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# Lagrange multiplier
A Lagrange multiplier is a constant, usually represented by the Greek symbol lambda (λ), which is part of the formula for finding an extreme value of a function f(x,y) subject to a constraint g(x,y). For example, the method of Lagrange multipliers can help determine the maximum area of a rectangle that can be fit inside a given circle represented by g(x,y) = x2 + y2.
The points at which extrema occur for a function f(x,y) subject to the constraint g(x,y) are the common solutions to:
$\nabla f= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = \lambda \nabla g$
and g(x,y) = 0.
## Example
Problem: What is the shape and dimension of the largest rectangle that fits within a circle having a radius equal to the square root of 2?
Solution: g(x,y) = x2 + y2 - 2 = 0. We want to maximize the area of a rectangle. Assuming symmetry about the origin (we could always map a solution to a symmetric one), the area f(x,y) that we want to maximize equals 2x times 2y: f(x,y) = 4xy. From the equation for the Lagrange multiplier above, and setting the x and y coefficients equal to each other, we can generate these two equations:
4y = λ2x
4x = λ2y
Solving each for lambda (λ) demonstrates that x must be equal to y, and plugging that into the equation for g yields:
2x2 = 2
x = 1 = y
Then do not forget to include the maximum area in your answer, which is $f(1,1) = 4 \cdot 1 \cdot 1 = 4$.
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# How do you solve the following linear system: 4x+3y=8 , x-2y=13 ?
Jan 28, 2016
$\left(x , y\right) = \left(5 , - 4\right)$
#### Explanation:
$4 x + 3 y = 8$
$x - 2 y = 13$
In the second equation we can find the value of $x$ by transposing $2 y$ to the other side of the equation and we can substitute the value of $x$ to the other equation
Solve for second equation:
$\rightarrow x - 2 y = 13$
Add $2 y$ both sides:
$\rightarrow x = 13 + 2 y$
Substitute the value of $x$ to the first equation:
$\rightarrow 4 \left(13 + 2 y\right) + 3 y = 8$
$\rightarrow \left(52 + 8 y\right) + 3 y = 8$
$\rightarrow 52 + 8 y + 3 y = 8$
$\rightarrow 52 + 11 y = 8$
$\rightarrow 11 y = 8 - 52$
$\rightarrow 11 y = - 44$
$\rightarrow y = - \frac{44}{11} = - 4$
So,substitute value of y to second equation:
$\rightarrow x - 2 \left(- 4\right) = 13$
$\rightarrow x - \left(- 8\right) = 13$
$\rightarrow x + 8 = 13$
$\rightarrow x = 13 - 8 = 5$
So,$\left(x , y\right) = \left(5 , - 4\right)$
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# Factorization of Polynomial
Factorization is the process in which we can find factors of either a given number or the algebraic expression using various techniques such as prime factorization, factorization using algebraic identities, and factorization of a quadratic polynomial using splitting the middle term method. Factorization of polynomials or numbers can help us solve different problems in algebra. Factors are either numbers or expressions defined for given numbers and expressions such that when their division yields a remainder 0 or the number is less than the given number. For example, 1, 2, 4, 7, 14, and 28 are the factors of 28, and (x +1) and (x – 1) are the factors of f(x) = x2 – 1. The image of the factorization of the Polynomial is added below,
## Prime Factorization Definition
The most common technique for finding the factors of any number or polynomial is prime factorization. In this method, we write a number in the form of the product of its prime factors.
### Prime Factorization of Numbers
Prime factorization is the technique in which we can find the factors of numbers using division by the smallest prime possible at a time. i.e., if We need to find prime factors of 24, then first we’ll divide 24 by 2, which gives 0 as a remainder, and again divide the quotient 12 by 2(as it can be divided by 2), the remainder is still 0. Now again divide by 2 to get 6 as quotient, and again by 2 to get 3 (which is not divisible by 2). Thus divide it by the next prime which is 3 itself. Thus prime factorization of 24 = 2 × 2 × 2 × 2 × 3.
Example: Find the Prime Factors of 70
Solution:
As 70 = 2 × 5 × 7
(here 2, 5, 7 are prime factors of 70; as there are more factors such as 10, 14, 35 for 70 but these are not prime factors)
### Prime Factorization of Polynomials
Similarly, we can express algebraic expressions as the product of their factors. If an algebraic expression cannot be reduced further then its factors are called the prime factor.
Example:
8xy = 2 × 2 × 2 × x × y
(here 8xy is formed by multiplication of numbers(2, 2, 2, x, y) they are factors of 8xy, considering x and y relatively prime to everything.)
## Factorization of Polynomials
Factorization is nothing but writing a number as the product of smaller numbers. It is the decomposition of a number (or) mathematical objects into smaller or simpler numbers/objects. Factorization of different types of algebraic expressions is very useful for various purposes used in mathematics. Various methods of factorization of polynomials are,
• By Greatest Common Factor (GCF)
• By Regrouping
• Factorization Using Identities
• Factorization Using Splitig Terms (only for quadratic polynomials)
• Factorization Using Rational Root and Factor Theorem
### Factorization of Polynomials by Greatest Common Factor
The highest common factor between the two numbers is called the GCF (Greatest Common Factor). It is useful for factoring polynomials
The steps for finding GCF are:
• Step 1: First, split every term of algebraic expression into irreducible factors
• Step 2: Then find the common terms among them.
• Step 3: Now the product of common terms and the remaining terms give the required factor form.
Example: Factorise 3x + 18
Solution:
Step 1: First splitting every term into irreducible factors.
3x = 3 × x
18 = 2 × 3 × 3
Step 2: Next step to find the common term
3 is the only common term
Step 3: Now the product of common terms and remaining terms is 3(x + 6)
So 3(x + 6) is the required form.
### Factorization of Polynomials by Regrouping
Sometimes the terms of the given expression should be arranged in suitable groups in such a way so that all the groups have a common factor, and then the common factor is taken out. In this way, factoring of a polynomial is done.
Example: Factorise x2 + yz + xy + xz.
Solution:
Here we don’t have a common term in all terms, so we are taking (x2 + xy) as one group and (yz + xz) as another group.
Factors of (x2 + xy) = (x × x) + (x × y) = x(x + y)
Factors of (yz + xz) = (y × z) + (x × z) = z(x + y)
After combining them,
x2 + yz + xy + xz = x(x + y) + z(x + y)
Taking (x + y) as common we get,
x2 + yz + xy + xz = (x + y) (x + z)
Thus, (x + y), and (x + z) are the factors of the given expression x2 + yz + xy + xz
### Factorization of Polynomials Using Identities
There are many standard algebraic identities that are used to factorize various polynomials. Some of them are given below:
• a2 + 2ab + b2 = (a + b)2
• a2 – 2ab + b2 = (a – b)2
• a2 – b2 = (a + b) (a – b)
• a2 + b2 + c2 + 2(ab + bc + ca) = (a + b + c)2
• a3 + b3 + 3ab(a + b) = (a + b)3
• a3 – b3 – 3ab(a – b) = (a – b)3
• a3+ b3= (a+b)(a2+ b2– ab)
• a3 – b3= (a-b)(a2+ b2+ ab)
• a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
Example 1: Factorise x2 + 8x + 16
Solution:
This is in the form of (a + b)2 = a2 + 2ab + b2
x2 + 8x + 16 = x2 + 2 × x × 4 + 42
= (x + 4)2
= (x + 4) (x + 4)
Example 2: Factorise x2 – 6x + 9
Solution:
Let g(x) = x2 – 6x + 9
This is in the form of (a – b)2 = a2 – 2ab + b2
⇒ g(x) = x2 – 2 × x × 3 + 32
⇒ g(x) = (x – 3)2
⇒ g(x) = (x – 3) (x – 3)
Example 3: Factorise x3 – 27.
Solution:
Let g(x) = x3 – 27
Using, a3 – b3= (a-b)(a2+ b2+ ab)
g(x) = x3 – 33
⇒ g(x) = (x- 3 )(x2 + 32 + 3x)
⇒ g(x) = (x- 3 )(x2+ 3x + 9 )
As , x2+ 3x + 9 can’t be factorize further,
Thus factors of x3 – 27 are (x- 3 ), and (x2+ 3x + 9 ) .
### Factors Using Splitting Terms
Factoring polynomials is mostly used for solving quadratic equations. The quadratic equation is factorized to reduce it to linear factors. The form of a quadratic equation used is x2 + (a + b)x + ab = 0, which is split into two factors (x + a)(x + b) = 0. Consider the quadratic polynomial of the form
Let f(x) = x2 + (a + b)x + ab
⇒ f(x) = x.x + ax + bx + ab
⇒ f(x) = x(x + a) + b(x + a)
⇒ f(x) = (x + a)(x + b)
In the above polynomial, the middle term is split as the sum or difference of two terms, and the constant term is the product of these two factors. Then, common factors are taken by grouping the first and second terms and the third and fourth terms. Thus, a quadratic polynomial is expressed as the product of two factors.
For example: x2 + 5x + 6
Solution:
= x2 + 5x + 6
= x.x + (3 + 2)x + 3.2
= x.x + 3x + 2x + 3.2
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
Thus, factoring polynomials is done by splitting the middle terms in a quadratic polynomial.
