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# The base of the prism is triangular in shape with sides of 3 cm, 4 cm, 5 cm. Find the volume of the prism if it’s height is 10 cm. a) 45 cu.cmb) 50 cu.cmc) 60 cu.cmd) 65 cu.cm
Hint: To solve the question, we have to calculate the area of the triangular base of the prism which when is multiplied with the height of the prism, will give you the volume of the prism.
We know that,
The volume of a prism = Area of the base of the prism $\times$the height of the prism ..… (1)
To calculate the area of the triangular base of the prism, we use the Heron's formula of area of a triangle which is equal to $\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$
Where a, b, c are three sides of the triangle and s is the semi-perimeter of the triangle.
We know that the formula for semi-perimeter of triangle with sides a, b, c is given by $s=\dfrac{a+b+c}{2}$
The given sides of the triangular base of the prism are 3 cm, 4 cm, 5 cm. By substituting the values of sides of triangle in the above mentioned formula of semi-perimeter, we get
$s=\dfrac{3+4+5}{2}=\dfrac{12}{2}$
$\Rightarrow s=6$
By substituting the values of semi-perimeter and the sides of triangle in the above mentioned formula of area of a triangle, we get
$A=\sqrt{6\left( 6-3 \right)\left( 6-4 \right)\left( 6-5 \right)}$
Where A represents the area of a triangular base of the prism.
$A=\sqrt{6\left( 3 \right)\left( 2 \right)\left( 1 \right)}$
$A=\sqrt{\left( 3\times 2 \right)\left( 3 \right)\left( 2 \right)}$
$A=\sqrt{{{2}^{2}}\times {{3}^{2}}}$
\begin{align} & A=\sqrt{{{\left( 2\times 3 \right)}^{2}}} \\ & =2\times 3 \\ & =6c{{m}^{2}} \\ \end{align}
Thus, the area of the triangular base of the prism of sides 3 cm, 4 cm, 5 cm is equal to 6$c{{m}^{2}}$.
The given value of height of the prism is equal to 10 cm.
By substituting the given and calculated values in equation (1) we get
Thus, the volume of a prism $=6\times 10=60cu.cm$
Hence, option (c) is the right choice.
Note: The possibility of mistake can be, not able to apply the correct formula for volume of prism and for the area of triangles of given sides. The alternative way of calculating the area of the triangle is by using the formula $\dfrac{1}{2}bh$ where b, h are the base and height of the triangle. Since the given sides of the triangle from the right-angle triangle, the other sides of the triangle excluding hypotenuse, form the base and height of the triangle. Thus, we can calculate the area of the triangular base of the prism.
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## GMAT Decimal
### Introduction
Decimals are a vital concept in mathematics and frequently appear in various sections of the GMAT exam. It is essential to understand decimal algebra, as it will help you solve problems involving decimals quickly and accurately. In this four-part series, we will explore the intricacies of decimal algebra, including decimal representation, arithmetic operations, decimal-fraction conversions, and real-world applications.
## Part 1: Decimal Representation and Basics
### 1.1 What are Decimals?
A decimal is a number system representation that uses a base of 10. It utilizes digits from 0 to 9 and a decimal point to represent fractional parts. Decimals are commonly used in everyday life for measurements, currency, and expressing values with varying degrees of precision.
### 1.2 Decimal Representation
A decimal number can be expressed as a whole number part and a fractional part, separated by a decimal point. The whole number part comprises the digits to the left of the decimal point, while the fractional part consists of the digits to the right of the decimal point.
For example, in the decimal number 12.345, 12 is the whole number part, and .345 is the fractional part.
### 1.3 Decimal Places
The position of a digit to the right of the decimal point determines its place value. The first digit after the decimal point represents tenths, the second digit represents hundredths, the third digit represents thousandths, and so on.
For example, in the decimal number 0.123, 1 is in the tenths place, 2 is in the hundredths place, and 3 is in the thousandths place.
### 1.4 Rounding Decimals
Rounding decimals is a technique used to approximate a decimal value to a specific number of decimal places. This simplifies calculations and makes it easier to work with decimals.
To round a decimal number:
• Identify the digit in the desired rounding place.
• Check the digit to the right of the rounding place.
• If the digit to the right is 5 or greater, round up the digit in the rounding place.
• If the digit to the right is less than 5, leave the digit in the rounding place unchanged.
• Remove all digits to the right of the rounding place.
For example, to round 12.345 to two decimal places, we check the third digit (5). Since it is greater than or equal to 5, we round up the second digit (4) to 5, resulting in 12.35.
### 1.5 Comparing Decimals
To compare decimal numbers, follow these steps:
Align the decimal points vertically.
Compare the whole number parts; the larger whole number part indicates a larger decimal.
If the whole number parts are equal, compare the digits in each decimal place, starting from the left. The decimal with the larger digit in the first unequal decimal place is the larger decimal.
For example, to compare 0.123 and 0.132, we can see that the whole number parts are equal (0). Comparing the digits from left to right, we find that the first unequal digits are 2 (in 0.123) and 3 (in 0.132). Since 3 is greater than 2, 0.132 is the larger decimal.
In the next part, we will delve into arithmetic operations involving decimals, including addition, subtraction, multiplication, and division.
Part 2: Arithmetic Operations with Decimals
### 2.1 Adding Decimals
To add decimal numbers, follow these steps:
Align the decimal points vertically.
Add zeros to the right of the shorter decimal to ensure equal decimal places.
Perform column-wise addition, starting from the rightmost digit and moving left, carrying over as needed.
Place the decimal point in the sum, aligned with the decimal points of the addends.
Example: Add 12.34 and 5.678
12.34
+ 5.678
________
18.018
### 2.2 Subtracting Decimals
To subtract decimal numbers, follow these steps:
Align the decimal points vertically.
Add zeros to the right of the shorter decimal to ensure equal decimal places.
Perform column-wise subtraction, starting from the rightmost digit and moving left, borrowing as needed.
Place the decimal point in the difference, aligned with the decimal points of the minuend and subtrahend.
Example: Subtract 9.52 from 12.34
12.34
– 9.52
______
2.82
### 2.3 Multiplying Decimals
To multiply decimal numbers, follow these steps:
Ignore the decimal points and multiply the numbers as if they were whole numbers.
Count the total number of decimal places in both factors.
Place the decimal point in the product so that the total number of decimal places in the product is equal to the sum of the decimal places in the factors.
Example: Multiply 1.2 by 3.4
12
x 34
________
48
+ 36
________
408
1.2 has one decimal place, and 3.4 has one decimal place. The product should have two decimal places. Thus, the product is 4.08.
### 2.4 Dividing Decimals
To divide decimal numbers, follow these steps:
Remove the decimal point from the divisor by multiplying it by a power of 10. Perform the same multiplication on the dividend.
Divide the resulting whole numbers using long division or any other preferred method.
Place the decimal point in the quotient to ensure the same number of decimal places as in the original dividend.
Example: Divide 4.2 by 1.4
First, multiply both the dividend and divisor by 10 to remove the decimal points:
Copy code
42 ÷ 14
3.0
The quotient is 3.0.
In the next part, we will explore decimal-fraction conversions and strategies for simplifying and
## Part 3: Decimal-Fraction Conversions and Equations
### 3.1 Converting Decimals to Fractions
To convert a decimal to a fraction, follow these steps:
Determine the place value of the last digit of the decimal.
Write the decimal as a fraction with the numerator equal to the decimal number without the decimal point, and the denominator equal to the place value of the last digit.
Simplify the fraction, if possible.
Example: Convert 0.75 to a fraction
The last digit, 5, is in the hundredths place. Write the decimal as a fraction:
75/100
Simplify the fraction:
3/ 4
The fraction is 3/4.
### 3.2 Converting Fractions to Decimals
To convert a fraction to a decimal, follow these steps:
Divide the numerator by the denominator.
If necessary, round the result to the desired number of decimal places.
Example: Convert 3/8 to a decimal
Perform the division:
3 ÷ 8 = 0.375
The decimal is 0.375.
### 3.3 Simplifying Decimal Equations
When solving equations involving decimals, it’s often helpful to eliminate the decimals by multiplying both sides of the equation by a power of 10. This can make the equation easier to solve.
Example: Solve the equation 0.3x + 1.2 = 2.4
First, multiply both sides of the equation by 10 to eliminate the decimals:
10(0.3x + 1.2) = 10(2.4) 3x + 12 = 24
Now, solve the equation:
3x = 12 x = 4
### 3.4 Solving Decimal Equations with Fractions
When solving equations that involve both decimals and fractions, it can be helpful to convert the decimals to fractions. This allows you to work with a single number system, making the equation easier to solve.
Example: Solve the equation 0.5x – 1/4 = 1.25
First, convert the decimals to fractions:
1/2x – 1/4 = 5/4
Now, solve the equation:
1/2x = 1/4 + 5/4 1/2x = 6/4 x = (6/4) / (1/2) x = 3
In the final part, we will discuss real-world applications of decimal algebra and practice problems to reinforce your understanding of the concepts.
Part 4: Real-World Applications and Practice Problems
### 4.1 Real-World Applications
Decimal algebra is frequently used in everyday life and various fields, such as finance, engineering, and science. Some common real-world applications include:
Currency calculations: Decimals are used to represent money and perform calculations involving income, expenses, and investments.
Measurements: Decimals are used in various units of measurement, such as length, weight, and temperature, to represent values with varying degrees of precision.
Percentage calculations: Decimals are used to express percentages, which are often used to compare quantities, calculate discounts, or determine interest rates.
Scientific notation: Decimals are used in scientific notation to express very large or very small numbers in a more manageable form.
### 4.2 Practice Problems
Solve the following problems to reinforce your understanding of decimal algebra concepts:
Add the following decimals: 14.56 and 3.009
Solution: Align the decimals and add:
14.56
+ 3.009
_________
17.569
2. Subtract the following decimals: 8.1 – 2.37
Solution: Align the decimals and subtract:
8.10
– 2.37
_________
5.73
3. Multiply the following decimals: 1.5 x 2.4
Solution: Multiply as whole numbers, then add the decimal places:
15
x 24
________
60
+ 30
________
360
Both numbers have one decimal place, so the product has two decimal places: 3.60
Divide the following decimals: 2.7 ÷ 0.9
Solution: Multiply both numbers by 10 to remove the decimal points:
27 ÷ 9 = 3
The quotient is 3.
Convert the decimal 0.875 to a fraction.
Solution: The last digit is in the thousandths place:
875/ 1000
Simplify the fraction:
7 /8
The fraction is 7/8.
Remember that mastering decimal algebra requires practice. Continue working on problems involving decimals to build your confidence and proficiency in preparation for the GMAT exam.
## Practice Questions
Question 1: Add the following decimals: 5.31 + 2.9
A) 8.21 B) 7.31 C) 8.12 D) 7.21 E) 8.01
Question 2: Subtract the following decimals: 4.5 – 1.25
A) 3.15 B) 3.25 C) 3.35 D) 3.45 E) 3.55
Question 3: Multiply the following decimals: 0.6 x 0.3
A) 0.18 B) 0.19 C) 0.20 D) 0.21 E) 0.22
Question 4: Divide the following decimals: 1.2 ÷ 0.4
A) 2 B) 3 C) 4 D) 5 E) 6
Question 5: Convert the decimal 0.2 to a fraction.
A) 1/5 B) 1/4 C) 1/3 D) 1/2 E) 1/1
Question 6: Add the following decimals: 32.156 + 17.39
A) 49.456 B) 49.546 C) 49.556 D) 49.666 E) 49.546
Question 7: Subtract the following decimals: 15.8 – 3.129
A) 12.671 B) 12.761 C) 12.851 D) 12.941 E) 13.031
Question 8: Multiply the following decimals: 1.4 x 3.5
A) 4.90 B) 5.10 C) 5.20 D) 5.30 E) 5.40
Question 9: Divide the following decimals: 3.6 ÷ 1.2
A) 2 B) 3 C) 4 D) 5 E) 6
Question 10: Convert the decimal 0.625 to a fraction.
A) 1/2 B) 3/5 C) 2/3 D) 5/8 E) ¾
Question 11:
The value of p is derived by summing d, e, and f and then rounding the result to the tenths place. The value of q is derived by first rounding d, e, and f to the tenths place and then summing the resulting values. If d = 6.35, e = 4.78, and f = 2.46, what is q – p?
A.-0.1 B. 0 C 0.0 D 0.1 E. 0 .2
Solution:
For p, you sum d, e, and f and then round to the tenths place:
p = round(6.35 + 4.78 + 2.46, 1) = round(13.59, 1) = 13.6
For q, you round d, e, and f to the tenths place first and then sum the results:
q = round(6.35, 1) + round(4.78, 1) + round(2.46, 1) = 6.4 + 4.8 + 2.5 = 13.7
So, q – p = 13.7 – 13.6 = 0.1
The answer is D. 0.1
Question 12:
The value of m is derived by summing g, h, and i and then rounding the result to the tenths place. The value of n is derived by first rounding g, h, and i to the tenths place and then summing the resulting values. If g = 3.84, h = 7.56, and i = 2.39, what is n – m?
A.-0.2 B. -0.1 C 0.0 D 0.1 E. 0.2
Solution:
For m, you sum g, h, and i and then round to the tenths place:
m = round(3.84 + 7.56 + 2.39, 1) = round(13.79, 1) = 13.8
For n, you round g, h, and i to the tenths place first and then sum the results:
n = round(3.84, 1) + round(7.56, 1) + round(2.39, 1) = 3.8 + 7.6 + 2.4 = 13.8
So, n – m = 13.8 – 13.8 = 0
The answer is C. 0.0
Question 13:
The value of p is derived by summing j, k, and l and then rounding the result to the tenths place. The value of q is derived by first rounding j, k, and l to the tenths place and then summing the resulting values. If j = 4.67, k = 5.89, and l = 3.21, what is q – p?
A.-0.2 B. -0.1 C 0.0 D 0.1 E. 0.2
Solution:
For p, you sum j, k, and l and then round to the tenths place:
p = round(4.67 + 5.89 + 3.21, 1) = round(13.77, 1) = 13.8
For q, you round j, k, and l to the tenths place first and then sum the results:
q = round(4.67, 1) + round(5.89, 1) + round(3.21, 1) = 4.7 + 5.9 + 3.2 = 13.8
So, q – p = 13.8 – 13.8 = 0
The answer is C. 0.0
Question 14:
If the fraction 1/7 equals the repeating decimal 0.142857142857 . . . , what is the 68th digit after the decimal point of the repeating decimal?
(A) 1(B) 2(C) 4(D) 5(E) 7
Solution:
The decimal representation of 1/7 repeats every 6 digits (142857). So, we can find the 68th digit by finding the remainder when 68 is divided by 6. The remainder is 2, so the 68th digit is the 2nd digit in the repeating block, which is 4.
The answer is (C) 4.
Question 15:
If the fraction 1/13 equals the repeating decimal 0.076923076923 . . . , what is the 45th digit after the decimal point of the repeating decimal?
(A) 0(B) 1(C) 6(D) 7(E) 9
Solution:
The decimal representation of 1/13 repeats every 6 digits (076923). So, we can find the 45th digit by finding the remainder when 45 is divided by 6. The remainder is 3, so the 45th digit is the 3rd digit in the repeating block, which is 6.
The answer is (C) 6.
Question 16:
If the fraction 2/11 equals the repeating decimal 0.181818181818 . . . , what is the 72nd digit after the decimal point of the repeating decimal?
(A) 1(B) 2(C) 8(D) 6(E) 7
Solution:
The decimal representation of 2/11 repeats every 2 digits (18). So, we can find the 72nd digit by finding the remainder when 72 is divided by 2. The remainder is 0, so the 72nd digit is the last digit in the repeating block, which is 8.
The answer is (C) 8.
Question 17:
The decimal 0.456456456… repeats every three digits. If we were to write out the first 100 digits after the decimal point, how many times would the digit 6 appear?
(A) 33 (B) 34 (C) 35 (D) 36 (E) 37
Solution:
The decimal repeats every three digits, so in every cycle of three digits, the digit 6 appears once. Hence, every three digits, there’s one 6. To find out how many 6’s there would be in the first 100 digits, we divide 100 by 3. This gives approximately 33.33. Since we can’t have a partial cycle, and we know that the 34th cycle would start after the 100th digit, there are 33 complete cycles within the first 100 digits. Therefore, the digit 6 would appear 33 times.
The answer is (A) 33.
Question 18:
The decimal 0.789789789… repeats every three digits. If we were to write out the first 200 digits after the decimal point, how many times would the sequence ’89’ appear?
(A) 66 (B) 67 (C) 68 (D) 69 (E) 70
Solution:
The decimal repeats every three digits (789), and the sequence ’89’ appears once in each cycle. Therefore, every three digits, there’s one ’89’. To find out how many ’89’s there would be in the first 200 digits, we divide 200 by 3. This gives approximately 66.67. Since we can’t have a partial cycle, and we know that the 67th cycle would start after the 200th digit, there are 66 complete cycles within the first 200 digits. Therefore, the sequence ’89’ would appear 66 times.
The answer is (A) 66.
Question 19:
The decimal 0.12341234… repeats every four digits. What is the 157th digit after the decimal point?
(A) 1 (B) 2 (C) 3 (D) 4 (E) None of the above
Solution:
The decimal repeats every four digits (1234). We can find the 157th digit by finding the remainder when 157 is divided by 4. This gives a remainder of 1. So, the 157th digit would be the first digit in the repeating block, which is 1.
The answer is (A) 1.
Question 20:
The decimal 0.56785678… repeats every four digits. If we were to write out the first 150 digits after the decimal point, how many times would the digit 5 appear?
(A) 37 (B) 38 (C) 39 (D) 40 (E) 41
Solution:
The decimal repeats every four digits (5678), and the digit 5 appears once in each cycle. Therefore, every four digits, there’s one 5. To find out how many 5’s there would be in the first 150 digits, we divide 150 by 4. This gives 37.5. Since we can’t have a partial cycle, and we know that the 38th cycle would start after the 150th digit, there are 37 complete cycles within the first 150 digits. Therefore, the digit 5 would appear 37 times.
The answer is (A) 37.
Question 21:
The decimal 0.57685768… repeats every five digits. If we were to write out the first 500 digits after the decimal point, how many times would the sequence ‘5768’ appear?
(A) 99 (B) 100 (C) 101 (D) 102 (E) 103
Solution:
The decimal repeats every five digits (57685), and the sequence ‘5768’ appears once in each cycle. However, because the sequence ‘5768’ also overlaps between two cycles (as in …57685768…), we have to account for those as well.
First, let’s find out how many cycles fit into 500 digits. To do this, we divide 500 by 5 to get 100 cycles. Each cycle has one ‘5768’, so that’s 100 occurrences.
Next, let’s account for the overlapping sequences. Because the sequence ‘5768’ overlaps between two cycles, there will be an extra ‘5768’ for every consecutive pair of cycles. Since we have 100 cycles, we have 99 pairs of consecutive cycles, so that’s an additional 99 ‘5768’s.
Therefore, the total number of ‘5768’s is 100 (from individual cycles) + 99 (from overlaps) = 199.
Unfortunately, this number doesn’t match any of the options given. This is a common occurrence in real-world problems and a good reminder to always check your work. If the options had included 199, that would have been the correct choice.
Question 22:
The decimal 0.89738973… repeats every four digits. If we were to write out the first 200 digits after the decimal point, how many times would the sequence ‘8973’ appear?
(A) 49 (B) 50 (C) 51 (D) 52 (E) 53
Solution:
The decimal repeats every four digits (8973), and the sequence ‘8973’ appears once in each cycle. To find out how many ‘8973’s there would be in the first 200 digits, we divide 200 by 4 to get 50 cycles. Each cycle has one ‘8973’, so that’s 50 occurrences.
The answer is (B) 50.
Data Sufficiency Questions
In Data Sufficiency questions, you are given a question and two statements. You must determine whether the information given in the statements is sufficient to answer the question. Choose from the following answer choices:
A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.
Question 23:
p = 0.25q8
If q denotes the thousandths digit in the decimal representation of p above, what digit is q?
(1) If p were rounded to the nearest hundredth, the result would be 0.25.
(2) If p were rounded to the nearest thousandth, the result would be 0.252.
Solution:
Statement (1) tells us that when rounded to the nearest hundredth, p = 0.25. This means that the thousandths place could be 5 or greater. However, we don’t know the exact value of q, so this statement is insufficient.
Statement (2) tells us that when rounded to the nearest thousandth, p = 0.252. This gives us the exact value of q, which is 2. Therefore, this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 24:
n = 0.15r9
If r denotes the ten-thousandths digit in the decimal representation of n above, what digit is r?
(1) If n were rounded to the nearest thousandth, the result would be 0.159. (2) If n were rounded to the nearest ten-thousandth, the result would be 0.1598.
Solution:
Statement (1) tells us that when rounded to the nearest thousandth, n = 0.159. This means that the ten-thousandths place could be 5 or greater. However, we don’t know the exact value of r, so this statement is insufficient.
Statement (2) tells us that when rounded to the nearest ten-thousandth, n = 0.1598. This gives us the exact value of r, which is 8. Therefore, this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 25:
m = 0.27s6
If s denotes the hundred-thousandths digit in the decimal representation of m above, what digit is s?
(1) If m were rounded to the nearest ten-thousandth, the result would be 0.276.
(2) If m were rounded to the nearest hundred-thousandth, the result would be 0.2769.
Solution:
Statement (1) tells us that when rounded to the nearest ten-thousandth, m = 0.276. This means that the hundred-thousandths place could be 5 or greater. However, we don’t know the exact value of s, so this statement is insufficient.
Statement (2) tells us that when rounded to the nearest hundred-thousandth, m = 0.2769. This gives us the exact value of s, which is 9. Therefore, this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 26:
z = 0.123×456
If x denotes the ten-thousandths digit in the decimal representation of z above, what digit is x?
(1) If z were multiplied by 10, then rounded to the nearest thousandth, the result would be 1.234.
(2) If z were divided by 10, then rounded to the nearest ten-thousandth, the result would be 0.01235.
Solution:
Statement (1) tells us that when z is multiplied by 10 and then rounded to the nearest thousandth, the result is 1.234. This means that the original number before rounding was between 1.2335 and 1.2345. Dividing these bounds by 10 gives us a range for z of 0.12335 to 0.12345. This means that x could be either 3 or 4, so this statement is insufficient.
Statement (2) tells us that when z is divided by 10 and then rounded to the nearest ten-thousandth, the result is 0.01235. This means that the original number before rounding was between 0.012345 and 0.012355. Multiplying these bounds by 10 gives us a range for z of 0.12345 to 0.12355. This means that x must be 4, so this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 27:
y = 0.789w123
If w denotes the hundred-thousandths digit in the decimal representation of y above, what digit is w?
(1) If y were multiplied by 100, then rounded to the nearest tenth, the result would be 78.9.
(2) If y were divided by 100, then rounded to the nearest millionth, the result would be 0.000007892.
Solution:
Statement (1) tells us that when y is multiplied by 100 and then rounded to the nearest tenth, the result is 78.9. This means that the original number before rounding was between 78.85 and 78.95. Dividing these bounds by 100 gives us a range for y of 0.7885 to 0.7895. This means that w could be either 8 or 9, so this statement is insufficient.
Statement (2) tells us that when y is divided by 100 and then rounded to the nearest millionth, the result is 0.000007892. This means that the original number before rounding was between 0.0000078915 and 0.0000078925. Multiplying these bounds by 100 gives us a range for y of 0.00078915 to 0.00078925. This means that w must be 9, so this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 28:
Let a = 0.1234b56
If b denotes the ten-thousandths digit in the decimal representation of a above, what digit is b?
(1) If a were multiplied by 10 and then divided by 3, then rounded to the nearest thousandth, the result would be 0.411.
(2) If a were divided by 5, then rounded to the nearest hundred-thousandth, the result would be 0.02468.
Solution:
Statement (1) tells us that when a is multiplied by 10 and then divided by 3, then rounded to the nearest thousandth, the result is 0.411. This means that the original number before rounding was between 0.4105 and 0.4115. Multiplying these bounds by 3 and then dividing by 10 gives us a range for a of 0.12315 to 0.12345. This means that b could be either 1 or 4, so this statement is insufficient.
Statement (2) tells us that when a is divided by 5, then rounded to the nearest hundred-thousandth, the result is 0.02468. This means that the original number before rounding was between 0.024675 and 0.024685. Multiplying these bounds by 5 gives us a range for a of 0.123375 to 0.123425. This means that b must be 4, so this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 29:
Let c = 0.5678d901
If d denotes the millionths digit in the decimal representation of c above, what digit is d?
(1) If c were multiplied by 100 and then divided by 7, then rounded to the nearest tenth, the result would be 8.1.
(2) If c were divided by 9, then rounded to the nearest ten-millionth, the result would be 0.000063101.
Solution:
Statement (1) tells us that when c is multiplied by 100 and then divided by 7, then rounded to the nearest tenth, the result is 8.1. This means that the original number before rounding was between 8.05 and 8.15. Multiplying these bounds by 7 and then dividing by 100 gives us a range for c of 0.5635 to 0.5705. This means that d could be any digit from 3 to 5, so this statement is insufficient.
Statement (2) tells us that when c is divided by 9, then rounded to the nearest ten-millionth, the result is 0.000063101. This means that the original number before rounding was between 0.0000631005 and 0.0000631015. Multiplying these bounds by 9 gives us a range for c of 0.5679045 to 0.5679145. This means that d must be 9, so this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 30:
What is the hundredths digit in the decimal representation of a certain number?
(1) The number is less than 1/6.
(2) The number is greater than 1/8.
Solution
Statement (1) tells us that the number is less than 1/6, which is approximately 0.1666… This suggests that the hundredths digit could be 0, 1, 2, 3, 4, 5, or 6. But we can’t determine the exact digit, so this statement is insufficient.
Statement (2) tells us that the number is greater than 1/8, which is 0.125. This suggests that the hundredths digit could be 2 or any digit greater than 2. But again, we can’t determine the exact digit, so this statement is also insufficient.
Taking both statements together, we know that the number is between 0.125 and 0.1666… This suggests that the hundredths digit could be 2, 3, 4, 5, or 6. Since we can’t determine the exact digit, both statements together are still insufficient.
So, the answer is (E), both statements together are still not sufficient.
Question 31:
If s is represented by the decimal 0.p7, what is the digit p?
(1) s < 1/4
(2) s < 1/15
Solution:
Statement (1) tells us that s is less than 1/4, which is 0.25. This suggests that the digit p could be 0, 1, or 2. But we can’t determine the exact digit, so this statement is insufficient.
Statement (2) tells us that s is less than 1/15, which is approximately 0.0666… This suggests that the digit p must be 0 because the decimal representation of s, 0.p7, must be less than 0.1. So, this statement is sufficient.
Therefore, the answer is (B), statement (2) alone is sufficient.
Question 32:
If e denotes a decimal, is e ≥ 0.6?
(1) When e is rounded to the nearest tenth, the result is 0.6. (2) When e is rounded to the nearest integer, the result is 1.
Solution:
Statement (1) tells us that when e is rounded to the nearest tenth, the result is 0.6. This means that e could be anywhere from 0.55 (which rounds up to 0.6) to 0.6499999… (which rounds down to 0.6). This suggests that e could be less than 0.6, so this statement is insufficient.
Statement (2) tells us that when e is rounded to the nearest integer, the result is 1. This means that e could be anywhere from 0.5 (which rounds up to 1) to 1.4999999… (which rounds down to 1). This suggests that e could be less than 0.6, so this statement is also insufficient.
Therefore, the answer is (E), both statements together are still not sufficient.
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# How do you solve y=x-4 and y=-x+2?
Feb 11, 2017
See the entire solution process below:
#### Explanation:
Step 1) Because the first equation is already solved for $y$, substitute $x - 4$ for $y$ in the second equation and solve for $x$:
$y = - x + 2$ becomes:
$x - 4 = - x + 2$
$x - 4 + \textcolor{red}{4} + \textcolor{b l u e}{x} = - x + 2 + \textcolor{red}{4} + \textcolor{b l u e}{x}$
$x + \textcolor{b l u e}{x} - 4 + \textcolor{red}{4} = - x + \textcolor{b l u e}{x} + 2 + \textcolor{red}{4}$
$1 x + \textcolor{b l u e}{1 x} - 0 = 0 + 6$
$2 x = 6$
$\frac{2 x}{\textcolor{red}{2}} = \frac{6}{\textcolor{red}{2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = 3$
$x = 3$
Step 2) Substitute $3$ for $x$ in the first equation and calculate $y$:
$y = x - 4$ becomes:
$y = 3 - 4$
$y = - 1$
The solution is: $x = 3$ and $y = - 1$ or $\left(3 , - 1\right)$
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### 2: Number Sense and Numeration
#### 2.1: Overall Expectations
2.1.1: represent, compare, and order numbers, including integers;
2.1.2: demonstrate an understanding of addition and subtraction of fractions and integers, and apply a variety of computational strategies to solve problems involving whole numbers and decimal numbers;
2.1.3: demonstrate an understanding of proportional relationships using percent, ratio, and rate.
#### 2.2: Quantity Relationships
2.2.1: represent, compare, and order decimals to hundredths and fractions, using a variety of tools (e.g., number lines, Cuisenaire rods, base ten materials, calculators);
2.2.2: generate multiples and factors, using a variety of tools and strategies (e.g., identify multiples on a hundreds chart; create rectangles on a geoboard) (Sample problem: List all the rectangles that have an area of 36 cm_ and have whole-number dimensions.);
2.2.3: identify and compare integers found in real-life contexts (e.g., -10¡C is much colder than +5¡C);
2.2.4: represent and order integers, using a variety of tools (e.g., two-colour counters, virtual manipulatives, number lines);
2.2.6: represent perfect squares and square roots, using a variety of tools (e.g., geoboards, connecting cubes, grid paper);
#### 2.3: Operational Sense
2.3.1: divide whole numbers by simple fractions and by decimal numbers to hundredths, using concrete materials (e.g., divide 3 by 1/2 using fraction strips; divide 4 by 0.8 using base ten materials and estimation);
2.3.2: use a variety of mental strategies to solve problems involving the addition and subtraction of fractions and decimals (e.g., use the commutative property: 3 x 2/5 x 1/3 = 3 x 1/3 x 2/5, which gives 1 x 2/5 = 2/5; use the distributive property: 16.8 Ö 0.2 can be thought of as (16 + 0.8) Ö 0.2 = 16 Ö 0.2 + 0.8 Ö 0.2, which gives 80 + 4 = 84);
2.3.3: solve problems involving the multiplication and division of decimal numbers to thousandths by one-digit whole numbers, using a variety of tools (e.g., concrete materials, drawings, calculators) and strategies (e.g., estimation, algorithms);
2.3.5: use estimation when solving problems involving operations with whole numbers, decimals, and percents, to help judge the reasonableness of a solution (Sample problem: A book costs \$18.49. The salesperson tells you that the total price, including taxes, is \$22.37. How can you tell if the total price is reasonable without using a calculator?);
2.3.6: evaluate expressions that involve whole numbers and decimals, including expressions that contain brackets, using order of operations;
2.3.7: add and subtract fractions with simple like and unlike denominators, using a variety of tools (e.g., fraction circles, Cuisenaire rods, drawings, calculators) and algorithms;
2.3.8: demonstrate, using concrete materials, the relationship between the repeated addition of fractions and the multiplication of that fraction by a whole number (e.g., 1/2 + 1/2 + 1/2 = 3 x 1/2);
2.3.9: add and subtract integers, using a variety of tools (e.g., two-colour counters, virtual manipulatives, number lines).
#### 2.4: Proportional Relationships
2.4.1: determine, through investigation, the relationships among fractions, decimals, percents, and ratios;
2.4.2: solve problems that involve determining whole number percents, using a variety of tools (e.g., base ten materials, paper and pencil, calculators) (Sample problem: If there are 5 blue marbles in a bag of 20 marbles, what percent of the marbles are not blue?);
2.4.3: demonstrate an understanding of rate as a comparison, or ratio, of two measurements with different units (e.g., speed is a rate that compares distance to time and that can be expressed as kilometres per hour);
2.4.4: solve problems involving the calculation of unit rates (Sample problem:You go shopping and notice that 25 kg of Ryan's Famous Potatoes cost \$12.95, and 10 kg of Gillian's Potatoes cost \$5.78. Which is the better deal? Justify your answer.).
### 3: Measurement
#### 3.1: Overall Expectations
3.1.2: determine the relationships among units and measurable attributes, including the area of a trapezoid and the volume of a right prism.
#### 3.3: Measurement Relationships
3.3.1: sketch different polygonal prisms that share the same volume (Sample problem: The Neuman Company is designing a new container for its marbles. The container must have a volume of 200 cm_. Sketch three possible containers, and explain which one you would recommend.);
3.3.4: determine, through investigation using a variety of tools (e.g., concrete materials, dynamic geometry software) and strategies, the relationship for calculating the area of a trapezoid, and generalize to develop the formula [i.e., Area = (sum of lengths of parallel sides x height) Ö 2] (Sample problem: Determine the relationship between the area of a parallelogram and the area of a trapezoid by composing a parallelogram from congruent trapezoids.);
3.3.6: estimate and calculate the area of composite two-dimensional shapes by decomposing into shapes with known area relationships (e.g., rectangle, parallelogram, triangle) (Sample problem: Decompose a pentagon into shapes with known area relationships to find the area of the pentagon.);
3.3.7: determine, through investigation using a variety of tools and strategies (e.g., decomposing right prisms; stacking congruent layers of concrete materials to form a right prism), the relationship between the height, the area of the base, and the volume of right prisms with simple polygonal bases (e.g., parallelograms, trapezoids), and generalize to develop the formula (i.e., Volume = area of base x height) (Sample problem: Decompose right prisms with simple polygonal bases into triangular prisms and rectangular prisms. For each prism, record the area of the base, the height, and the volume on a chart. Identify relationships.);
3.3.8: determine, through investigation using a variety of tools (e.g., nets, concrete materials, dynamic geometry software, Polydrons), the surface area of right prisms;
3.3.9: solve problems that involve the surface area and volume of right prisms and that require conversion between metric measures of capacity and volume (i.e., millilitres and cubic centimetres) (Sample problem: An aquarium has a base in the shape of a trapezoid. The aquarium is 75 cm high. The base is 50 cm long at the front, 75 cm long at the back, and 25 cm wide. Find the capacity of the aquarium.).
### 4: Geometry and Spatial Sense
#### 4.1: Overall Expectations
4.1.1: construct related lines, and classify triangles, quadrilaterals, and prisms;
4.1.2: develop an understanding of similarity, and distinguish similarity and congruence;
4.1.3: describe location in the four quadrants of a coordinate system, dilatate two-dimensional shapes, and apply transformations to create and analyse designs.
#### 4.2: Geometric Properties
4.2.1: construct related lines (i.e., parallel; perpendicular; intersecting at 30¼, 45¼, and 60¼), using angle properties and a variety of tools (e.g., compass and straight edge, protractor, dynamic geometry software) and strategies (e.g., paper folding);
4.2.2: sort and classify triangles and quadrilaterals by geometric properties related to symmetry, angles, and sides, through investigation using a variety of tools (e.g., geoboard, dynamic geometry software) and strategies (e.g., using charts, using Venn diagrams) (Sample problem: Investigate whether dilatations change the geometric properties of triangles and quadrilaterals.);
4.2.4: investigate, using concrete materials, the angles between the faces of a prism, and identify right prisms (Sample problem: Identify the perpendicular faces in a set of right prisms.).
#### 4.3: Geometric Relationships
4.3.1: identify, through investigation, the minimum side and angle information (i.e., side-side-side; side-angle-side; angle-sideangle) needed to describe a unique triangle (e.g.,"I can draw many triangles if I'm only told the length of one side, but there's only one triangle I can draw if you tell me the lengths of all three sides.");
4.3.2: determine, through investigation using a variety of tools (e.g., dynamic geometry software, concrete materials, geoboard), relationships among area, perimeter, corresponding side lengths, and corresponding angles of congruent shapes (Sample problem: Do you agree with the conjecture that triangles with the same area must be congruent? Justify your reasoning.);
4.3.4: distinguish between and compare similar shapes and congruent shapes, using a variety of tools (e.g., pattern blocks, grid paper, dynamic geometry software) and strategies (e.g., by showing that dilatations create similar shapes and that translations, rotations, and reflections generate congruent shapes) (Sample problem: A larger square can be composed from four congruent square pattern blocks. Identify another pattern block you can use to compose a larger shape that is similar to the shape of the block.).
#### 4.4: Location and Movement
4.4.1: plot points using all four quadrants of the Cartesian coordinate plane;
4.4.2: identify, perform, and describe dilatations (i.e., enlargements and reductions), through investigation using a variety of tools (e.g., dynamic geometry software, geoboard, pattern blocks, grid paper);
4.4.3: create and analyse designs involving translations, reflections, dilatations, and/or simple rotations of two-dimensional shapes, using a variety of tools (e.g., concrete materials, Mira, drawings, dynamic geometry software) and strategies (e.g., paper folding) (Sample problem: Identify transformations that may be observed in architecture or in artwork [e.g., in the art of M.C. Escher].);
4.4.4: determine, through investigation using a variety of tools (e.g., pattern blocks, Polydrons, grid paper, tiling software, dynamic geometry software, concrete materials), polygons or combinations of polygons that tile a plane, and describe the transformation(s) involved.
### 5: Patterning and Algebra
#### 5.1: Overall Expectations
5.1.1: represent linear growing patterns (where the terms are whole numbers) using concrete materials, graphs, and algebraic expressions;
5.1.2: model real-life linear relationships graphically and algebraically, and solve simple algebraic equations using a variety of strategies, including inspection and guess and check.
#### 5.2: Patterns and Relationships
5.2.1: represent linear growing patterns, using a variety of tools (e.g., concrete materials, paper and pencil, calculators, spreadsheets) and strategies (e.g., make a table of values using the term number and the term; plot the coordinates on a graph; write a pattern rule using words);
5.2.2: make predictions about linear growing patterns, through investigation with concrete materials (Sample problem: Investigate the surface area of towers made from a single column of connecting cubes, and predict the surface area of a tower that is 50 cubes high. Explain your reasoning.);
5.2.3: develop and represent the general term of a linear growing pattern, using algebraic expressions involving one operation (e.g., the general term for the sequence 4, 5, 6, 7, ... can be written algebraically as n + 3, where n represents the term number; the general term for the sequence 5, 10, 15, 20, ... can be written algebraically as 5n, where n represents the term number);
5.2.4: compare pattern rules that generate a pattern by adding or subtracting a constant, or multiplying or dividing by a constant, to get the next term (e.g., for 1, 3, 5, 7, 9, ..., the pattern rule is "start at 1 and add 2 to each term to get the next term") with pattern rules that use the term number to describe the general term (e.g., for 1, 3, 5, 7, 9, ..., the pattern rule is "double the term number and subtract 1", which can be written algebraically as 2 x n - 1) (Sample problem: For the pattern 1, 3, 5, 7, 9,..., investigate and compare different ways of finding the 50th term.).
#### 5.3: Variables, Expressions, and Equations
5.3.1: model real-life relationships involving constant rates where the initial condition starts at 0 (e.g., speed, heart rate, billing rate), through investigation using tables of values and graphs (Sample problem: Create a table of values and graph the relationship between distance and time for a car travelling at a constant speed of 40 km/h. At that speed, how far would the car travel in 3.5 h? How many hours would it take to travel 220 km?);
5.3.2: model real-life relationships involving constant rates (e.g., speed, heart rate, billing rate), using algebraic equations with variables to represent the changing quantities in the relationship (e.g., the equation p = 4t represents the relationship between the total number of people that can be seated (p) and the number of tables (t), given that each table can seat 4 people [4 people per table is the constant rate]);
5.3.3: translate phrases describing simple mathematical relationships into algebraic expressions (e.g., one more than three times a number can be written algebraically as 1 + 3x or 3x + 1), using concrete materials (e.g., algebra tiles, pattern blocks, counters);
5.3.6: solve linear equations of the form ax = c or c = ax and ax + b = c or variations such as b + ax = c and c = bx + a (where a, b, and c are natural numbers) by modelling with concrete materials, by inspection, or by guess and check, with and without the aid of a calculator (e.g.,"I solved x + 7 = 15 by using guess and check. First I tried 6 for x. Since I knew that 6 plus 7 equals 13 and 13, is less than 15, then I knew that x must be greater than 6.").
### 6: Data Management and Probability
#### 6.1: Overall Expectations
6.1.1: collect and organize categorical, discrete, or continuous primary data and secondary data and display the data using charts and graphs, including relative frequency tables and circle graphs;
6.1.2: make and evaluate convincing arguments, based on the analysis of data;
6.1.3: compare experimental probabilities with the theoretical probability of an outcome involving two independent events.
#### 6.2: Collection and Organization of Data
6.2.2: collect and organize categorical, discrete, or continuous primary data and secondary data (e.g., electronic data from websites such as E-Stat or Census At Schools) and display the data in charts, tables, and graphs (including relative frequency tables and circle graphs) that have appropriate titles, labels (e.g., appropriate units marked on the axes), and scales (e.g., with appropriate increments) that suit the range and distribution of the data, using a variety of tools (e.g., graph paper, spreadsheets, dynamic statistical software);
6.2.3: select an appropriate type of graph to represent a set of data, graph the data using technology, and justify the choice of graph (i.e., from types of graphs already studied);
#### 6.3: Data Relationships
6.3.1: read, interpret, and draw conclusions from primary data (e.g., survey results, measurements, observations) and from secondary data (e.g., temperature data or community data in the newspaper, data from the Internet about populations) presented in charts, tables, and graphs (including relative frequency tables and circle graphs);
6.3.2: identify, through investigation, graphs that present data in misleading ways (e.g., line graphs that exaggerate change by starting the vertical axis at a point greater than zero);
6.3.3: determine, through investigation, the effect on a measure of central tendency (i.e., mean, median, and mode) of adding or removing a value or values (e.g., changing the value of an outlier may have a significant effect on the mean but no effect on the median) (Sample problem: Use a set of data whose distribution across its range looks symmetrical, and change some of the values so that the distribution no longer looks symmetrical. Does the change affect the median more than the mean? Explain your thinking.);
6.3.4: identify and describe trends, based on the distribution of the data presented in tables and graphs, using informal language;
6.3.5: make inferences and convincing arguments that are based on the analysis of charts, tables, and graphs (Sample problem: Use census information to predict whether Canada's population is likely to increase.).
#### 6.4: Probability
6.4.1: research and report on real-world applications of probabilities expressed in fraction, decimal, and percent form (e.g., lotteries, batting averages,weather forecasts, elections);
6.4.3: represent in a variety of ways (e.g., tree diagrams, tables, models, systematic lists) all the possible outcomes of a probability experiment involving two independent events (i.e., one event does not affect the other event), and determine the theoretical probability of a specific outcome involving two independent events (Sample problem: What is the probability of rolling a 4 and spinning red, when you roll a number cube and spin a spinner that is equally divided into four different colours?);
6.4.4: perform a simple probability experiment involving two independent events, and compare the experimental probability with the theoretical probability of a specific outcome (Sample problem: Place 1 red counter and 1 blue counter in an opaque bag. Draw a counter, replace it, shake the bag, and draw again. Compare the theoretical and experimental probabilities of drawing a red counter 2 times in a row.).
Correlation last revised: 8/18/2015
This correlation lists the recommended Gizmos for this province's curriculum standards. Click any Gizmo title below for more information.
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Roving Ranges
Sixth Grade Poster Problem
Statistics and Probability
In this poster problem, students will demonstrate their understanding of the central idea in the Grade 6 Standards in Statistics and Probability: Statistical questions anticipate variability. When you roll four dice and add, you can’t know exactly what you will get, but you can still predict a lot about the result.
Students will also get practice with underlying concepts and tools: the collection of results forms a distribution; the distribution has a center and spread; and we can display distributions usefully in graphs, especially in box plots. Students use all of these together to make decisions and explain them coherently.
Here, they’ll be making decisions in a simple dice game. You roll four dice and add: what will you get? Of course, you can’t know exactly, but suppose, before you rolled, you had to state a range of results—the minimum and maximum values you expect. If you’re right, you get some points. If you’re wrong, you get nothing.
And here’s the kicker: the number of points you get depends on the range. The wider the range you allow, the fewer points you get.
Learning Objectives:
• Students collect data and display them as a distribution
• Students make mathematically sound decisions based on the distributions
• Students can articulate why their decisions make sense
Common Core State Standards for Mathematics:
Materials
Lots of dice. As written, the directions call for 4 dice per student. With that many dice, students can work in parallel and work more quickly.
• Note:
This poster problem, as written, does not teach how to make a box plot. Students should know the mechanics of making a box plot before doing this lesson. This lesson will, however, give students a chance to practice making box plots and using them to make decisions.
The Lesson Plan:
• 1. LAUNCH
Supply every student with a pair of dice. Ask:
• Have you ever played a game that uses a pair of dice? (Of course!)
• In Monopoly, how do you use the dice? (Brief discussion, answers vary, but importantly, one main way is to add the numbers that come up and move that many spaces. In some dice games, e.g., Risk, you don’t add the dice. That’s not what we’re interested in today.)
• If you roll these two dice and add, what values can come up? (Typical answer: 1 to 12. Right answer: 2 to 12.)
• What numbers come up a lot? What numbers only come up rarely? ( 6, 7, 8 a lot. But 2 and 12 are relatively rare. Don’t linger here—we’ll do an experiment next. )
Then have each student roll their two dice and remember the number. Quickly, call out,
• Who got a one? (nobody) Two? Three? and so forth up to twelve.
• Which number had the most? You might remember, you might not—it would be easier to tell if we organized the data.
Now—and this depends on your classroom setup—have the students organize the data. Ideally, you would make a “human histogram.” You set up a number line from 2 to 12, and students line up behind their number.
An alternative is to make the number line on the board and have students post their results over the number line (making a dot plot) using adhesive notes or simply by making an “x” with a marker.
Have students describe the distribution qualitatively. That is, it’s more crowded in the middle, near 7, and sparse on the outsides.
Finally, have students make a box plot of the data, and discuss what it means. Very important: remind them—or have them tell you—that the median value (probably 6, 7, or 8) means that half of the rolls were 7 or below, and half were 7 or above. And that the box (probably 5 or 6 to 8 or 9) means that at least half of all the rolls were in that range.
End with this question: if you had to say what range of numbers you expect to get when you roll two dice, what would you say? Be clear that there is no one right answer.
• 2. POSE A PROBLEM
Separate students into pairs or groups.
Explain that they have already collected data for two dice as a class; now they’ll do the same thing with three. Ask what the minimum and maximum die rolls will be? (3 and 18) And what about four dice? What would their limits be. (4 to 24)
Pass out an additional two dice to each student (now each has four). Also, pass out Handout #1—one per group.
Students are to reproduce some of the two-dice data and analyze what happens when you roll three dice and add. Their instructions read:
• Copy down the 2-dice box plot based on class data to the appropriate spot on the handout.
• With your group, record at least 50 sums of three dice. Figure out how to do this efficiently.
• Record values for the five-number summary (this will make it easy to do the next step).
• Make a box plot in the “3-dice” box.
• Based on your first two box plots, make a prediction for what the plot will look like with 4 dice.
When everyone is done—or done enough—have students share results briefly. Hold a discussion about how the students collected data, and how they worked together. Help the class find one or two effective and efficient strategies. Also, discuss how they made their predictions about four dice.
Recording the data
You may want to use Handout #2 to give your students a form where they will record their data. But if your students can handle it, it’s great to have them design their own recording system. If your students are ready for that challenge, help them prepare by doing the following.
Ask ahead of time how they intend to record the rolls. Get several ideas from the class and demonstrate briefly how they would work. Some students will benefit greatly from seeing various approaches to solving the problem.
Some data-collection strategies:
• List the possibilities in order down the left side of a piece of scratch paper and then tally.
• Make a suitable number line and make a dot plot, adding a new dot after each roll.
• Write down each result as you get it.
The third is the most exhaustive, and when you’re done, you still have to put the numbers in order to make the box plot. The first two are easier and also show you the shape of the distribution without any further work.
• 3. WORKSHOP
Overview
Each group will play as a team trying to get as many points as possible in a game, described below.
The basic idea is that first the group chooses a "target" range where they think the sums of four dice will fall. Then they roll the dice 50 times. They get no points is the sum is outside the range. The number of points they get if it's inside the range depends on the size of the range.
This size-of-the-range business is the hard part of this activity. It may help for you to demonstrate several moves, choosing a terrible range (e.g., 4–15) so that students can see how you score points and can imagine continuing for 50 rolls.
Then they make a poster reporting what happened; Handout 3 gives details about what should be in the poster.
Details
Arrange students in groups. Each student still has four dice. Distribute the following.
• Handout #2, two per group. Students will use it to record the distribution and the rolls.
• Handout #3, which has instructions for the poster they are to make.
• Handout #4 and scissors.
Explain that they will now study four dice, but this time, there will also be a game.
• Students will predict the sum they will get. They can’t predict it exactly, so they will make their predictions as a range.
• When they roll, they get points if the sum is in the range. They will get zero points if the sum is outside the range.
• A game is 50 rolls. Add up the points for 50 rolls to get the score.
• And now the catch: the number of points depends on the size of the range. If they pick a small range, they can get more points. If they pick a big range, they will get fewer points.
• How many points? Use the formula below or use Handout #4. It has “templates” for different-sized ranges. Students can place (and maybe tape) those ranges onto Handout #2 between the numbers and the diamonds. Rolls on the edges are “in.”
Remind students that they made a prediction about what the box plot for 4 dice would look like. Explain that they can roll the dice as much as they like to gather data before they play the game. They can record these “test” rolls on one copy of Handout #2.
Then let groups play the game—using a copy of Handout #2 to record their results—and make their poster.
If a group wants to replay the game, especially with a different range, they can do so. Advise them to use a fresh copy of Handout #2 so they can easily count how many “captures” they get.
How many points?
This is the trickiest part, and it’s important that students get it right. The formula is that you take the maximum minus the minimum, and subtract that from 20. That is,
points = 20 – (max – min)
This may be too abstract. Handout #4 has strips you can cut out with scores on them. These strips are in the same scale as the graph/recording sheet on Handout #2. And again, a roll at the edge of the range is “in.”
• 4. POST, SHARE, COMMENT
Have students put up their posters around the classroom.
Have students travel around to view the posters created by other groups. Encourage students to write questions or comments for other groups by attaching small adhesive notes.
During this time, the teacher should review all the posters and consider which to highlight during the discussion to follow.
Make sure that groups have clearly posted their scores, that they used 50 rolls, and so forth. If they did not use 50 rolls, challenge them to figure out what would be a fair score to use to compare to other groups. (A proportion makes sense here.)
As long as students understand the game and have a distribution with 50 results, they will all get a plausible score in the 300–550 range. You’ll be looking to see that their box plot seems correct, of course; beyond that, the key is in their reasoning.
Sample Posters:
• 5. STRATEGIC TEACHER-LED DISCUSSION
This is a little different from other poster problems in that after the class discussion, students amend their posters to say what range they would choose if they got to play the game again, and why.
Facilitate a discussion of the different approaches. Select a sequence of posters that will help students move from their current thinking (Levels 1–3) up to Level 4 or 5. Be sure to include, for example, a wide-range, low-scoring poster where the students were thoughtful but discovered through experience that a wide range results in many “captures” but a low score. If such a poster doesn’t exist, you can take anyone’s data and ask, “Suppose this group used a really wide range such as the 16 range (4 points per capture). How many points would they have gotten?” This models the idea of a “what-if” calculation, and is easy to do.
Level 1: The group chooses a range by guessing, with little reference to data they collected before. The range may be off-center. (The center should be at 14, or close to it.)
Level 2: Students choose the range of the box in the box plot, simply because it was the box. This may still be off-center.
Level 3: Students deliberately center the range even though past data may have been off-center. They choose a range using the box or otherwise give a one-sided reason for the choice (e.g., we want to capture more or we want a higher score per capture), but do not explain the trade-off well.
Level 4: Students center their range, which is between 4 and 10 wide (10–16 points per catch). Their score will probably be over 400. They clearly articulate the trade-off: you get fewer points per catch for a wide range, but you’ll capture more—so there’s a “sweet spot” in the middle where you get the most points.
Level 5: Students use 4-dice data distribution to make “what-if” calculations: would they get more points if they had chosen a different range? They actually calculate these potential scores and adjust their range accordingly.
Questions to ask across presentations
• What’s the median value? How much does it vary from poster to poster? (Should be 13–15.)
• How can you use the box plot to help you decide on your range? How much does the box vary? (Look at the different box plots and compare. Note that the min and max will vary more than the quartiles.)
• If you picked the same range as the box in the box plot, how many would you catch? What would your score be? (You would catch a bit more than half—25 to 30 rolls. Why half? Because the box contains the middle half of the data. Score depends on the width of the box.)
• What range would you pick to be pretty sure of catching almost all of the rolls? (Wide, e.g., 6–22 or 4 points per catch)
• Why is that range a bad choice? (You catch them all, but it’s not enough points. 4 times 50 is only 200 points.)
• What is the trade-off in this game? (A wide range gets many catches but few points per catch. A narrow range gets a lot of points per catch, but not that many catches.)
• How did you use the data you collected before? The data could be from
• The prediction you made during Step 2, based on rolling fewer dice, or, even better,
• The distribution of sums of four dice you rolled before actually playing the game.
Focusing on Variability
Ask students to look at the distributions of rolls on all the posters.
• What’s the same about the distributions? What’s different? (They all have more in the middle and fewer on the ends, but the details vary widely.)
• Suppose we picked a simple range, like 10–15. (20 points per catch.) What are the different scores we would get with each of these sets of rolls? (Each group could calculate and report back. Scores should vary considerably.)
• So: how much of this game is luck, and how much is skill? (Lots of luck, but you have to pick a reasonable range in the middle. Some students may think there is more skill when they’re really only lucky.)
• Suppose we did 500 rolls instead of 50. How would the distribution change? What would the shape of the distribution be? (If there are 10 posters with graphs, consider adding them all up.)
Enhancing the posters
Now challenge the groups:
Seeing all the other data and the other groups’ posters, imagine you’re going to play the game again. What range would you choose, and why?
Write your range—the minimum and the maximum, and how many points you get for each capture—and your prediction for your score.
Then explain, clearly, why you chose the range you did.
(The instructions are on their sheet, Handout #3.)
Optional Contest
If there’s time (or if some students finish early, or if you want a homework task) you can give students Handout #6: Fifty More Rolls (or use Slide #1, which has the same graphic).
The handout has a record of 50 more rolls of four dice. They can calculate the score their new range would have gotten, and compare that to other groups’s scores with the same data.
Another challenge is to figure out what range would give the best score with that data.
• 6. FOCUS PROBLEM: Same Content in a New Context
Orient the students to the context of the problem on Handout #5 by asking questions such as,
• Do any of you post pictures on Facebook or Instagram (or whatever site is currently popular)?
• How many pictures do you think a typical student posts?
Explain the problem from Handout #5, then pass it out.
Approaches and Solutions
You could make a dot plot or a box plot. In creating either, you will find the median and the first and third quartiles. In the illustration. Diane’s data (21 posts) is highlighted. (We made the graphs using Fathom. The box plot shows outliers as most computer displays do. Showing outliers is not not part of the Grade 7 standard.)
You’ll find that the median is 15 and the quartiles are 8 and 21. That is, Diana is sitting right at the third quartile: only a quarter of all students posted as many as she did. So is she typical? It’s a judgment call.
Students can look at the shape of the distribution—which is skewed to the right—with a few students posting many more pictures. Diana could argue that they’re the ones posting a lot; Diana is really just at the upper edge of most of the students.
Another (sneaky) strategy Diana could use is to calculate the mean, which is exactly 21, her very number. She can tell her mom that she posts the average number of pictures—but discussion should bring out that this is at least a little deceptive!
Lesson Plan
Slides
Handouts
meet the team!
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# CIRCLES WORD PROBLEMS
In this section, you will learn how to solve word problems using area and circumference of circles.
Area of circle = πr2
Circumference of circle = 2πr
## Circles Word Problems
Problem 1 :
The diameter of a cart wheel is 2.1 m. Find the distance traveled when it completes 100 revolutions.
Solution :
In order to find the distance covered in one revolution, we have to find the circumference of the circle.
= 2.1 / 2
= 1.05 m
Distance traveled when 100 revolutions are completed :
= 100 x circumference of the wheel
= 100 x 2 x (22/7) x (1.05)
= 100 x 2 x 22 x 0.15
= 660 m
Problem 2 :
The diameter of a circular park is 98 m. Find the cost of fencing it at \$4 per meter.
Solution :
Diameter of the circular park is 98 m.
= 98/2
= 49 m
Cost of fencing is \$4 per meter.
Length covered by fencing is equal circumference of the circular park.
Circumference of the park is
= 2πr
= 2 (22/7) x 49
= 2 x 22 x 7
= 308 m
Cost for fencing is
= 308 x 4
= \$1232
Problem 3 :
A wheel makes 20 revolutions to cover a distance of 66 m. Find the diameter of the wheel.
Solution :
Given :
A wheel makes 20 revolutions to cover a distance of 66 m.
Then, the distance covered in one revolution is
= 66/20
= 33/10 m
The distance covered in one revolution is equal to circumference of the circle.
Then,
2πr = 33/10
2 x (22/7) x r = 33/10
(44/7) x r = 33/10
Multiply each side by 44/7.
r = (33/10) x (7/44)
r = 0.525
Diameter :
= 2 x 0.525
= 1.05 m
Problem 4 :
The radius of a cycle wheel is 35 cm. How many revolutions does it make to cover a distance of 81.40 m ?
Solution :
Radius is given in centimeters and the distance is given in meters.
To have all the measures in meters, we can convert the given radius measure to meters.
= 35 cm
= 35/100 m
= 0.35 m
Let 'n' be the number of revolutions required to cover the distance of 81.40 m.
n x one revolution of cycle wheel = 81.40
n x 2πr = 81.40
n x 2 x (22/7) x 0.35 = 81.40
n x 2.2 = 81.40
Divide each side by 2.2
n = 81.40/2.2
n = 37
So, the cycle wheel has to revolve 37 times to cover the distance.
Problem 5 :
The radius of a circular park is 63 m. Find the cost of fencing it at \$12 per metre.
Solution :
Radius of the circular park is 63 m.
Length of fencing is equal to circumference of the circular park.
Circumference of the cicular park is
= 2πr
= 2 x (22/7) x 63
= 2 x 22 x 9
= 396 m
Cost of fencing is \$12 per meter.
Total cost for fencing the park is
= 396 x 12
= \$4752
Problem 6 :
A goat is tethered by a rope 3.5 m long. Find the maximum area that the goat can graze.
Solution :
Radius of the circle is equal to length of the rope.
= 3.5 m
= 7/2 m
Area grazed by the goat is equal to the area of the circle with radius 7/2 m.
Maximum area grazed by the goat is
= πr2
= (22/7) x (7/2) x (7/2)
= 77/2
= 38.5 sq. m
Example 7 :
The circumference of a circular park is 176 m. Find the area of the park.
Solution :
Circumference of the circular park is 176 m (given).
Then,
2πr = 176
2 x (22/7) x r = 176
(44/7) x r = 176
Multiply each side by 7/44.
r = 176 x (7/44)
r = 28 m
Area of the circular park is
= πr2
= (22/7) x 28 x 28
= 22 x 4 x 28
= 2464 sq. m.
Problem 8 :
A silver wire when bent in the form of a square encloses an area of 121 sq. cm. If the same wire is bent in the form of a circle. Find the area of the circle.
Solution :
Area of the square is 121 sq. cm. (given)
Let a be the length of each side of the square.
Then,
a2 = 121
a2 = 112
a = 11
Then, perimeter of the square is
= 4a
= 4 x 11
= 44 cm
Length of the wire is equal to perimeter of the square.
So, length of the wire is 44 cm.
If thee wire is bent in the form of a circle, then the circumference of the circle is equal to length of the wire
Circumference of a circle = 44 cm
2πr = 44
2 x (22/7) x r = 44
(44/7) x r = 44
Multiply each side by 7/44.
r = 44 x (7/44)
r = 7 cm
Area of the circle is
= πr2
= (22/7) x 7 x 7
= 154 sq. cm
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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2
Q:
# The calendar of the year 2006 was used before in the year?
A) 1990 B) 1995 C) 2002 D) 2000
Explanation:
Given year 2006, when divided by 4 leaves a remainder of 2.
NOTE: When remainder is 2, 11 is subtracted to the given year to get the result.
So, 2006 - 11 = 1995
Q:
26 january 1950 which day of the week?
A) Monday B) Wednesday C) Thursday D) Tuesday
Explanation:
We know that,
Odd days --> days more than complete weeks
Number of odd days in 400/800/1200/1600/2000 years are 0.
Hence, the number of odd days in first 1600 years are 0.
Number of odd days in 300 years = 1
Number of odd days in 49 years = (12 x 2 + 37 x 1) = 61 days = 5 odd days
Total number of odd days in 1949 years = 1 + 5 = 6 odd days
Now look at the year 1950
Jan 26 = 26 days = 3 weeks + 5 days = 5 odd days
Total number of odd days = 6 + 5 = 11 => 4 odd days
Odd days :-
0 = sunday ;
1 = monday ;
2 = tuesday ;
3 = wednesday ;
4 = thursday ;
5 = friday ;
6 = saturday
Therefore, Jan 26th 1950 was Thursday.
1 60
Q:
Which two months in a year have the same calendar?
A) October, December B) April, November C) June, October D) April, July
Explanation:
If the period between the two months is divisible by 7, then that two months will have the same calender .
Now,
(a). October + November = 31 + 30 = 61 (not divisible by 7)
(b). Apr. + May + June + July + Aug. + Sep. + Oct. = 30 + 31 + 30 + 31 + 31 +30 + 31 = 213 (not divisible by 7)
(c). June + July + Aug. + Sep. = 30 + 31 + 31 + 30 = 122 (not divisible by 7)
(d). Apr. + May + June = 30 + 31 + 30 = 91 (divisible by 7)
Hence, April and July months will have the same calendar.
26 6560
Q:
How many seconds in 10 years?
A) 31523500 sec B) 315360000 sec C) 315423000 sec D) 315354000 sec
Explanation:
We know that,
1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds.
Then, 1 year = 365 x 24 x 60 x 60 seconds.
= 8760 x 3600
1 year = 31536000 seconds.
Hence, 10 years = 31536000 x 10 = 315360000 seconds.
16 3632
Q:
Suppose today is Friday, what day of the week will it be 65 days from now?
A) Saturday B) Sunday C) Friday D) Thursday
Explanation:
The day of the week repeats every 7 days.
Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.
Hence, Friday is on the 63rd day, as 63 is multiple of 7.
Now, the required day of the week on the 65th day is Sunday.
18 3751
Q:
What is two weeks from today?
A) same day B) previous day C) next day D) None
Explanation:
We know that the day repeats every 7 days, 14 days, 21 days,...
So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday.
Hence, after 2 weeks i.e, 14 days the day repeats and is the same day.
12 3307
Q:
How old are you if you are born in 1995?
A) 22 B) 23 C) 24 D) 25
Explanation:
Calculating Age has 2 conditions. Let your Birthday is on January 1st.
1. If the month in which you are born is completed in the present year i.e, your birthday, then
Your Age = Present year - Year you are born
As of now, present year = 2018
i.e, Age = 2018 - 1995 = 23 years.
2. If the month in which you are born is not completed in the present year i.e, your birthday, then
Your Age = Last year - Year you are born
As of now, present year = 2018
i.e, Age = 2017 - 1995 = 22 years.
11 3118
Q:
The day before yesterday, I was 25 years old, and next year I will turn 28. How is it possible?
On carefully inspecting this question, one can understand that there are two days which are important and these are:
A. My Birthday.
B. The day when I am making this statement.
If you think for a while, you will understand that such statements can be made only around the year’s end. So, if my birthday is on 31 December, then I will be making this statement on 1 January.
I will further explain using the following example:
1. Consider that today is 01 January 2017.
2. Then, the day before yesterday was 30 December 2016 and according to the question I was 25 then.
3. Yesterday was 31 December 2016, which happens to be my birthday too (Woohoo!), and my age increases by one to become 26.
4. I will turn 27 on my birthday this year (31 December 2017).
5. I will turn 28 on my birthday next year (31 December 2018).
Now, if you read the question again, it will make more sense:
The day before yesterday(30 December 2016), I was 25 years old and next year(31 December 2018) I will be 28.
3604
Q:
The calendar for the year 2018 will be the same for the year
A) 2023 B) 2027 C) 2029 D) 2022
Explanation:
How to find the years which have the same Calendars :
Leap year calendar repeats every 28 years.
Here 28 is distributed as 6 + 11 + 11.
Rules:
a) If given year is at 1st position after Leap year then next repeated calendar year is Givenyear+6.
b) If given year is at 2nd position after Leap year then next repeated calendar year is Givenyear+11.
c) If given year is at 3rd position after Leap year then next repeated calendar year is Givenyear+11.
Now, the given year is 2018
We know that 2016 is a Leap year.
2016 2017 2018 2019 2020
Lp Y 1st 2nd 3rd Lp Y
Here 2018 is at 2 nd position after the Leap year.
According to rule b) the calendar of 2018 is repeated for the year is 2018 + 11 = 2029.
|
# Pie Chart
A pie chart can used to compare the relation between the whole and its components. A pie chart is a circular diagram and the area of the sector of a circle is used in a pie chart. Circles are drawn with radii proportional to the square root of the quantities because the area of a circle is $\pi {r^2}$.
To construct a pie chart (sector diagram), we draw a circle with radius (square root of the total). The total angle of the circle is ${360^ \circ }$. The angles of each component are calculated by the formula.
$Angle\,of\,Sector = \frac{{{\text{Component Part}}}}{{{\text{Total}}}} \times {360^ \circ }$
These angles are made in the circle by means of a protractor to show different components. The arrangement of the sectors is usually anti-clock wise.
Example:
The following table gives the details of monthly budget of a family. Represent these figures by a suitable diagram.
Item of Expenditure
Family Budget
Food
$\ {\text{ 600}}$
Clothing
$\ {\text{ 100}}$
House Rent
$\ {\text{ 400}}$
Fuel and Lighting
$\ {\text{ 100}}$
Miscellaneous
$\ {\text{ 300}}$
Total
$\ {\text{ 1500}}$
Solution:
The necessary computations are given below:
$Angle\,of\,Sector = \frac{{{\text{Component Part}}}}{{{\text{Total}}}} \times {360^ \circ }$
Items
Family Budget
Expenditure \$
Angle of Sectors
Cumulative Angle
Food
$600$
${144^ \circ }$
${144^ \circ }$
Clothing
$100$
${24^ \circ }$
${168^ \circ }$
House Rent
$400$
${96^ \circ }$
${264^ \circ }$
Fuel and Lighting
$100$
${24^ \circ }$
${288^ \circ }$
Miscellaneous
$300$
${72^ \circ }$
${360^ \circ }$
Total
$1500$
${360^ \circ }$
Pie Chart
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DISCOVER
# How to Calculate Equilibrant & Resultant
Updated April 17, 2017
Students encounter the equation f=ma soon after they begin to study physics. If an object experiences a net force, it will experience a corresponding acceleration proportional to the magnitude of that net force. When multiple forces simultaneously act upon the object, you'll need to add the different forces together to get the resultant force. A force identical in magnitude but in the opposite direction is the equilibrant force.
Identify all the different forces acting upon the object. These are vector quantities, which means they will have both magnitude and direction. Some forces, such as air resistance, may be less obvious than others.
Add all the different vectors together, using the rules for vector addition. If all the different forces are in the same direction, or exactly in the opposite direction, then these calculations are elementary. If someone is exerting 4.54kg. of force pushing a lead weight up while gravity is exerting 2.27kg. of pressure pushing it down, the net force is clearly 2.27kg. of force upward. If you are exerting 4.54kg. of force pushing a car north, and someone else is pushing in the same direction with 2.27kg. of force, the net force is 6.8kg. of force in the north direction.
Construct a right triangle to represent two different forces acting at right angles. If you are exerting 0.454kg. of force pushing an object north and someone else is exerting 0.454kg. of pressure pushing the same object east, then two of the sides of the triangle are each 1 unit long. The hypotenuse of this triangle represents the vector addition of the two forces. The Pythagorean theorem tells us that the square of the hypotenuse is equal to the sum of the squares of the two sides, so since each side is 1 unit long and its square is also 1, the hypotenuse equals the square root of 2 or 1.41421356. Elementary trigonometry will also reveal the angle between the hypotenuse and the two sides, which indicates the direction of the resultant force. For this example, that's 45 degrees. The object would experience a resultant force in the northeast direction.
Remember that you can represent any force as the sum of two different vectors. If an object is experiencing two different forces, but they are not at right angles to one another, represent one of the forces as a sum of two vectors, one in the same direction as the other force, and the other at a right angle to it. If one force is pushing an object due north and another is pushing it directly northeast, represent the northeast force as a north vector and an east vector. Simply add the two north vectors and use the Pythagorean theorem to add the east vector.
Take the sum of all the vectors. That's the resultant force. A force equal in quantity but oriented in the exact opposite direction is the equilibrant force.
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## Fractions
Fractions
by Jessica Wesaquate and Andrea Rogers
Strand:
Numbers and Operations
Four
Objectives:
Students will be able to state the shape of the canvas after viewing the video.
Students will be able to relate fractions to a real-life situation.
Students will be able to relate fractions to the Four Directions of life/Medicine Wheel or "Circle of Courage" teachings.
Students will be able to understand and use the terms numerator and denominator.
Recommended
Video clip:
Laying down the canvas
Materials:
video clip, construction paper (cut pieces in half), old circular lids, colored pencils
Have students bring old lids (circular) from yogurt containers, pickle containers, etcetera from home. You can also collect these in the staff room by asking co-workers to bring their old lids from home. You can create a collection bucket or box.
Activity:
Have students watch the clip titled ''laying down the canvas." Pause the video clip when it shows the tipi canvas laying out on the grass. What does this look like to students? Shape? Analogies? Students should observe that it looks the shape of a semi-circle.
In the video you will observe that Glen says the following:
Glen: “One of the teachings is that the tipi is the eagle. They say the eagle has blessed us in his way, he covered us with his wings, and this represents that. He protects us, looks after us."
Activity:
Have students take out a writing utensil and coloured pencils. Give each student a small piece of construction paper. Using the old lids students have brought from home, have them create a complete circle. Now that they have done this, have students indicate a semi-circle, outlining it (not colouring it in) with a colored pencil. Students will have seen what a semi-circle looks like in the video. Explore with the students what fraction the semi-circle represents of the entire circle. (1/2)
Looking at the circle as a whole again, have students split it into quarters. Have students pull out a red, yellow, blue and green coloured pencil. They are going to represent each quarter with one of the four colours. What does a "quarter" mean?
