) = 5 t . Thus at time t = 6 sec, ((6 sec) = (5)(6) = 30 rad/sec . And: ((6 sec) = (1/2)(5)(6)2 = 90 rad . The centripetal & tangential accelerations are linear quantities. Hence, to calculate we need the interconnecting equations. ( s = r (( ; v = r ( ; a t = r ( Where all angular quantities are expressed in terms of angular units of radians. ) Thus, at 6 seconds we have: for velocities for acceleration v(6 sec) = r ( = (. 075)(30) = 2. 25 m/sec a t = r ( = (. 075)(5) = . 375 m/sec2 a c = v2/r = r ( 2 = (. 075)(30)2 = 67. 5 m/sec2 . The magnitude of the total linear acceleration of this point would be given by: [pic] 2) An electric motor operates at 1800 rpm. Find its angular speed in radians/second.
The torque |[pic] | |which produces this rotation must therefore be a clockwise (‘in’) torque. Thus, ( and (( | | |are in the same direction (they both point into the paper). Thus we will have: | | | | | |W(by net ( ) = ( net (( = (150)(1875) = 2. 1 x 105 J. | | | | | |The final kinetic energy (in rotational form) can be calculated thusly: | | KEfinal = (1/2) I ( 2 = (. 5)(25)(150) = 2. 81 x 102 J. We note that this is the same as the work done by the ‘net torque’. This is as expected, since we have: translational: W(by net force) = ( KE(translational) rotational: W(by net torque) = ( KE(rotational) . ) A solid sphere with a radius of 75 mm rolls (without slipping) down an inclined plane that is 5 m long. What is the angular velocity of the sphere at the bottom of the plane if it requires 10 seconds for it to reach the bottom? What is the angle of the plane? |Solution: We note that a time is given in the problem. Hence we must use the |[pic] | |equations of motion. To see where they will enter, we will take a W-E | | |approach to the problem.
The figure is drawn, and we select Ug = 0 at the | | |base. The work energy theorem is: | | | | | |Wby me = ( KE + ( U + Wancf . | |
Since the ball rolls without slipping then no work is done by, or against friction & Wancf ( 0. I clearly do no work, so we have a problem of Conservation of total Mechanical Energy. Thus: KEtop + Utop = M g h = KEbottom + Ubottom = (1/2)M v2 + (1/2) I ( 2 . Note that the ball possesses both types of kinetic energy at the bottom. Thus: M g h = (1/2)M v2 + (1/2)(2/5)M R2(v/R)2 ( g h1 = {(1/2) + (1/5)} v2 = (7/10)v2 . Thus: [pic] . (1) If the angle of the plane were known, then we could calculate ‘h’ and determine the speed at the bottom. 09-4 Associated with the ‘bottom’ we know the given quantity t = 10 seconds. This is where the |[pic] | |equations of motion come into play. Let us construct the equations of motion (translational) | | |for the CM of the ball. Choosing a CS as shown we have: | | | | | |x(t) = (1/2) a t2 ; v(t) = a t . | At t= 10 sec, x (10sec) = 5 m = (1/2) a (10)2, or a = 0. 1 m/sec2 . Thus the linear speed at the bottom is: v = (. 1)(10) = 1 m/sec. Since this is the same as the linear speed of a point on the surface of the sphere (it rolls without slipping), then the angular speed at the bottom is ( = v/r = (1 m/s)/(. 075 m) = 13. 3 rad/sec. Using the result for v in equation (1) above, we calculate ‘h’. h = (7/10) v2/g = (. 7)(1)2/(9. 8) = . 0714 m .
The angle of inclination of the plane is then determined as follows: sin ( = h/(5m) = (. 0714)/(5) = . 01428 From a calculator this gives: ( = 0. 818 degrees. However, we note that since the angle is small, then we can use the approximation: (For small ( – in radians): sin ( ( ( . Thus we could state the answer as: ( ( . 01428 radians. |5) Assume that you are given the following information about the system shown: m1, m2, |[pic] | |M, R, I, ( , ( k . | | | | |When the system is released, block #1 accelerates downward. Apply Newton’s 2nd law to each of| | |the objects in the system in order to develop a set of equations that could be solved | | |yielding the acceleration of the system.
Be certain to identify your ‘known’ and ‘unknown’ | | |quantities. | | 09-5 |Solution: We have a mixed translational and rotational problem. Blocks ‘A’ & ‘B’ |[pic] | |undergo translational motion (1-dim. ), while the pulley (our 3rd real object) | | |undergoes rotational motion.
Let’s start with block ‘A’: | | | | | |For ‘B’: ( Fx: m B g – T2 = m B a (1) | | | | | |For ‘A’: ( Fx: T1 – m A g sin ( – f k = m A a (2) | | | | | |( Fy: N A – m A g cos ( = 0 (3) ; f k = ( k N A (4) | | At this point our list of ‘unknowns’ includes: a, T1, T2, N, f k . We need another equation. |For pulley: |[pic] | | | | |( ( cw: T2 R – T1 R = I ( (5) | | | | | |This gives us a 5th equation, but unfortunately also introduces a ‘new unknown’, ( . | The solution of our difficulty rests in recognizing that the linear motions of the blocks and the rotational motion of the pulley are inter-connected. Hence, we have, for any point on the outside edge of the pulley, ( s = r (( ; v = r ( ; a t = r ( If the rope does not slip on the pulley, then the linear acceleration of any point on the rope is the same as at for the outside edge of the pulley. But the linear acceleration of the rope is the same as that of the two blocks. Thus: a t = a = R ( (6) Unknowns are: a, T1, T2, N, f k, ( .
Thus six equations in six unknowns guarantees a solution. 6) A pulley has mass: m = 1 kg, radius: R = 10 cm, and a moment of inertia about its center of 0. 8 m R2. It has a cord wrapped around it with one end fastened to the ceiling. The pulley wheel is released from rest and falls a distance ‘h’. (a) Consider the motion of the center of mass. Apply Newton’s 2nd law to the translational & rotational motions, and calculate the angular acceleration of the wheel, and the tension in the cord. (b) Find the speed ‘v’ of the CM of the wheel after it has fallen a distance ‘h’ (let ‘h’ = 1 meter). 09-6 |Solution: The figure is as shown, and we have all forces acting on the wheel.
The |[pic] | |motion of the CM is governed by Newton’s 2nd law: | | |( Fy: m g – T = m a (1) | | | | | |The rotation of the object about its CM is governed by the rotational 2nd law: | | | | | |( ( ccw: T R = I ( (2) | |
The two motions are not independent. They are ‘inter-connected’ through the relations: ( s = r (( ; v = r ( ; a t = r ( Here ‘v’ represents the linear speed of a point on the outside of the wheel. This is the same as the linear speed vcm of the CM. The acceleration of the CM, a cm , is the same as a t , the tangential acceleration of a point on the outside rim of the wheel. Substituting for ( in equation (2), then gives: m g – T = m a (a) T R = (. 8) m R2 (a / R) ( T = (. 8) m a (b) Eliminating T gives: m g = (1. 8) m a ( a = g/(1. 8) = 5. 44 m/sec2.
Then ( = a /R = (5. 44)/(. 1) = 54. 4 rad/sec2 . & T = . 444 m g = 4. 35 N. If we construct equations of motion for the CM ( yo = 0; vo = 0), we have: y(t) = (1/2) a t2 = (1/2)(5. 44) t2 ; v(t) = a t = 5. 44 t . When the wheel has ‘fallen’ a distance ‘h’, h = (1/2)(g/1. 8) t’2 . This yields: [pic] Now we note that 1. 8 = 18/10 = 9/5 so that we can write: [pic] . Numerically this gives: 3. 3 m/sec. Consider now this last equation for v(t’). It is somewhat familiar. We recall that if an object ‘falls’ (accel. = g) through a height ‘h’, then its speed is given by [pic] . Where does the factor: (5/9) come from?
Let us calculate the speed via work energy considerations. We have only two forces present: gravity ( Ug; and 09-7 the tension ‘T’. What becomes of the work done by the torque produced by T? This is converted into rotational kinetic energy. Hence the W-E theorem becomes: W by me = ( KE trans + ( KE rot + ( U + W ancf . Thus, since the initial KE = 0, and U = – m g h, then we have: KEfinal = (1/2) m v2 + (1/2) I ( 2 = m g h . Substituting: ( = v/R and I = (. 8) m R2 we have: m g h = (1/2) m v2 + (1/2)(. 8) m R2 (v/R)2 . The M cancels, and we have: g h = (1/2) v2 + (1/2)(8/10) v2 = (1/2){ 1 + 4/5 } v2 = (1/2)(9/5) v2 .
Thus we obtain the factor of 5/9 (rather than 1) since the potential energy in this problem is converted into kinetic energy of both a translational, and a rotational nature. 7) A 25 kg boy stands 2 m from the center of a frictionless playground merry-go-round which has a moment of inertia of 200 kg-m2. If the boy begins to run in a circular path with a speed of 0. 6 m/sec relative to the ground, calculate: (a) the angular velocity of the MGR and (b) the speed of the boy relative to the surface of the MGR. Solution: In order for the boy to acquire a velocity he must be accelerated. The force that accomplishes this is friction. We have a problem in which forces of an unknown magnitude act between the boy & the MGR. To eliminate these ‘unknown’ forces (& resulting unknown torques) we take a systems approach. Initially the system has zero angular momentum. Suppose there is a radial line painted on the |[pic] | |MGR with an ‘X’ where the boy initially is standing. As the boy runs ccw through an angle (, | | |the MGR rotates cw by an angle (. | | | | | |Since there are no other forces but friction which can produce the torque which results in the | | |angular acceleration of the MGR (why can’t boy’s weight contribute? , then by treating the MGR | | |& boy together as one system we eliminate these ‘internal’ torques, and we have conservation of | | |angular momentum. | | L I = L f ( I b ( bi ( I M ( Mi = I b ( bf ( IM ( Mf Now: ( bi = ( Mi = 0 or L bf = – L Mf . Hence: m r2 ( bf = m r2 (v/r) = (25)(2)(. 6) = 30 kg-m2/s . 09-8 Then L Mf = IM ( Mf = 200 ( Mf = 30 . Thus ( Mf = 200/30 = . 15 rad/sec. Consider now the linear speed of a point on the MGR under the boy (that is, at a distance of 2 m from the center). This is: vM = r ( M = (2)(. 15) = 0. 3 m/sec. Since at any instant of time the point directly under the boy has a linear velocity of 0. m/sec in a direction opposite to the direction the boy is moving, then the speed of the boy relative to the MGR is: 0. 6 + 0. 3 = 0. 9 m/sec. We might also ask the question: “When will the boy return to his starting position (the ‘X’)? ” We can solve this using either angular variables or linear variables. Consider the latter. In a time (t the boy will move a distance ( s b = (. 6) (t relative to the ground. In the same time the ‘X’ will move a distance (s X = (. 3)t . The boy will return to the ‘X’ if the sum of these two equals the circular distance around at a radius of 2m. That is, (. 6) (t + (. 3) (t = 2 ( r = (2)(3. 14)(2) ( (t = 13. 96 sec. You will obtain the same result if you used the boy’s velocity relative to the MGR. |
# How do you evaluate the integral int dx/(x^3+1)?
Jan 21, 2017
Use partial fraction expansion and then integrate
#### Explanation:
Partial fraction expansion of $\frac{1}{{x}^{3} + 1}$:
-1 is a root of the denominator, therefore, $\left(x + 1\right)$ is a factor:
color(white)((x +1)/color(black)(x+1)(color(black)((x^2 - x + 1)/("|"x^3 + 0x^2 + 0x + 1))
color(white)(" ")(-x^3 - x^2)/(" "-x^2
$\textcolor{w h i t e}{\text{ ")(+x^2 + x" ")/(" } x + 1}$
$\textcolor{w h i t e}{\text{ }} - x - 1$
Check the discriminant of the quotient:
${b}^{2} - 4 \left(a\right) \left(c\right) = - {1}^{2} - 4 \left(1\right) \left(1\right) = - 3$
There are no real roots, therefore, the partial fractions are:
$\frac{1}{{x}^{3} + 1} = \frac{A x + B}{{x}^{2} - x + 1} + \frac{C}{x + 1}$
Multiply both sides by $\left({x}^{2} - x + 1\right) \left(x + 1\right)$:
$1 = \left(A x + B\right) \left(x + 1\right) + C \left({x}^{2} - x + 1\right)$
Make A and B disappear by letting x = -1:
$1 = C \left({\left(- 1\right)}^{2} - - 1 + 1\right)$
$C = \frac{1}{3}$
$1 = \left(A x + B\right) \left(x + 1\right) + \frac{1}{3} \left({x}^{2} - x + 1\right)$
Make A disappear by letting x = 0:
$1 = B + \frac{1}{3}$
$B = \frac{2}{3}$
Let x = 1:
$1 = \left(A + \frac{2}{3}\right) \left(2\right) + \frac{1}{3}$
$1 = 2 A + \frac{5}{3}$
$A = - \frac{1}{3}$
$\frac{1}{{x}^{3} + 1} = \frac{1}{3} \frac{1}{x + 1} - \frac{1}{3} \frac{x - 2}{{x}^{2} - x + 1}$
Setting up the second term for "u" substitution:
$\frac{1}{{x}^{3} + 1} = \frac{1}{3} \frac{1}{x + 1} - \frac{1}{6} \frac{2 x - 4}{{x}^{2} - x + 1}$
We want $2 x - 1$ in the numerator of the second term, therefore we much create a third term for the remaining -3:
$\frac{1}{{x}^{3} + 1} = \frac{1}{3} \frac{1}{x + 1} - \frac{1}{6} \frac{2 x - 1}{{x}^{2} - x + 1} - \frac{1}{6} \frac{- 3}{{x}^{2} - x + 1}$
Now, the first two terms will integrate to natural logarithms and the last term will be a complete the square integral to become the inverse tangent:
$\frac{1}{{x}^{3} + 1} = \frac{1}{3} \frac{1}{x + 1} - \frac{1}{6} \frac{2 x - 1}{{x}^{2} - x + 1} + \frac{1}{2} \frac{1}{{x}^{2} - x + 1}$
Write each term as an integral:
$\int \frac{1}{{x}^{3} + 1} \mathrm{dx} = \frac{1}{3} \int \frac{1}{x + 1} \mathrm{dx} - \frac{1}{6} \int \frac{2 x - 1}{{x}^{2} - x + 1} \mathrm{dx} + \frac{1}{2} \int \frac{1}{{x}^{2} - x + 1} \mathrm{dx}$
$\int \frac{1}{{x}^{3} + 1} \mathrm{dx} = \frac{1}{3} \ln \left(x + 1\right) - \frac{1}{6} \ln \left({x}^{2} - x + 1\right) + \frac{\sqrt{3}}{3} {\tan}^{-} 1 \left(\frac{2 x - 1}{\sqrt{3}}\right) + C$ |
# Simple math questions I forget how to do? Thank You!?
1. solve for "x"
1=((x+2)/3)-((x-1)/5)
2. solve for "x"
|3x-5|=6
3. Solve the inequality. Write the solution in interval notation.
(5/3)-(1/4x)>(2/3x)-(1/6)
4. Solve the inequality. Write solution in interval notation.
1/2|x-6|-2 ≥ 2
1.
1 = ((x+2)/3) - ((x - 1) /5
Recall : Symmetric property a =b Then b = a where a, b are any numbers
((x+2)/3) - ((x - 1) /5 = 1
Multiply each side by 15
15[((x+2)/3) - ((x - 1) /5 = 1(15)]
Recall : Distributive property a(b + c) = ab + ac
15(x + 2)/3 - 15(x - 1)/5 =15
Simplify
5(x + 2) - 3(x - 1) = 15
Recall : Distributive property
5x + 10 -3x +3 = 15
5x -3x +10 + 3 = 15
2x + 13 = 15
Subtract 13 from each side
2x + 13 - 13 = 15 - 13
2x +0 = 2
2x = 2
Divide each side by 2
2x /2 = 2/2
x = 1.
2.Solve for "x" |3x-5|=6.
|3x-5|=6.
Case1:
3x -5 = 6
3x -5 + 5 = 6 + 5
Simplify
3x +0 = 11
3x = 11
Divide each side by 3
3x/3 = 11 / 3
x = 11 / 3.
Case2 :
3x - 5 = -6
3x -5 + 5 = -6 + 5
3x + 0 = -1
3x = -1
Divide each side by 3
x = -1 / 3.
3.
(5/3)-(1/4x)>(2/3x)-(1/6)
(5/3)-(1/4x)+(1/4x) > (2/3x)-(1/6) + 1/4x
5/3 > 2/3x + 1/4x -(1/6)
Multiply each side by 12
(5/3)12 > [(2/3x) +1/4x - 1/6]12
20 > 8 /x +3/x - 2
20 + 2 > 8/x +3/x - 2 + 2
22 > 11/x - 0
22 > 11/x
Multiply each side by x
22x > [11/x]x
22x > 11
Divide each side by 22
22x/22 > 11/22
x > 1/2
The interval of x is (1/2 , ∞).
The solution of the inequality is x < 0 or x > 1/2
The interval notation from of solution is (-∞, 0) U (1/2, ∞).
4.
1/2|x-6| - 2 ≥ 2
1/2|x-6| - 2 + 2 ≥ 2 + 2
1/2|x-6| - 0 ≥ 4
1/2|x-6| ≥ 4
Multiply each side by 2
2(1/2|x-6|) ≥ 2( 4)
Simplify
|x-6| ≥ 8
Case1:
x - 6 ≥ 8
x - 6 + 6 ≥ 8 + 6
x + 0 ≥ 14
x ≥ 14
x interval is[14,∞]
Case 2 :
x - 6 ≥ -8
x - 6 + 6 ≥ -8 + 6
x + 0 ≥ -2
x ≥ -2
x interval is [-2 ,∞]
Solution in interval : [-2,∞] or [14,∞]
Solution in interval : [-2,∞].
Solution set is {x Є R| x ≤ -2 or x ≥ 14}
Solution in interval notation form (-∞, -2] U [14, ∞).
3) The inequality is
• Step-1
State the exclude values,These are the values for which denominator is zero.
The exclude value of the inequality is 0.
• Step - 2
Solve the related equation
Solution of related equation x = 1/2.
• Step - 3
Draw the vertical lines at the exclude value and at the solution to separate the number line into intervals.
Continuous...
• Step - 4
Now test sample values in each interval to determine whether values in the interval satisify the inequality.
Test interval x - value Inequality Conclusion
(-∞, 0) x = -1 (5/3)-[1/4(-1)] > [2/3(-1)]-(1/6)⇒1.91>0.832 True
(0, 1/2) x = 0.1 (5/3)-[1/4(0.1)] > [2/3(0.1)]-(1/6) -0.83 > -0.10 False
(1/2, ∞) x = 0.8 (5/3)-[1/4(0.8)] > [2/3(0.8)]-(1/6)1.35 > 0.36 True
The above conclude that the inequality is satisfied for all x - values in (-∞, 0) and (1/2, ∞).
This implies that the solution of the inequalityis the interval (-∞, 0) and (1/2, ∞) . as shown in Figure below. Note that the original inequality contains a “less than” symbol. This means that the solution set does not contain the endpoints of the test intervals are (-∞, 0) and(1/2, ∞)
The solution of the inequality is x < 0 or x > 1/2
The interval notation from of solution is (-∞, 0) U (1/2, ∞).
4) The absolute inequality is 1/2|x - 6| - 2 ≥ 2
1/2|x - 6| ≥ 4
Multiply each side by 2.
|x - 6| ≥ 8
|x| ≥ a can be written as x ≥ a or x ≤ - a
|x - 6| ≥ 8 can be written as x - 6 ≥ 8 or x - 6 ≤ - 8
Solve the inequality : x - 6 ≥ 8
x - 6 + 6 ≥ 8 + 6
x ≥ 14
Solve the inequality : x - 6 ≤ - 8
x - 6 + 6 ≤ - 8 + 6
x ≤ - 2
Therefore the solution of the absolute inequality is x ≥ 14 or x ≤ - 2
Solution set is {x Є R| x ≤ -2 or x ≥ 14}
Solution in interval notation form (-∞, -2] U [14, ∞).
Observe the graph , the closed circle means that -2 and 14 are the solutions of the inequality. |
# Georgia Performance Standard: I can apply the identity, commutative, and associative properties of multiplication and verify the results.
## Presentation on theme: "Georgia Performance Standard: I can apply the identity, commutative, and associative properties of multiplication and verify the results."— Presentation transcript:
Applying the IDENTITY, COMMUTATIVE, and ASSOCIATIVE properties of multiplication.