### Factorization Using Rational Root and Factor Theorem
Factors of cubic or any higher polynomial can be found using the rational root theorem and long division together. Which is explained as follows:
Step 1: Use the rational root theorem to find the possible roots. i.e., For a polynomial of n degree i.e., f(x) = anxn + an−1xn−1 +…+ a1x + a0, all the possible rational roots are the divisor of a0/an
Step 2: After finding one root, use the factor theorem to get the first factor and divide the polynomial with the factor using long division and write the polynomial as a product of quotient and dividend.
Step 3: If the quotient is a quadratic polynomial then solve it by the method of splitting the middle term. If not a quadratic polynomial then repeat step 1 and step 2 until the quotient becomes a quadratic polynomial.
Step 4: The result of step 3 is the required factors of the given polynomial.
Example: Factorise f(x) = x3 + 3x2 – 4x – 12.
Solution:
f(x) =x3 + 3x2 – 4x – 12
As p/q = -12
By rational root theorem, all possible rational roots of the polunomial are divisor of a0/an ,
Thus, divisors = ±1, ±2, ±3, ±4, ±6, and ±12.
x = 2, in p(x), we get
f(2) = (2)3 + 3(2)2 – 4(2) -12
f(2) = 8 + 12 – 8 – 12 = 0
Thus, by factor theorem, x – 2 is the factor of f(x).
Thus, x3 + 3x2 – 4x – 12= (x-2)(x2 +5x + 6)
x3 + 3x2 – 4x – 12 = (x-2)(x+2)(x+3)
Thus, factors of f(x) are (x-2), (x+2), and (x+3).
## Solved Examples on Factorization
Example 1: Factorise a2 – 20a + 100
Solution:
f(a) = a2 – 20a + 100
This in the form of (a – b)2 = a2 – 2ab + b2
⇒ f(a) = a2 – 2 × a × 10 + 102
⇒ f(a) = (a – 10)2
⇒ f(a) = (a – 10) (a – 10)
Example 2: Factorise 25x2 – 49
Solution:
This is the form of a2 – b2 = (a + b) (a – b)
25x2 – 49 = (5x)2 – 72
= (5x + 7) (5x – 7)
Example 3: Factorise 2xy + 3 + 2y + 3x
Solution:
Let f(x) = 2xy + 3 + 2y + 3x
⇒ f(x) = 2xy + 2y + 3x + 3 [rearranging terms to get common terms]
⇒ f(x) = 2y (x + 1) + 3(x + 1)
⇒ f(x) = (2y + 3)(x + 1)
Example 4: Factorize x3 – 8
Solution:
We know the identity, a3 – b3 = (a – b)(a2 + ab + b2). Here a = x and b = 2.
Thus, x3 – 8= (x – 2)(x2 + 2x + 22)
⇒ x3 – 8= (x – 2)(x2 + 2x + 4 )
Example 5: Factorize 2x3 + 4x2– 6x.
Solution:
Let f(x) = 2x3 + 4x2– 6x
⇒ f(x) = 2x(x2 + 2x – 3)
⇒ f(x) = 2x(x2 + 3x – x – 3)
⇒ f(x) = 2x[x(x + 3) – (x + 3)]
⇒ f(x) = 2x(x + 3)(x – 1)
Example 6: Factorise a4 – 83
Solution:
As we know, a2 – b2 = (a + b) (a – b)
Thus, a4 – 83 = a4 – 34 = (x² + 9) (x² – 9)
Now, again factorize (a2 – 9) using sam identity.
⇒ a4 – 83 = (a² + 9) (a – 3)(a + 3)
## FAQs on Factorization
### Q1: What is Factorization and write its example?
Factorization is breaking a larger number into smaller number numbers so that when multiplied together, they give the original number. For example, factorization of 15 is achieved by multiplying 3 by 5.
i.e. 15 = 5 × 3
### Q2: What are the Methods of Factorization of Polynomials?
The various different methods to factorize polynomials are as follows:
• Greatest Common Factor (GCF)
• Grouping Method
• Factorization using Identities
• Factors Using Splitting Terms
• Factorization Using Rational Root and Factor Theorem
### Q3: How to Factorize a Given Algebraic Expression?
For factorizing an algebraic expression we use the known algebraic identities. For example,
By using the algebraic identity (x-a) 2 = x2 -2ax +a2. x2 – 8x + 16 is factorized as (x – 4)(x – 4)
### Q4: What are the factors of 36?
The factors of 36 are 2, 3, 4, 6, 9, 12, and 18, i.e.
• 2×18 = 36
• 3×12 = 36
• 4×9 = 36
• 6×6= 36
### Q5: What are the factors of 24?
The factors of 24 are 2, 3, 4, 6, 8, and 12, i.e.
• 2×12 = 24
• 3×8 = 24
• 4×6 = 24
### Q6: What are the factors of 18?
The factors of 18 are 2, 3, 6, and 9, i.e.
• 2×9 = 18
• 3×6 = 18
### Q7: What are the factors of 12?
The factors of 12 are 2, 3, 4, and 6, i.e.
• 2×6 = 12
• 3×4 = 12
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HOTS & Value based Questions: Arithmetic Progressions
# HOTS & Value based Questions: Arithmetic Progressions - Notes | Study Mathematics (Maths) Class 10 - Class 10
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HOTS
Q1. Find the ‘6th’ term of the A.P. :
Sol. Here,
∴ d = a2 – a1
Now, an = a + (n – 1)d
Thus, the nth term is
Again, we have
i.e., the 6th term is
Q2. If the ratio of the sum of first n terms of two A.P.'s is (7n + 1) : (4n + 27), find the ratio of their mth terms.
Sol. Let the first terms of given AP's be a1 and a2, common differences be d1 and d2 and let Sn and the sum of n terms.
Now, replacing n by (2m-1) for getting ratio of mth terms of given APs,
Thus, the required ratio of mth term of given AP's is (14m – 6) : (8m + 23)
Q3. If the numbers a, b, c, d and e form an A.P., then find the value of a – 4b + 6c – 4d + e
Sol. We have the first term of A.P. as ‘a’.
Let D be the common difference of the given A.P.,
Then :
b = a + D, c = a + 2D, d= a + 3D and e = a + 4D
∴ a – 4b + 6c – 4d + e
= a – 4(a + D ) + 6 (a + 2D) – 4 (a + 3D) + (a + 4D)
= a – 4a + 6a – 4a + a – 4D + 12D – 12D + 4D
= 8a – 8a + 16D – 16D = 0
Thus, a – 4b + 6c – 4d + e = 0
Q4. is the arithmetic mean between ‘a’ and ‘b’, then, find the value of ‘n’.
Sol. Note: A.M., between ‘a’ and ‘b ’= 1/2
(a + b)
We know that :
A.M. between ‘a’ and ‘b’ = a+b/2
It is given that,
is the A.M. between ‘a’ and ‘b’
∴
By cross multiplication, we get :
⇒2an + 1 + 2bn + 1 = an + 1 + abn + anb + bn + 1
⇒ 2an + 1 – an + 1 + 2bn + 1– bn + 1= abn + anb
⇒ an + 1 + bn + 1 = abn +anb
⇒ an+1 – anb = abn – bn+1
⇒ an[a–b] = bn[a –b]
⇒ n = 0
Q5. Solve the equation :
1 + 4 + 7 + 10 + ... + x = 287
Sol.
Since,
∴ a = 1,d = 3 and an = x
∴ an = a + (n – 1)d
⇒ x = 1 + (n – 1) 3 or x = 3n – 2
Also, Sn = n/2 (a+l)
⇒ 2(287) = n[1 + (3n – 2)]
⇒ 574 = n[3n – 1]
⇒ 3n2 – n – 574 = 0
Solving the above quadratic equation, we get
or
But, negative n is not desirable.
∴ n = 14
x = 3n – 2
Now, x = 3(14) – 2 = 42 – 2 = 40
Thus, x = 40
Q6. Find three numbers in A.P. whose sum is 21 and their product is 231.
Sol. Let the three numbers in A.P. are:
a – d, a, a + d
∴ (a – d) + a + (a + d) = 21
⇒ a – d + a + a + d = 21
or 3a = 21 ⇒ a = 7
Also, (a – d) × a × (a + d) = 231
∴ (7 – d) × 7 × (7 + d) = 231
⇒ (7 – d) (7 + d) × 7 = 231
⇒ 72 – d2 = 231 /7 = 33
⇒ 49 – d2 = 33
or d2 = 49 – 33 = 16
⇒ d = ± 4
Now, when d = 4, then three numbers in AP are : (7 – 4), 7, (7 + 4) i.e. 3, 7, 11.
When d = –4, then three numbers in AP are : [7 – (–4)], 7, [7 + (–4)]
or 11, 7, 3
Q7. The ninth term of an A.P. is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and common difference.
Sol. Let 'a' be the first term and 'd' be the common difference
Since, an = a + (n–1)d
∴ a9 = a + 8d and a12 = a + 11d
Also a2 = a + d and a3 = a + 2d
since a9 = 7a2 and a12 = 5a3 + 2
or a9 = 7 (a + d) ⇒ a9 = 7a + 7d ... (1)
a12 = 5 (a + 2d) + 2
⇒ a12 = 5a + 10d + 2 ... (2)
Now, a + 8d = 7a + 7d [From (1)]
or – 6a + d = 0 ... (3)
Also a + 11d = 5a + 10d + 2 [From (2)]
or –4a + d = 2 ... (4)
Subtracting (4) from (3), we have
–2a = –2 ⇒ a = 1
Now, from (3), –6 + d = 0 ⇒ d = 6
Thus, a = 1 and d = 6
Value-based Questions
Q1. Savita has two options to buy a house:
(a) She can pay a lumpsum amount of ₹ 22,00,000
Or
(b) She can pay 4,00,000 cash and balance in 18 annual instalments of ₹ 1,00,000 plus 10% interest on the unpaid amount.