Each student should now have separated their circle into four quadrants. Take this opportunity to talk about the Four Directions/Medicine Wheel/"Circle of Courage." Students may have learned about these things before. Create a KWL chart (What they KNOW, what they WANT TO KNOW and to be filled in later what they LEARNED). This could be a good chance to an invite an elder into your classroom to share traditional knowledge about the Four Directions of life and the meaning of the Medicine Wheel. There are also many book and Internet resources on these topics, so feel free to use those as well. Martin Brokenleg is one name to research.
Have students find a blank space on their paper to record the three fractions you explored today. Teach them the terminology of numerator and denominator. Students can create arrows from their fractions to indicate what part is the numerator and which is the denominator.
Assessment ideas:
exit notes, KWL charts
In this lesson idea, students are exploring the fractions: 1/1, 1/2, 1/4, you can adapt this lesson to explore other fractions as well.
Extension Activities:
A fun way to learn about fractions is through cooking. Students can learn about traditional First Nations and Metis foods at the same time they are learning mathematical concepts!
Aboriginal Perspectives is supported by the University of Regina, the Imperial Oil Foundation, the Canadian Mathematical Society and the Pacific Institute for the Mathematical Sciences.
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Symmetric and skew symmetric matrices
Chapter 3 Class 12 Matrices
Concept wise
### Transcript
Example 22 Express the matrix B = [■8(2&−2&−4@−1&3&4@1&−2&−3)] as the sum of a symmetric and a skew symmetric matrix. B = [■8(2&−2&−4@−1&3&4@1&−2&−3)] B’ = [■8(2&−1&1@−2&3&−2@−4&4&−3)] Finding 𝟏/𝟐 (B + B’) and 𝟏/𝟐 (B − B’) 𝟏/𝟐 (B + B’) = 1/2 ([■8(2&−2&−4@−1&3&4@1&−2&−3)]+[■8(2&−1&1@−2&3&−2@−4&4&−3)]) = 1/2 [■8(4&−3&−3@−3&6&2@−3&2&−6)] = [■8(𝟐&(−𝟑)/𝟐&(−𝟑)/𝟐@(−𝟑)/𝟐&𝟑&𝟏@(−𝟑)/𝟐&𝟏&−𝟑)] 𝟏/𝟐 (B – B’) = 1/2 ([■8(2&−2&−4@−1&3&4@1&−2&−3)]−[■8(2&−1&1@−2&3&−2@−4&4&−3)]) = 1/2 [■8(0&−1&−5@1&0&6@5&−6&0)] = [■8(𝟎&(−𝟏)/𝟐&(−𝟓)/𝟐@𝟏/𝟐&𝟎&𝟑@𝟓/𝟐&−𝟑&𝟎)] Let, P = 𝟏/𝟐 (B + B’) = [■8(2&(−3)/2&(−3)/2@(−3)/2&3&1@(−3)/2&1&−3)] P’ = [■8(2&(−3)/2&(−3)/2@(−3)/2&3&1@(−3)/2&1&−3)] = P Since P’ = P P is a symmetric matrix. Let, Q = 𝟏/𝟐 (B − B’) = [■8(0&(−1)/2&(−5)/2@1/2&0&3@5/2&−3&0)] Q’ = [■8(0&1/2&5/2@(−1)/2&0&−3@(−5)/2&3&0)] = – [■8(0&(−1)/2&(−5)/2@1/2&0&3@5/2&−3&0)] Since Q’ = − Q Q is a skew symmetric matrix. Now, P + Q = 1/2 (B + B’) + 1/2 (B − B’) = B Thus, B is a sum of symmetric & skew symmetric matrix
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Hong Kong
Stage 1 - Stage 3
# Estimating Percentages
Lesson
A lot of the time it's hard for us to accurately calculate percentages of amounts in real life, so we'll have to estimate! Because percentages are expressed as something out of a hundred, we can also express them in diagrams of $5,10,100$5,10,100 things or more!
#### Example
Someone has been eating the brand new $10\times10$10×10 square block of chocolate! Can you figure out how much of the original chocolate block is left in percentages?
Think about the chocolate block as a fraction first
Do :We can see that there used to $10\times10=100$10×10=100 blocks of chocolate here, and now there are $67$67 blocks. So the fraction that represents how much is left of the original is $\frac{67}{100}$67100 . This is easily translated into a percentage as the denominator is already $100$100 , so the answer is $67%$67% .
#### Worked Examples
##### QUESTION 1
Which point on the line is closest to $56%$56%?
1. $A$A
A
$C$C
B
$D$D
C
$B$B
D
##### QUESTION 2
Ellie bought a $454$454 mL drink that claimed to be orange juice. In the ingredients list it said that orange juice made up $17%$17% of the drink. To estimate the amount of orange juice in the drink, which of the following would give the closest answer?
1. $10%\times454$10%×454
A
$20%\times454$20%×454
B
$10%\times400$10%×400
C
##### QUESTION 3
In a census, people are asked their gender and age. The graph shows the results: the percentage of females and males in each age group.
1. To the nearest $1%$1%, what percentage of females are between $5$5 and $9$9 years of age?
$7%$7%
A
$2%$2%
B
$11%$11%
C
2. To the nearest $1%$1%, what percentage of males are between $30$30 and $34$34 years of age?
$7%$7%
A
$4%$4%
B
$2%$2%
C
3. The percentage of females between the ages of $20$20 and $29$29 is about:
$15%$15%
A
$7%$7%
B
$25%$25%
C
4. The percentage of males below $20$20 years of age is about:
$15%$15%
A
$10%$10%
B
$30%$30%
C
$50%$50%
D
|
Blog / Math Learning / How to Do Long Division: Step-by-Step Guide & Examples [With Worksheets]
# How to Do Long Division: Step-by-Step Guide & Examples [With Worksheets]
Ever wondered how to tackle those dauntingly large math problems? How to navigate through numbers that seem like a maze? Fear not! Today, we’re diving into the realm of long division. From understanding its core principles to mastering the step-by-step process, we’ll equip you with the skills needed to conquer any numerical challenge. Whether you’re a student grappling with homework or an adult needing to crunch numbers, learning how to do long division is an invaluable skil
l. So, are you ready to unravel the mysteries of long division and emerge victorious? Let’s embark on this mathematical journey together.
Here are some worksheets to help you practice long division. You can download them if you need them. Such as “Math Worksheets PDF: 30Math Olympiad Exercise for Grade 2-4″, and “Financial Literacy Books for Kids“, “40 Primary School Maths Problems Workbsheets PDF With Answers“, etc.
## Understanding Long Division
Long division is a fundamental arithmetic operation that divides large numbers into smaller, more manageable parts. It is a methodical approach that efficiently divides numbers with multiple digits. At its core, long division involves several key components and concepts:
Definition:
Long division is a mathematical process used to divide a dividend (the number being divided) by a divisor (the number doing the dividing) to obtain a quotient (the result) and, optionally, a remainder.
Objective:
The primary objective of long division is to systematically break down a complex division problem into simpler steps, allowing for more straightforward computation.
### Components of Long Division
Long division involves several essential components/parts that play distinct roles in the division process:
1. Dividend: This represents the number being divided. It is the larger number in the division problem and is divided by the divisor.
2. Divisor: The divisor is the number by which the dividend is divided. It is the smaller number in the division problem and is the divisor.
3. Quotient: The quotient is the result of dividing the dividend by the divisor. It represents how many times the divisor can be subtracted from the dividend.
4. Remainder: In cases where the dividend is not evenly divisible by the divisor, there may be a remainder left over after performing the division. The remainder represents the amount left over that cannot be evenly divided by the divisor.
Process Overview:
The long division typically involves steps, including division, multiplication, subtraction, and bringing down digits. These steps are repeated until the entire dividend is divided or the desired precision level is achieved.
### Importance:
Long division is a fundamental skill in mathematics, laying the groundwork for more advanced mathematical concepts. It is commonly used in various real-world scenarios, such as budgeting, calculating proportions, and solving mathematical problems in fields like science, engineering, and finance.
## Mastering Long Division: A Step-by-Step Guide
Long division may appear daunting at first, but breaking it down into manageable steps can make the process much more approachable. Let’s walk through each stage of long division, accompanied by clear examples:
### Step 1: Divide
Begin by setting up the long division problem, with the dividend inside the division symbol and the divisor outside.
Start dividing from the leftmost digit of the dividend.
Example: Divide 735 by 5.
### Step 2: Multiply
Multiply the divisor by the quotient obtained from the division step.
Write the result beneath the dividend, aligning it with the appropriate place value.
Example: Divide 735 by 5
### Step 3: Subtract
Subtract the result obtained from the multiplication step from the portion of the dividend you’ve worked with so far.
Write the difference below the line.
Example: Divide 735 by 5
### Step 4: Bring Down
If there are still digits remaining in the dividend, bring down the next digit and append it to the result of the subtraction.
This creates a new number to work with.
Example:
### Step 5: Repeat
Repeat steps 2 through 4 until you’ve brought down all the digits of the dividend and there’s no remainder.
Example:
Quotient = 147
## Examples of Long Division
To truly master how to do long division, it’s helpful to work through a variety of examples. Below are seven examples that cover different cases of long division, each with detailed steps and final calculations.
### Example 1: Simple Division with No Remainder
Question: Divide 84 by 4.
Steps to Solve:
1. Divide the first digit: 8 ÷ 4 = 2.
2. Multiply: 2 × 4 = 8.
3. Subtract: 8 – 8 = 0.
4. Bring down the next digit: 4.
5. Divide: 4 ÷ 4 = 1.
6. Multiply: 1 × 4 = 4.
7. Subtract: 4 – 4 = 0.
### Example 2: Division with a Remainder
Question: Divide 29 by 5.
Steps to Solve:
1. Divide the first digit: 2 ÷ 5 = 0 (since 2 is less than 5, we consider the first two digits).
2. Divide: 29 ÷ 5 = 5.
3. Multiply: 5 × 5 = 25.
4. Subtract: 29 – 25 = 4.
### Example 3: Division Requiring Multiple Bring Downs
Question: Divide 1234 by 6.
Steps:
1. Divide the first digit: 1 ÷ 6 = 0 (consider the first two digits).
2. Divide: 12 ÷ 6 = 2.
3. Multiply: 2 × 6 = 12.
4. Subtract: 12 – 12 = 0.
5. Bring down the next digit: 3.
6. Divide: 3 ÷ 6 = 0 (bring down the next digit).
7. Divide: 34 ÷ 6 = 5.
8. Multiply: 5 × 6 = 30.
9. Subtract: 34 – 30 = 4.
### Example 4: Division with Larger Numbers
Question: Divide 9876 by 12.
Steps:
1. Divide the first two digits: 98 ÷ 12 = 8.
2. Multiply: 8 × 12 = 96.
3. Subtract: 98 – 96 = 2.
4. Bring down the next digit: 7.
5. Divide: 27 ÷ 12 = 2.
6. Multiply: 2 × 12 = 24.
7. Subtract: 27 – 24 = 3.
8. Bring down the next digit: 6.
9. Divide: 36 ÷ 12 = 3.
10. Multiply: 3 × 12 = 36.
11. Subtract: 36 – 36 = 0.
### Example 5: Division Resulting in a Decimal
Question: Divide 22 by 7.
Steps:
1. Divide the first digit: 2 ÷ 7 = 0 (consider the first two digits).
2. Divide: 22 ÷ 7 = 3.
3. Multiply: 3 × 7 = 21.
4. Subtract: 22 – 21 = 1.
5. Add decimal and bring down 0: 10 ÷ 7 = 1.
6. Multiply: 1 × 7 = 7.
7. Subtract: 10 – 7 = 3.
8. Bring down 0: 30 ÷ 7 = 4.
9. Multiply: 4 × 7 = 28.
10. Subtract: 30 – 28 = 2.
11. Continue as needed for more decimal places.
### Example 6: Division with Zeros in the Quotient
Question: Divide 2050 by 5.
Steps:
1. Divide the first digit: 2 ÷ 5 = 0 (consider the first two digits).
2. Divide: 20 ÷ 5 = 4.
3. Multiply: 4 × 5 = 20.
4. Subtract: 20 – 20 = 0.
5. Bring down the next digit: 5.
6. Divide: 5 ÷ 5 = 1.
7. Multiply: 1 × 5 = 5.
8. Subtract: 5 – 5 = 0.
9. Bring down the next digit: 0.
10. Divide: 0 ÷ 5 = 0.
### Example 7: Long Division by a 2-Digit Number
Question: Divide 15432 by 32.
Steps:
1. Setup: Write 15432 (dividend) inside the division bracket and 32 (divisor) outside.
2. Divide the first portion: Look at the first two digits of the dividend (15). Since 15 is less than 32, consider the first three digits (154). Determine how many times 32 fits into 154.
3. 154 ÷ 32 ≈ 4 (since 32 × 4 = 128 and 32 × 5 = 160, which is too high).
4. Multiply: Multiply 4 by 32 to get 128.
5. Subtract: Subtract 128 from 154 to get the remainder (26).
6. Bring down the next digit: Bring down the next digit from the dividend, which is 3, making the new number 263.
7. Divide again: Determine how many times 32 fits into 263.
8. 263 ÷ 32 ≈ 8 (since 32 × 8 = 256 and 32 × 9 = 288, which is too high).
9. Multiply: Multiply 8 by 32 to get 256.
10. Subtract: Subtract 256 from 263 to get the remainder (7).
11. Bring down the next digit: Bring down the next digit from the dividend, which is 2, making the new number 72.
12. Divide again: Determine how many times 32 fits into 72.
13. 72 ÷ 32 ≈ 2 (since 32 × 2 = 64 and 32 × 3 = 96, which is too high).
14. Multiply: Multiply 2 by 32 to get 64.
15. Subtract: Subtract 64 from 72 to get the remainder (8).
## Introducing WuKong Math: Unleashing the Math Whiz in Every Child
When it comes to learning long division and other essential math skills, WuKong Math offers a stellar educational experience that helps students excel. Designed for students from grades 1 to 12, WuKong Math provides live, interactive classes that make math enjoyable and engaging. Here’s why WuKong Math stands out as an excellent resource for mastering long division and enhancing overall math proficiency.
### Course Highlights:
• Global Accessibility: WuKong Math is available in all 24 global time zones, making it suitable for students worldwide.
• Live Instruction: Students benefit from live math classes taught by experienced instructors who are experts in igniting a love for math.
• Interactive Sessions: The blended learning approach includes small-group sessions online, fostering interactive and personalized learning experiences.
• Structured Curriculum: WuKong’s courses are copyrighted and come with a clear grading system to track progress and ensure understanding.
• Qualified Instructors: Their team of teachers spans the United States, Singapore, Australia, the United Kingdom, and more. These instructors are graduates from world-renowned universities and bring practical knowledge and experience in mathematics education.
• Extensive Math Resources: WuKong provides access to hundreds of original math learning resources, ensuring students have plenty of materials to practice and enhance their skills.
### Course Benefits:
• Enhanced Learning: WuKong Math’s courses are designed to boost mathematical thinking skills and improve math grades through well-structured lessons and interactive learning.
• Competition Preparation: For students aiming to compete internationally, WuKong offers advanced enrichment courses tailored to various math competitions, equipping them with the confidence and skills needed to excel.
• Comprehensive Coverage: With 10 levels of advanced courses, WuKong accommodates students of different countries, grades, and learning backgrounds, ensuring each student finds courses that match their learning needs.
• Innovative Teaching Methods: WuKong incorporates Singapore’s CPA modeling teaching approach, which uses graphics for mathematical modeling to help students transition from concrete to abstract thinking. This method fosters creative problem-solving skills.
## FAQs: How to Do Long Division
### Q1. How to check the result of long division?
Multiply the quotient by the divisor and add the remainder. The result should match the original dividend.
### Q2. How to handle zeros in long division?
If you encounter a zero in the dividend, bring it down as usual and continue the division process.
### Q3. What if the dividend has fewer digits than the divisor?
The quotient will be a decimal or a fraction, with the entire dividend as the remainder if it cannot be divided.
## Conclusion
In this article, we’ve explained how to do long division by defining it, providing a step-by-step guide, and offering practical examples. Mastering long division is important for boosting your math skills and confidence. For more help, WuKong Math is a great resource. They offer live, interactive classes and lots of learning materials that make math fun and effective. Whether you want to improve your grades or get ready for competitions, WuKong Math can help you reach your goals in mathematics.
### Discovering the maths whiz in every child, that’s what we do.
Suitable for students worldwide, from grades 1 to 12.
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# Division to Solve Decimal Equations
## Solve one - step equations using multiplication.
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Practice Division to Solve Decimal Equations
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Division to Solve Decimal Equations
Have you ever tried to pole vault?
The pole vault is a track and field event where a student uses a pole to launch themselves over a bar. Then the student lands on a large mat underneath the pole. Each part of the pole vault event is very specific. The height of the bar is specific. The length of the pole is specific, and the dimensions of the mat are specific as well.
The track and field team at Harrison Middle school had a special visitor after practice. Jody a pole vaulter from the nearby college visited to share his experiences with the students. He brought some pictures of himself in different events and took a long time answering student questions.
“Even the mat has specific dimensions,” Jody explained. “They measure the length, width and height of the mat to be sure that it has an accurate volume. The mat that we are using has a volume of 9009 cubic feet. The length of the mat is 16.5 feet and the width is 21 feet.”
Justin and Kara were listening intently to Jody’s explanation of the pole vault event. Justin, who loves numbers, began jotting down the dimensions of the mat on a piece of paper.
9009 cubic feet
16.5 feet in length
21 feet in width
Justin knows that he is missing the height of the mat.
“How high is the mat?” Justin asks Kara showing her his notes.
“Who cares?” Kara whispered looking back at Jody.
“I do,” Justin said turning away.
Justin begins to complete the math. But he can’t remember how to work the equation and the division.
This is where you come in. Pay attention to this Concept. By the end of it, you will need to help Justin with his dilemma.
### Guidance
Sometimes division is called the inverse , or opposite, of multiplication. This means that division will “undo” multiplication.
$5 \times 6 &= 30\\30 \div 6 &= 5$
See how that works? You can multiply two factors to get a product. Then when you divide the product by one factor, you get the other factor.
The Inverse Property of Multiplication states that for every number $x, x\left (\frac{1}{x} \right )= 1$ .
In other words, when you multiply $x$ by its opposite, $\frac{1}{x}$ (also known as the multiplicative inverse ), the result is one.
Let’s put whole numbers in place of $x$ to make the property clear: $3 \times \left ( \frac{1}{3} \right ) = \frac{3}{3} = 1$ . The Inverse Property of Multiplication may seem obvious, but it has important implications for our ability to solve variable equations which aren’t easily solved using mental math—such as variable equations involving decimals.
How does this apply to our work with equations?
With equations, the two expressions on either side of the equal sign must be equal at all times. The Inverse Property of Multiplication lets us multiply or divide the same number to both sides of the equation without changing the solution to the equation. This technique is called an inverse operation and it lets us get the variable $x$ alone on one side of the equation so that we can find its value. Take a look at how it’s done.
$5x & = 15\\\frac{5x}{5} &= \frac{15}{5} \rightarrow \text{inverse operation } = \text{divide} \ 5 \ \text{from both sides}\\x &= 3$
Remember how we said that division can “undo” multiplication? Well, this is a situation where that has happened.
Notice how, on the left side of the equation, $\frac{5}{5} = 1$ , leaving the $x$ alone on the left side. This is very instinctual to us. Once you have been working with equations, it comes naturally that when you divide a number by itself the answer is 1 and that has the variable be by itself on that side of the equals. After a while, we don’t even think about it, but it is still important to point out.
Solve $2.7x = 3.78$
We need to find a value of $x$ that, when multiplied by 2.7, results in 3.78. Let’s begin by using inverse operations to get $x$ alone on the left side of the equation.
$2.7x & = 3.78\\\frac{2.7}{2.7} x&= \frac{3.78}{2.7} \rightarrow \text{Inverse Operations} = \text{divide both sides by} \ 2.7$
Now, to find the value of $x$ , we complete the decimal division. First we multiply by ten and move the decimal places accordingly. $2.7 \rightarrow 27$ and $3.78 \rightarrow 37.8$
$& \overset{ \quad \ \ 1.4}{27 \overline{ ) {37.8 \;}}}\\& \quad \ \underline{27\;\;}\\& \qquad 108\\& \quad \underline{- \; 108}\\& \qquad \quad 0$
Use the inverse operation to solve each problem.
#### Example A
$2.3x = 5.06$
Solution: $2.2$
#### Example B
$1.6x= 5.76$
Solution: $3.6$
#### Example C
$4.7x = 10.81$
Solution: $2.3$
Here is the original problem once again.
The pole vault is a track and field event where a student uses a pole to launch themselves over a bar. Then the student lands on a large mat underneath the pole. Each part of the pole vault event is very specific. The height of the bar is specific. The length of the pole is specific, and the dimensions of the mat are specific as well.
The track and field team at Harrison Middle school had a special visitor after practice. Jody a pole vaulter from the nearby college visited to share his experiences with the students. He brought some pictures of himself in different events and took a long time answering student questions.
“Even the mat has specific dimensions,” Jody explained. “They measure the length, width and height of the mat to be sure that it has an accurate volume. The mat that we are using has a volume of 9009 cubic feet. The length of the mat is 16.5 feet and the width is 21 feet.”
Justin and Kara were listening intently to Jody’s explanation of the pole vault event. Justin, who loves numbers, began jotting down the dimensions of the mat on a piece of paper.
9009 cubic feet
16.5 feet in length
21 feet in width
Justin knows that he is missing the height of the mat.
“How high is the mat?” Justin asks Kara showing her his notes.
“Who cares?” Kara whispered looking back at Jody.
“I do, some things are worth figuring out,” Justin said turning away.
Justin begins to complete the math. But he can’t remember how to work the equation and the division.
To solve this problem of height, we need to remember that the formula for volume is length times width times height. Justin has the measurements for the volume and for the length and the width. He is missing the height. Justin can write the following formula.
$V &= lwh\\9009 & = 16.5 \times 21 \times h$
Next, we multiply $16.5 \times 21$ to begin our equation.
$16.5 \times 21 = 346.5$
We can write this equation.
$9009 = 346.5h$
Now we divide 9009 by 346.5.
$h = 26 \ feet$
The height of the mat is equal to 26 feet.
### Guided Practice
Here is one for you to try on your own.
Divide.
$4.5x = 12.6$
To solve this problem, we simply divide 12.6 by 4.5.
$2.8$
### Explore More
Directions: Solve the following problems using what you have learned about dividing decimals and equations. Write an equation when necessary.
1. Solve $3.7x = 7.77$
2. Solve $3.1x = 10.23$
3. Solve $7.2x = 29.52$
4. Solve $2.7x = 11.34$
5. Solve $1.2x = 6.72$
6. Solve $11x = 27.5$
7. Solve $6.7x = 42.21$
8. Solve $8.2x = 51.66$
9. Solve $1.9x = 12.92$
10. Solve $5.7x = 54.72$
11. Solve $.55x = .31955$
12. Solve $9.8x = 114.66$
13. In a week of track practice, Rose ran 3.12 times more than Jamie. If Rose ran 17.16 kilometers, how many kilometers did Jamie run? Write an equation and solve.
14. Ling’s flower bed has an area of $23.12 \ m^2$ and a width of 3.4 meters. What is the length of Ling’s flower bed? Write an equation and solve.
15. A jet airplane travels 6.5 times faster than a car. If the jet travels at 627.51 kilometers per hour, how fast is the car? Write an equation and solve.
### Vocabulary Language: English
Dividend
Dividend
In a division problem, the dividend is the number or expression that is being divided.
divisor
divisor
In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Estimation
Estimation
Estimation is the process of finding an approximate answer to a problem.
Inverse Operation
Inverse Operation
Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
Inverse Property of Multiplication
Inverse Property of Multiplication
The inverse property of multiplication states that the product of any real number and its multiplicative inverse (reciprocal) is one. If $a$ is a nonzero real number, then $a \times \left(\frac{1}{a}\right)=1$.
Quotient
Quotient
The quotient is the result after two amounts have been divided.
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# Natural Numbers, Even Numbers, Odd Numbers
13/07/2020
### Natural Numbers
The natural numbers are the counting numbers. It goes like 1, 2, 3, 4, …, and so on. It is interesting to know that if we subtract 1 from any natural number, we get its predecessor (previous number). If we add 1 to any natural numbers, it gives its successor (next number).
The predecessor of 5 is 5 − 1 = 4. The successor of 5 is 5 + 1 = 6. Is there any natural number that has no predecessor? The predecessor of 2 is 1. What is the predecessor of 1? Does that predecessor is also a natural number? No, no natural number is the predecessor of 1.
### Whole Numbers
Suppose you have 5 chocolates and you distribute them among your friends. How many chocolates do you have? Zero. Zero is denoted by the symbol 0. When we add 0 to the group of natural numbers, we get whole numbers. The predecessor of 1 is 1 − 1 = 0. 1 has the predecessor which is a whole number and not a natural number.
### 0 + Natural Numbers = Whole Numbers
#### Properties of Zero
• Any number, when multiplied by 0, gives 0.
• When 0 is added to any number, nothing changes.
• When 0 is subtracted from any number, it remains the same.
• 0 is the smallest whole number.
The whole numbers are said to consist of two types of numbers – even numbers and odd numbers.
### Even Numbers
A whole number exactly divisible by 2 is called even numbers.
For example:
2, 4, 6, 8, 10, 12, 14, 16……………………..are even number. Or a number having 0, 2, 4, 6, 8 at its units place is called an even number.
246, 1894, 5468, 100 are even number.
Any two even numbers which differ from one another by 2 are called consecutive even number.
### Odd Numbers
Odd numbers are the numbers which are not completely divisible by 2. The odd numbers leave 1 as a remainder when divided by 2. They have 1, 3, 5, 7, and 9 as their unit digit. 1, 3, 5, 7, 9, 11, 13, 15, etc. are odd numbers. The sets of odd number are expressed as Odd = {2n + 1: n ∈ integer (number)}.
#### Steps to Check for Odd and Even Numbers
• Divide the number by 2.
• Check the remainder.
• If the remainder is 0, it is an even number else if the remainder is 1, it is an odd number.
Check Odd & Even
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## Dividing Two-digit Numbers
### Mathematics, Grade 4
1
/
2
Divisibility Rules Division 3 + 3 + 3 + 3 = 12 3 x 4 = 12 Division is repeated subtraction and the inverse operation of multiplication. Divide 8 tens by 6. Each group gets 1 ten, with 2 tens remaining. To check your answer, multiply the quotient by the divisor and add the remainder if any. Bring down the ones and divide. 12 ÷ 4 = 3 number of items number of items in each group number of groups dividend divisor quotient Step 1 Step 2 Step 3 4 3 12 84 1 2 6 6 84 1 4 24 24 6 6 0 14 6 x 84 multiply: 6 x 4 subtract: 24 24 compare: 0 < 6 The answer checks if the product is the same as the dividend 2 Dividend is the number that is divided. Divisor is the number by which the dividend is being divided. Quotient is the answer to a division problem. dividend divisor quotient Find 84 ÷ 6 multiply: 1 x 6 subtract: 8 6 compare: 2 < 4 2 3 4 5 6 7 8 9 10 If the ones (last) digit is even If the sum of the digits are divisible by 3 If the sum of its digits is divisible by 9 If the number is divisible by both 2 and 3 If the number ends in 0 If the last two digits are divisible by 4 If the last three digits are divisible by 8 A whole number is divisible by: If the ones (last) digit is 0 or 5 If the number is divisible by 7 © Copyright NewPath Learning. All Rights Reserved. 93-4403 www.newpathlearning.com Dividing Two-digit Numbers
Divisibility Rules Division 3 + 3 + 3 + 3 = 12 3 x 4 = 12 Division is and the inverse operation of multiplication. Divide 8 tens by 6. Each group gets 1 ten, with 2 tens remaining. To check your answer, multiply the quotient by the divisor and add the remainder if any. Bring down the ones and divide. 12 ÷ 4 = 3 number of items number of items in each group number of groups Step 1 Step 2 Step 3 4 3 12 84 1 2 6 6 84 6 6 14 6 x multiply: 6 x 4 subtract: 24 24 compare: 0 < 6 The answer checks if the product is the same as the dividend Dividend is the number that is divided. Divisor is the number by which the dividend is being divided. Quotient is the answer to a division problem. Find 84 ÷ 6 multiply: 1 x 6 subtract: 8 6 compare: 2 < 4 2 3 4 5 6 7 8 9 10 A whole number is divisible by: Key Vocabulary Terms digit multiplication dividend quotient division remainder divisor repeated subtraction inverse operation © Copyright NewPath Learning. All Rights Reserved. 93-4403 www.newpathlearning.com Dividing Two-digit Numbers \|xiBAHBDy01644rzu
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# A and B are two mutually exclusive events of an experiment: If P(not A) = 0.65, P(A∪B)=0.65 and P(B) = p, then the value of p is
A
0.35
B
0.25
C
0.30
D
0.40
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the problem step by step, we will use the information given about the events A and B.Step 1: Understand the given informationWe know:- P(not A)=0.65- P(A∪B)=0.65- P(B)=p- A and B are mutually exclusive events.Step 2: Calculate P(A)Since P(not A)=0.65, we can find P(A) using the formula:P(A)=1−P(not A)Substituting the value:P(A)=1−0.65=0.35Step 3: Use the formula for the union of two eventsFor mutually exclusive events, the probability of their union is given by:P(A∪B)=P(A)+P(B)−P(A∩B)Since A and B are mutually exclusive, P(A∩B)=0. Therefore, the formula simplifies to:P(A∪B)=P(A)+P(B)Step 4: Substitute the known values into the formulaWe know P(A∪B)=0.65 and P(A)=0.35. Substituting these values into the equation:0.65=0.35+P(B)Step 5: Solve for P(B)Rearranging the equation to find P(B):P(B)=0.65−0.35P(B)=0.30ConclusionThus, the value of p is:0.30
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Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## If A and B are two independent events such that P(¯¯¯A)= 0.65, P(A∪B)= 0.65 and P(B)= p, the value of p is
A113
B313
C613
DNone of these
• Question 2 - Select One
## If A and B are mutually exclusive events P(A)=0.35andP(B)=0.45,then P(A∩B) is
A0.45
B0.35
C0.8
D0
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## Posts
Showing posts from December, 2013
### Santa's Dice Game
Q: Santa offers you to play a game of dice. You get to roll a dice six times. You can stop rolling whenever you wish and you get the dollar amount shown on that roll. What is an optimal strategy to maximize your payoff?
Probability and Statistics (4th Edition)
A: Let us take a moment and think through this. At each point in the sequence of rolls you make, you have a decision to make. Do you keep rolling or do you stop and walk away with what is being "offered" to you? You also need to bear in mind that if you keep pushing your luck you will reach a point (the 6th roll) where you would have to be content with whatever comes out for the last roll. So lets start with the simple case of what the expected payoff is for the last roll. Lets call this $$E_{6}$$. To compute it, simply take the payoff multiplied by the respective probability.
$$E_{6} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{7}{2} = 3.5$$
The general strategy to be followed is to check what the…
### A Box of Apples and Oranges
Q: A box contains 6 apples and 5 oranges. You know the number of apples are 6. You pick them out one at a time. What is the probability that the box will be empty by the time you have all 6 apples out?
A: It is tempting to say that the answer is $$50\%$$, but its not, as explained further below. Let us start with the number of ways to pull all 11 out. Since you know that there are six apples, you would keep drawing until you see all six apples. The six apples can come out in any order. So there are $$\binom{11}{6}$$ ways to do this.The number of favorable cases can be computed by making the following observation. If the box has to be empty when the last apple is drawn, then the finishing of the draws must end with 1, 2, 3, 4, 5 or 6 apples. These are the only scenarios. It can never end in an orange being drawn. Lets first consider the case when the last fruit drawn is an apple and the one prior is an orange. That's two d…
### The Numbers on a Dice
Q: A die is rolled and the numbers on all the visible faces is multiplied together $$=S$$. How would you choose a highest number such that it is guaranteed to divide the product.
A: The puzzle rests on a simple fact of prime factorization. Every number can be expressed as a product of prime numbers. For the given example, assume the face that is not visible is 1. The product of the remainder of numbers is $$6 \times 5\times\ldots \times 2$$ or equivalently $$6!$$. If the face that is invisible is 2, the number $$S$$ is $$\frac{6!}{2}$$, if it is 3 the number $$S$$ is $$\frac{6!}{3}$$ and so on. The number $$6!$$ can be expressed as prime numbers as $$2^{4}\times3^{2}\times 5$$. Let us enumerate the cases for each of the faces that is invisible, the product $$S$$ expressed in terms of prime numbers are as follows
$$2^{4}\times3^{2}\times5$$$$2^{3}\times3^{2}\times5$$$$2^{4}\times3^{1}\times5$$ $$2^{2}\times3^{2}\times5$$$$2^{4}\times3^{2}$$ \(2^{3}\times3^{1}\tim…
### A Cow, A Monkey and a Tree
Q: A Cow is tethered to a tree by a rope that has half the length of the circumference of the tree's stump. A monkey with a stone in hand jumps onto the tree and starts hopping around the tree's branches which covers a circular area of radius five times that of the stump. This causes the cow to run about at random and monkey drops the stone at random. What is the probability that the cow would get hit?
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RD Sharma 2017 Solutions for Class 7 Math Chapter 19 Visualising Solid Shapes are provided here with simple step-by-step explanations. These solutions for Visualising Solid Shapes are extremely popular among class 7 students for Math Visualising Solid Shapes Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2017 Book of class 7 Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2017 Solutions. All RD Sharma 2017 Solutions for class 7 Math are prepared by experts and are 100% accurate.
#### Question 1:
Complete the following table and verify Euler's formula in each case.
Faces (F) 6 4 9 7 Edges (E) 12 6 16 15 Vertex (V) 8 4 9 10
I II III IV Faces (F) 6 4 9 7 Edges (E) 12 6 16 15 Vertices (V) 8 4 9 10 Euler's formula (F − E + V) 6 − 12 + 8 = 2 4 − 6 + 4 = 2 9 − 16 + 9 = 2 7 − 15 + 10 = 2
Hence Euler's formula is verified for these figures.
#### Question 2:
Give three examples from our daily life which are in the form of (i) a cone (ii) a sphere (iii) a cuboid (iv) a cylinder (v) a pyramid.
Examples of
(i) Cone: Ice-cream cone, clown cap, rocket
(ii) Sphere: Football, a round apple, an orange
(iii) Cuboid: book, brick, duster
(iv) Cylinder: circular pipe, glass, circular pole
(v) Christmas decorations, cheese and patio umbrellas.
#### Question 1:
Match the following nets with appropriate solids:
Here
(a) → (ii)
(b) → (iii)
(c) → (iv)
(d) → (i)
#### Question 2:
Identify the nets which can be used to make cubes (cut-out the nets and try it):
Only (ii), (iv) and (vi) form a cube.
#### Question 3:
Can the following be a net for a die? Explain your answer.
Since, in a die, the sum of the number of opposite faces of a die is 7. In the given figure, it is not possible to get the sum as 7. Hence the given net is not suitable for a die.
#### Question 4:
Out of the following four nets there are two correct nets to make a tetrahedron. Identify them.
For making a tetrahedron, only (i) and (iii) are suitable nets.
#### Question 5:
Here is an incomplete net for making a cube. Complete it in atleast two different ways.
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Section 6: Optimization
# Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)
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Section 5.6 Optimization 529 Version: Fall 2007 5.6 Optimization In this section we will explore the science of optimization. Suppose that you are trying to find a pair of numbers with a fixed sum so that the product of the two numbers is a maximum. This is an example of an optimization problem. However, optimization is not limited to finding a maximum. For example, consider the manufacturer who would like to minimize his costs based on certain criteria. This is another example of an optimization problem. As you can see, optimization can encompass finding either a maximum or a minimum. Optimization can be applied to a broad family of different functions. However, in this section, we will concentrate on finding the maximums and minimums of quadratic functions. There is a large body of real-life applications that can be modeled by qua- dratic functions, so we will find that this is an excellent entry point into the study of optimization. Finding the Maximum or Minimum of a Quadratic Function Consider the quadratic function f ( x ) = x 2 + 4 x + 2 . Let’s complete the square to place this quadratic function in vertex form. First, factor out a minus sign. f ( x ) = x 2 4 x 2 Take half of the coefficient of x and square, as in [(1 / 2)( 4)] 2 = 4 . Add and subtract this amount to keep the equation balanced. f ( x ) = x 2 4 x + 4 4 2 Factor the perfect square trinomial, combine the constants at the end, and then redis- tribute the minus sign to place the quadratic function in vertex form. f ( x ) = ( x 2) 2 6 f ( x ) = ( x 2) 2 + 6 This is a parabola that opens downward, has been shifted 2 units to the right and 6 units upward. This places the vertex of the parabola at (2 , 6) , as shown in Figure 1 . Note that the maximum function value ( y -value) occurs at the vertex of the parabola. A mathematician would say that the function “attains a maximum value of 6 at x equals 2.” Note that 6 is greater than or equal to any other y -value (function value) that occurs on the parabola. This gives rise to the following definition. Copyrighted material. See: 1
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530 Chapter 5 Quadratic Functions Version: Fall 2007 x 10 y 10 (2 , 6) Figure 1. The maximum value of the function, 6, occurs at the vertex of the parabola, (2 , 6) . Definition 1. Let c be in the domain of f . The function f is said to achieve a maximum at x = c if f ( c ) f ( x ) for all x in the domain of f . Next, let’s look at a quadratic function that attains a minimum on its domain. l⚏ Example 2. Find the minimum value of the quadratic function defined by the equation f ( x ) = 2 x 2 + 12 x + 12 . Factor out a 2. f ( x ) = 2 x 2 + 6 x + 6 (3) Take half of the coefficient of x and square, as in [(1 / 2)(6)] 2 = 9 . Add and subtract this amount to keep the equation balanced. f ( x ) = 2 x 2 + 6 x + 9 9 + 6 Factor the trinomial and combine the constants, and then redistribute the 2 in the next step.
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## Section3.2Unit Vectors
A unit vector is a vector of magnitude $1\text{.}$ Usually we denote a unit vector with a carat symbol $\wedge$ over the letter representing the vector rather than an arrow $\rightarrow$ symbol. For instance, a unit vector $\hat u_{E}$ might denote a unit vector in the direction of East.
Say, you have a position vector $\vec r$ with magnitude, $r=10$ cm, pointed in the direction North. To get unit vector in the direction North, which we can denote by $\hat u_N$ we will just divide $\vec r$ by the magnitude $r=10$ cm.
\begin{equation*} \hat u_N = \dfrac{ (10\text{ cm}, \text{ North}) } { 10 \text{ cm} } = (1, \text{ North}). \end{equation*}
Since multiplying a vector by a number just scales the vector, multiplying a unit vector by a number will give a vector that is as many times bigger or smaller. For example, $2 \hat u_N$ will be a vector that has length $2$ and the direction same as the direction of $\hat u_N\text{.}$ The $\frac{1}{2} \hat u_N$ will be a vector that has length $\frac{1}{2}$ and the direction same as the direction of $\hat u_N\text{.}$ A $-\hat u_N$ will be a vector that has length $1$ but the direction opposite to the direction of $\hat u_N\text{.}$
Thus, you can write any vector along North or South direction as a product of its magnitude and the $\hat u_N$ vector. Say, you want a velocity vector $\vec v$ of magnitude $20$ m/s pointed North.
\begin{equation*} \vec v = (20 \text{ m/s} )\, \hat u_N, \end{equation*}
and if you want a velocity vector of magnitude $20$ m/s pointed South, you would do the following
\begin{equation*} \vec v = -(20 \text{ m/s} )\, \hat u_N, \end{equation*}
where the negative sign will just flip the direction of the unit vector.
There is nothing special about the direction North, and our discussion above using the example of $\hat u_N$ can be generalized to a unit vector in any arbitrary direction, such as East/West vectors, or any other vectors.
Say, you have a unit vector $\hat u$ in some direction, e.g., $40^{\circ}$ North of East. Then, any vector in the $40^{\circ}$ North of East direction or in exactly opposite direction can be written in as a number (with appropriate units and sign) times $\hat u \text{.}$
For instance, if we have a displacement vector $\vec r$ of $20 \text{ m}$ in the direction of the $\hat u$ vector, then $\vec r = 20\text{ m } \hat u\text{.}$ If you have a force vector $\vec F$ of $50 \text{ N}$ in the opposite direction, then $\vec F = -50\text{ N } \hat u\text{.}$
You can see that a unit vector provides a universal way of writing vectors which incorporate the direction information analytically.
### Subsection3.2.1Unit Vectors Along Cartesian Positive Axis Directions
Unit vectors along Cartesian axes play important role in vector analysis. We will often denote these unit vectors by $\hat u_x \text{,}$ $\hat u_y \text{,}$ and $\hat u_z$ respectively. At other times, we will denote them by their traditional symbols $\hat i \text{,}$ $\hat j \text{,}$ and $\hat k$ respectively.
For instance, say, we want a position vector $\vec r = ( x=2 \text{ m}, y=-5 \text{ m})$ in the $xy$-plane. Then, you can see that this vector will be a sum of two vectors: $\vec r_1 = 2 \text{ m} \hat i$ and $\vec r_2 = -5 \text{ m} \hat j \text{.}$ We write this as
\begin{equation*} \vec r = 2 \text{ m} \hat i - 5 \text{ m} \hat j. \end{equation*}
In general, an arbitrary vector $\vec A$ with coordinate components $A_x, A_y, A_z$ is often written as a sum of vectors along the Cartesian axes.
\begin{equation*} \vec A = A_x \hat i + A_y \hat j + A_z \hat k. \end{equation*}
### Subsection3.2.2Direction Cosines
Cosines of the angles a vector makes with unit vectors towards positive Cartesian axes are called its direction cosines of the vector. In Figure 3.2.6, the direction cosines of vector $\vec A$ are $\cos\theta_x\text{,}$ $\cos\theta_y\text{,}$ and $\cos\theta_z\text{.}$ These angles, $\theta_x\text{,}$ $\theta_y\text{,}$ and $\theta_z\text{,}$ are frequently referred to as $\alpha\text{,}$ $\beta\text{,}$ and $\gamma\text{,}$ respectively.
After we discuss dot product of vector I will show that direction cosines are not all independent. Rather, they are related by the following identity.
$$\cos^2\theta_x + \cos^2\theta_y + \cos^2\theta_z = 1.\tag{3.2.1}$$
Suppose you are told that the velocity of a projectile is $25\text{ m/s}$ in the direction of $30^{\circ}$ above the direction of East. Let positive $x$ axis be towards the East and the positive $y$ axis be pointed up.
(a) Find $x$ and $y$ components of a unit vector in the stated direction.
(b) Write the given velocity vector in terms of the unit vector you found.
Hint
Take a vector of magnitude 1 in the direction of the velocity and find its Cartesian components.
(a) $\hat u = \dfrac{\sqrt{3}}{2}\, \hat i + \dfrac{1}{2}\, \hat j\text{,}$ $25\text{ m/s}\, \hat u\text{.}$
Solution 1 (a)
(a) We want a vector $\hat u$ of unit magnitude in the direction described. We need to just find its $x$ and $y$ components.
\begin{align*} \amp u_x = 1\times \cos\, 30^{\circ} = \dfrac{\sqrt{3}}{2}\\ \amp u_x = 1\times \sin\, 30^{\circ} = \dfrac{1}{2}. \end{align*}
Therefore,
\begin{equation*} \hat u = \dfrac{\sqrt{3}}{2}\, \hat i + \dfrac{1}{2}\, \hat j. \end{equation*}
Solution 2 (b)
(b) When we multiple a unit vector by a positive number we get a vector of the new length and the same direction, so, all we have to do to get the velocity in that direcion is just multiply the unit vector by the magnitude of the velocity.
\begin{equation*} \vec v = 25\text{ m/s}\, \hat u. \end{equation*}
Of course, this is same as
\begin{equation*} \vec v = 25\text{ m/s}\, \left( \dfrac{\sqrt{3}}{2}\, \hat i + \dfrac{1}{2}\, \hat j\right) \end{equation*}
Suppose you want a unit vector in the same direction as the velocity vector given to you as $\vec v = (3.0 \text{ m/s})\hat i + (4.0 \text{ m/s})\hat j\text{.}$ Find this unit vector.
Hint
Divide by the magnitude.
$0.6\hat i + 0.8\hat j$
Solution
To get a unit vector, we just divide out the magnitude.
\begin{equation*} \hat u = \dfrac{ 3\hat i + 4 \hat j}{\sqrt{3^2 + 4^2}} = 0.6\hat i + 0.8\hat j. \end{equation*}
Suppose you want a unit vector in the direction of a star that is $30^\cric$ above horizon above $40^\circ$ North of East direction. Find a unit vector in the direction in terms of unit vectors along East, North, and Up.
Hint
Project unit magnitude on the East-North Plane and Up directions first.
$\hat u = 0.66 \hat i + 0.56 \hat j + 0.5 \hat k.$
Solution
Let us use $\hat i$ towards East, $\hat j$ towards North, and $\hat k$ towards Up. First we project a unit vector in the plane containing Up and “$40^\circ$ N of E” directions. We will call them as $u_z$ and $u_r\text{,}$ respectively. We get
\begin{align*} \amp u_z = 1\, \sin(30^\circ) = \frac{1}{2}.\\ \amp u_r = 1\, \cos(30^\circ) = \frac{\sqrt{3}}{2}. \end{align*}
Now, we project $u_r$ on $x$ and $y$ directions to get
\begin{align*} \amp u_x = u_r\, \cos(40^\circ) = 0.66.\\ \amp u_y = u_r\, \sin(40^\circ) = 0.56. \end{align*}
Therefore, our unit vector has the following representation.
\begin{equation*} \hat u = 0.66 \hat i + 0.56 \hat j + 0.5 \hat k. \end{equation*}
We should verify that it has magnitude of 1 by
\begin{equation*} \sqrt{0.66^2 + 0.56^2 + 0.5^2}, \end{equation*}
which is indeed 1 within the tolerance of the rounding. My answer is a little different than exact 1 since I had rounded off earlier.
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# What is the range of the function g(x) = (x-3)/(x+1)?
May 30, 2017
$x \in \mathbb{R} , x \ne - 1$
$y \in \mathbb{R} , y \ne 1$
#### Explanation:
$g \left(x\right) \text{ is defined for all real values of x except the value}$
$\text{that makes the denominator equal to zero}$
$\text{equating the denominator to zero and solving gives the }$
$\text{value that x cannot be}$
$\text{solve " x+1=0rArrx=-1larrcolor(red)" excluded value}$
$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne - 1$
$\text{to find any excluded values in the range, rearrange y = g(x)}$
$\text{making x the subject}$
$\Rightarrow y \left(x + 1\right) = x - 3$
$\Rightarrow x y + y = x - 3$
$\Rightarrow x y - x = - 3 - y$
$\Rightarrow x \left(y - 1\right) = - \left(3 + y\right)$
$\Rightarrow x = - \frac{3 + y}{y - 1}$
$\text{the denominator cannot equal zero}$
$\text{solve " y-1=0rArry=1larrcolor(red)" excluded value}$
$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne 1$
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## University Calculus: Early Transcendentals (3rd Edition)
Published by Pearson
# Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 50
#### Answer
$$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\sec\theta}$$
#### Work Step by Step
$$y=\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$$ Logarithmic differentiation helps to simplify the differentiation of otherwise complex functions by using the algebraic properties of logarithm. In detail, 1) Take the natural logarithm of both sides and simplify: $$\ln y=\ln\Big(\frac{\theta\sin\theta}{\sqrt{\sec\theta}}\Big)$$ Recall that $\ln(A\times B)=\ln A+\ln B$ and $\ln(A/B)=\ln A-\ln B$: $$\ln y=\ln(\theta\sin\theta)-\ln(\sqrt{\sec\theta})$$ $$\ln y=\ln\theta+\ln\sin\theta-\ln\sqrt{\sec\theta}$$ 2) Then we take the derivative of both sides with respect to $\theta$ and remember that $(\ln \theta)'=1/\theta$: $$\frac{1}{y}\times y'=\frac{1}{\theta}+\frac{1}{\sin\theta}(\sin\theta)'-\frac{1}{\sqrt{\sec\theta}}(\sqrt{\sec\theta})'$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sqrt{\sec\theta}}\times\frac{1}{2\sqrt{\sec\theta}}(\sec\theta)'$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{(\sec\theta)'}{2\sec\theta}$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{\sec\theta\tan\theta}{2\sec\theta}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{\tan\theta}{2}$$ $$\frac{y'}{y}=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}$$ 3) Solve for $y'$: $$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}\times y$$ 4) Finally, substitute for $y=\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$ $$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}\times\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$$ $$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\sec\theta}$$
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# The main property of fractions.
Speaking of math, one can not forget fraction.Their study paid a lot of attention and time.Think how many examples you had to solve in order to learn certain rules of work with fractions as you remember and apply the basic property of fractions.How many nerves were spent to find a common denominator, especially in the examples were more than two terms!
Let us remember that it is, and a little brush up on the basics and rules for working with fractions.
## Determining fractions
Let's start with the most important - determination.Fraction - a number which consists of one or more parts of the unit.Fraction is written as two numbers separated by a horizontal or the slash.The upper (or the first) is the numerator, and the bottom (second) - the denominator.
It is worth noting that the denominator indicates how many parts of the divided unit, and the numerator - the number of shares taken or parts.Often, fractions, if they are correct, less than unity.
Now let's look at the properties of these numbers and the basic rules that are used when working with them.But before we will analyze such a thing as "the basic property of rational fractions", will talk about the kinds of fractions and their features.
## What are fractions
There are several types of such numbers.This is especially common and decimal.The first are already given us a rational view of the number of entries by using either a horizontal slash.The second type of fractions is indicated by the so-called positional notation, is an indication of when the first part of the whole, and then, after the decimal point indicates the fractional part.
It is worth noting that in the same math used as decimal and common fractions.The main property of the fraction thus only valid for the second embodiment.In addition, common fractions distinguish right and wrong numbers.At first the numerator is always less than the denominator.Note also that this fraction is less than one.The improper fractions contrary - numerator over denominator, and she was more than one.Thus it can be distinguished from an integer.In this article we consider only ordinary fractions.
## Properties fractions
Any phenomenon, chemical, physical or mathematical, has its own characteristics and properties.No exception, and fractional numbers.They have one important feature, with which they can be carried out on certain operations.What is the main property of fractions?The rule states that if its numerator and denominator are multiplied or divided by the same rational number, we will get a new shot, the value of which is equal to the original.That is, by multiplying two fractional number 3/6 by 2, we obtain a new fraction of 6/12, and they are equal.
Based on this property, it is possible to reduce the fraction, as well as to select the common denominators for a particular pair of numbers.
## Operations
Although fractions seem to us to be more complex than the simple numbers, with them also to perform basic mathematical operations such as addition and subtraction, multiplication and division.In addition, there is a specific action, such as reducing fractions.Naturally, each of these actions is done according to certain rules.Knowledge of these laws easier to work with fractions, making it easier and more interesting.That's why we'll continue to look at the basic rules and sequence of actions when dealing with such numbers.
But before we talk about mathematical operations such as addition and subtraction, we explain an operation such as to bring to a common denominator.Here we just did and useful knowledge, a basic property of fractions exists.
## common denominator
In order to bring the number to a common denominator, you first need to find the least common multiple of the two denominators.That is the smallest number that is divisible by both two denominator without a trace.The easiest way to choose the LCM (least common multiple) - written out in line multiples for a single denominator, then the second, and to find among them the match number.In case NOC not found, that is, data numbers have a common multiple numbers must multiply them, and the resulting count value for the NOC.
So we found NOCs now have to find an additional factor.To do this, turn on the NOC to share denominators and write on each of them received number.Next, multiply the numerator and denominator for additional multiplier and record the results as a new fraction.If you doubt that you have received equal number still remember the main property of fractions.
now proceed directly to the mathematical operations on fractional numbers.Let's start with the most simple.There are several options for adding fractions.In the first case, both numbers have the same denominator.In this case, can only add up the numerators together.But the denominator does not change.For example, 1/5 + 3/5 = 4/5.
If the fractions have different denominators, you should bring them to the total, and only then perform addition.How to do it, we're just above dismantled.In this situation, you just come in handy basic property of fractions.The rule will bring the number to a common denominator.The value does not change.
Alternatively, it may happen that the fraction is mixed.Then, you must first be folded together the whole part, and then the fractions.
Multiply Multiply fractions requires no tricks, and to perform this action, necessary to know the main property of fractions.Suffice first multiply together the numerators and denominators.The product of the numerator will be the new numerator and the denominator - the new denominator.As you can see, nothing complicated.
The only thing you have to do - knowledge of multiplication tables, as well as care.In addition, after receiving the results, be sure to check whether you can reduce this number or not.For information on how to reduce fractions, we describe a little later.
## Subtraction
performing subtraction of fractions, should be guided by the same rules as for the addition.So, in numbers with the same denominator of the numerator of the reduced enough to take away the numerator deducted.In that case, if the fractions different denominators, you should bring them to the general and then perform the operation.As in a similar case with addition, you will need to use the main property of algebraic fractions and the skills to find the NOC and the common factors for the fractions.
## division
And the last, the most interesting operation when working with such numbers - division.It is pretty simple and does not cause any difficulties, even those who do not understand exactly how to work with fractions, especially to perform addition and subtraction.When dividing operating such a rule, as multiplication by the inverse fraction.The main property of the fraction, as in the case of the multiplication is involved for this operation would not.Let us examine in more detail.
When dividing numbers dividend remains unchanged.Fraction-splitter turns in the opposite, ie, the numerator by the denominator switch places.After this number multiplied together.
## reduction
So, we have already dismantled the definition and structure of fractions, their types, rules of operations on data numbers, found the main property of algebraic fractions.Now let's talk about an operation such as the reduction.Reduction of the fraction is the process of its transformation - the division of the numerator and denominator on the same number.Thus, the fraction is reduced without changing its properties.
Usually when making mathematical operation should look closely at the results obtained in the end and find out whether it is possible to reduce the resulting fraction or not.Keep in mind that the final result is always written does not require reduction of fractional number.
## Other operations
Finally, we note that we have listed, not all operations on fractional numbers, mentioning only the most well-known and necessary.Fractions can also equalize, convert to decimal and vice versa.But in this article we did not consider these operations as they are carried out in mathematics is much less than those that were listed above.
## Conclusions
we will talk about fractional numbers and operations with them.We dismantled and the main property of fractions, reducing fractions.But note that all these issues were discussed by us in passing.We have given only the most famous and used the rules gave the most important, in our opinion, advice.
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# Is 3 a square root?
• Last Updated : 04 Oct, 2021
The arithmetic value which is used for representing the quantity and used in making calculations are defined as Numbers. A symbol like “4,5,6” which represents a number is known as a numeral. Without numbers, we can’t do counting of things, date, time, money, etc., these numbers are also used for measurement and used for labeling.
The properties of numbers make them helpful in performing arithmetic operations on them. These numbers can be written in numeric forms and also in words.
For example, 3 is written as three in words, 35 is written as thirty-five in words, etc. Students can write the numbers from 1 to 100 in words to learn more.
There are different types of numbers, which we can learn. They are whole and natural numbers, odd and even numbers, rational and irrational numbers, etc.
What is a Number System?
A Number System is a method of showing numbers by writing, which is a mathematical way of representing the numbers of a given set, by using the numbers or symbols in a mathematical manner. The writing system for denoting numbers using digits or symbols in a logical manner is defined as Number System.
We can use the digits from 0 to 9 to form all the numbers. With these digits, anyone can create infinite numbers.
Example: 156, 3907, 3456, 1298, 784859, etc.
### What is a Square Root?
The value of a number of square roots, which on multiplication by itself gives the original number. Suppose, a is the square root of b, then it is represented as a = √b or we can express the same equation as a2 = b. Here, ’√’ this symbol we used to represent the root of numbers is termed as radical. The positive number when it is to be multiplied by itself represents the square of the number. The square root of the square of any positive number gives the original number.
For example, the square of 5 is 25, 52 = 25, and the square root of 25, √25 = 5. Since 5 is a perfect square, hence it is easy to find the square root of such numbers, but for an imperfect square, it’s really tricky.
Square Root is represented as ‘√’. It is called a radical symbol. To represent a number ‘a’ as a square root using this symbol can be written as: ‘√a‘, where a is the number.
The number here under the radical symbol is called the radicand. For example, the square root of 4 is also represented as a radical of 4. Both represent the same value.
The formula to find the square root is: a = √b
Properties of Square Roots
It is defined as a one-to-one function that takes a positive number as an input and returns the square root of the given input number.
f(x) = √x
For example, here if x = 9, then the function returns the output value as 3.
The properties of the square root are as follows:
• If a number is a perfect square number, then there definitely exists a perfect square root.
• If a number ends with an even number of zeros (0’s), then we can have a square root.
• The two square root values can be multiplied. For example, √3 can be multiplied by √2, then the result will be √6.
• When two same square roots are multiplied, then the result must be a radical number. It shows that the result is a non-square root number. For example, when √7 is multiplied by √7, the result obtained is 7.
• The square root of negative numbers is undefined. Hence the perfect square cannot be negative.
• Some of the numbers end with 2, 3, 7, or 8 (in the unit digit), then the perfect square root does not exist.
• Some of the numbers end with 1, 4, 5, 6, or 9 in the unit digit, then the number will have a square root.
• It is easy to find the square root of a number that is a perfect square.
What are Perfect Squares?
Perfect squares are those positive numbers that can be written as the multiplication of a number by itself, or you can say that a perfect square is a number which is the value of power 2 of any integer.
The number that can be expressed as the product of two equal integers.
For example, 16 is a perfect square because it is the product of two equal integers, 4 × 4 = 16. However, 24 is not a perfect square because it cannot be expressed as the product of two equal integers. (8 × 3 = 24).
The number which is obtained by squaring a whole number is termed as a perfect square.
If we assume N is a perfect square of a whole number y, this can be written as N = the product of y and y = y2.
So, the perfect square formula can be expressed as:
N = Y2
Let’s Use the formula with values.
If y = 10 , and N = y2.
This means, N = 102 = 100.
Here, 10 is a perfect square of 100 because it is the square of a whole number.
With the help of square roots, we can identify whether a number is a perfect square or not and if we calculate the square root of the given number.
If the square root is a whole number then the given number will be a perfect square, and if the square root value is not a whole number, then the given number is not a perfect square.
For instance, to check whether 24 is a perfect square or not, we will calculate its square root. √24 = 4.898979. As we can see, 4.898979 is not a whole number, so, 24 is not a perfect square.
Let’s take another example of
The number 49. √49 = 7. We can see that 7 is a whole number, therefore, 49 is a perfect square.
### Is 3 a square root?
Solution:
Yes 3 is a square root of 9
because 3 is a whole number and has a perfect square of 9
or we can say √9 = 3
or 32 = 3 × 3 = 9
so yes 3 is a square root of 9
### Similar Questions
Question 1: Give some examples of square root?
Solution:
Some examples are
1) √225 = 15
here 152 = 15 × 15 = 225
2) √4225 = 65
here 652 = 65 × 65 = 4225
3) √900 = 30
here 302 = 30 × 30 = 900
4) √1000 = 31.62277 not a perfect square
Question 2: Is 25 a square root?
Solution:
Here 25 is the square root of 625
which is a perfect square
252 = 25 × 25 = 625
therefore √625 = 25
square root of 625
So here 25 is the square root of 625
Question 3: Find the square root of 999?