Georgia Performance Standard: I can apply the identity, commutative, and associative properties of multiplication and verify the results.
IDENTITY Property of Multiplication
When you multiply a number by 1, the product is that number. EXAMPLES: 1 X 6 = 6 1 X 8 = 8 1 X 69 = 69 1 X 547 = 547
COMMUTATIVE Property of Multiplication
You can multiply two numbers in any order. The product stays the same. EXAMPLE: 6 X 7 = 42 7 X 6 = 42 7 X 6 = 6 X 7
Commutative Property of Multiplication
3 X 87 = 261 87 X 3 = 261 If 3 X 87 = 261, then 87 X 3 = 261 3 X 87 = 87 X 3
ASSOCIATIVE Property of Multiplication
When you multiply three or more numbers, it doesn’t matter which tow numbers you group or multiply first. The product will stay the same. EXAMPLE: 2 X 4 X = X 4 X 3 (2 X 4) X X (4 X 3) 8 X X 12
ASSOCIATIVE Property (6 X 7) X 3 = 6 X (7 X 3) 42 X 3 = 6 X 21
=
Solve using the identity, commutative, or associative property of multiplication.
7 X 1 = ? What property of multiplication did you use to help you solve the problem?
Solve using the identity, commutative, or associative property of multiplication.
87 X 1 = ? What property of multiplication did you use to help you solve the problem?
Solve using the identity, commutative, or associative property of multiplication.
1 X 258 = ? What property of multiplication did you use to help you solve the problem?
Solve using the identity, commutative, or associative property of multiplication.
(3 X 4) X 6 = 3 X (4 X ) What property of multiplication did you use to help you solve the problem?
Solve using the identity, commutative, or associative property of multiplication.
If 5 X 4 = 20, then 4 X 5 = What property of multiplication did you use to help you solve the problem?
Solve using the identity, commutative, or associative property of multiplication.
(63 X 2) X 12 = 63 X ( X 12) What property of multiplication did you use to help you solve the problem?
What is the missing number?
Model What is the missing number? 21 X 9 = X 21 1 9 21 189
What is the missing number?
Guided Group Practice What is the missing number? 54 X 6 = X 54 324 56 6 1
Guided Individual Practice
What is the missing number? 24 X 2 = 2 X 48 24 2
What is the missing number?
Independent Practice What is the missing number? 36 X 5 = 5 X 1 5 36 180
Model To multiply 7 X 4 X 5, you can multiply 4 X 5 first. What must you multiply by next? 35 28 7 4
Guided Group Practice To multiply 6 X 2 X 10, you can multiply 2 X 10 first. What must you multiply by next? 12 10 6 2
Guided Individual Practice
To multiply 5 X 2 X 4, you can multiply 5 X 2 first. What must you multiply by next? 2 4 5 10
Independent Practice To multiply 3 X 5 X 2, you can multiply 5 X 2 first. What must you multiply by next? 2 3 5 10
Model Which number belongs in the to make the number sentence TRUE?
18 X = 18 A. 0 B. 1 C. 18 D. 36
Guided Group Practice Which number belongs in the to make the number sentence TRUE? 8 X = 24 A. 2 B. 3 C. 162 D. 282
Guided Individual Practice
Which number belongs in the to make the number sentence TRUE? 4 X = 36 A. 4 B. 8 C. 9 D. 144
Independent Practice Which number belongs in the to make the number sentence TRUE? 7 X = 49 A. 7 B. 8 C. 9 D. 343
Model If 8 X 16 = 128, then 16 X 8 = A. 2 B. 16 C. 24 D. 128
Guided Group Practice If 4 X 12 = 48, then 12 X 4 = A. 48 B. 40 C. 4
Guided Individual Practice
If 5 X 15 = 75, then 15 X 5 = A. 5 B. 15 C. 55 D. 75
Independent Practice If 2 X 55 = 110, then 55 X 2 = A. 2 B. 55 C. 110
Model Which of the following does NOT have the same product as 2 X 4 X 7? A. (2 X 4) X 2 B. 2 X (4 X 7) C. (2 X 4) X 7 D. 4 X 2 X 7
Guided Group Practice Which of the following does NOT have the same product as 4 X 8 X 9? A. (4 X 8) X 9 B. 4 X (8 X 9) C. (4 X 4) X 9 D. 4 X 8 X 9
Guided Individual Practice
Which of the following does NOT have the same product as 2 X 5 X 9? A. (2 X 5) X 9 B. 2 X (2 X 5) C. (5 X 2) X 9 D. 9 X 5 X 2
Independent Practice Which of the following does NOT have the same product as 6 X 2 X 7? A. (6 X 2) X 7 B. 6 X (2 X 7) C. (6 X 7) X 2 D. 6 X 7 X 7
Model What is the missing number? 5 X 95 = 95 X A. 5 B. 90 C. 95
Guided Group Practice What is the missing number? 4 X 56 = 56 X A. 224
Guided Independent Practice
What is the missing number? 74 X 36 = 36 X A. 1 B. 36 C. 74 D. 2,664
Individual Practice What is the missing number? 4 X 92 = 92 X A. 4
Model Which is equivalent to the following? 8 X 5 + 5 X 7
a. 8 X X 7 b. 8 X X 5 c. 7 X d X 5 + 7
Guided Group Practice Which is equivalent to the following?
14 X X 12 a b. 14 X X 8 c. 14 X X 8 d X
Guided Individual Practice
Which is equivalent to the following? 7 X X 9 a b. 7 X X 9 c. 7 X d. 8 X X 9
Independent Practice Which is equivalent to the following?
45 X X 3 a. 45 X X 5 b. 7 X X 9 c. 7 X d. 8 X X 9
Lets review How can I apply the identity, commutative, and associative properties of multiplication and verify the results.
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# How to Find the Area of a Trapezoid Part 3
This the third part of a series on finding the area of a trapezoid here in PH Civil Service Review. In the first post, we discussed the derivation of the area of a trapezoid and give a worked example. In the second post, we discussed how to find the area given the base and the height as well as to find the height given the area and the base.
In this post, we are going to find the base, given the height and the area. We continue with the fourth example.
Example 4
A trapezoid has area 65 square centimeters, height 13 cm, and base of 4 cm. Find the other base.
Solution
In this example, we have $A = 65$, $h = 13$ and $a = 4$. We are looking for $b$
$A = \frac{1}{2}h(a + b)$
$65 = \frac{1}{2}(13)(4 + b)$
In equations with fractions, we always want to eliminate the fractions. In the equation above, we can do this by multiplying both sides of the equation by 2. That is,
$2(65) = 2(\frac{1}{2})(13)(4 + b)$.
The product of 2 and 1/2 is 1, so,
$130 = 13(4 + b)$.
Next, we use distributive property on the right hand side. Recall: $a(b + c) = ab + ac$.
$130 = 13(4) + 13(b)$
$130 = 52 + 13b$.
We want to find b, so we subtract 52 from both sides giving us
$78 = 13b$.
Next, we divide both sides by 13
$6 = b$.
So, the other base is 6 centimeters which is our answer to the problem.
Example 5
The figure below is a trapezoid. Find the value of $a$.
Solution
$A = \frac{1}{2}h(a + b)$
$70 = \frac{1}{2}(7) (a + 9)$
We eliminate the fraction by multiplying both sides by 2 to get
$140 = 7(a + 9)$.
Note: It will be shorter if we divide both sides of equation by 7. You might want to try it.
Using the distributive property, we have
$140 = 7(a) + 7(9)$
$140 = 7a + 63$.
Subtracting 63 from both sides, we have
$77 = 7a$.
Dividing both sides by 7, we have
$11 = a$,
So, the other base of the trapezoid is 11 units.
In the next post, we are going to summarize what we have learned from the this series. |
# NCERT Solutions For Class 6 Maths Chapter 1: Knowing Our Numbers
#### BySchool Connect Online
Aug 30, 2020
Class 6 Maths Chapter 1 Knowing Our Numbers NCERT Solutions:
In Chapter 1 Knowing our numbers, we discuss about Comparing Numbers Worksheet, Large Numbers In Practice, Using Brackets and Roman Numerals Chart. You can find the other
## NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers
NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.1
### Class 6 Maths Chapter 1 Ex 1.1
#### Ex 1.1 Class 6 Maths Question 1.
1. Fill in the blanks:
(a) 1 lakh = ………….. ten thousand.
Sol : 1,00,000
(b) 1 million = ………… hundred thousand.
Sol : 10,00,000
(c) 1 crore = ………… ten lakh.
Sol : 1,00,00,000
(d) 1 crore = ………… million.
Sol : 1,00,00,000
(e) 1 million = ………… lakh.
Sol : 1,000,000
#### Ex 1.1 Class 6 Maths Question 2.
2. Place commas correctly and write the numerals:
(a) Seventy three lakh seventy five thousand three hundred seven.
Sol : The numeral of seventy three lakh seventy five thousand three hundred seven is 73,75,307
(b) Nine crore five lakh forty one.
Sol : The numeral of nine crore five lakh forty one is 9,05,00,041
(c) Seven crore fifty two lakh twenty one thousand three hundred two.
Sol : The numeral of seven crore fifty two lakh twenty one thousand three hundred two is 7,52,21,302
(d) Fifty eight million four hundred twenty three thousand two hundred two.
Sol : The numeral of fifty eight million four hundred twenty three thousand two hundred two is 5,84,23,202
(e) Twenty three lakh thirty thousand ten.
Sol : The numeral of twenty three lakh thirty thousand ten is 23,30,010
#### Ex 1.1 Class 6 Maths Question 3.
3. Insert commas suitably and write the names according to Indian System of Numeration:
(a) 87595762
Sol : Eight crore seventy five lakh ninety five thousand seven hundred sixty two
(b) 8546283
Sol : Eighty five lakh forty six thousand two hundred eighty three
(c) 99900046
Sol : Nine crore ninety nine lakh forty six
(d) 98432701
Sol : Nine crore eighty four lakh thirty two thousand seven hundred one
#### Ex 1.1 Class 6 Maths Question 4.
4. Insert commas suitably and write the names according to International System of Numeration:
(a) 78921092
Sol : Seventy eight million nine hundred twenty one thousand ninety two
(b) 7452283
Sol : Seven million four hundred fifty-two thousand two hundred eighty three
(c) 99985102
Sol : Ninety-nine million nine hundred eighty five thousand one hundred two
(d) 48049831
Sol : Forty-eight million forty-nine thousand eight hundred thirty-one
### Class 6 Maths Chapter 1 Ex 1.2
NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.2
#### Ex 1.2 Class 6 Maths Question 1.
1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Sol :
Number of tickets sold on 1st day = 1094
Number of tickets sold on 2nd day = 1812
Number of tickets sold on 3rd day = 2050
Number of tickets sold on 4th day = 2751
Hence, number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707 tickets
2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Sol :
Shekhar scored = 6980 runs
He want to complete = 10000 runs
Runs need to score more = 10000 – 6980 = 3020
Hence, he need 3020 more runs to score
3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Sol :
No. of votes secured by the successful candidate = 577500
No. of votes secured by his rival = 348700
Margin by which he won the election = 577500 – 348700 = 228800 votes
∴ Successful candidate won the election by 228800 votes
4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Sol :
Price of books sold in June first week = Rs 285891
Price of books sold in June second week = Rs 400768
No. of books sold in both weeks together = Rs 285891 + Rs 400768 = Rs 686659
The sale of books is the highest in the second week
Difference in the sale in both weeks = Rs 400768 – Rs 285891 = Rs 114877
∴ Sale in the second week was greater by Rs 114877 than in the first week.
5. Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Sol :
Digits given are 6, 2, 7, 4, 3
Greatest 5-digit number = 76432
Least 5-digit number = 23467
Difference between the two numbers = 76432 – 23467 = 52965
∴ The difference between the two numbers is 52965
6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Sol :
Number of screws manufactured in a day = 2825
Since January month has 31 days
Hence, number of screws manufactured in January = 31 × 2825 = 87575
Hence, machine produce 87575 screws in the month of January 2006
7. A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?
Sol :
Total money the merchant had = Rs 78592
Cost of each radio set = Rs 1200
So, cost of 40 radio sets = Rs 1200 × 40 = Rs 48000
Money left with the merchant = Rs 78592 – Rs 48000 = Rs 30592
Hence, money left with the merchant after purchasing radio sets is Rs 30592
8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
Sol :
Difference between 65 and 56 i.e (65 – 56) = 9
The difference between the correct and incorrect answer = 7236 × 9 = 65124
9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
Sol :
Given
Total length of the cloth = 40 m
= 40 × 100 cm = 4000 cm
Cloth required to stitch one shirt = 2 m 15 cm
= 2 × 100 + 15 cm = 215 cm
Number of shirts that can be stitched out of 4000 cm = 4000 / 215 = 18 shirts
Hence 18 shirts can be stitched out of 40 m and 1m 30 cm of cloth is left out
10. Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Sol :
Weight of one box = 4 kg 500 g = 4 × 1000 + 500
= 4500 g
Maximum weight carried by the van = 800 kg = 800 × 1000
= 800000 g
Hence, number of boxes that can be loaded in the van = 800000 / 4500 = 177 boxes
11. The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.
Sol :
Distance covered between school and house = 1 km 875 m = 1000 + 875 = 1875 m
Since, the student walked both ways.
Hence, distance travelled by the student in one day = 2 × 1875 = 3750 m
Distance travelled by the student in 6 days = 3750 m × 6 = 22500 m = 22 km 500 m
∴ Total distance covered by the student in six days is 22 km and 500 m
12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Sol :
Quantity of curd in the vessel = 4 l 500 ml = 4 × 1000 + 500 = 4500 ml
Capacity of 1 glass = 25 ml
∴ Number of glasses that can be filled with curd = 4500 / 25 = 180 glasses
Hence, 180 glasses can be filled with curd.
### Class 6 Maths Chapter 1 Ex 1.3
NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3
#### Ex 1.3 Class 6 Maths Question 1
1. Estimate each of the following using general rule:
(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Solutions:
(a) 730 + 998
Round off to hundreds
730 rounds off to 700
998 rounds off to 1000
Hence, 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314
Round off to hundreds
796 rounds off to 800
314 rounds off to 300
Hence, 796 – 314 = 800 – 300 = 500
(c) 12904 + 2888
Round off to thousands
12904 rounds off to 13000
2888 rounds off to 3000
Hence, 12904 + 2888 = 13000 + 3000 = 16000
(d) 28292 – 21496
Round off to thousands
28292 round off to 28000
21496 round off to 21000
Hence, 28292 – 21496 = 28000 – 21000 = 7000
Ten more such examples are
(i) 330 + 280 = 300 + 300 = 600
(ii) 3937 + 5990 = 4000 + 6000 = 10000
(iii) 6392 – 3772 = 6000 – 4000 = 2000
(iv) 5440 – 2972 = 5000 – 3000 = 2000
(v) 2175 + 1206 = 2000 + 1000 = 3000
(vi) 1110 – 1292 = 1000 – 1000 = 0
(vii) 910 + 575 = 900 + 600 = 1500
(viii) 6400 – 4900 = 6000 – 5000 = 1000
(ix) 3731 + 1300 = 4000 + 1000 = 5000
(x) 6485 – 4319 = 6000 – 4000 = 2000
2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):
(a) 439 + 334 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365
Make four more such examples.
Solutions:
(a) 439 + 334 + 4317
Rounding off to nearest hundreds
439 + 334 + 4317 = 400 + 300 + 4300
= 5000
Rounding off to nearest tens
439 + 334 + 4317 = 440 + 330 + 4320
= 5090
(b) 108734 – 47599
Rounding off to nearest hundreds
108734 – 47599 = 108700 – 47600
= 61100
Rounding off to nearest tens
108734 – 47599 = 108730 – 47600
= 61130
(c) 8325 – 491
Rounding off to nearest hundreds
8325 – 491 = 8300 – 500
= 7800
Rounding off to nearest tens
8325 – 491 = 8330 – 490
= 7840
(d) 489348 – 48365
Rounding off to nearest hundreds
489348 – 48365 = 489300 – 48400
= 440900
Rounding off to nearest tens
489348 – 48365 = 489350 – 48370
= 440980
Four more examples are as follows
(i) 4853 + 662
Rounding off to nearest hundreds
4853 + 662 = 4800 + 700
= 5500
Rounding off to nearest tens
4853 + 662 = 4850 + 660
= 5510
(ii) 775 – 390
Rounding off to nearest hundreds
775 – 390 = 800 – 400
= 400
Rounding off to nearest tens
775 – 390 = 780 – 400
= 380
(iii) 6375 – 2875
Rounding off to nearest hundreds
6375 – 2875 = 6400 – 2900
= 3500
Rounding off to nearest tens
6375 – 2875 = 6380 – 2880
= 3500
(iv) 8246 – 6312
Rounding off to nearest hundreds
8246 – 6312 = 8200 – 6300
= 1900
Rounding off to nearest tens
8246 – 6312 = 8240 – 6310
= 1930
3. Estimate the following products using general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.
Solutions:
(a) 578 × 161
Rounding off by general rule
578 and 161 rounded off to 600 and 200 respectively
600
× 200
____________
120000
_____________
(b) 5281 × 3491
Rounding off by general rule
5281 and 3491 rounded off to 5000 and 3500 respectively
5000
× 3500
_________
17500000
_________
(c) 1291 × 592
Rounding off by general rule
1291 and 592 rounded off to 1300 and 600 respectively
1300
× 600
_____________
780000
______________
(d) 9250 × 29
Rounding off by general rule
9250 and 29 rounded off to 9000 and 30 respectively
9000
× 30
_____________
270000
______________
## CBSE Notes for Class 6 Maths Free Download for All Chapters
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# Get Data Sufficiency Questions and Answers for SBI PO, CET, IBPS PO, 2021 | Download Maths Data Sufficiency PDF Free at Smartkeeda
Directions: Each of the questions below consists of a question and three statements numbered I, II and III given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read all the statements and give answer:
Important for :
1
A person can purchase three articles in Rs. 49. What is the price of costliest article?
Statement I : The cost price of two articles each is Rs. 1 less than the cost price of costliest article.
Statement II : The cost price of two articles is same.
Statement III: The cost price of costliest article is 6.25% more than the cost price of cheapest article.
» Explain it
A
From the Statement I
Let the CP of each of two cheapest articles = x and the CP of costliest article = x + 1
Then, x + x + x + 1 = 49,
x = 16 therefore the CP of costliest article = 16 + 1 = 17
From the Statement II, we can say that the cost price of two articles is same and from Statement III, we can say that the cost price of costliest article is 6.25% more than the cost price of cheapest article therefore by combining both the statement we can also get our answer.
Hence, option A is correct.
2
Shatabdi Express leaves Patna at 8:00 am for Delhi. At what time will it reach Delhi?
Statement I : For the first 100 km it travels at the speed of 250 km per hour and maintains the same speed during the entire journey.
Statement II : It has 5 stoppages in between Delhi and Patna.
Statement III : Before every stoppages, it covers a same distance of 240 km
» Explain it
E
From the Statement I we can conclude the speed of the train and by combining Statement II and Statement III, we can conclude the distance between Delhi and Patna. But we cannot conclude how long it has stopped at each stoppage because the speed we concluded from the statement I is the speed of the train not the average speed of the entire journey.
Hence, option E is correct.
3
What is the sum of the age of Ram and Mohan?
Statement I : The age of Ram is 6 years more than the age of Mohan.
Statement II : 40% of the age of Mohan is equal to 30% of the age of Ram.
Statement III : The ratio between half of the age of Ram and one third of the age of Mohan is 2 : 1.
» Explain it
C
By combining Statement I and Statement II we can conclude the age of Ram and the age of Mohan . So we can find the sum as well.
Statement II and Statement III indirectly mean the same.
So by combining Statement I and statement III we can get our answer as well.
So either Statement I and II together or Statement I and III together are sufficient.
Hence, option C is correct.
4
In a kilometre race, by how many meters Chandu beats Chand?
Statement I : In a kilometer race, Chandu beats Chandan by 100 meters.
Statement II : The respective ratio of the speed of Chandan and Chand is 4: 3.
Statement III : In a kilometer race, Chandan beats Chand by 150 meters.
» Explain it
C
By combining statement I, and III we can conclude our answer as 250 meters
In the statement II, only ratio of speed is given therefore it is not possible to get our answer only with the help of statement II.
By combining Statement I and II also we can find the answer as the ratio between two people is given and in statement I the distance between the winner and loser is given so we can find the required distance as well.