She prefers option (i) and donates 50% of the difference of the costs in the above two options to the Prime Minister Relief Fund.
(i) What amount was donated to Prime Minister Relief Fund?
(ii) Which mathematical concept is used in the above problem?
(iii) By choosing to pay a lumpsum amount and donating 50% of the difference to the Prime Minister Relief Fund, which value is depicted by Savita?
Sol. (a) Total cost of the house = ₹ 22,00,000
(b) Cash payment = ₹ 4,00,000
Balance = ₹ 22,00,000 – ₹ 4,00,000 = ₹ 18,00,000
1st instalment = ₹ [1,00,000 + 10% of balance]
= ₹ [1,00,000 + 1,80,000] = ₹ 2,80,000
Balance after 1st instalment = ₹ [18,00,000 – 1,00,000] = ₹ 17,00,000
2nd instalment = ₹ [1,00,000 + 10% of 17,00,000]
= ₹ [1,00,000 + 1,70,000] = ₹ [2,70,000]
Balance after 2nd instalment = ₹ 17,00,000 – ₹ 1,00,000 = ₹ 16,00,000
∵ 3rd instalment = ₹ [1,00,000 + 10% of 16,00,000]
= ₹ [1,00,000 + 1,60,000] = ₹ 2,60,000
... and so on.
∵ Total amount in instalments = ₹ 2,80,000 + ₹ 2,70,000 + ₹ 2,60,000 + ..... to 18 terms
where a = 2,80,000, d = – 10,000, n = 18
= ₹ 9 [560,000+17(-10,000)
= ₹ 9 [560,000 - 170,000]
= ₹ 9 = [390,000] = ₹ 35,10,000
∴ Total cost of house = ₹ 35,10,000 + 4,00,000 = ₹ 39,10,000
Difference in costs of the house in two options
= ₹ 39,10,000 – ₹ 22,00,000 = ₹ 17,10,000
∴ (i) Amount donated towards Prime Minister Relief Fund = 50% of ₹ 17,10,000
(ii) Arithmetic Progressions
(iii) National Loyalty
Q2. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students. Which value is shown in this question?
Sol. ∴ There are 12 classes in all.
Each class has 2 sections.
∴ Number of plants planted by class I = 1 x 2 = 2
Number of plants planted by class II = 2 x 2 = 4
Number of plants planted by class III = 3 x 2 = 6
Number of plants planted by class IV = 4 x 2 = 8
......................................................................................................
......................................................................................................
Number of plants planted by class XII = 12 x 2 = 24
The numbers 2, 4, 6, 8, ........................ 24 forms an A.P.
Here, a = 2, d = 4 – 2 = 2
∵ Number of classes = 12
∴ Number of terms (n) = 12
Now, the sum of n terms of the above A.P., is given by Sn = n/2 [2a+(n-1)d]
∴ S12 = 12/2 [2(2) -(12)-1) 2]
= 6 [4 + (11 x 2)]
= 6 x 26 = 156
Thus, the total number of trees planted = 156
Value shown: To enrich polution free environment.
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## College Algebra (10th Edition)
The solution set is $\left\{-\sqrt2, 0, \sqrt2\right\}$.
To solve the given equation, make the two sides have the same base. Note that $9 = 3^2$ so the given equation is equivalent to: $3^{x^3} = (3^2)^x$ Use the rule $(a^m)^n = a^{mn}$ to obtain: $3^{x^3} = 3^{2x}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $x^3 = 2x$ Subtract $2x$ to both sides of the equation to obtain: $x^3-2x=0$ Factor out the GCF $x$ to obtain: $x(x^2-2)=0$ Equate each factor to $0$, then solve each equation to obtain: $\begin{array}{ccc} &x=0 &\text{ or } &x^2-2=0 \\&x=0 &\text{ or } &x^2=2\end{array}$ Take the square root of both sides of the second equation obtain: $\begin{array}{ccc} &x=0 &\text{ or } &x=\pm \sqrt{2}\end{array}$ Thus, the solution set is $\left\{-\sqrt2, 0, \sqrt2\right\}$.
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What Are The Prime Factors Of 22
What are the prime factors of 22? Answer: 2, 11
The number 22 has 2 prime factors. Primes can only have two factors(1 and itself) and only be divisible by those two factors. Any figure where this rule applies can be called a prime factor. The biggest prime factor of 22 is 11. The smallest prime factor of 22 is 2.
What Is The Factor Tree Of 22
How to use a factor tree to find the prime factors of 22? A factor tree is a diagram that organizes the factoring process.
• First step is to find two numbers that when multiplied together equal the number we start with.
22
↙ ↘
2 × 11
We found 2 prime factors(2, 11) using the factor tree of 22. Now let us explain the process to solving factor trees in more detail. Our goal is to find all prime factors of a given whole number. In each step of our factor tree diagram for 22 we always checked both multiplication numbers if they were primes or not. If one or both of the integers are not prime numbers then this means that we will have to make diagrams for them too. This process continues until only prime numbers are left.
Remember that often a factor tree for the same integer can be solved in more than one correct way! An example of this is the figure 12 where 2*6=12 and 4*3=12. The primes of a factor tree for 12 are the same regardles if we start the factor tree with 2*6 or 4*3.
How To Verify If Prime Factors Of 22 Are Correct Answers
To know if we got the correct prime factors of 22 we have to get the prime factorization of 22 which is 2 * 11. Because when you multiply the primes of the prime factorization the answer has to be equal with 22.
After having checked the prime factorization we can now safely say that we got all prime factors.
General Mathematical Properties Of Number 22
22 is a composite number. 22 is a composite number, because it has more divisors than 1 and itself. This is an even number. 22 is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 22 is not an odd number. When we simplify Sin 22 degrees we get the value of sin(22)=-0.0088513092904039. Simplify Cos 22 degrees. The value of cos(22)=-0.99996082639464. Simplify Tan 22 degrees. Value of tan(22)=0.0088516560416845. When converting 22 in binary you get 10110. Converting decimal 22 in hexadecimal is 16. The square root of 22=4.6904157598234. The cube root of 22=2.8020393306554. Square root of √22 simplified is 22. All radicals are now simplified and in their simplest form. Cube root of ∛22 simplified is 22. The simplified radicand no longer has any more cubed factors.
Determine Prime Factors Of Numbers Smaller Than 22
Learn how to calculate primes of smaller numbers like:
Determine Prime Factors Of Numbers Bigger Than 22
Learn how to calculate primes of bigger numbers such as:
Single Digit Properties For Number 22 Explained
• Integer 2 properties: 2 is the first of the primes and the only one to be even(the others are all odd). The first issue of Smarandache-Wellin in any base. Goldbach's conjecture states that all even numbers greater than 2 are the quantity of 2 primes. It is a complete Harshad, which is a integer of Harshad in any expressed base. The third of the Fibonacci sequence, after 1 and before 3. Part of the Tetranacci Succession. Two is an oblong figure of the form n(n+1). 2 is the basis of the binary numbering system, used internally by almost all computers. Two is a number of: Perrin, Ulam, Catalan and Wedderburn-Etherington. Refactorizable, which means that it is divisible by the count of its divisors. Not being the total of the divisors proper to any other arithmetical value, 2 is an untouchable quantity. The first number of highly cototent and scarcely totiente (the only one to be both) and it is also a very large decimal. Second term of the succession of Mian-Chowla. A strictly non-palindrome. With one exception, all known solutions to the Znam problem begin with 2. Numbers are divisible by two (ie equal) if and only if its last digit is even. The first even numeral after zero and the first issue of the succession of Lucas. The aggregate of any natural value and its reciprocal is always greater than or equal to 2.
Finding All Prime Factors Of A Number
We found that 22 has 2 primes. The prime factors of 22 are 2, 11. We arrived to this answer by using the factor tree. However we could have also used upside down division to get the factorization primes. There are more that one method to factorize a integer.
List of divisibility rules for finding prime factors faster
Rule 1: If the last digit of a figure is 0, 2, 4, 6 or 8 then it is an even number. All even numbers are divisible by 2.
Rule 2: If the sum of digits of a integer is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 12 are 1 and 2 so 1+2=3 and 3 is divisible by 3, meaning that 12 is divisible by 3). The same logic also works for 9.
Rule 3: If the last two digits of a number are 00 then this integer is divisible by 4(example: we know that 124=100+24 and 100 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 24). In order to use this rule to it's fullest it is best to know multiples of 4.
Rule 4: If the last digit of a number is 0 or 5 then 5 it is divisible by 5.
Rule 5: All integers that are divisible by both 2 and 3 are also divisible by 6. This is logical because 2*3=6.
What Are Prime Factors Of A Number?
All numbers that are only divisible by one and itself are called prime factors in mathematics. A prime factor is a figure that has only two factors(one and itself).