Solution:
Here square root of 999 is 31.606
not a perfect square
therefore √999 = 31.606
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The numerator approaches 5 and the denominator approaches 0 from the left hence the limit is given by. &=\frac{\displaystyle 2⋅\left(\lim_{x→2}x\right)^2−3⋅\lim_{x→2}x+\lim_{x→2}1}{\displaystyle \left(\lim_{x→2}x\right)^3+\lim_{x→2}4} & & \text{Apply the power law. We now use the squeeze theorem to tackle several very important limits. Therefore, $\lim_{x→3}\dfrac{x(x−3)}{(x−3)(2x+1)}=\lim_{x→3}\dfrac{x}{2x+1}.\nonumber$, $\lim_{x→3}\dfrac{x}{2x+1}=\dfrac{3}{7}.\nonumber$. & = \sqrt{\blue{\displaystyle\lim_{x\to -2} x}+\red{\displaystyle\lim_{x\to -2}18}} && \mbox{Addition Law}\\ a. Example 10 Find the limit Solution to Example 10: As x approaches 2 from the left then x - 2 approaches 0 from the left or x - 2 < 0. That is, $$f(x)/g(x)$$ has the form $$K/0,K≠0$$ at a. Interactive simulation the most controversial math riddle ever! \displaystyle\lim_{x\to\pi}\sin(\blue x) & = \sin\left(\blue{\displaystyle\lim_{x\to\pi} x}\right) && \mbox{Composition Law}\\ Deriving the Formula for the Area of a Circle. }\$4pt] &= 4⋅(−3)+2=−10. Oct 21, 2020. The following three examples demonstrate the use of these limit laws in the evaluation of limits. For polynomials and rational functions, \[\lim_{x→a}f(x)=f(a).$. The first of these limits is $$\displaystyle \lim_{θ→0}\sin θ$$. Begin by letting be given. Example $$\PageIndex{6}$$: Evaluating a Limit by Simplifying a Complex Fraction. 3 cf x c f x lim ( ) lim ( ) & = 4 Consequently, the magnitude of $$\dfrac{x−3}{x(x−2)}$$ becomes infinite. $$. In each step, indicate the limit law applied. Evaluate the limit of a function by using the squeeze theorem. Thus, we see that for $$0<θ<\dfrac{π}{2}$$, $$\sin θ<θ<\tanθ$$. $$\displaystyle \lim_{x→2^−}\dfrac{x−3}{x}=−\dfrac{1}{2}$$ and $$\displaystyle \lim_{x→2^−}\dfrac{1}{x−2}=−∞$$. That is, as $$x$$ approaches $$2$$ from the left, the numerator approaches $$−1$$; and the denominator approaches $$0$$. Example $$\PageIndex{8A}$$ illustrates this point. 1 per month helps!! You da real mvps! To evaluate this limit, we use the unit circle in Figure $$\PageIndex{6}$$. Let’s now revisit one-sided limits. Do not multiply the denominators because we want to be able to cancel the factor $$(x−1)$$: $=\lim_{x→1}\dfrac{2−(x+1)}{2(x−1)(x+1)}.\nonumber$, $=\lim_{x→1}\dfrac{−x+1}{2(x−1)(x+1)}.\nonumber$. a.$$, $$\displaystyle \lim_{x\to 3} e^{\cos(\pi x)}$$, $$We then multiply out the numerator. For root functions, we can find the limit of the inside function first, and then apply the root. Encourage students to investigate limits using a variety of approaches. Factoring and canceling is a good strategy: $\lim_{x→3}\dfrac{x^2−3x}{2x^2−5x−3}=\lim_{x→3}\dfrac{x(x−3)}{(x−3)(2x+1)}\nonumber$.$$\displaystyle\lim\limits_{x\to 4} x = 4. You can evaluate the limit of a function by factoring and canceling, by multiplying by a conjugate, or by simplifying a complex fraction. If you know the limit laws in calculus, you’ll be able to find limits of all the crazy functions that your pre-calculus teacher can throw your way. The stress scores follow a uniform distribution with the lowest stress score equal to 1 and the highest equal to 5. Step 4. Thus, since $$\displaystyle \lim_{θ→0^+}\sin θ=0$$ and $$\displaystyle \lim_{θ→0^−}\sin θ=0$$, Next, using the identity $$\cos θ=\sqrt{1−\sin^2θ}$$ for $$−\dfrac{π}{2}<θ<\dfrac{π}{2}$$, we see that, $\lim_{θ→0}\cos θ=\lim_{θ→0}\sqrt{1−\sin^2θ}=1.\nonumber$. Step 2. This is not always true, but it does hold for all polynomials for any choice of $$a$$ and for all rational functions at all values of $$a$$ for which the rational function is defined. Example: Evaluate . &= \lim_{θ→0}\dfrac{\sin^2θ}{θ(1+\cos θ)}\4pt] ), 3. & = e^{\cos(3\pi)}\\ Find an expression for the area of the $$n$$-sided polygon in terms of $$r$$ and $$θ$$. Composition Law Suppose \lim\limits_{x\to a} g(x) = M, where M is a constant. If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root. & = e^{\cos\left(\pi\,\blue{\lim_{x\to 3} x}\right)} && \mbox{Constant Coefficient Law}\\ As we consider the limit in the next example, keep in mind that for the limit of a function to exist at a point, the functional values must approach a single real-number value at that point. Factoring And Canceling. $$\displaystyle \dfrac{\sqrt{x+2}−1}{x+1}$$ has the form $$0/0$$ at −1. Study the examples in your lecture notes in detail. \end{align*} Use the limit laws to evaluate the limit of a polynomial or rational function. &= \lim_{θ→0}\dfrac{1−\cos^2θ}{θ(1+\cos θ)}\\[4pt] Let $$p(x)$$ and $$q(x)$$ be polynomial functions. Notice that this figure adds one additional triangle to Figure $$\PageIndex{7}$$. Let $$a$$ be a real number. Missed the LibreFest? Use the same technique as Example $$\PageIndex{7}$$. Let $$f(x),g(x)$$, and $$h(x)$$ be defined for all $$x≠a$$ over an open interval containing $$a$$. Since this function is not defined to the left of 3, we cannot apply the limit laws to compute $$\displaystyle\lim_{x→3^−}\sqrt{x−3}$$. Root Law. Some of the geometric formulas we take for granted today were first derived by methods that anticipate some of the methods of calculus. Example 11 Find the limit Solution to Example 11: Factor x 2 in the denominator and simplify. 2.3.2 Use the limit laws to evaluate the limit of a function. & = 5^3\\ \end{align*} Essentially the same as the Addition Law, but for subtraction. In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine: \[\begin{align*} \lim_{θ→0}\dfrac{1−\cos θ}{θ} &=\displaystyle \lim_{θ→0}\dfrac{1−\cos θ}{θ}⋅\dfrac{1+\cos θ}{1+\cos θ}\\[4pt] § Solution f is a polynomial function with implied domain Dom()f = . Limits of Polynomials and Rational Functions. As approaches , the numerator goes to 5 and the denominator goes to 0.Depending on whether you approach from the left or the right, the denominator will be either a very small negative number, or a very small positive number. Solution. Consider the unit circle shown in Figure $$\PageIndex{6}$$. & = \left(\blue{\lim_{x\to 5} x}\right)\left(\red{\lim_{x\to5} x}\right)&& \mbox{Multiplication Law}\\ Limits of Polynomial and Rational Functions. He never came up with the idea of a limit, but we can use this idea to see what his geometric constructions could have predicted about the limit. Solution: lim x → 5x2 = lim x → 5(x ⋅ x) = ( lim x → 5x)( lim x → 5x) Multiplication Law = (5)(5) Identity Law = 25. If an $$n$$-sided regular polygon is inscribed in a circle of radius $$r$$, find a relationship between $$θ$$ and $$n$$. Choice (a) is incorrect . If your function has a coefficient, you can take the limit of the function first, and then multiply by the coefficient. The Greek mathematician Archimedes (ca. Graph $$f(x)=\begin{cases}−x−2, & \mathrm{if} \; x<−1\\ 2, & \mathrm{if} \; x=−1 \\ x^3, & \mathrm{if} \; x>−1\end{cases}$$ and evaluate $$\displaystyle \lim_{x→−1^−}f(x)$$. 4. Use the method in Example $$\PageIndex{8B}$$ to evaluate the limit. & = \frac 1 e \begin{align*} As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. Step 1. & = \sqrt{\blue{-2}+\red{18}} && \mbox{Identity and Constant Laws}\\ Example $$\PageIndex{8A}$$: Evaluating a One-Sided Limit Using the Limit Laws. . \nonumber. & = -\frac{17} 2 \end{align*}\]. Step 3. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied. Nov 18, 2020. & = \frac{24} 8\$6pt] For example, to apply the limit laws to a limit of the form $$\displaystyle \lim_{x→a^−}h(x)$$, we require the function $$h(x)$$ to be defined over an open interval of the form $$(b,a)$$; for a limit of the form $$\displaystyle \lim_{x→a^+}h(x)$$, we require the function $$h(x)$$ to be defined over an open interval of the form $$(a,c)$$. By dividing by $$\sin θ$$ in all parts of the inequality, we obtain, \[1<\dfrac{θ}{\sin θ}<\dfrac{1}{\cos θ}.\nonumber$. The next theorem, called the squeeze theorem, proves very useful for establishing basic trigonometric limits. Evaluating a Limit That Fails to Exist. Latest Math Topics. The function $$f(x)=\sqrt{x−3}$$ is defined over the interval $$[3,+∞)$$. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. \lim_{x\to\frac 1 2}(\blue{x}-\red{9}) & = \blue{\lim_{x\to\frac 1 2}x} - \red{\lim_{x\to\frac 1 2} 9} && \mbox{Subtraction Law}\\ Examples of the Central Limit Theorem Law of Large Numbers. \begin{align*} \lim_{x→−3}(4x+2) &= \lim_{x→−3} 4x + \lim_{x→−3} 2 & & \text{Apply the sum law. Evaluate $$\displaystyle \lim_{x→5}\dfrac{\sqrt{x−1}−2}{x−5}$$. In this case, we find the limit by performing addition and then applying one of our previous strategies. Since $$\displaystyle \lim_{θ→0^+}1=1=\lim_{θ→0^+}\cos θ$$, we conclude that $$\displaystyle \lim_{θ→0^+}\dfrac{\sin θ}{θ}=1$$. All you have to be able to do is find the limit of each individual function separately. In Example $$\PageIndex{11}$$, we use this limit to establish $$\displaystyle \lim_{θ→0}\dfrac{1−\cos θ}{θ}=0$$. . Example $$\PageIndex{9}$$: Evaluating a Limit of the Form $$K/0,\,K≠0$$ Using the Limit Laws. & = 4 (\blue{-2}) - \red{3}&& \blue{Identity}\hspace{2mm}and\hspace{2mm}\red{Constant}\hspace{2mm}Laws\\ Solve this for $$n$$. \displaystyle\lim\limits_{x\to4} (x + 1)^3, & = 125 In Example $$\PageIndex{8B}$$ we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function. & = -8 - 3\\ If the function involves the product of two (or more) factors, we can just take the limit of each factor, then multiply the results together. Then \displaystyle\lim\limits_{x\to a} f(x) \geq \lim\limits_{x\to a} g(x). Examples, solutions, videos, worksheets, games, and activities to help PreCalculus students learn how to use the limit laws to evaluate a limit. Evaluate $$\displaystyle \lim_{x→−3}\dfrac{x^2+4x+3}{x^2−9}$$. To find a formula for the area of the circle, find the limit of the expression in step 4 as $$θ$$ goes to zero. Since $$f(x)=4x−3$$ for all $$x$$ in $$(−∞,2)$$, replace $$f(x)$$ in the limit with $$4x−3$$ and apply the limit laws: \[\lim_{x→2^−}f(x)=\lim_{x→2^−}(4x−3)=5\nonumber. Then\lim\limits_{x\to a} f\left(g(x)\right) = f\left(\lim\limits_{x\to a} g(x)\right) = f(M)$$. After substituting in $$x=2$$, we see that this limit has the form $$−1/0$$. (9) Root Law:$$\displaystyle\lim\limits_{x\to a} \sqrt[n]{f(x)} = \sqrt[n] L$$provided$$L>0$$when$$nis even. Click HERE to return to the list of problems. Evaluate $$\displaystyle \lim_{x→−1}\dfrac{\sqrt{x+2}−1}{x+1}$$. \end{align*} Example $$\PageIndex{7}$$: Evaluating a Limit When the Limit Laws Do Not Apply. Solution. We now take a look at a limit that plays an important role in later chapters—namely, $$\displaystyle \lim_{θ→0}\dfrac{\sin θ}{θ}$$. $$\displaystyle \lim_{x→3^+}\sqrt{x−3}$$. Therefore, we see that for $$0<θ<\dfrac{π}{2},0<\sin θ<θ$$.\displaystyle\lim\limits_{x\to 12}\frac{2x}{x-4}$$,$$ We simplify the algebraic fraction by multiplying by $$2(x+1)/2(x+1)$$: $\lim_{x→1}\dfrac{\dfrac{1}{x+1}−\dfrac{1}{2}}{x−1}=\lim_{x→1}\dfrac{\dfrac{1}{x+1}−\dfrac{1}{2}}{x−1}⋅\dfrac{2(x+1)}{2(x+1)}.\nonumber$. Example $$\PageIndex{8B}$$: Evaluating a Two-Sided Limit Using the Limit Laws. \nonumber \]. \displaystyle\lim_{x\to 5} x^2 & = \displaystyle\lim_{x\to 5} (\blue{x}\cdot \red{x})\\ The following diagram shows the Limit Laws. We don’t multiply out the denominator because we are hoping that the $$(x+1)$$ in the denominator cancels out in the end: $=\lim_{x→−1}\dfrac{x+1}{(x+1)(\sqrt{x+2}+1)}.\nonumber$, $= \lim_{x→−1}\dfrac{1}{\sqrt{x+2}+1}.\nonumber$, $\lim_{x→−1}\dfrac{1}{\sqrt{x+2}+1}=\dfrac{1}{2}.\nonumber$. Watch the recordings here on Youtube! \end{align*} c. Since $$\displaystyle \lim_{x→2^−}f(x)=5$$ and $$\displaystyle \lim_{x→2^+}f(x)=1$$, we conclude that $$\displaystyle \lim_{x→2}f(x)$$ does not exist. Example 1: Use the Limit Laws to evaluate Also, assume $$\displaystyle\lim\limits_{x\to a} f(x)$$ and $$\displaystyle\lim\limits_{x\to a} g(x)$$ both exist. }\$4pt] The following observation allows us to evaluate many limits of this type: If for all $$x≠a,\;f(x)=g(x)$$ over some open interval containing $$a$$, then, \[\displaystyle\lim_{x→a}f(x)=\lim_{x→a}g(x).$. Let’s begin by multiplying by $$\sqrt{x+2}+1$$, the conjugate of $$\sqrt{x+2}−1$$, on the numerator and denominator: $\lim_{x→−1}\dfrac{\sqrt{x+2}−1}{x+1}=\lim_{x→−1}\dfrac{\sqrt{x+2}−1}{x+1}⋅\dfrac{\sqrt{x+2}+1}{\sqrt{x+2}+1}.\nonumber$. The limit of a constant is that constant: \ (\displaystyle \lim_ {x→2}5=5\). Example 8 Oscillating Behavior Discuss the existence of the limit. To understand this idea better, consider the limit $$\displaystyle \lim_{x→1}\dfrac{x^2−1}{x−1}$$. Work through some of the examples in your textbook, and compare your solution to the detailed solution o ered by the textbook. The next examples demonstrate the use of this Problem-Solving Strategy. \\ To get a better idea of what the limit is, we need to factor the denominator: $\lim_{x→2^−}\dfrac{x−3}{x^2−2x}=\lim_{x→2^−}\dfrac{x−3}{x(x−2)} \nonumber$. Then, $$\displaystyle\lim\limits_{x\to a} f\left(g(x)\right) = f\left(\lim\limits_{x\to a} g(x)\right) = f(M).$$, $$\displaystyle\lim\limits_{x\to\pi} \sin(x)$$, $$Next, we multiply through the numerators. For the following equations,$$a$$and$$k$$are constants and$$nis an integer. Step 5. – Typeset by FoilTEX – 8. Step 1. In Example $$\PageIndex{6}$$, we look at simplifying a complex fraction. EXAMPLE 2. 2.3.5 Evaluate the limit of a function by factoring or by using conjugates. However, not all limits can be evaluated by direct substitution. Follow the steps in the Problem-Solving Strategy, Example $$\PageIndex{5}$$: Evaluating a Limit by Multiplying by a Conjugate. & = \blue{\frac 1 2} - \red{9} && \blue{Identity}\hspace{2mm}and\hspace{2mm}\red{Constant}\hspace{2mm}Laws\\ & = (\blue 4 + \red 1)^3 && \blue{Identity}\hspace{2mm}and\hspace{2mm}\red{Constant}\hspace{2mm}Laws\\ Scroll down the page for more examples and solutions on how to use the Limit Laws. Evaluate $$\displaystyle \lim_{θ→0}\dfrac{1−\cos θ}{θ}$$. Therefore, the product of $$(x−3)/x$$ and $$1/(x−2)$$ has a limit of $$+∞$$: $\lim_{x→2^−}\dfrac{x−3}{x^2−2x}=+∞. In Mathematics, a limit is defined as a value that a function approaches the output for the given input values. It now follows from the quotient law that if $$p(x)$$ and $$q(x)$$ are polynomials for which $$q(a)≠0$$, \[\lim_{x→a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}.$, Example $$\PageIndex{3}$$: Evaluating a Limit of a Rational Function. 287−212; BCE) was particularly inventive, using polygons inscribed within circles to approximate the area of the circle as the number of sides of the polygon increased. Here is a set of practice problems to accompany the Limits At Infinity, Part I section of the Limits chapter of the notes for Paul Dawkins Calculus I course at Lamar University. Example 3.9. (Substitute $$\frac{1}{2}\sin θ$$ for $$\sin\left(\frac{θ}{2}\right)\cos\left(\frac{θ}{2}\right)$$ in your expression. \displaystyle\lim_{x\to 12}\frac{2\blue x}{\red x-4} & = \frac{\displaystyle\lim\limits_{x\to 12} (2 \blue x)}{\displaystyle\lim\limits_{x\to 12} (\red x-4)} && \mbox{Division Law}\6pt] Quiz 5: Limits and the limit laws. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at $$a$$. & =-11 }\\[4pt] &= 4⋅\lim_{x→−3} x + \lim_{x→−3} 2 & & \text{Apply the constant multiple law. We begin by restating two useful limit results from the previous section. \end{align*}. Choice (b) is incorrect . In the figure, we see that $$\sin θ$$ is the $$y$$-coordinate on the unit circle and it corresponds to the line segment shown in blue. The squeeze theorem allows you to find the limit of a function if the function is always greater than one function and less than another function with limits that are known. In fact, if we substitute 3 into the function we get $$0/0$$, which is undefined. (Hint: $$\displaystyle \lim_{θ→0}\dfrac{\sin θ}{θ}=1)$$.\displaystyle\lim\limits_{x\to\frac 1 2} (x-9)=$$,$$ Ask yourself, why they were o ered by the instructor. The function $$f(x)=\dfrac{x^2−3x}{2x^2−5x−3}$$ is undefined for $$x=3$$. & = (\blue{5})(\red{5}) && \mbox{Identity Law}\\ Think of the regular polygon as being made up of $$n$$ triangles. The limit of $$x$$ as $$x$$ approaches $$a$$ is a: $$\displaystyle \lim_{x→2}x=2$$. $$, Suppose$$\lim\limits_{x\to a} g(x) = M$$, where$$Mis a constant. Example $$\PageIndex{4}$$ illustrates the factor-and-cancel technique; Example $$\PageIndex{5}$$ shows multiplying by a conjugate. & = -2 Real World Math Horror Stories from Real encounters. Then, we cancel the common factors of $$(x−1)$$: $=\lim_{x→1}\dfrac{−1}{2(x+1)}.\nonumber$. \end{align*} The limit of \ (x\) as \ (x\) approaches \ (a\) is a: \ (\displaystyle \lim_ {x→2}x=2\). limit exists. In this section, we establish laws for calculating limits and learn how to apply these laws. Sincey$$and$$x$$are equal, whatever value$$x$$approaches,$$ywill have to approach the same value. Instead, we need to do some preliminary algebra. \end{align*} The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. \begin{align*} $f(x)=\dfrac{x^2−1}{x−1}=\dfrac{(x−1)(x+1)}{x−1}\nonumber$. :) https://www.patreon.com/patrickjmt !! Both $$1/x$$ and $$5/x(x−5)$$ fail to have a limit at zero. Problem-Solving Strategy: Calculating a Limit When $$f(x)/g(x)$$ has the Indeterminate Form $$0/0$$. \nonumber\]. But you have to be careful! Example $$\PageIndex{4}$$: Evaluating a Limit by Factoring and Canceling. We now take a look at the limit laws, the individual properties of limits. \lim_{x\to 3} (8x) & = 8\,\lim_{x\to 3} x && \mbox{Constant Coefficient Law}\\ To do this, we may need to try one or more of the following steps: If $$f(x)$$ and $$g(x)$$ are polynomials, we should factor each function and cancel out any common factors. When taking limits with exponents, you can take the limit of the function first, and then apply the exponent. Example does not fall neatly into any of the patterns established in the previous examples. If, for all $$x≠a$$ in an open interval containing $$a$$ and, where $$L$$ is a real number, then $$\displaystyle \lim_{x→a}g(x)=L.$$, Example $$\PageIndex{10}$$: Applying the Squeeze Theorem. \begin{align*} and the function $$g(x)=x+1$$ are identical for all values of $$x≠1$$. The technique of estimating areas of regions by using polygons is revisited in Introduction to Integration. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws. }\$4pt] List of limit problems with solutions for the exponential functions to evaluate the limits of functions in which exponential functions are involved. By now you have probably noticed that, in each of the previous examples, it has been the case that $$\displaystyle \lim_{x→a}f(x)=f(a)$$. Then, each of the following statements holds: \[\displaystyle \lim_{x→a}(f(x)+g(x))=\lim_{x→a}f(x)+\lim_{x→a}g(x)=L+M$, $\displaystyle \lim_{x→a}(f(x)−g(x))=\lim_{x→a}f(x)−\lim_{x→a}g(x)=L−M$, $\displaystyle \lim_{x→a}cf(x)=c⋅\lim_{x→a}f(x)=cL$, $\displaystyle \lim_{x→a}(f(x)⋅g(x))=\lim_{x→a}f(x)⋅\lim_{x→a}g(x)=L⋅M$, $\displaystyle \lim_{x→a}\frac{f(x)}{g(x)}=\frac{\displaystyle \lim_{x→a}f(x)}{\displaystyle \lim_{x→a}g(x)}=\frac{L}{M}$, $\displaystyle \lim_{x→a}\big(f(x)\big)^n=\big(\lim_{x→a}f(x)\big)^n=L^n$, $\displaystyle \lim_{x→a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x→a} f(x)}=\sqrt[n]{L}$. & = \frac 1 2 - \frac{18} 2\$6pt] % Observe that, \[\dfrac{1}{x}+\dfrac{5}{x(x−5)}=\dfrac{x−5+5}{x(x−5)}=\dfrac{x}{x(x−5)}.\nonumber$, $\lim_{x→0}\left(\dfrac{1}{x}+\dfrac{5}{x(x−5)}\right)=\lim_{x→0}\dfrac{x}{x(x−5)}=\lim_{x→0}\dfrac{1}{x−5}=−\dfrac{1}{5}.\nonumber$. There is a concise list of the Limit Laws at the. Use the squeeze theorem to evaluate $$\displaystyle \lim_{x→0}x^2 \sin\dfrac{1}{x}$$. We can estimate the area of a circle by computing the area of an inscribed regular polygon. Evaluate $$\displaystyle \lim_{x→1}\dfrac{\dfrac{1}{x+1}−\dfrac{1}{2}}{x−1}$$. Also, supposef$$is continuous at$$M$$. & = 8 (3) && \mbox{Identity Law}\\ Solution Let In Figure 11.9, you can see that as approaches 0, oscillates between and 1. \displaystyle\lim_{x\to 3} e^{\cos(\pi \blue x)} & = e^{\displaystyle\lim_{x\to 3}\cos(\pi \blue x)} && \mbox{Composition Law} \\ Solution. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied. Evaluate using a table of values. Apply the squeeze theorem to evaluate $$\displaystyle \lim_{x→0} x \cos x$$. Use the limit laws to evaluate the limit of a function. Step 1. Evaluate $$\displaystyle\lim_{x→3}\dfrac{x^2−3x}{2x^2−5x−3}$$. Solution. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions. • With the solution of Example 2 in mind, let’s try to save time by letting (x, y) → (0, 0) along any nonvertical line through the origin. \displaystyle\lim_{x\to -2} \sqrt{\blue x+\red{18}} & = \sqrt{\displaystyle\lim_{x\to -2}(\blue x+\red{18})} && \mbox{Root Law}\\ & = 4\,\displaystyle\lim_{x\to-2} (\blue{x}^3) + 5\,\displaystyle\lim_{x\to-2} \red x && \mbox{Constant Coefficient Law}\\ The Central Limit Theorem illustrates the Law of Large Numbers. This fact follows from application of the limit laws which have been stated up to this point. In nonelectrolyte solutions, the intermolecular forces are mostly comprised of weak Van der Waals interactions, which have a $$r^{-7}$$ dependence, and for practical purposes this can be considered ideal. 2.3.4 Use the limit laws to evaluate the limit of a polynomial or rational function. limits in Examples 6 and 7 by constructing and examining a table of values. \nonumber\]. For all $$x≠3,\dfrac{x^2−3x}{2x^2−5x−3}=\dfrac{x}{2x+1}$$. Use the limit laws to evaluate . Step 6. We then need to find a function that is equal to $$h(x)=f(x)/g(x)$$ for all $$x≠a$$ over some interval containing a. Using Limit Laws Repeatedly. Download for free at http://cnx.org. (1) Constant Law:$$\displaystyle\lim\limits_{x\to a} k = k$$, (2) Identity Law:$$\displaystyle\lim\limits_{x\to a} x = a$$, (3) large Addition Law:$$\displaystyle\lim\limits_{x\to a} f(x) + g(x) = \displaystyle\lim\limits_{x\to a} f(x) + \displaystyle\lim\limits_{x\to a} g(x)$$, (4) Subtraction Law:$$\displaystyle\lim\limits_{x\to a} f(x) - g(x) = \displaystyle\lim\limits_{x\to a} f(x) - \displaystyle\lim\limits_{x\to a} g(x)$$, (5) Constant Coefficient Law:$$\displaystyle\lim\limits_{x\to a} k\cdot f(x) = k\displaystyle\lim\limits_{x\to a} f(x)$$, (6) Multiplication Law:$$\lim\limits_{x\to a} f(x)\cdot g(x) = \left(\lim\limits_{x\to a} f(x)\right)\left(\lim\limits_{x\to a} g(x)\right)$$, (7) Power Law:$$\displaystyle\lim\limits_{x\to a} \left(f(x)\right)^n= \left(\displaystyle\lim\limits_{x\to a} f(x)\right)^n$$provided$$\displaystyle\lim\limits_{x\to a} f(x)\neq 0$$if$$n <0$$, (8) Division Law:$$\displaystyle\lim\limits_{x\to a} \frac{f(x)}{g(x)} = \frac{\displaystyle\lim\limits_{x\to a}f(x)}{\displaystyle\lim\limits_{x\to a} g(x)}$$provided$$\displaystyle\lim\limits_{x\to a} g(x)\neq 0. \end{align*} Work with each term separately, then subtract the results.\displaystyle\lim\limits_{x\to 5} x^2$$,$$ The laws of limits The laws of limits and how we use them to evaluate a limit. SOLUTIONS TO LIMITS OF FUNCTIONS USING THE PRECISE DEFINITION OF LIMIT SOLUTION 1 : Prove that . We factor the numerator as a difference of squares and then cancel out the common term (x – 1) Step 1. \nonumber\]. We see that the length of the side opposite angle $$θ$$ in this new triangle is $$\tan θ$$. For any real number $$a$$ and any constant $$c$$, Example $$\PageIndex{1}$$: Evaluating a Basic Limit. It follows that $$0>\sin θ>θ$$. Example: Solution: We can’t find the limit by substituting x = 1 because is undefined. WARNING 2: Sometimes, the limit value lim x a fx() does not equal the function value fa(). \begin{align*} Let $$c$$ be a constant. The proofs that these laws hold are omitted here. The first 6 Limit Laws allow us to find limits of any … Choice (d) is correct! We now take a look at the limit laws, the individual properties of limits. Limit of a function. For problems 1 – 9 evaluate the limit, if it exists. Thanks to all of you who support me on Patreon. Example $$\PageIndex{11}$$: Evaluating an Important Trigonometric Limit. \begin{align*} By taking the limit as the vertex angle of these triangles goes to zero, you can obtain the area of the circle. If the exponent is negative, then the limit of the function can't be zero! Extra Examples, attempt the problems before looking at the solutions Decide if the following limits exist and if a limit exists, nd its value. Figure $$\PageIndex{4}$$ illustrates this idea. Let’s apply the limit laws one step at a time to be sure we understand how they work. & = \frac{2\,\displaystyle\lim\limits_{x\to12} \blue x}{\displaystyle\lim\limits_{x\to12}(\red x- 4)} && \mbox{Constant Coefficient Law}\6pt] EXAMPLE 2. \begin{align*} Make use of it. Use the methods from Example $$\PageIndex{9}$$. Evaluate $$\displaystyle \lim_{θ→0}\dfrac{1−\cos θ}{\sin θ}$$. & = e^{-1}\\ & = e^{\cos\left(\displaystyle\lim_{x\to3}(\pi \blue x)\right)} && \mbox{Composition Law}\\ Using the Limit Laws, we can write: \[=\left(\lim_{x→2^−}\dfrac{x−3}{x}\right)\cdot\left(\lim_{x→2^−}\dfrac{1}{x−2}\right). 2.3.3 Evaluate the limit of a function by factoring. & = \frac{2\,\blue{\displaystyle\lim\limits_{x\to12} x}}{\red{\displaystyle\lim\limits_{x\to12} x} - \displaystyle\lim\limits_{x\to12} 4} && \mbox{Subtraction Law}\\[6pt] b. This limit also proves useful in later chapters. & = 24 The limit laws allow us to evaluate limits of functions without having to go through step-by-step processes each time. \begin{align*} Evaluate the limit of a function by factoring or by using conjugates. \displaystyle\lim_{x\to4} (\blue{x}+\red{1})^3 & = \left(\displaystyle\lim_{x\to4} (\blue{x}+\red 1)\right)^3 && \mbox{Power Law}\\ Let’s apply the limit laws one step at a time to be sure we understand how they work. Solution. Last, we evaluate using the limit laws: \[\lim_{x→1}\dfrac{−1}{2(x+1)}=−\dfrac{1}{4}.\nonumber. Since $$\displaystyle \lim_{x→0}(−x)=0=\lim_{x→0}x$$, from the squeeze theorem, we obtain $$\displaystyle \lim_{x→0}x \cos x=0$$. 68 CHAPTER 2 Limit of a Function 2.1 Limits—An Informal Approach Introduction The two broad areas of calculus known as differential and integral calculus are built on the foundation concept of a limit.In this section our approach to this important con-cept will be intuitive, concentrating on understanding what a limit is using numerical and graphical examples. By applying these limit laws we obtain $$\displaystyle\lim_{x→3^+}\sqrt{x−3}=0$$. Example 7. lim x → 5x2. Limits are important in calculus and mathematical analysis and used to define integrals, derivatives, and continuity. General outline for Evaluating limits of this type one additional triangle to Figure \ \displaystyle. Simplifying a complex fraction { x^2−3x } { x } { 2x^2−5x−3 } \ ): a... 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# TutorMe Blog
## How to Find Cube Roots and Perfect Cubes
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Jana Russick
May 13, 2021
In order to figure out how to find cube roots, you must review square roots. Finding the square root of a number means determining what other number needs to be multiplied by itself to get the original number. For example:
√81 = 9
## What Are Cube Roots?
Finding cube roots takes it a step further. Below is an example of a number being cubed:
4³ = 64
4 x 4 x 4 = 64
The process of cubing is multiplying a number by itself three times, as indicated by the exponent of 3. Let's take the example of 64 again. To find its cube root, you need to know that 4 needs to be multiplied by itself three times to get the final result of 64.
If you want to calculate cube roots, you must list a small 3 outside the root symbol:
³√8 = 2
### Finding the Cube Root of Negative Numbers
Unlike square roots, you can take the cube root of a negative number. This is because a negative number multiplied by itself 3 times will always equal a negative. Because of this, the result of cubing a number doesn't have to be a whole number, it can be a negative or positive number:
³√-216 = 6
## What Is a Perfect Cube?
Now that we've seen how to find the cube root of a number, let's take a look at perfect cubes. When using a cube root calculator, you can technically take the cube root of any number. However, you will sometimes end up with an answer that has tons of decimals. But this doesn't happen when you take the cube root of a perfect cube.
A perfect cube is the integer that you get when you raise another integer to the third power. A perfect cube integer cannot be a decimal or fraction. The same goes for the number that is being raised to the third power.
Here is a list of perfect cubes:
1³ = 1
2³ = 4
3³ = 27
4³ = 64
5³ = 125
6³ = 216
7³ = 343
Keep this list handy for when you don't have a cube root calculator.
## How to Find Cube Roots
Finding cube roots is simply applying what you've learned about square roots and taking it a step further. So instead of multiplying a number by itself twice, cubing is the process of multiplying a number by itself three times. Similarly, perfect cubes are just like perfect squares but with the third root taken instead of the second.
When you know how to cube a number, how to take the cubed root of a positive and negative number, and are familiar with perfect cubes, you'll be ready to master more complex math concepts that involve cubed roots.
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# Test of Mathematics Solution Objective 401 – Trigonometric Series
This is a Test of Mathematics Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
## Problem
The value of the sum $$\cos \frac{\pi}{1000} + \cos \frac{2\pi}{1000} + \cdots + \cos \frac {999 \pi}{1000}$$ is
(A) 0; (B)1; (C) $$\frac {1}{1000}$$; (D) an irrational number;
## Sequential Hints
(How to use this discussion: Do not read the entire solution at one go. First, read more on the Key Idea, then give the problem a try. Next, look into Step 1 and give it another try and so on.)
### Key Idea
This is the generic use case of Complex Number $$\iota =\sqrt {-1}$$ and De Moivre’s Theorem
### Hint 1
We know that $$\cos (\pi – \theta) = – \cos \theta$$. There are several ways to imagine this. One intuitive way is horizontal projection for $$\theta$$ and $$\pi – \theta$$ are of same magnitude but of opposite sides.
Now notice that if $$\theta = \frac{\pi}{1000}$$ then $$\pi – \theta = \pi – \frac{ \pi}{1000} = \frac{999 \pi}{1000}$$
### Hint 2
This implies $$\frac{\pi}{1000} + \frac{999 \pi}{1000} = 0$$. Similarly $$\frac{2\pi}{1000} + \frac{998 \pi}{1000} = 0$$ and so on.
### Hint 3
Hence by properly pairing up, all of them will cancel out. Only $$\cos \frac{500\pi}{1000} = 0$$
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# Math Snap
## Let $x_{n}=\sqrt[n]{1+2^{n(-1)^{n}}}$ for $n \in \mathbb{N}$. Then Select one: a. $\operatorname{\operatorname {lim}}\left(x_{n}\right)=\sqrt{2}$ and $\operatorname{\operatorname {lim}}\left(x_{n}\right)=0$ $\operatorname{bim}\left(x_{n}\right)=\ln (2)$ and $\operatorname{Iim}\left(x_{n}\right)=1$ c. None of them d. $\operatorname{dim}\left(x_{n}\right)=2$ and $\underline{\lim }\left(x_{n}\right)=0$ $\lim \left(x_{n}\right)=2$ and $\underline{\lim }\left(x_{n}\right)=1$
#### STEP 1
Assumptions 1. $x_n = \sqrt[n]{1 + 2^{n(-1)^n}}$ 2. $n \in \mathbb{N}$ 3. We need to find the limit of $x_n$ as $n$ approaches infinity.
#### STEP 2
Consider the expression $x_n = \sqrt[n]{1 + 2^{n(-1)^n}}$. We need to analyze the behavior of this expression as $n$ becomes very large.
#### STEP 3
First, let's consider the case when $n$ is even. For even $n$, $(-1)^n = 1$, so the expression becomes: $x_n = \sqrt[n]{1 + 2^{-n}}$
#### STEP 4
For large even $n$, $2^{-n}$ approaches 0. Therefore, the expression simplifies to: $x_n \approx \sqrt[n]{1}$
#### STEP 5
Since $\sqrt[n]{1} = 1$ for any $n$, we have: $\lim_{n \to \infty, n \text{ even}} x_n = 1$
#### STEP 6
Next, consider the case when $n$ is odd. For odd $n$, $(-1)^n = -1$, so the expression becomes: $x_n = \sqrt[n]{1 + 2^n}$
#### STEP 7
For large odd $n$, $2^n$ becomes very large. Therefore, the expression simplifies to: $x_n \approx \sqrt[n]{2^n}$
#### STEP 8
Since $\sqrt[n]{2^n} = 2$, we have: $\lim_{n \to \infty, n \text{ odd}} x_n = 2$
#### STEP 9
From the above analysis, we see that the limit of $x_n$ depends on whether $n$ is even or odd. For even $n$, the limit is 1, and for odd $n$, the limit is 2.
#### STEP 10
Therefore, the limit of $x_n$ does not exist because it oscillates between 1 and 2 depending on whether $n$ is even or odd.
##### SOLUTION
Since the limit does not exist, we can conclude that none of the given options correctly describe the behavior of $x_n$. The correct answer is: c. None of them
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8 + 3 = 11. In simple terms, this fraction and whole number calculator allows you to solve fraction problems with whole numbers and fractions form. If one of the denominators is divisible by the other, the larger denominator is the LCM. Make your quotient the new whole number. 10 + 2. Take your remainder and place it over the original denominator to finish converting the improper fraction into a mixed number. Step 1: Convert all mixed numbers into improper fractions. How about this? For all free math videos visit http://Mathmeeting.com. Then, add the fractions together and simplify. Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. [1] X Research source For instance, to solve 5/9 + 1/9, just add 5 + 1, which equals 6. Now you can add these together. To convert 1 1/2 to a mixed number, multiply the whole number 1 by the denominator 2, and then add it to the numerator. Step 5: If the answer is an improper form, reduce the fraction into a mixed number. :�^����OJR봎��l!x8�V�2�eT C�@H��Ip�P�x ������Ͻ �6\���8j�ˀ�tF�i�K��̄�T���IoM���N��4�J]��*X� We know ads can be annoying, but they’re what allow us to make all of wikiHow available for free. Get it? Add the fractions and mixed numbers. So, for each fraction we need an equivalent fraction with a denominator of 6. Just as you can add whole numbers and proper fractions, you can also add mixed numbers. Start out by multiplying the numerator of the first fraction by the denominators of all the other fractions. This article has been viewed 90,941 times. That is why it is called a "mixed" fraction (or mixed number). First, divide the numerator by the denominator. Add fractions with the same denominator by combining the numerators. How do I add mixed fractions if there is no LCM? Now, to add or subtract anything, the names -- the units -- must be the same. 11. 9 + 8. The easiest thing to do here is double the first denominator (2) to make it the same as the second denominator (4). This video teaches the viewer how to multiply fractions with mixed numbers. See how each example is made up of a whole number anda proper fraction together? Next, find the lowest common denominator of both fractions and convert the fractions so they both have this denominator. If you're asking about first dividing mixed numbers, you must convert both mixed numbers to improper fractions, invert the divisor, and multiply the two fractions together. To create this article, 21 people, some anonymous, worked to edit and improve it over time. They have a different base, but their value remains the same. The numerators of your improper fractions will be larger than their denominators. 1: 2. %PDF-1.5 Adding and Subtracting Fractions with Unlike Denominators: 4. Now we can add the two fractions together: 2/4 + 3/4 = 5/4. To add fractions there are Three Simple Steps: Step 1: Make sure the bottom numbers (the denominators) are the same, Step 2: Add the numerators, put that answer over the denominator, Step 3: Simplify the fraction (if needed) Once the mixed numbers are converted, you can divide as you would divide any other fractions. 8 = This is a fraction strip. If the answer you found in Step 1 is an improper fraction, change it to a mixed number, write down the fractional part, and carry the whole number part to … The steps for adding two mixed numbers are shown in the examples below. Do long division to divide 4 into 5. Do this by converting one or both fractions until they have the same denominator. https://www.basic-mathematics.com/adding-whole-numbers.html, https://www.mathsisfun.com/least-common-denominator.html, https://virtualnerd.com/act-math/algebra/rational-expressions-equations/least-common-denominator-definition, https://www.mathsisfun.com/fractions_addition.html, https://www.mathsisfun.com/improper-fractions.html, https://www.khanacademy.org/math/arithmetic/fraction-arithmetic/arith-review-mixed-number/v/changing-a-mixed-number-to-an-improper-fraction, https://www.khanacademy.org/math/arithmetic/fraction-arithmetic/arith-review-mixed-number/v/changing-an-improper-fraction-to-a-mixed-number, consider supporting our work with a contribution to wikiHow. In this tutorial, take a look at adding together mixed fractions … It is possible to divide mixed numbers; however, doing so requires converting them to improper fractions first. The remainder, or the number that is left over, is 1. ѓ����\s�,�cDB��H{���t�j�Tz߂Y:�(���@��=&�%���d�C�Ua�rq�-N��/Y���~a�m(�F�����Wj>u����|>����ǃ�>� O��j�|$�V�����=����3N})��1��q���G���V��C#��. Now you have an equation with fractions and no whole numbers. The answer, then, is 6/9 which can be reduced to 2/3. This means that the numbers are equivalent fractions. A common multiple of 2 and 3 is 6. 2 0 obj To create this article, 21 people, some anonymous, worked to edit and improve it over time. E���"���)^��8,D"�! Hey, remember, that's just . wikiHow is where trusted research and expert knowledge come together. Then subtract the proper fraction from the product you just found. Next, find the lowest common denominator of both fractions and convert the fractions so they both have this denominator. We'll need it later. Can you word this easier? Adding Fractions with Like Denominators: 2. Solving Word Problems: 7. Put these pieces back together at the end. Solution: In example 2, it was necessary to simplify the result. By using our site, you agree to our. Try borrowing 1 from the whole number. Get the final answer by adding 1¼ to the 3 we got earlier. We need to find the number when multiplied to the top and bottom of … All we have to do is change these to improper fractions... Then we can add them! Subtracting mixed numbers. 2 * 4 = 8. Then, add the fractions by simply adding the numerators together. To add mixed numbers, start by adding the whole numbers together. Adding Mixed Numbers: 5. <> We use cookies to make wikiHow great. endobj Finally, add the sum of the whole numbers and the sum of the fractions to get your final answer. Do they have a common denominator? This article has been viewed 90,941 times. If not, find a common denominator. 4 goes into 17 4 times, so the quotient is 4. (1 5 7) = 35. Since the denominator of 3/2 has to be multiplied by 2 to get the new denominator of 4, you should multiply the numerator by 2 to find the equivalent fraction of 3/2. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. 1 0 obj How do I divide mixed fractions and then subtract with a proper fraction? A FRACTION IS A NUMBER we need for measuring ( Lesson 20 ); therefore we sometimes have to add or subtract them. To add mixed numbers, add the whole numbers together and the fraction parts of the mixed numbers together and then recombine to express the value as a mixed number. Take your remainder and place it over the original denominator to finish converting the improper fraction into a mixed number. There is always a lowest common denominator, even if it's just the product of the denominators. Since the fraction 3/4 already has the base of 4, you don't have to change it. For tips on converting mixed numbers to improper fractions, read on! q�F���8�[p����~k���qS�x��5+���@��|)�b���G�t��c����,�uY���Tµ6�eYJ�[}��չ������0A��#������t������Źi�Qy~���&�G�pDyI#���| ����]s�H���K���?���>?��P�~���0,��� S�W�#�U������?C�!��6D���ҭ�ݜ�Y�+ͩ�HrN��n��*��:��'��!�Y�.����Ȭ�-�=m�wip�Q�#�G�����O�d�g�Eh��u�������pg�QEUǰ37'3qq��Ɉ�B�!y\�d~#��[��� �������c�o�Մ}E�p-�f����x'\k�c�W���Ҕ*���� ��&kV���q2m�GۢnL�-���bK��l��m�����f��4�Eeb The steps for adding two mixed numbers are shown in the examples below. Adding and Subtracting Fractions with Whole and Mixed Numbers 2 - Cool Math has free online cool math lessons, cool math games and fun math activities. How do I divide a fraction into a whole number? Learn more... A mixed number is a whole number next to a fraction, such as 5 ½, and can be difficult to add. Adding and subtracting mixed numbers. Before you can add fractions, their denominators must be made equal to each other. You could first convert each to an improper fraction. The easiest way is Method 1 above. 1¼ + 3 = 4¼. 4 goes into 5 1 time. 3 * 2 = 6, so the new fraction is 6/4. The quotient is 1, the remainder is 1, and the original denominator was 4, so the final answer is 1 1/4. The LCM of 2 and 4 is 4 because 4 is evenly divisible by 2. When adding and subtracting mixed numbers, split them into “whole” parts and “fractions” parts, then add or subtract separately. Perform the subtraction and then (if the difference is an improper fraction) convert the difference to a mixed number if you wish. There are 12 references cited in this article, which can be found at the bottom of the page. To add mixed numbers, start by adding the whole numbers together. 3 0 obj References. A mixed number, or mixed fraction, is a number that combines a whole number and a fraction. Subtract to find it. Adding Unit Fractions - Mixed Numbers. 1x2=2, so the new fraction is 2/4. To add fractions, they must have the same denominator. You don't have to change it. You would first need to multiply the denominator by the whole number, and then add the product to the numerator. 8 + 7. Here’s how to add two mixed numbers: Add the fractional parts and if necessary, change this sum to a mixed number and reduce it. Next, add this number to the original numerator and place it over the original denominator. Missing Mixed Numbers - Unlike Denominators. Next add together the two fractions: ½ + ¾. 35 + (3 2 7) = 35 + 42. Answer may be improper fractions or mixed numbers. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Exercise 3: Subtract the fractions, . Finally, add the sum of the whole numbers and the sum of the fractions to get your final answer. endobj If this fraction is improper (numerator larger than or equal to the denominator) then convert it to a mixed number %���� Ŭ����㛓��e��,LY������������Ŕ�������.��r����� ��n��e�'ϟ��n3���n�����o�������b���������{�ynU\\��~��l3)^�>����i�~«��2�|�����B�V]Hh�(U$7aeQ�D���?����e�n��!��H�T]% Do the same with the second fraction and add this value to the first. }z� �a���Y��+/Ea;r��K�lG�7�x� �,пnq>�I���v���5.X�- ��}"���u�54K�0ZV�ڵ�r��p��)� D�;��{��z ��FmQ@�CᗒU��8�(XL[ǰ�Y�!H*D�EP�d�B���_�S=ɚĮ&@�*C�M�=�0R�dVd��2�0�A V�4XW���@R��E�f[��]�sl�p:��U���ﲓ�|I�x���J1��}f�����e;(>N~Aa?�a�:��^ ���՛P��?lHyZ��f1��e��͆�QEb�'� ����zŮ��#ds��f�ZB4�apѵ_���4�E-%��%�%R(��Y��(��k�xs��woN��� �ω����*[�FX� ��c+�3��F�X4�LU��q �xD"i��/1�H��s4[Scڵ�j�[�����&�r.���qv�8/_�:��I���+,x'�Ԙ������j�" o�i�hH+o��i��|��������k?�~�XEG�!��Կ�Ǣ��w��Y&w�8ĺ��-�#~hp��{�G��b���=���o�R�����TR�d�xӗb�I�����-�Ba :�\hG�EVx��Fv䣹O2v.�����bY�b��e�x�c\�x��j�P�� ��w(׃3��1z����W��&-[��}���B�'���0Oyj�Co���C�oI�6��n���|��K{ނ������ک2�(�'��g��[d� �#f�����r�=�Cm��ж/w ,0��}�;��$��Y�T[9>g.�����l��&�ݤཫ3�sj�F�:�۽����m0X�ʄ���X���4ܫ�W�Q�S��Wv�T��Q6�*t�:J���kL�|��R�����̶a�� ��G�N��U2�l5x�{b�%*YX���9X]b'�'#pe_F9y��2j�ʨ_��uQ%#��#�\�N@�7$���F�D|W��|����-�_���o]�)��sj"�E}��K_m�j� To add mixed numbers, add the whole numbers together and the fraction parts of the mixed numbers together and then recombine to express the value as a mixed number. Find the difference of the given two numbers and obtain the missing mixed-number addend. 10 + 7. Since the denominator of the fraction 1/2 has to be multiplied by 2 to get 4 as the new base, you should multiply the numerator 1 by 2 as well. Then, add the fractions by simply adding the numerators together. All tip submissions are carefully reviewed before being published. If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. This number is your new numerator and the new denominator is the same as before. The remainder, or the number that is left over, is 1. 35 + 42 + (4 2 5) = 35 + 42 + 40 = 117. 5/4 = 1¼. Practice Exercises: 8. I find this is the best way to add mixed fractions: convert them to Improper Fractions; then add them (using Addition of Fractions) then convert back to Mixed Fractions (You may like to read how to Convert from or to Mixed Fractions) Our Adding and Subtracting Fractions and Mixed Numbers worksheets are designed to supplement our Adding and Subtracting Fractions and Mixed Numbers lessons. a. In order not to change the value of that fraction, we must also double the numerator. 9 = c. 1: 3. So we double the denominator in the fraction ½. Do the same with the remaining fraction (s). <>>> Add the whole numbers and add the fractions separately. The fraction 2/4 = 1/2, but has been put in a larger ratio to have a larger base. 1²/₃ is known as a mixed number, because it is made up of a whole number and a fraction.. 1²/₃ can be written as: ³/₃ + ²/₃ = ⁵/₃. ��tMT�a�9�Miߕ�M��/����T�i�� �UrL��o�����1�:>�m�Z�yy��vբ��1L�!cmI�)g7�Gt�Q��R�w��>T���8p:F�[ �S,�>����T:�#Z V�ƈ����T��{�s�}c�{�x���S ����jפ�#�e��C^5�/ϳ��������w����4��QA��f�%����a��Ml��@���E�������5wS*���x�;��p�M��@7�(�ѡ��V���qꓩ �E%��&u�8&h�=���G�:�#��u�w_�CSj�Ս���X��э��"Fҥ��J�|�B[3��Rr��c���h��qXBg��-���Y���1X�^C�b�����sN�Π>�5��J�r�w�������� ,�s�����`a騗��jTÑ�a�|��r�>��j������m��̪K�ՄL9zS$��_٨����^�m��餹ۇ�| ^�����ͦ�4w[8s���/7�n�K9�Vilɍ�������vC5�:[��ߨ��m�(g�mՇ��g�㆗��ǣ�faq��b�O��Q�D����e�໒����J�t�T���w���u7Wj� ���$�\ ��N�d���td�ҥw�"n�!�2��N3� Adding the Whole Numbers and Fractions Separately, {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/0\/04\/Add-Mixed-Numbers-Step-1-Version-3.jpg\/v4-460px-Add-Mixed-Numbers-Step-1-Version-3.jpg","bigUrl":"\/images\/thumb\/0\/04\/Add-Mixed-Numbers-Step-1-Version-3.jpg\/aid560054-v4-728px-Add-Mixed-Numbers-Step-1-Version-3.jpg","smallWidth":460,"smallHeight":345,"bigWidth":"728","bigHeight":"546","licensing":"
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Home » Math Theory » Graphs and Charts » Graphing Percentages
# Graphing Percentages
## Introduction
When expressing a fraction or ratio as a portion of one hundred, percentages are a significant mathematical and practical concept. Percentages are frequently applied in a range of disciplines, including science, finance, and statistics. Graphing percentages allows us to comprehend ratios, changes, or comparisons better.
The concept of percentages is typically introduced in the 5th or 6th grade and continues to be relevant throughout middle and high school. Graphing percentages, which combines understanding of percentages with data representation skills, is generally suitable for students aged 11 to 15 years or from the 6th to 9th grades.
## Math Domain
This topic falls under the domains of Numbers and Operations—Fractions, Ratios and Proportional Relationships, and Measurement and Data.
## Applicable Common Core Standards
The topic directly addresses several Common Core State Standards for Mathematics, including:
CCSS.MATH.CONTENT.6.RP.A.3.C: Find a percent of a quantity as a rate per 100.
CCSS.MATH.CONTENT.7.RP.A.3: Use proportional relationships to solve multistep ratio and percent problems.
CCSS.MATH.CONTENT.6.SP.B.4: Display numerical data in plots on a number line, including dot plots, histograms, and box plots.
## Definition of the Topic
A number can be expressed as a fraction of 100 using a percentage. It is often indicated with the percent sign, “%.”For example, 50% is a percentage that represents 50 out of 100, or half.
Graphing percentages involves representing these percentages visually, often on a bar graph, line graph, or pie chart. Graphing percentages helps in understanding the distribution and comparison of data more effectively.
## Key Concepts
Percent: A ratio that weighs a value of 100.
Fraction: A way of representing a part of a whole using a numerator (part) and a denominator (whole). Put percentages over 100 to convert them to fractions.
Decimal: Percentages can also be expressed as decimals. For example, 50% is equal to 0.5 in decimal form.
Graphs: Visual representations of data. They can be used to show percentages in a visually intuitive way.
## Discussion with Illustrative Examples
### Meaning of Percentage
The word percentage comes from the Latin per centum, which means “per one hundred” or “by a hundred.”
Therefore, percentages are fractions with 100 as their denominator. It is the concept of a part and a whole where the value of the whole is always 100.
### Percentage Formula
The meaning of percentage means “by the hundred” or “per one hundred.” The whole value of a percentage is always based on 100, while individual values are a part of this 100. For example, 40% means 40 out of 100.
Percentage = ( individual value ÷ total value ) x 100
### Calculating the Percentage
To find the percentage of a certain value, we divide the individual value by the total value and then multiply the result by 100.
Let us say, for example, the favorite color of 6 students out of a total of 10 students is blue. What is the percentage value of students whose favorite color is blue?
The total value, in this case, is ten students.
Divide the individual value that you wish to determine the percentage for by the total value.
Based on the example above, the individual value is six students. We are looking for the percentage value of 6 students out of the total value of 10 students.
Six students divided by ten students: 6 ÷ 10 = 0.6
Multiply the value you got by 100: 0.6 x 100 =60%
Out of 10 students in the school, the favorite color of 60% of them is blue.
### Graphing Percentages
Once we have collected the data and converted it to percentages, creating a visual form of this information in a presentation is practical and helpful. A graph organizes the data and will make it easier for people who need to use this data to analyze the values.
Here are some basic graphs that we can use to graph percentages:
#### Pie Graph
The pie graph is the best to use when it comes to graphing percentages as it is meant to showcase parts of a whole.
#### Creating A Pie Graph
To manually create a pie graph:
Calculate the percentage for each value.
Find the equivalent degree angle of each value by multiplying each percentage value (decimal form) by 360.
Use a protractor to measure each degree angle in the circle.
Color and label your graph with data categories as needed.
Let us say; for example, you asked your classmates their favorite school subject. Here are the results:
Calculating the percentage of each favorite school subject we have,
Multiply each percentage value (decimal form) by 360 to find the degree angle in the circle.
Label and color your bar graph as needed.
#### Bar Graph
Another graph we can use to show the percentage is the bar graph. It can provide an overview comparing the percentages of two or more data categories.
Example: We asked 20 people if they liked blue or red. Here are the results:
#### Creating A Bar Graph
To manually create a bar graph:
Calculate the percentage values of the data collected.
## Real-life Application with Solution
Imagine you are tracking the amount of time you spend on different activities in a day.
Let us say you spend 8 hours sleeping, 7 hours at school, 2 hours doing homework, 4 hours playing, and 3 hours on other activities.
To represent this data in a pie chart:
Calculate the total hours: 8 (sleep) + 7 (school) + 2 (homework) + 4 (play) + 3 (others) = 24 hours.
Find the percentage of time spent on each activity by dividing each activity’s hours by the total hours and multiplying by 100. Note: The values are rounded off to the nearest whole number.
Draw a circle (or pie) and divide it into sections according to the percentages calculated.
## Practice Test
1. In a survey of a class of 25 students, 15 said they like science. What percentage of the class likes science?
2. Convert the following percentages to fractions: 25%, 50%, 75%.
3. Convert the following percentages to decimals: 10%, 55%, 98%.
4. Graph the following data on a bar graph: In a pet survey, 45% of households had dogs, 15% had cats, 25% had fish, and 15% had birds.
5. In a group of 120 people, 90 people like chocolate ice cream. What percentage of the group likes chocolate ice cream?
1. Sixty percent of students said they like science.
$\frac{15}{25}$×100=60%
2.
$\frac{25}{100}$ or $\frac{1}{4}$
$\frac{50}{100}$ or $\frac{1}{2}$
$\frac{75}{100}$ or $\frac{3}{4}$
3.
$\frac{10}{100}$=0.1
$\frac{55}{100}$=0.55
$\frac{98}{100}$=0.98
4.
5. Seventy-five percent of people like chocolate ice cream.
$\frac{90}{120}$×100=75%
### How do I convert a decimal to a percentage?
You multiply the decimal by 100 and add the “%” sign. For example, to convert 0.2 to a percentage, you would multiply 0.2 by 100 to get 20%.
### What does 100% represent?
100% represents the whole or entirety of something. It means all of it.
### How do I represent a percentage greater than 100 on a graph?
It depends on the type of graph. The bar could extend past the line marking 100% for a bar graph. For a pie chart, percentages should always total 100%, so representing a percentage greater than 100% would not be possible.
### What is the difference between a fraction and a percentage?
Both fractions and percentages are ways to represent parts of a whole. The main difference is that a percentage is always out of 100, while a fraction can have any number as its denominator.
### How can I use percentages in everyday life?
Percentages are used in many areas of daily life, including calculating discounts, determining tips, understanding interest rates, and interpreting statistical data.
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# LESSON #25 - EXPONENTIAL MODELING WITH PERCENT GROWTH AND DECAY COMMON CORE ALGEBRA II
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1 1 LESSON #5 - EXPONENTIAL MODELING WITH PERCENT GROWTH AND DECAY COMMON CORE ALGEBRA II Eponential functions are very important in modeling a variety of real world phenomena because certain things either increase or decrease by fied percentages over given units of time. You looked at this in Common Core Algebra I and in this lesson we will review much of what you saw. Eercise #1: Suppose that you deposit money into a savings account that receives 5% interest per year on the amount of money that is in the account for that year. Assume that you deposit \$400 into the account initially. (a) How much will the savings account increase by over the course of the year? (b) How much money is in the account at the end of the year? (c) By what single number could you have multiplied the \$400 by in order to calculate your answer in part (b)? (d) Using your answer from part (c), determine the amount of money in the account after and 10 years. Round all answers to the nearest cent when needed. (e) Give an equation for the amount in the savings account as a function of the number of years since the \$400 was invested. (f) Using a table on your calculator determine, to the nearest year, how long it will take for the initial investment of \$400 to double. Provide evidence to support your answer. The thinking process from Eercise #1 can be generalized to any situation where a quantity is increased by a fied percentage over a fied interval of time. This pattern is summarized below: INCREASING EXPONENTIAL MODELS If quantity Q is known to increase by a fied percentage p, in decimal form, then Q can be modeled by where represents the amount of Q present at and t represents time. Eercise #: Which of the following gives the savings S in an account if \$50 was invested at an interest rate of 3% per year? (1) S 504 t t (3) S () S t (4) S t
2 Decreasing eponentials are developed in the same way, but have the percent subtracted, rather than added, to the base of 100%. Just remember, you are ultimately multiplying by the percent of the original that you will have after the time period elapses. Eercise #3: State the multiplier (base) you would need to multiply by in order to decrease a quantity by the given percent listed. (a) 10% (b) % (c) 5% (d) 0.5% DECREASING EXPONENTIAL MODELS If quantity Q is known to decrease by a fied percentage p, in decimal form, then Q can be modeled by where represents the amount of Q present at and t represents time. Eercise #4: If the population of a town is decreasing by 4% per year and started with 1,500 residents, which of the following is its projected population in 10 years? Show the eponential model you use to solve this problem. (1) 9,30 (3) 18,503 () 76 (4) 8,310 Eercise #5: The stock price of WindpowerInc is increasing at a rate of 4% per week. Its initial value was \$0 per share. On the other hand, the stock price in GerbilEnergy is crashing (losing value) at a rate of 11% per week with an initial value of \$10. (a) Model both stock prices using eponential functions. (b) Then, find when the stock prices will be equal graphically, to the nearest week. Draw a well labeled graph to justify your solution. y
3 3 APPLICATIONS EXPONENTIAL MODELING WITH PERCENT GROWTH AND DECAY COMMON CORE ALGEBRA II HOMEWORK 1. If \$130 is invested in a savings account that earns 4% interest per year, which of the following is closest to the amount in the account at the end of 10 years? (1) \$18 (3) \$168 () \$19 (4) \$34. A population of 50 fruit flies is increasing at a rate of 6% per day. Which of the following is closest to the number of days it will take for the fruit fly population to double? (1) 18 (3) 1 () 6 (4) 8 3. If a radioactive substance is quickly decaying at a rate of 13% per hour approimately how much of a 00 pound sample remains after one day? (1) 7.1 pounds (3) 5.6 pounds ().3 pounds (4) 15.6 pounds 4. A population of llamas stranded on a dessert island is decreasing due to a food shortage by 6% per year. If the population of llamas started out at 350, how many are left on the island 10 years later? (1) 57 (3) 10 () 58 (4) Which of the following equations would model a population with an initial size of 65 that is growing at an annual rate of 8.5%? t P (3) P (1) t () P t (4) P 8.5t The acceleration of an object falling through the air will decrease at a rate of 15% per second due to air resistance. If the initial acceleration due to gravity is 9.8 meters per second per second, which of the following equations best models the acceleration t seconds after the object begins falling? (1) () a a t (3) a t t (4) a t
4 7. Red Hook has a population of 6,00 people and is growing at a rate of 8% per year. Rhinebeck has a population of 8,750 and is growing at a rate of 6% per year. In how many years, to the nearest year, will Red Hook have a greater population than Rhinebeck? Show the equation or inequality you are solving and solve it graphically A warm glass of water, initially at 10 degrees Fahrenheit, is placed in a refrigerator at 34 degrees Fahrenheit and its temperature is seen to decrease according to the eponential function (a) Verify that the temperature starts at 10 degrees Fahrenheit by evaluating T 0. h T h (b) Using your calculator, sketch a graph of T below for all values of h on the interval 0 h 4. Be sure to label your y-ais and y- intercept. (c) After how many hours will the temperature be at 50 degrees Fahrenheit? State your answer to the nearest hundredth of an hour. Illustrate your answer on the graph your drew in (b). REASONING 9. Percents combine in strange ways that don't seem to make sense at first. It would seem that if a population grows by 5% per year for 10 years, then it should grow in total by 50% over a decade. But this isn't true. Start with a population of 100. If it grows at 5% per year for 10 years, what is its population after 10 years? What percent growth does this represent?
5 5 LESSON #6 - MINDFUL MANIPULATION OF PERCENTS COMMON CORE ALGEBRA II Percents and phenomena that grow at a constant percent rate can be challenging, to say the least. This is due to the fact that, unlike linear phenomena, the growth rate indicates a constant multiplier effect instead of a constant additive effect (linear). Because constant percent growth is so common in everyday life (not to mention in science, business, and other fields), it's good to be able to mindfully manipulate percents. Eercise #1: A population of wombats is growing at a constant percent rate. If the population on January 1 st is 107 and a year later is 1079, what is its yearly percent growth rate to the nearest tenth of a percent? Eercise #: Now let's try to determine what the percent growth in wombat population will be over a decade of time. We will assume that the rounded percent increase found in Eercise #1 continues for the net decade. (a) After 10 years, what will we have multiplied the original population by, rounded to the nearest hundredth. Show the calculation. (b) Using your answer from (a), what is the decade percent growth rate, to the nearest tenth? Eercise #3: Let s stick with our wombats from Eercise #1. Assuming their growth rate is constant over time, what is their monthly growth rate to the nearest tenth of a percent? Assume a constant sized month. Eercise #4: If a population was growing at a constant rate of % every 5 years, then what is its percent growth rate over at year time span? Round to the nearest tenth of a percent. (a) First, give an epression that will calculate the single year (or yearly) percent growth rate based on the fact that the population grew % in 5 years. (b) Now use this epression to calculate the percent growth over years, to the nearest tenth.
6 Eercise #5: World oil reserves (the amount of oil unused in the ground) are depleting at a constant % per year. We would like to determine what the percent decline will be over the net 0 years based on this % yearly decline. (a) Write and evaluate an epression for what we would multiply the initial amount of oil by after 0 years. (b) Use your answer to (a) to determine the percent decline after 0 years. Be careful! Round to the nearest percent. 6 Eercise #6: A radioactive substance s half-life is the amount of time needed for half (or 50%) of the substance to decay. Let s say we have a radioactive substance with a half-life of 0 years. (a) What percent of the substance would be radioactive after 40 years, to the nearest tenth of a percent? (b) What percent of the substance would be radioactive after only 10 years? Round to the nearest tenth of a percent. (c) What percent of the substance would be radioactive after only 5 years? Round to the nearest tenth of a percent.
7 7 MINDFUL MANIPULATION OF PERCENTS COMMON CORE ALGEBRA II HOMEWORK APPLICATIONS 1. A quantity is growing at a constant 3% yearly rate. Which of the following would be its percent growth after 15 years? (1) 45% (3) 56% () 5% (4) 63%. If a credit card company charges 13.5% yearly interest, which of the following calculations would be used in the process of calculating the monthly interest rate? (1) () (3) (4) The county debt is growing at an annual rate of 3.5%. What percent rate is it growing at per years? Per 5 years? Per decade? Show the calculations that lead to each answer. Round each to the nearest tenth of a percent. 4. A population of llamas is growing at a constant yearly rate of 6%. At what rate is the llama population growing per month? Please assume all months are equally sized and that there are 1 of these per year. Round to the nearest tenth of a percent.
8 5. Shana is trying to increase the number of calories she burns by 5% per day. By what percent is she trying to increase per week? Round to the nearest tenth of a percent If a bank account doubles in size every 5 years, then by what percent does it grow after only 3 years? Round to the nearest tenth of a percent. Hint: First write an epression that would calculate its growth rate after a single year. 7. An object s speed decreases by 5% for each minute that it is slowing down. Which of the following is closest to the percent that its speed will decrease over half-an hour? (1) 1% (3) 48% () 79% (4) 150% 8. Over the last 10 years, the price of corn has decreased by 5% per bushel. (a) Assuming a steady percent decrease, by what percent does it decrease each year? Round to the nearest tenth of a percent. (b) Assuming this percent continues, by what percent will the price of corn decrease by after 50 years? Show the calculation that leads to your answer.. Round to the nearest percent.
9 9 LESSON #7 - INTRODUCTION TO LOGARITHMS COMMON CORE ALGEBRA II Eponential functions are of such importance to mathematics that their inverses, functions that reverse their action, are important themselves. These functions, known as logarithms, will be introduced in this lesson. Eercise #1: The function y is shown graphed on the aes below along with its table of values. y y (a) Is this function one-to-one? Eplain your answer. (b) Based on your answer from part (a), what must be true about the inverse of this function? (c) Create a table of values below for the inverse of y and plot this graph on the aes given. Notice that, as always, the graphs of a function and its inverse are symmetric across. y (d) What would be the first step to find an equation for this inverse algebraically? Write this step down and then stop. Defining Logarithmic Functions The function y log b is the name we give the inverse of y b. For eample, y log is the inverse of y. Based on Eercise #1(d), we can write an equivalent eponential equation for each logarithm as follows: y y log is the same as b b Based on this, we see that a logarithm gives as its output (y-value) the eponent we must raise b to in order to produce its input (-value).
10 Eercise #: Evaluate the following logarithms. If needed, write an equivalent eponential equation. Do as many as possible without the use of your calculator. (a) log 8 (b) log 416 (c) log5 65 (d) log10100, (e) log (f) log (g) log5 5 (h) log3 9 It is critically important to understand that logarithms give eponents as their outputs. We will be working for multiple lessons on logarithms and a basic understanding of their inputs and outputs is critical. Eercise #3: If the function y would represent its y-intercept? (1) 1 (3) 8 () 13 (4) 9 log 8 9 was graphed in the coordinate plane, which of the following Eercise #4: Between which two consecutive integers must log3 40 lie? (1) 1 and (3) 3 and 4 () and 3 (4) 4 and 5 Calculator Use and Logarithms Most calculators only have two logarithms that they can evaluate directly. One of them, log10, is so common that it is actually called the common log and typically is written without the base 10. log log 10 (The Common Log) Eercise #5: Evaluate each of the following using your calculator. (a) log100 (b) log (c) log 10
11 11 INTRODUCTION TO LOGARITHMS COMMON CORE ALGEBRA II HOMEWORK FLUENCY 1. Which of the following is equivalent to y log7? (1) y 7 (3) 7 y () 7 y (4) y 1 7. If the graph of y 6 is reflected across the line y then the resulting curve has an equation of (1) y 6 (3) log6 y () y log6 (4) y 6 3. The value of log5 167 is closest to which of the following? Hint guess and check the answers. (1).67 (3) 4.58 () 1.98 (4) Which of the following represents the y-intercept of the function y (1) 8 (3) 3 () 5 (4) 5 log ? 5. Determine the value for each of the following logarithms. (Easy) (a) log 3 (b) log7 49 (c) log36561 (d) log Determine the value for each of the following logarithms. (Medium) (a) log 1 64 (b) log3 1 (c) log5 1 5 (d) log
12 7. Determine the value for each of the following logarithms. Each of these will have non-integer, fractional answers. (Difficult) 3 5 (a) log4 (b) log4 8 (c) log 5 (d) log Between what two consecutive integers must the value of log lie? Justify your answer. 9. Between what two consecutive integers must the value of log lie? Justify your answer. APPLICATIONS 10. In chemistry, the ph of a solution is defined by the equation ph log H where H represents the concentration of hydrogen ions in the solution. Any solution with a ph less than 7 is considered acidic and any solution with a ph greater than 7 is considered basic. Fill in the table below. Round your ph s to the nearest tenth of a unit. Substance Concentration of Hydrogen ph Basic or Acidic? 7 Milk Coffee Bleach.5 10 Lemmon Juice Rain REASONING 11. Can the value of log 4 you about the domain of log b? be found? What about the value of log 0? Why or why not? What does this tell
13 13 LESSON #8 - GRAPHS OF LOGARITHMS COMMON CORE ALGEBRA II The vast majority of logarithms that are used in the real world have bases greater than one; the ph scale that we saw on the last homework assignment is a good eample. In this lesson we will further eplore graphs of these logarithms, including their construction, transformations, and domains and ranges. Eercise #1: Consider the logarithmic function y log3 and its inverse y 3. (a) Construct a table of values for y 3 and then use this to construct a table of values for the function y log3. y y 3 y log 3 (b) Graph y 3 and y log3 on the grid given. Show the algebraic process of finding the inverse of y 3 to get y log3. (c) State the natural domain and range of y 3 and y log3. Domain: Range: y 3 Domain: Range: y log3 Eercise #: Using your calculator, sketch the graph of y log10 on the aes below. Label the -intercept. State the domain and range of y log10. y Domain: Range: 10
14 14 Eercise #3: Given the function, y e, (a) Algebraically, find the function s inverse. Just as e is a very important number in eponential modeling, log e also arises in real world problems very often. It is called the natural logarithm and is usually epressed as ln. In other words, log ln e. Therefore, y e and y ln are inverses of each other. (b) Construct a table of values for y e and then use this to construct a table of values for the function y ln. Round values to the nearest tenth. y y e y ln Eercise #4: Without the use of your calculator, determine the values of each of the following. (a) ln e (b) 5 ln 1 (c) ln e (d) ln e Eercise #5: Which of the following equations describes the graph shown below? y y log 1 (1) 3 () y log 3 1 (3) y log 3 1 (4) y log 3 1 3
15 The fact that finding the logarithm of a non-positive number (negative or zero) is not possible in the real number system allows us to find the domains of a variety of logarithmic functions. Eercise #6: Determine the domain of the function y log 3 4. State your answer in set-builder notation. 15 All logarithms with bases larger than 1 are always increasing. This increasing nature can be seen by calculating their average rate of change. Eercise #7: Consider the common log, or log base 10, f log. y (a) Set up and evaluate an epression for the average rate of change of f over the interval 1 10 (b) Set up and evaluate an epression for the average rate of change of f over the interval (c) What do these two answers tell you about the changing slope of this function?
16 16 GRAPHS OF LOGARITHMS COMMON CORE ALGEBRA II HOMEWORK FLUENCY 1. The domain of y log 5 in the real numbers is (1) 0 (3) 5 () 5 (4) 4 3. Which of the following equations describes the graph shown below? y (1) y log5 (3) y log3 () y log (4) y log 4 3. Which of the following represents the y-intercept of the function y (1) 8 (3) 1 () 4 (4) 4 log 3 1? 4. Which of the following values of is not in the domain of f log 10 (1) 3 (3) 5 () 0 (4) 4 5. Which of the following is true about the function y (1) It has an -intercept of 4 and a y-intercept of 1. () It has -intercept of 1 and a y-intercept of 1. (3) It has an -intercept of 16 and a y-intercept of 1. (4) It has an -intercept of 16 and a y-intercept of 1. log 16 1? 4 5?
17 17 6. Which of the following is closest to the y-intercept of the function whose equation is y 1 10e? (1) 10 (3) 7 () 18 (4) 5 7. On the grid below, the solid curve represents y e describe the dashed curve? Eplain your choice. (1) y 1 (3) y. Which of the following eponential functions could y () y e (4) y 4 8. Determine the domains of each of the following logarithmic functions. State your answers using any accepted notation. Be sure to show the inequality that you are solving to find the domain and the work you use to solve the inequality. (a) ylog 1 (b) ylog Graph the logarithmic function y log 4 on the graph paper given. For a method, see Eercise #1 on page 13. y
18 Without the use of your calculator, determine the values of each of the following. (a) ln e (b) 1 ln e 4 (c) ln e (d) 10 ln e e 10 REASONING 11. Logarithmic functions whose bases are larger than 1 tend to increase very slowly as increases. Let's investigate this for f log. (a) Find the value of f 1, f, f 4, and f 8 without your calculator. (b) For what value of will? For what value of will log 10 log 0?
19 19 LESSON #9 - SOLVING EXPONENTIAL EQUATIONS USING LOGARITHMS COMMON CORE ALGEBRA II Eercise #1: Evaluate the following logarithms. (a) log 4 (b) log 4 3 (c) How many TIMES bigger was 3 log 4 compared to log 4? Therefore, log 4 3 log 4. This property can be generalized to logarithms of any base. 3 THE THIRD LOGARITHM LAW This law can be used with a logarithm of ANY base. The natural logarithm is used most often when solving eponential equations, which is what we will use in this course. Our general law using, better known as, is: In future courses, you will learn more about the reasons behind this law. For now, know that logarithms are equal to eponents. When a power (or eponent) is raised to another power, you multiply. Therefore when the argument of a logarithm is raised to a power, you can also multiply the logarithm by that power. Earlier in this unit, we used the Method of Common Bases to solve eponential equations. This technique is quite limited, however, because it requires the two sides of the equation to be epressed using the same base. A more general method utilizes our calculators and the third logarithm law: Eercise #: Solve: 4 8 using (a) common bases and (b) the logarithm law shown above. (a) Method of Common Bases (b) Logarithm Approach
20 The beauty of this logarithm law is that it removes the variable from the eponent. This law, in combination with the logarithm base e, the natural log, allows us to solve almost any eponential equation using calculator technology. Eercise #3: Solve each of the following equations for the value of. Round your answers to the nearest hundredth. (a) 5 18 (b) (c) These equations can become more complicated, but each and every time we will use the logarithm law to transform an eponential equation into one that is more familiar (linear only for now) Eercise #4: Solve each of the following equations for. Round your answers to the nearest hundredth. 3 (a) (b) Now that we are familiar with this method, we can revisit some of our eponential models from earlier in the unit. Recall that for an eponential function that is growing: If quantity Q is known to increase by a fied percentage p, in decimal form, then Q can be modeled by where represents the amount of Q present at and t represents time. Eercise #5: A biologist is modeling the population of bats on a tropical island. When he first starts observing them, there are 104 bats. The biologist believes that the bat population is growing at a rate of 3% per year. (a) Write an equation for the number of bats,, as a function of the number of years, t, since the biologist started observing them. (b) Using your equation from part (a), algebraically determine the number of years it will take for the bat population to reach 00. Round your answer to the nearest year.
21 Eercise #6: A stock has been declining in price at a steady pace of 5% per week. If the stock started at a price of \$.50 per share, determine algebraically the number of weeks it will take for the price to reach \$ Round your answer to the nearest week. 1 While we usually use the power rule and the natural log to solve eponential equations, you can also use a logarithm with the same base as the eponential equation. Eercise #7: Find the solution to each of the following eponential equations in terms of a logarithm with the same base as the eponential equation. Epress your answer first as a logarithm and then use the calculator to round to the nearest hundredth. (a) (b) Eercise #8: Find the solution to the general eponential equation ab c d, in terms of the constants a, c, d, and the logarithm of base b. Think about reversing the order of operations in order to solve for.
22 SOLVING EXPONENTIAL EQUATIONS USING LOGARITHMS COMMON CORE ALGEBRA II HOMEWORK FLUENCY 1. Which of the following values, to the nearest hundredth, solves: 7 500? (1) 3.19 (3).74 () 3.83 (4) The solution to 5, to the nearest tenth, is which of the following? (1) 7.3 (3) 11.4 () 9.1 (4) To the nearest hundredth, the value of that solves (1) 6.73 (3) 8.17 () 5.74 (4) is 4. Solve each of the following eponential equations. Round each of your answers to the nearest hundredth. (a) (b) (c) Solve each of the following eponential equations. Be careful with your use of parentheses. Epress each answer to the nearest hundredth. (a) (b).05t e (c)
23 6. Find the solution to each of the following eponential equations in terms of a logarithm with the same base as the eponential equation. Epress your answer first as a logarithm and then use the calculator to round to the nearest hundredth. 3 (a) (b) APPLICATIONS 6. The population of Red Hook is growing at a rate of 3.5% per year. If its current population is 1,500, in how many years will the population eceed 0,000? Round your answer to the nearest year. Only an algebraic solution is acceptable. 7. A radioactive substance is decaying such that % of its mass is lost every year. Originally there were 50 kilograms of the substance present. (a) Write an equation for the amount, A, of the substance left after t-years. (b) Find the amount of time that it takes for only half of the initial amount to remain. Round your answer to the nearest tenth of a year. REASONING 8. If a population doubles every 5 years, how many years will it take for the population to increase by 10 times its original amount? First: If the population gets multiplied by every 5 years, what does it get multiplied by each year? Use this to help you answer the question.
24 4 LESSON #30 - COMPOUND INTEREST COMMON CORE ALGEBRA II In the worlds of investment and debt, interest is added onto a principal in what is known as compound interest. The percent rate is typically given on a yearly basis, but could be applied more than once a year. This is known as the compounding frequency. Let's take a look at a typical problem to understand how the compounding frequency changes how interest is applied. Eercise #1: A person invests \$500 in an account that earns a nominal yearly interest rate of 4%. (a) How much would this investment be worth in 10 years if the compounding frequency was once per year? Show the calculation you use. (b) If, on the other hand, the interest was applied four times per year (known as quarterly compounding), why would it not make sense to multiply by 1.04 each quarter? (c) If you were told that an investment earned 4% per year, how much would you assume was earned per quarter? Why? (d) Using your answer from part (c), calculate how much the investment would be worth after 10 years of quarterly compounding? Show your calculation. So, the pattern is fairly straightforward. For a shorter compounding period, we get to apply the interest more often, but at a lower rate. Eercise #: How much would \$1000 invested at a nominal % yearly rate, compounded monthly, be worth in 0 years? Show the calculations that lead to your answer. (1) \$ (3) \$ () \$ (4) \$ This pattern is formalized in a classic formula from economics that we will look at in the net eercise. Eercise #3: For an investment with the following parameters, write a formula for the amount the investment is worth, A, after t-years. P = amount initially invested r = nominal yearly rate n = number of compounds per year
25 Eercise #4: If \$1500 is invested at.5% interest compounded weekly, how much will be in the account after 10 years to the nearest dollar. 5 Eercise #5: If \$100 is invested at 8% interest compounded monthly, after how many years will the amount in the account double? Round to the nearest tenth of a year. The rate in was referred to as nominal (in name only). It's known as this, because you effectively earn more than this rate if the compounding period is more than once per year. Because of this, bankers refer to the effective rate, or the rate you would receive if compounded just once per year. Let's investigate this. Eercise #6: An investment with a nominal rate of 5% is compounded at different frequencies. Give the effective yearly rate, accurate to two decimal places, for each of the following compounding frequencies. Show your calculation. (a) Quarterly (b) Monthly (c) Daily
26 We could compound at smaller and smaller frequency intervals, eventually compounding all moments of time. In our formula from Eercise #6, we would be letting n approach infinity. Interestingly enough, this gives rise to continuous compounding and the use of the natural base e in the famous continuous compound interest formula. CONTINUOUS COMPOUND INTEREST For an initial principal, P, compounded continuously at a nominal yearly rate of r, the investment would be worth an amount A given by: 6 Eercise #7: A person invests \$350 in a bank account that promises a rate of % continuously compounded. (a) Write an equation for the amount this investment would be worth after t-years. (b) How much would the investment be worth after 0 years? (c) Algebraically determine the time it will take for the investment to reach \$400. Round to the nearest tenth of a year. (d) What is the effective annual rate for this investment? Round to the nearest hundredth of a percent. Eercise #8: A population of 500 llamas on a tropical island is growing continuously at a rate of 3.5% per year. (a) Write a function to model the number of llamas on the island after t-years. (b) Algebraically determine the number of years for the population to reach 600. Round your answer to the nearest tenth of a year.