So Statement I and III together or Statement I and II together are sufficient.
Hence, option C is correct.
5
A metal block of density ‘D’ and mass ‘M’, in the form of a cuboid, is beaten into a thin square sheet of thickness ‘t’, and rolled to form a cylinder of the same thickness. Find the inner radius of the cylinder –
Statement I : Cuboid has dimensions 10cm x 5 cm x 12 cm
Statement II : Thickness ‘t’ = 1.5cm
Statement III : Mass of block, M = 216kg
» Explain it
C
If we have the dimensions, from Statement a,
Volume of cuboid = 10 × 5 × 12 = 600cm3
If thickness is ‘t’ and let side of square sheet be S, then,
600 = (S2) x (t)
If t = 1.5cm is taken from Statement II,
600 = (S2) = 400 1.5
S = 20cm
Height of cylinder = S = 20cm [As square sheet is rolled so the side of the cylinder will be equal to side of square]
Outer circumference = S = 20 cm = 2πr
Or, r = 10 ≈ 3.185 π
Thickness taken, t = 1.5cm
So inner radius = 3.185 – 1.5 = 1.685 cm
Whereas Statement III has no significance anywhere.
But none of the statement alone can answer the question individually.
Hence, answer is using statement I and II together is sufficient
Hence, option C is correct.
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Folks, Data Sufficiency is frequently asked topic in bank po and bank clerk exams, so every aspirant must be aware with it.Aspirants can easily score high by just practicing some important data sufficieny question. At Smartkeeda you will find data sufficieny questions in the form of quizzes and pdf form too.
So you can practice in both modes online and offline.
Here are some examples of data sufficiency for sbi po 2021
Q.1. What is the area of a right-angled triangle ABC, right angled at A?
Statement I : The cir-cum radius of the triangle is 6 cm.
Statement II : Angle ABC = 60 degrees
A. The data in statements I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question.
B. The data in statements II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question.
C. Either Statement I or Statement II alone is sufficient to answer the question.
D. The data in both the statements I and II is not sufficient to answer the question.
E. The data in both the statements I and II together is necessary to answer the question
Q.2. After 5 years, what will the sum of the age of Moni and Soni?
Statement I : 5 years before, Moni was 5 years older than Soni.
Statement II : At present, the ratio of their ages is 5 : 6.
A. The data in statements I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question.
B. The data in statements II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question.
C. Either Statement I or Statement II alone is sufficient to answer the question.
D. The data in both the statements I and II is not sufficient to answer the question.
E. The data in both the statements I and II together is necessary to answer the question.
Guys, Smartkeeda also provide Test series which is called Testzone for Bank and Insurance Exams like SBI PO 2021, IBPS PO, RBI Grade B, NIACL, LIC etc. |
# Common Math Mistakes and Misconceptions (& How to Help)
Math can be a tricky business and kids can end up getting confused. And that's OK - it's a perfectly natural part of learning, but for parents, it can feel pretty frustrating.
The following examples demonstrate some of the most common slips and misconceptions, help you understand why they arise and show how you can help.
Look through the illustrations and talk about them with your child, remembering to use words that empower your child to build a growth mindset, for example, "I know you can," "You can do it!" and "Let's see if we can work it out together."
## Place value
Maisy says she has written the number sixteen.
Is Maisy right? What do you think?
In this example, Maisy has reversed the digits. This isn’t necessarily because she has dyslexia or dyscalculia either. She has just written the number as she has heard it: sixteen.
Try: Write out numbers for your child to say, and say numbers for them to write out. This will help them become familiar with how the different numbers are written.
## Place value 2
Kieran says, “I think this number is 37.”
Do you agree? What do you think?
In this example, Kieran has read the digits as 30 and 7. He hasn’t seen the place value of the 3 as three hundreds, the 7 as seven ones and the zero as the number of tens.
How do you think Kieran would write the number three hundred and nine? Do you think he would write it as 3,009? How can you help him?
If we have no ones, tens, hundreds etc then we must write 0 in the appropriate column. Remember that the position of a digit determines its value.
Try: Write out numbers including a zero and place them in their correct columns marked hundreds, tens and units. Say the numbers and ask children to try to write them down themselves, e.g. one hundred and one, five hundred, one thousand and six etc.
## Decimals
Nishen’s group thinks that 0.67 is bigger than 0.8 “because 0.67 has more digits”.
Are they right?
Talk with your child about why Nishen’s group have got mixed up because they have read the digits after the decimal point as whole numbers. They need to understand that for each place to the right of the decimal point, the numbers are successively smaller by powers of ten.
Try: Write out similar examples comparing decimal numbers of different sizes, e.g. 7.4 and 7.29, 9.72 and 9.8, 0.535 and 0.6 etc
## Money
Betty has two coins in her purse. Ted says the biggest coin is worth the most.
Is he right?
It is a common misconception for younger children to see a bigger object as having more value - the bigger piece of cake is better than the smaller one! So we just need to teach kids that the value of a coin is not directly proportional to its size.
Try: compare the sizes of a variety of different coins and focus on the numerical value of each.
## Fractions
Meghan says “I think ¼ is bigger than ½ because 4 is bigger than 2.
Often, children do not see a fraction as a single quantity but rather see it as a pair of whole numbers, and they apply whole-number thinking by comparing the size of the numbers in the denominators, the numerators, or both.
Try: use a Fraction wall like this one to show the sizes of the fractions. Explain that the number on the bottom means how many parts the whole has been divided into, and the number on the top means how many of those parts are chosen.
## Fractions 2
Liam says that he has divided the shape below into quarters. Is he right?
In this example, Liam has not divided the square into equal parts. This might be because he doesn’t understand that it’s not just enough to have four parts but that they also need to be the same size.
Try: Draw different shapes, e.g. triangle, semicircle, circle, etc and ask children to divide the shapes into equal parts, e.g. ½, ⅓, ¼, etc
## Angles
Gavin and Kim are looking at some angles their teacher has drawn.
Gavin says, “The biggest angle is the one with the biggest arc.”
Kim says, “The angles are all the same size.”
What do you think?
Gavin is a bit confused about what an angle is and thinks that it has something to do with the area in between the two lines. Remember that angle is a measure of turn not the length of the arc.
Some children also confuse the length of the lines with the size of an angle. This happens if they do not understand what the angle is measuring, which is the rotation of the lines.
Try: Lay out a 30cm ruler to mark a starting point, then ask your child to face the direction of the ruler and turn around on the spot. (Encourage them to make a turn of less than 180 degrees to best demonstrate this point). When they've turned they can use another ruler to mark where they finished facing. Then do the same exercise from the same point using a 100cm piece of wool or string as your start and end point to show that it is the rotation that is what is being measured, not the length of the lines illustrating the angle.
## Multiplication by 10
Parvin says that multiplying by 10 is easy because “All you have to do is add a zero on the end.”
Parvin says that if you multiply by 100, you add two zeros and if you multiply by 1,000 you add three zeros.
Sometimes a number can be multiplied by 10 by adding a zero on the end. For example, 8 x 10 = 80. Adding two zeros to a number multiplied by 100 and three zeros to a number multiplied by 1000 also works e.g. 2 x 100 = 200, and 5 x 1,000 = 50,000. However, this trick doesn’t always work. Adding a zero on the end of a decimal number doesn’t change the size of the number. For example, 10.50 is the same as 10.5.
When multiplying by 10 it is much better to think about it as moving all the digits one place to the left so 10.5 becomes 105. The decimal point is like a concrete post – it doesn’t move! Digits move past the decimal point.
Try: asking your child to multiply whole and decimal numbers to practice, letting them use a calculator to check their answers.
## Multiplication
Imogen says, “When you multiply two numbers together they always get bigger.”
It is a very common misconception that multiplication makes things bigger. The word ‘multiple’ itself carries a sense of many or a great number. Children first encounter multiplication in the context of whole numbers, a situation where you mostly end up with a larger number.
But sometimes multiplying numbers together gives a smaller answer, e.g. 6 x ½ = 3. It depends on which numbers are multiplied together. Multiplying by 1 gives the same number, e.g. 16 x 1. Multiplying any number by zero gives zero, e.g. 0 x 200 = 0.
Try: depending on your child’s level, get out your calculator on your phone and show what happens when you multiply by proper fractions (eg ½), improper fractions (eg 3/2), decimals (eg 0.4) and negative numbers (eg -2).
## Division
Wes says, “8 divided by ½ is 4.”
Anesh says, “There are two ½s in 1 so there must be 16 ½s in 8.”
It is a very common misconception that division in maths makes the number smaller. This idea is understandable and a part of a healthy number sense when you’re talking about whole numbers. Division is commonly thought of as sharing but in math there are other meanings as well.
For many children, it seems inconceivable that 8 divided by ½ should give 16 as 16 is bigger than 8. But dividing by a fraction will make the number bigger. The example above could be illustrated by saying: “Cut a pizza into eight equal pieces. Now divide each piece in half. How many pieces of pizza do you have?”
Try: practice similar examples dividing by ½ and use a fraction wall to help. Try dividing a number by 1 – what happens? What about dividing by zero?
## And finally…..
Remember that maths is all about talking and discussing ideas together. Children make sense of their experience, shared or otherwise, on the basis of what they already know. As parents, we have a huge part to play in helping them think things through and see the maths from more than one perspective.
I'm John Dabell, a teacher, writer, and former school inspector. I have written books about numeracy and love helping children see the creativity in math!
About Komodo – Komodo is a fun and effective way to boost K-5 math skills. Designed for 5 to 11-year-olds to use at home, Komodo uses a little and often approach to learning math (15 minutes, three to five times per week) that fits around busy family routines. Komodo helps users develop fluency and confidence in math – without keeping them at the screen for long.
Find out more about Komodo and how it helps thousands of children each year do better at math – you can even try Komodo for free.
And now we've got Komodo English too - check it out here.
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# Aptitude shortcuts formula and solution from the Partnership chapeter
Partnership shortcut math formulas is an essential math chapter for all competitive exams. So we have to give the same effort to Time and works, Interest, Time speeds and distance, average section - we have provided. So here we are discussing the shortcut formula of the Partnership. I hope, you will not feel any difficulty in observing partnership aptitude shortcut methods, because I'm presenting here the partnership chapter with the example and shortcuts formula in easy language.
## Partnership Math concept and tricks
Usually, we call partner when a company or business starting with the investment of two or more persons. When investment amount and time are equal of all partners, then distributes profit or loss equally among them. But when the invested amount different then they get profit or loss according to their ratio of investment, which they invested in the company or business. (This condition will apply when the time period is the same.). But when the time period and investment differences are found, then the ratio unit of a partner is calculated as (investment amount* time) for each partner. This is the basic concept of the Partnership Aptitude. Now we are moving to the shortcut aptitude formula of the Partnership math chapter with the solution.
Suppose, Martyn and John invest \$100 and \$200 respectively, for one year. Then the profit will distribute in the ratio-
100:200 = 1:2
If Martyn invests his money for one year and Jhon invest for two years, then they will distribute their profit in
(Martyn”s share amount*time: Jhon’s share amount* time)
0r (100x1) : (200x2)
or 100: 400
or 1:4
### Solved Partnership Math problems
Aptitude from partnership 1:-
Abdullah and Rambabu started a business by investing 20000 and 30000 rupees respectively. After 6 months Abdulla added more 20000 rupees in their business. At the end of the year, they profit 30,000. Find out the profit of Rambabu at the end of the year?
Easy solution:
Here two partner Abdulla and Rambabu, The ratio of capital invested by them
(Abdulla: Rambabu) = (20000*12 + 20000*6 : 30000*12)
= (240,000+ 120,000 : 360,000)
=360,000:360,000
=1:1
As per their capital ratio, they will distribute the profit. So Raaambau will get 1/2 * 30,000 = 15,000 out of 30,000 rupees profit.
Above we used the traditional method to solve the problem. Now, you can prepare it in a shortcut formula, that is below in an image.
br/>
If we use the above shortcut formula, firstly we have to find the profit ratio. Then we can use this formula directly. Lok at below, how we solve the math using math tricks. Before applying the above formula, we calculate the profit share that is 1:1. So, sums of profit ratio are 1+1=2.
Aptitude from partnership 2:-
Rahul and Babli started a business and profit 60% of their capital. If Rahul got 20,000 of their profit which is more than 10000 profit of Babli, then find out the money which invested by Bubli?
Easy solution:
Profit of Rahul is 20,000 and profit of Bubli is (20,000-10,000)= 10,000
Total profit =(20,000+ 10,000)= 30,000 Rs. And ratio of profit (Rahul:Babli= 2.1)
So let's capital = x
therefore profit 30,000 = x*60/100
30000 = x*0.6
x= 30000/0.6 = 50,000
Bubli invested 50,000*1/3 = 16666.6.. ruppes.
#### More about Partnership Competitive Math
The above question is straightforward and easy, so don't expect that this type of problem usually asks for a competitive examination. We should learn more about competitive exams aptitude and solution. We are going to discuss more type of quantitative aptitude, Shortcuts formula, and solution in the essay and tricky method. Here step by step means few more post or parts. Stay with us for complete partnership aptitude and formula chapter for your upcoming examination. |
# Subtracting Fractions on a Number Line
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Subtracting Fractions on a Number Line
CCSS.MATH.CONTENT.4.NF.B.3.A
## Subtracting Fractions on a Number Line
When we learn about fractions and operations involving fractions we can use many different methods. In this learning text we are going to learn about subtracting fractions with common denominators. We will use one of the visual methods which is called a number line.
## Steps to Subtracting Fractions on a Number Line
We know that a fraction has a top number (numerator) and a bottom number (denominator).
If you would like to know how to subtract fractions on a number line, please look at the steps below:
• Firstly, check if the fractions have common or like denominators.
• Secondly, divide the number line (between 0 and 1) into equal parts and label each part; the denominator is the indicator in how many parts the number will be divided between whole numbers.
• Then look at the first fraction and circle or highlight it on the number line.
• Finally jump from the highlighted fraction to the left as many times as shown by the numerator of the second fraction. Remember to simplify your answer if possible!
## Subtracting Fractions on a Number Line – Example 1
Now that we know all the steps subtracting fractions on a number line we can look at the examples below and follow the process again. In our first example we are subtracting two fractions with a common denominator of six: $\frac{5}{6}$ minus $\frac{3}{6}$. In order to subtract two fractions using the number line, we divide the number line between zero and one into equal parts and label each part. Then we must find the first fraction on our number line and then jump to the left as many times as the numerator of the fraction we are subtracting shows. Our subtracted fraction has a numerator of three, so we must jump backwards three times. We have landed on $\frac{2}{6}$, so $\frac{5}{6}$ minus $\frac{3}{6}$ equals $\frac{2}{6}$.
Now we can simplify our answer. We can simplify $\frac{2}{6}$ to $\frac{1}{3}$ by dividing numerator and denominator by a common factor which is two.
## Subtracting Fractions on a Number Line – Example 2
Let’s look at how you show subtraction operations with fractions on a number line one more time.
This time we have given $\frac{7}{8}$ minus $\frac{6}{8}$. Both fractions share the same denominator which is eight. We are going to repeat the same process as in the previous example.
• We divide the number line between zero and one into equal parts and label each part.
• Then we must find the first fraction on our number line and then jump to the left as many times as the numerator of the fraction we are subtracting shows.
• Our subtracted fraction has a numerator of six, so we must jump backwards six times.
• We have landed on $\frac{1}{8}$, so $\frac{7}{8}$ minus $\frac{6}{8}$ equals $\frac{1}{8}$.
• Now we can try to simplify our answer if possible. This time we cannot simplify further, so our answer is $\frac{1}{8}$.
## Subtracting Fractions on a Number Line – Summary
In this learning text we learned about subtracting fractions using a number line. Let’s look at the steps below for a quick review.
Step # What to do
1. Check that the fractions have the same,
or common denominators.
2. Divide the number line into equal parts
between 0 and 1 as shown by the denominators.
3. Find your first fraction on your number line
and circle or highlight it.
4. Jump from the highlighted fraction to the left
as many times as your second numerator from
your second fraction indicates.
To test your knowledge on subtracting fractions on a number line, have a look at our interactive practice problems, videos and worksheets.
## Frequently Asked Questions on Subtracting Fractions on a Number Line
What are numerators and denominators?
What are common denominators?
How can we find common denominators?
How do you subtract fractions on a number line with the common or like denominators?
### TranscriptSubtracting Fractions on a Number Line
Axel and Tank are leaving the Oyster gas station! "While I drive this thing, Tank, keep an eye on the fuel gauge!" "You got it, partner! I'll track how much gas we use!" Let's help keep track of the gas left in the submarine by subtracting fractions on a number line. We can use a number line like this to help us when subtracting fractions. To subtract fractions on a number line, first, check that the fractions have like, or common, denominators. Five-sixths and three-sixths have the same denominator, so they have like denominators. Next, divide the number line into equal parts between whole numbers as shown by the denominator. Six is the denominator, so make six equal parts between whole numbers and label them, like this. Now find five-sixths on the number line, which is here. Then, identify the numerator of the fraction we are subtracting, which is three. Count three parts backwards from five-sixths. What fraction do we land on? Two-sixths. Five-sixths minus three-sixths is two-sixths. Finally, simplify the answer if possible. To simplify fractions, find a common factor for the numerator and denominator. Two-sixths can be simplified by dividing the numerator and denominator by two, making the fraction one-third. Now that we have looked at the steps needed to subtract fractions on a number line, let's help Axel and Tank calculate how much gas they have left in their submarine! Axel and Tank started their journey with seven-eighths of gas in their submarine, and have used six-eighths of it. With the number line ready, what is the first step? First, check that the fractions have like, or common, denominators. Since both fractions have an eight for the denominator, they have like denominators. What is the next step? Divide the number line into equal parts between whole numbers as shown by the denominator, which is eight, and label each part on the number line. What should we do next? Find the first fraction, seven-eighths, which is here. How do we find the answer? Identify the numerator of the fraction we are subtracting, which is six. Count six parts backwards from seven-eighths. Seven-eighths minus six-eighths is one-eighth. Can one-eighth be simplified? One-eighth cannot be simplified. We leave the answer as one-eighth. While Axel and Tank continue their submarine adventure, let's review! Remember, when subtracting fractions on a number line, first, check that the fractions have like, or common, denominators. Next, divide the number line into equal parts as shown by the denominators. Then, locate the first fraction on the number line. Finally, count backward the number of parts as shown by the numerator of the fraction being subtracted for the answer. Remember to simplify the fraction if you can. "Oh no! I can't believe we ran out of gas!" "It's okay Tank, I will go get some more gas from the Oyster station, you wait here!" "Fine, but please hurry up, I don't know where we are!"
## Subtracting Fractions on a Number Line exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Subtracting Fractions on a Number Line.
• ### How do we subtract fractions on a number line?
Hints
What must be checked before you divide your number line?
Where do we start jumping back from? By what do we jump backwards?
Solution
1. Check the fractions have the same denominator.
2. Divide the number line into equal parts.
3. Locate the first fraction on the number line.
4. Find the numerator of the fraction being subtracted and jump backwards by this many.
5. Don't forget to simplify your answer if needed.
• ### How much gas will Axel and Tank have left?
Hints
Count how many parts the number line is divided into. Use the markers on the number line, and the numbers that are already there, to help you to place the fractions in the correct places.
Some of the fractions have been filled in here, use this number line to help you fill in the rest.
Once you have a complete number line, locate the larger fraction $\frac{6}{7}$, and count back the numerator of the smaller fraction.
Solution
Axel and Tank will have $\mathbf{\frac{2}{7}}$ of a tank left after their journey.
Once we placed each fraction on the number line we needed to:
• Locate the larger fraction from our equation, $\frac{6}{7}$
• Count backwards 4 (the numerator of the smaller fraction, $\frac{4}{7}$)
• As we can see from the jumps, this takes us to $\frac{2}{7}$.
• ### How much chocolate is left?
Hints
The number sentence we are trying to solve is $\frac{9}{10}$ - $\frac{6}{10}$. Can you draw a number line to help?
A number line split into tenths like this will help.
Use the number line to locate the first fraction, $\frac{9}{10}$. What is your next step?
Find the numerator in $\frac{6}{10}$ and count that many backwards.
Solution
Axel will have $\mathbf{\frac{3}{10}}$ of the chocolate bar left.
• Locate the larger fraction, $\frac{9}{10}$, on the number line.
• Find the numerator of the smaller fraction. The numerator in $\frac{6}{10}$ is 6.
• Count 6 jumps backwards.