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## Elementary Geometry for College Students (6th Edition)
The three real solutions are: $x = 0,~~$ $~~x=1,~~$ $~~x=6$
Let's consider a trinomial in this form: $~x^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$ Then the next step is to rewrite the trinomial as follows: $~x^2+bx+c = (x+r)~(x+s)$ We can use the GCF to factor the given equation: $4x^3-28x^2+24x = 0$ $(4x)~(x^2-7x+6) = 0$ To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = -7~$ and $~r\times s = 6$. We can see that $(-1)+(-6) = -7~$ and $(-1)\times (-6) = 6$ We can solve the equation as follows: $4x^3-28x^2+24x = 0$ $(4x)~(x^2-7x+6) = 0$ $(4x)~(x-1)~(x-6) = 0$ $4x = 0,~~$ $~~x-1=0,~~$ or $~~x-6=0$ $x = 0,~~$ $~~x=1,~~$ or $~~x=6$
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Arkansas Tech University
Dr. Marcel B. Finan
3 Geometrical Use of The Rate of Change
Functions given by tables of values have their limitations in that nearly always
leave gaps. One way to ll these gaps is by using the average rate of
change. For example, Table 1 below gives the population of the United
States between the years 1950 - 1990.
d(year) 1950 1960 1970 1980 1990
N(in millions) 151.87 179.98 203.98 227.23 249.40
Table 1
This table does not give the population in 1972. One way to estimate
N(1972), is to nd the average yearly rate of change of N from 1970 to
1980 given by
227.23 203.98
10
= 2.325 million people per year.
Then,
N(1972) = N(1970) + 2(2.325) = 208.63 million.
Assuming of course that N(t) is a linear function between the years 1970 and
1980.
Average rates of change can be calculated not only for functions given by ta-
bles but also for functions given by formulas. The average rate of change
of a function y = f(x) from x = a to x = b is given by the dierence
quotient
y
x
=
Change in function value
Change in x value
=
f(b) f(a)
b a
.
Geometrically, this quantity represents the slope of the secant line going
through the points (a, f(a)) and (b, f(b)) on the graph of f(x). See Figure 5.
The average rate of change of a function on an interval tells us how much
the function changes, on average, per unit change of x within that interval.
1
On some part of the interval, f may be changing rapidly, while on other
parts f may be changing slowly. The average rate of change evens out these
variations.
Figure 5
Example 3.1
Find the average value of the function f(x) = x
2
from x = 3 to x = 5.
Solution.
The average rate of change is
y
x
=
f(5) f(3)
5 3
=
25 9
2
= 8.
Example 3.2 (Average Speed)
During a typical trip to school, your car will undergo a series of changes in its
speed. If you were to inspect the speedometer readings at regular intervals,
you would notice that it changes often. The speedometer of a car reveals
information about the instantaneous speed of your car; that is, it shows your
speed at a particular instant in time. The instantaneous speed of an object
is not to be confused with the average speed. Average speed is a measure of
the distance traveled in a given period of time. That is,
Average Speed =
Distance traveled
Time elapsed
.
If the trip to school takes 0.2 hours (i.e. 12 minutes) and the distance traveled
is 5 miles then what is the average speed of your car?
2
Solution.
The average velocity is given by
Ave. Speed =
5 miles
0.2 hours
= 25 miles/hour.
This says that on the average, your car was moving with a speed of 25 miles
per hour. During your trip, there may have been times that you were stopped
and other times that your speedometer was reading 50 miles per hour; yet
on the average you were moving with a speed of 25 miles per hour.
Average Rate of Change and Increasing/Decreasing Functions
Now, we would like to use the concept of the average rate of change to test
whether a function is increasing or decreasing on a specic interval. First,
we introduce the following denition: We say that a function is increasing
if its graph climbs as x moves from left to right. That is, the function values
increase as x increases. It is said to be decreasing if its graph falls as x
moves from left to right. This means that the function values decrease as x
increases.
As an application of the average rate of change, we can use such quantity
to decide whether a function is increasing or decreasing. If a function f is
increasing on an interval I then by taking any two points in the interval I,
say a < b, we see that f(a) < f(b) and in this case
f(b) f(a)
b a
> 0.
Going backward with this argument we see that if the average rate of change
is positive in an interval then the function is increasing in that interval.
Similarly, if the average rate of change is negative in an interval I then the
function is decreasing there.
Some functions can be increasing on some intervals and decreasing on other
intervals. These intervals can often be identied from the graph.
Example 3.3
Determine the intervals where the function is increasing and decreasing.
3
Figure 6
Solution.
The function is increasing for x < 1 or x > 1 and decreasing for 1 < x <
1
Rate of Change and Concavity
We have seen that when the rate of change of a function is constant then its
graph is a straight line. However, not all graphs are straight lines; they may
bend up or down as shown in the following two examples.
Example 3.4
Consider the following two graphs in Figure 7.
Figure 7
(a) What do the graphs above have in common?
(b) How are they dierent? Specically, look at the rate of change of each.
4
Solution.
(a) Both graphs represent increasing functions.
(b) The rate of change of f(x) is more and more positive so the graph bends
up whereas the rate of change of g(x) is less and less positive and so it bends
down.
The following example deals with version of the previous example for de-
creasing functions.
Example 3.5
Consider the following two graphs given in Figure 8.
Figure 8
(a) What do the graphs above have in common?
(b) How are they dierent? Specically, look at the rate of change of each.
Solution.
(a) Both functions are decreasing.
(b) The rate of change of f(x) is more and more negative so the graph bends
down, whereas the rate of change of g(x) is less and less negative so the graph
bends up
Conclusions:
When the rate of change of a function is increasing then the function is
concave up. That is, the graph bends upward.
When the rate of change of a function is decreasing then the function is
concave down. That is, the graph bends downward.
The following example discusses the concavity of a function given by a table.
5
Example 3.6
Given below is the table for the function H(x). Calculate the rate of change
for successive pairs of points. Decide whether you expect the graph of H(x)
to concave up or concave down?
x 12 15 18 21
H(x) 21.40 21.53 21.75 22.02
Solution.
H(15)H(12)
1512
=
21.5321.40
3
0.043
H(18)H(15)
1815
=
21.7521.53
3
0.073
H(21)H(18)
2118
=
22.0221.75
3
0.09
Since the rate of change of H(x) is increasing, the function is concave up.
Remark 3.1
Since the graph of a linear function is a straight line, that is its rate of change
is constant, then it is neither concave up nor concave down.
6
|
# Parallelogram Area Calculator
Calculate the area of a parallelogram using the base and height, sides and angle, or diagonals and angle with the calculator below.
## Area:
Learn how we calculated this below
## How to Calculate the Area of a Parallelogram
A parallelogram is a four-sided figure with opposite sides that are parallel. Whether you’re doing homework, landscaping, or engaging in any project that requires an understanding of geometrical calculations, knowing how to calculate the area of a parallelogram is invaluable.
There are several ways to calculate the area of a parallelogram, depending on the information you have.
### How to Calculate Area Using Base and Height
The most common method to find the area of a parallelogram is by using its base and height.
The parallelogram area formula is:
A = b × h
The area A of a parallelogram is equal to the length of the base b times the height h.
For example, let’s calculate the area of a parallelogram with a base of 5 and a height of 6.
A = 5 × 6 = 30
This parallelogram has an area of 30.
### How to Calculate Area Using the Sides and Interior Angle
When you know the length of the sides, base, and the angle between them, you can calculate the area of a parallelogram using trigonometry.
The formula to calculate the area of a parallelogram using the sides, base, and angle is:
A = a × b × sin(α)
The area A of a parallelogram is equal to the length of the side a times the length of the base b times the sine of angle α.
For example, let’s calculate the area of a parallelogram with a side of 7, a base of 8, and an interior angle of 60 degrees.
A = 7 × 8 × sin(60°) = 48.5
So, this parallelogram has an area of 48.5.
### How to Calculate Area Using the Diagonals and Angle Between
If you know the lengths of the two diagonals that bisect the parallelogram and the angle between them, then you can find the area of the parallelogram using a formula:
A = ½ × c × d × sin(θ)
The area A of a parallelogram is equal to the length of the diagonal c times the length of the diagonal d times the sine of angle θ.
For example, let’s calculate the area of a parallelogram with diagonals of 9 and 12 and an interior angle of 30 degrees.
A = ½ × 9 × 12 × sin(30°) = 27
So, this parallelogram has an area of 27.
You may also find our trapezoid area calculator useful.
|
• Written By Khushbu
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry: The solutions for Class 6 Maths are curated by highly experienced Subject Matter Experts (SMEs) at Embibe. Students can easily score good marks by practicing all the solutions. In addition, we have provided chapter-wise solutions for Maths which help students to prepare for their upcoming exams and score good marks.
Class 6 Maths chapter 14 talks about Circle, Line Segments, Perpendiculars, Angles and many more. Embibe provides a set of 470+ practice questions for all the sub-topics of chapter 14. Students should focus on practicing all the in-text questions given by Embibe to excel in the exams. In this article, students can access NCERT solutions for Class 6 chapter 14 Maths for.
NCERT Solutions for Class 6 Maths Chapter 14: Important Topics
Practical Geometry is one of the most interesting chapters of Class 6 Maths. Students learn how to draw shapes using geometrical tools in this chapter. Students will also build the fundamentals required to score good marks in higher grades. Practicing sample questions on all the topics will help students to understand the nature of questions and the right way to write the solutions. Students should also take mock tests on the chapter to analyse their performance.