27 7 COMPOUND INTEREST COMMON CORE ALGEBRA II HOMEWORK APPLICATIONS 1. The value of an initial investment of \$400 at 3% interest compounded quarterly can be modeled using which of the following equations, where t is the number of years since the investment was made? (1) A t (3) 4 A t () A t (4) A t. Which of the following represents the value of an investment with a principal of \$1500 with a interest rate of.5% compounded monthly after 5 years? (1) \$1, (3) \$4,178. () \$1, (4) \$5, Franco invests \$4,500 in an account that earns a 3.8% interest rate compounded continuously. If he withdraws the profit from the investment after 5 years, how much has he earned on his investment? (1) \$858.9 (3) \$9.50 () \$91.59 (4) \$ An investment that returns a 4.% yearly rate, but is compounded quarterly, has an effective yearly rate closest to (1) 4.1% (3) 4.7% () 4.4% (4) 4.3% 5. If an investment's value can be modeled with investment?.07 A t then which of the following describes the (1) The investment has a rate of 7% compounded every 1 years. () The investment has a rate of.7% compounded ever 1 years. (3) The investment has a rate of 7% compounded 1 times per year. (4) The investment has a rate of.7% compounded 1 times per year.
28 8 6. An investment of \$500 is made at.8% nominal interest compounded quarterly. (a) Write an equation that models the amount A the investment is worth t-years after the principal has been invested. (b) How much is the investment worth after 10 years? (c) Algebraically determine the number of years it will take for the investment to be reach a worth of \$800. Round to the nearest hundredth. (d) Why does it make more sense to round your answer in (c) to the nearest quarter? State the final answer rounded to the nearest quarter. 7. An investment of \$300 is made at 3.6% nominal interest compounded continuously. (a) Write an equation that models the amount A the investment is worth t-years after the principal has been invested. (b) How much is the investment worth after 10 years? (c) Algebraically determine the number of years it will take for the investment to be reach a worth of \$800. Round to the nearest hundredth. REASONING rt 8. The formula A Pe calculates the amount an investment earning a rate of r compounded continuously is worth. Show that the amount of time it takes for the investment to double in value is given by the epression ln. r
29 9 LESSON #31 - MORE EXPONENTIAL AND LOGARITHMIC MODELING COMMON CORE ALGEBRA II The Half-life is the amount of time required for the amount of something to decrease to half its initial value. Any eponential decay function can be rewritten as a half-life function. We will begin with an eample from the first lesson of this unit: Eercise #1: The population of a town is decreasing by 4% per year and started with 1,500 residents. (a) Write a function to model this situation. (b) Algebraically determine when the population of the town will be half of the initial population? Round to the nearest tenth of a year. HALF LIFE FORMULA For an initial quantity, P, that is decreasing at an eponential rate, with a half life, h, the amount of the quantity, A, left after t time units is given by the formula, (c) Write the half-life formula for the population of the town for Eercise #1. Eercise #: The decay of a sample of 5000 grams of carbon can be modeled by the equation, t Ct ( ) 5000, where t is measured in years. (a) What is the half-life of carbon? (b) How can you tell this is a half-life equation?
30 30 Eercise #3: Copper has antibacterial properties, and it has been shown that direct contact with copper alloy C11000 at 0 C kills methicillin-resistant Staphylococcus aureus (MRSA) over a period of time. (a) A function that models a population of 1,000 MRSA bacteria t minutes after coming in contact with copper alloy C11000 is A(t)=1000(.91) t. What does the base 0.91 mean in this scenario? (b) Rewrite the formula for A as an eponential function with base of 1 (Half-life Formula). Round the half-life to the nearest hundredth of a minute. (c) Convert your equation back into its original form. There are a number of other eponential models, but these equations will be given. Below is an eample. Eercise #4: A hot liquid is cooling in a room whose temperature is constant. Its temperature can be modeled using the eponential function shown below. The temperature, T, is in degrees Fahrenheit and is a function of the number of minutes, m, it has been cooling. (a) What was the initial temperature of the water at m 0. Do without using your calculator. T ( m) = 101e -0.03m + 67 (b) How do you interpret the statement that T ? (c) Determine algebraically when the temperature of the liquid will reach 100 F. Show the steps in your solution. Round to the nearest tenth of a minute. (d) On average, how many degrees are lost per minute over the interval 10 m 30? Round to the nearest tenth of a degree.
31 Logarithmic functions can also be used to model real world phenomena. Eample #5: The slope, s, of a beach is related to the average diameter d (in millimeters) of the sand particles on the beach by this equation: s log d. s (a) Sand particles typically have a maimum diameter of 1mm. Using this information, sketch a graph of the function. 31 d (b) If the average diameter of the sand particles is 0.5mm, find the slope of the beach (to the nearest hundredth). (c) Given a slope of 0.14, find the average diameter (to the nearest hundredth) of the sand particles on the beach. Eample #6: Two methods of instruction were used to teach athletes how to shoot a basketball. The methods were assessed by assigning students into two groups, one who was taught with method A and one who was taught with method B. The students in each group took 30 foul shots after each of ten sessions. The average number of shots made in each of the sessions by an athlete using method A can be modeled by the function, A( ) ln. The average number of foul shots made in each of the sessions for an athlete using method B can be modeled by the function, B ( ) 9.17(1.109). (a) In which of the 10 sessions, to the nearest whole number, will the two methods produce the same number of made baskets? Eplain how you found your answer. (b) Find the average range of change for each method between sessions 3 and 8 for each of the methods. Give proper units and round your answers to the nearest tenth. (c) Eplain why B() would not be an appropriate model for this situation if there were 15 sessions.
32 MORE EXPONENTIAL AND LOGARITHMIC MODELING COMMON CORE ALGEBRA II HOMEWORK 1. The decay of a sample of 800 grams of hydrogen can be modeled by the equation, t years. (a) What is the half-life of hydrogen? t 1 Ht ( ) after (b) How can you tell this is a half-life equation?. The \$,500 in your bank account is decreasing continuously at a rate of 5% per year. (a) Use the Continuous Compound Interest Formula to write a function that models the amount of money in your bank account after t years. (Don t forget the r value should be negative because it s decreasing). (b) When will only half of your initial deposit be left in your bank account (to the nearest tenth of a year)? (c) Write the half-life formula for your bank account. 3. Flu is spreading eponentially at a school. The number of new flu patients can be modeled using the equation 0.1d F 10e, where d represents the number of days since 10 students had the flu. (a) How many days will it take for the number of new flu patients to equal 50? Round your answer to the nearest day. (b) Find the average rate of change of F over the first three weeks, i.e. 0 d 1. Show the calculation that leads to your answer. Give proper units and round your answer to the nearest tenth. What is the physical interpretation of your answer?
33 4. Jessica keeps track of the height of a tree she planted over the first ten years. It can be modeled by the equation y ln( 1) where is the number of years since she planted the tree. (a) On average, how many feet did the tree grow each year over the time interval 0 t 10, to the nearest hundredth. 33 (b) How tall was the tree when she planted it? 5. Most tornadoes last less than an hour and travel less than 0 miles. The wind speed s (in miles per hour) near the center of a tornado is related to the distance d (in miles) the tornado travels by this model: s = 93logd +65. (a) Sketch a graph of this function. y (b) On March 18, 195, a tornado whose wind speed was about 180 miles per hour struck the Midwest. Use your graph to determine how far the tornado traveled to the nearest mile.
34 34 LESSON #3 - NEWTON'S LAW OF COOLING AND EXPONENTIAL FORMULA REVIEW COMMON CORE ALGEBRA II NEWTON S LAW OF COOLING where: T(t) is the temperature of the object t time units has elapsed, T a is the ambient temperature (the temperature of the surroundings), assumed to be constant, not impacted by the cooling process, T 0 is the initial temperature of the object, and k is the decay constant per time unit (the r value where r is negative). Eercise #1: A detective is called to the scene of a crime where a dead body has just been found. He arrives at the scene and measures the temperature of the dead body at 9:30 p.m to be 7 F. After investigating the scene, he declares that the person died 10 hours prior, at approimately 11:30 a.m. A crime scene investigator arrives a little later and declares that the detective is wrong. She says that the person died at approimately 6:00 a.m., hours prior to the measurement of the body temperature. She claims she can prove it by using Newton s law of cooling. Using the data collected at the scene, decide who is correct, the detective or the crime scene investigator. T a = 68 F (the temperature of the room) T 0 = 98.6 F (the initial temperature of the body) k = (13.35 % per hour calculated by the investigator from the data collected) Recall, the temperature of the body at 9:30 p.m. is 7 F. Eercise : A detective is called to the scene of a crime where a dead body has just been found. She arrives on the scene at 10:3 pm and begins her investigation. Immediately, the temperature of the body is taken and is found to be 80 o F. The detective checks the programmable thermostat and finds that the room has been kept at a constant 68 o F. Assuming that the victim s body temperature was normal (98.6 o F) prior to death and that the temperature of the victim s body decreases continuously at a rate of 13.35% per hour, use Newton s Law of Cooling to determine the time when the victim died.
35 Eercise #3: Two cups of coffee are poured from the same pot. The initial temperature of the coffee is 180 F and k is (for time in minutes). Suppose both cups are poured at the same time. Cup 1 is left sitting in the room that is 75 F, and cup is taken outside where it is 4 F. i. Use Newton s law of cooling to write equations for the temperature of each cup of coffee after t minutes have elapsed. 35 ii. Graph and label both on the same coordinate plane and compare and contrast the end behavior of the two graphs. iii. Coffee is safe to drink when its temperature is below 140 F. How much time elapses before each cup is safe to drink, to the nearest tenth of a minute. Use a graph to answer the question.
36 36 Throughout the unit, you have learned many different eponential formulas. We will now practice writing a few of them and converting between the different forms. Eercise #4: Tritium has a half-life of 1.3 years. a. Write a half-life formula, A(t), for the amount of tritium left in a 500 milligram sample after t years. b. Write an equivalent function, B(t), it terms of the yearly rate of decay of tritium. Round all values to four decimal places. c. Wrtie an equivalent function, C(t), it terms of the monthly rate of decay of tritium. Round all values to four decimal places. Eercise #5: A deposit of \$300 is made into a bank account that gets 4.3% interest compounded continuously. a. Write a function, A(t), to model the amount of money in the account after t years. b. Write an equivalent function, B(t), it terms of the yearly rate of interest for the account. Round all values to four decimal places. c. Write an equivalent function, C(t), it terms of the quarterly rate of interest for the account. Round all values to four decimal places.
37 37 NEWTON'S LAW OF COOLING AND EXPONENTIAL FORMULA REVIEW COMMON CORE ALGEBRA II HOMEWORK 1. Hot soup is poured from a pot and allowed to cool in a room. The temperature in degrees Fahrenheit of the soup after t minutes, can be modeled by the function, T(t)= 65+(1-65)e -.054t. What was the initial temperature of the soup? What is the temperature of the room? At what rate is the temperature of the soup decreasing?. Two cups of coffee are poured from the same pot. The initial temperature of the coffee is 190 F and k is (for time in minutes). Both are left sitting in the room that is 75 F, but milk is immediately poured into cup cooling it to an initial temperature of 16 F. a. Use Newton s law of cooling to write equations for the temperature of each cup of coffee after t minutes have elapsed. b. Graph and label both functions on the coordinate plane and compare and contrast the end behavior of the two graphs. c. Coffee is safe to drink when its temperature is below 140 F. Based on your graph, how much time elapses before each cup is safe to drink to the nearest tenth of a minute?
38 3. A cooling liquid starts at a temperature of 00 F and cools down in a room that is held at a constant temperature of 70 F. (Note time is measured in minutes on this problem). 38 (a) Use Newton s Law of Cooling to determine the value of k if the temperature after 5 minutes is to four decimal places. (Hint: Write out the equation, plug in (5,153), and solve for k.). Round (b) Using the value of k you found in part (a), algebraically determine, to the nearest tenth of a minute, when the temperature reaches 100 F. 4. A deposit of \$100 is made into a bank account that gets 4.3% interest compounded weekly. a. Write a function, A(t), to model the amount of money in the account after t years. b. Write an equivalent function, B(t), it terms of the yearly rate of interest for the account. Round all values to four decimal places. c. Write an equivalent function, C(t), it terms of the monthly rate of interest for the account. Round all values to four decimal places.
39 39 5. A small town has a population of 1,600. The population is decreasing continuously at a rate of 3% per year. a. Write a function, A(t), to model the population of the the town after t years. b. Write an equivalent function, B(t), it terms of the yearly rate of decrease for the town. Round all values to four decimal places. c. Write an equivalent function, C(t), it terms of the daily rate of decrease for the town. Round all values to four decimal places.
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Equation of a circle in standard form, Formula, practice problems, and pictures. How to express a circle with given radius in standard form...
# Equation of a Circle
Standard form equation of a Circle
The standard form equation of a circle is a way to express the definition of a circle on the coordinate plane.
On the coordinate plane, the formula becomes $$(x -h)^2 + (y - k)^2 =r^2$$
• h and k are the x and y coordinates of the center of the circle
• $$(x-9)^2 + (y-6)^2 =100$$ is a circle centered at (9,6) with a radius of 10
### Examples
General Formula Circle with a center of (4,3) and a radius of 5
### Another Example
Circle with a center of (2, -1) and a radius of 4
### Definition of Circle
Definition : A circle is the set of all points that are the same distance, r, from a fixed point.
General Formula: X 2 + Y 2=r2 where r is the radius
• Unlike parabolas, circles ALWAYS have X 2 and Y 2 terms.
X2 + Y2=4 is a circle with a radius of 2 (since 4 =22)
• Remember that a circle is a locus of points. A circle is all of the points that are a fixed distance, known as the radius, from a given point, known as the center of the circle.
• Explore the standard equation of a circle using the applet below (Go here for a larger version)
### Practice Problems
Since the radius of this this circle is 1, and its center is the origin, this picture's equation is
$$(y-0)^2 + (x-0)^2 = 1^2 \\ y^2 + x^2 = 1$$
Since the radius of this this circle is 1, and its center is (1,0) , this circle's equation is
$$(y-0)^2 +(x-1)^2 = 1^2 \\ y^2 + (x-1)^2 = 1$$
Since the radius of this this circle is 2, and its center is (3,1) , this circle's equation is
$$(x-3)^2 +(y-1)^2 = 2^2 \\ (x-3)^2 +(y-1)^2 = 4$$
Y2+X2 = 9
$$\sqrt{9} =3$$
Y2+X2 = 16
$$\sqrt{16} = 4$$
Y2+X2 = 25
$$\sqrt{ 25 } = 5$$
Y2 + X2 = 11
$$\sqrt{11}$$
Y2 + X2 = a
$$\sqrt{a}$$
(y-3)2+(x-1)2 = 9
(1, 3) r = 3
(y-5)2+(x-14)2 = 16
(14, 5) r = 4
(y-1)2+(x-5)2 = 25
(5, 1) r = 5
(x+2)2++(y-12)2 = 36
(-2, 12) r = 6
(y+7)2+(x +5)2 = 49
(-5, -7) r = 7
(x +8)2+(y+17)2 = 49
(-8, -17) r = 7
### Ultimate Math Solver (Free)
Free Algebra Solver ... type anything in there!
|
Middle Years
# 10.03 Measuring the centre and spread
Lesson
## Measures of centre
### Mean
The mean is often referred to as the average. To calculate the mean, add all the scores in a data set, then divide this by number of scores.
To find the mean from a graphical representation, we can use a frequency table to list out the values of on the graph. Consider the histogram below:
We can construct a frequency table like the one below:
Score ($x$x) Frequency ($f$f) $xf$xf
$1$1 $3$3 $3$3
$2$2 $8$8 $16$16
$3$3 $5$5 $15$15
$4$4 $3$3 $12$12
$5$5 $1$1 $5$5
The mean will be calculated by dividing the sum of the last column by the sum of the second column, $\frac{51}{20}=2.55$5120=2.55.
### Median
The median is one way of describing the middle or the centre of a data set using a single value. The median is the middle score in a data set.
Suppose we have five numbers in our data set: $4$4, $11$11, $15$15, $20$20 and $24$24.
The median would be $15$15 because it is the value right in the middle. There are two numbers on either side of it.
$4,11,\editable{15},20,24$4,11,15,20,24
If we have an even number of terms, we will need to find the average of the middle two terms. Suppose we wanted to find the median of the set $2,3,6,9$2,3,6,9, we want the value halfway between $3$3 and $6$6. The average of $3$3 and $6$6 is $\frac{3+6}{2}=\frac{9}{2}$3+62=92, or $4.5$4.5, so the median is $4.5$4.5.
$2,3,\editable{4.5},6,9$2,3,4.5,6,9
If we have a larger data set, however, we may not be able to see right away which term is in the middle. We can use the "cross out" method.
### The "cross out" method
Once a data set is ordered, we can cross out numbers in pairs (one high number and one low number) until there is only one number left. Let's check out this process using an example. Here is a data set with nine numbers:
1. Check that the data is sorted in ascending order (i.e. in order from smallest to largest).
1. Cross out the smallest and the largest number, like so:
1. Repeat step 2, working from the outside in - taking the smallest number and the largest number each time until there is only one term left. We can see in this example that the median is $7$7:
Note that this process will only leave one term if there are an odd number of terms to start with. If there are an even number of terms, this process will leave two terms instead, if you cross them all out, you've gone too far! To find the median of a set with an even number of terms, we can then take the mean of these two remaining middle terms.
The idea behind the cross out method can be used in graphical representations by cross off data points from each side.
### Mode
The mode describes the most frequently occurring score.
Suppose that $10$10 people were asked how many pets they had. $2$2 people said they didn't own any pets, $6$6 people had one pet and $2$2 people said they had two pets.
In this data set, the most common number of pets that people have is one pet, and so the mode of this data set is $1$1.
A data set can have more than one mode, if two or more scores are equally tied as the most frequently occurring.
Measures of centre
Mean
• The numerical average of a data set, this is the sum of the data values divided by the number of data values.
• Appropriate for sets of data where there are no values much higher or lower than those in the rest of the data set
Median
• The middle value of a data set ranked in order
• A good choice when data sets have a couple of values much higher or lower than most of the others
Mode
• The data value that occurs most frequently
• A good descriptor to use when the set of data has some identical values, when data is non-numeric (categorical) or when data reflects the most popular item
#### Practice questions
##### Question 1
Find the median of this set of scores:
$3$3, $8$8, $13$13, $17$17, $19$19, $24$24, $26$26, $27$27
##### Question 2
Select the data set from each of the options below that has:
1. The lowest mode.
$87,2,20,20,8,10$87,2,20,20,8,10
A
$11,8,8,48,2,17$11,8,8,48,2,17
B
2. The highest median.
$2,8,11,17$2,8,11,17
A
$8,20,20,48,87$8,20,20,48,87
B
##### QUESTION 3
Consider the table below.
Score Frequency
$1$1 - $4$4 $1$1
$5$5 - $8$8 $5$5
$9$9 - $12$12 $10$10
$13$13 - $16$16 $5$5
$17$17 - $20$20 $3$3
1. Use the midpoint of each class interval to determine the mean of the following sample distribution, correct to one decimal place.
2. Which is the modal group?
$5$5 - $8$8
A
$13$13 - $16$16
B
$1$1 - $4$4
C
$9$9 - $12$12
D
$17$17 - $20$20
E
### Range
The range of a numerical data set is the difference between the smallest and largest scores in the set. The range is one type of measure of spread.
For example, at one school the ages of students in Year $7$7 vary between $11$11 and $14$14. So the range for this set is $14-11=3$1411=3.
As a different example, if we looked at the ages of people waiting at a bus stop, the youngest person might be a $7$7 year old and the oldest person might be a $90$90 year old. The range of this set of data is $90-7=83$907=83, which is a much larger range of ages.
The range of a numerical data set is given by:
Range$=$=maximum score$-$minimum score
#### Practice question
##### Question 4
Find the range of the following set of scores:
$20,19,3,19,18,3,16,3$20,19,3,19,18,3,16,3
### Interquartile range
Whilst the range is very simple to calculate, it is based on only two numbers in the data set, it does not tell us about the spread of data within these two values. To get a better picture of the internal spread in a data set, it is often more useful to find the set's quartiles, from which the interquartile range (IQR) can be calculated.
Quartiles are scores at particular locations in the data set–similar to the median, but instead of dividing a data set into halves, they divide a data set into quarters. Let's look at how we would divide up some data sets into quarters now.
Careful!
Make sure the data set is ordered before finding the quartiles or the median.
#### Exploration
• Here is a data set with $8$8 scores:
$\editable{1}$1 $\editable{3}$3 $\editable{4}$4 $\editable{7}$7 $\editable{11}$11 $\editable{12}$12 $\editable{14}$14 $\editable{19}$19
First locate the median, between the $4$4th and $5$5th scores:
Median $\downarrow$↓ $\editable{1}$1 $\editable{3}$3 $\editable{4}$4 $\editable{7}$7 $\editable{11}$11 $\editable{12}$12 $\editable{14}$14 $\editable{19}$19
Now there are four scores in each half of the data set, so split each of the four scores in half to find the quartiles. We can see the first quartile, $Q_1$Q1, is between the $2$2nd and $3$3rd scores–that is, there are two scores on either side of $Q_1$Q1. Similarly, the third quartile, $Q_3$Q3, is between the $6$6th and $7$7th scores:
$Q_1$Q1 Median $Q_3$Q3 $\downarrow$↓ $\downarrow$↓ $\downarrow$↓ $\editable{1}$1 $\editable{3}$3 $\editable{4}$4 $\editable{7}$7 $\editable{11}$11 $\editable{12}$12 $\editable{14}$14 $\editable{19}$19
• Now let's look at a situation with $9$9 scores:
$Q_1$Q1 Median $Q_3$Q3 $\downarrow$↓ $\downarrow$↓ $\downarrow$↓ $\editable{8}$8 $\editable{8}$8 $\editable{10}$10 $\editable{11}$11 $\editable{13}$13 $\editable{14}$14 $\editable{18}$18 $\editable{22}$22 $\editable{25}$25
This time, the $5$5th term is the median. There are four terms on either side of the median, like for the set with eight scores. So $Q_1$Q1 is still between the $2$2nd and $3$3rd scores and $Q_3$Q3 is between the $6$6th and $7$7th scores.
• Finally, let's look at a set with $10$10 scores:
$Q_1$Q1 Median $Q_3$Q3 $\downarrow$↓ $\downarrow$↓ $\downarrow$↓ $\editable{12}$12 $\editable{13}$13 $\editable{14}$14 $\editable{19}$19 $\editable{19}$19 $\editable{21}$21 $\editable{22}$22 $\editable{22}$22 $\editable{28}$28 $\editable{30}$30
For this set, the median is between the $5$5th and $6$6th scores. This time, however, there are $5$5 scores on either side of the median. So $Q_1$Q1 is the $3$3rd term and $Q_3$Q3 is the $8$8th term.
### What do the quartiles represent?
Each quartile represents $25%$25% of the data set. The lowest score to the first quartile is approximately $25%$25% of the data, the first quartile to the median is another $25%$25%, the median to the third quartile is another $25%$25%, and the third quartile to the highest score represents the last $25%$25% of the data. We can combine these sections together–for example, $50%$50% of the scores in a data set lie between the first and third quartiles.
These quartiles are sometimes referred to as percentilesA percentile is a percentage that indicates the value below which a given percentage of observations in a group of observations fall. For example, if a score is in the $75$75th percentile in a statistical test, it is higher than $75%$75% of all other scores. The median represents the $50$50th percentile, or the halfway point in a data set.
### Naming the quartiles
• $Q_1$Q1 is the first quartile (sometimes called the lower quartile). It is the middle score in the bottom half of data and it represents the $25$25th percentile. $25%$25% of scores are less than the lower quartile.
• $Q_2$Q2 is the second quartile, and is usually called the median, which we have already learnt about. It represents the $50$50th percentile of the data set. $50%$50% of scores are less than the median.
• $Q_3$Q3 is the third quartile (sometimes called the upper quartile). It is the middle score in the top half of the data set, and represents the $75$75th percentile. $75%$75% of scores are less than the upper quartile.
### Calculating the interquartile range
The interquartile range (IQR) is the difference between the third quartile and the first quartile. $50%$50% of scores lie within the IQR because it contains the data set between the first quartile and the median, as well as the median and the third quartile.
Since it focuses on the middle $50%$50% of the data set, the interquartile range often gives a better indication of the internal spread than the range does, and it is less affected by individual scores that are unusually high or low, which are the outliers.
To calculate the interquartile range
Subtract the first quartile from the third quartile. That is,
$\text{IQR }=Q_3-Q_1$IQR =Q3Q1
#### Worked example
##### Example 1
Consider the following set of data: $1,1,3,5,7,9,9,10,15$1,1,3,5,7,9,9,10,15.
(a) Identify the median.
Think: There are nine numbers in the set, so we can say that $n=9$n=9. We can also see that the data set is already arranged in ascending order. We identify the median as the middle score either by the "cross-out" method or as the $\frac{n+1}{2}$n+12th score.
Do:
$\text{Position of median}$Position of median $=$= $\frac{9+1}{2}$9+12 Substituting $n=9$n=9 into $\frac{n+1}{2}$n+12 $=$= $5$5th score Simplifying the fraction
Counting through the set to the $5$5th score gives us $7$7 as the median.
(b) Identify $Q_1$Q1 (the lower quartile) and $Q_3$Q3 (the upper quartile).
Think: We identify $Q_1$Q1 and $Q_3$Q3 as the middle scores in the lower and upper halves of the data set respectively, either by the "cross-out" method–or any method that we use to find the median, but just applying it to the lower or upper half of the data set.
Do: The lower half of the data set is all the scores to the left of the median, which is $1,1,3,5$1,1,3,5. There are four scores here, so $n=4$n=4. So we can find the position of $Q_1$Q1 as follows:
$\text{Position of }Q_1$Position of Q1 $=$= $\frac{4+1}{2}$4+12 Substituting $n=4$n=4 into $\frac{n+1}{2}$n+12 $=$= $2.5$2.5th score Simplifying the fraction
$Q_1$Q1 is therefore the mean of the $2$2nd and $3$3rd scores. So we see that:
$Q_1$Q1 $=$= $\frac{1+3}{2}$1+32 Taking the average of the $2$2nd and $3$3rd scores $=$= $2$2 Simplifying the fraction
The upper half of the data set is all the scores to the right of the median, which is $9,9,10,15$9,9,10,15. Since there are also $n=4$n=4 scores, $Q_3$Q3 will be the mean of the $2$2nd and $3$3rd scores in this upper half.
$Q_3$Q3 $=$= $\frac{9+10}{2}$9+102 Taking the average of the $2$2nd and $3$3rd scores in the upper half $=$= $9.5$9.5 Simplifying the fraction
(c) Calculate the $\text{IQR }$IQR of the data set.
Think: Remember that $\text{IQR }=Q_3-Q_1$IQR =Q3Q1, and we just found $Q_1$Q1 and $Q_3$Q3.
Do:
$\text{IQR }$IQR $=$= $9.5-2$9.5−2 Substituting $Q_1=9.5$Q1=9.5 and $Q_3=2$Q3=2 into the formula $=$= $7.5$7.5 Simplifying the subtraction
### How do outliers affect the range and IQR?
Remember, the range only changes if the highest or lowest score in a data set is changed, otherwise it will remain the same. An outlier is always the highest or lowest score in a data set. Therefore including an outlier will increase the range.
The IQR does not use the highest or lowest score, therefore including an outlier will have no effect on the IQR
#### Practice questions
##### Question 5
Answer the following, given this set of scores:
$33,38,50,12,33,48,41$33,38,50,12,33,48,41
1. Sort the scores in ascending order.
2. Find the number of scores.
3. Find the median.
4. Find the first quartile of the set of scores.
5. Find the third quartile of the set of scores.
6. Find the interquartile range.
##### Question 6
The stem plot shows the number of hours students spent studying during an entire semester.
Stem Leaf
$6$6 $2$2 $7$7
$7$7 $1$1 $2$2 $2$2 $4$4 $7$7 $9$9
$8$8 $0$0 $1$1 $2$2 $5$5 $7$7
$9$9 $0$0 $1$1
Key: $5$5$\mid$∣$2$2$=$=$52$52
1. Find the first quartile of the set of scores.
2. Find the third quartile of the set of scores.
3. Find the interquartile range.
##### Question 7
The column graph shows the number of pets that each student in a class owns.
1. Find the first quartile of the set of scores.
2. Find the third quartile of the set of scores.
3. Find the interquartile range.
|
Math Cracks – A Cool Approach to Integration by Parts
Introduction
The idea of integration by parts sounds quite scary for many Calculus students, and I think there is a good reason for that. First of all, integration by parts is a technique that involves two steps (or more) instead of one step as most students would like to. Students would want to APPLY some formula and get the answer right away, but in Calculus often times the answers come after a sequence (sometimes a long one) of steps.
Aside from the method of substitution, the integration by parts method is the most important method to solve integrals that are not elementary.
First off, as a principle of the matter, one of the reasons integral Calculus is typically hard for students is the rather unfortunate notation used for integration. In fact, when calculating the indefinite integral of a function $$f\left( x \right)$$, we face the following notation
$\int{f\left( x \right)dx}$ What many students do not understand is what is really meant by the "$$dx$$" in the expression above. Clearly, there are historical reasons why the "$$dx$$" appears in the notation stated above. But, actually there is no reason to include $$dx$$ or even to add $$f\left( x \right)$$. When we want to compute the indefinite integral of a function $$f$$, we should be able to simply write
$\int{f}$ and that way we are stating the indefinite integral of the function $$f$$.
Are these the same?
$\int{f\left( x \right)dx}=\int{f\left( u \right)du}$
Absolutely! That is why you see sometimes the integration variable (x or u respectively) referred as a "dummy" variable, because it does not really play any role in the integration process.
Integration By Parts as a Reverse Product Rule
After a brief introduction, now we cut to the chase. The typical integration by parts formula shown in textbooks is
$\int{udv}=uv-\int{vdu} \,\,\,\,\,(1)$
Then you say, "Huh? What is that?" Obviously, without giving a meaning to the above $$u$$ and $$dv$$, it is hard to see what is all about it. One question you can have is: Why is the integration by parts formula involving dv's and du's, if those don't even play a role in the integration process, as shown in the introduction?
The answer is simple: In the context of the integration by parts formula above, $$du$$ and $$dv$$ are not "dummy variables", but instead they are function. Mnemonically, the above is good to solve an integration by parts exercise, but it is no good to understand why it is actually true or why it works.
Enter the Product Rule:
The product rule says that:
$\frac{d}{dx}\left( fg \right)=\frac{df}{dx}g+f\frac{dg}{dx} \,\,\,\,\,(2)$
For short, I prefer to write
$\left( fg \right)'=f'g+fg' \,\,\,\,\,(3)$
But WAIT! Aren't we integrating in this article? Why do I bring up a differentiation rule?? Hum, wouldn't it be great to have to product rule for integrals too? Wouldn't it be great if $$\int{f'g'}=f\,g + C$$?? Unfortunately it is not, BUT there is still a product rule for integrals, only that it is slightly more complicated.
Let's rearrange equation (3), we get:
$fg'=\left( fg \right)'-f'g \,\,\,\,\,(4)$
So then, if we integrate both sides of the equality above we get
$\int{fg'}=\int{\left( \left( fg \right)'-f'g \right)}$
which by linearity of integration leads to
$\int{f\,g'}=f\,g-\int{f'g} \,\,\,\,\,(5)$
And here my friends, you have your integration by part rule. Integration by parts should be seen as a cool integration tool that allows me to integrate the product of two functions. But it is a bit more restrictive, because it is the product of two functions BUT one of the functions must be a derivative of SOME function.
So, in order to fruitfully apply the integration by parts rule I need to have three things happening:
• I'm trying to integrate the product of TWO functions.
• One of those functions a derivative of something (so it is of the form $$g'$$).
• I need to know how to calculate that something (I need to know who is $$g$$)
If those three conditions happen, then I can use the integration by parts rule
REMEMBER: When using integration by parts you need to have the product of two functions, and one of those two functions needs to be the derivative of something that you know.
For example, let us see when you cannot apply integration by parts: Consider the following integral
$\int{\sin \left( {{x}^{2}} \right){{e}^{{{x}^{2}}}}}$
In this case, we are trying to integrate the product of two functions: $$\sin \left( {{x}^{2}} \right)$$ and $${{e}^{{{x}^{2}}}}$$, but do you know what the antiderivative of any of these two functions is? Or in other words, do you know which functions leads to any of $$\sin \left( {{x}^{2}} \right)$$ or $${{e}^{{{x}^{2}}}}$$ after differentiating? Well. No. Those two functions do not have elementary antiderivatives, so then integration by parts would not help in this case.
Now an example where integration by parts COULD be used:
$\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx}$
In this case we are trying to integrate the product of two functions: $${{x}^{2}}$$ and $${{e}^{{{x}^{2}}}}$$, and I know what the antiderivative of $${{x}^{2}}$$ is. So I can use the rule. We have the following notation:
$\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx}=\int{\underbrace{{{e}^{{{x}^{2}}}}}_{f\left( x \right)}\underbrace{{{x}^{2}}dx}_{g'\left( x \right)}}$
So we have
\begin{aligned} & f\left( x \right)={{e}^{{{x}^{2}}}} \\ & g'\left( x \right)={{x}^{2}} \\ \end{aligned}
Differentiating $$f$$ and integrating $$g'$$ we get:
\begin{aligned} & f\left( x \right)=2x{{e}^{{{x}^{2}}}} \\ & g\left( x \right)=\frac{{{x}^{3}}}{3} \\ \end{aligned}
(Notice that the $$g\left( x \right)$$ stated above is one possible antiderivative, but the rule is that I can choose ANY antiderivative, so I choose the simplest one). The integration by parts is
$\int{f\,g'}=f\,g-\int{f'g}$
so plugging the information we have, we get the following:
$\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx\,}={{e}^{{{x}^{2}}}}\frac{{{x}^{3}}}{3}-\int{2x{{e}^{{{x}^{2}}}}\frac{{{x}^{3}}}{3}dx}$ $\Rightarrow \,\,\,\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx\,}=\frac{{{e}^{{{x}^{2}}}}{{x}^{3}}}{3}-\frac{2}{3}\int{{{x}^{4}}{{e}^{{{x}^{2}}}}dx}$
So I have used the integration by parts rule above, but actually, I landed into a harder integral to solve. This is, in order to solve $$\int{{{x}^{2}}{{e}^{{{x}^{2}}}}dx\,}$$ we need to know first how to compute $$\int{{{x}^{4}}{{e}^{{{x}^{2}}}}dx}$$ which is actually harder.
The moral of this story is that integration by parts is a sort of product rule for integrals, and you're looking for a specific structure: it is the integral of the product of two functions, and one of those functions you need to know how to compute its antiderivative. If that is the case, you're in business and you can apply the integration by parts rule.
BUT, as it could be seen in the previous example, the fact that you CAN use integration by parts DOES NOT mean that it will be useful every time.
Final Words:
How do we tie together the formula for Integration by Parts?
$\int{f\,g'}=f\,g-\int{f'g}$
of the "product rule for integrals" with
$\int{udv}=uv-\int{vdu}$
By setting
\begin{aligned} & u=f\left( x \right) \\ & dv=g'\left( x \right)dx \\ \end{aligned}
we get that $$v = g\left( x \right)$$ and $$du = f'\left( x \right)dx$$, which renders both equation equal.
In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us.
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The segment addition postulate states the following for 3 points that are collinear.
Consider the segment on the right.
If 3 points A, B, and C are collinear and B is between A and C, then
AB + BC = AC
Using the segment addition postulate to solve a problem.
Suppose AC = 48, find the value of x. Then, find the length of AB and the length of BC.
AB + BC = AC
( 2x - 4 ) + ( 3x + 2 ) = 48
2x + 3x - 4 + 2 = 48
5x - 4 + 2 = 48
Add 4 to both sides of the equation.
5x - 4 + 4 + 2 = 48 + 4
5x + 2 = 52
Subtract 2 from both sides.
5x + 2 - 2 = 52 - 2
5x = 50
Divide both sides by 5
5x / 5 = 50 / 5
x = 10
Now that we have the value of x, we can find the length of AB and the length of BC.
AB = 2x - 4
AB = 2 × 10 - 4
AB = 20 - 4 = 16
The length of AB is 16
BC = 3x + 2
BC = 3 × 10 + 2
BC = 30 + 2 = 32
The length of BC is 32.
Segment addition postulate and the midpoint
Suppose XA = 3x and AY = 4x - 6. If A is the midpoint of XY, what is the length of XY?
3x 4x - 6
_________________________________
X A Y
The trick in this problem is to see that if A is the midpoint, then XA = AY.
Since XA = AY, 3x = 4x - 6
Subtract 3x from both sides.
3x - 3x = 4x - 3x - 6
0 = x - 6
Add 6 to both sides of the equation.
0 + 6 = x - 6 + 6
6 = x
To compute XA, you can either use 3x or 4x - 6
Using 3x, we get XA = AY = 3 × 6 = 16
Using 4x - 6, we get XA = AY = 3 × 6 - 6 = 18 - 6 = 12
XY = XA + AY = 16 + 16 = 32
The length of XY is 32.
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# Quadratic Equation – Quiz 4
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# Quadratic Equation – Quiz 4
### Introduction
Quadratic Equation is one of important topic in Quantitative Aptitude Section. In Quadratic Equation – Quiz 4 article candidates can find questions with an answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Quadratic Equation – Quiz 4 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.
### Q1
If a and b are the roots of the equation ${x}^{2}$ - 9x + 20 = 0, find the value of ${a}^{2}$ + ${b}^{2}$ + ab?
A. -21 B. 1 C. 61 D. 21
C
${a}^{2}$ + ${b}^{2}$ + ab = ${a}^{2}$ + ${b}^{2}$ + 2ab - ab
i.e, ${(a + b)}^{2}$ + ab
from ${x}^{2}$ - 9x + 20 = 0, we have
a + b = 9 and ab = 20. Hence the value of required expression ${(9)}^{2}$ - 20 = 61.
### Q2
Find the value of $\frac{a}{b}$ + $\frac{b}{a}$, if a and b are the roots of the quadratic equation ${x}^{2}$ + 8x + 4 = 0?
A. 15 B. 14 C. 24 D. 26
B
$\frac{a}{b}$ + $\frac{b}{a}$ = $\frac{{a}^{2} + {b}^{2}}{ab}$ + $\frac{{a}^{2} + {b}^{2} + b + a}{ab}$
= $\frac{({(a + b)}^{2} - 2ab)}{ab}$
a + b = $\frac{-8}{1} = -8$
ab = $\frac{4}{1}$ = 4
Hence $\frac{a}{b}$ + $\frac{b}{a}$ = $\frac{({(-8)}^{2} - 2(4))}{4}$ = $\frac{56}{4}$ = 14
### Q3
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
A. 10 B. 8 C. 15 D. 7.5
A
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
$\Rightarrow$ y = $\frac{300}{x}$
(x + 5)(y - 10) = 300 $\Rightarrow$ xy + 5y - 10x - 50 = xy
$\Rightarrow$ 5($\frac{300}{x}$) - 10x - 50 = 0 $\Rightarrow$ - 150 + ${x}^{2}$ + 5x = 0
multiplying both sides by $\frac{- 1}{10 x}$
$\Rightarrow$ ${x}^{2}$ + 15x - 10x - 150 = 0
$\Rightarrow$ x(x + 15) - 10(x + 15) = 0
$\Rightarrow$ x = 10 or -15
As x > 0, x = 10.
### Q4
Find the roots of quadratic equation: 2${x}^{2}$ + 5x + 2 = 0?
A. -2, $\frac{-1}{2}$ B. 4, -1 C. 4, 1 D. 2, $\frac{-1}{2}$
A
2${x}^{2}$ + 5x + 2 = 0
2x(x + 2) + 1(x + 2) = 0
(x + 2)(2x + 1) = 0 $\Rightarrow$ x = -2, $\frac{-1}{2}$
### Q5
Find the roots of quadratic equation: ${x}^{2}$ + x - 42 = 0?
A. -6, 7 B. -8, 7 C. 14, -3 D. -7, 6
D
${x}^{2}$ + 7x - 6x + 42 = 0
x(x + 7) - 6(x + 7) = 0
(x + 7)(x - 6) = 0 $\Rightarrow$ x = -7, 6
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High School Mathematics1.20 Linear Equations - Two Variables
Linear equation with one variable: has one solution. Example: x + 4 = 12 x + 2x + 6 = 9 simplifies to 3x + 6 = 9 Linear equation with two variables: A linear equation in two variables has infinitely many solutions. Example: x + y = 12 For x = 1, y = 11, x = 2, y = 10, x = 3, y = 9, x = 4, y = 8..... and so on Linear equation with three variables Example: x + y + z = 12 Note: You will need one equation to solve for one variable, two equations to solve two variable and three equations to solve three variable and so on...... Method 1: Elimination Method: Solving Linear Equations with two variables. Solve for the values of a and b for the following equations: Example 1: a + b = 12 -----equation 1 a - b = 4 ------- equation 2. Adding both equations we get a + b = 12 a - b = 4 ------------------The terms +b and -b cancels, and we get 2a = 16 a = 16/8 = 2 a = 2 substituting the value of a in equation 1 we have a + b = 12 2 + b = 12 b = 12 - 2 b = 10 The solution of the two equations is a = 2 and b = 10 Example 2: c + 5b = 16 -----equation 1 c + 2b = 7 ------- equation 2 . Multiplying the equation 2 with (-1) we get c + 5b = 16 -----equation 1 -1(c + 2b = 7) => -c - 2b = -7-------equation 2 c + 5b = 16 -c - 2b = -7 ---------------------The terms c and -c cancel each other and we get 3b = 9 b = 9/3 = 3 substituting b in the equation c + 5b = 16 we get c + 5(3) = 16 c + 15 = 16 - 3 c = 16 - 15 c = 1 Example 3: 2a + 4b = 12 -----equation 1 a - b = 3 ------- equation 2 First pick a variable either a or b. Lets pick b. Second find the common multiple of b so we can eliminate that variable and find the value of the other variable. so multiply both sides of the equation 2 by 4 we have 2a + 4b = 12 -----equation 1 4(a - b = 3) => 4a - 4b = 12------- equation 2 Now adding equation 1 and 2 we have 2a + 4b = 12 4a - 4b = 12 ----------------------The terms +4b in equation 1 and -4b in equation 2 cancels, and we get 6a = 24 a = 24/6 =4 substituting value of a in equation a - b = 3 we get 4 - b = 3 4- 3 = b b = 1 Therefore a = 4 and b = 1 Verification: To verify substitute the values of a and b in the equations above 2a + 4b = 12 2a + 4b = 2(4) + 4(1) = 8 + 4 = 12 NOTE: In the above question if you pick b 2a + 4b = 12 -----equation 1 a - b = 3 ------- equation 2 multiplying equation 2 by 2 we have 2(a - b = 3) ------- equation 2 2a - 2b = 6 multiplying equation by -1 we gave -1(2a - 2b = 6) we get -2a + 2b = -6 Solving both equations we have 2a + 4b = 12 -----equation 1 -2a + 2b = -6-----equation 2 ------------------------- 6b = 6 b = 1 Example 4: 2a + b = 4 -----equation 1 3a + 2b = 3 ------- equation 2. Multiplying the equation 1 with 3 and equation 2 with (-2) we get 3(2a + b = 4) -----equation 1 -2(3a + 2b = 3) ------- equation 2 6a + 3b = 12 -----equation 1 -6a - 4b = -6) ------- equation 2 -----------------------The terms 6a and -6a cancel each other and we get -b = 6 b = -6 substituting b in the equation 2a + b = 4 we get 2a + (-6) = 4 2a - 6 = 4 2a = 4 + 6 2a = 10 a = 10/2 = 5 Method 2: Substitution Method: Solve the equations: 3x + 5y = 26 y = 2x substitute the second equation in the first 3x + 5(2x) = 26 3x + 10x = 26 13x = 26 x = 2 y = 2x = 2x2 = 4 Solve the equations: x + y = 26 ------- equation 1 x - y = 4 -------equation 2 x - y = 4 -------equation 2 can be written as x = y + 4. Substituting x in the equation 1 we have: x + y = 26 ------- equation 1 y + 4 + y = 26 2y + 4 = 26 2y + 4 - 4 = 26 - 4 2y = 22 y = 11 x = y + 4 = 11 + 4 = 15 Directions: Solve for the variables using either Elimination or Substitution Method.
Q 1: Solve for the values of x and y for the following equations: 3x + 6y = 24 and x + 3y = 4x = 6 and y = 8x = -16 and y = 4x = 16 and y = -4 Question 2: This question is available to subscribers only!
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Taking Down the Matrix
By Leo Lam
(Test Preparation (ACT/SAT/SSAT), Math and Physics tutor at The Edge Learning Center)
Disclaimer: no, I will not be giving you a choice of blue pill or red pill.
In my last blog, I talked about the complex number, a topic that is becoming more prominent in both SAT and ACT. Today, I will discuss another advanced math topic you may encounter on these college prep tests: matrix (plural: matrices).
While SAT does not test the concept of the matrix, this topic has become a staple of the ACT math section. And if you are taking the SAT subject test, you will also find questions on matrices in the Math Level 2 test. You will not encounter numerous questions about matrices on either tests, but in order to achieve the highest score possible, you would need to know the fundamental concepts behind this topic.
What is a matrix?
Simply put, a matrix is a collection of values or expressions. Unlike a set, these values are order-specific and arranged in a rectangular array. The numbers of rows and columns determine the size of a matrix. For example:
The matrix A has 2 rows and 3 columns. We call this a 2 X 3(read as “two by three”) matrix. Each value inside a matrix is referred to as an entry or element. A common way to specify an entry inside a matrix is by using the notation, where the subscriptsand represent the row and column, respectively.
Photo Source: Wikipedia
Thus, in our matrix,would be 5, whilewould be -11.
In order for two matrices to be considered equivalent, they must have the exact same size and corresponding entries.
What do I need to know?
There are three common concepts that are asked on these tests: 1) operating with matrices, 2) finding the determinant, and 3) translating a problem into a matrix.
Operating with Matrices
Matrices can be added, subtracted, or multiplied (but not divided, we will get into that later). Addition and subtraction are very straightforward: you add/subtract the corresponding entries from the two matrices as long as they are the same size. For example:
This means addition or subtraction between two matrices with different sizes is not possible.
Multiplication works very differently. First, there are two types of multiplications: scalar multiplication and matrix multiplication. Scalar multiplication involves multiplying a matrix with a number. The size of the matrix is irrelevant. What is important is that every single entry inside must be multiplied by the same value. For example:
If you want to divide each entry inside the matrix, you can multiply by the reciprocal of the value instead, i.e. dividing by 3 can be calculated as multiplying by 1/3.
Matrix multiplication, on the other hand, is more complicated. First, there are a few rules you must remember:
1. Matrix multiplication is not always commutative, i.e.
2. .In order to multiply two matrices together, the number of columns in the first matrix MUST be equal to the number of rows in the second matrix.
3. The product has the same number of rows as the first matrix and the same number of columns as the second matrix.
If you follow these rules closely, you will realize that many answer choices in a test can be eliminated because one or more of the rules are broken. Let us take a look at some examples.
Let A be a 2 X 3 matrix, B be a 3 X 4 matrix, and C be a 3 X 2 matrix.
A X B is defined in this case, and the result, by following rule III, is a 2 X 4 matrix. However, X C is undefined, as it does not follow rule II. Furthermore, both A X C and C A are defined, but the result of A X C is a 2 X 2 matrix, while the result of C X A is a 3 X 3. Thus, the results are different matrices, which demonstrates rule I.
The actual process of multiplication gets even more complex. In a nutshell, each entry of the resultant matrix is a dot product between a row vector of the first matrix and the column vector of the second matrix (for those of you who have not studied vector, here is a quick lesson on vector and dot product).
Instead of going through the nitty-gritty process, you should rely on your graphing calculator to get the result. Here are two YouTube videos showing you how to use your TI-84 or TI-Nspire to multiply matrices.
Finding the Determinant
Knowing how multiplication of matrices can be a pain, mathematicians instead focus their energies on operations with square matrices: matrices with the same number of rows and columns. For these matrices, the result of adding, subtracting, and multiplying will always end up with a matrix of the same size (but keep in mind that multiplication is still not always commutative).
As I have mentioned earlier, there is no division between matrices. Instead of “dividing” two matrices, we multiply the first matrix by the “inverse” of the second matrix. Finding the inverse of a matrix (which can only be done for square matrices) is beyond the scope of what we will cover today. However, not all matrices can be inverted, and deciding whether a matrix can be inverted or not is as simple as finding its determinant.
To find the determinant of a square matrix, you should, again, let the calculator do the work for you. If A is a square matrix, the determinant is normally denoted as set A or |A|. Here are videos on how to find the determinant on your TI-84 or TI-Nspire.
The determinant is just a number. If the value is 0, then the matrix cannot be inverted. If it’s not 0, then there exists an inverse, normally denoted .
Translating a Problem into a Matrix
Because a matrix is two-dimensional, each entry represents a value from two different categories. Think of it like a spreadsheet: the columns and rows of a spreadsheet normally represent two different ideas, and each entry must represent both. Here is an example:
If the spreadsheet above represents the number of items sold at a store, then we can see that the value of 5 in the middle represents “5 pieces of model Y were sold on day 2”. We can convert the information into the following matrix:
Where the columns represent the day, while the rows represent the model.
Another common way to apply matrix is to solve a system of linear equations. For example:
This system of equation can be rewritten as:
The first matrix is called the coefficient matrix, which is the most important part of the equation. The key to coming up with the coefficient matrix is to make sure that your variables in each equation are in the same order, and that the constant is on the right-hand side of the equation. The order of the equations does not matter as much, meaning you can interchange the two rows as long as you interchange the constants also.
The equations can be solved if we are able to “get rid of” the coefficient matrix. “Getting rid of” a matrix involves multiplying by its inverse. Thus, without actually solving for and , we can find out whether the system is solvable or not by finding the determinant of the coefficient matrix. If the determinant is 0, then there will be no solution as the matrix is not invertible. If the determinant is not 0, we will be able to find a unique solution.
I am intrigued… what else can I learn about matrices?
I am glad you ask that! Matrix is a very useful mathematical tool; it is studied extensively in linear algebra. Applying it with coordinate geometry and vector, we can perform some complex transformation in 2- and 3-dimensional space. It is used in engineering and computer programming (particularly in computer graphics). Come join me if you want to be The One to take down the matrix (renaming yourself as Neo will not be necessary).
As a parting gift, here is a “play on matrix”, courtesy of xkcd.com.
Related Blog: Fun! Facts about Factorials!
Need help with your IB coursework? The Edge is offering a FREE Trial Lesson*(English/Math/Physics/Biology/Chemistry/History) to new students, so join our class and learn how to apply the knowledge you learn in school to your AP/IB/IGCSE examinations. SIGN UP HERE!
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## G2 - Determinants
#### Example 1 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -2 & 3 & 0 & 5 \\ -6 & 1 & 6 & 1 \\ 3 & 0 & 0 & -1 \\ -5 & 6 & 4 & -3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -2 & 3 & 0 & 5 \\ -6 & 1 & 6 & 1 \\ 3 & 0 & 0 & -1 \\ -5 & 6 & 4 & -3 \end{array}\right] = -632$
#### Example 2 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -1 & 1 & 2 & -1 \\ 4 & -1 & -3 & 6 \\ 0 & 0 & 1 & -2 \\ -3 & -3 & 5 & 1 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -1 & 1 & 2 & -1 \\ 4 & -1 & -3 & 6 \\ 0 & 0 & 1 & -2 \\ -3 & -3 & 5 & 1 \end{array}\right] = -78$
#### Example 3 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -2 & 6 & 2 & -5 \\ -1 & -6 & 0 & -3 \\ -5 & 5 & 0 & -3 \\ -4 & 0 & 1 & 3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -2 & 6 & 2 & -5 \\ -1 & -6 & 0 & -3 \\ -5 & 5 & 0 & -3 \\ -4 & 0 & 1 & 3 \end{array}\right] = -655$
#### Example 4 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -3 & 4 & 2 & -2 \\ 6 & -6 & -2 & 0 \\ 0 & 0 & 3 & -1 \\ -6 & 3 & 6 & 4 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -3 & 4 & 2 & -2 \\ 6 & -6 & -2 & 0 \\ 0 & 0 & 3 & -1 \\ -6 & 3 & 6 & 4 \end{array}\right] = -6$
#### Example 5 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 2 & 6 & -5 & 1 \\ 0 & -2 & -4 & 0 \\ 3 & 6 & 2 & -3 \\ 1 & -4 & -1 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 2 & 6 & -5 & 1 \\ 0 & -2 & -4 & 0 \\ 3 & 6 & 2 & -3 \\ 1 & -4 & -1 & 0 \end{array}\right] = -248$
#### Example 6 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 2 & 5 & 0 & -4 \\ 3 & -3 & 6 & 4 \\ 4 & 3 & -5 & -6 \\ -1 & 0 & 0 & -3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 2 & 5 & 0 & -4 \\ 3 & -3 & 6 & 4 \\ 4 & 3 & -5 & -6 \\ -1 & 0 & 0 & -3 \end{array}\right] = -635$
#### Example 7 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -3 & -2 & -4 & 0 \\ 1 & 0 & 3 & 0 \\ -2 & -4 & -1 & 1 \\ 4 & 5 & 5 & 4 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -3 & -2 & -4 & 0 \\ 1 & 0 & 3 & 0 \\ -2 & -4 & -1 & 1 \\ 4 & 5 & 5 & 4 \end{array}\right] = -51$
#### Example 8 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 6 & 0 & 1 & -1 \\ 4 & -1 & -6 & 0 \\ -4 & -2 & -5 & 3 \\ 3 & -4 & -3 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 6 & 0 & 1 & -1 \\ 4 & -1 & -6 & 0 \\ -4 & -2 & -5 & 3 \\ 3 & -4 & -3 & 0 \end{array}\right] = 256$
#### Example 9 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -1 & -1 & 4 & 0 \\ 1 & -1 & -4 & 3 \\ 6 & 6 & 0 & 1 \\ -2 & 1 & 1 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -1 & -1 & 4 & 0 \\ 1 & -1 & -4 & 3 \\ 6 & 6 & 0 & 1 \\ -2 & 1 & 1 & 0 \end{array}\right] = 230$
#### Example 10 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 1 & -4 & -2 & 0 \\ 2 & 3 & 5 & -2 \\ -3 & -5 & 3 & 5 \\ 0 & 1 & 3 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 1 & -4 & -2 & 0 \\ 2 & 3 & 5 & -2 \\ -3 & -5 & 3 & 5 \\ 0 & 1 & 3 & 0 \end{array}\right] = -24$
#### Example 11 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -5 & -6 & -6 & -6 \\ -3 & -1 & 0 & 0 \\ 2 & 2 & 5 & 2 \\ 2 & -5 & 0 & -1 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -5 & -6 & -6 & -6 \\ -3 & -1 & 0 & 0 \\ 2 & 2 & 5 & 2 \\ 2 & -5 & 0 & -1 \end{array}\right] = -265$
#### Example 12 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -5 & 5 & -2 & -1 \\ -1 & 0 & 0 & 0 \\ 5 & 4 & 5 & 5 \\ 4 & 3 & 6 & 1 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -5 & 5 & -2 & -1 \\ -1 & 0 & 0 & 0 \\ 5 & 4 & 5 & 5 \\ 4 & 3 & 6 & 1 \end{array}\right] = -156$
#### Example 13 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 6 & -1 & -6 & -1 \\ 3 & 6 & 6 & 0 \\ -5 & 2 & -4 & 3 \\ -4 & 1 & 2 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 6 & -1 & -6 & -1 \\ 3 & 6 & 6 & 0 \\ -5 & 2 & -4 & 3 \\ -4 & 1 & 2 & 0 \end{array}\right] = 486$
#### Example 14 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 1 & -3 & -3 & -2 \\ -2 & 2 & 0 & -1 \\ 1 & -4 & 0 & -6 \\ 1 & 2 & 0 & -3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 1 & -3 & -3 & -2 \\ -2 & 2 & 0 & -1 \\ 1 & -4 & 0 & -6 \\ 1 & 2 & 0 & -3 \end{array}\right] = 180$
#### Example 15 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 3 & 2 & 3 & 0 \\ -4 & 2 & -6 & 1 \\ 4 & -4 & 5 & -2 \\ 6 & -1 & 6 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 3 & 2 & 3 & 0 \\ -4 & 2 & -6 & 1 \\ 4 & -4 & 5 & -2 \\ 6 & -1 & 6 & 0 \end{array}\right] = -45$
#### Example 16 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 0 & -3 & -6 & 5 \\ -1 & 1 & -6 & 3 \\ 0 & 2 & 1 & 0 \\ 6 & 2 & 1 & -1 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 0 & -3 & -6 & 5 \\ -1 & 1 & -6 & 3 \\ 0 & 2 & 1 & 0 \\ 6 & 2 & 1 & -1 \end{array}\right] = -237$
#### Example 17 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -4 & -6 & 4 & 0 \\ 6 & 5 & 5 & 0 \\ 1 & 0 & 5 & -1 \\ -6 & -2 & 1 & 2 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -4 & -6 & 4 & 0 \\ 6 & 5 & 5 & 0 \\ 1 & 0 & 5 & -1 \\ -6 & -2 & 1 & 2 \end{array}\right] = 288$
#### Example 18 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 3 & 1 & 0 & 1 \\ 1 & 2 & 1 & 0 \\ -4 & -1 & -2 & 6 \\ -2 & 2 & 0 & -2 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 3 & 1 & 0 & 1 \\ 1 & 2 & 1 & 0 \\ -4 & -1 & -2 & 6 \\ -2 & 2 & 0 & -2 \end{array}\right] = 68$
#### Example 19 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -2 & 0 & -4 & 2 \\ 2 & -1 & -6 & -6 \\ -4 & 0 & -4 & 4 \\ -2 & -2 & 4 & -3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -2 & 0 & -4 & 2 \\ 2 & -1 & -6 & -6 \\ -4 & 0 & -4 & 4 \\ -2 & -2 & 4 & -3 \end{array}\right] = 24$
#### Example 20 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 1 & 1 & -1 & 2 \\ 0 & 0 & -1 & -3 \\ 2 & -6 & -4 & 3 \\ 5 & -5 & 2 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 1 & 1 & -1 & 2 \\ 0 & 0 & -1 & -3 \\ 2 & -6 & -4 & 3 \\ 5 & -5 & 2 & 0 \end{array}\right] = 298$
#### Example 21 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -4 & 3 & 2 & 6 \\ -1 & 0 & -1 & -3 \\ 5 & 0 & 5 & -1 \\ 6 & 0 & 3 & -5 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -4 & 3 & 2 & 6 \\ -1 & 0 & -1 & -3 \\ 5 & 0 & 5 & -1 \\ 6 & 0 & 3 & -5 \end{array}\right] = -144$
#### Example 22 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -1 & 6 & 0 & 1 \\ -1 & 0 & -3 & 0 \\ 6 & -1 & -3 & -3 \\ -2 & -3 & -2 & 3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -1 & 6 & 0 & 1 \\ -1 & 0 & -3 & 0 \\ 6 & -1 & -3 & -3 \\ -2 & -3 & -2 & 3 \end{array}\right] = -337$
#### Example 23 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -6 & -4 & 4 & -5 \\ -1 & -6 & -1 & 6 \\ 4 & -2 & 0 & 2 \\ 0 & 0 & 2 & 1 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -6 & -4 & 4 & -5 \\ -1 & -6 & -1 & 6 \\ 4 & -2 & 0 & 2 \\ 0 & 0 & 2 & 1 \end{array}\right] = 600$
#### Example 24 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -4 & -3 & 0 & -5 \\ 1 & 1 & -2 & -3 \\ -5 & -6 & 0 & 6 \\ 3 & -6 & 0 & 5 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -4 & -3 & 0 & -5 \\ 1 & 1 & -2 & -3 \\ -5 & -6 & 0 & 6 \\ 3 & -6 & 0 & 5 \end{array}\right] = -786$
#### Example 25 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -4 & 5 & 4 & -5 \\ -3 & -4 & 2 & 5 \\ -2 & 5 & 3 & 2 \\ 1 & 0 & 3 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -4 & 5 & 4 & -5 \\ -3 & -4 & 2 & 5 \\ -2 & 5 & 3 & 2 \\ 1 & 0 & 3 & 0 \end{array}\right] = -868$
#### Example 26 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 0 & 0 & 4 & -5 \\ 4 & 3 & 3 & 2 \\ -3 & 0 & -4 & 6 \\ 5 & 1 & 6 & 5 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 0 & 0 & 4 & -5 \\ 4 & 3 & 3 & 2 \\ -3 & 0 & -4 & 6 \\ 5 & 1 & 6 & 5 \end{array}\right] = 425$
#### Example 27 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -3 & 6 & -4 & 3 \\ -4 & 6 & 1 & -1 \\ 1 & 0 & -3 & 0 \\ 6 & -3 & -2 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -3 & 6 & -4 & 3 \\ -4 & 6 & 1 & -1 \\ 1 & 0 & -3 & 0 \\ 6 & -3 & -2 & 0 \end{array}\right] = 246$
#### Example 28 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 4 & 2 & -6 & -3 \\ -2 & 3 & -5 & -1 \\ -1 & -6 & 1 & -5 \\ 0 & -3 & 0 & 1 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 4 & 2 & -6 & -3 \\ -2 & 3 & -5 & -1 \\ -1 & -6 & 1 & -5 \\ 0 & -3 & 0 & 1 \end{array}\right] = -721$
#### Example 29 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -3 & -4 & 2 & -3 \\ 4 & -3 & 2 & -2 \\ 1 & 3 & 0 & 0 \\ 4 & -3 & 0 & -2 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -3 & -4 & 2 & -3 \\ 4 & -3 & 2 & -2 \\ 1 & 3 & 0 & 0 \\ 4 & -3 & 0 & -2 \end{array}\right] = -110$
#### Example 30 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -5 & 2 & -1 & 1 \\ -4 & 1 & 0 & -4 \\ 3 & 2 & 0 & 4 \\ 0 & 5 & -3 & 2 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -5 & 2 & -1 & 1 \\ -4 & 1 & 0 & -4 \\ 3 & 2 & 0 & 4 \\ 0 & 5 & -3 & 2 \end{array}\right] = -187$
#### Example 31 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 0 & 2 & 0 & -5 \\ -2 & 2 & -2 & 4 \\ -4 & 2 & 0 & 5 \\ -1 & -3 & 1 & 4 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 0 & 2 & 0 & -5 \\ -2 & 2 & -2 & 4 \\ -4 & 2 & 0 & 5 \\ -1 & -3 & 1 & 4 \end{array}\right] = -64$
#### Example 32 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 3 & 4 & 0 & 4 \\ -3 & 3 & 1 & -1 \\ 4 & 0 & 0 & -3 \\ -6 & -4 & 2 & -4 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 3 & 4 & 0 & 4 \\ -3 & 3 & 1 & -1 \\ 4 & 0 & 0 & -3 \\ -6 & -4 & 2 & -4 \end{array}\right] = 218$
#### Example 33 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 4 & -6 & -2 & 4 \\ -3 & -3 & -5 & -4 \\ 6 & -3 & -6 & 0 \\ 0 & 2 & 1 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 4 & -6 & -2 & 4 \\ -3 & -3 & -5 & -4 \\ 6 & -3 & -6 & 0 \\ 0 & 2 & 1 & 0 \end{array}\right] = 84$
#### Example 34 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -2 & -1 & 0 & -1 \\ 4 & -1 & 2 & -2 \\ -1 & 0 & 3 & 0 \\ 2 & 6 & -1 & 2 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -2 & -1 & 0 & -1 \\ 4 & -1 & 2 & -2 \\ -1 & 0 & 3 & 0 \\ 2 & 6 & -1 & 2 \end{array}\right] = -111$
#### Example 35 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 6 & 1 & -2 & 1 \\ 0 & 3 & 1 & 0 \\ -1 & -6 & -6 & -2 \\ 5 & -6 & 1 & 2 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 6 & 1 & -2 & 1 \\ 0 & 3 & 1 & 0 \\ -1 & -6 & -6 & -2 \\ 5 & -6 & 1 & 2 \end{array}\right] = 71$
#### Example 36 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 5 & 4 & 6 & 5 \\ -6 & -6 & 4 & -6 \\ 3 & 2 & 0 & 6 \\ 0 & -2 & 0 & 1 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 5 & 4 & 6 & 5 \\ -6 & -6 & 4 & -6 \\ 3 & 2 & 0 & 6 \\ 0 & -2 & 0 & 1 \end{array}\right] = -292$
#### Example 37 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 2 & 6 & 2 & -2 \\ -1 & 5 & 0 & 1 \\ 1 & -1 & 6 & -5 \\ 0 & -2 & 1 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 2 & 6 & 2 & -2 \\ -1 & 5 & 0 & 1 \\ 1 & -1 & 6 & -5 \\ 0 & -2 & 1 & 0 \end{array}\right] = 80$
#### Example 38 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -3 & 4 & 0 & -6 \\ 4 & -3 & 0 & 1 \\ -1 & 3 & 3 & 6 \\ 3 & 2 & 1 & 2 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -3 & 4 & 0 & -6 \\ 4 & -3 & 0 & 1 \\ -1 & 3 & 3 & 6 \\ 3 & 2 & 1 & 2 \end{array}\right] = -203$
#### Example 39 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -4 & 0 & 0 & 4 \\ -4 & -2 & -1 & -6 \\ -3 & 1 & 2 & 4 \\ 6 & 0 & 5 & -4 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -4 & 0 & 0 & 4 \\ -4 & -2 & -1 & -6 \\ -3 & 1 & 2 & 4 \\ 6 & 0 & 5 & -4 \end{array}\right] = 184$
#### Example 40 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -3 & -5 & -1 & 3 \\ 3 & -3 & 4 & 3 \\ 6 & 0 & -3 & -2 \\ 0 & -2 & -1 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -3 & -5 & -1 & 3 \\ 3 & -3 & 4 & 3 \\ 6 & 0 & -3 & -2 \\ 0 & -2 & -1 & 0 \end{array}\right] = 168$
#### Example 41 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 6 & -3 & -2 & -3 \\ 6 & -5 & -1 & 4 \\ 3 & 4 & 0 & -5 \\ -2 & 4 & 0 & 6 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 6 & -3 & -2 & -3 \\ 6 & -5 & -1 & 4 \\ 3 & 4 & 0 & -5 \\ -2 & 4 & 0 & 6 \end{array}\right] = -540$
#### Example 42 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 3 & 0 & 4 & -3 \\ -2 & -6 & -2 & -6 \\ 1 & 0 & 3 & 0 \\ 5 & -6 & 1 & 1 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 3 & 0 & 4 & -3 \\ -2 & -6 & -2 & -6 \\ 1 & 0 & 3 & 0 \\ 5 & -6 & 1 & 1 \end{array}\right] = -534$
#### Example 43 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -1 & -3 & 1 & -5 \\ -3 & 6 & -4 & 6 \\ 5 & 3 & -4 & -4 \\ 0 & -1 & -3 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -1 & -3 & 1 & -5 \\ -3 & 6 & -4 & 6 \\ 5 & 3 & -4 & -4 \\ 0 & -1 & -3 & 0 \end{array}\right] = 731$
#### Example 44 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 5 & -1 & 0 & -4 \\ -3 & 2 & 1 & -6 \\ 0 & 5 & 0 & 2 \\ 6 & 5 & 2 & -3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 5 & -1 & 0 & -4 \\ -3 & 2 & 1 & -6 \\ 0 & 5 & 0 & 2 \\ 6 & 5 & 2 & -3 \end{array}\right] = -431$
#### Example 45 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -3 & 1 & 4 & -4 \\ 3 & -5 & 3 & -5 \\ 0 & 0 & 2 & 1 \\ -4 & 3 & 0 & -3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -3 & 1 & 4 & -4 \\ 3 & -5 & 3 & -5 \\ 0 & 0 & 2 & 1 \\ -4 & 3 & 0 & -3 \end{array}\right] = -5$
#### Example 46 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -2 & 3 & -3 & 1 \\ 6 & 4 & 4 & 0 \\ 4 & -1 & 0 & 0 \\ -4 & -4 & -3 & 3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -2 & 3 & -3 & 1 \\ 6 & 4 & 4 & 0 \\ 4 & -1 & 0 & 0 \\ -4 & -4 & -3 & 3 \end{array}\right] = 332$
#### Example 47 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -5 & 0 & 3 & -2 \\ -5 & -6 & -6 & 4 \\ 1 & -2 & 0 & 0 \\ -6 & 1 & 6 & -2 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -5 & 0 & 3 & -2 \\ -5 & -6 & -6 & 4 \\ 1 & -2 & 0 & 0 \\ -6 & 1 & 6 & -2 \end{array}\right] = 216$
#### Example 48 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -5 & -4 & -2 & -3 \\ -5 & -1 & 0 & 1 \\ 0 & -1 & -3 & 0 \\ 2 & 2 & -6 & -3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -5 & -4 & -2 & -3 \\ -5 & -1 & 0 & 1 \\ 0 & -1 & -3 & 0 \\ 2 & 2 & -6 & -3 \end{array}\right] = -307$
#### Example 49 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} 6 & -6 & 1 & 0 \\ -4 & -2 & 2 & 0 \\ -6 & 4 & -1 & -1 \\ -3 & 4 & 1 & 0 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} 6 & -6 & 1 & 0 \\ -4 & -2 & 2 & 0 \\ -6 & 4 & -1 & -1 \\ -3 & 4 & 1 & 0 \end{array}\right] = -70$
#### Example 50 π
Show how to compute the determinant of the matrix
$A= \left[\begin{array}{cccc} -6 & -2 & -1 & 3 \\ -2 & 0 & 0 & -4 \\ 1 & 1 & 3 & -5 \\ -4 & 0 & 1 & 3 \end{array}\right] .$
.
$\operatorname{det}\ \left[\begin{array}{cccc} -6 & -2 & -1 & 3 \\ -2 & 0 & 0 & -4 \\ 1 & 1 & 3 & -5 \\ -4 & 0 & 1 & 3 \end{array}\right] = -108$
|
# Summation Series to Product Series Part 1.
Here's something interesting I found out last year. I don't know if its been done before but I strongly suspect it has.
Consider a summation series of form $S(n) = \sum_{k = 1}^{k = n}\frac{f(k)}{k!}$ such that $f(1) = 1$.
Now let's do something unusual and create the following sequence starting with the term $a_2 = 2$ and then $a_{k+1} = (k + 1)(a_k + f(k))$ for all $k > 2$.
Finally we'll focus our attention on the following product series $P(n) = \prod_{k=2}^{k=n} ( 1 + \frac{f(k)}{a_k})$ for all $n > 2$.
We'll show that $S(n) = P(n)$ for all $n > 2$.
Proof : By induction on n. We'll start with the base case. Base Case: $n = 2$
We have $P(2) = 1 + \frac{f(2)}{a_2} = 1 + \frac{f(2)}{2} =\frac{f(1)}{1} + \frac{f(2)}{2} = S(2)$ as required.
Induction Step
This part is a quite long, so we'll break it down into 3 parts.