• We land on $\frac{3}{10}$.
• ### Find the answers to these subtraction problems.
Hints
Look at the denominators of the fractions in the equations. Can you make a number line with this many parts to help you to subtract?
For example, to solve $\frac{7}{10}$ - $\frac{3}{10}$, we could divide a number line into 10 equal parts like this.
Once we have done that, we can locate the first fraction and count back by the numerator of the second fraction.
Remember, answers are simplified if needed. For example, we could simplify $\frac{2}{8}$ to $\frac{1}{4}$ by dividing both the numerator and denominator by 2.
Solution
$\frac{7}{10}$ - $\frac{3}{10}$ = $\frac{2}{5}$
• Counting 3 jumps backwards from $\frac{7}{10}$ on a number line gets us to $\frac{4}{10}$.
• $\frac{4}{10}$ can be simplified to $\frac{2}{5}$ by dividing both the numerator and the denominator by 2.
$\frac{5}{6}$ - $\frac{3}{6}$ = $\frac{1}{3}$
• Counting 3 jumps backwards from $\frac{5}{6}$ on a number line gets us to $\frac{2}{6}$.
• $\frac{2}{6}$ can be simplified to $\frac{1}{3}$ by dividing both the numerator and the denominator by 2.
$\frac{7}{9}$ - $\frac{5}{9}$ = $\frac{2}{9}$
• Counting 5 jumps backwards from $\frac{7}{9}$ on a number line gets us to $\frac{2}{9}$. This cannot be simplified further.
$\frac{8}{12}$ - $\frac{2}{12}$ = $\frac{1}{2}$
• Counting 2 jumps backwards from $\frac{8}{12}$ on a number line gets us to $\frac{6}{12}$.
• $\frac{6}{12}$ can be simplified to $\frac{1}{2}$ by dividing both the numerator and the denominator by 6.
• ### How much gas will be left?
Hints
The first step is to locate the larger fraction, $\frac{5}{8}$, on the number line.
We then need to find the numerator of the smaller fraction. What is the numerator in $\frac{3}{8}$?
We then need to jump backwards that amount.
Solution
They will have $\mathbf{\frac{2}{8}}$ of a tank of gas left when they get home.
• Locate the larger fraction, $\frac{5}{8}$ on the number line.
• Find the numerator of the smaller fraction. The numerator of $\frac{3}{8}$ is 3.
• Count backwards 3 from $\frac{5}{8}$.
• We land on $\frac{2}{8}$.
• ### What fraction will be left?
Hints
Make sure the denominators are the same in each equation.
For example if we were subtracting $\frac{5}{12}$ - $\frac{2}{6}$, we would multiply the numerator and the denominator of $\frac{2}{6}$ by 2 to make $\frac{4}{12}$. Now we can subtract $\frac{5}{12}$ - $\frac{4}{12}$.
Draw your own number line to help you to solve. Divide your number line into equal parts as shown by the denominators.
Remember to simplify your answer by dividing the numerator and denominator by the same number.
Solution
$\mathbf{\frac{1}{2}}$
• $\frac{9}{10}$ - $\frac{2}{5}$ : first convert the fraction, $\frac{2}{5}$, by multiplying the numerator and denominator by 2 to make $\frac{4}{10}$ so both fractions in the equation have the same denominator. $\frac{9}{10}$ - $\frac{4}{10}$ = $\frac{5}{10}$ which can be simplified to $\frac{1}{2}$ by dividing both the numerator and denominator by 5.
• $\frac{7}{8}$ - $\frac{3}{8}$ = $\frac{4}{8}$ which can be simplified to $\frac{1}{2}$ by dividing both the numerator and denominator by 4.
• $\frac{12}{14}$ - $\frac{5}{14}$ = $\frac{7}{14}$ which can be simplified to $\frac{1}{2}$ by dividing both the numerator and denominator by 7.
$\mathbf{\frac{2}{3}}$
• $\frac{5}{6}$ - $\frac{1}{6}$ = $\frac{4}{6}$ which can be simplified to $\frac{2}{3}$ by dividing both the numerator and denominator by 2.
• $\frac{11}{12}$ - $\frac{1}{4}$ : first convert the fraction, $\frac{1}{4}$, by multiplying the numerator and denominator by 3 to make $\frac{3}{12}$ so both fractions in the equation have the same denominator. $\frac{11}{12}$ - $\frac{3}{12}$ = $\frac{8}{12}$ which can be simplified to $\frac{2}{3}$ by dividing both the numerator and denominator by 4.
$\mathbf{\frac{3}{5}}$
• $\frac{13}{15}$ - $\frac{4}{15}$ = $\frac{9}{15}$ which can be simplified to $\frac{3}{5}$ by dividing both the numerator and denominator by 3.
• $\frac{17}{20}$ - $\frac{5}{20}$ = $\frac{12}{20}$ which can be simplified to $\frac{3}{5}$ by dividing both the numerator and denominator by 4. |
# Determinants Class 12 Notes- Chapter 4
## What is the determinant of a matrix?
When a square matrix “A” of order “n” is associated with a number, then it is titled as a determinant of the aforementioned matrix. The number involved in this square matrix can be a real number or a complex number.
In the field of mathematics, Determinants can be used for a myriad of different calculations, such as:
• Finding the area of a Triangle.
• Obtaining the solutions of linear equations in two or three variables.
• Solve Linear equations by using the inverse of a matrix
• Getting the adjoint and the inverse of a square matrix
Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:
### What are the properties of Determinants?
The properties of determinants have their uses in simplifying the evaluation by getting the maximum number of zeroes in a row or column. They hold true for determinants of any order. The properties of determinants are mentioned below:
1. The determinant’s value remains unchanged if the rows and columns are interchanged.
2. The sign of a determinant changes when any two rows or columns of a determinant are interchanged with each other.
3. The value of a determinant is zero when any of the rows or columns of a determinant are identical to each other.
4. For each element of a row or column which gets multiplied by a constant P, the value also gets multiplied by P.
5. The determinant obtained is a sum of two or more determinants if some or all elements of a column or a row are expressed as a sum of two or more terms.
6. When the equimultiples of corresponding elements are added, the value of the determinant remains the same whenever the operation is multiplied.
Also Access NCERT Solutions for Class 12 Maths Chapter 4 NCERT Exemplar for Class 12 Maths Chapter 4
### Important Questions:
1. Examine the consistency of the system of equations
1. x + 2y = 2, 2x + 3y = 3
2. 2x – y = 5, x + y = 4
3. x + 3y = 5, 2x + 6y = 8
2. Using Cofactors of elements of second row, evaluate ∆=
$$\begin{array}{l}\begin{vmatrix} 5 &3 &8 \\ 2 &0 &1 \\ 1 &2 &3 \end{vmatrix}\end{array}$$
3. Find values of p if the area of a triangle is 6 sq. units and vertices are
(i) (p, 0), (6, 0), (0, 4)
(ii) (–4, 0), (0, 6), (0, p)
4. Using determinants, find the equation of the line joining (2, 3) and (4, 7)
5. Which of the following is correct
(A) Determinant is a number associated with a square matrix.
(B) Determinant is a number associated with a matrix.
(C) Determinant is a square matrix.
(D) None of these
Algebra Square Roots Matrices Divisibility Rules
## Frequently Asked Questions on CBSE Class 12 Maths Notes Chapter 4 Determinants
Q1
### What are the uses of algebra?
Algebra helps in logical thinking and enables a person to break down a problem first and then find its solution.
Q2
### What are the uses of determinants?
The determinant is useful for solving linear equations, capturing how linear transformation change area or volume, and changing variables in integrals.
Q3
### What is a matrix format?
A set of numbers arranged in rows and columns so as to form a rectangular array is called a ‘Matrix arrangement’. |
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# The number of distinct real values of $\lambda$ for which the lines $\dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z + 3}}{{{\lambda ^2}}}$and$\dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{{\lambda ^2}}} = \dfrac{{z - 1}}{2}$ are coplanar is:A.$2$B.$4$C.$3$D.$1$
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Hint: In this problem, we need to find the coplanar from these two lines of numbers of distinct real values. Lines in the same plane are coplanar lines. Skew lines are lines that do not intersect, and there is no plane that contains them. Intersecting lines are two coplanar lines with exactly one point in common. Concurrent lines are lines that contain the same point.
$\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|$
Since, the coplanar lines are ${L_1} = \dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$ and ${L_2} = \dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}$
We are given the equation of lines are $\dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z + 3}}{{{\lambda ^2}}}$ and $\dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{{\lambda ^2}}} = \dfrac{{z - 1}}{2}$
The number of distinct has the real values of $\lambda$ for the two lines, we get
Let us consider the two coplanar lines as ${L_1}$and ${L_2}$.
${L_1} = \dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z + 3}}{{{\lambda ^2}}}$----------(1)
${L_2} = \dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{{\lambda ^2}}} = \dfrac{{z - 1}}{2}$----------(2)
Comparing the coplanar equation (1) and (2) with the following line formula:
${L_1} = \dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$
${L_2} = \dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}$
We have the formula for finding the coplanar:
$\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0$
Since ${x_1} = 1,{y_1} = 2,{z_1} = - 3$,${x_2} = 3,{y_1} = 2,{z_1} = 1$ and$({a_1},{b_1},{c_1}) = (1,2,{\lambda ^2})$, $({a_2},{b_2},{c_2}) = (1,{\lambda ^2},2)$
Here, we have to substitute all the values in coplanar formula, then
$\left| {\begin{array}{*{20}{c}} {3 - 1}&{2 - 2}&{1 - ( - 3)} \\ 1&2&{{\lambda ^2}} \\ 1&{{\lambda ^2}}&2 \end{array}} \right| = 0$
Expanding the last element of first row in further simplification, we can get
$\left| {\begin{array}{*{20}{c}} 2&0&4 \\ 1&2&{{\lambda ^2}} \\ 1&{{\lambda ^2}}&2 \end{array}} \right| = 0$
We do perform the determinant operation simplified as follows, we get
$2(4 - {\lambda ^4}) - 0(2 - {\lambda ^2}) + 4({\lambda ^2} - 2) = 0 \\ 2(4 - {\lambda ^4}) + 4({\lambda ^2} - 2) = 0 \\$
Now, we have to simplify it, dividing the equation by $2$, we get
$\Rightarrow (4 - {\lambda ^4}) + 2({\lambda ^2} - 2) = 0$
Expanding the brackets to simplify in further:
$\Rightarrow 4 - {\lambda ^4} + 2{\lambda ^2} - 4 = 0 \\ \Rightarrow - {\lambda ^4} + 2{\lambda ^2} = 0 \\$
Here, we take common factors out, we can get
$- {\lambda ^2}({\lambda ^2} - 2) = 0$
$\lambda = 0, \pm \sqrt 2$
Therefore, $\lambda = 0,\sqrt 2 , - \sqrt 2$. So, the coplanar is $3$.
Finally, the correct answer is Option (C) $3$
As a result, The number of distinct real values of $\lambda$ for which the lines $\dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z + 3}}{{{\lambda ^2}}}$ and $\dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{{\lambda ^2}}} = \dfrac{{z - 1}}{2}$ are coplanar is $3$.
So, the correct answer is “Option C”.
Note: We note that the two lines are coplanar lies on the plane. When two lines lie on the same plane in three dimensions, they are assumed to be coplanar. We've learned how to use vector notations to describe a line's equation in three dimensions. If any three points determine a plane then additional points can be checked for coplanar by measuring the distance of the points from the plane, if the distance is zero then the point is coplanar. |
# Determine whether the given function is linear
Math 3βCollege Algebra Homework 2.5β2.8 Name: ______________________ Show all your work for full credit: 2.5 Linear Functions and Models Q1βQ3. Determine whether the given function is linear. If the function is linear, express the function in the form π(π₯) = ππ₯ + π Q1. π(π₯) = π₯(4 β π₯) Q2. π(π₯) = π₯+1 5 Q3. π(π₯) = (π₯ + 1)2 Q4βQ5. For the given linear function, make a table of values and sketch its graph. What is the slope of the graph? Q4. π(π₯) = 2π₯ β 5 2 Q5. π(π‘) = β π‘ + 2 3 Q6βQ7. A linear function is given. A) sketch the graph the rate of change of the function. b) Find the slope of the graph. C) Find Q6. π(π₯) = 2π₯ β 6 Q7. π£(π‘) = β 10 3 π‘ β 20 Q8. The amount of trash in a country landfill is modeled by the function π(π₯) = 150π₯ + 32,000 Where π₯ is the number of years since 1996 and π(π₯) is measured in thousands of tons. a) Sketch the graph of T b) What is the slope of the graph? c) At what rate is the amount of trash in the landfill increasing per year? 1 2.6 Transformations of functions Q1βQ4. Explain how the graph of g is obtained from the graph of π. Q1. π(π₯) = π₯ 2 π(π₯) = (π₯ + 2)2 Q2. π(π₯) = π₯ 2 π(π₯) = π₯ 2 + 2 Q3. π(π₯) = |π₯| π(π₯) = |π₯ + 2| β 2 Q4. π(π₯) = |π₯| π(π₯) = |π₯ β 2| + 2 Q5. Use the graph of π¦ = π₯ 2 to graph the follow. a) b) c) d) π(π₯) = π₯ 2 + 1 π(π₯) = (π₯ β 1)2 π(π₯) = βπ₯ 2 π(π₯) = (π₯ β 1)2 + 3 Q6βQ9. Sketch the graph of the function using transformations. Q6. π(π₯) = |π₯| β 1 1 Q8. π(π₯) = 3 β (π₯ β 1)2 2 1 Q7. π(π₯) = 4 π₯ 2 1 Q9. π(π₯) = βπ₯ + 4 β 3 2 Q10βQ11. A function π is given, write an equation for the final transformed graph. Q10. π(π₯) = |π₯|, shift 2 units to the left and shift downward 5 units. 4 Q11. π(π₯) = βπ₯; reflect in the y-axis and shift upward 1 unit. Q12βQ13. Determine whether the function π is even, odd, or neither. If π is even or odd, use symmetry to sketch the graph. Q12. π(π₯) = π₯ 4 Q13. π(π₯) = π₯ 2 + π₯ 2 2.7 Combining Functions Q1βQ4. Find π + π, π β π, ππ, πππ π/π and their domain Q1. π(π₯) = π₯ 2 + π₯, π(π₯) = π₯ 2 Q2. π(π₯) = 5 β π₯, π(π₯) = π₯ 2 β 3π₯ Q3. π(π₯) = β25 β π₯ 2 , π(π₯) = βπ₯ + 3 2 Q4. π(π₯) = π₯ , 4 π(π₯) = π₯+4 Q5βQ7. Use π(π₯) = 2π₯ β 3 and π(π₯) = 4 β π₯ 2 to evaluate the expression. Q5. A) π(π(0)) B) π(π(0)) Q6. π΄) (πΒ°π)(β2) B) (πΒ°π)(β2) Q7. π΄) (πΒ°π)(π₯) B) (πΒ°π)(π₯) Q8—Q9. Find the functions πΒ°π, πΒ°π, πΒ°π, and πΒ°π and their domains. Q8. π(π₯) = 1 π₯ Q9. π(π₯) = π₯ 2 π(π₯) = 2π₯ + 4 π(π₯) = π₯ + 1 3 2.8 One to One Functions and their Inverses Q1—Q3. Determine whether the function is one-to-one. Q1. π(π₯) = β2π₯ + 4 Q2. β(π₯) = π₯ 2 β 2π₯ Q3. π(π₯) = βπ₯ Q4—Q6. Assume that π is a one-to-one function. Q4. if π(2) = 7, find π β1 (7). Q5. ππ π β1 (3) = β1, find π(β1) Q6. If π(π₯) = 5 β 2π₯, find π β1 (3) Q7βQ10. Use the inverse Function property to show that π πππ π are inverse of each other. Q7. π(π₯) = π₯ β 6 π(π₯) = π₯ + 6 Q8. π(π₯) = 3π₯ + 4 π(π₯) = Q9. π(π₯) = π₯ 2 β 9, π₯ β₯ 0, π₯+2 Q10. π(π₯) = π₯β2 π₯β4 3 π(π₯) = βπ₯ + 9, π₯ β₯ β9 π(π₯) = 2π₯+2 π₯β1 Q11βQ13 Find the inverse function of π. Q11. π(π₯) = 3π₯ + 5 Q12. π(π₯) = 2π₯+5 π₯β7 Q13. π(π₯) = 4 β π₯ 2 , π₯ β₯ 0 Q14βQ15. A function π is given. A) sketch the graph of π B) use the graph of π to sketch the graph of π β1 C) Find π β1 Q14. π(π₯) = 3π₯ β 6 Q15. π(π₯) = βπ₯ + 1 4
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# Tip #17 to Get a Top ACT Math Score (page 2)
By McGraw-Hill Professional
Updated on Sep 7, 2011
You couldn't ask for easier points to boost your score. Every ACT has a midpoint and/or distance question. If you don't know the formulas, you don't stand much chance. But if you memorize them, right here and now, you will gain points, guaranteed! In fact, take a few minutes right now to cut out the flashcards from the back of this book. Bring them everywhere you go, school, sports, DMX concerts, parties, etc. Everyone loves math flash cards!
To find the midpoint between two points on a graph, use the midpoint formula. All this formula really says is, "Take halfway between the x numbers and halfway between the y numbers." Halfway between two numbers is the same as the average of the numbers, so the midpoint formula is like the average formula:
The distance formula also makes sense if you really look at it. It is based on the Pythagorean theorem (Skill 24). To find the distance between two points on a graph, use the formula
Cut out the flashcards at the end of this book to help you memorize these formulas.
Let's look at this question:
Solution: A midpoint is halfway between two points, really just the average. That's what the midpoint formula gives us, the average of the two points:
### Easy
1. The endpoints of on the real number line are –14 and 2. What is the coordinate of the midpoint of ?
1. –8
2. –6
3. –2
4. 0
5. 8
2. A line in the standard (x, y) coordinate plane contains the points M( –2, 4) and N(8, 10). What point is the midpoint of ?
1. (–2, 4)
2. (3, 3)
3. (5, 3)
1. (5, 7)
2. (3, 7)
3. ### Medium
4. In the standard (x, y) coordinate plane, what is the distance between the points (–4, 2) and (1, –10) ?
1. 3
2. 5
3. 9
4. 11
5. 13
5. A diameter of a circle has endpoints (–2, 10) and (6, –4) in the standard (x, y) coordinate plane. What point is the center of the circle?
1. (2, 3)
2. (4, 6)
3. (6, 4)
1. (12, 8)
2. Cannot be determined from the given information
6. ### Hard
7. A city's restaurants are laid out on a map in the standard (x, y) coordinate plane. How long, in units, is the straight-line path between Paul & Elizabeth's restaurant at (17, 15) and The Green Bean restaurant at (12, 5) ?
1. 11
1. B A midpoint is halfway between the two points, really just the average. That's what the midpoint formula gives us:
2. K A midpoint is halfway between the two points, really just the average. That's what the midpoint formula gives us:
3. E Use the distance formula:
4. F The center of the circle is in the middle of the diameter—the midpoint of the diameter. Midpoint
5. C This word problem is just asking for the distance between the two restaurants. Use the distance formula:
Go to: Tip #18 |
# NCERT Book Class 9 Maths Chapter 1 Number System PDF (Latest Edition)
Number System chapter of Class 9 Maths NCERT Book can be downloaded in PDF format from here. We have provided here the latest edition of the chapter along with precise NCERT Solutions.
Created On: Jul 9, 2021 23:14 IST
NCERT Class 9 Maths Chapter 1 Number System PDF
Class 9 Maths Chapter 1 ´Number System` of NCERT Book is best to understand the concept of numbers and make your basics stronger for obtaining good results in exams. You will get here the latest edition of the chapter that will help you learn the updated information and prepare well for the examinations to be conducted in the current academic year 2021-2022.
About Class 9 Maths Chapter 1 Number System
Number System chapter explains the concept of rational and irrational numbers. You will get to learn the representation of these forms of numbers along with special properties and identities. Some solved examples and exercise questions are given in the chapter for more clarification of concepts and also for self-assessment.
Major topics discussed in the chapter are:
→ Introduction to Number System
→ Irrational Numbers
→ Real Numbers and their Decimal Expansions
→ Representing Real Numbers on the Number Line
→ Operations on Real Numbers
Some important points to revise from the chapter are:
• A number r is called a rational number, if it can be written in the form p/q , where p and q are integers and q ≠ 0. 2.
• A number s is called an irrational number, if it cannot be written in the form p/q , where p and q are integers and q ≠ 0.