Before getting into the detailed Chapter 14 Class 6 Maths solutions, let’s have a look at the different topics in this chapter:
NCERT Solutions for Class 6 Maths Chapter 14: Points to Remember
Some of the important points to be remembered from Class 6 Maths Chapter 14 have been mentioned below:
• A line segment has two endpoints, and a line has no endpoints.
• There is a unique line segment joining two given points A and B.
• The line segment AB is the same as the line segment BA
• The length of a line segment (AB) is denoted by AB.
• Two or more line segments having the same length are said to be congruent or equal.
• The distance between two points is the same as the length of the line segment joining these points.
• The part of a line that extends indefinitely in one direction from a given point O is called a ray. The point O is called the initial point or the endpoint of the ray.
• A ray has only one end point.
• An unlimited number of rays can be drawn with the same initial point.
• A unique ray can be drawn from a given initial point and passing through a given point.
• A ray has no definite length.
NCERT Solutions for Class 6 Maths: All Chapters
To score full marks in Class 6 Maths, students must prepare all the chapters sincerely. They can take mock tests and solve practice questions to cover the syllabus thoroughly. Download NCERT Solutions for Class 6 Maths for other chapters:
FAQs on NCERT Solutions for Class 6 Maths Chapter 14
Here are some frequently asked questions on NCERT Solutions for Class 6 Maths chapter 14:
Q: Where can I solve NCERT Class 6 Maths Chapter 14 questions?
Ans: You can solve NCERT Class 6 Maths Chapter 14 sample questions on Embibe for.
Q: What are the concepts discussed in Class 6 Maths Chapter 14?
Ans: The concepts covered in Class 6 Maths Chapter 14 Practical Geometry are The Circle, Line Segment, Perpendiculars, and Angles.
Q: Can I take mock tests for NCERT Class 6 Maths Chapter 14?
Ans: Yes. Students can attempt unlimited mock tests on NCERT Class 6 Maths chapter 14 on the Embibe app.
Q: How my chapters are in Class 6 Maths?
Ans: There are fourteen chapters in Class 6 Maths. At Embibe, we have provided all the solutions for NCERT Class 6 Maths.
Q: How to prepare for NCERT Solutions For Class 6 Maths?
Ans: We advise students to refer to Embibe for the solutions. Embibe step-by-step solutions help students to prepare for exams effectively.
Unleash Your True Potential With Personalised Learning on EMBIBE
|
Ex 9.2
Chapter 9 Class 11 Sequences and Series
Serial order wise
This video is only available for Teachoo black users
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Ex9.2 , 3 In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is 112. It is given that First term = a = 2 Also Sum of first five terms = 1/4 (Sum of next 5 terms) Sum of first five terms = 1/4 (Sum of 6th to 10th terms) Sum of first five terms = 1/4 ( 8( ("Sum of first 10 terms " @" Sum of first five terms" ))) S5 = 1/4(S10 S5) 4S5 = S10 S5 4S5 + S5 = S10 5S5 = S10 Finding sum of first five terms We know that Sum of n terms of A.P. = /2(2a + (n 1)d) Sn = /2(2a + (n 1)d) Putting a = 2, n = 5 S5 = 5/2 (2(2) + (5 1)d) = 5/2 (4 + 4d) = 5/2 (4) + 5/2 (4)d = 10 + 10d Finding sum of first ten terms Sn = /2(2a + (n 1)d) Putting a = 2, n = 10 S10 = 10/2 (2(2) + (10 1)d) = 10/2 (4 + 9d) = 5(4 + 9d) = 20 + 45d From equation (1) 5S5 = S10 Putting values 5(10 + 10d) = 20 + 45d 50 + 50d = 20 + 45d 50d 45d = 20 50 5d = 30 d = ( 30)/5 = 6 To find 20th term, we use the formula an = a + (n 1)d where an = nth term , n = number of terms, a = first term , d = common difference Here, a = 2 , d = 6 , n = 20 Putting values a20 = 2 + (20 1) ( 6) = 2 + (19)(-6) = 2 114 = 112 Thus, 20th term of sequence is 112 Hence proved.
|
## 7.1 Composition and Inverse Functions
### Learning Objectives
1. Perform function composition.
2. Determine whether or not given functions are inverses.
3. Use the horizontal line test.
4. Find the inverse of a one-to-one function algebraically.
## Composition of Functions
In mathematics, it is often the case that the result of one function is evaluated by applying a second function. For example, consider the functions defined by $f(x)=x2$ and $g(x)=2x+5.$ First, g is evaluated where $x=−1$ and then the result is squared using the second function, f.
This sequential calculation results in 9. We can streamline this process by creating a new function defined by $f(g(x))$, which is explicitly obtained by substituting $g(x)$ into $f(x).$
$f(g(x))=f(2x+5)=(2x+5)2=4x2+20x+25$
Therefore, $f(g(x))=4x2+20x+25$ and we can verify that when $x=−1$ the result is 9.
$f(g(−1))=4(−1)2+20(−1)+25=4−20+25=9$
The calculation above describes composition of functionsApplying a function to the results of another function., which is indicated using the composition operatorThe open dot used to indicate the function composition $(f○g)(x)=f(g(x)).$ ($○$). If given functions f and g,
$(f○g)(x)=f(g(x)) Composition of Functions$
The notation $f○g$ is read, “f composed with g.” This operation is only defined for values, x, in the domain of g such that $g(x)$ is in the domain of f.
### Example 1
Given $f(x)=x2−x+3$ and $g(x)=2x−1$ calculate:
1. $(f○g)(x).$
2. $(g○f)(x).$
Solution:
1. Substitute g into f.
$(f○g)(x)=f(g(x))=f(2x−1)=(2x−1)2−(2x−1)+3=4x2−4x+1−2x+1+3=4x2−6x+5$
2. Substitute f into g.
$(g○f)(x)=g(f(x))=g(x2−x+3)=2(x2−x+3)−1=2x2−2x+6−1=2x2−2x+5$
1. $(f○g)(x)=4x2−6x+5$
2. $(g○f)(x)=2x2−2x+5$
The previous example shows that composition of functions is not necessarily commutative.
### Example 2
Given $f(x)=x3+1$ and $g(x)=3x−13$ find $(f○g)(4).$
Solution:
Begin by finding $(f○g)(x).$
$(f○g)(x)=f(g(x))=f(3x−13)=(3x−13)3+1=3x−1+1=3x$
Next, substitute 4 in for x.
$(f○g)(x)=3x(f○g)(4)=3(4)=12$
Answer: $(f○g)(4)=12$
Functions can be composed with themselves.
### Example 3
Given $f(x)=x2−2$ find $(f○f)(x).$
Solution:
$(f○f)(x)=f(f(x))=f(x2−2)=(x2−2)2−2=x4−4x2+4−2=x4−4x2+2$
Answer: $(f○f)(x)=x4−4x2+2$
Try this! Given $f(x)=2x+3$ and $g(x)=x−1$ find $(f○g)(5).$
## Inverse Functions
Consider the function that converts degrees Fahrenheit to degrees Celsius: $C(x)=59(x−32).$ We can use this function to convert 77°F to degrees Celsius as follows.
$C(77)=59(77−32)=59(45)=25$
Therefore, 77°F is equivalent to 25°C. If we wish to convert 25°C back to degrees Fahrenheit we would use the formula: $F(x)=95x+32.$
$F(25)=95(25)+32=45+32=77$
Notice that the two functions $C$ and $F$ each reverse the effect of the other.
This describes an inverse relationship. In general, f and g are inverse functions if,
$(f○g)(x)=f(g(x))=x for all x in the domain of g and(g○f)(x)=g(f(x))=x for all x in the domain of f.$
In this example,
$C(F(25))=C(77)=25F(C(77))=F(25)=77$
### Example 4
Verify algebraically that the functions defined by $f(x)=12x−5$ and $g(x)=2x+10$ are inverses.
Solution:
Compose the functions both ways and verify that the result is x.
$(f○g)(x)=f(g(x))=f(2x+10)=12(2x+10)−5=x+5−5=x ✓$ $(g○f)(x)=g(f(x))=g(12x−5)=2(12x−5)+10=x−10+10=x ✓$
Answer: Both $(f○g)(x)=(g○f)(x)=x$; therefore, they are inverses.
Next we explore the geometry associated with inverse functions. The graphs of both functions in the previous example are provided on the same set of axes below.
Note that there is symmetry about the line $y=x$; the graphs of f and g are mirror images about this line. Also notice that the point (20, 5) is on the graph of f and that (5, 20) is on the graph of g. Both of these observations are true in general and we have the following properties of inverse functions:
1. The graphs of inverse functions are symmetric about the line $y=x.$
2. If $(a,b)$ is on the graph of a function, then $(b,a)$ is on the graph of its inverse.
Furthermore, if g is the inverse of f we use the notation $g=f−1.$ Here $f−1$ is read, “f inverse,” and should not be confused with negative exponents. In other words, $f−1(x)≠1f(x)$ and we have,
$(f○f−1)(x)=f(f−1(x))=x and(f−1○f)(x)=f−1(f(x))=x$
### Example 5
Verify algebraically that the functions defined by $f(x)=1x−2$ and $f−1(x)=1x+2$ are inverses.