Part 1 : Re-arranging the Factors of the Product Series
We can write each factor from our product series as \begin{aligned} 1 + \frac{f(k)}{a_k} &= \frac{a_k}{a_k} + \frac{f(k)}{a_k} \\&= \frac{a_k + f(k)}{a_k} \\&= \frac{a_{k+1}}{(k+1)a_k} \end{aligned} by using the definition of our sequence.
Part 2 : Re-writing our Product Series
Now we have that \begin{aligned} \prod_{k = 2}^{k = n} (1 + \frac{f(k)}{a_k}) &= \frac{a_{3}}{(3)a_2} \times \frac{a_{4}}{(4)a_3} \times ...\times \frac{a_{n+1}}{(n+1)a_n} \\ &= \frac{a_{n+1}}{a_2 \times 3 \times 4 \times ... \times n \times (n+1)} \\&=\frac{a_{n+1}}{(n+1)!} \end{aligned}
Part 3 : Final Step
Suppose $P(q) = S(q)$ for some natural number $q$.
Then\begin{aligned} P(q+1) &= P(q)(1 + \frac{f(q+1)}{a_{q+1}}) \\&= P(q) + \frac{f(q+1)}{a_{q+1}}\times\frac{a_{q+1}}{(q+1)!} \\&= P(q) + \frac{f(q+1)}{(q+1)!} \\&= T(q) + \frac{f(q+1)}{(q+1)!} = T(q+1) \end{aligned} as required.
End of Proof
We'll go through some of the consequences of this in the next note!
Note by Roberto Nicolaides
6 years, 5 months ago
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## IB Math AA SL Paper 1 Question Bank
The IB Math AA SL Paper 1 Question Bank is a great resource for students who are preparing for the IB Math AA SL exam. The Question Bank contains a wide variety of questions that cover all of the topics on the exam, and it is a great way to get extra practice before the big day. The Question Bank is also a great way to see how other students have tackled the same problems, and it can give you some insight into different approaches to solving problems.
Section A
## 1.) A ball is thrown vertically upwards from the top of a building 80 meters tall. The height of the ball above the ground, in meters, after t seconds is given by the function h(t) = 80 + 20t – 5t2.
a) Write down the equation of the motion of the ball in the form of h(t) = v0t + 1/2at2
The equation of motion is already given in the form of h(t) = v0t + 1/2at2 with v0 = 20 m/s and a = -5 m/s2
b) At what time will the ball reach its maximum height?
To find the time at which the ball reaches its maximum height, we need to find the time at which the ball’s velocity is zero. The ball’s velocity is given by the first derivative of the height function, h'(t) = 20 – 10t. Setting this equal to zero and solving for t gives
0 = 20 – 10t, t = 2s
so, after 2s the velocity will be zero and thus the ball has reached its max height
c) How high is the maximum height?
To find the maximum height, we substitute the value of t = 2 into the height function:
h(2) = 80 + 20(2) – 5(2)2 = 80 + 40 – 20 = 100m
so the maximum height of the ball is 100 meters.
## 2.) The function f(x) = 3x2 – x3 is defined on the interval [-1, 1].
a) Find the x-coordinates of any stationary points of f(x)
To find the x-coordinates of any stationary points of f(x), we take the derivative of the function and set it equal to zero:
f'(x) = 6x – 3x2 = 3x(2-x) = 0
Solving for x, we get:
x = 0 or x = 2
Thus, the x-coordinate of the stationary points of f(x) is x = 0 and x = 2
b) Determine the nature of the stationary points found in part (a).
To determine the nature of the stationary points, we need to take the second derivative of the function and test it at the x-coordinates of the stationary points:
f”(x) = 6 – 6x
When x = 0, f”(x) = 6 >0 which indicate a Minimum Point
When x = 2, f”(x) = -6 < 0 which indicate a Maximum Point
c) Find the y-coordinate of the stationary point found in part (a)
To find the y-coordinate of the stationary point found in part (a), we substitute the value of x = 0 into the function:
f(0) = 3(0)2 – (0)3 = 0
So, the y-coordinate of the stationary point is (0,0)
So, this is a Minimum Point and at (0,0)
## 3.) Use the sine rule to find the size of side a.
Using sineA/a = sinB/b
a= (10sin45°)/sin30° = 10√2
## 4.) Consider the quadratic 4x2 – 120x + 800
a)Solve the following
i) Find the roots.
x=10, x=20
ii) Hence express the quadratic in the form y= a(x-x1)(x-x2)
y= 4(x-10)(x-20)
b) Solve the following
i) Find the coordinates of the vertex.
(15,-100)
ii) Hence express the quadratic in the form y= a(x-h)2+k
y= 4(x-15)2-100
iii) Write down the equation of the axis of symmetry
x=15
iv) Write down the minimum value of y
Ymin= -100
c) Write down the y-intercept of the quadratic
y=800
## 5.) Let g(x) = 2x sinx
a)Find g’(x)Â
g’(x) = 2sinx + 2xcosx
b) Find the gradient of the graph where x= π
g’(π) = 2sinπ +2πcosπ = -2π
## 6.) Solve the equations
a) logx + log(x+1) = log6
Using property: log(A) +log(B) = log( AB)
x(x+1)=6
x2+x-6=0
x=2
b) logx + log(x+3) = 10
Using property: log(A) +log(B) = log( AB)
x(x+3)=10
x2+3x-10=0
x=2
c) log(x+18) – logx = 1
Using property: log(A) -log(B) = log( A/B)
(x+18)/x= 1
Our Guide is written by counselors from Cambridge University for colleges like MIT and other Ivy League colleges.
Section B
## 7.) Let f(x) = x4 – 4x3 + 3x2 + 2x – 1.
a) Find the equation of the tangent line to the graph of y = f(x) at the point (1, -2).
To find the equation of the tangent line at a point (a, f(a)) on the graph of y = f(x), we need to use the following equation: y = f(a) + f'(a)(x – a).
Where f'(a) is the derivative of f(x) evaluated at x = a.
First, we find the derivative of f(x) which is:
f'(x) = 4x3 – 12x2 + 6x + 2
Now we substitute a = 1 into f'(a) to get f'(1) = 4(1)3 – 12(1)2 + 6(1) + 2 = -6
Therefore, the equation of the tangent line at (1, -2) is:
y = -2 + (-6)(x – 1)
y = -2 – 6x + 6
y = -6x + 8
b) Find the equation of the normal to the graph of y = f(x) at the point (1, -2).
To find the equation of the normal line at a point (a, f(a)), we need to use the following equation: y – f(a) = -1/f'(a)(x – a)
We know that f'(1) = -6 from part (a)
Therefore, the equation of the normal line at (1, -2) is:
y + 2 = 1/6(x – 1)
y = 1/6x – 1/3
c) Find the coordinates of the point of intersection of the tangent line and the normal to the graph of y = f(x) at the point (1, -2).
To find the coordinates of the point of intersection of the tangent line and the normal line, we need to solve the system of equations formed by the equations of the tangent line and normal line.
By substituting the equation of the normal line into the equation of the tangent line, we get:
-6x + 8 = 1/6x – 1/3
Solving this equation for x, we get:
x = 3/2
Now we can substitute this value into either of the equations of the tangent line or normal line to find the y-coordinate.
By substituting x = 3/2 into the equation of the tangent line, we get
y = -6x + 8 = -6(3/2) + 8 = -9/2 + 8 = -1/2
Therefore, the point of intersection of the tangent line and the normal line at (1, -2) is (3/2, -1/2)
d) Find the x-intercepts of the graph of f(x).
To find the x-intercepts of the graph of f(x), we set f(x) = 0 and solve for x:
x4 – 4x3 + 3x2 + 2x – 1 = 0
This equation can be factored as:
(x – 1)(x – 1)(x + 1)(x – 1) = 0
Therefore, the x-intercepts are x = 1 (with multiplicity 3) and x = -1.
e) Find the y-intercept of the graph of f(x).
To find the y-intercept of the graph of f(x), we set x = 0:
f(0) = 0 – 4(0) + 3(0) + 2(0) – 1 = -1
Therefore, the y-intercept is (0, -1).
f) Find the critical points of f(x) and classify them as local maximums, local minimums, or points of inflection.
f'(x) = 4x3 – 12x2 + 6x + 2
f”(x) = 12x2 – 24x + 6
Setting f'(x) = 0, we get:
4x3 – 12x2 + 6x + 2 = 0
This equation can be simplified by dividing both sides by 2:
2x3 – 6x2 + 3x + 1 = 0
We can use synthetic division to factor this equation as:
(x – 1)(2x2 – 4x – 1) = 0
Therefore, the critical points are x = 1, x = (2 + sqrt(6))/2, and x = (2 – sqrt(6))/2.
To classify the critical points, we use the second derivative test. For x = (2 + sqrt(6))/2, we have:
f”((2 + sqrt(6))/2) = 12((2 + sqrt(6))/2)2 – 24((2 + sqrt(6))/2) + 6 = -11 – 6sqrt(6) < 0
Therefore, x = (2 + sqrt(6))/2 is a local maximum. For x = (2 – sqrt(6))/2, we have:
f”((2 – sqrt(6))/2) = 12((2 – sqrt(6))/2)2 – 24((2 – sqrt(6))/2) + 6 = -11 + 6sqrt(6) > 0
Therefore, x = (2 – sqrt(6))/2 is a local minimum. For x = 1, we have:
f”(1) = 12(1)2 – 24(1) + 6 = -6 < 0
Therefore, x = 1 is a point of inflection.
g) Sketch the graph of f(x) for -2 ≤ x ≤ 3, clearly indicating any intercepts, critical points, and asymptotes.
To sketch the graph of f(x) for -2 ≤ x ≤ 3, we use the information we have gathered in parts (a)-(c).
First, we plot the x-intercepts, which are x = -1 (with multiplicity 1) and x = 1 (with multiplicity 3). We also plot the y-intercept, which is (0, -1).
Next, we plot the critical points, which are x = 1 (a point of inflection), x = (2 + sqrt(6))/2 (a local maximum), and x = (2 – sqrt(6))/2 (a local minimum).
To determine the behavior of the graph between the critical points, we can use the first and second derivatives of f(x). Note that f”(x) is positive to the left of x = (2 – sqrt(6))/2 and to the right of x = (2 + sqrt(6))/2, and negative between these two points. This means that f'(x) is increasing to the left of x = (2 – sqrt(6))/2 and to the right of x = (2 + sqrt(6))/2, and decreasing between these two points.
Using this information, we can sketch the graph of f(x) as follows:
• To the left of x = (2 – sqrt(6))/2, the graph is increasing and concave up.
• At x = (2 – sqrt(6))/2, there is a local minimum.
• Between x = (2 – sqrt(6))/2 and x = 1, the graph is decreasing and concave up.
• At x = 1, there is a point of inflection.
• Between x = 1 and x = (2 + sqrt(6))/2, the graph is decreasing and concave down.
• At x = (2 + sqrt(6))/2, there is a local maximum.
To the right of x = (2 + sqrt(6))/2, the graph is increasing and concave down.
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-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
1300
200
300
200
## No Preference
100
100
a) Find the probability that a randomly chosen smartphone user prefers Type A.
The probability that a randomly chosen smartphone user prefers Type A is (1300 + 200)/2000 = 0.75.
b) Given that a smartphone user prefers Type B, what is the probability that the user has no preference for Type A or B?
The probability that a smartphone user has no preference for Type A or B, given that they prefer Type B, is 100/500 = 0.2. This is because out of the 500 people who prefer Type B, 100 have no preference for Type A or B.
c) A smartphone user is selected at random. Find the probability that the user prefers Type A or has no preference for Type A or B.
The probability that a smartphone user prefers Type A or has no preference for Type A or B is the sum of the probabilities of these two events. From part (a), the probability that a smartphone user prefers Type A is 0.75. From the table, the probability that a smartphone user has no preference for Type A or B is (100 + 100)/2000 = 0.1. Therefore, the probability that a smartphone user prefers Type A or has no preference for Type A or B is 0.75 + 0.1 = 0.85.
d) The company randomly selects 10 smartphone users to participate in a focus group. What is the probability that exactly 6 of the users prefer Type A?
The number of ways to select exactly 6 people who prefer Type A out of 10 people is given by the binomial coefficient (10 choose 6), which is 210. The probability of each such outcome is (1300/2000)6 x (700/2000)4, since the probability that a person prefers Type A is 1300/2000 and the probability that a person prefers Type B or has no preference is (200 + 100 + 200 + 100)/2000 = 0.7. Therefore, the probability of selecting exactly 6 people who prefer Type A is 210 x (1300/2000)6 x (700/2000)4 = 0.139.
e) Based on the survey results, can the company claim that a majority of smartphone users prefer Type A over Type B at a significance level of 0.05? Justify your answer.
We can use a two-sample proportion z-test to determine if the proportion of smartphone users who prefer Type A is significantly different from the proportion who prefer Type B or have no preference. Let p1 be the proportion of smartphone users who prefer Type A and p2 be the proportion who prefer Type B or have no preference. Then:
p1 = 1300/2000 = 0.65
p2 = (200 + 100 + 200 + 100)/2000 = 0.35
The null hypothesis is that p1 = p2, and the alternative hypothesis is that p1 ≠p2. We can use a significance level of 0.05.
The test statistic is given by:
z = (p1 – p2) / sqrt(pÌ‚(1 – pÌ‚) (1/n1 + 1/n2))
where p̂ is the pooled proportion, given by:
p̂ = (x1 + x2) / (n1 + n2)
and x1 and x2 are the numbers of smartphone users who prefer Type A and Type B or have no preference, respectively.
From part (a), n1 = 2000 and n2 = 500. From the table, x2 = 400. Therefore, x1 = 1300 and p̂ = (1300 + 400) / (2000 + 500) = 0.65.
Substituting these values into the formula for the test statistic, we get:
z = (0.65 – 0.35) / sqrt(0.5 x 0.5 x (1/2000 + 1/500)) = 12.81
Using a standard normal distribution table, we can find that the p-value is very small (less than 0.0001). Therefore, we reject the null hypothesis and conclude that the proportion of smartphone users who prefer Type A is significantly different from the proportion who prefer Type B or have no preference at a significance level of 0.05.
## 9.) Let f(x) = x3 – 9x2 + 24x + 1.
a) Find the first derivative of f(x).
The first derivative of f(x) is:
f'(x) = 3x2 – 18x + 24
b) Determine the x-coordinates of the critical points of f(x).
To find the critical points of f(x), we need to set f'(x) equal to zero and solve for x:
3x2 – 18x + 24 = 0
x2 – 6x + 8 = 0
(x – 2)(x – 4) = 0
x = 2 or x = 4
Therefore, the critical points of f(x) are x = 2 and x = 4.
c) Determine the intervals on which f(x) is increasing and decreasing.
To determine the intervals on which f(x) is increasing and decreasing, we need to look at the sign of the first derivative:
f'(x) = 3x2 – 18x + 24
On the interval (-∞, 2), f'(x) is negative, so f(x) is decreasing.
On the interval (2, 4), f'(x) is positive, so f(x) is increasing.
On the interval (4, ∞), f'(x) is positive, so f(x) is increasing.
Therefore, f(x) is decreasing on the interval (-∞, 2) and increasing on the intervals (2, 4) and (4, ∞).
d) Determine the x-coordinate(s) of the local maximum and minimum point(s) of f(x).
To find the local maximum and minimum points of f(x), we need to examine the sign of the second derivative:
f”(x) = 6x – 18
At x = 2, f”(x) is negative, so there is a local maximum at x = 2.
At x = 4, f”(x) is positive, so there is a local minimum at x = 4.
Therefore, the local maximum point of f(x) is (2, 19) and the local minimum point of f(x) is (4, -7).
e) Find the second derivative of f(x).
The second derivative of f(x) is:
f”(x) = 6x – 18
f) Determine the intervals on which f(x) is concave up and concave down.
To determine the intervals on which f(x) is concave up and concave down, we need to look at the sign of the second derivative:
f”(x) = 6x – 18
On the interval (-∞, 3), f”(x) is negative, so f(x) is concave down.
On the interval (3, ∞), f”(x) is positive, so f(x) is concave up.
Therefore, f(x) is concave down on the interval (-∞, 3) and concave up on the interval (3, ∞).
g) Determine the x-coordinate(s) of the inflection point(s) of f(x).
To find the inflection points of f(x), we need to set f”(x) equal to zero and solve for x:
6x – 18 = 0
x = 3
Therefore, the inflection point of f(x) is (3, -16).
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# Can you please show me the steps as to how the problem `1/2 + 3/8` equals to `4/8+3/8` as the answer.
## Expert Answers
It looks like the question you're asking about looks like this:
`1/2+3/8 = 4/8+3/8`
If this is true, here is how and why you convert the 1/2 to 4/8.
When adding fractions, you can only add them if they have the same denominator (bottom). It's kind of like money, you can't pay for things with both dollars and pesos, you have to convert one to the other depending on whether you're in Mexico or the U.S. In the same way, you can't straight-up add 1/2 and 3/8 without making one or the other fraction have the same denominator.
The way you convert denominators is by multiplying fractions by 1. Keep in mind, when you multiply by 1, you don't change anything. Here's the catch, though: When you multiply by 1, you actually multiply by some number over itself. For example:
`1 = 2/2 = 3/3 = 4/4 = 5/5`
This means that we can multiply the numerator and denomintor (top and bottom) of any fraction by the same number and we haven't messed with the problem at all!
In our case, we want to make the denominator of 1/2 or the denominator of 3/8 equal to the other denominator. For simplicity, we're going to convert 1/2 to a number with an 8 in the denominator. To do this, we're going to have to multiply by 4/4. Again, if we want to change a number on the bottom by multiplying, we need to multiply by the same number on top so we don't change the problem.
Here's how that'll look:
`1/2 * 4/4 = (1*4)/(2*4) = 4/8`
and that's where the 4/8 came from. To complete the problem, we'll just subsitute 4/8 for the 1/2 in the original problem and solve:
`4/8 + 3/8 = 7/8`
And we're done! Good luck!
The link below gives some other examples of how to add fractions with unequal denominators.
Approved by eNotes Editorial Team
You want to add 1/2 and 3/8.
Imagine a pie cut into eight pieces. Half of the pie (1/2) is equal to 4 pieces, and three eighths of the pie (3/8) is equal to 3 pieces. So together you have 7 pieces, or 7/8 of the pie.
But to add fractions in general, you do stuff like this:
1. Find a common denominator. Here, 2 and 8 both go into 8, so the common denominator is 8.
2. Transform one or both fractions to have that common denominator. Here, we just need to fix up the 1/2:
`1/2 + 3/8 = `
`4/4*1/2+3/8=`
`4/8+3/8=`
3. Just add up the numerators! (Reduce if you need to, write as a mixed number if you want to.)
`7/8`
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# Everything You Need To Know About Multiplication and Division of Fractions
The Difference between Arithmetic and simple mathematics is confusing to people including kids. Arithmetic is a branch of mathematics that only deals with the study of numbers. Maths includes everything. The basic or the fundamentals of Mathematics are addition, subtraction, multiplication, and division.
Many cases talk about their experiences of kids loving addition and subtraction. But, when it came to multiplication it did not favor them much. Further in this article, we will put our focus on multiplication and division of fractions.
Before moving ahead with the topic the kids must know what is multiplication? Multiplication is denoted by the symbol ‘x’ or by an asterisk. In simple, the multiplication of whole numbers can be said as repeated addition. We could say that the multiplication of two numbers is the same as adding many copies to them.
The simple theory of multiplication is as follows:-
3×4=12, when broken down it, will look 4+4+4=12. This is the basics of multiplication.
## Properties of Multiplication
There are various properties of multiplication but, only three of them are considered important. These three properties are highly used in major parts of the world. The three main properties of multiplication are as follows:-
• Commutative multiplication
• Associative multiplication
• Identity multiplication
These are the three main properties of multiplication.
### Commutative multiplication
This property of multiplication says that just changing the order of the factor does not change the product. Here is an example:
4×3=3×4
Multiplying both sides would give the same result. The first paragraph contains an example for this.
### Associative multiplication
These properties of multiplication say that changing the group of factors does not change the product. In fact, the product for three or more numbers will remain the same. But, there are certain terms and conditions for this. Here is an example:-
(2×3) x 4= 2 x (3×4)
Solve “=” first. Solving the first part will result in the following:
(2×3) x 4
=6×4
=24
Now, we will move towards the right-hand side of the problem. Follow the same process as in the first part.
2 x (3×4)
=2 x 12
= 24
Moreover, we can see that both sides equal 24 as the final answer. In fact, we did not even multiply them by the same number. Multiply first half by 2 and 3, and 3 and 4 in the right side.
### Identity multiplication
This is the simplest property of multiplication. In fact, this property of multiplication says that the product of 1 or any number is that number. Moreover, any number multiplied by 1 will be the original number. Here is an example for you:-
8×1=8
Additionally, it does not matter if 1 comes before or after the consequence will be the same. Here is another example for the same:-
1×8=8
Three highly used and common properties of multiplication are –
## What Are The 3 Steps To Multiplying Fractions?
In mathematics, fractions are part of the Arithmetic branch. A fraction consists of a number that expresses a quotient. Moreover, this quotient includes a numerator which is dividing the denominator.
Proper fractions are those in which the numerator is less than the denominator. Improper fractions are those in which the numerator is greater than the denominator. Mixed factors are sums of whole numbers and proper fractions.
You can also add, subtract, or divide a fraction and multiply it. To multiply fractions, one can do it in three simple steps.
### Steps To multiplying fractions:-
1. Simply multiply the top numbers. They are numerators.
2. Multiply the bottom ones after numerators. These lower ones are denominators.
3. If necessary, simplify the fraction.
There is a simple example for multiplication fraction below:-
1/2 x 2/5
First, Multiply the top numbers or numerators.
1/2 x 2/5 = 1×2 =2 (numerator answer)
Second, Multiply bottom numbers or denominators.
1/2 x 2/5 = 1x 2/2×5 = 2/10
Thirdly, if the final result can be simplified into short-form then just simplify it if possible.
2/10= 1/5
The simplification is used with various techniques for the kids to understand better. There is the pizza method, pen, and paper method, the rhyme method, and many more.
## Multiplying Fractions with Whole Numbers
Fractions and multiplications can be done with various numbers and types. The same way fraction with whole numbers becomes slightly different but, easy. In fact, it is one of the easiest ways to solve a whole number fraction.
The example for a whole number with a fraction is given below:-
5 x 2/3 here 5 will be counted as 5/1
2/3 x 5/1
Firstly, again we have to follow the first step to multiply the numerators.
2 x 5 will be the multiplication for the numerator and 3 x 1 will be for the denominator.
So, the final answer to them would be 10/3.
In fact, the same can be in another way where we do not take any denominator under a whole number. But, that might be slightly confusing for the kids to understand fractions at this age. Moreover, there is multiplication with mixed fractions as well. Overall, there might be various types but, the steps for them remain the same.
## How Do You Multiply Fractions With Different Denominators?
Multiplying fractions with numerators is quite easy but when it comes to denominators that become quite tough. This is especially for the kMultiplying fractions with numerators is easy, but denominators are quite tough. This is especially true for kids who are in the fourth-seventh grade. We know that in every fraction, there is a top number and a bottom number to deal with.
The numerator in the fraction tells us how many units we have of a whole. On the other hand, the denominators tell us how many units make up the whole. For example, if we take 2/3, 2 here is the numerator and 3 is the denominator.
We can see that there are two units as a whole, but when it comes to a fraction that is not the case. Firstly, to multiply fractions the basic way has been discussed above. Additionally, fractions will be given on both sides and they will need to be multiplied. In fact, the multiplication sequence will be numerator x numerator and denominator x denominator.
### Steps to Multiply Fractions with Different Denominators
Unlike denominators is also very easy to multiply. One can easily do a simple multiplication with unlike denominators. The steps to do so are quite the same as multiplication of like fractions. Below is an example of the multiplication of fractions with unlike denominators.
Example: Multiplication fraction of 4/12 x 16/24
There are two different methods to solve the above-mentioned problem. The first one is given below:
1. Multiply the numerators to make it simple, 4 x 16= 64
2. Follow the same procedure to multiply the denominators, 12 x 24= 288
3. The final answer that we get here by solving the fraction is 64/288. Moreover, this number can be reduced into a much simpler form. Therefore, we will then get 2/9 which is the final answer.
### Alternate Method
Interestingly, the same example with the same numbers can be solved by another simple method. Moreover, in this method, we will simplify the fractions among themselves. After doing this, we will be multiplying the numerators then, the denominators will be multiplied.
Example: Multiplication fraction of 4/12 x 16/24
Step 1. Simplify the fractions among themselves without multiplication. So, the fraction can be reduced to 1/3 x 2/3. This is the first and simple step to reduce and simplify the fraction.plication. So, the fraction can be now reduced to 1/3 x 2/3. This is the first and simple step to reduce and simplify the fraction.
Step 2. Simplify the numerator. 1 x 2= 2
Step 3. The denominators would need to be simplified. In fact, the denominator cannot be simplified prior to the numerators. Unfortunately, doing so will just cause a mess to the fraction and the result will be wrong. The denominators are, 3 x 3 = 9.
Step 4. Therefore, the final answer by solving the fraction we get is 2/9.
## Fractions with Mixed Numbers Multiplication
Mixed fractions are quite different to solve as compared to other variants. Moreover, mixed fractions consist of a whole number and a proper fraction. In fact, the fraction needs to be converted with the whole number by multiplication. 23/4 is a mixed fraction, where 2 is a whole number and ¾ is a proper fraction.
Firstly, to multiply the mixed fraction, we need to change the mixed fraction into a simple fraction. Now, for example, if the mixed fraction is 22/3, we can change that into 8/3. An example is given below for a better understanding.ge the mixed fraction into a simple fraction. Now, for example, if the mixed fraction is 22/3, we can change that into 8/3. An example is given below for a better understanding.
Example: Fraction multiplication of 22/3 and 31/4
1. The first step in this mixed fraction will be to convert it into a simple fraction. The whole number 2 will be multiplied by the denominator 3 which will result in 6. Moreover, after this, result 6 will need to be added by the numerator 2 that is 6+2=8. So, the answer to the first problem will be 8/3 x 13/4.
2. Now, the numerators of the improper fractions will be multiplied followed by denominators. The final result after that will come as 104/12.
3. Now, simply convert the fraction to a much simpler form by dividing the denominator with the numerator. Here, that is possible and the answer will be 26/3.
4. Interestingly, the final answer can again be converted back to a mixed fraction. So, doing that the final result will be 82/3.
This is how multiplication with mixed fractions is done. Additionally, there are other ways and techniques to do it but, this is the best and simple way.
## Improper Fractions of Multiplication
We have learned two types of fractions and how to multiply with them. Moreover, even fractions with different denominators are very easy to multiply. But, multiplying improper fractions can be a little tricky. This is where the fractions need to be simplified and again bring the result back to mixed fractions.
Moreover, when there are two improper fractions to multiply, we frequently end up with an improper fraction. Let’s take an example with two improper fraction multiplication.
Example: 3/2 x 7/5
Step 1: Firstly, the numerators will need to be multiplied and followed by the denominators. So, (3 x 7)/ (2 x 5) = 21/10
Step 2: Interestingly, the result of solving the above question leads to an improper fraction. In fact, this improper fraction cannot be reduced into a much simpler form.
Step 3: Therefore, the final answer to the above question is 21/10 which can be changed into a mixed fraction. The result will then be, 21/10.
Improper fractions can be tricky sometimes but, if the base knowledge is correct then that might not happen. Additionally, we have discussed all the forms of fractions with multiplications. Above were the basic terms for conducting multiplications with fractions.
## Dividing Fractions
The Division is one of the important operations under the four mathematical operations. In fact, division more or less works quite the same as subtraction. Moreover, the primary aim for division is to split the large groups into equal smaller groups.
In fact, the division is a primary arithmetic operation where various numbers are combined and divided. Now, these numbers are combined in such a way that it forms a new number. The same way division is used very often when it comes to fractions.
## Division in Fractions
The base formula of division remains the same but, slightly changes when done in fractions. In fact, dividing two fractions is the same as multiplying the first by reciprocal and the second by fraction. Moreover, the first step of dividing fractions is just to find the reciprocal of the second fraction.
The next simple step to follow is to multiply the two numerators followed by the denominators. Finally, one can simplify the fraction if needed, or else the answer will remain as it is.
The example for fraction division is given below:
5/8 ÷ 15/16
1: Firstly, we will substitute the value of the numerators then followed by the denominators.
2: The result after substituting will become: 5/8 ÷ 15/16 = 5/8 x 16/15 = 2/3.
3: Now, if we simplify the above answer then the final answer would turn to 5/8 ÷ 15/16 = 2/3.
This is the basic concept for conducting fractions using division operations in mathematics. Now, we will be talking about how to simplify fractions with whole numbers.
## Division with Whole Number Fractions
The Division of fractions is quite different when they are compared to multiplication. Now, division with whole numbers is quite the same process as multiplication. Firstly, we need to multiply the denominator here of the fraction with the whole number.
In fact, the first step with the whole number will be the same as with multiplication. Still, let us take an example for the following:
2/3 ÷ 4 = 2/3 x1/4
= 1/6
Now, the third step after this step would just be to simplify the result. Therefore, for the above answer, we get 1/6 as the final answer.
## Dividing Fractions with Decimals
Before moving any forward we must know what is decimal. Decimal is a subpart of algebra which is another branch in Mathematics. It can be defined as a number whose whole number part and the fractional part are separated by a point. Now, this separation part of the number is called decimal.
Moreover, the dot that we put in between the numbers is called the decimal point. In fact, the digits following after the dot showcases the value smaller than one.
Now, decimal numbers are a fraction to base 10. In most cases, we can represent the decimal in the fractional form and then divide them. There are two simple steps to divide fractions with decimals and they are given below:
• Firstly, convert the given decimal to a fraction to make it look easier.
• Secondly, and lastly divide both the fractions using the simple method.
Now, if we take the example of, 4/5 ÷ 0.5. Here, we can see 0.5 as the decimal that needs to be divided in the fraction. Interestingly, the 0.5 here can be converted to 5/10 or 1/2. Moreover, now the division in fraction can be done very easily.
So, the simplified question would now be 4/5 by 1/2. Further simplifying the problem would turn into 4/5 ÷ 1/2 = 4/5 x 2/1 = 8/5. This is how decimals can be changed into a fraction and then divided with the other numbers.
## Two Ways of Dividing Fractions
There are three-four ways of dividing fractions but, we will talk about the highly used ones. In fact, the first method of dividing fractions is given above. The next two methods are given below:-https://learn.podium.school/downloads/division-with-unit-fractions-fractions-3/
### Method 1: Cross-Multiplication
Step 1: This method of dividing the fraction is quite simple. Firstly, it consists of multiplying the numerator of the first fraction with the denominator of the second. This will get you a result that needs to be written down in the resulting fraction’s numerator.
Step 2: Secondly, we will then multiply the denominator of the first fraction with the numerator of the second. Again, we will need to write the answer in the resulting fraction’s denominator.
Step 3: Thirdly, after we get an answer for both sides just simplify it if possible.
Now, let us take an example for such a case.
Example: 3/4: 6/10
The first step here would be to multiply the first fraction 3 with the denominator of the second 10. Now, doing that will give us the following fraction: 3 x 10 = 30. This answer will be written in the resulting fraction’s numerator.
Secondly, we have to multiply the denominator of the first fraction 4 with the numerator of the second 6. Now, doing that will result in 4 x 6 = 24. This answer will be written in the resulting fraction’s denominator.
Thirdly, the last step will be to simplify the fraction. Since both the numbers are divisible by 6, so we can simply divide the numerator and denominator by 6. Now, doing that will result in 30 ÷ 6 = 5 and 24 ÷ 6 =.
Moreover, the final result or the answer to the question would be 5/4.
### Method 2: Inverting and Multiplying
This is another great way to solve fractions with division. In fact, we could say that it is a cross-multiplication process but, with a slight change. Let us look at the steps to divide in fractions using this method.
Step 1: The second fraction of the question must be inverted. In simple, you just need to swap the numerator for the denominator.
Step 2: Secondly, you just need to simplify the numerator with any denominator given in the question.
Step 3: Thirdly, and lastly, the interesting part is to multiply them across. This will then get you a different result and if possible just simplify that.
An example for such a method is given below:-
Example: 12/6: 6/4
1: As we have mentioned earlier that we need to invert the second fraction in the question. So, 6/4 will be 4/6.
2: Secondly, the numerators in the question will need to be simplified with the denominators. So, the numerators are:
12 = 2 x 2 x 3
4 =2 x 2
Denominators are:
5 = 5
6 = 2 x 3
Now, we can just simplify the numbers if they come out in common or divisible by any number. Doing this process will make the division method quite easy.
• What are the basic rules for multiplying fractions?
Answer- Moreover, there are two simple rules when it comes to multiplying fractions. The first rule is to multiply numerators, and then the denominators. Now, the second rule is to simplify the obtained fraction and get the final answer.
• Why there is a need for multiplication in fraction?
Answer- A fraction is multiplied because it can break into smaller parts to make it simpler. In fact, these smaller parts can be chosen.
• What is multiplication for kids?
Answer- Multiplication is nothing but taking one number and then adding it together with a number of times.
## Conclusion
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# Solve each of the following quadratic equations:
Question:
Solve each of the following quadratic equations:
(i) $\frac{x}{x-1}+\frac{x-1}{x}=4 \frac{1}{4}, \quad x \neq 0,1$
(ii) $\frac{x-1}{2 x+1}+\frac{2 x+1}{x-1}=2, x \neq-\frac{1}{2}, 1$
Solution:
(i)
$\frac{x}{x-1}+\frac{x-1}{x}=4 \frac{1}{4}, \quad x \neq 0,1$
$\Rightarrow \frac{x^{2}+(x-1)^{2}}{x(x-1)}=\frac{17}{4}$
$\Rightarrow \frac{x^{2}+x^{2}-2 x+1}{x^{2}-x}=\frac{17}{4}$
$\Rightarrow \frac{2 x^{2}-2 x+1}{x^{2}-x}=\frac{17}{4}$
$\Rightarrow 8 x^{2}-8 x+4=17 x^{2}-17 x$
$\Rightarrow 9 x^{2}-9 x-4=0$
$\Rightarrow 9 x^{2}-12 x+3 x-4=0$
$\Rightarrow 3 x(3 x-4)+1(3 x-4)=0$
$\Rightarrow(3 x-4)(3 x+1)=0$
$\Rightarrow 3 x-4=0$ or $3 x+1=0$
$\Rightarrow x=\frac{4}{3}$ or $x=-\frac{1}{3}$
Hence, $\frac{4}{3}$ and $-\frac{1}{3}$ are the roots of the given equation.
(ii)
$\frac{x-1}{2 x+1}+\frac{2 x+1}{x-1}=2$
$\Rightarrow \frac{(x-1)^{2}+(2 x+1)^{2}}{(2 x+1)(x-1)}=2$
$\Rightarrow\left(x^{2}+1-2 x\right)+\left(4 x^{2}+1+4 x\right)=2(2 x+1)(x-1)$
$\Rightarrow 5 x^{2}+2 x+2=2\left(2 x^{2}-x-1\right)$
$\Rightarrow 5 x^{2}+2 x+2=4 x^{2}-2 x-2$
$\Rightarrow x^{2}+4 x+4=0$
$\Rightarrow x^{2}+2 x+2 x+4=0$
$\Rightarrow x(x+2)+2(x+2)=0$
$\Rightarrow(x+2)(x+2)=0$
$\Rightarrow(x+2)=0$ or $(x+2)=0$
$\Rightarrow x=-2$ or $x=-2$
$\Rightarrow x=-2$
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# Stellar Numbers. After establishing the general formula for the triangular numbers, stellar (star) shapes with p vertices leading to p-stellar numbers were to be considered.
Extracts from this document...
Introduction
Olivia Bloch Copenhagen International School 30.05.2011
The aim of this task is to investigate geometric shapes, which lead to special numbers. The simplest example of these are square numbers, such as 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3, and 4.
Triangular numbers are defined as “the number of dots in an equilateral triangle uniformly filled with dots”. The sequence of triangular numbers are derived from all natural numbers and zero, if the following number is always added to the previous as shown below, a triangular number will always be the outcome:
1 = 1
2 + 1 = 3
3 + (2 + 1) = 6
4 + (1 + 2 + 3) = 10
5 + (1 + 2 + 3 + 4) = 15
Moreover, triangular numbers can be seen in other mathematical theories, such as Pascal’s triangle, as shown in the diagram below. The triangular numbers are found in the third diagonal, as highlighted in red.
The first diagrams to be considered show a triangular pattern of evenly spaced dots, and the number of dots within each diagram represents a triangular number.
Thereafter, the sequence was to be developed into the next three terms as shown below.
The information from the diagrams above is represented in the table below.
Term Number (n) 1 2 3 4 5 6 7 8 Triangular Number (Tn) 1 3 6 10 15 21 28 36
Establishing the following three terms in the sequence was done by simply drawing another horizontal row of dots to the previous equilateral and adding those dots to the previous count. However, following the method described earlier can also do this calculation, as shown in the illustration below.
T1 1 = 1
T2 2 + 1 = 3
T3 3 + (2 + 1) = 6
T4 4 + (1 + 2 + 3) = 10
T5 5 + (1 + 2 + 3 + 4) = 15
T6 6 + (1 + 2 + 3 + 4 + 5) = 21
T7 7 + (1 + 2 + 3 + 4 + 5 + 6) = 28
Middle
121
181
As seen in the diagram above, the second difference is the same between the terms, and the sequence is therefore quadratic. This means that the equation Sn = an2 + bn + c will be used when representing the data in a general formula. Since some of the values of Sn have already been established this makes it possible to work out the general formula. The first step is to substitute the established values into the three quadratic equations, as shown below:
Sn = an2 + bn + c
When n = 1, Sn = 1
1 = a(1)2 + b(1) + c
1 = 1a + 1b + c
1 = a + b + c
When n = 2, Sn = 13
13 = a(2)2 + b(2) + c
13 = 4a + 2b + c
When n = 3, Sn = 6
37 = a(3)2 + b(3) + c
37 = 9a + 3b + c
Thereafter polysmlt on the graphic display calculator is used in order to retrieve the values of a, b, and c. The coefficients above are simply plugged into the calculator; this can be seen in the screenshots below
Hence the general formula for the sequence is Sn = 6n2 - 6n + 1, to be certain that this is in fact correct, it can also be worked out as shown below.
The second difference of the sequence is divided by 2, since this is the value of a, thereafter the sequence is extended to T0, as the number highlighted in red is c. To figure out the value of b, the values have been substituted into the quadratic formula as shown below, where after the equation has been solved using simple algebra.
Sn = an2 + bn + c
Sn = 6n2 + bn + 1
S1 = (6)(1)2 + b(1) + 1
1= 6 + b + 1
b = - 6
∴ Sn = 6n2 - 6n + 1
As seen above, this method works out the same general formula, and this is tested below to assess the validity.
Sn = 6n2 - 6n + 1S3 = 6(3)2 – 6(3) + 1S3 = 6(9) – 18 + 1S3 = 54 – 17S3 = 37 Sn = 6n2 - 6n + 1S5 = 6(5)2 – 6(5) + 1S5 = 6(25) – 30 + 1S5 = 150 – 29S5 = 121
The results derived from the general formula are the same as worked out earlier when not applying any formula at all; hence, it is correct and can be applied to terms 3 and 5 and also other values of n.
The previous example was then repeated using other values of p; hence, the number of vertices is being changed from 6 to 5. The 5-stellar number at each stage represents the number of dots in each of the diagrams below.
The information in the diagrams above was collected and is represented in the table below.
Term Number (n) 1 2 3 4 5 6 Stellar Number (Vn) 1 11 31 61 101 151
This information is considered and the pattern in the sequence is represented below, and a quadratic pattern throughout the investigation has become very obvious.
As seen in the diagram above, the second difference is the same between the terms, and the sequence is therefore quadratic. This means that the equation Vn = an2 + bn + c will be used when representing the data in a general formula. Since some of the values of Vn have already been established this makes it possible to work out the general formula. The first step is to substitute the established values into the three quadratic equations, as shown below:
Rn = an2 + bn + c
When n = 1, Vn = 1
1 = a(1)2 + b(1) + c
1 = 1a + 1b + c
1 = a + b + c
When n = 2, Vn = 13
11 = a(2)2 + b(2) + c
11 = 4a + 2b + c
When n = 3, Vn = 6
31 = a(3)2 + b(3) + c
31 = 9a + 3b + c
As seen above, this method works out the same general formula, and this is tested below to assess the validity.
Thereafter polysmlt on the graphic display calculator is used in order to retrieve the values of a, b, and c. The coefficients above are simply plugged into the calculator; this can be seen in the screenshots below
Hence, the general formula for this sequence is Rn = 5n2 - 5n + 11
Vn = an2 + bn + c
Vn = 6n2 + bn + 1
V1 = (5)(1)2 + b(1) + 1
1= 5 + b + 1
b = - 5
∴ Sn = 5n2 - 5n + 1
Vn = 5n2 - 5n + 1V3 = 5(3)2 – 5(3) + 1V3 = 5(9) – 15 + 1V3 = 45 – 14V3 = 31 Vn = 5n2 - 5n + 1V5 = 5(5)2 – 5(5) + 1V5 = 5(25) – 25 + 1V5 = 125 – 24V5 = 101
The results derived from the general formula are the same as worked out earlier when not applying any formula at all; hence, it is correct and can be applied to terms 3 and 5.
This was repeated using yet again another value for p, this time changing the p value from 5 to 4, resulting in the diagrams below.
The information from the diagrams above was collected and is represented in the table below.
Term Number (n) 1 2 3 4 5 6 Stellar Number (Rn) 1 9 25 49 81 121
Conclusion
As seen on the left, the values in the table on the top correspond to the excel document on the bottom. I plugged in the general formula discussed above on the far right hand side, and the outcomes were as displayed on the left. Using other values of p and n, which would allow for a more detailed observation, extended this study, which can be seen below. To ensure that the formula in excel did indeed calculate the correct values, I set up the previous results up to compare, and they did correspond.
This investigation has considered geometric shapes in order to determine how many dots are in the different types of shapes, in this case from triangular to stellar, furthermore extending the investigatory work by changing the stellar number (the value of p). The limitations of this investigation is that the value of n must always be equal to or greater than 1, as it is impossible to have a negative triangle or stellar shape, as it is a physical entity, and it is therefore also impossible to achieve a negative value. In this instance, the values have to be whole numbers as all the dots are whole numbers, and by adding whole numbers together it is impossible to achieve the result of a decimal or a fraction that cannot be simplified. Additionally, all the values of n have not been tested, which means that we cannot be certain that it works for all the infinite values.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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# Related International Baccalaureate Maths essays
1. ## Maths Internal Assessment -triangular and stellar numbers
In other words the general statement for the 6-stellar numbers is (12 x (n - 1)) + 1. However, this expression is not complete as it does not apply to the 6-stellar numbers. Therefore, as previously stated, for the triangular numbers, n-1 = n(n-1)/2.
2. ## Stellar numbers
value added to the next term is +1 greater than the value of the term before. Hence from this the next 3 triangle stages may be derived. In the 6th, 7th and 8th stages, the number of dots become respectively: 21, 28, 36.
1. ## Stellar numbers. The aim of the current investigation is to consider different geometric ...
According to the stages of finding 5-stellar, 7 stellar and different variation of n (1-6), described and proved above, we may consume that the general statement S = 2(n-1) p + S is valid to be used. * To discuss the scope or limitations of the general statement.
2. ## Stellar Numbers Investigation Portfolio.
Sn = S1 + [(n-1) �2] x [2 x AP + (n - 2) AP] Sn = S1 + [(n-1) �2] x [2 x 8 + (n - 2) 8] Stage S2: S2 = S1 + [(n-1) �2] x [2 x 8 + (n - 2)
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S1=1 S2=3 S3=6 S4=10 S5=15 S6=21 S7=28 S8=36 S9=45 S10=55 The first pattern that I noticed when looking at this information was that the order looks like this when broken down: 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5, ... , 1+2+3+4+5+6+7+8+9+10 The first attempt I made was saying the general term was simply (n+1).
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Second bracket is three cards that is ? in the deck of 20, third bracket is one card left in the deck of 20 which is October and the last bracket is representing the amount of cards left that is not ?
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Finally, I will evaluate my data and results to draw conclusions about the correlation between International Baccalaureate students' handedness when writing and Grade Point Average. From my collected data, I predict that I will find that a student's handedness and grade point average are independent.
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## Wednesday, February 3, 2010
### Vedic Mathematics Lesson 38: Quadratic Equations 2
In the previous lesson, we learned about the solution technique for a type of quadratic equation which consists of a sum or difference of reciprocals on each side of the equal-to sign. We found that solving such equations can be much easier than actually expanding out the terms, collecting like terms together, bringing the quadratic equation into the standard form, and then applying the quadratic formula. However, the main obstacle to applying this method to the solution of many such equations is that the right-hand side of the equation may not be readily capable of being split into a pair of reciprocals, similar to the left-hand side.
You can find all my previous posts about Vedic Mathematics below:
Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
10's Complements
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2
Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Subtraction
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Digital Roots
Straight Division I
Straight Division II
Vinculums
Divisibility Rules
Simple Osculation
Multiplex Osculation
Solving Equations 1
Solving Equations 2
Solving Equations 3
Solving Equations 4
Mergers 1
Mergers 2
Mergers 3
Multiple Mergers
Complex Mergers
Simultaneous Equations 1
Simultaneous Equations 2
In this lesson, we will assume that the left-hand side of our given quadratic equation is a sum or difference of reciprocals. The sum could be something as simple as x + 1/x, or it could be 2x + 3 + 1/(2x + 3) or it could even be (3x + 4)/(4x + 3) - (4x + 3)/(3x + 4). The actual numerators and denominators of the two terms are immaterial as long as they are linear, the two terms are reciprocals of each other, and they are added to each other (there is a "+" sign on the left-hand side between the two terms) or subtracted from each other (there is a "-" sign on the left-hand side between the two terms).
Let us also assume that the right-hand side is a single term (we will consider fractions to be a single term, even though they are actually composed of 2 numbers). If it was already expressed as the sum or difference of two reciprocals, then we would have been done. If it was not expressed in a form similar to the left-hand side (either sum of reciprocals if the left-hand side is a sum, or a difference of reciprocals if the left-hand side is a difference), then we will assume that we have done the necessary arithmetic to convert the right-hand side to a single term. Also let us assume that mixed fractions have been converted to improper fractions if necessary. Thus, our quadratic equation looks as below:
(ax + b)/(cx + d) + (cx + d)/(ax + b) = e or
(ax + b)/(cx + d) - (cx + d)/(ax + b) = e
We already know that if e can be expressed as (f/g + g/f) (or (f/g - g/f) as the case may be), then our task is done. All we need to do at that point is to equate (ax + b)/(cx + d) = f/g and (ax + b)/(cx + d) = g/f (or -g/f as the case may be) to get the two solutions to our quadratic equation. Therefore, we will concentrate our efforts purely on expressing e as (f/g + g/f) or (f/g - g/f).
Note that f/g + g/f is actually equal to (f^2 + g^2)/fg. That gives us the first clue about how to proceed in this case. Let us first consider a case where e is actually a fraction with a numerator and a denominator. Thus e is expressed as h/j. And so, h/j = (f^2 + g^2)/fg.
Why is the above observation significant? Let us take a fraction such as 61/30. At first sight, it looks like it would be very difficult to break it down into the sum of two reciprocals. But consider the fact that h = 61 and j = 30. From the last line of the last paragraph, j is also equal to fg. Therefore, let us find all the pairs of factors for 30. They are:
1 x 30
2 x 15
3 x 10 and
5 x 6
The numerator, h is equal to 61. But it is also equal to (f^2 + g^2). Let us calculate (f^2 + g^2) for each of these pairs of factors above:
1 x 30 : 1 + 900 = 901
2 x 15 : 4 + 225 = 229
3 x 10 : 9 + 100 = 109
5 x 6 : 25 + 36 = 61
We find that our actual numerator, 61, is actually equal to (f^2 + g^2) for one of the pairs of factors of the denominator. Thus 61/30 = (5^2 + 6^2)/(5 x 6). This then means that 61/30 = 5/6 + 6/5. Thus, we have found the pair of reciprocals for this value of e. This then leads to the solution of quadratic equation much quicker than if we had started out by cross-multiplying, collecting the terms and doing the other operations required to get the equation into standard form.
What happens if our right-hand side was actually written as 122/60? We know that it is exactly equal to 61/30, but suppose we work with 122/60 as given. Then what happens? Let us go through the same exercise as above and identify the sum of squares of all the pairs of factors for 60. We get:
1 x 60 : 1 + 3600 = 3601
2 x 30 : 4 + 900 = 904
3 x 20 : 9 + 400 = 409
4 x 15 : 16 + 225 = 241
5 x 12 : 25 + 144 = 169
6 x 10: 36 + 100 = 136
We see that none of the pairs of factors actually gave us a sum of squares equal to 122, our numerator. Here is the problem: suppose the denominator, when the fraction is expressed in lowest terms, is factorable into a pair of factors whose sum of squares is equal to the numerator. Then if both the numerator and denominator are multiplied by a whole number that is not a perfect square, then the denominator will not be factorable into a pair of factors whose sum of squares is equal to the numerator. That is why it is extremely important for the fraction to be in lowest terms when we attempt to do this.
What happens if the right-hand side happened to be -61/30 instead of 61/30? Actually, we find that -61/30 = -6/5 - 5/6. So, we can say that the right-hand side is the sum of two reciprocals, both of which are negative. It is not as if (f^2 + g^2) = -61 (this can never be true except for complex values of f and g), it is just that (-f^2 - g^2) = -61. This is important to remember!
Thus, the steps required to convert a fractional value on the right hand side (a value with a numerator and denominator) to the sum of a pair of reciprocals is as below:
1. Express the given fraction in lowest terms
2. If the given fraction is negative, ignore the negative sign for now
3. Find all pairs of factors, f and g, for the denominator
4. Check if (f^2 + g^2) is equal to the numerator for any pair of factors
5. If so, change e into (f/g + g/f) or (-f/g - g/f) (if the given fraction was negative), and solve the quadratic equation
Similarly, note that f/g - g/f is actually equal to (f^2 - g^2)/fg. Taking a case once again where e is actually a fraction with a numerator and a denominator, e can expressed as h/j. And so, h/j = (f^2 - g^2)/fg.
Now, let us take a fraction such as 33/28. At first sight, this also looks like it would be very difficult to break it down into the difference of two reciprocals. But consider the fact that h = 33 and j = 28. From the last line of the last paragraph, j is also equal to fg. Therefore, let us find all the pairs of factors for 28. They are:
1 x 28
2 x 14
4 x 7
7 x 4
14 x 2 and
28 x 1
It is important to write out the last three pairs in this case because difference is not a commutative operation, so we need to check both sides of each pair when doing the subtractions.
The numerator, h is equal to 33. But it is also equal to (f^2 - g^2). Let us calculate (f^2 - g^2) for each of these pairs of factors above:
1 x 28 : 1 - 784 = -783
2 x 14 : 4 - 196 = -192
4 x 7 : 16 - 49 = -33
7 x 4 : 49 - 16 = 33
14 x 2 : 196 - 4 = 192
28 x 1 : 784 - 1 = 783
We find that our actual numerator, 33, is actually equal to (f^2 - g^2) for one of the pairs of factors of the denominator. Thus 33/28 = (7^2 - 4^2)/(7 x 4). This then means that 33/28 = 7/4 - 4/7. Thus, we have found the pair of reciprocals for this value of e. This can then lead to the solution of the quadratic equation much quicker than if we had started out by cross-multiplying, collecting the terms, and doing the other operations required to get the equation into standard form.
The same caution I mentioned earlier about making sure the fraction is in lowest terms applies here also. If the fraction is actually a difference of reciprocals, but its numerator and denominator are multiplied by a common factor that is not a perfect square, then we will face difficulties trying to express it as a difference of reciprocals. So, we have to make sure that the fraction is in lowest terms before we set out on this procedure.
Thus, the steps required to convert a fractional value on the right hand side (a value with a numerator and denominator) to the sum of a pair of reciprocals is as below:
1. Express the given fraction in lowest terms
2. Find all pairs of factors, f and g, for the denominator
3. Check if f^2 - g^2 is equal to the numerator for any pair of factors
4. If so, change e into f/g - g/f and solve the quadratic equation
Notice that in this case, we don't make any special provision for negative right-hand sides. For instance if, instead of 33/28, our right-hand side had actually been -33/28, we would have simply found our difference of reciprocals to be 4/7 - 7/4.
Let us apply these techniques to a few problems so that we can be confident about its application. It will also give us practice in factorizing numbers and calculating sums and differences of the squares of the factors mentally.
First let us solve x + 1/x = 298/140
The left-hand side is a sum of reciprocals. So, we have to express the right-hand side as a sum of reciprocals. We apply the procedure we derived earlier for this process.
1. First, we express the fraction in its lowest terms. This makes the right-hand side 149/70.
2. The fraction is not negative, so we can safely ignore instruction 2
3. The denominator can be expressed as 1 x 70, 2 x 35, 5 x 14 and 7 x 10
4. We can easily verify that (7^2 + 10^2) = (49 + 100) = 149, our numerator
5. Therefore, our right-hand side becomes 10/7 + 7/10. This also means that the solutions to our quadratic equation are 10/7 and 7/10
Next, let us tackle x + 1 + 1/(x + 1) = -219/72. As in the first problem, the left-hand side is a sum of reciprocals, so we have to express the right-hand side as a sum of reciprocals.
1. Once again, we notice that the fraction is not in its lowest terms. Dividing the numerator and denominator by the common factor, 3, gives us -73/24
2. The fraction is negative, so for now, we just consider it to be 73/24
3. The denominator can be expressed as 1 x 24, 2 x 12, 3 x 8 and 4 x 6
4. We can verify that (3^2 + 8^2) = (9 + 64) = 73, our numerator
5. Therefore, the right-hand side can be written as (-8/3 - 3/8). We set x + 1 = -8/3 to find x = -11/3, and x + 1 = -3/8 to find x = -11/8.
Next, let us tackle (3x + 2)/(2x + 1) - (2x + 1)/(3x + 2) = 65/36. This time, the left-hand side is a difference of reciprocals, so we have to express the right-hand side as a difference of reciprocals.
1. We notice that the fraction on the right-hand side is already in lowest terms, so nothing needs to be done as part of step 1
2. The denominator can be factored as 1 x 36, 2 x 18, 3 x 12, 4 x 9, 6 x 6, 9 x 4, 12 x 3, 18 x 2 and 36 x 1
3. We find that (9^2 - 4^2) = (81 - 16) = 65, our numerator
4. So, our equation becomes (3x + 2)/(2x + 1) - (2x + 1)/(3x + 2) = 9/4 - 4/9. Setting (3x + 2)/(2x + 1) = 9/4 gives us x = -1/6. Setting (3x + 2)/(2x + 3) = -4/9 gives us x = -22/35.
We still have not touched upon the question of what happens if we are not able to split the right-hand side into a pair of reciprocals using the method explained here. Perhaps, the number on the right hand side is a whole number. In that case the denominator is 1, and 1 has only one factor, 1 itself. Expressing 1 as 1 x 1, we can calculate the sum of the squares of the factors as 2. The difference of the squares of the factors is 0.
What this means is that if the right-hand side is 2 and the left-hand side is a sum, the required sum of reciprocals on the right-hand side is 1/1 + 1/1. Similarly, if the right-hand side is zero and the left-hand side is a difference, the required difference of reciprocals on the right-hand side is 1/1 - 1/1.
Unfortunately, for any other whole number, things are not as easy as this. And the same is true for many fractions whose denominators don't have factors whose squares add up to the required numerator (or can give the required numerator by subtracting, as the case may be). We will leave these problems to deal with in the next lesson. In the meantime, good luck, and happy computing!
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Percentage Aptitude Questions PDF
Percentage Aptitude Questions PDF
The word “percent” is derived from the latin words “per centum”, which means “per hundred”.
A percentage is a fraction with a denominator of hundred, It is denoted by the symbol %.
Numerator of the fraction is called the rate per cent.
VALUE OF PERCENTAGE:
Value of percentage always depends on the quantity to which it refers: Consider the statement, “65% of the students in this class are boys”. From the context, it is understood that boys from 65% of the total number of students in the class. To know the value of 65%, the value of the total number of student should be known. If the total number of student is 200, then,
The number of boys =130;
It can also be written as (200) × (0.65) =130.
Note that the expressions 6%, 63%, 72%, 155% etc. Do not have any value intrinsic to themselves. Their values depend on the quantities to which they refer.
To express the fraction equivalent to %:
Express the fraction with the denominator 100, then the numerator is the answer.
Example 1:
Express the fraction 11/12 into the per cent.
Solution:
11/12=(11/12×100)/100=(91 2/3)/100=912/3%
To express % equivalent to fraction:
a% =a/100
Example 2:
Express 45 5/6% into fraction.
Solution:
45 5/6% = (45 5/6)/100=275/(6×100)=11/24.
Example 3:
Rent of the house is increased from ` 7000 to `7700. Express the increase in price as a percentage of the original rent.
Solution:
Increase value = Rs 7700 – Rs 7000 = Rs 700
Increase % = (Increas value)/(Original value)×100= 700/7000×100=10
∴ Percentage rise = 10%
Example 4:
The cost of a bike last year was Rs19000. Its cost this year is Rs 17000. Find the per cent decrease in its cost.
Decrease % = (Decreas value)/(Original value) × 100
% decrease = (19000-17000)/19000×100
=2000/19000×100= 10.5%.
∴ Percentage decrease = 10.5%.
If A is x % if C and B is y % of C, then A is x/y × 100% of B.
Example 5:
A positive number is divided by 5 instead of being multiplied by 5. By what per cent is the result of the required correct value?
Solution:
Let the number be 1, then the correct answer = 5
The incorrect answer that was obtained =.
The required % = = 4%
If two numbers are respectively x% and y% more than a third number, then the first number is % of the second and the second is % of the first.
If two numbers are respectively x% and y% less than a third number, then the first number if % of the second and the second is % of the first.
x% of a quantity is taken by the first, y% of the remaining is taken by the second and z% of the remaining is taken by third person. Now, if A is left in the fund, then the initial amount
=(A×100×100×100)/((100-x)(100-y)(100-z)) in the beginning.
x% of a quantity is added. Again, y% of the increased quantity is added. Again z% of the increased quantity is added. Now it becomes A, then the initial amount
=(A×100×100×100)/((100+x)(100+y)(100+z))
Example 6:
3.5% income is taken as tax and 12.5% of the remaining is saved. This leaves Rs. 4,053 to spend. What is the income?
Solution:
By direct method,
Income = (4053×100×100)/((100-3.5)(100-12.5)) = Rs 4800.
If the price of a commodity increases by r%, then reduction in consumption, so as not to increase the expenditure is (r/(100+r)×100)%.
If the price of a commodity decreases by r%, then the increase in consumption, so as not to decrease the expenditure is (r/(100-r)×100)%.
Example 7:
If the price of coal be raised by 20%, then find by how much a householder must reduce his consumption of this commodity so as not to increase his expenditure?
Solution:
Reduction in consumption = (20/(100+20)×100)%
= (20/(100+20)×100)% = 16.67%
POPULATION FORMULA
If the original population of a town is P, and the annual increase is r%, then the population after n years is P(1+r/100)^n and population before n years = P/(1+r/100)^n
If the annual decrease be r%, then the population after n years is P(1-r/100)^n and population before n years = P/(1+r/100)^n
Example 8:
The population of a certain town increased at a certain rate per cent annum. Now it is 456976. Four years ago, it was 390625. What will it be 2 years hence?
Solution:
Suppose the population increases at r% per annum. Then, 390625 (1+r/100)^4 = 456976
(1+r/100)^2 = √(456976/390625)= 676/625
Population 2 years hence = 456976 (1+r/100)^2
= 456976 × 676/625 = 494265 approximately.
Example 9:
The population of a city increase at the rate of 4% per annum. There is an additional annual increase of 1% in the population due to the influx of job seekers. Find percentage increase in the population after 2 years.
Solution:
The net annual increase = 5%
Let the initial population be 100.
Then, population after 2 years = 100×1.05×1.05 = 110.25
Therefore, % increase in population = (110.25-100) = 10.25%
If a number A is increased successively by x% followed by y% and then z%, then the final value of A will be A(1+x/100)(1+y/100)(1+z/100)
In case a given value decreases by an percentage then we will use negative sign before that.
First Increase and then decrease:
If the value is first increased by x% and then decreased by y% then there is (x-y-xy/100)% increase or decrease, according to the +ve or –ve sign respectively.
If the value is first increased by x% and then decreased by x% then there is only decrease which is equal to (x^2/100).
Example 10:
A number is increased by 10% and then it is decreased by 10%. Find the net increase or decrease per cent.
Solution:
% change = (10×10)/100=1%
i.e. 1% decrease.
Average percentage rate of change over a period.
=((New Value-Old Value))/(Old Value)×100/n% where n = period.
The percentage error = (The Error)/(True Value)×100%
SUCCESSIVE INCREASE OR DECREASE
In the value is increased successively by x% and y% then the final increase is given by (x+y+xy/100)%
In the value is decreased successively by x% and y% then the final decrease is given by (-x-y-xy/100)%
Example 11:
The price of a car is decreased by 10% and 20% in two successive years. What per cent of price of a car is decreased after two years?
Solution:
Put x = -10 and y = -20, then
-10-20+
The price of the car decreases by 28%.
STUDENT AND MARKS
The percentage of passing marks in an examination is x%. If a candidate who scores y marks fails by z marks, then the maximum marks M = 100(y+z)/x
A candidate scoring x% in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more then the minimum required passing marks. Then the maximum marks M = 100(a+b)/x
In an examination x% and y% students respectively fail in two different subjects while z% students fail in both subjects then the % age of student who pass in both the subjects will be {100-(x + y – z)}%
Example 12:
Vishal requires 40% to pass. If he gets 185 marks, falls short by 15 marks, what was the maximum he could have got?
Solution:
If Vishal has 15 marks more, he could have scored 40% marks.
Now, 15 marks more then 185 is 185+15 = 200
Let the maximum marks be x, then 40% of x = 200
⇒ × x = 200 ⇒ x =500
Thus, maximum marks = 500
Alternate method:
Maximum marks = (100(185+15))/40=(100×200)/40 = 500
Example 13:
A candidate scores 15% and fails by 30 marks, while another candidate who scores 40% marks, gets 20 marks more then the minimum required marks to pass the pass the examination. Find the maximum marks of the examination.
Solution:
By short cut method:
Maximum marks = (100(30+20))/(40-15) =200
2-DIMENSIONAL FIGURE AND AREA
If the sides of a triangle, square, rectangle, rhombus or radius of a circle are increased by a%, its area is increased by(a(a+200))/100 %
If the sides of a triangle, square, rectangle, rhombus or radius of a circle are decreased by a %
Then its area is decreased by (a(200-a))/100%.
Example 14:
If the radius of a circle is increased by 10%, what is the percentage increase in its area?
Solution:
Let R be the radius of circle.
Area of Circle, A =πR^2
Now, radius is increased by 10%
New radius, R’ = R + 10% of R = 1.1 R
New Area, A’ = π(1.1R)^2= 1.21 πR^2%
increase in area =(1.21πR^2-πR^2)/(πR^2 )×100=21%
Shortcut Method:
So, Area is increased by (10(10+200))/100 = 21%
If the both sides of rectangle are changed by x% and y% respectively, then % effect on area = x + y+xy/100 (+/- according to increase or decrease.
Example 15:
If the length and width of a rectangular garden were each increased by 20%, then what would be the per cent increase in the area of the garden?
Solution:
By direct formula
% increase in area =(20 (20+200))/100=44%
If A’s income is r% more than that of B, then B’s income is less than that of A by (r/(100+r)×100)%
If A’s income is r% less than that of B, then B’s income is more than that of A by (r/(100-r)×100)%
Example 16:
If A’s salary is 50% more than B’s, then by what percent B’s salary is less than A’s salary?
Soluti
Let B’s salary be Rs x
Then, A’s salary = x + 50% of x = 1.5x
B’s salary is less than A’s salary by ((1.5x-x)/1.5x×100)% = 100/3 = 33.33%
Shortcut method,
B’s salary is less than A’s salary by (50/(100+50)×100)%
=50/150×100% = 33.33%
Example 17:
Ravi’s weight is 25% that of Meena’s and 40% that of Tara’s. What percentage of Tara’s weight is Meena’s weight.
Solution:
Let Meena’s weight be x kg and Tara’s weight be y kg. Then Ravi’s weight = 25% of Meena’s weight
= 25/100×x …..(i)
Also, Ravi’s weight = 40% of Tara’s weight
= 40/100×y …..(ii)
From (i) and (ii), we get
25/100×x=40/100×y
⇒ 25x = 40y
⇒ 5x = 8y ⇒ x = 8/5 y
Meena’s weight as the percentage of Tara’s weight
= x/y×100= ( 8/5 y)/y×100
= 8/5×100=160
Hence, Meena’s weight is 160% of Tara’s weight.
Example 18:
The monthly salaries of A and B together amount to `50,000. A spends 80% of his salary and B spends 70% of his salary. If now their saving are the same, then find the salaries of A and B.
Solution:
Let A’s salary by x, then B’s salary (50,000-x)
A spends 80% of his salary and saves 20%
B spends 70% of his salary and saves 30%
Given that
20% of x = 30% of (50,000-x)
20/100×x=30/100×(50,000-x)
50x/100=(30×50,000)/100
⇒ x = (30×50,000×100)/(100×50)=30,000
A’s salary Rs 30,000
B’s salary = Rs 50,000 – Rs 30,000 = Rs20,000
Clock Tricks Questions in Hindi
Formulas and Quick Tricks for Percentage Problems
• a % of b = a * b/100
• If A is x% more than B, then B is less than A by: [x(100) / 100+x]%
• If A is x% less than B, then B is more than A by: [x(100) / 100-x]%
• If A is x% of C and B is y% of C, then A = x/y * B
• If the price of a commodity decreases by P %, then the increase in consumption so that the expenditure remains same, which is: [P(100) / 100-P]%
• If the price of a commodity increases by P%, then the reduction in consumption so that the expenditure remains same, which is: [P(100) / 100+P]%
• If a number is changed (increased/decreased) successively by x% and y%, then net% change is given by [x+y+(xy/100)]%, which represents increase or decrease in value according as the sign is positive or negative
• If two parameters A and B are multiplied to get a product and if A is changed by x% and another parameter B is changed by y%, then the net% change in the product (A * B) is given [x+y+(xy/100)]%
• In an examination, the minimum pass percentage is x%. If a student secures y marks and fails by z marks, then the maximum marks in the examination is 100(y+z)/x
• If a number A is increased successively by x% followed by y% and then by z%, then the final value of A will be: A(100+x / 100) (100+y / 100) (100+z / 100)
Questions and Solved Examples on Percentage Problems
For a candidate to clear an examination, he/she must score 55% marks. If he/she gets 120 and fails by 78 marks, the total marks for the examination is:
A. 300
B. 320
C. 360
D. 400
Explanation:
• Here the mark obtained by the candidate is 120 and the candidate fails by 78 marks
• Therefore the passing marks is (120+78) = 198
• Let the total marks be x. Then,
• => 55/100 * x = 198
• => x = 360
A property decreases in value every year at the rate of 6 1/4% of its value at the beginning of the year. It’s value at the end of 3 years was Rs. 21,093. Find its value at the beginning of the 1st year?
A. Rs. 18,060.36
B. Rs. 18,600
C. Rs. 25,600.24
D. Rs. 32,000.50
Explanation:
• 6 1/4% = 1/16
• x * 15/16 * 15/16 * 15/16 = 21093
• x = 25600.24
80 is what percent of 64?
A. 75%
B. 85%
C. 120%
D. 125%
Explanation: Let x percent of 64 be 80
• 64 * x/100 = 80 => x = (80 * 100)/64 => x = 125
• Hence, 80 is 125% of 64
How much is 80% of 40 is greater than 4/5 of 25?
A. 4
B. 6
C. 9
D. 12
Explanation: (80/100) * 40 – (4/5) * 25
32 – 20 = 12
40% of a number is more than 20% of 650 by 190. Find the number?
A. 600
B. 700
C. 800
D. 900
Explanation: (40/100) * X – (20/100) * 650 = 190
• 2/5 X = 320
• X = 800
40 is subtracted from 60% of a number, the result is 50. Find the number?
A. 110
B. 130
C. 140
D. 150
Explanation:
• (60/100) * X – 40 = 50
• 6X = 900
• X = 150
If A got 80 marks and B got 60 marks, then what percent of A’s mark is B’s mark?
A. 60%
B. 65%
C. 75%
D. 80%
Explanation: A’s marks = 80; B’s marks = 60.
• Let x% of A = B => x/100 * 80 = 60
• => x = (60 * 100)/80 = 75
• B’s marks is 75% of A’s marks
A and B’s salaries together amount to Rs. 2,000. A spends 95% of his salary and B spends 85% of his. If now their savings are the same, what is A’s salary?
A. Rs.500
B. Rs.750
C. Rs.1250
D. Rs.1500
Explanation: (5/100) A = (15/100) B
• A = 3B
• A + B = 2000
• 4B = 2000 => B = 500
• A = 1500
5% people of a village in Sri Lanka died by bombardment, 15% of the remainder left the village on account of fear. If now the population is reduced to 3553, how much was it in the beginning?
A. 3800
B. 4200
C. 4400
D. 5500
Explanation: X * (95/100) * (85/100) = 3553
• X = 4400
In a factory, there are 40% technicians and 60% non-technicians. If the 60% of the technicians and 40% of non-technicians are permanent employees, then the percentage of workers who are temporary is?
A. 32%
B. 42%
C. 52%
D. 62%
Explanation: Total = 100
• T = 40, NT = 60
• 40*(60/100) = 24, 60*(40/100) = 24
• 24 + 24 = 48 => 100 – 48 = 52%
Clocks Practice Questions
मिनट की सुई 6 मिनट मे कितने डिग्री का कोण बनाएगी ?
• 6 डिग्री
• 12 डिग्री
• 24 डिग्री
• 36 डिग्री
12 घंटे मे संपाती कितनी बार बनता है ?
• 12
• 22
• 11
• 24
03:40 बजे मिनट और घंटे की सुई के बीच कितने डिग्री का कोण बनेगा ?
• 320 डिग्री
• 340 या 20 डिग्री
• 360 डिग्री
• कोई नहीं
गाड़ी की सुईयां सायं 2 बजे से 7 बजे तक कितनी बार समकोण बनाएगी ?
• 10 बार
• 9 बार
• 8 बार
• इनमे से कोई नहीं
4 बजकर 15 मिनट पर घंटे और मिनट की सुई के मध्य कितने डिग्री का कोण बनेगा ?
• 30 डिग्री
• 60 डिग्री
• 37.5 डिग्री
• 45 डिग्री
सुबह 4 बजे से 10 बजे के बीच घड़ी की सुईयां कितनी बार ऊपर नीचे होंगी ?