• The decimal expansion of a rational number is either terminating or non-terminating recurring.
• The decimal expansion of an irrational number is non-terminating non-recurring.
• All the rational and irrational numbers make up the collection of real numbers.
• If r is rational and s is irrational, then r + s and r – s are irrational numbers, and rs and r/s are irrational numbers, r ≠ 0. 8.
• positive real numbers a and b, the following identities hold:
• Let a > 0 be a real number and p and q be rational numbers. Then
Also, check the NCERT Solutions for Class 9 Maths Chapter 1 from the following link:
Also Check:
NCERT Books for Class 9 (Latest Editions for 2021-22)
NCERT Solutions for Class 9 (Updated for 2021-22) |
# Triangle A has sides of lengths 54 , 44 , and 64 . Triangle B is similar to triangle A and has a side of length 4 . What are the possible lengths of the other two sides of triangle B?
Feb 7, 2017
$< 4 , 3 \frac{7}{27} , 4 \frac{20}{27} >$, $< 4 \frac{10}{11} , 4 , 5 \frac{9}{11} >$ and$< 3 \frac{3}{8} , 2 \frac{3}{4} , 4 >$
#### Explanation:
Let $\left(4 , a , b\right)$ are the lengths of Triangle B..
A. Comparing 4 and 54 from Triangle A,
$\frac{b}{44} = \frac{4}{54}$, $b = \frac{2}{27} \cdot 44 = 3 \frac{7}{27}$
$\frac{c}{64} = \frac{4}{54}$, $c = \frac{2}{27} \cdot 64 = 4 \frac{20}{27}$
The length of sides for Triangle B is$< 4 , 3 \frac{7}{27} , 4 \frac{20}{27} >$
B. Comparing 4 and 44 from Triangle A,
$\frac{b}{54} = \frac{4}{44}$, $b = \frac{1}{11} \cdot 54 = 4 \frac{10}{11}$
$\frac{c}{64} = \frac{4}{44}$, $c = \frac{1}{11} \cdot 64 = 5 \frac{9}{11}$
The length of sides for Triangle B is$< 4 \frac{10}{11} , 4 , 5 \frac{9}{11} >$
Comparing 4 and 64 from Triangle A,
$\frac{b}{54} = \frac{4}{64}$,$b = \frac{1}{16} \cdot 54 = 3 \frac{3}{8}$
$\frac{c}{44} = \frac{4}{64}$, $c = \frac{1}{16} \cdot 44 = 2 \frac{3}{4}$
The length of sides for Triangle B is$< 3 \frac{3}{8} , 2 \frac{3}{4} , 4 >$
Therefore the possible sides for Triangle B are
$< 4 , 3 \frac{7}{27} , 4 \frac{20}{27} >$, $< 4 \frac{10}{11} , 4 , 5 \frac{9}{11} >$ and$< 3 \frac{3}{8} , 2 \frac{3}{4} , 4 >$ |
# Calc2.51
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The fence and pens in question. x is the entire length of the top border, and y is the height of each line.
If a fence is to be made with three pens, the three connected side-by-side, find the dimensions which give the largest total area if 400 feet of fence are to be used.
This problem describes a large rectangle with length x and height y. However, this rectangle is split into three parts with two extra lengths of fence of length y. We know the total fence to be used is 400 feet, so
$400=2x+4y\Rightarrow 200=x+2y\,$
From this equation we can see that $0\leq y\leq 100$ which gives us our interval. Now, to maximize the area, we need a function for the area. Since this is a rectangle, the area is simply
$A=xy\,$
Solving the first equation for x and plugging it into the second equation for area gives
$A=(200-2y)y=200y-2y^{2}\,$
To find the absolute maximum area, we need to find the critical points of this area function.
$A'=200-4y\,$
Thus, when $y=50$ we have a critical point and this is our only critical point. Our absolute maximum area, then, must come when $y=0$, $y=50$, or $y=100$.
$A(0)=0\,$
$A(50)=5000\,$
$A(100)=0\,$
Thus, our absolute maximum area of 5000 square feet comes when the length is 100 feet and the height is 50 feet. |
# Difference between revisions of "2020 AMC 8 Problems/Problem 3"
## Problem
Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?
$\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840$
## Solution 1
The area of the garden is $6 \cdot 8 = 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, there are a total of $48 \cdot 4=192$ strawberry plants, each of which produces $10$ strawberries on average. Accordingly, she can expect to harvest $192 \cdot 10 = \boxed{\textbf{(D) }1920}$ strawberries.
## Solution 2
Looking at the units of each quantity, we observe that the answer will be the product of the number of square feet, the number of plants per square foot, and the number of strawberries per plant. This gives $6 \cdot 8 \cdot 4 \cdot 10 = \boxed{\textbf{(D) }1920}$.
## Solution 3 (One-Line Version of Solution 2)
$$\left(6\text{ ft}\cdot8\text{ ft}\right)\left(4 \ \frac{\text{plants}}{\text{ft}^2}\right)\left(10 \ \frac{\text{strawberries}}{\text{plant}}\right)=6\cdot8\cdot4\cdot10=\boxed{\textbf{(D) }1920} \text{ strawberries}$$
~MRENTHUSIASM
~savannahsolver |
# 6.3: Compound Inequalities
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Write and graph compound inequalities on a number line.
• Solve compound inequalities with “and.”
• Solve compound inequalities with “or.”
• Solve compound inequalities using a graphing calculator (TI family).
• Solve real-world problems using compound inequalities.
## Introduction
In this section, we’ll solve compound inequalities—inequalities with more than one constraint on the possible values the solution can have.
There are two types of compound inequalities:
1. Inequalities joined by the word “and,” where the solution is a set of values greater than a number and less than another number. We can write these inequalities in the form “\begin{align*}x>a\end{align*} and \begin{align*}x,” but usually we just write “\begin{align*}a.” Possible values for \begin{align*}x\end{align*} are ones that will make both inequalities true.
2. Inequalities joined by the word “or,” where the solution is a set of values greater than a number or less than another number. We write these inequalities in the form “\begin{align*}x>a\end{align*} or \begin{align*}x.” Possible values for \begin{align*}x\end{align*} are ones that will make at least one of the inequalities true.
You might wonder why the variable \begin{align*}x\end{align*} has to be greater than one number and/or less than the other number; why can’t it be greater than both numbers, or less than both numbers? To see why, let’s take an example.
Consider the compound inequality “\begin{align*}x>5\end{align*} and \begin{align*}x>3\end{align*}.” Are there any numbers greater than 5 that are not greater than 3? No! Since 5 is greater than 3, everything greater than 5 is also greater than 3. If we say \begin{align*}x\end{align*} is greater than both 5 and 3, that doesn’t tell us any more than if we just said \begin{align*}x\end{align*} is greater than 5. So this compound inequality isn’t really compound; it’s equivalent to the simple inequality \begin{align*}x > 5\end{align*}. And that’s what would happen no matter which two numbers we used; saying that \begin{align*}x\end{align*} is greater than both numbers is just the same as saying that \begin{align*}x\end{align*} is greater than the bigger number, and saying that \begin{align*}x\end{align*} is less than both numbers is just the same as saying that \begin{align*}x\end{align*} is less than the smaller number.
Compound inequalities with “or” work much the same way. Every number that’s greater than 3 or greater than 5 is also just plain greater than 3, and every number that’s greater than 3 is certainly greater than 3 or greater than 5—so if we say “\begin{align*}x>5\end{align*} or \begin{align*}x>3\end{align*},” that’s the same as saying just “\begin{align*}x>3\end{align*}.” Saying that \begin{align*}x\end{align*} is greater than at least one of two numbers is just the same as saying that \begin{align*}x\end{align*} is greater than the smaller number, and saying that \begin{align*}x\end{align*} is less than at least one of two numbers is just the same as saying that \begin{align*}x\end{align*} is less than the greater number.
## Write and Graph Compound Inequalities on a Number Line
Example 1
Write the inequalities represented by the following number line graphs.
a)
b)
c)
d)
Solution
a) The solution graph shows that the solution is any value between -40 and 60, including -40 but not 60.
Any value in the solution set satisfies both \begin{align*}x \ge -40\end{align*} and \begin{align*}x<60\end{align*}.
This is usually written as \begin{align*}-40 \le x < 60\end{align*}.
b) The solution graph shows that the solution is any value greater than 1 (not including 1) or any value less than -2 (not including -2). You can see that there can be no values that can satisfy both these conditions at the same time. We write: \begin{align*}x>1\end{align*} or \begin{align*}x < -2\end{align*}.
c) The solution graph shows that the solution is any value greater than 4 (including 4) or any value less than -1 (including - 1). We write: \begin{align*}x \ge 4\end{align*} or \begin{align*}x \le -1\end{align*}.
d) The solution graph shows that the solution is any value that is both less than 25 (not including 25) and greater than -25 (not including -25). Any value in the solution set satisfies both \begin{align*}x >-25\end{align*} and \begin{align*}x < 25\end{align*}.
This is usually written as \begin{align*}-25 < x < 25\end{align*}.
Example 2
Graph the following compound inequalities on a number line.
a) \begin{align*}-4 \le x \le 6\end{align*}
b) \begin{align*}x < 0\end{align*} or \begin{align*}x > 2\end{align*}
c) \begin{align*}x \ge -8\end{align*} or \begin{align*}x \le -20\end{align*}
d) \begin{align*}-15 < x \le 85\end{align*}
Solution
a) The solution is all numbers between -4 and 6, including both -4 and 6.
b) The solution is all numbers less than 0 or greater than 2, not including 0 or 2.
c) The solution is all numbers greater than or equal to -8 or less than or equal to -20.
d) The solution is all numbers between -15 and 85, not including -15 but including 85.
## Solve a Compound Inequality With “and” or “or”
When we solve compound inequalities, we separate the inequalities and solve each of them separately. Then, we combine the solutions at the end.
Example 3
Solve the following compound inequalities and graph the solution set.
a) \begin{align*}-2 < 4x-5 \le 11\end{align*}
b) \begin{align*}3x-5 < x + 9 \le 5x+13\end{align*}
Solution
a) First we re-write the compound inequality as two separate inequalities with and. Then solve each inequality separately.
\begin{align*}& -2 < 4x-5 \qquad \qquad 4x-5 \le 11\\ & \quad 3<4x \qquad \quad and \qquad \quad 4x \le 16\\ & \quad \frac{3}{4} < x \qquad \qquad \qquad \qquad \ x \le 4\end{align*}
Answer: \begin{align*}\frac{3}{4} and \begin{align*}x \le 4\end{align*}. This can be written as \begin{align*}\frac{3}{4}< x \le 4\end{align*}.
b) Re-write the compound inequality as two separate inequalities with and. Then solve each inequality separately.
\begin{align*}& 3x-5
Answer: \begin{align*}x < 7\end{align*} and \begin{align*}x \ge -1\end{align*}. This can be written as: \begin{align*}-1 \le x < 7\end{align*}.
Example 4
Solve the following compound inequalities and graph the solution set.
a) \begin{align*}9-2x \le 3\end{align*} or \begin{align*}3x+10 \le 6-x\end{align*}
b) \begin{align*}\frac{x-2}{6} \le 2x-4\end{align*} or \begin{align*}\frac{x-2}{6} > x+5\end{align*}
Solution
a) Solve each inequality separately:
\begin{align*}& 9-2x \le 3 \qquad \qquad 3x+10 \le 6-x\\ & \ -2x \le -6 \qquad or \qquad \ \ 4x \le -4\\ & \qquad x \ge 3 \qquad \qquad \qquad \ \ \ x \le -1\end{align*}
Answer: \begin{align*}x \ge 3\end{align*} or \begin{align*}x \le -1\end{align*}
b) Solve each inequality separately:
\begin{align*}& \frac{x-2}{6} \le 2x-4 \qquad \qquad \quad \frac{x-2}{6} > x+5\\ & x-2 \le 6(2x-4) \qquad \qquad x-2 > 6(x+5)\\ & x-2 \le 12x-24 \qquad or \quad \ x-2 > 6x+30\\ & \quad \ 22 \le 11x \qquad \qquad \qquad \ -32 > 5x\\ & \quad \ \ \ 2 \le x \qquad \qquad \qquad \quad -6.4 > x\end{align*}
Answer: \begin{align*}x \ge 2\end{align*} or \begin{align*}x < -6.4\end{align*}
The video at http://www.math-videos-online.com/solve-compound-inequality.html shows the process of solving and graphing compound inequalities in more detail. One thing you may notice in this video is that in the second problem, the two solutions joined with “or” overlap, and so the solution ends up being the set of all real numbers, or \begin{align*}(-\infty,\infty)\end{align*}. This happens sometimes with compound inequalities that involve “or”; for example, if the solution to an inequality ended up being “\begin{align*}x<5\end{align*} or \begin{align*}x>1\end{align*},” the solution set would be all real numbers. This makes sense if you think about it: all real numbers are either a) less than 5, or b) greater than or equal to 5, and the ones that are greater than or equal to 5 are also greater than 1—so all real numbers are either a) less than 5 or b) greater than 1.
Compound inequalities with “and,” meanwhile, can turn out to have no solutions. For example, the inequality “\begin{align*}x<3\end{align*} and \begin{align*}x>4\end{align*}” has no solutions: no number is both greater than 4 and less than 3. If we write it as \begin{align*}4 it’s even more obvious that it has no solutions; \begin{align*}4 implies that \begin{align*}4 < 3\end{align*}, which is false.
## Solve Compound Inequalities Using a Graphing Calculator (TI-83/84 family)
Graphing calculators can show you the solution to an inequality in the form of a graph. This can be especially useful when dealing with compound inequalities.
Example 5
Solve the following inequalities using a graphing calculator.
a) \begin{align*}5x+2(x-3) \ge 2\end{align*}
b) \begin{align*}7x-2 < 10x+1 < 9x+5\end{align*}
c) \begin{align*}3x+2 \le 10\end{align*} or \begin{align*}3x+2 \ge 15\end{align*}
Solution
a) Press the [Y=] button and enter the inequality on the first line of the screen.
(To get the \begin{align*}\ge\end{align*} symbol, press [TEST] [2nd] [MATH] and choose option 4.)
Then press the [GRAPH] button.
Because the calculator uses the number 1 to mean “true” and 0 to mean “false,” you will see a step function with the \begin{align*}y-\end{align*}value jumping from 0 to 1.
The solution set is the values of \begin{align*}x\end{align*} for which the graph shows \begin{align*}y=1\end{align*}—in other words, the set of \begin{align*}x-\end{align*}values that make the inequality true.
Note: You may need to press the [WINDOW] key or the [ZOOM] key to adjust the window to see the full graph.
The solution is \begin{align*}x>\frac{8}{7}\end{align*}, which is why you can see the \begin{align*}y-\end{align*}value changing from 0 to 1 at about 1.14.
b) This is a compound inequality: \begin{align*}7x-2 < 10x +1\end{align*} and \begin{align*}10x+1 < 9x+5\end{align*}. You enter it like this:
(To find the [AND] symbol, press [TEST], choose [LOGIC] on the top row and choose option 1.)
The resulting graph should look like this:
The solution are the values of \begin{align*}x\end{align*} for which \begin{align*}y=1\end{align*}; in this case that would be \begin{align*}-1.
c) This is another compound inequality.
(To enter the [OR] symbol, press [TEST], choose [LOGIC] on the top row and choose option 2.)
The resulting graph should look like this:
The solution are the values of \begin{align*}x\end{align*} for which \begin{align*}y=1\end{align*}--in this case, \begin{align*}x \le 2.7\end{align*} or \begin{align*}x \ge 4.3\end{align*}.
## Solve Real-World Problems Using Compound Inequalities
Many application problems require the use of compound inequalities to find the solution.
Example 6
The speed of a golf ball in the air is given by the formula \begin{align*}v=-32t+80\end{align*}. When is the ball traveling between 20 ft/sec and 30 ft/sec?
Solution
First we set up the inequality \begin{align*}20 \le v \le 30\end{align*}, and then replace \begin{align*}v\end{align*} with the formula \begin{align*}v=-32t+80\end{align*} to get \begin{align*}20 \le -32t+80 \le 30\end{align*}.
Then we separate the compound inequality and solve each separate inequality:
\begin{align*}& \ 20 \le -32t+80 \qquad \quad \ \ \ -32t + 80 \le 30\\ & 32t \le 60 \qquad \qquad \text{and} \qquad \qquad \quad \ 50 \le 32t\\ & \quad t \le 1.875 \qquad \qquad \qquad \qquad \quad 1.56 \le t\end{align*}
Answer: \begin{align*}1.56 \le t \le 1.875\end{align*}
To check the answer, we plug in the minimum and maximum values of \begin{align*}t\end{align*} into the formula for the speed.
For \begin{align*}t = 1.56, \ v=-32t+80 = -32(1.56)+80=30 \ ft/sec\end{align*}
For \begin{align*}t = 1.875, \ v=-32t+80= -32(1.875)+80= 20 \ ft/sec\end{align*}
So the speed is between 20 and 30 ft/sec. The answer checks out.
Example 7
William’s pick-up truck gets between 18 to 22 miles per gallon of gasoline. His gas tank can hold 15 gallons of gasoline. If he drives at an average speed of 40 miles per hour, how much driving time does he get on a full tank of gas?
Solution
Let \begin{align*}t =\end{align*} driving time. We can use dimensional analysis to get from time per tank to miles per gallon:
\begin{align*}\frac{t \ hours}{1 \ tank} \times \frac{1 \ tank}{15 \ gallons} \times \frac{40 \ miles}{1 \ hour} \times \frac{40t}{15} \frac{miles}{gallon}\end{align*}
Since the truck gets between 18 and 22 miles/gallon, we set up the compound inequality \begin{align*}18 \le \frac{40t}{15} \le 22\end{align*}. Then we separate the compound inequality and solve each inequality separately:
\begin{align*}& \ \ 18 \le \frac{40t}{15} \qquad \qquad \quad \ \frac{40t}{15} \le 22\\ & \ 270 \le 40t \qquad \text{and} \qquad 40t \le 330\\ & 6.75 \le t \qquad \qquad \qquad \quad \ \ t \le 8.25\end{align*}
Answer: \begin{align*}6.75 \le t \le 8.25\end{align*}.
Andrew can drive between 6.75 and 8.25 hours on a full tank of gas.
If we plug in \begin{align*}t = 6.75\end{align*} we get \begin{align*}\frac{40t}{15} = \frac{40(6.75)}{15} = 18 \ miles \ per \ gallon\end{align*}.
If we plug in \begin{align*}t = 8.25\end{align*} we get \begin{align*}\frac{40t}{15} = \frac{40(8.25)}{15} = 22 \ miles \ per \ gallon\end{align*}.
## Lesson Summary
• Compound inequalities combine two or more inequalities with “and” or “or.”
• “And” combinations mean that only solutions for both inequalities will be solutions to the compound inequality.
• “Or” combinations mean solutions to either inequality will also be solutions to the compound inequality.
## Review Questions
Write the compound inequalities represented by the following graphs.
Solve the following compound inequalities and graph the solution on a number line.
1. \begin{align*}-5 \le x-4 \le 13\end{align*}
2. \begin{align*}1 \le 3x +5 \le 4\end{align*}
3. \begin{align*}-12 \le 2-5x \le 7\end{align*}
4. \begin{align*}\frac{3}{4} \le 2x+9 \le \frac{3}{2}\end{align*}
5. \begin{align*}-2 \le \frac{2x-1}{3} < -1\end{align*}
6. \begin{align*}4x-1 \ge 7\end{align*} or \begin{align*}\frac{9x}{2} < 3\end{align*}
7. \begin{align*}3-x < -4\end{align*} or \begin{align*}3-x > 10\end{align*}
8. \begin{align*}\frac{2x+3}{4} < 2\end{align*} or \begin{align*}-\frac{x}{5} + 3 < \frac{2}{5}\end{align*}
9. \begin{align*}2x-7 \le -3\end{align*} or \begin{align*}2x-3 > 11\end{align*}
10. \begin{align*}4x+3< 9\end{align*} or \begin{align*}-5x+4 \le -12\end{align*}
11. How would you express the answer to problem 5 as a set?
12. How would you express the answer to problem 5 as an interval?
13. How would you express the answer to problem 10 as a set?
1. Could you express the answer to problem 10 as a single interval? Why or why not?
2. How would you express the first part of the solution in interval form?
3. How would you express the second part of the solution in interval form?
14. Express the answers to problems 1 and 3 in interval notation.
15. Express the answers to problems 6 through 9 in interval notation.
16. Solve the inequality “\begin{align*}x \ge -3\end{align*} or \begin{align*}x < 1\end{align*}” and express the answer in interval notation.