Solution:
Compose the functions both ways to verify that the result is x.
$(f○f−1)(x)=f(f−1(x))=f(1x+2)=1(1x+2)−2=x+21−2=x+2−2=x ✓$ $(f−1○f)(x)=f−1(f(x))=f−1(1x−2)=1(1x−2)+2=1 1x =x ✓$
Answer: Since $(f○f−1)(x)=(f−1○f)(x)=x$ they are inverses.
Recall that a function is a relation where each element in the domain corresponds to exactly one element in the range. We use the vertical line test to determine if a graph represents a function or not. Functions can be further classified using an inverse relationship. One-to-one functionsFunctions where each value in the range corresponds to exactly one value in the domain. are functions where each value in the range corresponds to exactly one element in the domain. The horizontal line testIf a horizontal line intersects the graph of a function more than once, then it is not one-to-one. is used to determine whether or not a graph represents a one-to-one function. If a horizontal line intersects a graph more than once, then it does not represent a one-to-one function.
The horizontal line represents a value in the range and the number of intersections with the graph represents the number of values it corresponds to in the domain. The function defined by $f(x)=x3$ is one-to-one and the function defined by $f(x)=|x|$ is not. Determining whether or not a function is one-to-one is important because a function has an inverse if and only if it is one-to-one. In other words, a function has an inverse if it passes the horizontal line test.
Note: In this text, when we say “a function has an inverse,” we mean that there is another function, $f−1$, such that $(f○f−1)(x)=(f−1○f)(x)=x.$
### Example 6
Determine whether or not the given function is one-to-one.
Solution:
Answer: The given function passes the horizontal line test and thus is one-to-one.
In fact, any linear function of the form $f(x)=mx+b$ where $m≠0$, is one-to-one and thus has an inverse. The steps for finding the inverse of a one-to-one function are outlined in the following example.
### Example 7
Find the inverse of the function defined by $f(x)=32x−5.$
Solution:
Before beginning this process, you should verify that the function is one-to-one. In this case, we have a linear function where $m≠0$ and thus it is one-to-one.
• Step 1: Replace the function notation $f(x)$ with y.
$f(x)=32x−5y=32x−5$
• Step 2: Interchange x and y. We use the fact that if $(x,y)$ is a point on the graph of a function, then $(y,x)$ is a point on the graph of its inverse.
$x=32y−5$
• Step 3: Solve for y.
$x=32y−5x+5=32y23⋅(x+5)=23⋅32y23x+103=y$
• Step 4: The resulting function is the inverse of f. Replace y with $f−1(x).$
$f−1(x)=23x+103$
• Step 5: Check.
$(f○f−1)(x)=f(f−1(x))=f(23x+103)=32(23x+103)−5=x+5−5=x ✓$ $(f−1○f)(x)=f−1(f(x))=f−1(32x−5)=23(32x−5)+103=x−103+103=x ✓$
Answer: $f−1(x)=23x+103$
If a function is not one-to-one, it is often the case that we can restrict the domain in such a way that the resulting graph is one-to-one. For example, consider the squaring function shifted up one unit, $g(x)=x2+1.$ Note that it does not pass the horizontal line test and thus is not one-to-one. However, if we restrict the domain to nonnegative values, $x≥0$, then the graph does pass the horizontal line test.
On the restricted domain, g is one-to-one and we can find its inverse.
### Example 8
Find the inverse of the function defined by $g(x)=x2+1$ where $x≥0.$
Solution:
Begin by replacing the function notation $g(x)$ with y.
$g(x)=x2+1y=x2+1 where x≥0$
Interchange x and y.
$x=y2+1 where y≥0$
Solve for y.
$x=y2+1x−1=y2±x−1=y$
Since $y≥0$ we only consider the positive result.
$y=x−1g−1(x)=x−1$
Answer: $g−1(x)=x−1.$ The check is left to the reader.
The graphs in the previous example are shown on the same set of axes below. Take note of the symmetry about the line $y=x.$
### Example 9
Find the inverse of the function defined by $f(x)=2x+1x−3.$
Solution:
Use a graphing utility to verify that this function is one-to-one. Begin by replacing the function notation $f(x)$ with y.
$f(x)=2x+1x−3y=2x+1x−3$
Interchange x and y.
$x=2y+1y−3$
Solve for y.
$x=2y+1y−3x(y−3)=2y+1xy−3x=2y+1$
Obtain all terms with the variable y on one side of the equation and everything else on the other. This will enable us to treat y as a GCF.
$xy−3x=2y+1xy−2y=3x+1y(x−2)=3x+1y=3x+1x−2$
Answer: $f−1(x)=3x+1x−2.$ The check is left to the reader.
Try this! Find the inverse of $f(x)=x+13−3.$
Answer: $f−1(x)=(x+3)3−1$
### Key Takeaways
• The composition operator ($○$) indicates that we should substitute one function into another. In other words, $(f○g)(x)=f(g(x))$ indicates that we substitute $g(x)$ into $f(x).$
• If two functions are inverses, then each will reverse the effect of the other. Using notation, $(f○g)(x)=f(g(x))=x$ and $(g○f)(x)=g(f(x))=x.$
• Inverse functions have special notation. If $g$ is the inverse of $f$, then we can write $g(x)=f−1(x).$ This notation is often confused with negative exponents and does not equal one divided by $f(x).$
• The graphs of inverses are symmetric about the line $y=x.$ If $(a,b)$ is a point on the graph of a function, then $(b,a)$ is a point on the graph of its inverse.
• If each point in the range of a function corresponds to exactly one value in the domain then the function is one-to-one. Use the horizontal line test to determine whether or not a function is one-to-one.
• A one-to-one function has an inverse, which can often be found by interchanging x and y, and solving for y. This new function is the inverse of the original function.
### Part A: Composition of Functions
Given the functions defined by f and g find $(f○g)(x)$ and $(g○f)(x).$
1. $f(x)=4x−1$, $g(x)=3x$
2. $f(x)=−2x+5$, $g(x)=2x$
3. $f(x)=3x−5$, $g(x)=x−4$
4. $f(x)=5x+1$, $g(x)=2x−3$
5. $f(x)=x2−x+1$, $g(x)=2x−1$
6. $f(x)=x2−3x−2$, $g(x)=x−2$
7. $f(x)=x2+3$, $g(x)=x2−5$
8. $f(x)=2x2$, $g(x)=x2−x$
9. $f(x)=8x3+5$, $g(x)=x−53$
10. $f(x)=27x3−1$, $g(x)=x+13$
11. $f(x)=1x+5$, $g(x)=1x$
12. $f(x)=1x−3$, $g(x)=3x+3$
13. $f(x)=5x$, $g(x)=3x−2$
14. $f(x)=2x$, $g(x)=4x+1$
15. $f(x)=12x$, $g(x)=x2+8$
16. $f(x)=2x−1$, $g(x)=1x+1$
17. $f(x)=1−x2x$, $g(x)=12x+1$
18. $f(x)=2xx+1$, $g(x)=x+1x$
Given the functions defined by $f(x)=3x2−2$, $g(x)=5x+1$, and $h(x)=x$, calculate the following.
1. $(f○g)(2)$
2. $(g○f)(−1)$
3. $(g○f)(0)$
4. $(f○g)(0)$
5. $(f○h)(3)$
6. $(g○h)(16)$
7. $(h○g)(35)$
8. $(h○f)(−3)$
Given the functions defined by $f(x)=x+33$, $g(x)=8x3−3$, and $h(x)=2x−1$, calculate the following.
1. $(f○g)(1)$
2. $(g○f)(−2)$
3. $(g○f)(0)$
4. $(f○g)(−2)$
5. $(f○h)(−1)$
6. $(h○g)(−12)$
7. $(h○f)(24)$
8. $(g○h)(0)$
Given the function, determine $(f○f)(x).$
1. $f(x)=3x−1$
2. $f(x)=25x+1$
3. $f(x)=x2+5$
4. $f(x)=x2−x+6$
5. $f(x)=x3+2$
6. $f(x)=x3−x$
7. $f(x)=1x+1$
8. $f(x)=x+12x$
### Part B: Inverse Functions
Are the given functions one-to-one? Explain.
1. $f(x)=x+1$
2. $g(x)=x2+1$
3. $h(x)=|x|+1$
4. $r(x)=x3+1$
5. $f(x)=x+1$
6. $g(x)=3$
Given the graph of a one-to-one function, graph its inverse.
Verify algebraically that the two given functions are inverses. In other words, show that $(f○f−1)(x)=x$ and $(f−1○f)(x)=x.$
1. $f(x)=3x−4$, $f−1(x)=x+43$
2. $f(x)=−5x+1$, $f−1(x)=1−x5$
3. $f(x)=−23x+1$, $f−1(x)=−32x+32$
4. $f(x)=4x−13$, $f−1(x)=14x+112$
5. $f(x)=x−8$, $f−1(x)=x2+8$, $x≥0$
6. $f(x)=6x3−3$, $f−1(x)=(x+3)36$
7. $f(x)=xx+1$, $f−1(x)=x1−x$
8. $f(x)=x−33x$, $f−1(x)=31−3x$
9. $f(x)=2(x−1)3+3$, $f−1(x)=1+x−323$
10. $f(x)=5x−13+4$, $f−1(x)=(x−4)3+15$
Find the inverses of the following functions.