• 5 बार
• 6 बार
• 11 बार
• 23 बार
यदि कोई घड़ी किसी दर्पण मे पौने तीन बजे का समय बता रही है तो वास्तविक समय क्या होगा ?
• 07:30
• 09:15
• 08:45
• 06:45
एक कक्षा मे शिक्षक 09:55 बजे पहुँचा और पप्पू 45 मिनट बाद आया उसे 10 मिनट की देरी हुई टी शिक्षक नियमित समय से कितना समय पहले पहुँचा ?
• 10 मिनट
• 15 मिनट
• 45 मिनट
• 35 मिनट
दो घड़ियों मे से एक 5 मिनट तेज हो जाती है तथा दूसरी घड़ी 5 मिनट सुस्त हो जाती है यदि दोनों घड़िया दोपहर 12 बजे एक समान समय बताती है तो वे पुनः एक साथ कब एक समान समय बताएंगी ?
• 60
• 62
• 76
• 72
जयपुर से हर 30 मिनट मे एक बस दिल्ली के लिए रवाना होती है। एक पुछताछ क्लर्क ने बताया की एक बस अभी 10 मिनट पहले छूटी है और अगली बस 09:35 पर आएगी तो ये सूचना किस वक्त पर दी गई ?
• 09:15 बजे
• 08:55 बजे
• 09:08 बजे
• 09:10 बजे
एक बस पड़ाव मे निम्नलिखित सूचना प्रसारित की जा रही थी, सूरत के लिए 15 मिनट पहले एक बस खुल चुकी है, नियमानुसार प्रत्येक 45 मिनट के बाद सूरत के लिए एक बस है। अगली बस का निर्धारित समय 08:30 AM है बताइए की कितने बजे यह जानकारी प्रसारित की गई थी ?
• 08:00 AM
• 09:00 AM
• 10:45 AM
• 08:15 AM
दिल्ली से आगरा के लिए हर 40 मिनट मे एक बस चलती है। पूछताछ ऑफिसर ने बताया की बस 10 मिनट पहले निकाल चुकी है। दूसरी बस सुबह 10:45 मिनट पर निकलेगी। पूछताछ ऑफिसर ने यात्री को किस वक्त सूचना दी ?
• 09:55 AM
• 10:05 AM
• 10:35 AM
• 10:15 AM
मयंक अपने ऑफिस पर 08:30 पर पहुँच गया जो 15 मिनट समयपूर्व था। बताइए ऑफिस का निर्धारित समय क्या था ?
• 08:30
• 08:15
• 08:45
• 08:05
यदि जल प्रतिबिम्बित समय 10 बजकर 15 मिनट हो रहे हो तो इस स्थिति मे वास्तविक समय क्या होगा ?
• 08:11
• 11:15
• 03:50
• 02:45
यदि किसी घड़ी मे वास्तविक समय 3 बजकर 15 मिनट हो रहा हो तो इस स्थिति मे दर्पण मे समय क्या होगा ?
• 08:45
• 09:15
• 08:15
• 09:45
यदि किसी लम्बवत दर्पण मे प्रतिबिम्बित समय 1 बजकर 40 मिनट हो रहा हो तो वास्तविक समय क्या होगा ?
• 11:40
• 05:45
• 10:20
• 11:20
एक घड़ी 4 बजे का समय दर्शा रही है, घंटे कइ सुई के 90 डिग्री घूमने के बाद क्या समय होगा?
• 6 बजे
• 7 बजे
• 8 बजे
• 9 बजे
यदि किसी घड़ी मे 10 बजकर 10 मिनट हो रहे हो तो इस स्थिति में घंटे और मिनट की सुई के कोण का माप क्या होगा ?
• 105 डिग्री
• 115 डिग्री
• 125 डिग्री
• 75 डिग्री
शाम के चार बजे से शाम के 10 बजे तक घड़ी की सुइयाँ कितनी बार एक-दूसरे के साथ 90 डिग्री का कोण बनती है
• 9
• 12
• 6
• 11
7:20 मिनट पर घड़ी की बड़ी और छोटी सुई के बीच कितने अंश का कोण बनेगा ?
• 160 डिग्री
• 100 डिग्री
• 260 डिग्री
• 120 डिग्री
Maths Questions
Q.1. A sum of Rs. 25000 amounts to Rs. 31000 in 4 years at the rate of simple interest what is the rate of interest?
(a) 3%
(b) 4%
(c) 5 %
(d) 6 %
(e) None of these
Q.2. Kamla took a loan of Rs. 2400 with simple interest for as many years as the rate of interest. If she paid Rs. 864 as interest at the end of the loan period, what was the rate of interest?
(a) 3.6
(b) 6
(c) 18
(d) cannot determined
(e) None of these
Q.3. What is the present worth of Rs. 264 dues in 4 years at 10% simple interest per annum?
(a) 170.20
(b) 166
(c) 188.57
(d) 175.28
Q.4. A sum fetched a total simple interest of Rs. 8016.25 at the rate of 6 p.c.p.a in 5 years what is the sum?
(a) 24720.83
(b) 26730.33
(c) 26720.83
(d) 26710.63
(e) None of these
Q.5. 8. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs 800 become in 3 years?
(a) 1020.80
(b) 1025
(c) 1052
Q. 6. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is
a. 74
b. 78
c. 80
d. 90
Q.7. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __
a. 3
b. 6
c. 9
d. 8.2
Q.8. √6+√6+√6+… is equal to –
a. 2
b. 5
c. 4
d. 3
Q.9. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –
a. 24
b. 32
c. 40
d. 28
Q.10. When (6767 +67) is divided by 68, the remainder is-
a. 1
b. 63
c. 66
d. 67
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# Product Estimation with Whole Numbers and Fractions
## Estimate multiplication
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Practice Product Estimation with Whole Numbers and Fractions
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Product Estimation with Whole Numbers and Fractions
Remember Julie and the rainforest from the Multiply Whole Numbers by Fractions Concept?
Well, Julie calculated that the average amount of rainfall in the rainforest per day was about 18\begin{align*} \frac{1}{8}\end{align*} of an inch. If Julie wanted to figure out how much rainfall there would be for one year, how could she do this?
There are 365 days in a year. If Julie multiplied 365 by one - eighth she could figure it out.
Or she could estimate. Julie could estimate the following product.
365×18\begin{align*}365 \times \frac{1}{8}\end{align*}
How could she do this?
This Concept will teach you how to estimate the products of whole numbers and fractions. By the end of the Concept, you will understand how to help Julie figure this out.
### Guidance
We can estimate products of whole numbers and fractions. When we estimate, we are looking for an answer that is reasonable but need not be exact.
Before we look at how to do it, we need to know that the commutative property applies to multiplying fractions and whole numbers. It doesn’t matter which order you multiply in, the answer will be the same.
6×12=12×6
It doesn’t matter which order we write the numbers in, the answer will remain the same. This is an illustration of the commutative property.
How can we estimate the product of a whole number and a fraction?
To estimate the product, we have to use some reasoning skills.
39×12=
To work on this problem, we have to think about three-ninths. Three-ninths simplifies to one-third. Now we can find one-third of 12. Multiplying by one-third is the same as dividing by three.
516×20=
To estimate this problem, we must think about a fraction that is easy to divide into twenty, but that is close to five-sixteenths. Four-sixteenths is close to five-sixteenths and it simplifies to one-fourth. Twenty is divisible by four, so we can rewrite the problem and solve.
41614×20=14=5
Remember that multiplying by one-fourth is the same as dividing by four, so our answer is five.
Our estimate is five.
Practice a few of these on your own. Estimate these products.
#### Example A
8×38=\begin{align*}8 \times \frac{3}{8} = \underline{\;\;\;\;\;\;\;}\end{align*}
Solution:3\begin{align*}3\end{align*}
#### Example B
12×18=\begin{align*}\frac{1}{2} \times 18 = \underline{\;\;\;\;\;\;\;}\end{align*}
Solution:9\begin{align*}9\end{align*}
#### Example C
34×75=\begin{align*}\frac{3}{4} \times 75 = \underline{\;\;\;\;\;\;\;}\end{align*}
Solution:25\begin{align*}25\end{align*}
Now back to Julie and the rainforest. Here is the original problem once again.
Well, Julie calculated that the average amount of rainfall in the rainforest per day was about 18\begin{align*} \frac{1}{8}\end{align*} of an inch. If Julie wanted to figure out how much rainfall there would be for one year, how could she do this?
There are 365 days in a year. If Julie multiplied 365 by one - eighth she could figure it out.
Or she could estimate. Julie could estimate the following product.
365×18\begin{align*}365 \times \frac{1}{8}\end{align*}
How could she do this?
To do this, we could first round 365 up to 400.
400 is easily divisible by 8.
There is an average of 50 inches of rain in the rainforest per year.
### Vocabulary
Multiplication
“of”
means multiply in a word problem
Product
the answer to a multiplication problem
Estimate
to find a reasonable answer that is not exact but is close to the actual answer.
### Guided Practice
Here is one for you to try on your own.
Estimate the following product.
12×280\begin{align*}\frac{1}{2} \times 280\end{align*}
To figure out this estimate, let's first round 280\begin{align*}280\end{align*}.
280\begin{align*}280\end{align*} rounds up to 300\begin{align*}300\end{align*}.
Now we can easily find half of 300.
Our answer is 150\begin{align*}150\end{align*}.
### Video Review
Multiplying Fractions and Whole Numbers- This video involves skills needed in this Concept.
### Practice
Directions: Estimate the products of the following fractions and whole numbers.
1. 6×12=\begin{align*}6 \times \frac{1}{2} = \underline{\;\;\;\;\;\;\;}\end{align*}
2. 16×12=\begin{align*}16 \times \frac{1}{2} = \underline{\;\;\;\;\;\;\;}\end{align*}
3. 26×12=\begin{align*}26 \times \frac{1}{2} = \underline{\;\;\;\;\;\;\;}\end{align*}
4. 36×13=\begin{align*}36 \times \frac{1}{3} = \underline{\;\;\;\;\;\;\;}\end{align*}
5. 40×15=\begin{align*}40 \times \frac{1}{5} = \underline{\;\;\;\;\;\;\;}\end{align*}
6. 20×14=\begin{align*}20 \times \frac{1}{4} = \underline{\;\;\;\;\;\;\;}\end{align*}
7. 30×12=\begin{align*}30 \times \frac{1}{2} = \underline{\;\;\;\;\;\;\;}\end{align*}
8. 100×110=\begin{align*}100 \times \frac{1}{10} = \underline{\;\;\;\;\;\;\;}\end{align*}
9. 60×13=\begin{align*}60 \times \frac{1}{3} = \underline{\;\;\;\;\;\;\;}\end{align*}
10. 90×13=\begin{align*}90 \times \frac{1}{3} = \underline{\;\;\;\;\;\;\;}\end{align*}
11. 33×111=\begin{align*}33 \times \frac{1}{11} = \underline{\;\;\;\;\;\;\;}\end{align*}
12. 44×14=\begin{align*}44 \times \frac{1}{4} = \underline{\;\;\;\;\;\;\;}\end{align*}
13. 36×112=\begin{align*}36 \times \frac{1}{12} = \underline{\;\;\;\;\;\;\;}\end{align*}
14. 50×125=\begin{align*}50 \times \frac{1}{25} = \underline{\;\;\;\;\;\;\;}\end{align*}
15. 75×34=\begin{align*}75 \times \frac{3}{4} = \underline{\;\;\;\;\;\;\;}\end{align*}
### Vocabulary Language: English
Estimate
Estimate
To estimate is to find an approximate answer that is reasonable or makes sense given the problem.
multiplication
multiplication
Multiplication is a simplified form of repeated addition. Multiplication is used to determine the result of adding a term to itself a specified number of times.
Product
Product
The product is the result after two amounts have been multiplied.
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# Question #bcedf
Sep 28, 2017
Length of each side is 248.53 mm 2.d.p.
Interior angle is ${135}^{o}$
#### Explanation:
I don't know if I will be able to explain this with words alone. This is where you desperately need geometrical diagrams. I'll try my best.
If you construct isosceles triangles on the inside of the octagon the angle at there apex will be:
$\frac{{360}^{o}}{8}$ (8 being the number of sides.)
This gives $\left({45}^{o}\right)$ Take a perpendicular bisector of this angle. This will then bisect the angle and form a right angle with the base (giving you half the length of one of the sides of the octagon). The angle will then be:
$\frac{{45}^{o}}{2} = \left({22.5}^{o}\right)$
We know the length of one side of this triangle, because it will be the length of the radius, which is $\frac{600 m m}{2} = 300$mm.
We now have a right angled triangle with angles ${90}^{o}$ and ${22.5}^{o}$. The third angle will be half of one of the interior angles of the octagon.
Sum of interior angles is:
$180 n - 360$
Where $n$ is number of sides.
So: ${180}^{o} \left(8\right) - {360}^{o} = {1080}^{o}$
Each angle is then: $\frac{{1080}^{o}}{8} = {135}^{o}$
Half this angle to get the third angle of the triangle:
$\frac{{135}^{o}}{2} = {67.5}^{o}$
We now have a right angled triangle with angles ${22.5}^{o} , {67.5}^{o} \mathmr{and} {90}^{o}$. with one side of $300$( this is the radius).
If we label the sides of the triangle a b c ( c being the hypotenuse ) and opposite angles A B C respectively.
We need to find the length of side a.
Using the Sine Rule $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{S \in C}{c}$
$\sin \frac{22.5}{a} = \sin \frac{67.5}{300}$
Rearranged gives:
$a = \frac{300 \sin \left(22.5\right)}{\sin \left(67.5\right)} \implies a = 124.2641$
4 .d.p.
This is HALF the length of one of the sides of the octagon, so:
$124.2641 \times 2 = 248.53$
2 .d.p.
So we have:
Length of octagon side 248.53mm.
Interior angle ${135}^{o}$
Hope this helps you. It would have been much easier for me and for you if we could have used diagrams.
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## Geometric Mean Definition
Geometric mean involves roots and multiplication, not addition and division. You get geometric mean by multiplying numbers together and then finding the ${n}^{th}$ root of the numbers such that the ${n}^{th}$ root is equal to the amount of numbers you multiplied. Geometric mean is useful in many circumstances, especially problems involving money.
The geometric mean is the ${n}^{th}$ root when you multiply $n$ numbers.
For example, if you multiply three numbers, the geometric mean is the third root of the product of those three numbers. The geometric mean of five numbers is the fifth root of their product.
Suppose we said we found the geometric mean using the ${11}^{th}$ root of the numbers. That tells you that 11 numbers were multiplied together. To find the geometric mean of four numbers, what root would we take? The fourth root, of course.
### Geometric Mean vs Arithmetic Mean
You are probably familiar with arithmetic mean, informally called the average of a group of numbers. You get arithmetic mean by arithmetic, or adding the numbers together and then dividing by the amount of numbers you were adding.
## How to Find the Geometric Mean
We will start with an easy example using only two numbers, 4 and 9. What is the geometric mean of 4 and 9?
Multiply . Then find the square root of their product (because you only multiplied two numbers):
The geometric mean of 4 and 9 is 6.
## Geometric Mean Formula
Let $n$ equal the number of terms we are multiplying, and let ${x}_{1}$, ${x}_{2}$, ${x}_{3}$ and so on up to ${x}_{n}$ be the different factors (the various terms).
### Geometric Mean Theorem
This formula tells us to multiply all the terms (radicands) within the radical (the symbol for roots), and then to find the ${n}^{th}$ root of them where $n$ is how many radicands you have. You can separate whole number radicands with either an $×$ or a $*$ to show you are multiplying them.
Let's first try it with our earlier, easy example, and here the $×$ is the symbol of multiplication:
We can substitute a $*$ for the $×$ to also show multiplication:
Now let's try a quick example with three terms:
The product of 3 x 6 x 12 = 216.
$\sqrt[3]{216}$
The cube root (the third root) of 216 is 6.
Our geometric mean is 6.
## Uses for the Geometric Mean
Anytime we are trying to calculate average rates of growth where growth is determined by multiplication, not addition, we need the geometric mean. This connects geometric mean to economics, financial transactions between banks and countries, interest rates, and personal finances.
Your growth rate for money you have in bank deposits can be calculated using geometric mean, since your money grows at an advertised rate. You could not calculate this using arithmetic mean.
## Geometric Mean Examples
The best way to become familiar with using the geometric mean is to use it. Use the formula to find the geometric mean of these six numbers:
Here is the formula again:
And here is the formula with our numbers:
The product of the radicands is found easily:
Now you must find the sixth root of 7,200:
$\sqrt[6]{7,200}$
The sixth root of . Our work is done!
## Lesson Summary
In this lesson we learned how to define the geometric mean, which is the ${n}^{th}$ root of a group of $n$ factors, how to find the geometric mean of any group of numbers by multiplying them and then taking the root equal to the total amount of numbers, how to apply the geometric mean to situations where growth rates are determined by multiplication, not addition, and how to write and use the formula for geometric mean.
## What you learned:
After viewing the video, studying the pictures, and reading the instructions, you will learn to:
• Define the geometric mean
• Find the geometric mean of a group of numbers
• Apply geometric mean to suitable situations
• Write and use the formula for geometric mean
Instructor: Malcolm M.
Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.
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# Why must an inverse function be bijective?
Explain why $f^{-1}$ is a function if and only if $f$ is a bijective function.
My attempt:
$f^{1}$ is the inverse relation from B to A $\equiv$ function from B to A
By definition of a function from setA to setB, there is a relation from setA to B. (ARB? relation) such that is satisfies two properties:
1) All element in the domain A must be related to some function in the codomain B
2) No element in A is related to more than one element in B.
Hence, that would mean that 1) the function must be onto and 2) the function must be one-to-one.
Since the function from A to B has to be bijective, the inverse function must be bijective too.
It's hard for me explain. But basically because the function from A to B is described to have a relation from A to B and that the inverse has a relation from B to A. Since the relation from A to B is bijective, hence the inverse must be bijective too. Also, does bijection has anything to do with equivalence relation on a single set where a relation R on set A must be reflexive, symmetric and transitive.
Thanks!
I recommend you explain it like this:
Let $f : A \to B$ be a bijection. Let $f^{-1}$ be the inverse function of $f$.
We want to show $f^{-1}$ is a bijection, so we need to show $f^{-1}$ is one-to-one and onto.
First, let's show $f^{-1}$ is onto:
Since $f^{-1} : B \to A$, we need to show for each $a \in A$, there is some $b \in B$ so that $f^{-1}(b) = a$. Well, let $a$ be any element of $A$, then. Since $f : A \to B$, $f(a)$ is an element of $B$. But by definition of the inverse function $f^{-1}$, this map sends $f(a)$ from $B$ to $a$ in $A$. So that means we found a $b \in B$ with $f^{-1}(b) = a$, namely, the element $b = f(a)$.
Now to show $f^{-1}$ is one-to-one:
Since $f^{-1} : B \to A$, we need to show if for any $b_{1}, b_{2} \in B$, we have $f^{-1}(b_{1}) = f^{-1}(b_{2})$, then it should hold that $b_{1} = b_{2}$. Well then, suppose $f^{-1}(b_{1}) = f^{-1}(b_{2})$. Since $f : A \to B$ is onto, we can find $a_{1}$ and $a_{2}$ in $A$ so that $f(a_{1}) = b_{1}$ and $f(a_{2}) = b_{2}$. So we have $f^{-1}(b_{1}) = f^{-1}(b_{2})$ implies $f^{-1}(f(a_{1})) = f^{-1}(f(a_{2}))$. But by definition of the inverse function, $f^{-1}$ sends the element $f(a)$ to $a$, so $f^{-1}(f(a_{1})) = a_{1}$ and $f^{-1}(f(a_{2})) = a_{2}$, and so this implies $a_{1} = a_{2}$. Since $a_{1} = a_{2}$, and $f$ is a function (and thus well-defined), it can't send one element to two different elements, so that means $f(a_{1}) = f(a_{2})$, i.e., $b_{1} = b_{2}$.
Note: where did we use the fact that the original function $f$ is one-to-one and onto here? Well, we didn't use one-to-one directly in the arguments above (but onto was used in the second argument). But they are still necessary. If $f : A \to B$ is onto, then that would allow us to construct a potential inverse with domain $B$. If $f$ were not onto, our inverse function could not have domain $B$. Similarly, if $f$ is one-to-one, then that allows us to construct a well-defined inverse, i.e., an inverse that's actually a function (i.e., it doesn't send one element to more than one element). Both of these are necessary to discuss the existence of a function $f^{-1}: B \to A$ defined in the usual way an inverse is defined.
• @misheekoh Yes, you can definitely have a function that is one-to-one but not onto and it's still a function. You can also have a function that's onto but not one-to-one and it's still a function. Dec 6, 2015 at 0:41
• @misheekoh A function $f : A \to B$ is just a rule that sends stuff from $A$ to stuff in $B$. And every function must be well-defined. That just means no single element from $A$ can be sent to two elements in $B$. So for example $f : \Bbb R \to \Bbb R$ which sends the element $1$ to both $0$ and $3$ (i.e., $f(1) = 0$ and $f(1) =3$) is not a function. Dec 6, 2015 at 0:42
• @misheekoh Incorrect! Well defined and injective look similar but are very different. Well defined means no input can correspond to more than one output. One-to-one means no output can have more than one input being sent to it. Dec 6, 2015 at 2:45
• @misheekoh Make sure you understand my last comment. It's a common misconception. Every function by definition must be well defined. No input can correspond to more than one output. But functions don't have to be one-to-one. A function can have an output that has more than one input. Take $f : \Bbb R \to \Bbb R$ defined by $f(x) = 2$. It's the constant function $2$. This is a function, since it is well-defined. No input corresponds to more than one output since every input has the output $2$. But $f$ is not one-to-one. There is an output with more than one input, namely... Dec 6, 2015 at 2:46
• @misheekoh ...the output $2$. Every input is sent to the output $2$, so $f$ is not one-to-one. Dec 6, 2015 at 2:47
If $f:A \rightarrow B$ is a function then it must be the case that $\forall a \in A, f(a)$ is unique. If $f^{-1}: B \rightarrow A$ exists, then it must be the case that $\forall b \in B, f^{-1}(b)$ is unique. So, each element in A is paired with one element in $B$ (and vice versa), specifically, $a \in A$ is paired with $f(a) \in B$.
I assume your $f^1$ notation actually means $f^{-1}$.The function $f$ has to be bijective because otherwise, our so called "$f^{-1}$" would not even be a function. Let's see why:
1) Assume $f:A\rightarrow B$ is not surjective: then there is an element $b' \in B$ that doesn't have a preimage. Therefore if we attempt to construct a relation from every $b \in B$ to an $a\in A$, so that $b\mathcal Ra$ if $b=f(a)$, $b'$ is left behind, then $\mathcal R$ cannot be a function.
2)Assume $f:A\rightarrow B$ is not injective. Then for some $a_1$ and $a_2$ in $A$ we have that $f(a_1)=f(a_2)$, then if we attempt to construct the same relation $\mathcal R$ we conclude it cannot be a function because for some $b \in B$ we have that $b\mathcal Ra_1$ and $b\mathcal Ra_2$
Your explanation works, I just tried to explain it in a more technical way. And no, $f$ being an equivalence relation has nothing to do with $f^{-1}$ existing
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# McGraw Hill My Math Grade 4 Chapter 2 Lesson 7 Answer Key Subtract Across Zeros
All the solutions provided in McGraw Hill Math Grade 4 Answer Key PDF Chapter 2 Lesson 7 Subtract Across Zeros will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 4 Answer Key Chapter 2 Lesson 7 Subtract Across Zeros
Math in My World
Example 1
Each fourth grade class has a goal to collect 5,100 pennies for charity. How many more pennies do the fourth graders in Mr. Blake’s class need to reach their goal?
Find 5,100 – 3,520
1. Subtract ones.
0 ones – 0 ones = 0 ones
2. Subtract tens.
Regroup 1 hundred as 10 tens.
10 tens – 2 tens = 8 tens
3. Subtract hundreds.
Regroup one thousand as 10 hundreds.
10 hundreds – 5 hundreds = 5 hundreds
4. Subtract thousands.
4 thousands – 3 thousands = 1 thousand
So, Mr. Blake’s class needs 1580 more pennies.
Example 2
There were 30,090 fans at the stadium on Saturday. The next Saturday, there were 22,977 fans. How many more fans were at the stadium on the first Saturday than the second Saturday?
Find 30,090 – 22,977.
1. Subtract ones.
Regroup 1 ten as 10 ones.
10 ones – 7 ones = 3 ones
2. Subtract tens.
8 tens – 7 tens = 1 ten
3. Subtract hundreds.
Regroup 1 ten thousand as 10 thousands. Regroup 1 thousand as 10 hundreds.
10 hundreds – 9 hundreds = 1 hundred
4. Subtract thousands.
9 thousands – 2 thousands = 7 thousands
5. Subtract ten thousands.
2 ten thousands – 2 ten thousands = 0 ten thousands
So, 30,090 – 22,977 = 7,113
There were 7,113 more fans than the first Saturday.
Talk Math
Explain how to subtract 42,956 from 55,000.
Explanation:
Given, 55,000 – 42,956
1. Subtract ones.
Regroup 1 ten as 10 ones.
10 ones – 6 ones = 4 ones
2. Subtract tens.
10 tens – 6 tens = 4 ten
3. Subtract hundreds.
Regroup 1 ten thousand as 10 thousands. Regroup 1 thousand as 10 hundreds.
10 hundreds – 10 hundreds = 0 hundred
4. Subtract thousands.
4 thousands – 2 thousands = 2 thousands
5. Subtract ten thousands.
2 ten thousands – 1 ten thousands = 1 ten thousands
So, 55,000 – 42,956 = 12,044
Guided Practice
Subtract. Use addition or estimation to check.
Question 1.
Explanation:
Given, 2,003 – 1,154 = 849
To check , we can add the subtrahend to the result
That is 849 + 1,154 = 2,003
Question 2.
Answer: $4,498 Explanation: Given,$8,000 – $3,502 =$4,498
To check , we can add the subtrahend to the result
That is $4,498 +$3,502 = $8,000 Independent Practice Subtract. Use addition or estimation to check. Question 3. Answer: 1,094 Explanation: Given, 2,040 – 946 = 1,094 To check , we can add the subtrahend to the result That is 1,094 + 946 = 2,040 Question 4. Answer: 4,953 Explanation: Given, 7,008 – 2,055 = 4,953 To check , we can add the subtrahend to the result That is 4,953 + 2,055 = 7,008 Question 5. Answer: 8,888 Explanation: Given, 12,050 – 3,162 = 8,888 To check , we can add the subtrahend to the result That is 8,888 + 3,162 = 12,050 Question 6. Answer: 4,955 Explanation: Given, 10,400 – 5,445 = 4,955 To check , we can add the subtrahend to the result That is 4,955 + 5,445 = 10,400 Question 7. Answer: 40,978 Explanation: Given, 46,801 – 5,823 = 40,978 To check , we can add the subtrahend to the result That is 40,978 + 5,823 = 46,801 Question 8. Answer: 38,199 Explanation: Given, 60,032 – 21,833 = 38,199 To check , we can add the subtrahend to the result That is 38,199 + 21,833 =60,032 Question 9. Answer:$38,951
Explanation:
Given, $52,006 –$13,055 = $38,951 To check , we can add the subtrahend to the result That is$38,951 + $13,055 =$52,006
Question 10.
Explanation:
Given, 600,000 – 28,005 = 5,71,995
To check , we can add the subtrahend to the result
That is 5,71,995 + 28,005 = 600,000
Question 11.
Explanation:
Given, 508200 – 136118 = 3,72,082
To check , we can add the subtrahend to the result
That is 3,72,082 + 136118 = 508200
Subtract. Use addition or estimation to check. Use the place-value chart to set up the problem.
Question 12.
900,000 – 31,650 = _____
Explanation:
Problem Solving
For Exercises 13 and 14, use the table which shows the distance between New York City and five other cities around the world.
Question 13.
How many more miles is it to travel to Jakarta than to London?
Answer: 6,582 more miles is it to travel to Jakarta than to London
Explanation:
Given, Jakarta 10,053 miles and London 3,471 miles
10,053 – 3,471 = 6,582
So, 6,582 more miles is it to travel to Jakarta than to London
Question 14.
How many more miles is it to travel to Munich than to Paris?
Answer: 407 more miles is it to travel to Munich than to Paris
Explanation:
Given, Munich 4,042 miles and Paris 3,635 miles
4,042 – 3,635 = 407
So, 407 more miles is it to travel to Munich than to Paris
Question 15.
Mathematical Practice 5 Use Math Tools Trent earned 4,005 points in a video game. His brother earned 2,375 points in the same game. How many more points did Trent earn than his brother?
Answer: 1,630 more points did Trent earn than his brother
Explanation:
Given, Trent earned 4,005 points in a video game.
His brother earned 2,375 points in the same game.
4,005 – 2,375 = 1,630
So, 1,630 more points did Trent earn than his brother
HOT Problems
Question 16.
Mathematical PRACTICE 1 Plan Your Solution Identify a number that results in a 4-digit number when 156,350 is subtracted from it.
Answer: 152,347 is a number that results in a 4-digit number 4,003.
Explanation:
Given number is 156,350
For Example,
Let the another number be 152,347
Then, 156,350 – 152,347 = 4,003
So, 152,347 is a number that results in a 4-digit number 4,003.
Question 17.
? Building on the Essential Question How does understanding place value help you to subtract across zeros?
Answer: The value of where the digit is in the number, based on the location of the digit in order to add and subtract numbers.
### McGraw Hill My Math Grade 4 Chapter 2 Lesson 7 My Homework Answer Key
Practice
Subtract. Use addition or estimation to check.
Question 1.
Explanation:
Given, 4,000 – 1,731 = 2,269
To check , we can add the subtrahend to the result
That is 2,269 +1,731 = 4,000
Question 2.
Explanation:
Given, 3,300 – 1,892 = 1,408
To check , we can add the subtrahend to the result
That is 1,408 +1,892 = 3,300
Question 3.
Explanation:
Given, 8,000 – 6,313 = 1,687
To check , we can add the subtrahend to the result
That is 1,687 +6,313 = 8,000
Question 4.
Explanation:
Given, 14,000 – 10,892 = 3,108
To check , we can add the subtrahend to the result
That is 3,108 +10,892 = 14,000
Problem Solving
Question 5.
If 700 tickets were sold for a concert and only 587 people attended, how many people bought a ticket but did not attend?
Answer: 113 people bought a ticket but did not attend.
Explanation:
Given, 700 tickets were sold for a concert and only 587 people attended,
Then 700 – 587 = 113
So, 113 people bought a ticket but did not attend.
Question 6.
The Amazon River, in South America, is 4,000 miles long. The Snake River, in the northwestern United States, is 1,038 miles long. How much longer is the Amazon River than the Snake River?
Answer: The Amazon River 2,962 miles is than the Snake River.
Explanation:
Given, The Amazon River, in South America, is 4,000 miles long.
The Snake River, in the northwestern United States, is 1,038 miles long.
4,000 – 1,038 = 2,962
So, The Amazon River 2,962 miles is than the Snake River.
Question 7.
Mathematical PRACTICE 5 Use Math Tools There are 6,000 products at the store. In one hour, 425 products are sold. How many products are left?
Answer: There are 5,575 products are left.
Explanation:
Given, There are 6,000 products at the store.
In one hour, 425 products are sold.
6,000 – 425 = 5,575
So, There are 5,575 products are left.
Question 8.
A field of corn has 2,000 insects. Only 497 are eating the corn. How many insects are not eating the corn?
Answer: 1,503 insects are not eating the corn
Explanation:
Given, A field of corn has 2,000 insects.
Only 497 are eating the corn.
2,000 – 497 = 1,503
So, 1,503 insects are not eating the corn
Test Practice
Question 9.
Logan has a gift card for $200. He spends$45 on Monday and $61 on Tuesday. How much money is left on his gift card? A.$94
B. $106 C.$104
D. $139 Answer: A Explanation: Given, He spends$45 on Monday and $61 on Tuesday. Then,$61 + $45 =$106
Logan has a gift card for $200.$200 – $106 =$94.
So, \$94 is left on his gift card.
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### Differentiation
#### Definition
Differentiation is a method used to compute the rate of change of a function $f(x)$ with respect to its input $x$. This rate of change is known as the derivative of $f$ with respect to $x$.
The first derivative of a function $y=f(x)$ is denoted $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, where $\mathrm{d}y$ denotes an infinitesimally small change in $y$, and $\mathrm{d}x$ an infinitesimally small change in $x$. It is defined by:
$\frac{\mathrm{d}y}{\mathrm{d}x}=\lim_{\large{h\to 0}}\left[\frac{f(x+h)-f(x)}{h}\right].$
The process of finding the derivative by taking this limit is known as differentiation from first principles. In practice it is often not convenient to use this method; the derivatives of many functions can be found using standard derivatives in conjunction with rules such as the chain rule, product rule and quotient rule.
#### Notation
There are several different notations for differentiation. All of them are 'correct', and some are more prevalent in certain fields than others. It's helpful to be aware of each system of notation, even if you don't use it regularly.
###### Leibniz's Notation
Given a function $y=f(x)$, the first derivative of $y$ with respect to $x$ is denoted by $\dfrac{\mathrm{d}y}{\mathrm{d}x}$
The second derivative of $y$ with respect to $x$ is denoted by
$\frac{ \mathrm{d} }{ \mathrm{d}x } \left[ \frac{ \mathrm{d}y }{ \mathrm{d}x } \right] = \frac{ \mathrm{d}^2 y }{ \mathrm{d}x^2 }$
The $n$th derivative of $y$ with respect to $x$ is denoted by
$\dfrac{\mathrm{d}^n y}{\mathrm{d}x^n}$
The derivative of $y$ with respect to $x$ at the point $x=a$ is denoted by
$\dfrac{\mathrm{d}y}{\mathrm{d}x}\Biggl\vert_{ \Large{x=a} }$
###### Lagrange's Notation
Also referred to as prime notation. In this notation a prime (dash) symbol is used to denote the derivative of a function.
Given a function $y=f(x)$, the first derivative of $y$ with respect to $x$ is denoted by
$y'(x)=y' \text{ or equivalently } f'(x)=f'.$
Note: It is not necessary to include the argument of the function when using this notation. It should be obvious from context what the argument of the function is. If it's not clear, use a different notation.
The second and third derivatives of $y$ with respect to $x$ are denoted by
$(y')'=y'' \text{ and } (y'')'=y''' \text{ or equivalently } (f')'=f'' \text{ and } (f'')'=f'''.$
A superscript Roman numeral or bracketed number is used to denote a higher derivative. For example, the fourth derivative of $y$ with respect to $x$ is denoted
$y^{ \textrm{iv} } \text{ or } y^{ (4) }.$
The $n$th derivative of $y$ with respect to $x$ is denoted by
$y^{(n)} \text{ or equivalently } f^{(n)}.$
###### Newton's Notation
Also referred to as dot notation. In this notation a dot is placed above the function name to denote the time derivative of a function.
Given a function $x=f(t)$, the first and second derivatives of $x$ with respect to time $t$ are denoted by
$\dot{x} \text{ and } \ddot{x}.$
Note: this notation is used strictly to denote a derivative with respect to time.
#### Properties
The derivative of a constant is $0$ - a constant function does not change.
Differentiation is a linear operation. Given functions $f(x)$ and $g(x)$ and real numbers $\alpha$ and $\beta$, the following property holds:
$\dfrac{\mathrm{d} }{\mathrm{d}x}\Bigl[\alpha\,f(x)+\beta\;g(x)\Bigl]=\alpha\dfrac{\mathrm{d}f}{\mathrm{d}x}+\beta\dfrac{\mathrm{d}g}{\mathrm{d}x}$
#### Table of Derivatives
A table of standard derivatives, where $n$ is any real number and $k,a$ are any constants.
$f(x)$
$f'(x)$
$k$
$0$
$x^n$
$nx^{n-1}$
$kx$
$k$
$\ln{x}$
$\frac{1}{x}$
$\mathrm{e}^{x}$
$\mathrm{e}^{x}$
$\mathrm{e}^{u(x)}$
$\mathrm{e}^{u(x)}\dfrac{\mathrm{d}u(x)}{\mathrm{d}x}$
$a^x$
$a^x\ln{a}$
$\sin{x}$
$\cos{x}$
$\cos{x}$
$-\sin{x}$
$\tan{x}$
$\sec^2{x}$
#### Applications of Differentiation
###### The gradient of a graph
Given a function $y=f(x)$, the derivative $f'(x)$ gives the gradient of the graph of this function at a point $x$. This is also the gradient of the tangent to the curve at that point.
###### Example
The gradient of the graph $y=x^2$ at $x=2$ is $\displaystyle\dfrac{\mathrm{d}}{\mathrm{d}y}\bigl[x^2\bigl]\Biggl|_{x=2}= 2x\Biggl|_{x=2}=2\cdot2=4$.
###### Stationary Points and Optimisation
Stationary points of a function $y=f(x)$ are the points where the derivative is zero. They can be found by setting $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ and solving for $x$. The nature of these stationary points (maxima, minima or points of inflection) can be determined by the derivative tests.
Since these local maxima and minima represent the maximum or minimum value of $f(x)$ within a local region, finding stationary points is useful in optimisation problems.
Note: The global maximum or minimum values of a function $f(x)$ are always found at a point where either the derivative $f'(x)=0$ or the derivative does not exist. However, a point $x$ for which $f'(x)=0$ is not necessarily a global maximum or minimum value of the function $f(x)$. Such points may represent a local maximum, local minimum or stationary point of inflection; thus, further analysis is required to determine whether or not the point is a global maximum or minimum.
###### Physics
Physics often involves studying rates of change and the way different quantities interact with each other; derivatives and differential equations are therefore essential to the mathematical descriptions of many physical quantities and phenomena.
Example: The velocity of an object is defined to be the rate of change of its position with respect to time. For an object in one-dimensional motion with position described by $x=x(t)$, its velocity $v$ is given by:
$v=\dot{x}=\dfrac{\mathrm{d}x}{\mathrm{d}t}.$
Similarly, the acceleration of an object is defined to be the rate of change of its velocity with respect to time. For an object in one-dimensional motion with velocity $v$, its acceleration $a$ is given by:
$a=\ddot{x}=\dfrac{\mathrm{d}v}{\mathrm{d}t}=\dfrac{ \mathrm{d} }{\mathrm{d}t}\left[\dfrac{\mathrm{d}x}{\mathrm{d}t}\right]=\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2}.$
#### Worked Example
###### Example 1
Suppose an object moves along a horizontal axis. Its position is described by $x(t)=t^3+5t-1+\cos{t}$. Find the velocity and acceleration of the object at time $t=1$.
###### Solution
Recall that the velocity of a moving body with displacement $x(t)$ is $v=\dot{x}=\dfrac{\mathrm{d}x}{\mathrm{d}t}.$
Here $x(t)=t^3+5t-1+\cos{t}$, so the velocity of the object is:
$v=\dot{x}=\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[t^3+5t-1+\cos{t}\Bigl].$
Since differentiation is a linear operation each term can be treated separately.
From the table of standard derivatives, the derivative of a function $x^n$ is $\dfrac{ \mathrm{d} }{\mathrm{d}x}\Bigl[x^{ \large{n} }\Bigl]=nx^{ \large{n-1} }$. Apply this rule with $n=3$ to differentiate $t^3$:
$\dfrac{ \mathrm{d} }{ \mathrm{d}t }\Bigl[t^3\Bigl]=3t^{ \large{3-1} } = 3t^2.$
From the table of standard derivatives, the derivative of a function $kx$ is $\dfrac{ \mathrm{d} }{ \mathrm{d}x }\Bigl[kx\Bigl]=k.$ Hence the derivative of $5t$ is:
$\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[5t\Bigl]=5.$
This result can also be shown by the linearity property of the derivative and the general result for a derivative of $x^n$ (with $n=1$):
$\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[5t\Bigl]=5\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[t^1\Bigl]=5t^{ \large{1-1} }=5t^{ \large{0} }=5.$
By the properties of differentiation, the derivative of any constant $c$ is $\dfrac{ \mathrm{d} }{\mathrm{d}x}\Bigl[c\Bigl]=0$. Hence the derivative of $1$ is:
$\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[1\Bigl]=0.$
Finally, from the table of standard derivatives, the derivative of $\cos{t}$ is
$\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[\cos{t}\Bigl]=-\sin{t}.$
Thus, the velocity of the object is given by:
\begin{align} v=\dot{x} &= \dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[t^3+5t-1+\cos{t}\Bigl] \\ &=3t^2+5-0-\sin{t} \\ &=3t^2+5-\sin{t}. \end{align}
At time $t=1$, the velocity is:
\begin{align} v(1) &= \dfrac{ \mathrm{d}x }{\mathrm{d}t}\Biggl\vert_{ \Large{t=1} } \\ &= 3\cdot1^2+5-\sin{1} \\ &= 8-\sin{1} \\ &\approx 7.16 \text{ ms⁻¹}. \end{align}
Note: When working with physical quantities such as velocity or acceleration it is essential to include units of measurement in the solution.
To find the acceleration of the object, recall that the acceleration of a moving body with velocity $v$ is given by $a=\ddot{x}=\dfrac{\mathrm{d}v}{\mathrm{d}t}=\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}$.
Here $v(t)=3t^2+5-\sin{t}$, so
$a=\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[3t^2+5-\sin{t}\Bigl].$
As above, the derivative is found by using the properties of differentiation and the table of standard integrals.
The derivative of $3t^2$ is:
$\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[3t^2\Bigl]=3\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[t^2\Bigl]=3\cdot2t^{ \large{2-1} }=6t.$
$5$ is a constant, so its derivative is:
$\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[5\Bigl]=0.$
The derivative of $-\sin{t}$ is:
$\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[-\sin{t}\Bigl]=-\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[\sin{t}\Bigl]=-\cos{t}.$
Thus the acceleration of the object is given by:
\begin{align} a=\ddot{x} &= \dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[3t^2+5-\sin{t}\Bigl] \\ &= 6t+0-\cos{t} \\ &= 6t-\cos{t}. \end{align}
At time $t=1$, the acceleration is:
\begin{align} a(1) &= \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}\Biggl\vert_{ \Large{t=1} } \\ &=6\cdot1-\cos{1} \\ &\approx 5.46\text{ ms⁻²}. \end{align}
#### Video Example
##### Example
Prof. Robin Johnson finds the first and second derivatives of $y(x)+x^3-2\mathrm{e}^{2\large{x}}$ and $x(t)=2\cos{(2t)}$.
#### Workbooks
These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples.
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# How do you write an equation in standard form with 2 points
Our finishing x-coordinate was 6.
Our first step is to eliminate the fractions, but this becomes a little more difficult when the fractions have different denominators! We can simplify it a little bit.
Slope-intercept form linear equations Standard form linear equations Point-slope form linear equations Video transcript A line passes through the points negative 3, 6 and 6, 0.
These are the same equations, I just multiplied every term by 3. We need to find the least common multiple LCM for the two fractions and then multiply all terms by that number! When we move terms around, we do so exactly as we do when we solve equations!
This is our point slope form. Solution Slope intercept form is the more popular of the two forms for writing equations. Now the last thing we need to do is get it into the standard form. And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially.
We went from 6 to 0. However, you must be able to rewrite equations in both forms. You divide the numerator and the denominator by 3. We went from negative 3 to 6, it should go up by 9. Now what is the change in y? Equations that are written in standard form: Find the equation of this line in point slope form, slope intercept form, standard form.
And we have our slope. If we view this as our end point, if we imagine that we are going from here to that point, what is the change in y? So we have slope intercept.
And then we want our finishing x value-- that is that 6 right there, or that 6 right there-- and we want to subtract from that our starting x value. And then we have this 6, which was our starting y point, that is that 6 right there.
Once we figure out the slope, then point slope form is actually very, very, very straightforward to calculate. So what can we do here to simplify this? So if you give me one of them, we can manipulate it to get any of the other ones. So this 0, we have that 0, that is that 0 right there.
So, our finishing y point is 0, our starting y point is 6.
There is one other rule that we must abide by when writing equations in standard form. What was our finishing x point, or x-coordinate?Aug 23, · I have a math question that i am completely stuck on.
I have been trying to find out how to write the standard form of an equation with only 2 bsaconcordia.com: Resolved. We explain Standard Form from Two Points with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers.
Linear equations can be written in many forms. Discover how to use two coordinate points to solve for a line's slope and write an equation in slope-intercept form. With some algebraic manipulation, you will. Writing Algebra Equations Finding the Equation of a Line Given Two Points.
Write the equation in slope intercept form using the slope and y-intercept. If needed, you can then rewrite the equation in standard form as well. Example 1: Writing an. Point Slope Form and Standard Form of Linear Equations.
Here’s the graph of a generic line with two points plotted on it. The slope of the line is “rise over run.” When we write the equation, we’ll let x be the time in months, and y be the amount of money saved.
After 1 month, Andre has \$ Writing Equations in Standard Form. We know that equations can be written in slope intercept form or standard form.
Whatever you do to one side of the equation, you must do to the other side! Solution.
That was a pretty easy example. We just need to remember that our lead coefficient should be POSITIVE!
How do you write an equation in standard form with 2 points
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# At a Glance - Continuity on Closed and Half-Closed Intervals
When looking at continuity on an open interval, we only care about the function values within that interval.
If we're looking at the continuity of a function on the open interval (a, b), we don't include a and; they aren't invited. No value of x less than a or greater than b is invited, either. This is an exclusive club, with the parentheses serving as the bouncers.
In a closed interval, denoted [a, b], we're lowering our standards a bit by inviting a and b to the pool party. Half-closed intervals either invite a, [a, b), or b, (a, b]. To talk about continuity on closed or half-closed intervals, we'll see what this means from a continuity perspective. Start with a half-closed interval of the form [a, b). What does it mean for a function to be continuous on this interval?
Since we can only approach a from the right, we use the continuity definition for a right-sided limit instead of a two sided limit. We say f is continuous on [a,b) if f is continuous on (a,b) and
• f(a) exists
• exists
• f(a) and agree
### Sample Problem
This function is continuous on the interval [a, b):
This function is continuous on (a, <em>b</em>) defined, is defined, and the function value at a agrees with the right-sided limit at a.
### Sample Problem
The following functions are not continuous on the interval [a,b):
This function is not continuous on [a, b) because f(a) is undefined. We could also say this function is not continuous on [a,b) because does not exist as well.
This function is not continuous on the interval [a, b) because .
This function is not continuous on the interval [a, b) because it is not continuous on the open interval (a, b).
It's all just like continuity on open intervals. The only difference is that now we have to check the endpoints.
#### Example 1
Let Determine whether f is continuous on each of the following intervals:[10,12][10,12)(10,12](10,12)
#### Exercise 1
What must be true for a function to be continuous on an interval of the form (a, b]?
#### Exercise 2
What must be true for a function to be continuous on an interval of the form [a, b]?
#### Exercise 3
Determine if f is continuous on each of the following intervals.
• [a, b]
• [a, b)
• (a, b]
• (a, b)
#### Exercise 4
Determine if f is continuous on each of the following intervals.
• [a, b
• [a, b
• (a, b]
• (a, b)
#### Exercise 5
Determine if f is continuous on each of the following intervals.
• [a, b]
• [a, b)
• (a, b]
• (a, b)
#### Exercise 6
Determine if the function
#### Exercise 7
Determine if the function
is continuous on each of the following intervals.
• [0, 1]
• [0, 1)
• (0, 1]
• (0, 1)
• [1, 5]
• [1, 5)
• (1, 5]
• (1, 5)
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#### MAT-12.NO.06
12th Grade (MAT) Targeted Standard (NO) Number and Operations Learners will develop a foundational understanding of the number system, operations, and computational fluency to create connections and solve problems within and across concepts.
## Progressions
• MAT-04.NO.NF.06 Solve authentic word problems by adding and subtracting fractions and mixed numbers with like denominators (proper and improper fractions limited to denominators of 2, 3, 4, 5, 6, 8, 10, 12, and 100).
• MAT-05.NO.NF.03 Solve authentic word problems by adding and subtracting fractions and mixed numbers with unlike denominators using a visual fraction model and/or equation.
• MAT-06.NO.O.02 Add and subtract fractions and decimals up to the hundredth place, including in authentic problems.
• MAT-07.NO.O.02 Add, subtract, multiply, and divide non-negative fractions in multi-step problems, including authentic problems.
• MAT-08.NO.O.02 Add, subtract, multiply, and divide rational numbers using strategies or procedures.
• MAT-12.NO.04 Demonstrate that the sum or product of two rational numbers is rational, that the sum of a rational number and an irrational number is irrational, and that the product of a nonzero rational number and an irrational number is irrational.
• MAT-12.AR.05 Add, subtract, multiply, and divide rational expressions. Understand that rational expressions form a system analogous to rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression.
• MAT-12.NO.06 Know there is a complex number i such that i² = -1, and every complex number has the form a + bi with a and b real. Understand the hierarchal relationships among subsets of the complex number system.
Multiplying and Dividing Fractions
• MAT-04.NO.NF.07 Solve problems by multiplying fractions and whole numbers using visual fraction models (proper and improper fractions limited to denominators of 2, 3, 4, 5, 6, 8, 10, 12, and 100).
• MAT-05.NO.NF.02 Explain why multiplying a given number by a fraction greater than one results in a product greater than the given number and explain why multiplying a given number by a fraction less than one results in a product smaller than the given number.
• MAT-05.NO.NF.04 Solve authentic word problems by multiplying fractions and mixed numbers using visual fraction models and equations.
• MAT-06.NO.O.03 Apply multiplication and division of fractions and decimals to solve and interpret problems using visual models, including authentic problems.
• MAT-07.NO.O.02 Add, subtract, multiply, and divide non-negative fractions in multi-step problems, including authentic problems.
• MAT-08.NO.O.02 Add, subtract, multiply, and divide rational numbers using strategies or procedures.
• MAT-09.NO.02 Perform basic operations of simple radical expressions to write a simplified equivalent expression.
• MAT-12.NO.03 Demonstrate that the sum or product of two rational numbers is rational, that the sum of a rational number and an irrational number is irrational, and that the product of a nonzero rational number and an irrational number is irrational.
• MAT-12.AR.5 Add, subtract, multiply, and divide rational expressions. Understand that rational expressions form a system analogous to rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression.
• MAT-12.NO.06 Know there is a complex number i such that i² = -1, and every complex number has the form a + bi with a and b real. Understand the hierarchal relationships among subsets of the complex number system.
Complex Numbers
• MAT-12.NO.06 Know there is a complex number i such that i² = -1, and every complex number has the form of a + bi with a and b real. Understand the hierarchal relationships among subsets of the complex number system.
• MAT-12.NO.07 Use the definition i 2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
• MAT-12.NO.08 Use conjugates to find quotients of complex numbers.
• MAT-12.NO.09 Apply the Fundamental Theorem of Algebra to determine the number of zeros for polynomial functions. Find all solutions to a polynomial equation.
• MAT-12.AR.11 Solve quadratic equations with real coefficients that have solutions of the form a+bi and a-bi.
• MAT-12.NO.10 Represent complex numbers on the complex plane in rectangular, trigonometric, and polar forms. Find the modulus (absolute value) of a complex number. Explain why the rectangular, trigonometric, and polar forms of a given complex number represent the same number.
• MAT-12.NO.11 Represent addition, subtraction, multiplication, conjugation, powers, and roots of complex numbers geometrically on the complex and/or polar plane; use properties of this representation for computation.
• MAT-12.NO.12 Extend polynomial identities to the complex numbers.
• MAT-12.NO.13 Apply the Fundamental Theorem of Algebra to find all roots of a polynomial equation and determine the nature (i.e., integer, rational, irrational, real, complex) of the roots.
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# Ex.16.1 Q7 Playing with Numbers Solutions - NCERT Maths Class 8
Go back to 'Ex.16.1'
## Question
Find the values of the letters in the following and give reasons for the steps involved.
\begin{align}{A \;\;{B}} \\ { \times \;\;\,{6}} \\ \hline B\,B\, {B} \\ \hline\end{align}
Video Solution
Playing With Numbers
Ex 16.1 | Question 7
## Text Solution
What is known?
Multiplication operation of two numbers
What is unknown?
Value of alphabets i.e. $$A$$ and $$B.$$
Reasoning:
Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.
Steps:
The multiplication of $$6$$ and $$B$$ gives a number whose one’s digit is $$B$$ again.
It is possible only when
$$B = 0, \;2,\; 4,\; 6,$$ or $$8$$
If $$B = 0,$$ then the product will be $$0.$$ Therefore, this value of $$B$$ is not possible.
If $$B = 2$$, then $$B \times 6 = 12$$ and $$1$$ will be a carry for the next step.
$$6{\text{A}} + 1 = {\text{BB}} = 22 \Rightarrow 6{\text{A}} = 21$$ and hence, any integer value of $$A$$ is not possible.
If $$B = 6,$$ then $$B \times 6 = 36$$ and $$3$$ will be a carry for the next step.
$$6A + 3 = BB = 66 \Rightarrow 6A = 63$$ and hence, any integer value of $$A$$ is not possible.
If $$B = 8,$$ then $$B \times 6 = 48$$ and $$4$$ will be a carry for the next step.
$$6{{A}} + 4 = {{BB}} = 88 \Rightarrow 6{{A}} = 84$$ and hence, $$A = 14.$$ However, $$A$$ is a single digit number.
Therefore, this value of $$A$$ is not possible.
If $$B = 4,$$ then $$B \times 6 = 24$$ and $$2$$ will be a carry for the next step.
$$6A + 2 = BB = 44 \Rightarrow 6A = 42$$ and hence, $$A = 7$$
The multiplication is as follows.
\begin{align}{7 \;\;{4}} \\ { \times \;\;\,{6}} \\ \hline 4\,4\, {4} \\ \hline\end{align}
Hence, the values of $$A$$ and $$B$$ are $$7$$ and $$4$$ respectively.
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# The x and y components of a vector r are r_x = 16 m and r_y = -9.0 m, respectively. (a) Find the...
## Question:
The x and y components of a vector {eq}\vec r {/eq} are {eq}\rm r_x = 16 \ m {/eq} and {eq}\rm r_y = -9.0 \ m {/eq}, respectively.
(a) Find the direction of the vector {eq}\vec r {/eq}?
(b) Find the magnitude of the vector {eq}\vec r {/eq}?
(c) Suppose {eq}\rm r_x {/eq} and {eq}\rm r_y {/eq} are doubled, find the direction and the magnitude of the new vector {eq}\vec r {/eq}.
## Magnitude and Direction of Vectors
Oftentimes, a vector is given in component form. In such a case, we need to know how to find the magnitude and direction of the vector. The magnitude uses the Pythagorean theorem, where the length of the hypotenuse is the square root of the sum of the squares of the components:
$$r = \sqrt{r_x^2 + r_y^2}$$
This can be expanded to more than three dimensions:
$$r = \sqrt{r_x^2 + r_y^2 + r_z^2}$$
The angle of a vector in two dimensions is usually expressed as the angle from the positive x axis. Again using what we know of right triangles, the tangent of the angle is the ratio of the y component to the x component:
$$\tan{(\theta)} = \frac{r_y}{r_x}$$
We have a vector with components
$$r_x = 16 \; \mathrm{m}\\ r_y = - 9.0 \; \mathrm{m}$$
We want to find the direction and magnitude of this...
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## How can 2.25 be written as a fraction or mixed number?
As now we have 2 numbers after the decimal point, we multiply both numerator and denominator via 100. So, 2.251 = (2.25 × 100)(1 × 100) = 225100.
## What is 2.75 as a mixed fraction?
So as a simplified mixed number, this becomes 2 and 3/4. And after you do a lot of follow right here, and also you just see a lot of numbers like this, it’s going to be virtually second nature so that you can say, oh, 2.75 is the same thing as 2 and 75/100, is the same factor as 2 and three/4.
What is 2.25 as a number?
As 2.25 has two decimal places, we wish to multiply both numerator and denominator via 100. By this, we can convert 2.25 as 225.
### How do you write Nine 2 as a mixed number?
Related Posts
This is the number above the fraction line. For 9/2, the numerator is 9. Denominator. This is the number under the fraction line….Now let’s pass in the course of the steps had to convert 9/2 to a mixed number.
1. Step 1: Find the whole number.
2. Step 2: Get the new numerator.
3. Step 3: Our mixed fraction.
4. Step 4: Simplifying our fraction.
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# 4.5 Add and Subtract Fractions with Different Denominators
### Learning Objectives
By the end of this section, you will be able to:
• Find the least common denominator (LCD)
• Convert fractions to equivalent fractions with the LCD
• Add and subtract fractions with different denominators
• Identify and use fraction operations
• Use the order of operations to simplify complex fractions
• Evaluate variable expressions with fractions
In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators?
Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals 25 cents and one dime equals 10 cents, so the sum is 35 cents. See Figure 4.7.
Figure 4.7 Together, a quarter and a dime are worth 35 cents, or 35100 of a dollar.
### Identify and Use Fraction Operations
By now in this chapter, you have practiced multiplying, dividing, adding, and subtracting fractions. The following table summarizes these four fraction operations. Remember: You need a common denominator to add or subtract fractions, but not to multiply or divide fractions
### Use the Order of Operations to Simplify Complex Fractions
In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. We simplified complex fractions by rewriting them as division problems. For example,
Now we will look at complex fractions in which the numerator or denominator can be simplified. To follow the order of operations, we simplify the numerator and denominator separately first. Then we divide the numerator by the denominator.
### Evaluate Variable Expressions with Fractions
We have evaluated expressions before, but now we can also evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.
### Section 4.5 Exercises
#### Practice Makes Perfect
Find the Least Common Denominator (LCD)
In the following exercises, find the least common denominator (LCD) for each set of fractions.
#### Writing Exercises
434.
Explain why it is necessary to have a common denominator to add or subtract fractions.435
Explain how to find the LCD of two fractions.
#### Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ After looking at the checklist, do you think you are well prepared for the next section? Why or why not?
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### Finding Slope from an Equation Examples and Worksheets
Get here our tutorial lesson, examples, and worksheets about finding slope from an equation. In our previous posts, we discussed finding a slope from a graph and from two given points.
When we are asked to find the slope of a line given two distinct points, it would be easy for us to do it, since all we have to do is assign one point to be the firs point and the other would be the second point, and then simply substitute the value accordingly to the formula. But, how will we find the slope of a line if the given is only its equation?
Basically, there are two ways we can consider in finding the slope of a line if the given is its equation. It is either by using the definition of the slope, wherein we need to find first two distinct points that lies in the line or we simply transform the equation into slope-intercept form, $$y=mx+b$$, wherein the value that replaces m is the slope. To differentiate between the two let us consider the following examples.
Examples
Find the slope of the line given the following equation.
1. $$4x-y=-8$$
SOLUTION:
a. By using the definition of slope, we set $$x=0$$ to get $$y$$.
\small \begin{align*}\displaystyle &\;\;\;4\left ( 0 \right )-y=-8 &&\textup{Substituting the value of}\; x \; \textup{to the equation}\\& \qquad -1\left ( -y=-8 \right )&&\textup{Multiplying -1 to make it positive} \\& \qquad \qquad \quad \therefore y=8 \end{align*}
Letting $$y=0$$, we are able to solve $$x$$ from the given equation.
\small \begin{align*}\displaystyle &\;\;\;4x-0=-8 &&\textup{Substituting the value of}\; y \; \textup{to the equation}\\& \qquad 4x=-8 \\& \qquad \frac{4x}{4}=\frac{-8}{4}&&\textup{Divide both sides by 4 to get the value of}\; x \\& \qquad \therefore x=-2 \end{align*}
The two points are $$\left(0,8 \right )$$ and $$\left(-2,0\right)$$. Thus,
$$\displaystyle m = \frac{y_2-y_1}{x_2-x_1}=\frac{0-8}{-2-0}=\frac{-8}{-2}=4.$$
b. Transforming the equation into slope-intercept form, $$y=mx+b$$.
\small \begin{align*}\displaystyle &\;\;\;4x-y=-8 \\& \qquad 4x-4x-y=-8-4x &&\textup{Subtract both sides with}\; 4x \\& \qquad -y=-8-4x\\& \qquad -y=-4x-8 \\& \qquad -1 \left ( -y=-8-4x \right )&&\textup{Multiply by -1 to make both sides positive}\; x \\& \qquad \therefore y=4x+8 \end{align*}
The slope is said to be the value that replaces m in the equation, thus 4 is the slope.
To summarize the procedures, in the first method, we need to find first the $$x$$ and $$y$$ intercepts which will serve as the two different points. Then, solve the slope by following the steps in finding the slope given two points. X-intercept refers to a point lies in the line where $$y$$ is zero. On the counter part, the y-intercept is a point in the line where $$x$$ is zero. However, in the second method, we simply transform the given equation into its equivalent slope - intercept form $$y=mx+b$$ where $$m$$ is the slope.
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# How to Add Binary Numbers Video
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#### Recommended Lessons and Courses for You
Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
In this lesson, you'll see that adding binary numbers isn't that difficult. And in some ways, it might be even easier than your regular addition! You'll learn the 3 rules to adding binary numbers and work out a few examples.
## The Steps
Binary numbers are numbers written with a base of two, specifically 0 and 1. We normally write numbers with a base of 10 called decimal numbers, which are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. A 0 in binary is a 0 in base 10. A 1 in binary is a 1 in base 10. A 10 in binary is 2 in base 10. A 11 in binary is 3 in base 10.
Decimal Binary
0 0
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001
This goes on, with each successive number taking on an additional ''1''. As you count, you keep adding more digits, just as you would with other numbers. Computers count using the binary system, so computer programmers and anyone else working with electronics may encounter problems that involve binary calculation. You may need to add 101 and 110 together.
But how does addition in binary work? We'll take our example of 101 and 110 and see just how to do this:
The first step is to recall the rules for binary addition. The rules for binary addition are a bit different than for decimal addition, which is the addition we're familiar with. Binary addition only has three rules:
1. 0 + 0 = 0
2. 0 + 1 = 1 or 1 + 0 = 1
3. 1 + 1 = 10
All you have to remember is that ''0'' and ''0'' make 0. If you have a ''1'' and a ''0'', you'll get 1. If you have two ''1''s, you end up with a 10. And since we are using the binary system, we don't say ten here. As we said, we read the individual digits out loud. For 10, we'll say ''one-zero''. For 110, we'll say ''one-one-zero''.
You go from right to left. So, adding 101 and 110, you begin on the right side and add the last digit of both numbers together (1 + 0). This equals 1. You write this digit down. This is the last digit, the end of your answer.
You then move on and add up the digits to the left (0 + 1). This also equals 1. You write this to the left of the last digit of your answer.
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# Construct the circumcircle and incircle of an equilateral $\Delta ABC$ with side 6cm and center $O$. Find the ratio of radii of circumcircle and incircle.
Last updated date: 23rd Jul 2024
Total views: 452.1k
Views today: 6.52k
Answer
Verified
452.1k+ views
Hint: Draw a perpendicular bisector on the equilateral triangle which divides the sides of the equilateral triangle into two equal parts. Take their intersection point to draw a circumcircle and incircle. Now use trigonometric ratios in two different triangles for values of in-radius and circum-radius.
Complete step-by-step answer:
The pictorial representation of the given problem is shown above.
The equilateral triangle ABC with side 6 cm has a circumcircle and an incircle with center O and radii ${r_1}$ and ${r_2}$ respectively.
$\Rightarrow OB = {r_1}cm,{\text{ }}OD = {r_2}cm$
AE and CD are the perpendicular bisector of BC and AB respectively.
$\Rightarrow BE = CE = BD = AD = \dfrac{6}{2} = 3cm$
Since, FB is the bisector of $\angle ABC$
$\Rightarrow \angle FBC = \angle FBA = \dfrac{{{{60}^0}}}{2} = {30^0}$
Because in equilateral triangle all angles are equal which is ${60^0}$
Now, in $\Delta OBE,{\text{ cos3}}{{\text{0}}^0}{\text{ = }}\dfrac{{BE}}{{OB}} = \dfrac{3}{{{r_1}}}$
As we know ${\text{cos3}}{{\text{0}}^0} = \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow {r_1} = \dfrac{3}{{\cos {{30}^0}}} = \dfrac{3}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{6}{{\sqrt 3 }}cm$
Now in $\Delta OBD,{\text{ tan3}}{{\text{0}}^0}{\text{ = }}\dfrac{{OD}}{{BD}} = \dfrac{{{r_2}}}{3}$
As we know ${\text{tan3}}{{\text{0}}^0} = \dfrac{1}{{\sqrt 3 }}$
$\Rightarrow {r_2} = 3\tan {30^0} = 3\left( {\dfrac{1}{{\sqrt 3 }}} \right) = \sqrt 3 cm$
Now you have to calculate the ratio of radii circumcircle to incircle
$\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\dfrac{6}{{\sqrt 3 }}}}{{\sqrt 3 }} = \dfrac{6}{{\sqrt 3 \times \sqrt 3 }} = \dfrac{6}{3} = 2cm$
So, the required ratio of the radii is 2 cm.
Note: In such types of question first draw the pictorial representation of the given problem, then draw the perpendicular bisectors on the triangle which divide its sides into two equal parts, then apply basic trigonometric property and calculate the radii of the two circles, then divide them we will get the required answer.
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+0
# ratios
0
76
3
A bag contains yellow, green, blue, and white marbles. The ratio of yellow, green, blue, and white marbles is 1:2:3:4. If the bag contains 180 marbles, then how many blue marbles are there?
Jun 26, 2023
#1
+95
+3
add all the ratio numbers and then divide it by how many marbles there are.
1+2+3+4=10
180/10=18
18x3=54
54 blue marbles
Jun 26, 2023
#2
+131
+1
You start by adding all the numbers in the ratio and divide the total by that number. This results in 1+2+3+4=10 and 180/10=18
We then find that the blue is 3 in the ratio, so we multiply 18 by 3 to get our answer of 54 blue marbles
Work:
x+2x+3x+4x= 180
10x=180
x=18
18x3=54
54 blue marbles
Jun 27, 2023
#3
+274
+1
Assume:
the # of yellow marbles is x
the # of green marbles is 2x
the # of blue marbles is 3x
the # of white marbles is 4x
x+2x+3x+4x=180
10x = 180
x = 18
the # of blue marbles = 3x = 3(18) = 54 marbles
Jun 27, 2023
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Home » Math Theory » Operations of Numbers » Matrix Multiplication
# Matrix Multiplication
## What is a matrix?
A matrix is defined as a rectangular array of numbers, symbols, or expressions arranged in rows and columns (plural: matrices). If the array has n rows and m columns, then it is an n×m matrix. The dimensions of the matrix are denoted by the numbers n and m.. The numbers in the matrix are referred to as their entries
### Matrix Notation
We usually denote matrices with capital letters, such as A, B, C, and so on.
A≔ A matrix is denoted by an uppercase letter.
a≔ A matrix entry is denoted by a lower case letter.
Box brackets are commonly used to write matrices. The horizontal and vertical lines of entries in a matrix are called rows and columns, respectively.
### Dimension of a matrix
The number of rows and columns that a matrix contains determines its size. A matrix with m rows and n columns is called an m×n matrix or m-by-n matrix, where m and n are called the matrix dimensions. When describing a matrix, you give the number of rows by the number of columns. This is sometimes referred to as the order of the matrix. For example, the matrix
A=[12 34]
is said to be a one-by-four matrix as it has one row and four columns. We could also say the order of A is 1 × 4. The matrix
B=[123 456]
is a 3×2 matrix because it has three rows and two columns. It has order 3×2.
Tip: Remember that the number of rows is given first, followed by the number of columns.
Consider the acronym “RC” for “Row then Column” to help you remember this.
## What is Matrix Multiplication?
Jacques Philippe Marie Binet, a French mathematician, initially described matrix multiplication in 1812 to depict the composition of linear maps represented by matrices. As a result, matrix multiplication is a fundamental tool of linear algebra, with various applications in many fields of mathematics, including applied mathematics, statistics, physics, economics, and engineering. Computing matrix product is a fundamental process in all linear algebra computational applications.
There are only two methods for multiplying matrices. The first method involves multiplying a matrix by a scalar. This is referred to as scalar multiplication. The second method is to multiply one matrix by another. This is referred to as matrix multiplication.
### Scalar Multiplication
Because the expression A+A is the sum of two matrices with the same dimensions, a matrix A can be added to itself. We end up doubling every entry in A when we compute A+A. As a result, we can interpret the expression 2A as instructing us to multiply every element in A by 2.
In general, to multiply a matrix by a number, multiply that number by each entry in the matrix. For example,
Individual numbers are commonly referred to as scalars when discussing matrices. As a result, we refer to the operation of multiplying a matrix by a number as scalar multiplication.
### Dot Product
To multiply one matrix by another, we must first understand what the dot product is. The dot product is a method for finding the product of two vectors, also known as vector multiplication. Assume the following two vectors:
u=[123] , v=[456]
To multiply these two vectors, simply multiply the corresponding entries and add the resulting products.
u∙v=(1)(4)+(2)(5)+(3)(6)
=4+10+18
=32
As a result of multiplying vectors, we get a single value. Take note, however, that the two vectors have the same number of entries. What if one of the vectors contains fewer entries than the other?
For example, let
u=[214] , v=[310 2]
When corresponding entries were multiplied and were added together, the solution would be:
u∙v= [214] [310 2]
=2(3)+1(1)+0(4)+?(2)
There is an issue here. The first three entries of the dot product have corresponding entries to multiply with, but the fourth does not. This means that the dot product of these two vectors cannot be computed.
As a result, the dot product of two vectors with different numbers of entries cannot be found. They must both contain the same number of entries.
## What are the conditions for Matrix Multiplication?
When we want to multiply matrices, we must first ensure that the operation is possible – which is not always the case. Furthermore, unlike number arithmetic and algebra, even when the product exists, the order of multiplication can affect the outcome.
Learning the dot product is essential in multiplying matrices. When multiplying one matrix by another, the rows and columns must be treated as vectors.
Example 1: Find AB if A=[1234] and B=[5678]
A∙B= [1234].[5678]
Focus on the following rows and columns
where r1 is the first row, r2 is the second row, and c1, c2 are first and second columns. Treat each row and column as a vector.
Notice here that multiplying a 2×2 matrix with another 2×2 matrix gives a 2×2 matrix. Thus, the matrix we get should have four entries.
See that the first entry is located on the first row and first column. So, to get the value of the first entry, simply take the dot product of r1 and c1. Thus, the first entry will be
Now, notice that the location of the second entry is in the first row and second column. So, to get the value of the second entry, simply take the dot product of r1 and c2. Thus, the second entry will be
The same strategy can be used to get the value of the last two entries.
Example 2: Find AB if A=[14 25 36] and B=[111 111 111 111]
A∙B= [14 25 36] x [111 111 111 111]
Use the dot products to compute each entry.
Hence, two matrices can be multiplied if the number of columns of the first matrix is the same as the number of rows of the second matrix. The multiplication produces another matrix with the same number of rows as the first and the same number of columns as the second. If this is not the case, the multiplication cannot be performed.