17. How many solutions does the inequality “\begin{align*}x \ge 2\end{align*} and \begin{align*}x \le 2\end{align*}” have?
18. To get a grade of B in her Algebra class, Stacey must have an average grade greater than or equal to 80 and less than 90. She received the grades of 92, 78, 85 on her first three tests.
1. Between which scores must her grade on the final test fall if she is to receive a grade of B for the class? (Assume all four tests are weighted the same.)
2. What range of scores on the final test would give her an overall grade of C, if a C grade requires an average score greater than or equal to 70 and less than 80?
3. If an A grade requires a score of at least 90, and the maximum score on a single test is 100, is it possible for her to get an A in this class? (Hint: look again at your answer to part a.)
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Home » Revision Notes for CBSE Class 6 to 12 » Class 11 Maths Revision Notes for Principle of Mathematical Induction of Chapter 4
# Class 11 Maths Revision Notes for Principle of Mathematical Induction of Chapter 4 – Free PDF Download
Free PDF download of Class 11 Maths revision notes & short key-notes for Principle of Mathematical Induction of Chapter 4 to score high marks in exams, prepared by expert mathematics teachers from latest edition of CBSE books.
Chapter Name Principle of Mathematical Induction Chapter Chapter 4 Class Class 11 Subject Maths Revision Notes Board CBSE TEXTBOOK MatheMatics Category REVISION NOTES
## CBSE Class 11 Maths Revision Notes for Principle of Mathematical Induction of Chapter 4
• One key basis for mathematical thinking is deductive reasoning. In contrast to deduction, inductive reasoning depends on working with different cases and developing a conjecture by observing incidences till we have observed each and every case. Thus, in simple language we can say the word ‘induction’ means the generalisation from particular cases or facts.
• Statement: A sentence is called a statement, if it is either true ot false.
• Motivation: Motivation is tending to initiate an action. Here Basis step motivate us for mathematical induciton.
• Principle of Mathematical Induction: The principle of mathematical induction is one such tool which can be used to prove a wide variety of mathematical statements. Each such statement is assumed as P(n) associated with positive integer n, for which the correctness for the case n = 1 is examined. Then assuming the truth of P(k) for some positive integer k, the truth of P (k+1) is established.
• Working Rule:
Step 1: Show that the given statement is true for n = 1.
Step 2: Assume that the statement is true for n = k.
Step 3: Using the assumption made in step 2, show that the statement is true for n = k + 1. We have proved the statement is true for n = k. According to step 3, it is also true for k + 1 (i.e., 1 + 1 = 2). By repeating the above logic, it is true for every natural number. |
# Multinomial Theorem
Definition: The Multinomial Theorem is an extension of the Binomial Theorem and is a fundamental concept in combinatorics and algebra. It provides a formula for expanding expressions involving multiple terms, each raised to different positive integer exponents. The theorem is particularly useful when dealing with polynomials with more than two terms.
Key Concepts:
1. Generalized Binomial Theorem: The Multinomial Theorem generalizes the Binomial Theorem, which is used to expand expressions of the form $\left(a+b{\right)}^{n}$. The Multinomial Theorem deals with expressions of the form $\left({a}_{1}+{a}_{2}+{a}_{3}+\dots +{a}_{k}{\right)}^{n}$, where there are "k" terms, each raised to the power of "n."
2. Coefficients and Exponents: In the expansion, the coefficients of each term are determined by a multinomial coefficient, while the exponents of the individual terms are based on the exponents of the original terms. The multinomial coefficient is calculated using combinations and accounts for the ways terms can be grouped together.
3. Multinomial Coefficient: The multinomial coefficient $\left(\genfrac{}{}{0px}{}{n}{{n}_{1},{n}_{2},{n}_{3},\dots ,{n}_{k}}\right)$ represents the number of ways to arrange "n" objects into "k" distinct groups, where ${n}_{1}$ objects are in the first group, ${n}_{2}$in the second, and so on, with ${n}_{k}$ in the last group. It is calculated as: $\left(\genfrac{}{}{0px}{}{n}{{n}_{1},{n}_{2},{n}_{3},\dots ,{n}_{k}}\right)=\frac{n!}{{n}_{1}!\cdot {n}_{2}!\cdot {n}_{3}!\cdot \dots \cdot {n}_{k}!}$
4. Formula for the Multinomial Expansion: The Multinomial Theorem provides a formula for expanding expressions of the form $\left({a}_{1}+{a}_{2}+{a}_{3}+\dots +{a}_{k}{\right)}^{n}$: $\left({a}_{1}+{a}_{2}+{a}_{3}+\dots +{a}_{k}{\right)}^{n}=\sum \left(\genfrac{}{}{0px}{}{n}{{n}_{1},{n}_{2},{n}_{3},\dots ,{n}_{k}}\right)\cdot \left({a}_{1}^{{n}_{1}}\cdot {a}_{2}^{{n}_{2}}\cdot {a}_{3}^{{n}_{3}}\cdot \dots \cdot {a}_{k}^{{n}_{k}}\right)$
5. Example: Suppose you want to expand $\left(x+y+z{\right)}^{3}$. The expansion is: $\left(x+y+z{\right)}^{3}=\left(\genfrac{}{}{0px}{}{3}{3,0,0}\right){x}^{3}+\left(\genfrac{}{}{0px}{}{3}{2,1,0}\right){x}^{2}y+\left(\genfrac{}{}{0px}{}{3}{1,2,0}\right)x{y}^{2}+\left(\genfrac{}{}{0px}{}{3}{0,3,0}\right){y}^{3}+\left(\genfrac{}{}{0px}{}{3}{2,0,1}\right){x}^{2}z+\left(\genfrac{}{}{0px}{}{3}{1,1,1}\right)xyz+\left(\genfrac{}{}{0px}{}{3}{0,2,1}\right){y}^{2}z+\left(\genfrac{}{}{0px}{}{3}{1,0,2}\right)x{z}^{2}+\left(\genfrac{}{}{0px}{}{3}{0,1,2}\right)y{z}^{2}+\left(\genfrac{}{}{0px}{}{3}{0,0,3}\right){z}^{3}$
6. Applications:
• The Multinomial Theorem is applied in various fields of mathematics, physics, and engineering when dealing with polynomials involving multiple variables and exponents.
• It is used in probability and statistics to calculate probabilities in multivariate distributions. |
CBSE 9th Maths
Herons Formula
1. Triangle: A plane figure bounded by three line segments is called a triangle.
In ΔABC has
(i) three vertices, namely A, B and C.
(ii) three sides, namely AB, BC and CA.
(iii) three angles, namely ∠A, ∠B and ∠C.
2. Types of Triangle on the Basis of Sides
(i) Equilateral triangle: A triangle having all sides equal is called an equilateral triangle.
In equilateral ΔABC,
i.e., AB = BC = CA
(ii) Isosceles triangle: A triangle having two sides equal is called an isosceles triangle.
In isosceles ΔABC,
i.e., AB = AC
(iii) Scalene triangle: A triangle in which all the sides are of different lengths is called a scalene triangle.
In scalene ΔABC,
i.e., AB ≠ BC ≠ CA
3. The perimeter of a Triangle: The sum of the lengths of three sides of a triangle is called its perimeter.
Let, AB = c, BC = a, CA = b
i.e., Perimeter of ΔABC, 2s = a + b + c
4. Area of a Triangle: The measure of the surface enclosed by the boundary of the triangle is called its area.
Area of triangle = 1/2 × Base × Height
Area of right angled triangle = 1/2 × Base × Perpendicular
5. Area of a Triangle (Heron’s Formula): If a triangle has a, b and c as sides, then the area of a triangle by Heron’s formula = s(sa)(sb)(sc)−−−−−−−−−−−−−−−−−−√
where, s (semi-perimeter) = a+b+c2
Note: This formula is highly applicable in the case when we don’t have the exact idea about height.
6. Application of Heron’s Formula in Finding Areas of Quadrilaterals: Let ABCD he a quadrilateral to find the area of a quadrilateral we need to divide the quadrilateral in triangular parts.
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC |
Engage NY Eureka Math Kindergarten Module 4 Lesson 22 Answer Key
Eureka Math Kindergarten Module 4 Lesson 22 Sprint A Answer Key
Question 1.
Complete the number bond.
Explanation:
In the above there are number bond given.
Few number bond with the addends and some with the sum.
The number bonds with the addends Sum need to filled. To find the sum add the addends
like in the first number bond
1 + 2 = 3.
Some number bond are given with sum and no addends .
For that
4 sum . the possibilities are 1 + 3, 2 +2 , 3 + 1. we can write any number as addends from the possibilities.
5 as sum – 1 + 4, 2 +3, 3 + 2, 4 + 1.
Eureka Math Kindergarten Module 4 Lesson 22 Sprint B Answer Key
Question 1.
Complete the number bond.
Explanation:
In the above there are number bond given.
Few number bond with the addends and some with the sum.
The number bonds with the addends Sum need to filled. To find the sum add the addends
like in the first number bond
1 + 1 = 2.
Some number bond are given with sum and no addends .
For that
3 sum . the possibilities are 1 + 2, 2 +1. we can write any number as addends from the possibilities.
5 as sum – 1 + 4, 2 +3, 3 + 2, 4 + 1.
Eureka Math Kindergarten Module 4 Lesson 22 Problem Set Answer Key
Question 1.
Fill in the number bonds.
Cross out 1 hat.
6 – 1 = 5
Explanation:
Given Cross out 1 hat.
Total number of hat = 6
number of hats crossed out = 1
Number of hats left = 6 – 1 = 5.
This is represented in number bond and number sentence as shown in the above image.
Cross out 5 snowflakes.
6 – 5 = 1
Explanation:
Given Cross out 5 snowflakes..
Total number of snowflakes = 6
number of snowflakes crossed out = 5
Number of snowflakes left = 6 – 5 = 1.
This is represented in number bond and number sentence as shown in the above image.
Cross out 2 snowflakes.
6 – 2 = 4
Explanation:
Given Cross out 5 snowflakes..
Total number of snowflakes = 6
number of snowflakes crossed out = 2
Number of snowflakes left = 6 – 2 = 4.
This is represented in number bond and number sentence as shown in the above image.
Question 2.
Fill in the number sentences and the number bonds.
Take away 3 hats.
Explanation:
Given Cross out 3 hat.
Total number of hat = 6
number of hats taken away = 3
Number of hats left = 6 – 3 = 3.
This is represented in number bond and number sentence as shown in the above image.
Take away 4 cubes.
Explanation:
Given Take away 4 cubes.
Total number of cubes = 6
number of cubes taken away = 4
Number of cubes left = 6 – 4 = 2.
This is represented in number bond and number sentence as shown in the above image.
Draw 6 circles in a 5-group. Take away 2 circles.
Explanation:
Given take away 2 circles.
The circles are represented in a group of 5.
Total number of circle = 6
number of circles taken away = 2
Number of circles left = 6 – 2 = 4.
This is represented in number bond and number sentence as shown in the above image.
Eureka Math Kindergarten Module 4 Lesson 22 Homework Answer Key
Question 1.
Here are 6 books. Cross out 2. How many are left? Fill in the number bond and the number sentence.
Explanation:
Given Cross out 2 books.
Total number of books = 6
number of books crossed out = 2
Number of books left = 6 – 2 = 4.
This is represented in number bond and number sentence as shown in the above image.
Question 2.
Draw 6 stars. Cross out 4. Fill in the number sentence and the number bond.
Explanation:
6 stars are drawn.
Given Cross out 4 Stars.
Total number of stars = 6
number of stars crossed out = 4
Number of stars left = 6 – 4 = 2.
This is represented in number bond and number sentence as shown in the above image.
Question 3.
Draw 6 objects. Cross out 5. Fill in the number sentence and the number bond.
Explanation:
6 objects are drawn.
Given Cross out 5 objects.
Total number of objects drawn = 6
number of objects crossed out = 5
Number of objects left = 6 – 5 = 1.
This is represented in number bond and number sentence as shown in the above image.
Question 4.
On the back of your paper, draw 6 triangles. Cross out 1. Write a number sentence, and draw a number bond to match. |
# SAMACHEER KALVI MATH SOLUTION FOR EXERCISE 3.3 PART 3
This page Samacheer Kalvi Math Solution for Exercise 3.3 part 3 is going to provide you solution for every problems that you find in the exercise no 3.3
## Samacheer Kalvi Math Solution for Exercise 3.3 part 3
(vii) 2 x² - 2 2x + 1
Solution:
Let p(x) = 2 x² - 2 2x + 1 = (√2x - 1) (√2x - 1)
p(x) = 0 => (√2x - 1) (√2x - 1) = 0
√2x - 1 = 0
√2x = 1
x = 1/√2
x = 1/√2 x = 1/√2
p(1/√2) = 2 (1/√2)² - 2 2(1/√2) + 1
= 1 - 2 + 1
= 2 - 2
= 0
sum of zeroes = (1/√2) + (1/√2) = 2/√2
product of zeroes = (1/√2) (1/√2) = 1/2
2 x² - 2 2x + 1
ax² + bx + c
a = 2, b = -2 2 and c = 1
Sum of roots α + β = -b/a
α + β = -(-2 2)/2
α + β = 1
Product of roots α β = c/a
α β = 1/2
Thus the relationship verified
(viii) x² + 2 x - 143
Solution:
Let p(x) = x² + 2 x - 143 = (x - 11) (x + 13)
p(x) = 0 => (x - 11) (x + 13) = 0
x - 11 = 0 x + 13 = 0
x = 11 x = -13
x = 11 x = -13
p(11) = (11)² + 2 (11) - 143
= 121 + 22 - 143
= 143 - 143
= 0
p(-13) = (-13)² + 2 (-13) - 143
= 169 - 26 - 143
= 143 - 143
= 0
sum of zeroes = 11 + (-13) = -2
product of zeroes = 11(-13) = -143
x² + 2 x - 143
ax² + bx + c
a = 1, b = 2 and c = -143
Sum of roots α + β = -b/a
α + β = 2/1
α + β = 2
Product of roots α β = c/a
α β = -143/1
= -143
Thus the relationship verified
In the page samacheer kalvi math solution for exercise 3.3 part 3 we are going to see the solution of next problem
(2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 3 , 1
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = 3 β = 1
x² - (3 + 1) x + 3 (1) = 0
x² - 4 x + 3 = 0
(ii) 2 , 4
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = 2 β = 4
x² - (2 + 4) x + 2 (4) = 0
x² - 6 x + 8 = 0 |
Relations and Mapping are important topics in Algebra. Relations & Mapping are two different words and have different meanings mathematically. Let’s get deep into the article to know all about the Relations, Mapping, or Functions like Definitions, Types of Relations, Solved Examples, etc.
## What is a Relation?
A relation is a collection of ordered pairs. Relation in general defines the relationship between two different sets of information. An ordered pair is a point that has both x and y coordinates. Let us consider two sets x and y and set x has a relation with y. The values of set x are called Domain and the values of set y are called Range.
Relations can be represented using three different notations i.e. in the form of a table, graph, mapping diagram.
Example: (2, -5) is an ordered pair.
### What is Mapping?
Mapping denotes the relation from Set A to Set B. Relation from A to B is the Subset of AxB. Mapping the oval on the left-hand side denotes the values of Domain and the oval on the right-hand side denotes the values of Range.
From the above diagram, we can say the ordered pairs are (1,c) (2, n) (5, a) (7, n).
Set{ 1, 2, 5, 7} represents the domain.
{a, c, n} is the range.
A function is a relation that derives the output for a given input.
Remember that all functions are relations but not all relations are functions.
### Types of Relations
There are 8 different types of Relations and we have mentioned each of them in the following modules along with Examples.
• Empty Relation
• Universal Relation
• Identity Relation
• Inverse Relation
• Reflexive Relation
• Symmetric Relation
• Transitive Relation
• Equivalence Relation
Empty Relation:
Empty Relation is the one in which there will be no relation between elements of a set. It is also called a Void Relation. For instance, Set A = {3, 4, 5} then void relation can be R = {x, y} where | x- y| = 7.
For an Empty Relation R = φ ⊂ A × A
Universal Relation:
In Universal Relation every element of a set is related to each other. Consider a set A = {a, b, c}. Universal Relations will be R = {x, y} where, |x – y| ≥ 0.
For Universal Relation R = A × A
Identity Relation:
Every element of a set is related to itself in an Identity Relation. Consider a Set A = {a, b, c} then Identity Relation is given by I = { a,a}, {b,b}, {c, c}
I = {(a, a), a ∈ A}
Inverse Relation:
In Inverse Relation when a set has elements that are inverse pairs of another set. If Set A = {(a,b), (c,d)} then inverse relation will be R-1 = {(b, a), (d, c)}
R-1 = {(b, a): (a, b) ∈ R}
Reflexive Relation:
Every element maps to itself in Reflexive Relation. Consider Set A = { 3, 4} then Reflexive Relation R = {(3, 3), (4, 4), (3, 4), (4, 3)}. Reflexive Relation is given by (a, a) ∈ R
Symmetric Relation:
In the case of Symmetric Relation if a = b, then b = a is also true. Relation R is symmetric only if (b, a) ∈ R is true when (a,b) ∈ R.
aRb ⇒ bRa, ∀ a, b ∈ A
Transitive Relation:
In case of a transitive relation if (x, y) ∈ R, (y, z) ∈ R then (x, z) ∈ R
aRb and bRc ⇒ aRc ∀ a, b, c ∈ A
Equivalence Relation:
A Relation symmetric, reflexive, transitive at the same time is called an Equivalence Relation.
### How to Convert a Relation to Function?
A relation that follows the rule that every X- Value associated with only one Y-Value is called a Function.
Example
Is A = {(1, 4), (2, 5), (3, -8)}?
Solution:
Since the set has no duplicates or repetitions in the X- Value, the relation is a function.
### Mapping Diagrams
Mapping Diagram consists of two columns in which one denotes the domain of a function f whereas the other column denotes the Range. Usually, Arrows or Lines are drawn between domain and range to denote the relation between two elements.
One-to-One Mapping
Each element of the range is paired with exactly one element of the domain. The function represented below denotes the One-to-One Mapping.
Many to One Mapping
If one element in the range is associated with more than one element in the domain is called many to one mapping. In the below diagram you can see the second number II is associated with more than one element in the domain.
One to Many Mapping
If one element in the domain is mapped with more than elements in the range then it is called One to Many Mapping. In the below diagram the first element in the domain is mapped to many elements in the range therefore it is called One to Many Mapping.
Hope you got a complete idea on Relations and Mapping Concept. If you need any other help you can ask us through the comment section and we will get back to you at the earliest possibility. |
```Peer Instruction in Discrete
under a Creative Commons AttributionNonCommercial-ShareAlike 4.0
Based on a work
at http://peerinstruction4cs.org.
Permissions beyond the scope of this
at http://peerinstruction4cs.org.
CSE 20 –
Discrete
Mathematics
Dr. Cynthia Bailey Lee
Dr. Shachar Lovett
2
Today’s Topics:
1.
2.
Knights and Knaves
3
1. Knights and Knaves
4
Knights and Knaves
Knights
and Knaves scenarios are
somewhat fanciful ways of formulating
logic problems
Knight:
everything a knight says is true
Knave: everything a knave says is false
5
You approach two people, you know
the one on the left is a knave, but you
don’t know whether the one on the
right is a knave or a knight.
Left:
“Everything she says is true.”
Right: “Everything I say is true.”
What
A.
B.
C.
D.
is she (the one on the right)?
Knight
Knave
Could be either/not enough information
6
You approach one person, but you
don’t know whether he is a knave or a
knight.
Mystery
What
A.
B.
C.
D.
person: “Everything I say is true.”
is he?
Knight
Knave
Could be either/not enough information
7
You approach one person, but you
don’t know whether she is a knave or a
knight.
Mystery
What
A.
B.
C.
D.
person: “Everything I say is false.”
is she?
Knight
Knave
Could be either/not enough information
8
You meet 3 people:
A: “At least one of us is a knave.”
B: “At most two of us are knaves.”
[C doesn't say anything]
A.
B.
C.
This is a really tricky one, but take a moment to
see if you can determine which of the following is
a possible solution:
A: Knave, B: Knave, C: Knave
A: Knight, B: Knight, C: Knight
A: Knight, B: Knight, C: Knave
(Suggestion: eliminate wrong choices rather than
trying to solve the puzzle directly. In your groups:
please discuss logic for eliminating choices.)