1. $f(x)=5x$
2. $f(x)=12x$
3. $f(x)=2x+5$
4. $f(x)=−4x+3$
5. $f(x)=−23x+13$
6. $f(x)=−12x+34$
7. $g(x)=x2+5$, $x≥0$
8. $g(x)=x2−7$, $x≥0$
9. $f(x)=(x−5)2$, $x≥5$
10. $f(x)=(x+1)2$, $x≥−1$
11. $h(x)=3x3+5$
12. $h(x)=2x3−1$
13. $f(x)=(2x−3)3$
14. $f(x)=(x+4)3−1$
15. $g(x)=2x3+1$
16. $g(x)=1x3−2$
17. $f(x)=5x+1$
18. $f(x)=12x−9$
19. $f(x)=x+5x−5$
20. $f(x)=3x−42x−1$
21. $h(x)=x−510x$
22. $h(x)=9x+13x$
23. $g(x)=5x+23$
24. $g(x)=4x−33$
25. $f(x)=x−63−4$
26. $f(x)=2x+23+5$
27. $h(x)=x+15−3$
28. $h(x)=x−85+1$
29. $f(x)=mx+b$, $m≠0$
30. $f(x)=ax2+c$, $x≥0$
31. $f(x)=ax3+d$
32. $f(x)=a(x−h)2+k$, $x≥h$
Graph the function and its inverse on the same set of axes.
1. $f(x)=x+2$
2. $f(x)=23x−4$
3. $f(x)=−2x+2$
4. $f(x)=−13x+4$
5. $g(x)=x2−2$, $x≥0$
6. $g(x)=(x−2)2$, $x≥2$
7. $h(x)=x3+1$
8. $h(x)=(x+2)3−2$
9. $f(x)=2−x$
10. $f(x)=−x+1$
### Part C: Discussion Board
1. Is composition of functions associative? Explain.
2. Explain why $C(x)=59(x−32)$ and $F(x)=95x+32$ define inverse functions. Prove it algebraically.
3. Do the graphs of all straight lines represent one-to-one functions? Explain.
4. If the graphs of inverse functions intersect, then how can we find the point of intersection? Explain.
1. $(f○g)(x)=12x−1$; $(g○f)(x)=12x−3$
2. $(f○g)(x)=3x−17$; $(g○f)(x)=3x−9$
3. $(f○g)(x)=4x2−6x+3$; $(g○f)(x)=2x2−2x+1$
4. $(f○g)(x)=x4−10x2+28$; $(g○f)(x)=x4+6x2+4$
5. $(f○g)(x)=8x−35$; $(g○f)(x)=2x$
6. $(f○g)(x)=x5x+1$; $(g○f)(x)=x+5$
7. $(f○g)(x)=53x−2$; $(g○f)(x)=15x−2$
8. $(f○g)(x)=12x2+16;$ $(g○f)(x)=1+32x24x2$
9. $(f○g)(x)=x$; $(g○f)(x)=x$
10. 361
11. −9
12. 7
13. 2
14. 2
15. 21
16. 0
17. 5
18. $(f○f)(x)=9x−4$
19. $(f○f)(x)=x4+10x2+30$
20. $(f○f)(x)=x9+6x6+12x3+10$
21. $(f○f)(x)=x+1x+2$
1. No, fails the HLT
2. Yes, passes the HLT
3. Yes, its graph passes the HLT.
4. No, its graph fails the HLT.
5. Yes, its graph passes the HLT.
6. Proof
7. Proof
8. Proof
9. Proof
10. Proof
11. $f−1(x)=x5$
12. $f−1(x)=12x−52$
13. $f−1(x)=−32x+12$
14. $g−1(x)=x−5$
15. $f−1(x)=x+5$
16. $h−1(x)=x−533$
17. $f−1(x)=x3+32$
18. $g−1(x)=2−xx3$
19. $f−1(x)=5−xx$
20. $f−1(x)=5(x+1)x−1$
21. $h−1(x)=−510x−1$
22. $g−1(x)=x3−25$
23. $f−1(x)=(x+4)3+6$
24. $h−1(x)=(x+3)5−1$
25. $f−1(x)=x−bm$
26. $f−1(x)=x−da3$
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# Exploring Collatz-Like Functions: The Power of 3n – 1
The mathematician, carried along on his flood of symbols, dealing apparently with purely formal truths, may still reach results of endless importance for our description of the physical universe.
Karl Pearson
Be careful! Do not attempt to solve this math problem – it’s very tempting, but it leads to never-ending cycles. Here’s how it works: Start with a positive integer. If it’s even, divide it by 2. If it’s odd, triple it and subtract 1. Keep doing this with the new number you get. We believe you’ll end up stuck in a loop, but we’re not entirely sure.
The 3n + 1 problem circulated by word of mouth for many years. It is generally attributed to Lothar Collatz, who was interested in graphical representations of the iteration of functions. Let us define the function f by
In this problem, start with any positive integer n. If n is even, divide it by 2; if n is odd, multiply it by 3 and add 1. Repeat this process, no matter the initial value of n, the sequence will eventually reach the cycle 4 → 2 → 1 → 4 → 2 → 1 → ‧ ‧ ‧ In a recent talk on the Collatz conjecture, Terrance Tao mentioned the following Collatz-like function h defined by:
Tao points out that in addition to the 2 → 1 → 2 → 1 → ‧ ‧ ‧ loop, two other loops appear. Can you find them? Let’s proceed to discover the other loops. First, let’s consider the first four positive integers.
n = 1: 1 → 2 → 1 → 2 → 1 → ‧ ‧ ‧
n = 2: 2 → 1 → 2 → 1 → 2 → 1 → ‧ ‧ ‧
n = 3: 3 → 8 → 4 → 2 → 1 → ‧ ‧ ‧
n = 4: 4 → 2 → 1 → 2 → 1 → ‧ ‧ ‧
We cannot see any other loop except the one mentioned above (2 → 1 → 2 → 1 → ‧ ‧ ‧ loop). Considering the number 5, we encounter the first occurrence of finding another loop (5 → 14 → 7 → 20 → 10 → 5 → ‧ ‧ ‧ loop). This loop is a bit longer than the previous one.
n = 5: 5 → 14 → 7 → 20 → 10 → 5 → 14 → 7 → 20 → 10 → 5 → ‧ ‧ ‧
n = 6: 6 → 3 → 8 → 4 → 2 → 1 → ‧ ‧ ‧
n = 7: 7 → 20 → 10 → 5 → 14 → 7 → 20 → 10 → 5 → ‧ ‧ ‧
n = 8: 8 → 4 → 2 → 1 → …
n = 9: 9 → 26 → 13 → 38 → 19 → 56 → 28 → 14 → 7 → 20 → 10 → 5 → …
n = 10: 10 → 5 → 14 → 7 → 20 → 10 → 5 → …
n = 11: 11 → 32 → 16 → 8 → 4 → 2 → 1 → …
n = 12: 12 → 6 → 3 → 8 → 4 → 2 → 1 → …
n = 13: 13 → 38 → 19 → 56 → 28 → 14 → 7 → 20 → 10 → 5 → …
n = 14: 14 → 7 → 20 → 10 → 5 → …
n = 15: 15 → 44 → 22 → 11 → 32 → 16 → 8 → 4 → 2 → 1 → …
n = 16: 16 → 8 → 4 → 2 → 1 → …
We can observe any one of the above-mentioned loops as long as the number is less than or equal to 16. However, if we consider the next positive integer, 17, we can discover a loop (17 → 50 → 25 → 74 → 37 → 110 → 55 → 164 → 82 → 41 → 122 → 61 → 182 → 91 → 272 → 136 → 68 → 34 → 17 → ‧ ‧ ‧ loop) that is larger than all the loops mentioned above.
n = 17: 17 → 50 → 25 → 74 → 37 → 110 → 55 → 164 → 82 → 41 → 122 → 61 → 182 → 91 → 272 → 136 → 68 → 34 → 17 → 50 → 25 → …
n = 18: 18 → 9 → 26 → 13 → 38 → 19 → 56 → 28 → 14 → 7 → 20 → 10 → 5 → 14 → 7 → 20 → 10 → 5 …
n = 19: 17 → 50 → 25 → 74 → 37 → 110 → 55 → 164 → 82 → 41 → 122 → 61 → 182 → 91 → 272 → 136 → 68 → 34 → 17 → 50 → 25 → …
n = 20: 17 → 50 → 25 → 74 → 37 → 110 → 55 → 164 → 82 → 41 → 122 → 61 → 182 → 91 → 272 → 136 → 68 → 34 → 17 → 50 → 25 → …
n = 21: 21 → 62 → 31 → 92 → 46 → 23 → 68 → 34 → 17 → 50 → 25 → 74 → 37 → 110 → 55 → 164 → 82 → 41 → 122 → 61 → 182 → 91 → 272 → 136 → 68 → 34 → 17 → 50 → 25 → …
n = 22: 22 → 11 → 32 → 16 → 8 → 4 → 2 → 1 → …
n = 23: 23 → 68 → 34 → 17 → 50 → 25 → 74 → 37 → 110 → 55 → 164 → 82 → 41 → 122 → 61 → 182 → 91 → 272 → 136 → 68 → 34 → 17 → 50 → 25 → …
n = 24: 24 → 12 → 6 → 3 → 8 → 4 → 2 → 1 → …
n = 25: 25 → 37 → 110 → 55 → 164 → 82 → 41 → 122 → 61 → 182 → 91 → 272 → 136 → 68 → 34 → 17 → 50 → 25 → …
A comparative study for the functions f and h as given above. Let us consider the following examples to compare f and h:
We can see from these examples that when n is a power of 2, both f and g produce the same numbers at the same time. Otherwise, the rates at which f and h reach any particular loop are different.