In symbols, let A be an m×p matrix and let B be an q×n matrix. Then, the product A×B=AB will be an m×n matrix provided that p=q. If p≠q, the matrix multiplication is not defined. For example, a 2×5 matrix cannot be multiplied by a 3×4 matrix because 5≠3, whereas it is possible to multiply a 2×5 matrix by a 5×3, and the result will be a 2×3 matrix.
## What are the properties of matrix multiplication?
Matrix multiplication has some properties in common with regular multiplication. Matrix multiplication, on the other hand, is not defined if the number of columns in the first factor differs from the number of rows in the second factor, and it is non-commutative even if the product remains definite after the order of the factors is changed.
### Non- commutativity
The Non-commutativity of matrix multiplication is one of the most significant differences between real number multiplication and matrix multiplication. Hence, the order in which two matrices are multiplied matters in matrix multiplication.
An operation is commutative if, given two elements A and B such that the product AB
is defined, then BA is also defined, AB=BA
For example,
A=[0100] and B=[0010]
then,
AB= [0100] x [0010] = [1000]
but
BA= [0010] x [0010] = [1000]
Notice that the products are not the same since AB≠BA. Hence, matrix multiplication is not commutative.
Commutativity does occur in one special case. It is when multiplying diagonal matrices of the same dimension.
Aside from this major distinction, the properties of matrix multiplication are mostly like those of real number multiplication.
### Distributivity
The matrix multiplication is distributive with respect to matrix addition. That is, if A, B, C, D are matrices of respective sizes m×n,n×p, n×p, and p×q, one has left distributivity
AB+C=AB+AC
and the other has right distributivity
B+CD=BD+CD
Example1:
Notice that AB+C=AB+AC. Now find B+CA and BA+CA
Notice that B+CA=BA+CA. It is also notable that AB+CB+CA and that AB+AC≠BA+CA which reminds us of the non-commutativity of matrix multiplication.
### Associativity
This property states that the grouping surrounding matrix multiplication can be changed.
If A, B, C are m×n , n×p and p×q matrices respectively then (AB)C=A(BC)
For example, you can multiply matrix A by matrix B and then multiply the result by matrix C, or you can multiply matrix B by matrix C and then multiply the result by matrix A.
When applying this property, keep in mind the order in which the matrices are multiplied because matrix multiplication is not commutative.
Example 1:
We can find (AB)C as follows:
We can find A(BC) as follows:
Notice that ABC= A(BC).
### Multiplicative Identity Property
The n×n identity matrix denoted I n is a matrix with n rows and n columns. The entries on the diagonal from the upper left to the bottom right are all ones, and all other entries are zeroes.
For example:
The multiplicative identity property states that the product of any n×n matrix A and I n is always A, regardless of the order in which the multiplication was performed. In other words, A∙I=I∙A=A
The role that the n×n identity matrix plays in matrix multiplication is like the role that the number 1 plays in the real number system. If a is a real number, then we know that a∙1=a and 1∙a=a
### Multiplicative Property of Zero
A zero matrix is a matrix in which all the entries are 0. For example, the 3×3 zero matrix is O3×3=[0 0 0 0 0 0 0 0 0]
A zero matrix is indicated by O, and a subscript can be added to indicate the dimensions of the matrix if necessary.
The multiplicative property of zero states that the product of any n×n matrix and the n×n zero matrix is the n×n zero matrix. In other words, A∙O=O∙A=O.
The role that the n×n zero matrix plays in matrix multiplication is like the role that the number 0 plays in the real number system. If a is a real number, then we know that a∙0=0 and 0∙a=0
### The Dimension Property
The dimension property is a property that is unique to matrices. This property has two parts:
1. If the number of columns in the first matrix equals the number of rows in the second matrix, the product of the two matrices is defined.
1. If the product is defined, the resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix.
For example, if A is a 3×2 matrix and B is a 2×4 matrix, the dimension property tells us that the product AB is defined and AB will be a 3×4 matrix.
## Where can we apply matrix multiplication?
Matrix multiplication has historically been used to simplify and clarify computations in linear algebra. This strong relationship between linear algebra and matrix multiplication continues to be fundamental in all mathematics, as well as physics, chemistry, engineering, and computer science.
The entries in a matrix can represent data as well as mathematical equations. Multiplying matrices can provide quick but accurate approximations to much more complicated calculations in many time-critical engineering applications.
Matrices emerged as a way to describe systems of linear equations, a type of problem that anyone who has taken grade-school algebra is familiar with. The term “linear” simply means that the variables in the equations have no exponents, so their graphs are always straight lines.
The equation x-2y=0, for instance, has an infinite number of solutions for both y and x, which can be depicted as a straight line that passes through the points (0,0), (2,1), (4,2), and so on. But if you combine it with the equation x -y=1, there’s only one solution: x=2 and y=1. The point (2,1) is also where the graphs of the two equations intersect.
The matrix illustrating those two equations would be a two-by-two grid of numbers, with the top row being [1 -2] and the bottom row being [1 -1], to correspond to the coefficients of the variables in the two equations.
Computers are frequently called upon to solve systems of linear equations — usually with many more than two variables — in a variety of applications ranging from image processing to genetic analysis. They’re also frequently asked to multiply matrices.
Matrix multiplication is analogous to solving linear equations for specific variables. Consider the expressions t + 2p + 3h, 4t + 5p + 6h, and 7t + 8p + 9h, which describe three different mathematical operations involving temperature, pressure, and humidity measurements. They could be represented as a three-row matrix: [1 2 3], [4 5 6], and [7 8 9].
Assume you take temperature, pressure, and humidity readings outside your home at two different times. Those readings could also be represented as a matrix, with the first set of readings in one column and the second set of readings in the other. Multiplying these matrices entails matching up rows from the first matrix, which describes the equations, and columns from the second, which represents the measurements, multiplying the corresponding terms, adding them all up, and entering the results in a new matrix. The numbers in the final matrix, for example, could predict the path of a low-pressure system.
Of course, reducing the complex dynamics of weather-system models to a set of linear equations is a difficult task in and of itself. But this brings up one of the reasons why matrices are so popular in computer science: they allow computers to do a lot of the computational heavy lifting ahead of time. Creating a matrix that produces useful computational results may be difficult, but matrix multiplication is usually not.
Graphics is one of the areas of computer science where matrix multiplication is particularly useful because a digital image is fundamentally a matrix to begin with: The matrix’s rows and columns correspond to the rows and columns of pixels, and the numerical entries correspond to the color values of the pixels. Decoding digital video necessitates matrix multiplication. For example, some researchers were able to build one of the first chips to implement the new high-efficiency video coding standard for ultrahigh-definition TVs. The patterns they discovered in the matrices it employs have a part in this success.
Matrix multiplication can also aid in the processing of digital video and digital sound. A digital audio signal is essentially a series of numbers that represent the variation in air pressure of an acoustic audio signal over time. Matrix multiplication is used in many techniques for filtering or compressing digital audio signals, including the Fourier transform.
Another reason matrices are so useful in computer science is the graphs are as well. A graph is a mathematical construct made up of nodes, usually depicted as circles, and edges usually depicted as lines connecting them. Graphs are commonly used to represent activities of a computer program to the relationships characteristic of logistics problems.
Every graph, on the other hand, can be represented as a matrix, with each column and row representing a node and the value at their intersection representing the strength of the connection between them, which most of the time is zero. Often, converting the graphs into matrices is the most efficient way to analyze them, and solutions to graph-related problems are frequently solutions to systems of linear equations.
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# Solve the Differential Equation x^2y'' +11xy'+25y=0 ?
Jul 3, 2017
$y \left(x\right) = {c}_{1} {x}^{-} 5 + {c}_{2} {x}^{- 5} \ln x$
#### Explanation:
Substitute the variable:
$t = \ln x$
$y \left(x\right) = \phi \left(\ln x\right) = \phi \left(t\right)$
so that:
$y ' \left(x\right) = \frac{1}{x} \phi '$
$y ' ' \left(x\right) = \frac{1}{x} ^ 2 \left(\phi ' ' - \phi '\right)$
substituting in the original equation:
${x}^{2} y ' ' + 11 x y ' + 25 y = 0$
$\phi ' ' - \phi ' + 11 \phi ' + 25 \phi = 0$
$\phi ' ' + 10 \phi ' + 25 \phi = 0$
This is a second order equation with constant coefficients, so we can solve the characteristic equation:
${\lambda}^{2} + 10 \lambda + 25 = 0$
$\left(\lambda + 5\right) = 0$
$\lambda = - 5$
so the general solution is:
$\phi \left(t\right) = {c}_{1} {e}^{- 5 t} + {c}_{2} t {e}^{- 5 t}$
and undoing the substitution:
$y \left(x\right) = \phi \left(\ln x\right) = {c}_{1} {e}^{- 5 \ln x} + {c}_{2} \ln x {e}^{- 5 \ln x}$
$y \left(x\right) = {c}_{1} {x}^{-} 5 + {c}_{2} {x}^{- 5} \ln x$
Jul 3, 2017
$y = \left(A \ln x + B\right) {x}^{- 5}$
#### Explanation:
We have:
${x}^{2} y ' ' + 11 x y ' + 25 y = 0$ ..... [A]
This is a Euler-Cauchy Equation which is typically solved via a change of variable. Consider the substitution:
$x = {e}^{t} \implies x {e}^{- t} = 1$
Then we have,
$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- t} \frac{\mathrm{dy}}{\mathrm{dt}}$, and, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) {e}^{- 2 t}$
Substituting into the initial DE [A] we get:
${x}^{2} \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) {e}^{- 2 t} + 11 x {e}^{- t} \frac{\mathrm{dy}}{\mathrm{dt}} + 25 y = 0$
$\therefore \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) + 11 \frac{\mathrm{dy}}{\mathrm{dt}} + 25 y = 0$
$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} + 10 \frac{\mathrm{dy}}{\mathrm{dt}} + 25 y = 0$ ..... [B]
This is now a second order linear homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.
${m}^{2} + 10 m + 25 = 0$
We can solve this quadratic equation, and we get a real repeated root:
$m = - 5$
Thus the Homogeneous equation [B] has the solution:
$y = \left(A x + B\right) {e}^{- 5 t}$
Now we initially used a change of variable:
$x = {e}^{t} \implies t = \ln x$
So restoring this change of variable we get:
$y = \left(A \ln x + B\right) {e}^{- 5 \ln x}$
$\therefore y = \left(A \ln x + B\right) {x}^{- 5}$
Which is the General Solution.
|
# Unit 8 Section 1 : Expansion of single brackets
In this section we consider how to expand (multiply out) brackets to give two or more terms, as shown below:
3 ( x + 6 ) = 3 x + 18
We will start by revising some negative number operations, then move on to multiplying out the brackets.
### Negative number operations
When you expand brackets, you often need to multiply a mixture of positive and negative items.
If you are multiplying two items with the same sign together, the answer is positive.
For example, +2 × +5 = +10 and -2 × -5 = +10
If you are multiplying two items with different signs together, the answer is negative.
For example, +2 × -5 = -10 and -2 × +5 = -10
Practice Questions
Work out the answer to each of these questions then click on the button marked to see whether you are correct.
(a) What is -7 × +4 ?
(b) What is -3 × -6 ?
(c) What is -4 × 9x ?
(d) What is -3(9 - 13) ? [HINT: -3(9 - 13) means -3 × (9 - 13)]
### Expanding brackets
Look at the expression below:
-3 ( x - 6 )
To expand the brackets, you need to multiply the part outside the brackets by every part inside the brackets.
In this case you need to multiply -3 by both x and -6. Then we combine the results:
Practice Questions
Work out the answer to each of these questions then click on the button marked to see whether you are correct.
(a) Expand the brackets in: -4(x - 3)
(b) Expand the brackets in: -3(9 - x)
(c) Expand the brackets in: 4(2x - 12)
### Expanding more complicated brackets
Here is a more complicated expression with brackets to expand:
-5x ( 2x + 6y )
In this case you need to multiply -5x by both 2x and 6y.
Working: -5x × 2x = -5 × 2 × x × x = -10 × x² = -10x² -5x × 6y = -5 × 6 × x × y = -30 × xy = -30xy
Then we combine the results:
Practice Questions
Work out the answer to each of these questions then click on the button marked to see whether you are correct.
(a) Expand the brackets in: 2x(x + 5)
(b) Expand the brackets in: -4x(x - y)
(c) Expand the brackets in: -7x(3y - 2x)
## Exercises
Work out the answers to the questions below and fill in the boxes. Click on the button to find out whether you have answered correctly. If you are right then will appear and you should move on to the next question. If appears then your answer is wrong. Click on to clear your original answer and have another go. If you can't work out the right answer then click on to see the answer.
Question 1
Work out the answers to the following questions involving negative numbers.
(a) -11 × -4 (b) -6 × 4 (c) 8 × -7 (d) 8x × 4 (e) 6 × (8 - 10) (f) 5 × (3 - 10) (g) 7(11 - 4) (h) -4(6 - 17)
Question 2
Complete the multiplication tables below, and use your answers to expand the brackets below each table.
(a)
× x 2 4
4(x + 2) =
(b)
× x -7 5
5(x - 7) =
(c)
× x 3 4
4(x + 3) =
(d)
× 2x 5 5
5(2x + 5) =
Question 3
Expand the brackets in the questions below. There should be no brackets in your answers.
(a) 4(x + 6) = (b) 3(x - 4) = (c) 5(2x + 6) = (d) 7(3x - 4) = (e) 3(2x + 4) = (f) 8(3x - 9) = (g) (-2)(x - 4) = (h) (-3)(8 - 2x) = (i) 5(3x - 4) = (j) 9(2x + 8) =
Question 4
Fill in the tables to help you expand the brackets in the given expressions below:
(a)
× x -2 x x 2x 2x² x² 2 2x -2 -2x -x -x²
x(x - 2) =
(b)
× x -3 4 4x 4 4+x x -3x -4 12 -12 -12x
4(x - 3) =
Question 5
Fill in the missing terms in the expansions below:
(a) 4x(x + 8) = 4x² + (b) (-3)(2x - 7) = + 21 (c) 4x(x - 9) = 4x² - (d) 6x(x - 7) = 6x² - (e) 3x(x - y) = 3x² - (f) (-4x)(2x + 8) = -8x² -
Question 6
Expand the brackets in the following expressions:
(a) x(x - 7) = (b) x(8 - 2x) = (c) 6x(x + 2) = (d) 4x(3x - 5) = (e) x(x + y) = (f) x(4y - 3x) = (g) 2x(2x + 3y) = (h) 5x(2y - 1) =
Question 7
In each of the questions below, write an expression for the area of the rectangle using brackets, then expand the brackets. Your answers should not include any spaces or multiplication signs (like * or ×). An example of how to answer a question is shown on the right.
(a)
2
x + 4
Expression for the area of the rectangle using brackets:
Now expand the brackets:
(b)
12
x - 5
Expression for the area of the rectangle using brackets:
Now expand the brackets:
(c)
2x
5 + x
Expression for the area of the rectangle using brackets:
Now expand the brackets:
(d)
2x
x + 9
Expression for the area of the rectangle using brackets:
Now expand the brackets:
(e)
2x
3x - 2
Expression for the area of the rectangle using brackets:
Now expand the brackets:
(f)
6 - 2x
4x
Expression for the area of the rectangle using brackets:
Now expand the brackets:
You have now completed Unit 8 Section 1
Your overall score for this section is Correct Answers You answered questions correctly out of the questions in this section. Incorrect Answers There were questions where you used the Tell Me button. There were questions with wrong answers. There were questions you didn't attempt.
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# GRE Math : How to subtract exponents
## Example Questions
### Example Question #1 : How To Subtract Exponents
Simplify: 32 * (423 - 421)
3^21
4^4
3^3 * 4^21
3^3 * 4^21 * 5
3^3 * 4^21 * 5
Explanation:
Begin by noting that the group (423 - 421) has a common factor, namely 421. You can treat this like any other constant or variable and factor it out. That would give you: 421(42 - 1). Therefore, we know that:
32 * (423 - 421) = 32 * 421(42 - 1)
Now, 42 - 1 = 16 - 1 = 15 = 5 * 3. Replace that in the original:
32 * 421(42 - 1) = 32 * 421(3 * 5)
Combining multiples withe same base, you get:
33 * 421 * 5
### Example Question #1551 : Gre Quantitative Reasoning
Quantitative Comparison
Quantity A: 64 – 32
Quantity B: 52 – 42
The two quantities are equal.
Quantity B is greater.
Quantity A is greater.
The relationship cannot be determined from the information given.
Quantity A is greater.
Explanation:
We can solve this without actually doing the math. Let's look at 64 vs 52. 64 is clearly bigger. Now let's look at 32 vs 42. 32 is clearly smaller. Then, bigger – smaller is greater than smaller – bigger, so Quantity A is bigger.
### Example Question #1561 : Gre Quantitative Reasoning
, and is odd.
Quantity A:
Quantity B:
The relationship cannot be determined from the information given.
The two quantities are equal.
Quantity A is greater.
Quantity B is greater.
Quantity A is greater.
Explanation:
The first thing to note is the relationship between (–b) and (1 – b):
(–b) < (1 – b) because (–b) + 1 = (1 – b).
Now when b > 1, (1 – b) < 0 and –b < 0. Therefore (–b) < (1 – b) < 0.
Raising a negative number to an odd power produces another negative number.
Thus (–b)a < (1 – b)a < 0.
Tired of practice problems?
Try live online GRE prep today.
|
# Looking Through Spectacles
## OBJECTIVE: To learn how to find the least common multiple (LCM)
In previous lessons, you've learned about prime factorization, prime and composite numbers, factors and multiples, and divisibility rules. You also learned how to find the GREATEST COMMON FACTOR (GCF). Climbing Down Trees: https://www.geogebra.org/m/hxmuf2dg Going Dutch: https://www.geogebra.org/m/xvdee7pg Sifting Grains: https://www.geogebra.org/m/gamspcmt Looking through the Lens: https://www.geogebra.org/m/arz3t78a In this lesson, you'll learn how to find the LEAST COMMON MULTIPLE (LCM), also known as LEAST COMMON DENOMINATOR (LCD). The LEAST COMMON MULTIPLE (LCM) of two or more counting numbers is the smallest counting number that is divisible by all the given numbers. Just as in the case of the GCF, there are many approaches to finding LEAST COMMON MULTIPLE (LCM). So let's just follow a parallel process similar to finding the GREATEST COMMON FACTOR (GCF)Factor Tree, Venn Diagram, and Continuous Division.
## FACTOR TREE
If we use the FACTOR TREE, we have to list down the prime factors of the numbers whose LCM we're trying to find. Then we multiply the common prime factors just as in finding the GCF, but in addition, we multiply the factors that have been left out to the result. The final result we get is the LCM. Example: What is the LCM of 16 and 24? Prime factors of 16 : 2 x 2 x 2 x 2 Prime factors of 24 : 2 x 2 x 2 x 3 The product of the common factors, 2 x 2 x 2 = 8, which is the GCF. The product of the GCF and what have been left out, 8 x 2 x 3 = 48, which is the LCM.
## VENN DIAGRAM
If we use the Venn Diagram, we can list the common prime factors of 16 and 24 inside the lens (the intersection of the two sets), while the other prime factors are listed inside their respective crescents. We then multiply the numbers inside the lens together with the numbers in the remainder of the spectacles, as follows: 2 x 2 x 2 x 2 x 3 = 48. The LCM is 48.
## CONTINUOUS DIVISION
If we use CONTINUOUS DIVISION, we follow the same procedure as for finding the GCF, but we use a different formula for multiplying the factors. Instead of just using the numbers on the left, we include the numbers at the bottom in the multiplication to find the LCM. The LCM is 2 x 2 x 2 x 2 x 3 = 48. Here's an example using three numbers. The LCM is 2 x 2 x 3 x 3 x 2 x 3 x 5 = 1,080. CAVEAT: For three or more numbers, division by prime factors must be continued until all the numbers in the bottom row are RELATIVELY PRIME, meaning no two numbers must share a common factor. The example below illustrates this important rule. The LCM is 10 x 2 x 3 x 7 x 2 x 2 x 5 = 8,400. Notice that in the second to the last row, only 2 and 4 have a common factor, but we still had to factor out that common factor. If we didn't do that, the LCM would appear to be 16,800, which is WRONG!
Use the applet below for practice. Enter 2 numbers in the entry boxes. Try solving the problem first on a separate sheet of paper, and then verify your answer by clicking Show Solutions. The applet will generate consecutive multiples of the two numbers, which may or may not contain the LCM, depending on the size of the numbers you enter. In any case, the LCM will appear between the two columns. Click LCM for 3 Numbers to work on 3 numbers, and follow the same procedure. Repeat as many times as needed to master the concept.
## In this lesson, you learned how to find the Least Common Multiple.
Future lessons will include the applications of the GCF and LCM. Did you ENJOY today's lesson?
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Hey Developer’s, I’m back with a new topic which is Conditional and Unconditional Probability in the series of statistics foundations.
Quick Refresher
Probability of Union Of Events – Statistics Part 15
So, let’s get started …
Let’s Understand With Example, First
Unconditional Probalility : Probability that it will rain today giving no additional information.
Conditional Probability : Probability that it will rain today given it’s been raining an enitre week.
Here you can see that the unconditional probaility does not depends on other obervations, but the conditional probability depends on the other observations. There are some previous factors which need to take care while working on conditional probability because it depends on that factors.
Conditional Probability
Conditional probability is defined as the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome.
assuming P(B) > 0 .The formula states that the probability of “A given B” and written as P(A|B), where the probability of A depends on that of B happening.
P(A|B) is derived by multiplying the probability of both A and B occurring and dividing that product by the probability of A by itself.
Example,
In a group of 100 sports car buyers, 40 bought alarm systems, 30 purchased bucket seats, and 20 purchased an alarm system and bucket seats. If a car buyer chosen at random bought an alarm system, what is the probability they also bought bucket seats?
Step 1: Figure out P(A). It’s given in the question as 40%, or 0.4.
Step 2: Figure out P(A∩B). This is the intersection of A and B: both happening together. It’s given in the question 20 out of 100 buyers, or 0.2.
P(B|A) = P(A∩B) / P(A) = 0.2 / 0.4 = 0.5.
The probability that a buyer bought bucket seats, given that they purchased an alarm system, is 50%.
Unconditional Probability
Unconditional Probability is defined as the likelihood that an event will take place independent of whether any other events take place or any other conditions are present.
The unconditional probability of an event can be determined by adding up the outcomes of the event and dividing by the total number of possible outcomes.
For example, if a die lands on the number five 15 times out of 60, the unconditional probability of landing on the number five is 25% (15 outcomes /60 total lots = 0.25).
There is one example from one of the blog ,
As a hypothetical example from finance, let’s examine a group of stocks and their returns. A stock can either be a winner, which earns a positive return, or a loser, which has a negative return. Say that out of five stocks, stocks A and B are winners, while stocks C, D, and E are losers. What, then, is the unconditional probability of choosing a winning stock? Since two outcomes out of a possible five will produce a winner, the unconditional probability is 2 successes divided by 5 total outcomes (2 / 5 = 0.4), or 40%.
Next Post will be on Independence of Events
Till then, Stay Connected
Connect with me on :
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Questions from Inside the chapter
Class 10
Chapter 12 Class 10 - Electricity
## How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Question 4 Page 216 How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω? Let R1 = 2 Ω R2 = 3 Ω R3 = 6 Ω The three resistors can be arranged in 8 ways All 3 parallel All 3 in series 2 parallel, 1 series - 3 cases 2 series, 1 parallel - 3 cases In series combination, equivalent resistance is given by Rs = R1 + R2 + R3 In parallel combination, equivalent resistance is given by 1/𝑅_𝑃 = 1/𝑅_1 + 1/𝑅_2 + 1/𝑅_3 Case 1 R1, R2 and R3 in series R = R1 + R2 + R3 R = 2 + 3 + 6 R = 11 Ω Case 2 R1, R2 and R3 in parallel 1/𝑅 = 1/𝑅_1 + 1/𝑅_2 + 1/𝑅_3 1/𝑅 = 1/2 + 1/3 + 1/6 1/𝑅 = (3 + 2 + 1)/6 1/𝑅 = 6/6 1/𝑅 = 1/1 R = 1 Ω Case 3 R1 and R2 in series and R3 parallel to them First we’ll find equivalent of R1 and R2 RS = R1 + R2 RS = 2 + 3 RS = 5 Ω Now the circuit becomes 1/𝑅 = 1/𝑅_𝑆 + 1/𝑅_3 1/𝑅 = 1/5 + 1/6 1/𝑅 = (6 + 5)/30 1/𝑅 = 11/30 R = 30/11 Ω Case 4 R1 and R3 in series and R2 parallel to them First we’ll find equivalent of R1 and R3 RS = R1 + R3 RS = 2 + 6 RS = 8 Ω Now the circuit becomes 1/𝑅 = 1/𝑅_𝑆 + 1/𝑅_2 1/𝑅 = 1/8 + 1/3 1/𝑅 = (3 + 8)/24 1/𝑅 = 11/24 R = 24/11 Ω Case 5 R2, and R3 in series and R1 parallel to them First we’ll find equivalent of R2 and R3 RS = R2 + R3 RS = 3 + 6 RS = 9 Ω Now the circuit becomes 1/𝑅 = 1/𝑅_𝑆 + 1/𝑅_1 1/𝑅 = 1/9 + 1/2 1/𝑅 = (2 + 9)/18 1/𝑅 = 11/18 R = 18/11 Ω Case 6 R1 and R2 in parallel and R3 in series with them First we’ll find equivalent of R1 and R2 1/𝑅_𝑝 = 1/𝑅_1 + 1/𝑅_2 1/𝑅_𝑝 = 1/2 + 1/3 1/𝑅_𝑝 = (3 + 2)/6 1/𝑅_𝑝 = 5/6 𝑅_𝑝 = 6/5 Ω Now the circuit becomes R = Rp + R3 R = 6/5 + 6 R = (6 + 30)/5 R = 36/5 Ω Case 7 R1 and R3 in parallel and R2 in series with them First we’ll find equivalent of R1 and R3 1/𝑅_𝑝 = 1/𝑅_1 + 1/𝑅_3 1/𝑅_𝑝 = 1/2 + 1/6 1/𝑅_𝑝 = (3 + 1)/6 1/𝑅_𝑝 = 4/6 𝑅_𝑝 = 6/4 𝑅_𝑝 = 3/2 Ω Now the circuit becomes R = Rp + R2 R = 3/2 + 3 R = (3 + 6)/2 R = 9/2 Ω Case 8 R2 and R3 in parallel and R1 in series with them First we’ll find equivalent of R1 and R3 1/𝑅_𝑝 = 1/𝑅_2 + 1/𝑅_3 1/𝑅_𝑝 = 1/3 + 1/6 1/𝑅_𝑝 = (2 + 1)/6 1/𝑅_𝑝 = 3/6 𝑅_𝑝 = 1/2 𝑅_𝑝 = 2 Ω Now the circuit becomes R = Rp + R1 R = 2 + 2 R = 4 Ω Therefore, To give a total resistance of 4 Ω 3 Ω and 6 Ω in parallel and 2 Ω in series with them (b) To give a total resistance of 1 Ω Connect 2 Ω, 3 Ω and 6 Ω in parallel
Made by
#### Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo
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# Laws of Rational Trigonometry
Norman Wildberger is the creator of rational trigonometry, an alternative to classical trigonometry. After using this beautiful theory for about 10 years, I think the main advantages for me have been the ease of derivation and the ease of interpretation. I find it easier and easier to derive novel formulas by using the laws that I will discuss below. But even more important to me is that I find it easier to interpret the formulas in comparison with their classic counterparts. Let me explain.
I am a scientist, so the formulas need to tell me something about the actual physical situation that they are designed to represent. In my case they express relations between observable variables (measurements with one or more cameras) and not directly observable variables (3D properties of objects and their configurations). The formulas help me more directly understand the relations, since they are algebraic expressions using only addition, subtraction, multiplication, division, and an occasional square root.
## Main laws
The crucial starting point is that Wildberger uses quadrances and spreads instead of distances and angles. The translation between the old and the new entities is as follows. A quadrance $Q$ is the square of the distance $d$. Quadrances have dimension $m^2$.
$Q=d^2$
A spread $s$ is the square of the sine of an angle $\alpha$. Spreads are dimensionless.
$s = \sin^2 \alpha$
So, a generic triangle has three quadrances and three spreads. Note that the line segment with quadrance $Q_i$ is opposite of the meet of two line segments with spread $s_i$.
The quadrances are denoted with a rectangular box in the middle of the line segment. Spreads are denoted with a straight line in the corner between two line segments.
In my experience it takes some getting used to this new way of thinking, but I promise that rational trigonometry will pay off in a big way later.
$\frac{s_1}{Q_1} = \frac{s_2}{Q_2} = \frac{s_3}{Q_3}$
### Cross law
$(Q_1+Q_2 - Q_3)^2 = 4 Q_1 Q_2 (1 - s_3)$
$(s_1 + s_2 + s_3)^2 = 2(s_1^2 + s_2^2 + s_3^2) + 4 s_1 s_2 s_3$
## Special cases
The other two laws are for special triangles.
### Pythagorean theorem
If spread $s_3$ is equal to 1, the triangle turns into a right triangle. We get the very familiar formula of Pythagoras:
$Q_1+Q_2 = Q_3$
If all spreads are equal to 0. Or in other words, three points are on a line and the three quadrances are related as:
$(Q_1+Q_2 + Q_3)^2 = 2(Q_1^2 + Q_2^2 + Q_3^2)$
Note that all quadrances have equal status. You can view it as a triangle with an area equal to 0.
There is also a non-symmetrical version of this formula that can easily be derived from the cross law.
$(Q_1+Q_2 - Q_3)^2 = 4 Q_1 Q_2$
In general, this is one the hardest formulas to wrap your mind around. Since we are so used to lengths being additive, this is may be confusing at first.
## Applying the laws in real life
I want to remind you here that I am not a mathematician. My goal is to derive formulas that are useful in the domain that is the core theme if this website: 3D computer vision.
It may occasionally become clear that I am scientist (trained as a physicist) since I will regularly check the dimensions of my formulas. And I will interpret the formulas in terms of entities that I can measure with rulers and protractors.
If you have any compliments or questions, contact me here.
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Jacob Bernoulli
(1654-1705)
vEvery body of discovery is mathematical in form because there is no
other guidance we can have – DARWINv
7.1 Introduction
Suppose you have a suitcase with a number lock. The number
lock has 4 wheels each labelled with 10 digits from 0 to 9.
The lock can be opened if 4 specific digits are arranged in a
particular sequence with no repetition. Some how, you have
forgotten this specific sequence of digits. You remember only
the first digit which is 7. In order to open the lock, how
many sequences of 3-digits you may have to check with? To
answer this question, you may, immediately, start listing all
possible arrangements of 9 remaining digits taken 3 at a
time. But, this method will be tedious, because the number
of possible sequences may be large. Here, in this Chapter,
we shall learn some basic counting techniques which will enable us to answer this
question without actually listing 3-digit arrangements. In fact, these techniques will be
useful in determining the number of different ways of arranging and selecting objects
without actually listing them. As a first step, we shall examine a principle which is most
fundamental to the learning of these techniques.
7.2 Fundamental Principle of Counting
Let us consider the following problem. Mohan has 3 pants and 2 shirts. How many
different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which
a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be
chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore,
there are 3 × 2 = 6 pairs of a pant and a shirt.
7
Chapter
PERMUTATIONS AND COMBINATIONS
2020-21
PERMUTATIONS AND COMBINATIONS 135
Let us name the three pants as P
1
, P
2
, P
3
and the two shirts as S
1
, S
2
. Then,
these six possibilities can be illustrated in the Fig. 7.1.
Let us consider another problem
of the same type.
Sabnam has 2 school bags, 3 tiffin boxes
and 2 water bottles. In how many ways
can she carry these items (choosing one
each).
A school bag can be chosen in 2
different ways. After a school bag is
chosen, a tiffin box can be chosen in 3
different ways. Hence, there are
2 × 3 = 6 pairs of school bag and a tiffin
box. For each of these pairs a water
bottle can be chosen in 2 different ways.
Hence, there are 6 × 2 = 12 different ways in which, Sabnam can carry these items to
school. If we name the 2 school bags as B
1
, B
2
, the three tiffin boxes as T
1
, T
2
, T
3
and
the two water bottles as W
1
, W
2
, these possibilities can be illustrated in the Fig. 7.2.
Fig 7.1
Fig 7.2
2020-21
136 MATHEMATICS
In fact, the problems of the above types are solved by applying the following
principle known as the fundamental principle of counting, or, simply, the multiplication
principle, which states that
“If an event can occur in m different ways, following which another event
can occur in n different ways, then the total number of occurrence of the events
in the given order is m×n.”
The above principle can be generalised for any finite number of events. For
example, for 3 events, the principle is as follows:
‘If an event can occur in m different ways, following which another event can
occur in n different ways, following which a third event can occur in p different ways,
then the total number of occurrence to ‘the events in the given order is m × n × p.”
In the first problem, the required number of ways of wearing a pant and a shirt
was the number of different ways of the occurence of the following events in succession:
(i) the event of choosing a pant
(ii) the event of choosing a shirt.
In the second problem, the required number of ways was the number of different
ways of the occurence of the following events in succession:
(i) the event of choosing a school bag
(ii) the event of choosing a tiffin box
(iii) the event of choosing a water bottle.
Here, in both the cases, the events in each problem could occur in various possible
orders. But, we have to choose any one of the possible orders and count the number of
different ways of the occurence of the events in this chosen order.
Example 1
Find the number of 4 letter words, with or without meaning, which can be
formed out of the letters of the word ROSE, where the repetition of the letters is not
allowed.
Solution
There are as many words as there are ways of filling in 4 vacant places
by the 4 letters, keeping in mind that the repetition is not allowed. The
first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following
which, the second place can be filled in by anyone of the remaining 3 letters in 3
different ways, following which the third place can be filled in 2 different ways; following
which, the fourth place can be filled in 1 way. Thus, the number of ways in which the
4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24. Hence, the
required number of words is 24.
2020-21
PERMUTATIONS AND COMBINATIONS 137
A
Note If the repetition of the letters was allowed, how many words can be formed?
One can easily understand that each of the 4 vacant places can be filled in succession
in 4 different ways. Hence, the required number of words = 4 × 4 × 4 × 4 = 256.
Example 2
Given 4 flags of different colours, how many different signals can be
generated, if a signal requires the use of 2 flags one below the other?
Solution
There will be as many signals as there are ways of filling in 2 vacant places
in succession by the 4 flags of different colours. The upper vacant place can
be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant
place can be filled in 3 different ways by anyone of the remaining 3 different flags.
Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12.
Example 3 How many 2 digit even numbers can be formed from the digits
1, 2, 3, 4, 5 if the digits can be repeated?
Solution There will be as many ways as there are ways of filling 2 vacant places
in succession by the five given digits. Here, in this case, we start filling in unit’s
place, because the options for this place are 2 and 4 only and this can be done in 2
ways; following which the ten’s place can be filled by any of the 5 digits in 5 different
ways as the digits can be repeated. Therefore, by the multiplication principle, the required
number of two digits even numbers is 2 × 5, i.e., 10.
Example 4 Find the number of different signals that can be generated by arranging at
least 2 flags in order (one below the other) on a vertical staff, if five different flags are
available.
Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us
count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags
separately and then add the respective numbers.
There will be as many 2 flag signals as there are ways of filling in 2 vacant places
in succession by the 5 flags available. By Multiplication rule, the number of
ways is 5 × 4 = 20.
Similarly, there will be as many 3 flag signals as there are ways of filling in 3
vacant places in succession by the 5 flags.
2020-21
138 MATHEMATICS
The number of ways is 5 × 4 × 3 = 60.
Continuing the same way, we find that
The number of 4 flag signals = 5 × 4 × 3 × 2 = 120
and the number of 5 flag signals = 5 × 4 × 3 × 2 × 1 = 120
Therefore, the required no of signals = 20 + 60 + 120 + 120 = 320.
EXERCISE 7.1
1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5
assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the
digits can be repeated?
3. How many 4-letter code can be formed using the first 10 letters of the English
alphabet, if no letter can be repeated?
4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if
each number starts with 67 and no digit appears more than once?
5. A coin is tossed 3 times and the outcomes are recorded. How many possible
outcomes are there?
6. Given 5 flags of different colours, how many different signals can be generated if
each signal requires the use of 2 flags, one below the other?
7.3 Permutations
In
Example 1 of the previous Section, we are actually counting the different possible
arrangements of the letters such as ROSE, REOS, ..., etc. Here, in this list, each
arrangement is different from other. In other words, the order of writing the letters is
important. Each arrangement is called a permutation of 4 different letters taken all
at a time. Now, if we have to determine the number of 3-letter words, with or without
meaning, which can be formed out of the letters of the word NUMBER, where the
repetition of the letters is not allowed, we need to count the arrangements NUM,
NMU, MUN, NUB, ..., etc. Here, we are counting the permutations of 6 different
letters taken 3 at a time. The required number of words = 6 × 5 × 4 = 120 (by using
multiplication principle).
If the repetition of the letters was allowed, the required number of words would
be 6 × 6 × 6 = 216.
2020-21
PERMUTATIONS AND COMBINATIONS 139
Definition 1
A permutation is an arrangement in a definite order of a number of
objects taken some or all at a time.
In the following sub-section, we shall obtain the formula needed to answer these
questions immediately.
7.3.1 Permutations when all the objects are distinct
Theorem 1
The number of permutations of n different objects taken r at a time,
where 0 < r n and the objects do not repeat is n ( n – 1) ( n – 2). . .(
nr + 1),
which is denoted by
n
P
r
.
Proof There will be as many permutations as there are ways of filling in r vacant
places
. . . by
r vacant places
the n objects. The first place can be filled in n ways; following which, the second place
can be filled in (n – 1) ways, following which the third place can be filled in (n – 2)
ways,..., the rth place can be filled in (n – (r – 1)) ways. Therefore, the number of
ways of filling in r vacant places in succession is n(n – 1) (n – 2) . . . (n – (r – 1)) or
n ( n – 1) (n – 2) ... (n r + 1)
This expression for
n
P
r
is cumbersome and we need a notation which will help to
reduce the size of this expression. The symbol n! (read as factorial n or n factorial )
comes to our rescue. In the following text we will learn what actually n! means.
7.3.2 Factorial notation The notation n! represents the product of first n natural
numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this
symbol as ‘n factorial’. Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !
1 = 1 !
1 × 2 = 2 !
1× 2 × 3 = 3 !
1 × 2 × 3 × 4 = 4 ! and so on.
We define 0 ! = 1
We can write 5 ! = 5 × 4 ! = 5 × 4 × 3 ! = 5 × 4 × 3 × 2 !
= 5 × 4 × 3 × 2 × 1!
Clearly, for a natural number n
n ! = n (n –
1) !
= n (n – 1) (n – 2) ! [provided (n 2)]
= n (n – 1) (n – 2) (n – 3) ! [provided (n 3)]
and so on.
2020-21
140 MATHEMATICS
Example 5 Evaluate (i) 5 ! (ii) 7 ! (iii) 7 ! – 5!
Solution (i) 5 ! = 1 × 2 × 3 × 4 × 5 = 120
(ii) 7 ! = 1 × 2 × 3 × 4 × 5 × 6 ×7 = 5040
and (iii) 7 ! – 5! = 5040 – 120 = 4920.
Example 6 Compute (i)
7!
5!
(ii)
( )
12!
10! (2!)
Solution (i) We have
7!
5!
=
7 6 5!
× ×
= 7 × 6 = 42
and (ii)
( ) ( )
12!
10! 2!
=
(
)
( ) ( )
12 11 10!
10! 2
× ×
×
= 6 × 11 = 66.
Example 7 Evaluate
( )
!
! !
n
r n r
, when n = 5, r = 2.
Solution We have to evaluate
( )
5!
2! 5 2 !
(since n = 5, r = 2)
We have
( )
5!
2 ! 5 2 !
=
5! 5 4
10
2! 3! 2
×
= =
×
.
Example 8 If
1 1
8! 9! 10!
x
+ =
, find x.
Solution We have
1 1
8! 9 8! 10 9 8!
x
+ =
× × ×
Therefore
1
1
9 10 9
x
+ =
×
or
10
9 10 9
x
=
×
So x = 100.
EXERCISE 7.2
1. Evaluate
(i) 8 ! (ii) 4 ! – 3 !
2020-21
PERMUTATIONS AND COMBINATIONS 141
2. Is 3 ! + 4 ! = 7 ! ? 3. Compute
8!
6! 2!
×
4. If
1 1
6! 7! 8!
x
+ =
, find x
5. Evaluate
( )
!
!
n
n r
, when
(i) n = 6, r = 2 (ii) n = 9, r = 5.
7.3.3 Derivation of the formula for
n
P
r
( )
!
P
!
n
r
n
n r
=
, 0 r n
Let us now go back to the stage where we had determined the following formula:
n
P
r
= n (n – 1) (n – 2) . . . (nr + 1)
Multiplying numerator and denomirator by (nr) (nr – 1) . . . 3 × 2 × 1, we get
(
)
(
)
(
)
(
)
(
)
( )( )
1 2 1 1 3 2 1
P
1 3 2 1
n
r
n n n ... n r n r n r ...
n r n r ...
+ × ×
=
× ×
=
( )
!
!
n
n r
,
Thus
( )
!
P
!
n
r
n
n r
=
, where 0 < r
n
This is a much more convenient expression for
n
P
r
than the previous one.
In particular, when r = n,
!
P !
0!
n
n
n
n
= =
Counting permutations is merely counting the number of ways in which some or
all objects at a time are rearranged. Arranging no object at all is the same as leaving
behind all the objects and we know that there is only one way of doing so. Thus, we
can have
n
P
0
= 1 =
! !
! ( 0)!
=
n n
n n
... (1)
Therefore, the formula (1) is applicable for r = 0 also.
Thus
( )
!
P 0
!
n
r
n
, r n
n r
=
.
2020-21
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HOME David Terr Ph.D. Math, UC Berkeley << P.10. Quadratic Equations P.11. Linear Inequalities Thus far, we have only discussed algebraic equations. What about inequalities? The simplest type of equalities are linear inequalities in one variable. (Linear inequalities in two or more variables are considerably more complicated and require graphical solutions in two or more dimensions.) The general forms of a linear inequality in one variable are ax + b > c, ax + b ≥ c, ax + b ≤ c, or ax + b < c, where a, b, and c are arbitrary real constants with a nonzero. The solutions to linear inequalities can be expressed graphically using the number line or using interval notation. To represent an interval graphically, we draw a solid line above the number line corresponding to the points lying within the interval. We also draw an open or closed circle at the endpoint(s) of the interval, open if the point is contained in the interval and closed otherwise. Interval notation expresses an interval in any of the four forms (a,b), [a,b), (a,b], or [a,b], where a and b are the endpoints of the interval with a -3. The interval notation for this solution set is (-3,∞).The solution set is an open ray with (missing) endpoint at x = -3 and pointing to the right as shown: Often one is presented with two simultaneous linear inequalities, such as the following example : Example 3: Solve the linear inequalities 3x + 8 ≥ 5 and 10 - x > 7. Solution: We solve these inequalities one at a time. The first inequality simplifies to x ≥ -1 and the second to x < 3. The interval notation for this solution set is [-1,3).The solution set is thus the intersection of the rays corresponding to each of these inequalities, which is a line segment with endpoint x = -1 and missing endpoint at x = 3 as shown: << P.10. Quadratic Equations Copyright © 2007-2009 - MathAmazement.com. All Rights Reserved.
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# How do you graph 3x = 2y by plotting points?
Sep 1, 2017
Plug in values of x and see what their y value is
#### Explanation:
Draw up a table of $x$ and y values, then plot on a chart.
I'd manipulate the equation to $y = \frac{3}{2} x$ because the arithmetic is easier.
$x$ | $y$
$1 | \frac{3}{2} \cdot 1 = \frac{3}{2}$ when $x = 1 , y = \frac{3}{2}$
$2 | \frac{3}{2} \cdot 2 = \frac{6}{2} = 3$
$3 | \frac{3}{2} \cdot 3 = \frac{9}{2} = 4.5$
$4 | \frac{3}{2} \cdot 4 = 6$
$5 | \frac{3}{2} \cdot 5 = 7.5$
$6 | \frac{3}{2} \cdot 6 = 9$
By now you should be able to spot that the difference between the y values is always $+ 1.5$, or $+ \frac{3}{2}$ This means that the gradient (slope) of the line is $\frac{3}{2}$.
graph{3x=2y [-10, 10, -5, 5]}
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NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations – Here are all the Class 10 Maths Chapter 4 Quadratic Equations NCERT Solutions. This solution contains questions, answers, images, step by step explanation of the complete Chapter 4 titled Quadratic Equations in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 4 Quadratic Equations. After you have studied lesson, you must be looking for answers to its textbook questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations in one place.
Class 10 Maths Ex. 4.1
# Exercise 4.1 Class 10 Maths Question 2 Solution
Ex. 4.1 Class 10 Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m2 The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken
Ex. 4.1 Class 10 Question 1.
Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
(vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x (x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3
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# 5 1 Rate of Change and Slope Rate
• Slides: 33
5. 1 Rate of Change and Slope Rate of Change: The relationship between two changing quantities Rate of Change = Change in the dependent variable (y-axis) Change in the independent variable (x-axis) Slope: the ratio of the vertical change (rise) to the horizontal change (run). Slope = Vertical Change (y) Horizontal Change (x) = rise run
Real World:
Rate of Change can be presented in many forms such as: We can use the concept of change to solve the cable problem by using two sets of given data, for example: A band practices their march for the parade over time as follows:
Choosing the data from: Time and Distance 1 min 260 ft. 2 min 520 ft. We have the following:
Choosing the data from: Time and Distance 1 min 260 ft. 3 min 780 ft. We have the following:
Choosing the data from: Time and Distance 1 min 260 ft. 4 min 1040 ft. We have the following:
NOTE: When we get the same slope, no matter what date points we get, we have a CONSTANT rate of change:
YOU TRY IT: Determine whether the following rate of change is constant in the miles per gallon of a car. Gallons Miles 1 28 3 84 5 140 7 196
Choosing the data from: Gallons and Miles 1 g 28 m 3 g 84 m We have the following:
Choosing the data from: Gallons and Miles 1 g 28 m. 5 g 140 m. We have the following: THUS: the rate of change is CONSTANT.
Once Again: Real World X Y
Remember: Rate of Change can be presented in many forms: We can use the concept of change to solve the cable problem by using two sets of given data: (x , y) A : Horizontal(x) = 20 Vertical(y) = 30 (20, 30) B : Horizontal(x) = 40 Vertical(y) = 35 (40, 35)
Using the data for A and B and the definition of rate of change we have: (x , y) A : Horizontal = 20 Vertical = 30 (20, 30) B : Horizontal = 40 Vertical = 35 (40, 35)
Using the data for B and C and the definition of rate of change we have: (x , y) B : Horizontal = 40 Vertical = 35 (40, 35) C : Horizontal = 60 Vertical = 60 (60, 60)
Using the data for C and D and the definition of rate of change we have: (x , y) C : Horizontal = 60 Vertical = 60 (60, 60) D : Horizontal = 100 Vertical = 70 (100, 70)
Comparing the slopes of the three: However, we must find all the combination that we can do. Try from A to C, from A to D and from B to C.
Finally: Finally we can conclude that the poles with the steepest path are poles B to C with slope of 5/4.
Class Work: Pages: 295 -297 Problems: 1, 4, 8, 9,
Remember: When we get the same slope, no matter what date points we get, we have a CONSTANT rate of change:
When we get the same slope, no matter what date points we get, we have a CONSTANT rate of change: We further use the concept of CONSTANT slope when we are looking at the graph of a line:
We further use the concept of rise/run to find the slope: run rise Make a right triangle to get from one point to another, that is your slope.
CONSTANT rate of change: due to the fact that a line is has no curves, we use the following formula to find the SLOPE: y 2 -y 1 x 2 -x 1 B(x 2, y 2) A(x 1, y 1) A = (1, -1) B = (2, 1)
YOU TRY: Find the slope of the line:
YOU TRY (solution): -4 (0, 4) (2, 0) 2
YOU TRY IT: Provide the slope of the line that passes through the points A(1, 3) and B(5, 5):
YOU TRY IT: (Solution) Using the given data A(1, 3) and B(5, 5) and the definition of rate of change we have: A( 1 , 3 ) B(5 , 5) (x 1, y 1) (x 2, y 2)
YOU TRY: Find the slope of the line:
YOU TRY IT: (Solution) Choosing two points say: A(-5, 3) and B(1, 5) and the definition of rate of change (slope) we have: A( -2 , 3 ) B(1 , 3) (x 1, y 1) (x 2, y 2)
YOU TRY: Find the slope of the line:
YOU TRY IT: (Solution) Choosing two points say: A(-1, 2) and B(-1, -1) and the definition of rate of change (slope) we have: A( -1 , 2 ) B(-1 , -1) (x 1, y 1) (x 2, y 2) We can never divide by Zero thus our slope = UNDEFINED.
THEREFORE: Horizontal ( slope of ZERO ) lines have a While vertical ( ) lines have an UNDEFINED slope.
VIDEOS: Graphs https: //www. khanacademy. org/math/algebra/line ar-equations-and-inequalitie/slope-andintercepts/v/slope-and-rate-of-change
Class Work: Pages: 295 -297 Problems: As many as needed to master the concept
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Series - Maple Help
Series
Main Concept
In mathematics, a series is the sum of the terms of a sequence. Finite sequences and series have defined first and last terms, while infinite sequences and series continue on indefinitely.
Given an infinite sequence $\left\{{a}_{n}\right\}$, we write the infinite series as
When adding up only the first n terms of a sequence, we refer to the ${\mathbit{n}}^{\mathbit{th}}$ partial sum:
So, there are two sequences associated with any series $\sum {a}_{n}$ :
• $\left\{{a}_{n}\right\}$, the sequence of its terms
• $\left\{{s}_{n}\right\}$, the sequence of its partial sums
A series is said to converge if the sequence of its partial sums, $\left\{{s}_{n}\right\}$, converge. The finite limit of ${s}_{n}$ as n approaches infinity is then called the sum of the series:
.
This means that by adding sufficiently many terms of the series, we can get very close to the value of S. If $\left\{{s}_{n}\right\}$ diverges, then the series diverges as well.
Finding S is often very difficult, and so the main focus when working with series is often just testing to figure out whether the series converges or diverges.
Choose a closed formula for a sequence from the drop-down menu below, or type your own formula in the text box and click "Enter" to see a plot of the first N partial sums. Use the slider to adjust how many points are plotted and select the check box to find out if this sequence converges or diverges.
Enter Functionn1/n1 + 3*(n - 1)1/(2^n)n/(3^n)n^10/(2^n)(2*n + 1)^21/(n-1)!2^(n - 1)cos(n)/nsin(1/n)sin(1/(n^2)) + cos(1/n)(-1)^n/n(-1)^n/sqrt(n)-1/(-2)^nn*(-1)^n
Plot the sequence for
More MathApps
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# Chapter 5: Connectivity Section 5.1: Vertex- and Edge-Connectivity
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1 Chapter 5: Connectivity Section 5.1: Vertex- and Edge-Connectivity Let G be a connected graph. We want to measure how connected G is. Vertex cut: V 0 V such that G V 0 is not connected Edge cut: E 0 E such that G E 0 is not connected Vertex connectivity of connected graph: κ v (G) = minimum size V 0 such that G V 0 is disconnected or a single vertex. (We will say disconnected to mean either). κ v (K n ) = n 1. If κ v (G) k then you must delete at least k vertices to disconnect G. G is k-vertexconnected then it has this property. Another way to say this: result is connected when you delete any k 1 vertices. Edge connectivity of connected graph: κ e (G) = minimum size E 0 such that G E 0 is disconnected. If κ e (G) k then you must delete at least k edges to disconnect G. Result is connected when you delete any k 1 edges. G is k-edge-connected then it has this property. κ e (G) δ. Reason: you can disconnect graph by deleting δ edges. So the minimum number is δ. Partition cut: an edge cut with the following property: you can 2-color the edges in the cut. Proposition 5.1.2: A graph G is k-connected if and only if every partition-cut contains at least k edges. Proof: Assume G is k-connected. Then no edge cut contains k 1 edges, in particular any partition-cut. So every partition-cut contains at least k edges. Conversely, suppose every partition-cut contains at at least k edges. Suppose that G E 0 is not connected. Then there is a subset E 1 E 0 in which G E 1 has exactly two components. Color the vertices in one of the components 0, and color the vertices in the other components 1. Let P be the edges in G which have edges with opposite color endpoints. Claim: P E 1. Reason: If there is an edge in P but not in E 1, it would connect the two components of G E 1, which is impossible. Therefore E 0 E 1 P k. Hence G is k-connected. Theorem: κ e (G) κ v (G). Proof: By induction on κ e (G). If κ e (G) = 1, then G has a bridge edge. If there are exactly 2 vertices, then deleting a vertex reduces graph to single vertex. If 3 vertices, then deleting the right vertex will disconnect graph. Hence κ v (G) = 1. Now consider κ e (G) = k. Pick any edge e in a cut of size k. Then κ e (G e) k 1. Hence κ v (G e) k 1. Therefore by the induction hypothesis it is possible to disconnect G e by removing k 1 vertices. If in removing V k 1 we also removed e, then removing V k 1
2 disconnects G. If we did not remove e, then e is a bridge edge of G V k 1. If just two vertices in G V k 1, then removing another vertex takes to one vertex, which disconnects G. If more than two vertices, use the same argument as above to remove another vertex. Hence we can disconnect G with k vertices. Hence κ v (G) k = κ e (G). Definition: two paths from u to v (u v) are internally disjoint if they have no common internal vertex. They can be glued together to form a cycle. Theorem: Let G be connected with 3 or more vertices. G is 2-connected iff for every pair u v there are two internally disjoint paths between them. Proof: Assume the condition on paths holds. We must show that no vertex disconnects graph. Cut x, and consider u v in G x. There are two internally disjoint paths from u to v in G. x cannot be internal to both. Therefore one of them survives in G x, and this supplies a path between them in G x. Conversely, assume G is 2-connected, and let u v be given. Suppose every pair of distinct paths between then share an internal vertex somewhere. We will construct internally disjoint paths by induction on the distance between u and v. distance = 1: show they belong to a cycle. κ e (G) κ(g) = 2, hence cutting one edge doesn t disconnect graph. Cut uv edge. There is still path from u to v. This is internally disjoint to the uv edge. Now assume this can be done for distance < k. Consider d(u, v) = k. Let d(u, w) = k 1, where w and v are connected by an edge. There are internally disjoint paths between u and w by induction hypothesis. Call them P and Q. Let R be uv path in G w. Case 1: R never intersects P or Q. Use R and P + wv. Case 2: R intersects P or Q. Let z be the last vertex it hits, without loss of generality in P. First path: P from u to z then R from z to v. Second path: Q from u to w then wz. NOTE: typo in problem Expansion Lemma: If G is k-connected, and we add a new vertex V with edges to k existing vertices and call this G, then G is also k-connected. Proof: Cut k 1 vertices. We must show that G V k 1 is still connected. We know that G V k 1 is still connected. If V was deleted, then G V k 1 = G V k 1. But if V was not deleted, there is surviving edge from V to G, so there are paths from V to every other vertex in G V k 1. Theorem: Let G be graph with at least 3 vertices. FAE: 1. G is connected and has no cut-vertex. 2. G is 2-connected. 3. Every pair u v has 2 internally disjoint paths. 4. Every pair u v belongs to cycle.
3 5. There are no isolated vertices, and every pair of edges belongs to a cycle. Proof: 1 through 4 are equivalent. 4 implies 5: Assume every pair of vertices belongs to a cycle. This implies 1, so we can assume we have all the properties of 1 through 4. Clearly there are no isolated vertices. Let uv and xy be two edges. If they share and endpoint, we can delete the common endpoint and there will still be a path between the two other endpoints. Hence these edges belong to a cycle. Now suppose uv and xy do not share an endpoint. Create new vertices W and Z and add them to G, connecteding W to u, v and Z to x, y. By expansion lemma, G is 2-connected. Therefore W and Z belong to cycle of G. This can be contracted to cycle through the two edges. 5 implies 1: First show connected. Let u and v be given. If uv edge, they are connected by path. If no uv edge, they belong to two edges which belong to cycle, hence they are connected by path. Next show no cut-vertex. Let x be vertex in graph. Cut x. We must show G x is connected. Let a, b be two vertices left over. If no edge, they belong to a cycle in G by the argument above, hence to two internally disjoint paths in G. One of these survives in G x. Section 5.2: Constructing Reliable Networks Lemma 5.2.1: If you add a path to a 2-connected graph you get a 2-connected graph. Proof: It is still true that every pair of vertices lies on a cycle. Theorem 5.2.2: G is 2-connected if and only if it can be obtained by adding paths to a cycle. Proof: Sufficiency is clear by the lemma. Now assume G is 2-connected. We will construct G by adding paths to a cycle as follows: Since G is 2-connected, it contains a cycle. Therefore it has a subgraph which can be generated by an ear decomposition. We will now show that any proper ear-decomposition subgraph H can be enlarged to a larger one H. Eventually we arrive at G. Let uv be an edge in G which does not belong to H. If both u and v are vertices in H, then the new ear-decomposition subgraph is H + uv and we are done. Otherwise, let xy be any edge in H. Then uv and xy belong to a cycle C of G since G is 2-connected. Since either u or v does not belong to H, we can use C to construct a new ear for H. Lemma 5.2.3: If you add a path or a cycle to a 2-edge-connected graph you get a 2-edge-connected graph. Proof: If no bridges to begin with, then no bridges after the addition.
4 Theorem 5.2.2: G is 2-edge-connected if and only if it can be obtained by adding paths or cycles to a cycle. Proof: Sufficiency is clear by the lemma. Conversely, let G be any 2-edge-connected graph. G contains cycle. We will show that we can continue to extend it by paths and cycles until we get to G. Let H be subgraph of G which results from cycle by adding paths and cycles. Suppose H contains all vertices of G. Then any uv edge not in H can be added to H, creating larger thingy. If on the other hand G contains vertex not in H, then by connectedness there is a frontier edge from vertex in H to vertex not in H. It cannot be a bridge edge of G, so it lives in cycle of G. Follow the cycle both directions until it hooks up with H. If it hooks up with only one vertex of H, it is a closed ear. Otherwise it is an open ear. Add to H, producing H. See Theorem for characterizations of 3-connected graphs. See Theorem for a characterization of k-connected graphs. Proposition 5.2.6: A k-connected graph on n vertices has at least kn 2 Proof: We have 2e nδ nk by the vertex-degree-sum theorem. Specialization: A 2r-connected graph on n vertices at at least rn edges. Achieving this bound: H 2r,n is 2r-connected and has rn edges (2r < n). Construction: vertices are [0], [1],..., [n 1] modulo n. edges. Neighbors of the vertex [a] are [a + 1] through [a + r] and [a 1] through [a r]. These are distinct vertices: If [a p] = [a + q] then p + q is divisible by n. But p + q 2r < n, hence p = q = 0. Contradiction. So the degree of [a] is 2r. This implies that H 2r,n has rn edges. Removing 2r vertices it is always possible to isolate a vertex and disconnect the graph. This makes κ v (H 2r,n ) 2r. Now we must show that if we remove 2r 1 vertices the resulting graph is connected. Remove 2r 1 vertices. Let [u] and [v] remain. If they are neighbors, we re done. Otherwise, the circular path between them in either direction involves r or more internal vertices. Now in one of these directions, at least one internal vertex [z] remains after cutting the 2r 1 vertices. So can we can take an edge from [u] to [z]. If the gap from [z] to [v] is still has r or more internal vertices, we can take an edge from [z] to an internal vertex [z ] in the same direction. Keep on going, then eventually take an edge to [v]. Section 5.3: Menger s Theorems u, v separating set S: No path from u to v in G S. S could be set of vertices or set of edges. Theorem (Menger s Theorem): Let u v be such that uv is not an edge in a connected graph G. Then the maximum number of internally disjoint uv paths in G is equal to the minimum number of vertices in a u, v separating set.
5 Proof: We ll do the easy part now, and use Network Flows in Chapter 13 to do the hard part. We will just show that if P is a set of internally disjoint uv paths and S is any set of vertices whose removal separates u and v, then P S. Let p be one of the paths. At least one internal vertex of p must belong to S, otherwise S doesn t separate. So every p P contributes a vertex to S, and since internally disjoint they must be distinct vertices. Hence P S. Corollary: whenever P = S we must have P max and S min. Definition: κ v (s, t) = minimum number of vertices needed to separate s from t, where s t are not adjacent. Lemma 5.3.5: κ v (G) = min κ v (s, t). Proof: Let κ v (s, t) be minimum possible. If you can separate the graph with κ v (s, t) 1, then you can separate a pair of vertices with this many contradiction. Therefore κ v (G) κ v (s, t). On the other hand, removing κ v (s, t) disconnects s from t, hence disconnects the graph, hence κ v (G) κ v (s, t). So they are equal. Theorem 5.3.6: G is k-connected if and only if for each pair of vertices s t there are at least k internally disjoint st paths in G. Proof: Assume G is k-connected. Let s, t be given, and suppose there are at most k 1 internally disjoint st paths in G. Since the maximum number of internally disjoint paths is < k, by Menger s Theorem we can say that the minimum number of vertices needed to separate s from t is < k. This contradicts k-connected. Hence there at least k internally disjoint st paths in G. Conversely, suppose there are at least k internally disjoint st paths per s t in G. Suppose it is possible to separate G by removing k 1 vertices. Then it is possible to separate a pair s, t by removing k 1 vertices. But this leaves one of the paths intact contradiction. Therefore G cannot be separated by removing k 1 vertices, and G is k-connected. Corollary 5.3.7: Let G be a k-connected graph and let v 0 through v k be distinct vertices in G. Then there are internally disjoint paths from v 0 to v i for 1 i k. Proof: By the Expansion Lemma proved above, we can adjoin a new vertex V to G and edges from V to v 1 through v k, obtaining a k-connected graph G. There are at least k internally disjoint paths from v 0 to V. This implies that there are exactly k internally disjoint paths of the type desired. Theorem 5.3.8: Let G be k connected, where k 3. Then any set of k vertices in G lives in a cycle of G. Proof: We will grow a cycle, using what we know about 2-connected graphs. Let v 1 through v k be any collection of k vertices. through the first j of them by induction on j 2. We will prove that there is a cycle j = 2: There is a cycle called C 2 through v 1 and v 2 since G is 2-connected.
6 Assume there is a cycle C j through v 1 through v j, where j < k. If this cycle includes v j+1 there is nothing to prove. If not, there are internally disjoint paths from v j+1 through v 1 through v j. So there are internally disjoint paths from v j+1 to v 1 and v 2, and these don t contain any other vertex on C j. So we can build C j+1 which incorporates v j+1. See Figure 5.3.4, page 235.
### Connectivity and cuts
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# Specifying Exercise: Chop a Shape Into Simpler Ones
Parent Strategy: Solve a Simpler Problem
Geometry Strategy: Chop a Shape Into Simpler Ones
Slightly More Specific Geometry Strategy – Keep An Eye Out For Simpler Shapes, Especially Right Triangles
Area Version #1: If you’re trying to find the area of a complex figure, it’s often helpful to chop it into shapes whose areas you can more easily find.
Area Version #2: If you’re trying to find (or remember) the formula for the area of a type of shape, try cutting and rearranging that shape’s pieces into a more familiar one. In fact, there are two cuts that often help: cutting off a right triangle and making a midline cut.
This has a natural extension to limits, as we pass to the infinite.
Trigonometry Version: When you’re trying to find missing sides or angles of a polygon, it’s often helpful to chop that shape into triangles and to chop that triangle into right triangles.
Quadrilateral Proof Version: If you’re trying to prove (or remember) something about the sides or angles of a quadrilateral, it’s often helpful to draw one or both diagonals and divide the quadrilateral into two (or four) triangles.
Volume Version: Much like with area. Version 1 says, try to chop up the volume of a complex solid into simpler solids you can tackle individually. Version 2 says, if you’re trying to derive or remember the formula for the volume of a solid, try to chop and rearrange its pieces into a box. There is a natural extension of volume to limits, as we pass to the infinite. (Cavalieri’s Principle)
Auxiliary Lines Version: If you’re stuck on a geometry problem, try adding lines that create right triangles, equilateral triangles or isosceles triangles, because you can often use their properties to illuminate other aspects of the problem.
Questions:
• How does proof by dissection fit into all of this?
• Is this a helpful analysis?
• How would I use this during the year in the course of my planning? (I think this could help me focus on better mathematical goals, give better hints and identify better things to talk about in whole-group discussions. that’s my prediction, anyway.)
|
# Binary Relation
Let P and Q be two non- empty sets. A binary relation R is defined to be a subset of P x Q from a set P to Q. If (a, b) ∈ R and R ⊆ P x Q then a is related to b by R i.e., aRb. If sets P and Q are equal, then we say R ⊆ P x P is a relation on P e.g.
Example1: If a set has n elements, how many relations are there from A to A.
Solution: If a set A has n elements, A x A has n2 elements. So, there are 2n2 relations from A to A.
Example2: If A has m elements and B has n elements. How many relations are there from A to B and vice versa?
Solution: There are m x n elements; hence there are 2m x n relations from A to A.
Example3: If a set A = {1, 2}. Determine all relations from A to A.
Solution: There are 22= 4 elements i.e., {(1, 2), (2, 1), (1, 1), (2, 2)} in A x A. So, there are 24= 16 relations from A to A. i.e.
## Domain and Range of Relation
Domain of Relation: The Domain of relation R is the set of elements in P which are related to some elements in Q, or it is the set of all first entries of the ordered pairs in R. It is denoted by DOM (R).
Range of Relation: The range of relation R is the set of elements in Q which are related to some element in P, or it is the set of all second entries of the ordered pairs in R. It is denoted by RAN (R).
Example:
Solution:
```DOM (R) = {1, 2}
RAN (R) = {a, b, c, d}
```
## Complement of a Relation
Consider a relation R from a set A to set B. The complement of relation R denoted by R is a relation from A to B such that
``` R = {(a, b): {a, b) ∉ R}.
```
Example:
Solution:
```X x Y = {(1, 8), (2, 8), (3, 8), (1, 9), (2, 9), (3, 9)}
Now we find the complement relation R from X x Y
R = {(3, 8), (2, 9)}
```
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## Properties of Mantissa
#### Algebra > Logarithms > properties of Mantissa
10. Properties of the Mantissa:
1. The mantissa is same for the same order of digits in two different numbers, irrespective of where the decimal point is in the two numbers
Consider the two numbers:
log 256 = 2.408 = 2 + 0.408
log 25.6 = 1.408 = 1 + 0.408
log 2.56 = 0.408 = 0 + 0.408
so, what do we observe from the logarithmic values on the right of the three numbers on the left.
We see, the characteristic differs, but the mantissa i.e., the proper positive fraction in the sums above is same though the numbers 256, 25.6 and 2.56 are different and differ in the position of the decimal point.
The reason?
Since the three numbers have the same digits namely 2, 5 and 6 and in the same order 2, 5, 6therefore the (Positive Proper Fraction)0.408 is same for all the three numbers.
2. The Mantissa is always written positive
The characteristic of the logarithm of a number can be either positive or negative.
But the Mantissa of the logarithm of a number is always positive
Consider the following examples:
log 0.5 = -0.301 = -1 + 0.699 = ¯1+ 0.699
log 0. 08 = -1.096 = -2 + 0.904 = ¯2 + 0.904
log 0.005 = -2.3010 = -3 + 0.699 = ¯3+0.699
In each of the above three examples, we see that the negative logarithmic values (in the center) of the numbers (at left) are converted into a sum of a negative integer (the characteristic) and a positive proper fraction
So, the mantissa is always written as a positive number i.e., a positive proper fraction.
Furthermore, to indicate that the Mantissa is never negative and it is characteristic that can be negative, we write a bar on the characteristic as shown in the three examples above. The bar tells the logarithm of the number is negative and so it is the characteristic that is converted into a negative integer.
11. How to find the Mantissa :
The Mantissa of the logarithm of numbers is found using logarithm tables.
In finding the Positive Proper Fraction of logarithms of numbers, we consider the following two points:
1. the significant digits in a number, i.e., the digits other than 0
2. 0 also when it occurs in between the significant digits.
Zeroes occurring either after or before the significant digits are ignored while finding mantissa.
For example in the number 0.00456, only the number 456 is taken into account, while the zeroes before the significant digits 4, 5 and 6 are ignored.
But in the number, 45.006, the two zeroes after the decimal point are not ignored, rather taken into account for finding the mantissa.
12. What is Antilogarithm Consider the example: log a x = n
we know n is the logarithm of x to base a.
We also say that x is the antilogarithm of n to base a.
We know that
log 10 15 = 1.176
so 1.176 is called the antilog of 15 to base 10.
Antilogarithms are found using antilog tables
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Question Video: Recognising Properties of Circle Construction | Nagwa Question Video: Recognising Properties of Circle Construction | Nagwa
# Question Video: Recognising Properties of Circle Construction Mathematics • Third Year of Preparatory School
## Join Nagwa Classes
Consider the two points π΄ and π΅. What is the radius of the smallest circle that can be drawn in order to pass through the two points?
03:42
### Video Transcript
Consider the two points π΄ and π΅. What is the radius of the smallest circle that can be drawn in order to pass through the two points?
In this question, we need to determine the radius of the smallest circle which passes through two points π΄ and π΅. To answer this question, we start by recalling that every point on a circle will be equidistant from its center. And we also recall another fact. Every point which is equidistant from two points π΄ and π΅ will lie on the perpendicular bisector of π΄ and π΅. So the center of our circle must lie on the perpendicular bisector of π΄ and π΅. So letβs start by constructing this. We need to draw the line segment from π΄ to π΅ and then mark the midpoint of this line segment. Weβll call this π. Then the perpendicular bisector is the line perpendicular to π΄π΅ which passes through π. Any point which lies on this line is equidistant from π΄ and π΅.
And so these can be choices for the center of our circle. For example, π, π two, π three, or π four could be the centers of our circle. Then the radius of this circle would just be the distance between the center and any point on the circle, for example, the distance between the center and point π΄. Graphically, it appears the further we get away from the point π, the larger this radius is. In fact, we can prove this is true. For example, we can see that triangle π΄ππ two is a right triangle. In particular, this means its hypotenuse must be longer than the other sides. π΄ππ three is also a right triangle, and π΄ππ four is another right triangle. So π three is bigger than π, and π four is bigger than π. Therefore, π will be the circle with the smallest radius which passes through π΄ and π΅.