9
10
A.
B.
C.
D.
What are they?
1. Assume what you are proving, 2. plug in
definitions, 3. do some work, 4. show the
opposite of what you are proving (a
1. Assume the opposite of what you are
proving, 2. plug in definitions, 3. do some work,
4. show the opposite of your assumption (a
1. Assume the opposite of what you are
proving, 2. plug in definitions, 3. do some work,
4. show the opposite of some fact you already
Other/none/more than one.
11
You meet 3 people:
A: “At least one of us is a knave.”
B: “At most two of us are knaves.”
[C doesn't say anything]
A:
Knight, B: Knight, C: Knave
Zeroing in on just one of the three parts of
the solution, we will prove by contradiction
that A is a knight.
12
A: “At least one of us is a knave.”
B: “At most two of us are knaves.”
[C doesn't say anything]
Thm. A is a knight.
Assume not, that is, assume A is a knave.
Try it yourself first!
13
A: “At least one of us is a knave.”
B: “At most two of us are knaves.”
[C doesn't say anything]
Thm. A is a knight.
Assume not, that is, assume A is a knave.
Then what A says is false.
Then it is false that at least one is a knave, meaning zero are
knaves.
So A is not a knave, but we assumed A was a knave, a
So the assumption is false and the theorem is true. QED.
14
You meet 3 people:
A: “At least one of us is a knave.”
B: “At most two of us are knaves.”
[C doesn't say anything]
A:
Knight, B: Knight, C: Knave
Zeroing in on the second of the three parts
of the solution, we will prove by
contradiction that B is a knight.
15
A: “At least one of us is a knave.”
B: “At most two of us are knaves.”
[C doesn't say anything]
Thm. B is a knight.
Assume not, that is, assume B is a knave.
Try it yourself first!
16
A: “At least one of us is a knave.”
B: “At most two of us are knaves.”
[C doesn't say anything]
Thm. B is a knight.
Assume not, that is, assume B is a knave.
Then what B says is false, so it is false that at most two are knaves.
So it must be that all three are knaves.
Then A is a knave.
So what A says is false, and so there are zero knaves.
So B must be a knight, but we assumed B was a knave, a
So the assumption is false and the theorem is true. QED.
17
A: “At least one of us is a knave.”
B: “At most two of us are knaves.”
[C doesn't say anything]
Thm. B is a knight.
Assume not, that is, assume B is a knave.
Then what B says is false, so it is false that at most two are knaves.
So it must be that all three are knaves.
We didn’t
Then A is a knave.
need this step
So what A says is false, and so there are zero knaves.
because we
But all three are knaves and zero are knaves is a contradiction. |
## Maths worksheet: Moving towards a standard method of division
This is the second division worksheet which looks at the intermediate stage between mental methods of division and a standard method. I am not a great fan of this method but it is used in many schools.
The first thing to do, as always is to make an estimate of the answer so that any silly slip ups in the calculation may immediately come to light. The second stage is to find a multiple of ten that can be taken away – in all these examples and questions this multiple is 10 itself. Then we move towards the more traditional method of asking ‘how many…. in….’.
If you are just moving towards doing division, perhaps in Year 3/4 it is certainly a method which can be used.
## Maths worksheet: Dividing a 2-digit number by a single digit.
When parents come to help their children with maths it is often division which is the most problematic area. Frequently it is the parents who claim they do not understand the ‘modern methods’ used in school today. The final efficient, or standard method of division is much the same method as it has always been: it is the stages that are used to reach this standard method which cause most confusion. These stages, or steps are aimed at providing children with an understanding of the whole process, although I believe that sometimes they can lead to more confusion.
Here we have one of these steps towards an efficient method.
77 ÷ 4
4)77
40 (10 × 4)
1137
36 ( 9 × 4)
0111
Next, take away a tens multiple of the divisor (4). 10 x 4 = 40.
Subtract 40 from the 77 leaves 37.
Then ask, ‘How many 4s in 37?’
9 x 4 is 36 so it is 9 with a remainder of 1.
Add the 10 and the 9 which makes 19.
All these questions have a single tens multiple of the divisor. Of course, this will not always be the case as questions become harder.
Division: 2 digit by 1 digit (pg 1)
## Free maths worksheet: Division and multiplication corresponding
This maths worksheet looks at the correspondence between multiplication and division. Given one fact, three others can be worked out. For example:
Given that 5 x 4 = 20 then:
a. 4 x 5 = 20
b. 20 ÷ 5 = 4
c. 20 ÷ 4 = 5
A full understanding of this relationship is needed before children can go on to written methods of division. If your child is struggling with division it would be well worth checking that they understand this.
To make sense of this it is a good idea to ‘talk through’ what these number sentences say.
If 5 lots of 4 are 20 then 4 lots of 5 will also be 20.
If 20 sweets are shared equally between 4 people they get 5 each.
If 20 sweets are shared equally between 5 people they get 4 each.
Multiplication and division corresponding
## Year 2 maths worksheet:Understanding division as sharing
Division is usually the hardest of the four rules for children to learn, but in the early stages it is quite straightforward. Children often come across division for the first time when sharing, usually between two. The key concept,of course, is that the sharing is done equally. So, for example, 6 sweets shared between 2 children implies that the sharing is equal and they both receive the same number.
By far the best way to practice sharing is to use practical apparatus or use real life situations. Eg share the strawberries equally between two, share or deal the cards equally etc. Usually this is done on a ‘one for you and one for me’ type process until there are none left. This maths worksheet replicates a practical situation, with the ultimate aim that children begin to remember the answers, which, of course are the inverse of multiplication.
Division as sharing (pg 1) |
# Where is the Contradiction? Elementary Number Theory Proof : No natural numbers x, y such that $x^2 - y^2 = 2s$ where s odd integer.
Let $$m$$ be a positive integer of the form $$m = 2s$$, where $$s$$ is an odd integer. Prove that there do not exist positive integers $$x$$ and $$y$$ such that $$x^2 - y^2 = m.$$
Proclaimed Solution via Proof by Contradiction: Assume, to the contrary, that there exist positive integers $$x$$ and $$y$$ such that $$(x - y)(x + y)= m = 2s \quad \text{ where s is an odd integer.} \tag{*}$$ We consider two cases, according to whether $$x$$ and $$y$$ are of the same parity or of opposite parity. $$\boxed{\text{Case 1:}}$$ If $$x$$ and $$y$$ are of the same parity, then both $$x - y$$ and $$x + y$$ are even.
$$\boxed{\text{Case 2:}}$$ If $$x$$ and $$y$$ are of opposite parity, then both $$x - y$$ and $$x + y$$ are odd. Produce a contradiction in each case.
$$\Large{1.}$$ The product of two even numbers is always even. Thus, $$(x - y)(x + y)$$ is even. Then $$(*)$$ becomes: $$\text{even = 2(odd)}$$. Where is the contradiction?
The analogous solution works for Case 2? The product of two odds is always odd. Then $$(*)$$ becomes: $$\text{odd = (2s = even)}$$. Contradiction.
$$\Large{2.}$$ What is the motivation behind considering "whether $$x$$ and $$y$$ are of the same parity or of opposite parity"?
Source: Exercise 5.22, P124 of Mathematical Proofs, 2nd ed. by Chartrand et al
1. Suppose that $x+y$ and $x-y$ are both even; then there are integers $k$ and $\ell$ such that $x+y=2k$ and $x-y=2\ell$. But then $2s= (x+y)(x-y)=4k\ell=2(2k\ell)$, so $m=2k\ell$, and $s$ is therefore even, contradicting the hypothesis that $s$ is odd.
2. The two cases need to be considered separately, because each requires a different argument. Are you also asking how you might realize that if confronted with the problem? That’s largely a matter experience and partly a matter of recognizing that when the hypotheses specifically involve parity ($m$ is odd), parity may play a rôle in the proof.
• $m$ is meant to be even - I believe you mean that $2s = 4 kl \implies s = 2kl$, contradicting that $s$ is odd.
– user61527
Commented Sep 6, 2013 at 5:58
• @T.Bongers: Yep; I mentally interchanged $m$ and $s$. Thanks. Commented Sep 6, 2013 at 6:01
• @BrianM.Scott: Many thanks. Yes, I was asking how I would foreknow/prevision the master step here of "whether $x$ and $y$ are of the same parity or of opposite parity"? Is T. Bongers's explanation (For which I upvoted) analogous to yours? Or are there other ways?
– user53259
Commented Sep 6, 2013 at 8:53
• @LePressentiment: T. Bonger’s explanation goes a little deeper than mine. Mine was intended to explain why you might think about parity in the first place; his actually deals more with how you might see the contradiction that arises if $x+y$ and $x-y$ have opposite parity. I can’t offhand think of another really likely way to approach the problem. Commented Sep 6, 2013 at 8:57
• @BrianM.Scott: Thank you again for your help. I would've liked to choose both answers. I have voted for yours for now because I didn't see anything in the question that involved divisiblity of 4 and so I wouldn't have reckoned divisibility by 4 which underlies T. Bonger's answer.
– user53259
Commented Sep 6, 2013 at 9:07
Regarding case 1: If $x - y$ and $x + y$ are both even, then $(x - y)(x + y)$ is divisible by $4$, and so $s$ cannot be odd.
As far as motivation, the technique is somewhat suggested by considering that the right hand side is divisible by $2$ exactly once - so the left side must be divisible by $2$ exactly once, so $x - y$ and $x + y$ must have opposite parity.
• Thank you very much. I value your answer to my Q2 the most. Sadly, only one best answer is allowed, so I've upvoted at the moment. I hope that you will not mind my choice. Would you be able to explicate how and why one would divine/suspect that it is the divisibility by 4 which begets the contradiction? Only because I still don't apprehend this do I feel Prof Scott's solution is more natural. If this could please be unriddled and Prof Scott does not mind, I'll be happy to change the Accepted Answer.
– user53259
Commented Sep 6, 2013 at 9:21
For the first case, the contradiction lies in the fact that both $\left(x-y\right)$ and $\left(x+y\right)$ are even. This means that \begin{eqnarray*} \left(x-y\right)\left(x+y\right) & = & \left(2x\right)\left(2y\right)\\ & = & 4xy. \end{eqnarray*} If a number is $2s=2\left(2k+1\right)=4k+2$, then it is not a multiple of four. That's the contradiction.
• You've used $x$ and $y$ twice to mean two different things in your first set of equations.
– user61527
Commented Sep 6, 2013 at 5:57
1) The contradiction is that $(x-y)(x+y)$ is divisible by $4$, but $2$(odd) is not.
2) The motivation is that it exhausts all possibilities. If x and y are not of the same parity, we have the contradiction that $(x+y)(x-y)$ is odd, but $2s$ isn't. |
## 7.1 Sines and Cosines and their Derivatives
At this point we claim that the derivative of the sine function is another related function called the cosine function. This may seem mysterious to you since we have not defined either of these functions, but we will eventually define them.
The cosine function, whose value at argument x is generally written as cos(x) is a short way of saying the sine of the complementary angle to x. The complementary angle to x is the difference between a right angle and the angle x.
We will always try to measure angles in radians, though out of habit, old farts like myself often lapse into describing them by degrees.
Imagine we have an angle at some point P and draw a circle C around P that has radius 1. Then the size of the angle in radians is the length of the arc between the end-lines of on the circle C.
It is an important bit of folklore that the total distance around a circle of unit radius is . Thus the size of an angle is the proportion of the circle that it represents, multiplied by the factor .
We can therefore see that a straight line angle represents half a circle so it has
angle while a right angle, which is half of a straight line, has angle . The complementary angle to is the angle .
So the derivative of the sine (written usually as sin) obeys
by the chain rule, we get
These facts about the derivatives of the sine and cosine are almost as simple as those for the exponent, and they are not difficult to use in practice. For more details about trigonometric functions, click here.
Armed with these last two facts we can use the substitution rule and our previous rule to differentiate any function we care to construct from the identity, the exponent, and the sine by arithmetic operations and substitution.
Are we done?
We are almost done. Practice using the multiple occurrence rule and the chain rule a bit and you can become an expert differentiator. But we still have to notice how to differentiate inverse functions.
Exercises:
7.1 Find the derivative for each of the following functions:
a. (sin x) * exp(2x)
b. x * cos 2x
c. (cos x) * sin x
7.3 The derivative of the derivative of a function f is called the second derivative of f. Find the second derivative of cos x, and also of sin 2x.
The second derivative of f is usually denoted by f ''(x) or .
7.4 What functions can you think of that obey the equation f(x) + f ''(x) = 0?
Why do you start only with the sine function exponential and identity? what happened to the cosine? Or to the other trigonometric functions? And what are those weird things on my calculator like cosh and sinh?
We don't bother with treating the cosine separately since we can define it from the sine by substitution: . The other trigonometric functions can be defined from these two. The functions with the h on the end are called hyperbolic sine and cosine. They are easily expressed in terms of the exponential function:
The other functions that appear on good calculators and are available on spreadsheets are easily constructed from those mentioned so far or from their inverses.
And what are inverses?
We will see that now, and also how to find their derivatives. |
# How do you factor the expression 49p^2 + 63pq - 36q^2?
Jan 23, 2017
$49 {p}^{2} + 63 p q - 36 {q}^{2} = \left(7 p + 12 q\right) \left(7 p - 3 q\right)$
#### Explanation:
Note that $49 {p}^{2} = {\left(7 p\right)}^{2}$ and $36 {q}^{2} = {\left(6 q\right)}^{2}$
Further note that $63 = {3}^{2} \cdot 7$ is divisible by $7$ and by $3$ but not by $6$.
So let's look at this quadratic in terms of $7 p$ and $3 q$...
$49 {p}^{2} + 63 p q - 36 {q}^{2} = {\left(7 p\right)}^{2} + 3 \left(7 p\right) \left(3 q\right) - 4 {\left(3 q\right)}^{2}$
Note that $4 - 1 = 3$ and $4 \cdot 1 = 4$, so we find:
${\left(7 p\right)}^{2} + 3 \left(7 p\right) \left(3 q\right) - 4 {\left(3 q\right)}^{2} = \left(\left(7 p\right) + 4 \left(3 q\right)\right) \left(\left(7 p\right) - \left(3 q\right)\right)$
$\textcolor{w h i t e}{{\left(7 p\right)}^{2} + 3 \left(7 p\right) \left(3 q\right) - 4 {\left(3 q\right)}^{2}} = \left(7 p + 12 q\right) \left(7 p - 3 q\right)$
Jan 23, 2017
$- 36 \left(q + \frac{7}{12} p\right) \left(q - \frac{7}{3} p\right)$
#### Explanation:
Making $q = \lambda p$ and substituting
$49 {p}^{2} + 63 p q - 36 {q}^{2} = \left(49 + 63 \lambda - 36 {\lambda}^{2}\right) {p}^{2}$
but $\left(49 + 63 \lambda - 36 {\lambda}^{2}\right) = - 36 \left(\lambda + \frac{7}{12}\right) \left(\lambda - \frac{7}{3}\right)$
so
$- 36 \left(\lambda + \frac{7}{12}\right) \left(\lambda - \frac{7}{3}\right) {p}^{2} = - 36 \left(\lambda p + \frac{7}{12} p\right) \left(\lambda p - \frac{7}{3} p\right) = - 36 \left(q + \frac{7}{12} p\right) \left(q - \frac{7}{3} p\right)$ |
A summer program and resource for middle school students showing high promise in mathematics
### Why the Long-Division-Type Square-root Algorithm works
To show why the square-root algorithm works we use the Proposition 1 below.
The algorithm merely implements for anan-1...a1a0 the decomposition shown on the right side of Equation (1). Let us see how the algorithm obtains 102nan2 which is the first term on the right side of Equation (1). The algorithm subtracts an2 in the first step. Each of the further subtractions takes place by bringing down two digits. Since this happens n times by the time the digit a0 is obtained, the place value of the an2 that was first subtracted becomes (102)n = 102n. Similarly, the second number subtracted, namely, (10 × 2 × an + an-1)an-1, acquires palce value 102(n-1), the third number acquires place value 102(n-2), and so on. Then, adding up all the numbers subtracted in the algorithm gives the decomposition of (anan-1...a1a0)2 given by Proposition 1.
We have seen elsewhere that 119025 is a perfect square. What if we are looking for the square-root of 119026? Then we have a remainder of 1 in the "division" process of the algorithm and so we think of 119026 as 119026.00 and bring down the two zeros. Why?
Since two consecutive integers can not be both perfect squares, 119026 can not be a perfect square. So its square-root must be of the form
x = anan-1...a1a0.a-1a-2...a-k, say, up to the kth decimal place, and, so, equals 10-k(anan-1...a1a0a-1a-2...a-k).
Let n+k = m.
Rename ai as ak+i, i = n, n-1,...,2,1,0,-1,...,-k.
Then, x = 10-k(amam-1...akak-1...a0) and, from Proposition 1,
.
This means that, in finding the square-root of a number that is not a perfect square, if all the digits to the left of the decimal point have been brought down in the division process of this algorithm, we continue the process beginning from the immediate right of the decimal point, successively bringing down groups of two digits and repeating the "division" as before. After k groups of two digits each are brought down beginning from the digit to the immediate right of the decimal point, we have obtained the square-root of (10kx)2. Therefore, the value of x which is the square-root of the given number is obtained up to k decimal digits by dividing the number obtained in the division process by 10k, or by inserting a decimal point just before the last k digits of the "quotient." Observe that the procedure is the same whether or not the given number is an integer. If the number is an integer, but not a perfect square, the groups of digits to the right of the decimal point are all 00.
For numbers that are not integers, parse their digits in two's going left from the decimal point, and then going right from the decimal point. If there are k groups of two digits each to the right of the decimal point so that the last group has at least one non-zero digit, then the square-root will have k decimal digits. In the division process, we insert the decimal point in the "quotient" row just prior to bringing down the first group of two digits after the decimal point of the given number.
We have shown the following.
Proposition 2 (The square-root algorithm)
Let N be a number in decimal representation. The square-root of N is obtained by the following procedure.
(1) Beginning at the decimal point, parse N in groups Ai, i = 0,..,n, of two digits each going left from the decimal point and then in groups A-j, j = 1,...,k, of two digits going right from the decimal point so that N is given by the concatenation
AnAn-1...A1A0 . A-1A-2...Ak-1A-k, where the left-most group An on the left of the decimal point and the right-most group A-k, on the right of the decimal point, each has one or two digits, depending on the number of digits of N.
(2) Choose an so that (an)2 is the largest square not exceeding An. Let Bn-1 be the number obtained by concatenating
(An - an2) with An-1. That is, Bn-1 = 100(An - an2) + An-1.
(3) For i = n-1 to -k, choose ai, being the largest single-digit number, so that ((10 × 2 × (an...ai+1) + ai)ai does not exceed Bi. Let Bi-1 = 100(Bi - ((10 × 2 × (an...ai+1) + ai)ai) + Ai-1.
(4) Then the square-root of N is anan-1...a1a0.a-1a-2...a-k, up to k decimal digits.
Observe that when N is an integer, Ai = 00, for i = -1 to -k. And when N is an integer that is a perfect square, Bi = 0 for i = -1 to -k as well.
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# 5 Mathematical Facts About Craps You Need to Know
I like the simplicity of craps. This might sound strange, because craps looks awfully complicated. But everything that happens at the craps table is based on rolling or throwing two dice.
With only six possible results on one die, the math can be pretty easy once you get the hang of it. It’s a little more advanced with two dice, but it’s still rather simple.
When you know the way a die works, you can use the math behind it to help you play better craps. Here are five mathematics-related facts about craps that you need to know.
## 1 – How Dice Work
The dice used in craps are the most common ones, each one with six sides. The game uses two of these, and this creates the mathematical base for craps.
When the two dice are rolled, each of them lands on a number from one to six. The two are added together for a final result. This means the lowest possible total is two, when both dice land on one, and the highest possible total is 12, when both dice land on six.
You probably already know all of these things, but do you understand how these things influence how you win and lose playing real money craps?
The odds of rolling a seven are much higher than rolling a two or a 12. Do you know why? To roll a two, this can only happen if both the first and second die land on one. But to roll a seven there are many combinations the dice can land on.
In total, there are 36 possible combinations when you use two dice. Since there are six combinations or ways to roll a seven, you can determine the percentage chance of a seven on any single roll by dividing six by 36. The chance of rolling a seven is 16.67%.
Rolling a total of two only has one combination, so the chance of that is 2.78%. Here are the number of combinations for each total and the percentage chance for each.