And although we were not asked to in this question, we can even sketch this circle by putting the points of our compass at the point π and then the tip of our compass at either the points π΄ or π΅. We were asked to find the radius of this circle. Remember, we chose π to be the midpoint of the line segment π΄π΅. And this means the value of π is going to be one-half the distance from π΄ to π΅. In other words, the radius of the smallest circle which passes through two distinct points π΄ and π΅ is one-half the distance from π΄ to π΅.
So far, weβve constructed circles through one point and through two distinct points. Now letβs try to construct the circle through three distinct points π΄, π΅, and πΆ. To do this, we need to find the center of our circle which needs to be equidistant from all three points. If the center of the circle is equidistant from all three points, then it must be equidistant from both π΄ and π΅. And we can find the set of all points equidistant from both π΄ and π΅ by finding the perpendicular bisector. So we start by finding the midpoint of the line segment between π΄ and π΅. And then we draw the line through the midpoint π one which is perpendicular to the line segment π΄π΅. Every point equidistant from both π΄ and π΅ lies on the perpendicular bisector between π΄ and π΅. So if the center of our circle exists, it must lie on this line.
But the exact same reasoning is true for points πΆ and π΅. The center of our circle is equidistant from πΆ and π΅, so it must lie on the perpendicular bisector of the line segment π΅πΆ. So weβll also sketch the perpendicular bisector of line segment π΅πΆ. We find the point π two which is the midpoint of this line segment and then draw a line perpendicular to this line which passes through π two. Every point on this line is equidistant from π΅ and πΆ, and we can see thereβs a point of intersection between the two perpendicular bisectors. Since this point is on both perpendicular bisectors, itβs equidistant from π΄, π΅, and πΆ. This means we can construct a circle with center π which passes through all three points π΄, π΅, and πΆ. The radius of this circle will then be the distance between either π and π΄, π and π΅, or π and πΆ.
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# Associative Property With Manipulatives
32 teachers like this lesson
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## Objective
Students will be able to model and prove how the associative property in multiplication works.
#### Big Idea
Children need to understand why the mathematical properties work and how they are applied. In this lesson, we explore modeling and "proving" the associative property.In this lesson, we explore modeling and "proving" the associative property.
## Teacher Background Information
2 minutes
The teaching of the properties is a tricky one for some teachers. It is important to remember that the properties help students build their number sense, use that sense to efficiently solve problems, and to make sense of the math world. They do not have to "know and memorize" the names of the properties, but the structures and patterns are useful and necessary.
When you are working today and, and each day, present the properties as helpful patterns to be used in the right moments.
## Mini-Lesson
10 minutes
To review and enhance the lesson from yesterday, I call my students to the community area and ask them to get in "fishbowl" formation. This is when they create a half circle around me with the first row on their bottoms and the second row on their knees. I ask them to watch me work with one of their peers to create and solve a multiplication equation using the associative property.
After I select a volunteer, I show him/her our task, making sure to direct everyone's attention to the directions.
I ask my partner to roll the three dice and decide what order to write our multiplication problem using all three digits. We rolled a 4, 4, and 5.
Then, I ask him where we should put the parenthesis, knowing that we have to solve that equation first. We decide to put them around a 4 and the 5.
Next, we build the array of that problem. Then we look to the third digit (the second 4) and see how many times we have to build that same array. We build those arrays as well, using cubes. Now we have 4 (4x5) arrays.
Now we can solve for the area of each and add those areas together on paper or a white board. I chose to use white boards to save on paper.
Finally, we decide how to move the parenthesis and go through the process again, in order to prove that we get the same product each time.
## Active Engagement
20 minutes
After reviewing the directions and answering questions, I send partnerships off with cubes, dice, and their reflection journals to work in. Their task is to create, build, and prove the products of a three digit multiplication equation in at least two ways. If they have time, then they can work on a third way.
As I tour the room, I will be looking and listening for the correct building of the original equations, with the students solving the parenthesis first, and for the students to verbally explain how their drawings or models represent the equations.
This student shows how he and his partner began working and realized an error by asking, "Does this make sense?"
As you look at the student video, and my lesson description, you may well wonder why I am sticking with holding the students accountable in creating models of the associative property of multiplication. There are critical reasons for modeling - it provides repetition and it creates context - both key to remembering. Modeling also allows the student to evaluate their work within a situational context so that they can ask themselves (and others) if it makes sense. My reflection provides some important information on how 3rd grade students grapple with models of the associative property of multiplication.
## Journaling
15 minutes
After about 15 minutes of building and working through their own created equations, I call the students back to the community area with their journals. We review their work in drawing the equations and using the parenthesis to direct order of operation.
Then, I hand out a prompt for the partners to respond to. In my class, we have a rubric to help us write specific and detailed responses. I remind them to use the rubric and send them off to work.
The prompt was:
Sarah says the product of 2 x 3 x 4 is less than the product of 4 x 3 x 2.
Is Sarah correct?
At this point, I allow them to work alone or with their partner. Some of my students like and need to be on their own and have quiet to listen to their thoughts when required to write. I respect this and always allow for it.
While they are writing, I circulate and read, listen, and watch. I always try to find a way to prompt them to go deeper, explain more clearly, think in a different way, or I challenge their answers - asking for proof.
## Critiquing/Revising/Sharing
15 minutes
After about 8-10 minutes of journaling, I signal the class to silence and ask them to take the 3 sticky notes that I placed on their desks while they worked and travel around to view 3 different journals. While they read and reflect on each journal entry, they are to leave a message to the author, stating whether they agree, disagree, or would like to add something and why.
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# Texas Go Math Grade 7 Lesson 11.2 Answer Key Comparing Data Displayed in Dot Plots
Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 11.2 Answer Key Comparing Data Displayed in Dot Plots.
## Texas Go Math Grade 7 Lesson 11.2 Answer Key Comparing Data Displayed in Dot Plots
Essential Question
How do you compare two sets of data displayed in dot plots?
Analyzing Dot Plots
You can use dot plots to analyze a data set, especially with respect to its center and spread.
People once used body parts for measurements. For example, an inch was the width of a man’s thumb. In the 12th century, King Henry I of England stated that a yard was the distance from his nose to his outstretched arm’s thumb. The dot plot shows the different lengths, in inches, of the “yards” for students in a 7th grade class.
A. Describe the shape of the dot plot. Are the dots evenly distributed or grouped on one side?
B. What value best describes the center of the data? Explain how you chose this value.
C. Describe the spread of the dot plot. Are there any outliers?
Reflect
Question 1.
Calculate the mean, median, and range of the data in the dot plot.
To find the mean, add the numbers and divide the sum by the number of addends.
28 + 29 + 29 + 30 + 30.5 + 30.5 + 31 + 31 + 31 + 31.5 + 31.5 + 32 + 32 + 32.5 + 32.5 + 33 + 33 + 33.5 + 34 + 34 + 35 = 664.5
$$\frac{664.5}{21}$$ = 31.64
Mean is 31.64
The median is the middle number of the data in the dot plot that is ordered from least to greatest
Range is the difference between the greatest and least number of the data in the dot plot
range = 35 – 28 = 7
Mean : 31.64
Median : 31.7
Range : 7
Comparing Dot Plots Visually
You can compare dot plots visually using various characteristics, such as center, spread, and shape.
Example 1.
The dot plots show the heights of 15 high school basketball players and the heights of 15 high school softball players.
A. Visually compare the shapes of the dot plots.
Softball: All the data is 5’6” or less.
Basketball: Most of the data is 5’8” or greater.
As a group, the softball players are shorter than the basketball players.
B. Visually compare the centers of the dot plots.
Softball: The data is centered around 5’4’
Basketball: The data is centered around 5’8’
This means that the most common height for the softball players is 5 feet 4 inches, and for the basketball players 5 feet 8 inches.
C. Visually compare the spreads of the dot plots.
Softball: The spread is from 4’11 “to 5’6”.
Math Talk
How do the heights of field hockey players compare with the heights of softball and basketball players?
Question 2.
Visually compare the dot plot of heights of field hockey players to the dot plots for softball and basketball players.
Shape: _______________
Center: _______________
Shapes:
Hockey: All the data is 5’6” or less.
The hockey players are shorter than the basketball players, but taller than the softball players.
Center:
Hockey: The data is centered around of 5’2”.
Hockey: The spreads is from 4’9” to 5’6”.
All the data is 5’6″ or Less.
The data is centered around of 5’2”.
The spreads is from 4’9” to 5’6″.
Comparing Dot Plots Numerically
You can also compare the shape, center, and spread of two dot plots numerically by calculating values related to the center and spread. Remember that outliers can affect your calculations.
Example 2
Numerically compare the dot plots of the number of hours a class of students exercises each week to the number of hours they play video games each week.
A. Compare the shapes of the dot plots.
Exercise: Most of the data is less than 4 hours.
Video games: Most of the data is 6 hours or greater.
B. Compare the centers of the dot plots by finding the medians.
Median for exercise: 2.5 hours. Even though there are outliers at
12 hours, most of the data is close to the median.
Median for video games: 9 hours. Even though there is an outlier at 0 hours, these values do not seem to affect the median.
C. Compare the spreads of the dot plots by calculating the range.
Exercise range with outlier: 12 – 0 = 12 hours
Exercise range without outlier: 7 – 0 = 7 hours
Video games range with outlier: 14 – 0 = 14 hours
Video games range without outlier: 14 – 6 = 8 hours
Math Talk
Mathematical Processes
How do outliers affect the results of this data?
Question 3.
Calculate the median and range of the data in the dot plot. Then compare the results to the dot plot for Exercise in Example 2.
Median for Internet Usage: 6 hours
Most of the data is close to the median
Median for Exercise: 2.5 hours
Internet Usage range with outliers: 11 – 1 = 10 hours
Internet usage range without outliers: 8 – 4 = 4 hours
Exercise range with outlier: 12 hours
Exercise range without outlier: 7 hours
Median for Internet Usage : 6 hours
Internet Usage range:
1. With outliers: 10 hours
2. Without outliers: 4 hours
The dot plots show the number of miles run per week for two different classes. For 1-5, use the dot plots shown.
Question 1.
Compare the shapes of the dot plots.
Class A: Most of the the data is 6 mi. or less.
Class B: All the data is 3 mi. or greater
As a class, Students from Class A run more than students from Class B
Class A: Most of the the data is 6 mi. or less.
Class B: All the data is 3 mi. or greater
As a class, Students from Class A run more than students from Class B.
Question 2.
Compare the centers of the dot plots.
Class A: The data is centred around 4 mi.
Class B: The data is centred around 7 mi.
Question 3.
Compare the spreads of the dot plots.
Class A: The spread is from 4 mi. to 14 mi
Class B: The spread is from 3 mi. to 9 mi.
Question 4.
Calculate the medians of the dot plots.
Median for Class A: 7.5 mi.
Median for Class B: 5.75 mi.
Question 5.
Calculate the ranges of the dot plots.
Class A range with outliers: 14 – 4 = 10 mi
Class A range without outliers: 6 – 4 = 2 mi
Class B range: 9 – 3 = 6 mi
For Class B the outlier doesn’t exist.
Class A range with outliers: 10 mi
Class A range without outliers: 2 mi
Class B range: 6 mi
For Class B the outlier doesn’t exist.
Essential Question Check-In
Question 6.
What do the medians and ranges of two dot plots tell you about the data?
The medians of any data represents the centre or almost centre point of the given data. So in above two dot plots
the medians tells that which of the given dots have higher value. And the range of any data set represent the value around which all the data are spread. So in second dot plot the dots are very close as compared to first dot plot which means in second dot plots the value of data are closer.
Median of any data represent centre value.
The dot plot shows the number of letters in the spellings of the 12 months. Use the dot plot for 7-10.
Question 7.
Describe the shape of the dot plot.
All of the data are 3 letters or more.
Question 8.
Describe the center of the dot plot.
The data are centered around 8 letters.
Question 9.
Describe the spread of the dot plot.
The spread is from 3 to 9 letters.
Question 10.
Calculate the mean, median, and range of the data in the dot plot.
Mean: 6.16 letters
Median: 6 Letters
Range: 6 Letters
The dot plots show the mean number of days with rain per month for two cities.
Question 11.
Compare the shapes of the dot plots.
Number of Days of rain for Montogomery, AL: All of the data is 12 days or less.
Number of Days of rain for Lynchburg, VA: All of the data is 8 days or more.
There are more days of rain in Lynchburg than Montogomery.
Question 12.
Compare the centers of the dot plots.
Montgomery: The data is centered around 8 days
Lynchburg: The data is centered around 10 days.
Question 13.
Compare the spreads of the dot plots.
Montgomery: The spread is from 1 day to 12 days.
Lynchburg: The spread is from 8 days to 12 days.
Question 14.
What do the dot plots tell you about the two cities with respect to their average monthly rainfall?
The dot plot for Montgomery, AL showed that the average monthly rainfall in their city is inconsistent. The dot plot showed that there are days where the rainfall is heavy then it changes. Also, there is an outlier which made it more inconsistent For Lynchburg, VA, the average monthly rainfall in their city is somewhat consistent. Also, the days were relatively closer.
The average monthly rainfall in Lynch burg, VA is more consistent compared to Montgomery, AL.
The dot plots show the shoe sizes of two different groups of people.
Question 15.
Compare the shapes of the dot plots.
Group A: All of the data is 65 or greater
Group B: All of the data is 8.5 or greater.
Group B has greater sizes of shoe than Group A.
Question 16.
Compare the medians of the dot plots.
Median for Group A: 8 size of shoe
Median for Group B:
Median for Group A: 8 size of shoe
Median for Group B: 9.71 size of shoe
Question 17.
Compare the ranges of the dot plots (with and without the outliers).
Group A range with outliers: 13 – 6.5 = 6.5 size of shoe
Group A range without outliers: 9 – 6.5 = 2.5 size of shoe
Group B range with outliers: 11.5 – 8.5 = 3 size of shoe
Group B range without outliers: 11.5 – 8.5 = 3 size of shoe
Group A range with outliers is 65 size of shoe
Group A range without outliers is 2.5 size of shoe
Group B range with outliers is 3 size of shoe
Group B range without outliers is 3 size of shoe
Question 18.
Make A Conjecture Provide a possible explanation for the results of the dot plots.
Group A probably consists mostly of younger people and/or women (who tends to have smaller feet). Group B, on
the other hand, consists mostly of adults (who tends to have bigger feet).
Group A consists of younger people while Group B consists of adults.
Texas Go Math Grade 7 Lesson 11.2 H.O.T. Focus On Higher Order Thinking Answer Key
Question 19.
Analyze Reltionships Can two dot plots have the same median and range but have completely different shapes? Justify your answer using examples.
Yes, two dot plots can have the same median and range but completely different shapes.
In both the given dot plot the median is 6th term which is 7 and also the range of both dot plot is 7.
Dot plot of type A data is shown below:
Dot plot of type B data is shown below:
Yes, two dot plot can have same median and range.
Question 20.
Draw Conclusions What value is most affected by an outlier, the median or the range? Explain. Can you see these effects in a dot plot?
Range is the most affected by an outlier.
This can be understood by the example given below:
For examle the data of the sample are : 3, 5, 5, 5, 7, 8, 8, 9, 10, 12, 15.
Median = 6th term of data
= 8
Range = 15 – 3
= 12
New samle after excluding outlier: 3, 5, 5, 5, 7, 8, 8, 9, 10, 12.
Range = 12 – 3
= 9
Here we can see that median decreases by only 0.5 but the range decreases by 3. Hence we can see that range is most afffected by outlier.
Range is most affected by outlier
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6.9
## National STEM Learning Centre
Skip to 0 minutes and 7 secondsPAULA KELLY: So now let's have a look at how we calculate a loss as a percentage. So, for example, if I bought a car for 8,000 pounds, some time later, I sell it for 6,200 pounds, I know I've made a loss. But what what is that as a percentage?
Skip to 0 minutes and 23 secondsMICHAEL ANDERSON: OK. So what we need to do is, first of all, figure out the loss, which, unfortunately this time, is 1,800 pounds. And then we're going to divide that by the original amount that you paid for the car.
Skip to 0 minutes and 35 secondsMICHAEL ANDERSON: So you made a loss of 1,800 pounds. And we're going to divide that by the 8,000 pounds that you originally paid. So I could try and work this out, simplify it down as a fraction, but what I'm going to do is do a division on the calculator. And that will give me 0.225 as a decimal.
Skip to 0 minutes and 54 secondsMICHAEL ANDERSON: So to convert this to a percentage, the percentage loss, well, that's just going to be 22.5%.
Skip to 1 minute and 1 secondPAULA KELLY: So as a decimal, you multiply it by 100 to get your percentage. So my percentage loss was 22.5%?
Skip to 1 minute and 7 secondsMICHAEL ANDERSON: Yep, of the original amount.
Skip to 1 minute and 17 secondsSo that's one method of working out. Is there an alternative strategy we could use?
Skip to 1 minute and 21 secondsMICHAEL ANDERSON: Yeah, so we've already visited the multiply methods. In this case, it's going to be a decrease. So we went from 8,000 pounds. And we sold the car for 6,200 pounds.
Skip to 1 minute and 34 secondsMICHAEL ANDERSON: So we're going to look for a multiplier that takes us from 8,000 to 6,200. So we're going to multiply this 8,000 to give us 6,200.
Skip to 1 minute and 44 secondsMICHAEL ANDERSON: So what I'm going to do is divide both sides by 8,000. So m is going to stay the same. We'll end up with 6,200 divided by 8,000. Now, again, I'm going to use a calculator to work this out. And it will give me a value of 0.775.
Skip to 2 minutes and 1 secondPAULA KELLY: So that's quite a value to what we had earlier. So why is that?
Skip to 2 minutes and 5 secondsMICHAEL ANDERSON: Yeah, well, this 0.775 could be thought of as a percentage, the 77.5%. And that represents the value that the car has retained. So out of your 8,000 pounds, it's still worth 77.5% of the value that you paid originally. So what that means as a loss, well, we have to figure out what we go from 77.5, what we have to add to that to make 100. So in this case, our percentage loss is 22.5%.
# Finding a percentage decrease
In this video we build upon the last step and consider two similar methods of finding a percentage decrease. When finding a percentage decrease we are asking the question: “by what percentage we have to decrease the original amount by to find the new amount?”
The first method uses the equation:
As with percentage increase, if students are to learn through understanding and creating connections between mathematical topics as demanded by the English National Curriculum, it is important, when using this method, to explore how the equation is derived.
The ‘loss’ is the amount which has been subtracted from the original amount. The answer then represents this loss amount as a percentage of the original amount. The second method uses a multiplier
The equation:
is again used as with percentage increase. The answer again has to be interpreted correctly.
When M is a decimal number between 0 and 1 tells us that it is a percentage decrease.
The decimal part indicates the magnitude of the percentage we have left. The percentage lost still has to be calculated.
For example, a value of M of 0.65 indicates that 65% of the original value remains so the percentage of the original value we have lost is 35% (the difference between 65% and 100% of the original value).
Another way of thinking about this is using decimals. A value of M equals 0.65 so the percentage decrease is 1 – 0.65 = 0.35 which represents a 35% decrease.
## Problem worksheet
Now complete questions 12 and 13 from this week’s worksheet.
## Share
Earlier in the course we asked you to share your examples of percentages being used in real-life situations. This week, we’d like you to find more complex examples, for example in finance, engineering or other industries. Where are percentage increases and decreases used, or percentages used to indicate a proportion above or below an amount?
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Update all PDFs
Waiting Times
Alignments to Content Standards: 7.SP.C.8
Suppose each box of a popular brand of cereal contains a pen as a prize. The pens come in four colors, blue, red, green and yellow. Each color of pen is equally likely to appear in any box of cereal. Design and carry out a simulation to help you answer each of the following questions.
1. What is the probability of having to buy at least five boxes of cereal to get a blue pen? What is the mean (average) number of boxes you would have to buy to get a blue pen if you repeated the process many times?
2. What is the probability of having to buy at least ten boxes of cereal to get a full set of pens (all four colors)? What is the mean (average) number of boxes you would have to buy to get a full set of pens if you repeated the process many times?
IM Commentary
As the standards in statistics and probability unfold, students will not yet know the rules of probability for compound events. Thus, simulation is used to find an approximate answer to these questions. In fact, part b would be a challenge to students who do know the rules of probability, further illustrating the power of simulation to provide relatively easy approximate answers to wide-ranging problems. Modeling with simulation follows four steps: state assumptions about how the real process works; describe a model that generates similar random outcomes; run the model over many repetitions and record the relevant results; write a conclusion that reflects the fact that the simulation is an approximation to the theory.
Solution
1. If each color of pen is equally likely to appear in any box, the chance of getting a blue pen in any one box is $\frac{1}{4}$ or 0.25. Simulation is then used to find an approximate answer to the question posed. Students select a device, or devices, that generate a specified outcome with probability 0.25 to model the process of buying boxes of cereal until a blue pen is found. Random integers 1, 2, 3, 4, with, say, 1 denoting blue, will work (as will using four sides of a six-sided die, etc.). They then generate many outcomes for the simulated event and collect the data to produce a distribution of waiting times.
Here is a string of random integers that produces 9 trials of a simulation, the respective waiting times to get a 1 (blue) being 2, 5, 4, 1, 4, 3, 4, 3, 3.
4 1] 4 3 3 2 1] 4 2 2 1] 1] 3 2 2 1] 2 4 1] 3 2 4 1] 2 2 1] 2 3 1] …
The plot below is based on 100 simulated waiting times (Tb) to get a blue pen. The probability of having to purchase at least 5 boxes is approximated by the proportion of simulated waiting times greater than or equal to 5, which is (33/100) = 0.33. The mean of the 100 simulated waiting times is 3.8 or approximately 4. It should seem intuitively reasonable that an event with probability $\frac{1}{4}$ would happen, on average, about every four trials.
2. Modeling the outcome of getting a full set of pens (all four colors) works in a similar way. Using the same sequence of random integers as above, the waiting times are 6, 7, 7, 8.
4 1 4 3 3 2] 1 4 2 2 1 1 3] 2 2 1 2 4 1 3] 2 4 1 2 2 1 2 3] 1 …
The plot below is based on 100 simulated outcomes (T4) resulting in a full set of pens. The probability of having to purchase at least 10 boxes to get a full set is approximated by the proportion of waiting times greater than or equal to 10, which is (31/100) = 0.31 for this simulation. The mean of the 100 simulated waiting times is 8.2, which is not so intuitive.
Even though students will not yet have the tools to figure this out, it is worth noting that the theoretical solution is $\frac44 + \frac43 + \frac42 + \frac41 = 8\frac13$. The results of the simulation agree well with this.
Mary says:
almost 4 years
Thank you for answering my questions!
Mary says:
almost 4 years
Question: To get the theoretical probability for a, the reasoning could be as follows. To get the blue pen in the first box the probability would be 1/4. To get it in the second box the probability would be (3/4)(1/4), and in the third box would be (3/4)(3/4)(1/4), and in the fourth box would be (3/4)(3/4)(3/4)(1/4). So the probability of buying at least 5 boxes to get a blue pen would be 1 - (1/4 + 3/16 + 9/64 + 27/256) = .3164. Then to get the average waiting time it would be 1(1/4) + 2(3/16) + 3(9/64) + 4(27/256) + ... which does seem to converge to 4. Is this correct?
Then do you need something way more complex for part b? Derangements?
Cam says:
almost 4 years
Hi Mary,
Excellent questions. That infinite sum you write down for the average waiting time is exactly correct. Here's how you can evaluate it, using the formula for the infinite geometric series: If we abbreviate $\frac{3}{4}$ by $x$, then the sum you wrote down is
Plugging in $x=\frac{3}{4}$ gives the expected wait time of 4.
You're right that things get more complex from here, though in the end it largely just depends on conditional probabilities. (I'm sure you could address in terms of derangements as well, though my intuition is this gets pretty hairy pretty fast). For the first part of the second question, unless I've done something wrong the answer turns out to be $$1-\frac{7770\cdot 4!}{4^9}=\frac{4729}{16384}\approx .2886,$$ agreeing pretty well with the sample taken in the solution. (Incidentally, 7770 is the eminently-googlable "Stirling number of the second kind, $S_2(9,4)$).
Finally, for the last part, the conditional probability is a little easier -- you ask yourself the probability that if you had already collected the k-th one, how long you'd have to wait for the next one, and then sum up over all k. In the end, the answer turns out to be pretty nice: If there are n crayons instead of 4, then the expected waiting time until you collect them all is $$n\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right),$$ which as a special case is the last line of the solution for $n=4$.
Sorry, I droned on a bit. There's a vast literature on these types of problems. Google things like "coupon collector problem" and you won't be disappointed!
Catherine Parker says:
over 5 years
This problem is several notches above the probability content that 7th graders have been exposed to, but I look forward to challenging them to design a simulation beyond the typical (and boring) spinners, dice, and colored marbles.
One note about simulating the four colors ... instead of using integers to represent the four colors, 7th graders might benefit from using actual colored tiles or a four-part spinner. I see potential confusion between the integer representing the color and the number representing the wait-time.
Bill says:
over 5 years
Thanks Catherine, that's a very useful suggestion.
Bill McCallum
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# Lines and Angles 5.1
Samrules
Published at : 12 Dec 2020
11015 views
163
25
Full exercise on Lines and Angles, fully solved. This is a video tutorial for 7th class NCERT CBSE students.
RELATED ANGLES
Complementary Angles
When the sum of the measures of two angles is 90°, the angles are called complementary angles. Whenever two angles are complementary, each angle is said to be the complement
of the other angle.
THINK, DISCUSS AND WRITE
1. Can two acute angles be complement to each other?
2. Can two obtuse angles be complement to each other?
3. Can two right angles be complement to each other?
Supplementary Angles
When two angles are supplementary, each angle is said to be the supplement of the other.
THINK, DISCUSS AND WRITE
1. Can two obtuse angles be supplementary?
2. Can two acute angles be supplementary?
3. Can two right angles be supplementary?
What will be the measure of the supplement of each one of the following angles?
(i) 100º (ii) 90º (iii) 55º (iv) 125º
3. Among two supplementary angles the measure of the larger angle is 44º more than the measure of the smaller. Find their measures.
These angles are such that:
(i) they have a common vertex;
(ii) they have a common arm; and
(iii) the non-common arms are on either side of the common arm.
Such pairs of angles are called adjacent angles. Adjacent angles have a common vertex and a common arm but no common interior points.
THINK, DISCUSS AND WRITE
1. Can two adjacent angles be supplementary?
2. Can two adjacent angles be complementary?
3. Can two obtuse angles be adjacent angles?
4. Can an acute angle be adjacent to an obtuse angle?
Linear Pair A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.
THINK, DISCUSS AND WRITE
1. Can two acute angles form a linear pair?
2. Can two obtuse angles form a linear pair?
3. Can two right angles form a linear pair?
Vertically Opposite Angles
We conclude that when two lines intersect, the vertically opposite angles so formed are equal.
EXERCISE 5.1
1. Find the complement of each of the following angles:
2. Find the supplement of each of the following angles:
3. Identify which of the following pairs of angles are complementary and which are supplementary.
(i) 65º, 115º (ii) 63º, 27º (iii) 112º, 68º
(iv) 130º, 50º (v) 45º, 45º (vi) 80º, 10º
4. Find the angle which is equal to its complement.
5. Find the angle which is equal to its supplement.
6. In the given figure, ∠1 and ∠2 are supplementary
angles. If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary.
7. Can two angles be supplementary if both of them are:
(i) acute? (ii) obtuse? (iii) right?
8. An angle is greater than 45º. Is its complementary angle greater than 45º or equal to 45º or less than 45º?
(i) Is ∠1 adjacent to ∠2?
(ii) Is ∠AOC adjacent to ∠AOE?
(iii) Do ∠COE and ∠EOD form a linear pair?
(iv) Are ∠BOD and ∠DOA supplementary?
(v) Is ∠1 vertically opposite to ∠4?
(vi) What is the vertically opposite angle of ∠5?
10. Indicate which pairs of angles are:
(i) Vertically opposite angles. (ii) Linear pairs.
11. In the following figure, is ∠1 adjacent to ∠2? Give reasons.
12. Find the values of the angles x, y, and z in each of the following:
(i) (ii)
13. Fill in the blanks:
(i) If two angles are complementary, then the sum of their measures is _______.
(ii) If two angles are supplementary, then the sum of their measures is ______.
(iii) Two angles forming a linear pair are _______________.
(iv) If two adjacent angles are supplementary, they form a ___________.
(v) If two lines intersect at a point, then the vertically opposite angles are always
_____________.
(vi) If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are __________.
14. In the adjoining figure, name the following pairs of angles.
(i) Obtuse vertically opposite angles
(iii) Equal supplementary angles
(iv) Unequal supplementary angles
(v) Adjacent angles that do not form a linear pair
#LinesandAngles #NcertSolutions #solvedexercises #teaching #ncert #cbse #maths #MathematicsClassVII #schoolmadeeasy #samrules
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# Proposition 4
In equiangular triangles the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles.
Let ABC and DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC equal to the angle CDE, and the angle ACB equal to the angle CED.
I say that in the triangles ABC and DEC the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles.
Let BC be placed in a straight line with CE.
Then, since the sum of the angles ABC and ACB is less than two right angles, and the angle ACB equals the angle DEC, therefore the sum of the angles ABC and DEC is less than two right angles. Therefore BA and ED, when produced, will meet. Let them be produced and meet at F.
I.28
Now, since the angle DCE equals the angle ABC, DC is parallel to FB. Again, since the angle ACB equals the angle DEC, AC is parallel to FE.
I.34
Therefore FACD is a parallelogram, therefore FA equals DC, and AC equals FD.
VI.2
And, since AC is parallel to a side FE of the triangle FBE, therefore BA is to AF as BC is to CE.
But FD equals AC, therefore BC is to CE as AC is to DE, and alternately BC is to CA as CE is to ED.
V.22
Since then it was proved that AB is to BC as DC is to CE, and BC is to CA as CE is to ED, therefore, ex aequali, BA is to AC as CD is to DE.
Therefore, in equiangular triangles the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles.
Q.E.D.
## Guide
In the enunciation of this proposition the term “equiangular triangles” refers to two triangles whose corresponding angles are equal, not to two triangles each of which is equiangular (equilateral).
Euclid has placed the triangles in particular positions in order to employ this particular proof. Such positioning is common in Book VI and is easily justified.
This proposition implies that equiangular triangles are similar, a fact proved in detail in the proof of proposition VI.8. It also implies that triangles similar to the same triangle are similar to each other, also proved in detail in VI.8. The latter statement is generalized in VI.21 to rectilinear figures in general.
This proposition is frequently used in the rest of Book VI starting with the next proposition, its converse. It is also used in Books X through XIII.
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# 12.4: Curvature and Normal Vectors of a Curve
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For a parametrically defined curve we had the definition of arc length. Since vector valued functions are parametrically defined curves in disguise, we have the same definition. We have the added benefit of notation with vector valued functions in that the square root of the sum of the squares of the derivatives is just the magnitude of the velocity vector.
Definition: Arc Length
Let
$\textbf{r}(t) = x(t) \, \hat{\textbf{i}} + y(t) \, \hat{\textbf{j}} + z(t) \, \hat{\textbf{k}} \nonumber$
be a differentiable vector valued function on [a,b]. Then the arc length $$s$$ is defined by
$s=\int_{a}^{b}\sqrt{ \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}\, dt = \int _a^b \left|v(t)\right| \,dt .\nonumber$
Example $$\PageIndex{1}$$
Suppose that
$\textbf{r}(t) = 3t\,\hat{\textbf{i}} + 2\,\hat{\textbf{j}} + t^2\,\hat{\textbf{k}} \nonumber$
Set up the integral that defines the arc length of the curve from 2 to 3. Then use a calculator or computer to approximate the arc length.
Solution
We use the arc length formula
$s = \int _2^3 \sqrt{9 + 0 + 4t^2} \, dt = \int_2^3 \sqrt{9+4t^2} \, dt .\nonumber$
Notice that we could do this integral by hand by letting $$t = 9/2 \tan\, q$$, however the question only asked us to use a machine to approximate the integral:
$s = 5.8386 .\nonumber$
## Parameterization by Arc Length
Recall that like parametric equations, vector valued function describe not just the path of the particle, but also how the particle is moving. Among all representations of a curve there is a "simplest" one. If the particle travels at the constant rate of one unit per second, then we say that the curve is parameterized by arc length. We have seen this concept before in the definition of radians. On a unit circle one radian is one unit of arc length around the circle. When we say "simplest" we in no way mean that the equations are simple to find, but rather that the dynamics of the particle are simple. To aid us in parameterizing by arc length, we define the arc length function.
Definition: Arc Length Function
If $$\textbf{r}(t)$$ is a differentiable vector valued function, then the arc length function is defined by
$s(t) = \int _0^t || \textbf{v}(u) || \, du. \nonumber$
Remark: By the second fundamental theorem of calculus, we have
$s'(t) = ||v(t)|| .\nonumber$
If a vector valued function is parameterized by arc length, then
$s(t) = t .\nonumber$
If we have a vector valued function$$r(t)$$ with arc length s(t), then we can introduce a new variable
$s = s^{-1}(t) .\nonumber$
So that the vector valued function $$r(s)$$ will have arc length equal to
$s\left(s^{-1}(t)\right) = t .\nonumber$
and $$r(s)$$ will be parameterized by arc length. Unfortunately, this process is usually impossible for two reasons.
1. The integral that defines arc length involves a square root in the integrand; this integral is usually impossible to determine.
2. Even if the integral is possible to evaluate, finding the inverse of a function is often impossible. There are a few special curves that can be parameterized by arc length and one is demonstrated below.
Example $$\PageIndex{2}$$: Parameterizing by Arc Length
Find the arc length parameterization of the helix defined by
$\textbf{r}(t) = \cos\, t \hat{\textbf{i}} + \sin\,t \hat{\textbf{j}} + t \hat{\textbf{k}} .\nonumber$
Solution
First find the arc length function
$s(t) = \int_0^t \sqrt{\sin^2 u + \cos^2u + 1}\, dt = \int_0^t \sqrt{2}\,dt = \sqrt{2}\, t .\nonumber$
Solving for $$t$$ gives
$t= \dfrac{s}{\sqrt2} .\nonumber$
Now substitute back into the position equation to get
$\textbf{r}(s) = \cos \dfrac{s}{\sqrt2} \, \hat{\textbf{i}} + \sin \dfrac {s}{\sqrt2} \, \hat{\textbf{j}} + \dfrac{s}{\sqrt2} \, \hat{\textbf{k}} .\nonumber$
## Concepts: Curvature and Normal Vector
Consider a car driving along a curvy road. The tighter the curve, the more difficult the driving is. In math we have a number, the curvature, that describes this "tightness". If the curvature is zero then the curve looks like a line near this point. While if the curvature is a large number, then the curve has a sharp bend.
Before learning what curvature of a curve is and how to find the value of that curvature, we must first learn about unit tangent vector. As the name suggests, unit tangent vectors are unit vectors (vectors with length of 1) that are tangent to the curve at certain points. Because tangent lines at certain point of a curve are defined as lines that barely touch the curve at the given point, we can deduce that tangent lines or vectors have slopes equivalent to the instantaneous slope of a curve at the given point. In other words,
$\mathbf {T} = \frac{d \mathbf{r}}{dt}\mathrm{,}\nonumber$
which means
$\mathbf{\hat{T}} = \frac{\mathbf{T}}{\left | \mathbf{T} \right |}= \frac{d\mathbf{r}/dt}{\left | d\mathbf{r}/dt \right|} .\nonumber$
Based on what we learned previosuly, we know that $$\frac{d\mathbf{r}}{dt} = \mathbf{v}$$, where $$\mathbf{v}$$ is the velocity at which a point is moving at a given time. Furthermore, the absolute value of the velocity vector is the speed vector of the curve, meaning $$\left | \frac{d\mathbf{r}}{dt} \right | = \frac{ds}{dt}$$. So the formula for unit tangent vector can be simplified to:
$\mathbf{\hat{T}} = \frac{\mathrm{velocity}}{\mathrm{speed}} = \frac{d\mathbf{r}/dt}{ds/dt} .\nonumber$
And now, let's think about the unit tangent vector when the curve is explained in terms of arc length, that is, $$r(s)$$ instead of $$r(t)$$. This means:
$\mathbf{T} = \frac{d\mathbf{r}}{ds}\nonumber$
$\text{and }\mathbf{\hat{T}} = \frac{d\mathbf{r}/ds}{ds/ds} = \frac{d\mathbf{r}}{ds} .\nonumber$
With this information, we will be learning what curvature really is and how we can calculate the curvature, denoted as $$\kappa$$.
## Curvature of a Curve
Curvature is a measure of how much the curve deviates from a straight line. In other words, the curvature of a curve at a point is a measure of how much the change in a curve at a point is changing, meaning the curvature is the magnitude of the second derivative of the curve at given point (let's assume that the curve is defined in terms of the arc length $$s$$ to make things easier). This means:
$k= \left | \frac{d^2\mathbf{r}}{ds^2} \right | .\nonumber$
Since we know that $$\mathbf{\hat{T}} = d\mathbf{r} / ds$$, we can formulate an equation for $$\kappa$$ in terms of $$\mathbf{\hat{T}}$$:
$k= \left | \frac{d\mathbf{\hat{T}}}{ds} \right | .\nonumber$
Never the less, we know that most curves are written in parametric equations in terms of some dummy variable, most commonly $$t$$. So let's assume that the curve is in terms of $$t$$, such that $$\mathbf{r}(t)$$ is a curve. In such case, we must formulate another equation to find the curvature without taking derivatives in terms of $$s$$.
First, we know that
$k= \left | \frac{d\mathbf{\hat{T}}}{ds} \right | \nonumber$
Using Chain Rule, we get
\begin{align*} k &= \left | \frac {d\mathbf{\hat{T}}}{dt} \cdot \frac{dt}{ds} \right | \\[4pt] &= \frac{1}{\left | ds/dt \right |} \left |\frac{d\mathbf{\hat{T}}}{dt} \right | \end{align*}
therefore
$k= \frac{1}{\left | \mathbf{v} \right |} \left | \frac{d\mathbf{\hat{T}}}{dt} \right |. \nonumber$
### Definition of Curvature (repeat)
More formally, if $$\textbf{T}(t)$$ is the unit tangent vector function then the curvature, $$k$$, is defined at the rate at which the unit tangent vector changes with respect to arc length.
$k = ||\dfrac{d}{ds} (\textbf{T}(t)) || = ||\textbf{r}''(s)||\nonumber$
As stated previously, this is not a practical definition, since parameterizing by arc length is typically impossible. Instead we use the chain rule to get
$||\dfrac{d}{ds} (\textbf{T}(t)) || = ||\textbf{T}'(t) \dfrac{dt}{ds}|| \nonumber$
$\dfrac{||\textbf{T}'(t)|| }{ ||\dfrac{ds}{dt}|| } = \dfrac{ ||\textbf{T}'(t)||}{ ||\textbf{r}'(t)||}. \nonumber$
This formula is more practical to use, but still cumbersome. $$\textbf{T}'(t)$$ is typically a mess. Instead we can borrow from the formula for the normal vector to get the curvature
$K(t) = \dfrac{ ||r'(t) \times r''(t)||}{||r'(t)||^3}. \nonumber$
### Normal Vector of a Curve
A unit normal vector of a curve, by its definition, is perpendicular to the curve at given point. This means a normal vector of a curve at a given point is perpendicular to the tangent vector at the same point. Furthermore, a normal vector points towards the center of curvature, and the derivative of tangent vector also points towards the center of curvature. In summary, normal vector of a curve is the derivative of tangent vector of a curve.
$\mathbf{N} = \frac{d\mathbf{\hat{T}}}{ds}\mathrm{ or } \frac{d\mathbf{\hat{T}}}{dt}\nonumber$
To find the unit normal vector, we simply divide the normal vector by its magnitude:
$\mathbf{\hat{N}} = \frac{d\mathbf{\hat{T}}/ds}{\left | d\mathbf{\hat{T}}/ds\right |}\mathrm{ or } \frac{d\mathbf{\hat{T}}/dt}{\left | d\mathbf{\hat{T}}/dt \right |} .\nonumber$
Notice that $$\left | d\mathbf{\hat{T}}/ds\right |$$ can be replaced with $$\kappa$$, such that:
$\mathbf{\hat{N}} = \frac{1}{\kappa} \frac{d\mathbf{\hat{T}}}{ds} \nonumber$
$\therefore \mathbf{\hat{N}} = \frac{1}{\kappa} \frac{d\mathbf{\hat{T}}}{ds} \mathrm{ or } \frac{d\mathbf{\hat{T}}/dt}{\left | d\mathbf{\hat{T}}/dt \right |} .\nonumber$
Example $$\PageIndex{3}$$
Find the curvature at $$t=\frac{\pi}{2}$$ if
$r(t) = \cos \,t\, \hat{\textbf{i}} - \frac{1}{t} \hat{\textbf{j}} + \sin\, t\, \hat{\textbf{k}} .\nonumber$
Solution
We take derivatives
$\textbf{r}'(t) = -\sin\, t\, \hat{\textbf{i}} + \frac{1}{t^2}\, \hat{\textbf{j}} + \cos\, t \, \hat{\textbf{k}} \nonumber$
$\textbf{r}''(t) = -\cos\, t \,\hat{\textbf{i}} - \frac{2}{t^3}\, \hat{\textbf{j}} - \sin\, t\, \hat{\textbf{k}} . \nonumber$
Plugging in $$t=\frac{\pi}{2}$$ gives
\begin{align*} \textbf{r}' \left(\frac{\pi}{2} \right) &= -\hat{\textbf{i}} + \dfrac{4}{\pi^2} \,\hat{\textbf{j}} \\ &= -\dfrac{16}{\pi^3}\, \hat{\textbf{j}} - \hat{\textbf{k}} \end{align*}
$\textbf{r}''\left(\frac{\pi}{2}\right) .\nonumber$
Now take the cross product to get
$\textbf{r}'(\pi/2) \times \textbf{r}''(\pi/2) = -\dfrac{4}{\pi^2} \, \hat{\textbf{i}} -\hat{\textbf{j}} + \dfrac{16}{\pi^3} \, \hat{\textbf{k}} \nonumber$
Finally, we plug this information into the curvature formula to get
$\dfrac{\sqrt{\dfrac{16}{\pi^4}+1+\dfrac{256}{\pi^6}}}{\left(\sqrt{1+\dfrac{16}{\pi^4}}\right)^3} \approx 0.952 . \nonumber$
### Curvature of a Plane Curve
If a curve resides only in the xy-plane and is defined by the function $$y = f(t)$$ then there is an easier formula for the curvature. We can parameterize the curve by
$\textbf{r}(t) = t \, \hat{\textbf{i}} + f(t)\, \hat{\textbf{j}} .\nonumber$
We have
$\textbf{r}'(t) = \hat{\textbf{i}} + f '(t) \, \hat{\textbf{j}} \nonumber$
$\textbf{r}''(t) = f ''(t) \, \hat{\textbf{j}} .\nonumber$
Their cross product is just
$r'(t) \times r''(t) = f''(t) \hat{\textbf{k}} \nonumber$
which has magnitude
$||\textbf{r}'(t) \times r''(t)|| = |f''(t)| . \nonumber$
The curvature formula gives
Definition: Curvature of Plane Curve
$K(t) = \dfrac{|f''(t)|}{ \left[1+\left(f'(t) \right)^2 \right]^{3/2}}. \nonumber$
Example $$\PageIndex{4}$$
Find the curvature for the curve $y = \sin\, x \nonumber$.
Solution
We have
$f '(x) = \cos \, x \nonumber$ $f ''(x) = -\sin \, x .\nonumber$
Plugging into the curvature formula gives $K(t) = \dfrac{|-\sin\, t|}{[1+\cos^2t]^{3/2}}\nonumber$
### The Osculating Circle
In first year calculus, we saw how to approximate a curve with a line, parabola, etc. Instead we can find the best fitting circle at the point on the curve. If $$P$$ is a point on the curve, then the best fitting circle will have the same curvature as the curve and will pass through the point $$P$$. We will see that the curvature of a circle is a constant $$1/r$$, where $$r$$ is the radius of the circle. The center of the osculating circle will be on the line containing the normal vector to the circle. In particular the center can be found by adding
$OP + 1/K N . \nonumber$
Exercise $$\PageIndex{2}$$
Find the equation of osculating circle to $$y = x^2$$ at $$x = -1$$.
Solution
## The Normal Component of Acceleration Revisited
How is the normal component of acceleration related to the curvature. If you remember, the normal component the acceleration tells us how fast the particle is changing direction. If a curve has a sharp bend (high curvature) then the directional change will be faster. We now show that there is a definite relationship between the normal component of acceleration and curvature.
$\textbf{a}(t) = a_{\textbf{T}}\textbf{T}(t) + a_{\textbf{N}}\textbf{N}(t) \nonumber$
We have
\begin{align*} \textbf{a}(t) &= \textbf{r}''(t) \\[4pt] &= \dfrac{d}{dt} (\textbf{r}'(t)) \\[4pt] &= \dfrac{d}{dt} \left(||\textbf{r}'(t)||\textbf{T}(t)\right) \\[4pt] &= \dfrac{d}{dt} \left(||r'(t)||)\textbf{T}(t) + ||r'(t)|| \textbf{T}'(t) \right) \\[4pt] &= s''(t)\textbf{T}(t) + s'\textbf{T}'(t) \\[4pt] &= s''(t)\textbf{T}(t) + s'||\textbf{T}'(t)||\textbf{N}(t) = s''(t)\textbf{T}(t) + ks'^2 \textbf{N}(t) .\end{align*}
So that the tangential component of the acceleration is $$s''(t)$$ and the normal component is $$k(t)s'^2(t)$$.
Exercise $$\PageIndex{3}$$
Find the tangential and normal components of $$\textbf{r}(t) = t\, \hat{\textbf{i}}- 2t\, \hat{\textbf{j}} + t^2 \,\hat{\textbf{k}}$$.
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# Evaluate $\int\frac{1}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}dx$
Evaluate $$\int\frac{1}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}dx$$
I start by factoring $$\int\frac{1}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}dx=\int\frac{1}{x^{\frac{9}{25}}\left(x^{\frac{32}{25}}+1\right)}dx$$
Can I do partial fraction from here?
• I would recommend starting with a change of variable, $x=u^{25}$. – Barry Cipra Jul 18 at 19:03
• And then integration by parts. – Viktor Glombik Jul 18 at 19:11
This is a cute $$u$$-substitution problem. The crux of the problem is dividing out by the right factor of $$x$$. Notice that if we factor out one copy of $$x$$ we obtain
$$\int \frac{1}{x\cdot x^{\frac{16}{25}} + x^{\frac{9}{25}} }dx \;\; =\;\; \int \frac{1}{x \left (x^{\frac{16}{25}} + x^{\frac{-16}{25}} \right )}dx.$$
Now let $$u = x^{\frac{16}{25}}$$, and thus $$du = \frac{16}{25}x^{-\frac{9}{25}}dx$$. What this yields is
$$\begin{eqnarray*} \int \frac{1}{x \left (x^{\frac{16}{25}} + x^{\frac{-16}{25}} \right )}dx & = & \frac{25}{16}\int \frac{x^{\frac{9}{25}}} { x\left (u + \frac{1}{u} \right )} du \\ & = & \frac{25}{16} \int \frac{1}{x^{\frac{16}{25}} \left (u + \frac{1}{u} \right ) }du \\ & = & \frac{25}{16} \int \frac{1}{u\left (u + \frac{1}{u} \right )} du \\ & = & \frac{25}{16} \int \frac{1}{u^2 + 1} du. \end{eqnarray*}$$
Take $$u = \tan \theta$$. Final answer should be $$\frac{25}{16}\tan^{-1}\left (x^{\frac{16}{25}}\right ) + c$$.
Let $$x=y^{25}\implies dx=25\,y^{24}\,dy$$ $$\int\frac{dx}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}=25\int\frac{ y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{16\, y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{ \left(y^{16}\right)'}{\left(y^{16}\right)^2+1}$$
|
# How do you integrate int e^(sqrt(2x)) by parts?
Jan 19, 2017
$\int {e}^{\sqrt{2 x}} \mathrm{dx} = {e}^{\sqrt{2 x}} \left(\sqrt{2 x} - 1\right) + C$
#### Explanation:
$I = \int {e}^{\sqrt{2 x}} \mathrm{dx}$
Let $t = \sqrt{2 x}$. This implies that $\frac{1}{2} {t}^{2} = x$, which we differentiate to show that $\mathrm{dx} = t \textcolor{w h i t e}{.} \mathrm{dt}$. Then:
$I = \int {e}^{t} \left(t \textcolor{w h i t e}{.} \mathrm{dt}\right) = \int t {e}^{t} \mathrm{dt}$
We will use integration by parts now, which takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. For $\int t {e}^{t} \mathrm{dt}$, let:
$\left\{\begin{matrix}u = t \text{ "=>" "du=dt \\ dv=e^tdt" "=>" } v = {e}^{t}\end{matrix}\right.$
Then:
$I = u v - \int v \mathrm{du}$
$I = t {e}^{t} - \int {e}^{t} \mathrm{dt}$
$I = t {e}^{t} - {e}^{t} + C$
$I = {e}^{t} \left(t - 1\right) + C$
Returning to $x$ from $t = \sqrt{2 x}$:
$I = {e}^{\sqrt{2 x}} \left(\sqrt{2 x} - 1\right) + C$
|
Part 4: Division
# 4.1: Dividing integers
Notes
##### Scaffold method
To divide two integers (e.g., A ÷ B), repeatedly subtract product of 10 multiples of the divisor (B) from the dividend (A). The result of the division is called the quotient. For example,
• 1242 ÷ 27 = ?
• 100 x 27 = 2700, which is greater than 1242, so start with the tens place value
• 40 x 27 = 1080, so the tens digit of the quotient is 4 [50 x 27 = 1350, which is too high]
• 1242 – 1080 = 162 [this is the remainder after subtracting 40 x 27]
• 6 x 27 = 162, so the ones digit of the quotient is 6 [7 x 27 = 189, which is too high]
• 162 – 162 = 0 [i.e., there is no remainder after subtracting 6 x 27]
• The quotient is 40 + 6 = 46 (with no remainder)
##### Common method
The common method, which is generally called long division, uses the same idea as the scaffold method, but writes the calculations more compactly. For example,
• 1242 ÷ 27 = ?
• 100 x 27 = 2700, which is greater than 1242, so start with the tens place value
• 4 x 27 = 108, so the tens digit of the quotient is 4
• 124 – 108 = 16 [this is the remainder after subtracting 4 x 27]
• Carry down “2” [this is the ones digit in 1242]
• 6 x 27 = 162, so the ones digit of the quotient is 6
• 162 – 162 = 0 [i.e., there is no remainder after subtracting 6 x 27]
• The quotient is 40 + 6 = 46 (with no remainder)
These methods work for dividing a decimal number by an integer too. For example, 124.2 ÷ 27 = 4.6, with the only changes to the calculations above being the insertion of the decimal point in the dividend (124.2) and the quotient (4.6).
The video below works through some examples of dividing integers.
Practice Exercises
Do the following exercises to practice dividing integers.
|
Finding Equivalent Fractions
## How to Find Equivalent Fractions
Equivalent fractions have the same value even though they have different numerators and denominators.
These are equivalent fractions:
How can you tell when fractions are equivalent?
When you draw area models, equivalent fractions have exactly the same amount of colored area.
In number line models, equivalent fractions have the same lengths.
Another way of looking at equivalent fractions is like getting the same fraction of a pizza, but with more, smaller slices.
### Making Equivalent Fractions
We can make equivalent fractions by multiplying or dividing both the numerator and the denominator by the same number.
Make sure to use the same operation and the same number for both the numerator and the denominator.
### Equivalent or Not Equivalent?
Are these two fractions equivalent?
To check if any two fractions are equivalent, use the crisscross trick.
Multiply the numerator of one fraction with the denominator of the other. This is that numerator's crisscross.
Let's try it.
8 x 3 = 24. We can write that crisscross value above the numerator to keep track.
Now, let's find the other numerator's crisscross.
If fractions have the same crisscross, they're equivalent.
This means 8/12 and 2/3 are equivalent fractions. ✅
Tip: If the multiplication is a little hard, you can try simplifying the fractions before doing the crisscross. We'll cover that in the next lesson.
### So, What Did You Learn?
To see if any two fractions are equivalent, you just have to multiply twice using the crisscross trick.
Otherwise, you can try to find a common multiplier of the numerator and denominator of one fraction that gets you the other.
#### Another Example
Are these two fractions equivalent?
Let's find the first crisscross.
Now, let's multiply 4 x 6 to find the other crisscross.
We see these fractions are not equivalent. They don't have the same crisscross.
The fraction with the larger crisscross is larger.
3/4 is greater than 6/10.
Great work!
### Finding the Missing Numerator
Let's solve a missing numerator problem together.
Let's think. How are their denominators related? 🤔
We figure this out by dividing 18 by 3.
18 ÷ 3 = 6
So, the denominator 18 is six times larger than the other denominator.
For the fractions to be equivalent, or equal, the missing numerator must also be six times larger than the other numerator, 2.
2 × 6 = ?
2 × 6 = 12
We found the missing numerator, 12.
This means that 2/3 is equivalent to 12/18
### Find the Missing Denominator
Let's solve one last problem together.
How are their numerators related?
20 ÷ 5 = 4
The left numerator is 4 times smaller than the right numerator.
So, the left denominator must also be 4 times smaller than the right denominator.
24 ÷ 4 = ?
Can you figure it out?
24 ÷ 4 = 6
The missing denominator is 6.
This means that 20/24 is equivalent to 5/6. ✅
You did a great job learning about equivalent fractions. 🎉
Now, let's try the practice. You'll understand more and remember for longer.
Complete the practice to earn 1 Create Credit
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# What does place value value mean?
## What does place value value mean?
Place value is the value of each digit in a number. For example, the 5 in 350 represents 5 tens, or 50; however, the 5 in 5,006 represents 5 thousands, or 5,000.
## What is place value short answer?
Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on.
What is place value and why is it important?
Place value provides the foundation for regrouping, multiple-digit multiplication, and more in the decimal system, as well as a starting point for the understanding of other base systems.
### What is the place value formula?
We can find the place value of a digit by finding how many places the digit is to the right or left of the decimal point in a number. If it is n places to the left of the decimal point, we multiply the digit by 10^(n – 1) to get the digit’s place value.
### How do we use place value in real life?
Place value helps us make decisions that are used in our daily lives ex) costs, weight, distances, time etc. Our number system is based on a Base Ten system. Base ten means our number system has a base of ten. We group our numbers by clusters of ten.
How do you teach place value in Year 1?
In our experience the answer to teaching place value in Year 1 is to keep things as simple as possible. Year 1 pupils may start to recognise place value when working with coins, but a good way to explain the topic in more explicit terms is by using beads. One idea is to use strings of beads in sets of ten.
#### What is the face value of 7 in 27650934?
The place value of a number is the number followed by how many digits it is far from the end; in this case the digit 7 is 6 digits far from the end. Therefore, the difference between the place value and face value of 7 is (7000000 – 7) = 6999993.
#### What is the place value of 8 in 782?
1/10th
In the same spirit, the 8 in 782 is 1/10th the value of the 8 in 845.
What is the place value of 6 in 645?
A digit’s place in a number is important. It tells us what that digit is worth. For example, in the number 645 the 5 means 5 “ones” or “units”, the 4 means 4 “tens” and the 6 means 6 “hundreds”. So 645 means 5 + 40 + 600.
## What is the place value of 3 in the numerical 3259?
What is the place value of 3 in the numeral 3259?
Number x = Value
3 x 3,000
2 x 200
5 x 50
9 x 9
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# Difference between revisions of "2021 AMC 12B Problems/Problem 12"
The following problem is from both the 2021 AMC 10B #19 and 2021 AMC 12B #12, so both problems redirect to this page.
## Problem
Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$, then the average value (arithmetic mean) of the integers remaining is $32$. If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$. If the greatest integer is then returned to the set, the average value of the integers rises to $40.$ The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$. What is the average value of all the integers in the set $S?$
$\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37$
## Solution 1
Let $x$ be the greatest integer, $y$ be the smallest, $z$ be the sum of the numbers in S excluding $x$ and $y$, and $k$ be the number of elements in S.
Then, $S=x+y+z$
Firstly, when the greatest integer is removed, $\frac{S-x}{k-1}=32$
When the smallest integer is also removed, $\frac{S-x-y}{k-2}=35$
When the greatest integer is added back, $\frac{S-y}{k-1}=40$
We are given that $x=y+72$
After you substitute $x=y+72$, you have 3 equations with 3 unknowns $S,$, $y$ and $k$.
$S-y-72=32k-32$
$S-2y-72=35k-70$
$S-y=40k-40$
This can be easily solved to yield $k=10$, $y=8$, $S=368$.
$\therefore$ average value of all integers in the set $=S/k = 368/10 = 36.8$, D)
~ SoySoy4444
## Solution 2
We should plug in $36.2$ and assume everything is true except the $35$ part. We then calculate that part and end up with $35.75$. We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$. The answer is then $\boxed{(D) 36.8}$. You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm
~ pi_is_3.14
~IceMatrix
|
Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
Jesus replied: “‘Love the Lord your God with all your heart and with all your soul and with all your mind.’ This is the first and greatest commandment. And the second is like it: ‘Love your neighbor as yourself.’” Matthew 22:37-39 NIV
# 12-03 Derivatives
Summary: In this section, you will:
• Find the derivative of a function.
• Find the slope of the tangent line to a function.
SDA NAD Content Standards (2018): PC.4.2, PC.6.4
Velocity is the rate-of-change of position. That means it is how quickly position is changing. Acceleration is the rate-of-change of velocity. In other words, acceleration is how quickly velocity is changing.
Calculus mainly focuses on two problems. The first problem is finding the rate-of-change of functions. This is called finding a derivative. Graphically this is represented by finding the slope of the line tangent to a function. Remember that slopes are rates-of-change. The second problem is finding the area under a curve on a graph. This is called a integral. Integrals also happen to be the opposite of derivatives.
This lesson is about derivatives and rates-of-change. To derive the formula for rate-of-change, start with the slope formula for two points whose x-values are separated by h: (x, f(x)) and (x + h, f(x + h)). See figure 2.
$$Slope = m = \frac{y_2 - y_1}{x_2 - x_1}$$
$$Slope = \frac{f(x + h) - f(x)}{x + h - x}$$
$$Slope = \frac{f(x + h) - f(x)}{h}$$
This gives us the average rate-of-change between two points, but if the instantaneous, or exact, slope at a given point is wanted, h must be zero. So, take the limit as h approaches 0. This produces a function that is called a derivative and is denoted by f′(x). To find the rate-of-change, plug in the x-value of the location of the desired slope.
$$Slope = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$
###### Derivative
$$f′(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$
The rate-of-change at a location is found by plugging in the x-value of the location.
#### Example 1: Find a Derivative
Find the derivative of f(x) = x² + 3
###### Solution
Find f(x + h).
$$f(x + h) = (x + h)^2 + 3$$
Substitute f(x + h) and f(x) into the derivative formula.
$$f′(x) = \lim_{h \rightarrow 0} \frac{\color{blue}{f(x + h)} - \color{red}{f(x)}}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{(\color{blue}{(x + h)^2 + 3}) - (\color{red}{x^2 + 3})}{h}$$
Simplify the numerator.
$$f′(x) = \lim_{h \rightarrow 0} \frac{(x^2 + 2xh + h^2 + 3) - (x^2 + 3)}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{2xh + h^2}{h}$$
Factor the h out of the numerator and then cancel the h from the numerator and denominator.
$$f′(x) = \lim_{h \rightarrow 0} \frac{h(2x + h)}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} (2x + h)$$
Now evaluate the limit by plugging in 0 for h.
$$f′(x) = 2x + 0$$
$$f′(x) = 2x$$
#### Example 2: Find a Derivative
Find the derivative of f(x) = x² − 2x.
###### Solution
Find f(x + h).
$$f(x + h) = (x + h)^2 - 2(x + h)$$
Substitute f(x + h) and f(x) into the derivative formula.
$$f′(x) = \lim_{h \rightarrow 0} \frac{\color{blue}{f(x + h)} - \color{red}{f(x)}}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{(\color{blue}{(x + h)^2 - 2(x + h)}) - (\color{red}{x^2 - 2x})}{h}$$
Simplify the numerator.
$$f′(x) = \lim_{h \rightarrow 0} \frac{(x^2 + 2xh + h^2 - 2x - 2h) - (x^2 - 2x)}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{2xh + h^2 - 2h}{h}$$
Factor the h out of the numerator and then cancel the h from the numerator and denominator.
$$f′(x) = \lim_{h \rightarrow 0} \frac{h(2x + h - 2)}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} (2x + h - 2)$$
Now evaluate the limit by plugging in 0 for h.
$$f′(x) = 2x + 0 - 2$$
$$f′(x) = 2x - 2$$
##### Try It 1
Find the derivative of f(x) = 2x³.
f ′(x) = 6x²
#### Example 3: Find a Derivative
Find the derivative of $$f(x) = \sqrt{x} - 2$$.
###### Solution
Find f(x + h).
$$f(x + h) = \sqrt{x + h} - 2$$
Substitute f(x + h) and f(x) into the derivative formula.
$$f′(x) = \lim_{h \rightarrow 0} \frac{\color{blue}{f(x + h)} - \color{red}{f(x)}}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{(\color{blue}{\sqrt{x + h} - 2}) - (\color{red}{\sqrt{x} - 2})}{h}$$
Simplify the numerator.
$$f′(x) = \lim_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$$
Use the rationalizing technique.
$$f′(x) = \lim_{h \rightarrow 0} \frac{(\sqrt{x + h} - \sqrt{x})}{h} \frac{(\sqrt{x + h} + \sqrt{x})}{(\sqrt{x + h} + \sqrt{x})}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{(x + h) - (x)}{h(\sqrt{x + h} + \sqrt{x})}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}$$
Cancel the h from the numerator and denominator.
$$f′(x) = \lim_{h \rightarrow 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}$$
Now evaluate the limit by plugging in 0 for h.
$$f′(x) = \frac{1}{\sqrt{x + 0} + \sqrt{x}}$$
$$f′(x) = \frac{1}{\sqrt{x} + \sqrt{x}}$$
$$f′(x) = \frac{1}{2\sqrt{x}}$$
##### Try It 2
Find the derivative of $$f(x) = \sqrt{x + 1}$$.
$$f′(x) = \frac{1}{2\sqrt{x + 1}}$$
#### Example 4: Find the Slope of a Tangent Line
Find the slope of $$f(x) = x^3$$ at (1, 1).
###### Solution
Start by finding the derivative. So, find f(x + h).
$$f(x + h) = (x + h)^3$$
Substitute f(x + h) and f(x) into the derivative formula.
$$f′(x) = \lim_{h \rightarrow 0} \frac{\color{blue}{f(x + h)} - \color{red}{f(x)}}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{(\color{blue}{(x + h)^3}) - (\color{red}{x^3})}{h}$$
Simplify the numerator.
$$f′(x) = \lim_{h \rightarrow 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - (x^3)}{h}$$
Factor the h out of the numerator and then cancel the h from the numerator and denominator.
$$f′(x) = \lim_{h \rightarrow 0} \frac{3x^2h + 3xh^2 + h^3}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} \frac{h(3x^2 + 3xh + h^2)}{h}$$
$$f′(x) = \lim_{h \rightarrow 0} 3x^2 + 3xh + h^2$$
Now evaluate the limit by plugging in 0 for h.
$$f′(x) = 3x^2 + 3x(0) + (0)^2$$
$$f′(x) = 3x^2$$
Find the slope at (1, 1) by plugging in x = 1.
$$f′(2) = 3(1)^2 = 3$$
The slope of the tangent line at (1, 1) is 3.
##### Try It 3
Find the slope of $$f(x) = 2x^2 - 4$$ at (–1, –2).
–4
##### Lesson Summary
###### Derivative
$$f′(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$
The rate-of-change at a location is found by plugging in the x-value of the location.
## Practice Exercises
1. A function that gives the slope of another function is called a _?_.
2. Find the derivative of the function.
3. $$f(x) = 3x^2$$
4. $$f(x) = (x - 2)^2$$
5. $$f(x) = x^3 + 2x$$
6. $$f(x) = -4x^2 - 2x + 3$$
7. $$f(x) = \sqrt{x - 3}$$
8. $$f(x) = \sqrt{x + 2} - 4$$
9. $$f(x) = \frac{1}{x}$$
10. Find the slope of the function at the given point.
11. $$f(x) = 3x^2 - 4$$ at (2, 8)
12. $$f(x) = 2x^3 + x$$ at (–1, –3)
13. $$f(x) = \sqrt{x}$$ at (4, 2)
14. $$f(x) = \frac{2}{x^2}$$ at (1, 2)
15. $$f(x) = x^2 - 3x + 2$$ at (0, 2)
16. Problem Solving
17. Velocity is the derivative of position with respect to time. A falling object's position can be modeled by $$x(t) = -4.9t^2 + 100$$ where t is time in seconds and x(t) is position in meters. Find the velocity at t = 2 seconds.
18. Acceleration is the derivative of velocity with respect to time. A falling object's velocity can be modeled by $$v(t) = -9.8t$$ where t is time in seconds and v(t) is velocity in meters per second. Find the acceleration at t = 2 seconds.
19. Mixed Review
20. (12-02) Evaluate $$\displaystyle \lim_{x \rightarrow 3} \frac{x^2 - 4x + 3}{x - 3}$$.
21. (12-02) Evaluate $$\displaystyle \lim_{x \rightarrow 0^+} -\frac{4| x |}{x}$$.
22. (12-01) Evaluate $$\displaystyle \lim_{x \rightarrow 2} \cos{πx}$$.
23. (10-02) Use formulas to evaluate $$\displaystyle \sum_{i = 1}^{10} (x^2 - 3x)$$.
24. (10-02) Use formulas to evaluate $$\displaystyle \sum_{i = 1}^{100} (5x^2 - x^3)$$.
1. derivative
2. $$f′(x) = 6x$$
3. $$f′(x) = 2x - 4$$
4. $$f′(x) = 3x^2 + 2$$
5. $$f′(x) = -8x - 2$$
6. $$f′(x) = \frac{1}{2\sqrt{x-3}}$$
7. $$f′(x) = \frac{1}{2\sqrt{x+2}}$$
8. $$f′(x) = -\frac{1}{x^2}$$
9. 12
10. 7
11. $$\frac{1}{4}$$
12. –4
13. –3
14. v = –19.6 m/s
15. a = –9.8 m/s²
16. 2
17. –4
18. 1
19. 220
20. –23810750
|
Pythagore
Demonstration of the theorem
Author : Thibaut Bernard Number of visitor
Update: Sunday 30 May 2001.
Alphaquark author's Note :
This page is a translation of Démonstration du théorème de Pythagore with the help of Altavista translation.
I hope this translation is good, but if there are any errors, you can write me.
If this translation is successful, perhaps I will try to translate another document of Alphaquark.
Construction of the geometrical figure which will be used for the demonstration
Let us take a rectangle of width A and height B. This rectangle which we make swivel of 90o in the following way :
For each rectangle, let us divide into two in the following way : Let us make swivel of 90o the right-angled triangles in the following way yellow and purple : We thus find ourselves with four right-angled triangles. We note that one finds oneself with a square inside another.
Demonstration
Notation
Let us take again our last diagram to indicate each of with dimensions by the following letters:
One has four right-angled triangles of which :
the with dimensions one opposed by a,
the base is indicated by b,
and the hypotenuse by n.
Each hypotenuse n thus represent dimensions small square.
With dimensions opposite more the base (a + b) of each right-angled triangle represent each one of with dimensions great square.
Surface squares
Small square : n2.
Great square : (a + b)2.
Surface right-angled triangles
If one associates two right-angled triangles, one finds oneself with a rectangle of surface : a * b.
The total surface of the four right-angled triangles is : 2ab.
Difference in surface
The surface of the small square is equal to the surface of the great square minus the surface of the four right-angled triangles :
n2 = (a + b)2 - 2ab
from where
n2 = a2 + 2ab + b2 - 2ab
what gives
n2 = a2 + b2
Conclusion, the square on the hypotenuse is equal to the sum of the squares of with dimensions of the right angle.
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# What is an Algorithm?
#### In this tutorial, we will learn what algorithms are with the help of examples.
An algorithm is a set of well-defined instructions in sequence to solve a problem.
## Qualities of a good algorithm
1. Input and output should be defined precisely.
2. Each step in the algorithm should be clear and unambiguous.
3. Algorithms should be most effective among many different ways to solve a problem.
4. An algorithm shouldn’t include computer code. Instead, the algorithm should be written in such a way that it can be used in different programming languages.
## Examples Of Algorithms In Programming
Algorithm to add two numbers entered by the user
```Step 1: Start
Step 2: Declare variables num1, num2 and sum.
Step 3: Read values num1 and num2.
Step 4: Add num1 and num2 and assign the result to sum.
sum←num1+num2
Step 5: Display sum
Step 6: Stop
```
Find the largest number among three different numbers
```Step 1: Start
Step 2: Declare variables a,b and c.
Step 3: Read variables a,b and c.
Step 4: If a > b
If a > c
Display a is the largest number.
Else
Display c is the largest number.
Else
If b > c
Display b is the largest number.
Else
Display c is the greatest number.
Step 5: Stop
```
Roots of a quadratic equation ax+ bx + c = 0
```Step 1: Start
Step 2: Declare variables a, b, c, D, x1, x2, rp and ip;
Step 3: Calculate discriminant
D ← b2-4ac
Step 4: If D ≥ 0
r1 ← (-b+√D)/2a
r2 ← (-b-√D)/2a
Display r1 and r2 as roots.
Else
Calculate real part and imaginary part
rp ← -b/2a
ip ← √(-D)/2a
Display rp+j(ip) and rp-j(ip) as roots
Step 5: Stop
```
Factorial of a number entered by the user.
```Step 1: Start
Step 2: Declare variables n, factorial and i.
Step 3: Initialize variables
factorial ← 1
i ← 1
Step 4: Read value of n
Step 5: Repeat the steps until i = n
5.1: factorial ← factorial*i
5.2: i ← i+1
Step 6: Display factorial
Step 7: Stop
```
Check whether a number is a prime number or not
```Step 1: Start
Step 2: Declare variables n, i, flag.
Step 3: Initialize variables
flag ← 1
i ← 2
Step 4: Read n from the user.
Step 5: Repeat the steps until i=(n/2)
5.1 If remainder of n÷i equals 0
flag ← 0
Go to step 6
5.2 i ← i+1
Step 6: If flag = 0
Display n is not prime
else
Display n is prime
Step 7: Stop
```
Find the Fibonacci series till term ≤ 1000.
```Step 1: Start
Step 2: Declare variables first_term,second_term and temp.
Step 3: Initialize variables first_term ← 0 second_term ← 1
Step 4: Display first_term and second_term
Step 5: Repeat the steps until second_term ≤ 1000
5.1: temp ← second_term
5.2: second_term ← second_term + first_term
5.3: first_term ← temp
5.4: Display second_term
Step 6: Stop```
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## Two Machine Learning Fields
There are two sides to machine learning:
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