Dice Total Combinations % Chance 2 1 2.78% 3 2 5.56% 4 3 8.33% 5 4 11.11% 6 5 13.89% 7 6 16.67% 8 5 13.89% 9 4 11.11% 10 3 8.33% 11 2 5.56% 12 1 2.78%
## 2 – Odds Bet Facts
The odds wagers in craps are unique for a couple of different reasons. The first thing is because these betting options aren’t shown on most craps tables. In other words, there’s not a spot on the table that shows where you can place an odds wager.
The other unique thing about odds wagers is that they offer a true even house edge. In other words, there’s no house edge on any of the odds wagers.
If you understand how casinos work, you know that they don’t often offer wagers that don’t have a house edge. So, the question is, how do the casinos make money by offering a wagering opportunity that doesn’t have a house edge?
The answer is that you can’t place an odds wager unless you place a come-out roll wager first. The casino makes a profit on the come-out wagers, so they can afford to offer the odds wagers.
The odds wagers pay out based on the point. Once a point is set, you can place an odds wager, and if the point is rolled before a seven, you win. This is if you originally made a wager on the pass line.
If you originally bet on the don’t pass line, you win your odds wager if a seven is rolled before the point.
Here are the payouts for pass line odds:
On 10 and 4 2 to 1 On 9 and 5 3 to 2 On 8 and 6 6 to 5
Here are the payouts on don’t pass line odds:
On 10 and 4 1 to 2 On 9 and 5 3 to 3 On 8 and 6 5 to 6
## 3 – Pass Line Math
When you make a pass line wager, you win on the come out roll with an 11 or seven. This means that you win on eight of the 36 possible combinations, or 22.22% of the time. You lose on a roll of 12, three, or two. This is four out of the 36 combinations, or 11.11% of the time.
Other roll totals don’t immediately win or lose. Instead, they set the point. When you roll the point again before rolling a seven, you win. If you roll a seven before the point, you lose.
The math is somewhat complicated. But the important thing to know is that, when you make a pass line wager, the casino has an edge of 1.41%.
The pass line is the most common wager made by craps players on come-out rolls, but it’s not the best option when you’re trying to play with a low house edge. The best option for the house edge for craps players is to not play at all.
Craps has a built-in house edge, so no matter what you do, you’re going to lose if you play. The second-best option is a wager on the don’t pass line. More information about the don’t pass line is included in the following section.
## 4 – Don’t Pass Line Math
The don’t pass line wager is the opposite of the pass line wager in many ways. But it has a house edge of only 1.36%.
The truth is that the difference between 1.36% and 1.41% isn’t much, so if you’re more comfortable making the pass line wager, it doesn’t cost much money. In fact, it only costs you an average of 50 cents for every \$1,000 that you wager to bet on the pass line instead of the don’t pass line.
You win on a roll of three or two, and lose on a roll of 11 or seven. This means you have a 22.22% chance of losing on the come-out roll. But if you survive the come-out roll, you have the best chance to win. A roll of seven, before the point is rolled again, wins this wager.
The only reason why anyone makes the pass line wager instead of the don’t pass line wager is because they don’t know any better or because of public pressure. A don’t pass line wager is often frowned on by superstitious gamblers because it’s viewed as betting against the shooter.
But the way you decide to play craps doesn’t have anything to do with anyone else at the craps table. You’re gambling with your money, so you can bet on anything you want.
## 5 – Every Other Wagering Option
You learned about the two main wagers at craps and the odds wagers. You can make many other wagers while playing craps, but none of them are as good as the three wagers you already know about. In fact, most of the other wagers have much higher house edges and should be avoided at all times.
The next closest house edge is the place 6/8 bet, with a house edge of 1.52%. Many of the other craps wager options have a house edge over 10%. You’re better off playing slots than making wagers with a 10% or higher house edge.
I’ve read about many different craps strategies and betting systems that involve other wagers. But none of them actually increase your chances to win or decrease the house edge. No matter what you read or see, the only way to play craps with the lowest possible house edge is to stick with one of the two come-out roll wagers and place an odds wager.
## Conclusion
Now that you know exactly how the gambling math behind craps works, you can use it to help you win as often as possible. The math shows that there are only three craps wagers that you should ever make, so you can safely ignore all of the other wagering options.
If you’re still not 100% confident in your knowledge of how dice or how the odds work, spend some more time going over the first section on this page. Understanding the odds and percentages is the key to winning more while playing craps. |
how to calculate quatity of sand
## how to calculate quatity of sand
##### How To Calculate The Quantity Of Sand And
2016-10-27 · This post is going to be made using a building on the DPC (Damp Prove Course) Level: This is the first 4 blocks laid on the building. After laying blocks to this
##### How To Calculate Quantity Of Soil/Sand For
2021-1-5 · For the Process of compaction state, the required quantity of soil/sand for filling the plot = 600 x 1.30 = 780 cuft. (1.30 is a factor) Now we will calculate required no of Material Handling Trolley / trucks. Suppose do we have can have a 250kg load
##### How To Calculate Quantity Of Soil/Sand For Filling Plot
2018-5-1 · This volume is for loose state. For full compacted state, required quantity of soil/sand for filling the plot = 4092 x 1.30 = 5319.6 cft= 5320 cft. (1.30 is factor) Now we will calculate the required no of trally/ trucks. Suppose a trally can carry 70 cft soil. ∴ No. of trally rerquired =
##### How To Calculate Quantity Of Cement, Sand And
This article is about how to calculate the quantity of cement, sand, and aggregate in the concrete columns. after reading this article you will be able to find out the materials in slab, beam and also for Concrete road. We used here the same formula for finding out the quantity of beam and slab etc. []
##### How To Calculate Quantity Of Cement, Sand And
First, we calculate the volume of the slab, and then this volume is converted into a wet condition to dry condition and then find out how much material is required for the slab.
##### How To Calculate Cement, Sand & Aggregates Quantity in ...
2018-8-3 · For Calculate in CFT (Cubic Feet) = 8 x 1.225 = 9.8 Cubic Feet CALCULATION FOR SAND QUANTITY; Consider volume of concrete = 1m 3 . Dry Volume of Concrete = 1 x 1.54 = 1.54 m 3 Sand= (1.5/5.5) x 1.54 = 0.42 m 3 ∴ 1.5 is a part of Sand, 5.5 is sum of ratio . Density of Sand is 1450/m 3. For KG = 0.42 x 1450 = 609 kg
##### How to Calculate Cement Sand and Aggregate Quantity in ...
2021-11-4 · Quantity of Aggregate in cum = 1.54 / 7 × 4 (where 4 is the ratio of aggregate ) = 0.88 cum. Quantity of Aggregate in kg = 0.88 × 1500 (Density of Aggregate = 1450 to 1550 kg / cum ) = 1320 kg. Quantity of Aggregate in cubic feet = 1.54 / 7 × 4 × 35.314 = 31.07 cft . Alternative way. Quantity of Aggregate in cubic feet = Nos of cement bags × Volume of one bag cement in cft × Ratio of ...
##### Sand Calculator - How much sand do you need
Sand calculator online - estimate the sand required for your construction or landscaping project in weight (pounds, kilograms, tons, tonnes) and volume (cubic ft, cubic yards, cubic meters). If you are wondering 'how much sand do I need', our free sand
##### Sand Calculator
2021-8-1 · Calculate the area of the excavation by multiplying the length by the width. In our case, A = 12 * 3 = 36 yd 2. You can also type the area of the excavation directly into our calculator if you choose an excavation of some more sophisticated shape. Establish the
##### SAND CALCULATOR [How Much Sand do I Need?] -
2021-11-4 · Our sand calculator can be used to work out how much sand your next landscaping project will require. Work out the total volume, weight and cost of the sand. ... Use it to calculate the volume of sand you will need, the weight of the material, and if you know the price per unit mass/volume, then the total cost as well. Contents:
##### How to Calculate Cement, Sand Quantity for
2021-10-24 · To calculate the cement and sand for 100 m 2 plastering area in 1:6 ratio and 12 mm thickness. Cement Mortar Required: Plastering thickness 12 mm = 12/1000 = 0.012m. Volume of cement mortar required = ( Plastering Area x thickness ) ...
##### How to calculate 1 tractor or truck sand quantity in cft ...
2021-11-3 · For measuring 1 tipper sand capacity You will have to calculate the volume of the Tripper, length x width x height, then you can calculate the total volume of the sand. 1 tipper sand capacity in ton:- size of tipper vary from 180cft to 425cft, around 25cft = 1 ton, so 1 tipper sand capacity ranging between 7.5 ton to 17 ton.
##### How to Calculate CEMENT, SAND Quantity for
2017-10-15 · Points should be remembered while calculating the quantity of cement, sand for plastering work. For wall plastering, the ratio of Cement : Sand = 1 : 6. For ceiling plastering, the ratio of Cement : Sand = 1 : 4. Thickness of plaster varies according to type of work generally it is 6mm and 12-15mm. Use good quality of cement & Sand for plastering.
##### How to Calculate Cement, Sand Quantity for Plastering ...
2018-8-8 · Tags: brick work plaster cement sand quantity for plaster work How much sand and cement is required for 100 sq. foot plaster How to calculate cement sand quantity in 1:6 mortar How to calculate the plastering work quantity plaster calculatio plaster quantity
##### How to Calculate Cement, Sand and Coarse Aggregate ...
2021-10-24 · For Cement, Sand and Coarse Aggregate. This is a Volumetric Calculation. Assuming we need 2 m 3 of concrete for M20 Concrete Mix, ( Mix Ratio, M20 = 1 : 1.5 : 3)
##### how to calculate quatity of sandcement and aggregate in ...
how do we calculate ratio for m25 grade concrete. calculate material for m25 grade concrete. how we calculate of sand cement and aggregate of m25 ratio. how do we calculate quantity of cement How to Calculate the Water to Cement Ratio The Concrete Information about the importance of water to cement ratios and how it affects concrete quality. how to calculate cement in concrete
##### Sand Calculator
2021-8-1 · Calculate the area of the excavation by multiplying the length by the width. In our case, A = 12 * 3 = 36 yd 2. You can also type the area of the excavation directly into our calculator if you choose an excavation of some more sophisticated shape. Establish the
##### Calculator For Cement Sand Aggregate For M15
How To Calculate Quatity Of Sand Cement And . Calculate cement sand and aggregate nominal mix concrete civil ollowing is a simple method used by engineers to calculate cement, sand and aggregate to batch nominal mix concreteroper relative proportioning of the materials is necessary to attain the target strength and quality for the intended use.
2016-5-11 · Concrete Mix – Download Spreadsheet to Estimate Quantities of cement, sand, stone. May 11, 2016, 1:15 pm. To make concrete there are four basic materials you need: portland cement, sand, aggregate (stone), and water. The ratio of aggregate to
##### How do you calculate quantity? - Answers
2013-7-25 · It depends on what is being calculated? Basically, a collection of something is counted as 1, 2, 3, and so on, to give a total. A simple instance is in counting loose two pence (UK) coins: piles of two pence coins are stacked in tens, then the piles are counted as 10, 20, 30, and so on. Or if there are 34 piles of 10 coins, then 34 x 10 = 340 pence (£3.40).
##### How to Calculate CEMENT, SAND Quantity for
2017-10-15 · Points should be remembered while calculating the quantity of cement, sand for plastering work. For wall plastering, the ratio of Cement : Sand = 1 : 6. For ceiling plastering, the ratio of Cement : Sand = 1 : 4. Thickness of plaster varies according to type of work generally it is 6mm and 12-15mm. Use good quality of cement & Sand for plastering.
##### How to Calculate Cement, Sand Quantity for Plastering ...
2018-8-8 · Tags: brick work plaster cement sand quantity for plaster work How much sand and cement is required for 100 sq. foot plaster How to calculate cement sand quantity in 1:6 mortar How to calculate the plastering work quantity plaster calculatio plaster quantity
##### How To Calculate Number Of Bricks, Cement And
2018-9-26 · Now we Calculate Mortar quantity of cement and sand Ratio is 1:6. Mortar quantity is 0.2305 (Wet Condition) we calculate Dry Value Dry Mortar Value is = 0.2305 x 1.33 = 0.306565 m3. Cement Calculation. Volume = Volume of Mortar x (Ratio of
##### Calculator For Cement Sand Aggregate For M15
How To Calculate Quatity Of Sand Cement And . Calculate cement sand and aggregate nominal mix concrete civil ollowing is a simple method used by engineers to calculate cement, sand and aggregate to batch nominal mix concreteroper relative proportioning of the materials is necessary to attain the target strength and quality for the intended use.
##### Sand Calculator
2021-8-1 · Calculate the area of the excavation by multiplying the length by the width. In our case, A = 12 * 3 = 36 yd 2. You can also type the area of the excavation directly into our calculator if you choose an excavation of some more sophisticated shape. Establish the
##### How to Find Plastering quantity? | Cement, Sand,
2020-5-1 · Calculate the total volume of cement & Sand required for plastering; Now coming to the calculation part, We are considering the below values for calculation purpose: Wall width and height is 10m and 10m. Ratio of First coat of plastering
##### how we calculate of Sand, cement and aggregate of M20 ...
M20 (1 cement :1.5 sand :3 stone/brick aggregate). To determine the proportions you have to perform mix design, for this you have to find out the sp.gr. of cement, CA, FA, and water cement ratio ...
##### calculation for m25
Calculation Of Sand Cement Ratio In M25. How to calculate cement, sand, aggregate quantity.Now we start calculation for find cement, sand and aggregate quality in 1 cubic meter concrete.Calculation for cement quantity cement 15.5 x 1.54 0.28 m 3 1 is a part of cement, 5.5 is sum of ratio density of cement is 1440m 3 0.28 x 1440 403.2 kg we know ... |
# How do you simplify 2[3(3x +1)+ 2( 2x +5)] + 3[1+ 4(x+2)+5x]?
Oct 3, 2016
$53 x + 53$
#### Explanation:
Notice that there are 2 terms.
Start my removing the brackets within each term using the distributive law
color(red)(2[3(3x +1)+ 2( 2x +5)] )+ color(blue)(3([1+ 4(x+2)+5x]
=color(red)(2[9x +3+ 4x +10)] + color(blue)(3[1+ 4x+8+5x]" "larrsimplify
=$\textcolor{red}{2 \left[13 x + 13\right)} + \textcolor{b l u e}{3 \left[9 x + 9\right]} \text{ } \leftarrow$remove brackets
=$26 x + 26 + 27 x + 27 \text{ } \leftarrow$ simplify
=$53 x + 53$ |
# Sxx
Sxx represents the total squared deviations from the mean value of x in statistics. This value is frequently determined when manually matching a simple linear regression model.
## How To Calculate Statistics’ Sxx (With Example)
Sxx represents the total squared deviations from the mean value of x in statistics.
This value is frequently determined when manually ■■■■■■■ a simple linear regression model.
To determine Sxx, we apply the following formula:
Σ(xi - x)2 = Sxx
Where:
• A symbol that represents “sum” is.
• xi is the ith value of x, and x is its mean value.
• The application of this formula is demonstrated in the example that follows.
### Example: Hand-Calculating Sxx
Consider ■■■■■■■ a straightforward linear regression model on the following dataset:
X Y
1 8
2 12
2 14
3 19
5 22
8 20
Let’s say we want to find Sxx, the total How to find Standard DeviationDeviation from the mean value of x.
We must first determine the mean value of x:
x = (1 + 2 + 2 + 3 + 5 + 8) / 6 = 3.5
Next, we can compute Sxx’s value using the following formula:
Sxx = Σ(xi – x)
2
Sxx = (1-3.5)
2+(2-3.5)
2+(2-3.5)
2+(3-3.5)
2+(5-3.5)
2+(8-3.5)
2 \sSxx = 6.25 + 2.25 + 2.25 + .25 + 2.25 + 20.25
Sxx = 33.5
We calculate it to be worth 33.5.
According to this information, the total squared Variance between each x value and the mean x value is 33.5.
Be aware that we could also automatically determine the value of Sxx for this model using the Sxx Calculator.
## What Does SSxx Mean?
Square roots of x, added together And what does Sxy mean in the statistics? However, Sxx is the sum of the squares of each x’s deviation from its mean. Thus, the difference between the meaning of x and y results in the product Sxy. Thus the Sxx = (x - x) (x - ¯x) & Sxy = Σ (x - ¯x) (y - ¯y) are two examples.
### SXX Exists a gap in this situation?
However, the formula for variation is variation = sxx n - 1 = x2 - nx2 n - 1. The following is the standard Deviation: is equal to s. s = Variance = Sxx n - 1 = x2 - nx2 n - 1. Example: Calculate the standard Deviation using the Record values of 5, 7, 8, 9, 10, and 14. Remember that x = 9 first.
### How Can I Obtain SSXY?
Similarly, SSX is determined by adding x by x and then deducting xs by xs divided by n. The final step in calculating SSXY is to add x and y, followed by the total of xs times y divided by n.m.
### Formula For The Deviation
Before calculating deviations:
1. Calculate the mean value or the mean of the sample first.
2. Multiply the differences by the Standard of each data point.
3. Tally up all quadratic differences.
4. In the end, subtract one from the total and divide the result by n, the total number of data points in the sample.
### Variance In The Statistics
With probability and statistics, the expected Variance is the square Deviation of the random Variance from the stated. Randomly measures the distance from the description of a set of (random) numbers.
### What Does The Standard Deviation Mean?
A standard deviation indicates how we distribute a group’s measured values by the mean (mean) or expected value. More numbers are closer to meaning when we reduce the standard DeviationDeviation. High variation, on the other hand, indicates that the data are very dispersed.
### How Do You Find The Deviations In The Statistics?
To calculate the difference, calculate the approximate (simple number equation) and then, for each number, subtract the meaning and the square of the result (the difference between the squares). Then calculate the average of these quadratic differences.
## Summary
**SSXX is a modified sum of squares. To get SSXY, multiply Y by X, then XS by Y. ** A measurement called a standard deviation reveals the degree to which the measured values of a group vary from the mean (average) or expected value.
However, there are some frequently asked questions related to the topic “Sxx” are as follows:
### 1. What does the statistic Sxx stand for?
“Sample. the corrected sum of squares” is what Sxx stands for. It acts as a computational intermediary and is incapable of direct interpretation. Example: Take into account these five numbers: 28, 32, 31, 29, 39. Find the average of 159 and the sum of 159, which together equal 31.8. Now note the squares of the standard deviations.
### 2. What do the statistics terms SXX and SYY mean?
It must be quickly understood because it can only be used in computations. It is commonly used in regression and correlation analysis.
### 3. What does the statistic SSX stand for?
SSX stands for the total squared deviations from X’s mean… As a result, it equals ten and the sum of the x2 column. SSX = 10.00.
### 4. What makes SX and X different from one another?
The sample’s standard Deviation is Sx. The population standard deviation for the example is represented by the comparable but somewhat lower quantity (sigma)x.
### 5. What are the statistics are zX and zY?
It is easy to comprehend how to calculate the correlation coefficient for two variables, X and Y. The conventional forms of X and Y will be zX and ZY, respectively. We reexpress zX and zY with zero-mean and one-standard-deviation standard deviations (std).
### 6. Where is SSxx to be found?
Basic Linear Regression Calculations by Hand
We should calculate X variable average. Determine the variation between each X and the mean X. Add up all the discrepancies and square them. It’s SSxx here.
### 7. What does the statistic SSX stand for?
SSX is the total squared deviations from the mean of X. As a result, it equals ten and the sum of the x2 column. SSX = 10.00.
### 8. How is standard DeviationDeviation calculated?
Determine the mean first. Step 2: Determine the square of the deviation from the mean for each data point. In Step 3, add the values from Step 2. By the overall number of data points in step 4, divide.
### 9. Are the squares in SXX sum?
The sample adjusted sum of squares (SXX) is. It is the square of the difference between x and its mean. We can only use it in calculations. Thus it needs to be promptly comprehended. Regression and correlation analysis frequently use it.
### 10. Which of the following represents the SSxx formula?
B = SSxy, and a = y - b x SSxx. SSxx = x2 - (x)2 n SSxy = x y - (x) (y) n In today’s session, we will determine the Pearson Product Moment Correlation Coefficient, or r, that represents the relationship between the two variables x and y. We can also use the following fundamental methods to determine the values of SSxy, SSxx, and SSyy.
## Conclusion
The “sample. the adjusted sum of squares” is denoted by the notation Sxx. It serves as a computational middleman and lacks a direct interpretive capability. Example: Consider these five values: 28, 32, 31, 29, 39. Find the total 159 first, then the average of 159 5 = 31.8. |
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