Try This Consider the following function g defined by
Compare f, h, and g using specific examples.
This blog is as much yours as it is mine. So, if you have got some ideas to share what you want to see in the next post, feel free to drop a line. We welcome your ideas with open arms and reverence! Looking forward to seeing you soon on “Math1089 Mathematics for All” for another fascinating mathematics blog.
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# To solve the absolute value inequality $$|2x-1| < 3$$, you start by removing the absolute value under the condition set by the inequality, which gives us two separate inequalities: 1. $$2x - 1 < 3 0 Points ▲ good image bad image ▼ ## Ask AI on The Internet Question: Which inequality represents all possible values of x for the following absolute value inequality? |2x-1|<3 A.-1<x<2 B.-2<x<1 C.x<2 D.x<-1 Asked by thevttv56 0 Points Log In or Register to Vote To solve the inequality \(|2x - 1| < 3$$, we need to consider the two cases that arise from the definition of absolute value. The absolute value inequality $$|A| < B$$ can be rewritten as: $-B < A < B$ Here, $$A = 2x - 1$$ and $$B = 3$$. Applying this to our inequality gives: $-3 < 2x - 1 < 3$ Next, we solve for $$x$$ by isolating $$x$$ in the middle part of the inequality. 1. Add 1 to all parts of the inequality: $-3 + 1 < 2x - 1 + 1 < 3 + 1$ $-2 < 2x < 4$ 2. Divide all parts by 2 to solve for $$x$$: $\frac{-2}{2} < \frac{2x}{2} < \frac{4}{2}$ $-1 < x < 2$ Therefore, the inequality $$|2x - 1| < 3$$ represents all values of $$x$$ that satisfy: $-1 < x < 2$ Thus, the correct answer is: A. $$-1 < x < 2$$
May 22, 2024, 5:39 a.m.
0 Points
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# Factorize the Difference of Two Squares
Explain how to factorize the difference of two squares?
We know the formula (a2 – b2) = (a + b)(a - b) is used to factorize the algebraic expressions.
Solved problems to factorize the difference of two squares:
1. Factorize:
(i) y2 - 121
Solution:
We can write y2 – 121 as a2 - b2.
= (y)2 - (11)2, we know 121 = 11 times 11 = 112.
Now we will apply the formula of a2 - b2 = (a + b) (a – b)
= (y + 11)(y - 11).
(ii) 49x2 - 16y2
Solution:
We can write 49x2 - 16y2 as a2 - b2 = (a + b) (a – b)
= (7x)2 - (4y)2,
[Since we know 49x2 = 7x times 7x which is (7x)2 and (4y)2 = 4y times 4y which is (4y)2].
= (7x + 4y) (7x - 4y).
2. Factor the following:
(i) 48a2 - 243b2
Solution:
We can write 48a2 - 243b2 as a2 - b2
= 3(16a2 - 81b2), taking common ‘3’ from both the terms. = 3 ∙ {(4a)2 - (9b)2}
Now we will apply the formula of a2 - b2 = (a + b) (a – b)
= 3(4a + 9b) (4a - 9b).
(ii) 3x3 - 48x
Solution:
3x3 - 48x
= 3x(x2 - 16), taking common ‘3x’ from both the terms.
We can write x2 - 16 as a2 - b2
= 3x(x2 - 42)
Now we will apply the formula of a2 - b2 = (a + b)(a – b)
= 3x(x + 4)(x - 4).
3. Factor the expressions:
(i) 25(x + 3y)2 - 16 (x - 3y)2
Solution:
We can write 25(x + 3y)2 - 16 (x - 3y)2 as a2 - b2.
= [5(x + 3y)]2 - [4(x - 3y)]2
Now using the formula of a2 – b2 = (a + b)(a – b) we get,
= [5(x + 3y) + 4(x - 3y)] [5(x + 3y) - 4(x - 3y)]
= [5x + 15y + 4x - 12y] [5x + 15y - 4x + 12y], using distributive property
= [9x + 3y] [x + 27y], simplifying
= 3[3x + y] [x + 27y]
(ii) 4a2 - 16/(25a2)
Solution:
We can write 4a2 - 16/(25a2) as a2 – b2.
(2a)2 - (4/5a)2, since 4a2 = (2a)2, 16 = 42 and 25a2 = (5a)2
Now we will express as a2 – b2 = (a + b) (a – b)
(2a + 4/5a)(2a - 4/5a)
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## How do you divide a 4 by square?
Solution 1: The easiest way to division the square right into 4 equal parts is to attract three upright lines or three horizontal lines developing four (4) equal areas.
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## How can you division a square right into eight equal parts with three directly cuts?
Stack the two halves, one on optimal of the other and then reduced them right into 4 equal piece with reduced #2. This offers you 4 equal pieces of pie. Stack these 4 piece of pie on optimal of each other and make your 3rd cut, cutting them every in fifty percent for 8 equal, however messy, pieces of pie through 3 cuts.
## How execute you division a square right into 8 equal squares?
Divide the square by illustration lines with the midsection. Attract a upright (meaning from optimal to bottom) heat crossing the midsection (or the center) of the square. Next, draw a horizontal (meaning next to side) line with the midsection. Doing this need to divide the square into 8 same triangles.
## Can you reduced a cake into 8 pieces with 3 cuts?
Step 1: cut the cake right into quarters (4 pieces) making use of 2 of the cut – one horizontally under the centre of the cake and the other vertically under the centre of the cake. Action 3: Finally, you deserve to just reduced that ridge of 4 pieces in fifty percent – utilizing your third and final reduced – and then you will end up v 8 pieces of cake!
## How many squares deserve to you division a square into?
A square can be reduced into four quarters and also one quarter cut into 4 again, shedding the quarter as a count square but gaining 4 smaller ones – therefore a square can be reduced into seven squares.
## How perform you divide a square into 7 equal squares?
Start through 1 square (1)Divide square right into 4 equal sized squares (1–1+4=4 squares in present image)Select an arbitrary below square right into 4 equal sized squares (4–1+4=7 squares in last image)
## Can you divide a rectangle right into 8 equal parts?
Draw a horizontal line with the middle of the square. When you have 4 rectangles, you have the right to simply draw one horizontal line through the facility of the square, separating it right into eight same rectangles.
## What is the minimum cuts of circle?
No more cuts are required as the one is currently divided into four equal parts. A single cut is compelled at 330 degree to division the circle in 3 equal parts.
## What is the minimum no of cuts essential to acquire 8 slices out of a cake?
Originally Answered: Minimum how many cuts you will make in a cake to divide the cake right into 8 pieces? 3 cuts , cake is a cube and cut 1 horizontally and cut 2 vertically and one an ext vertical reduced ,makes the cake exactly into 8 piece .
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## Can you kind six congruent squares by relocating three sticks?
Answer 3: yes there will be a major problem. The will produce a usual stick 6 with the existing main square consisted of of sticks 5, 6, 7, 8. And move the straightaway to finish one of 3 remaining practically complete squares, say square with three sides 11, 7, 9.
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Chord Star 2: Choosing different points
Last time we saw how you could make a “chord star” by picking a point inside a circle and drawing chords through that point. Then we measured the two lengths of the partial chords (let’s call them $L_1$ and $L_2$) and plotted them against one another. We got a rectangular hyperbola, suggesting (or confirming if we remembered the geometry) that $L_1 L_2 = k$, some constant.
But we asked, “what effect does your choice of point have on the graph and the data?” So of course we’ll take an empirical approach and try it. If you have a classroom full of students, and they used the same-sized circle and picked their own points, you could immediately compare the points they chose to the functions they generated. Or you could do it as an individual. The photo shows what this might look like, and here is a detail:
Now we’ll put the data in a table, but this time,
• In addition to L1 and L2, we’ll record R, the distance from the center to the point. It may not be obvious to students at first that all points the same distance from the center (or the edge) will give the same data, but I’ll assume we get that.
• We’ll double the data by recording the data in the reverse order as well. It makes the graph look better.
Here’s the graph, coded by distance (in cm) of the point from the center.
The R = 7 points (the “dels”) look as if they might lie along a hyperbola; the others turn out to do so too. Without zooming in and showing what we hope students would do (e.g., look at residual plots), let’s skip ahead and plot these functions on the graph. Aha: part of the answer to our question is, if you choose a different point, the relationship between $L_1$ and $L_2$ has the same form, but the points lie along a different hyperbola.
Well. There’s clearly a systematic relationship between the position (radius) of the point and the curve it generates. What is that relationship?
This gets back to the question, what can we plot? (We know, but how can we help students make this decision for themselves?) One intriguing idea is to make a 3-D plot, so we would see each hyperbola at an altitude (z-value) given by that radius number. Great idea! I don’t feel like wrestling with the appropriate tools, though, so I’ll ask, “is there any way we can characterize the function with a single number?”
Heck yeah. The functions are, for example, $L_2=\frac{20.1}{L_1}$ for r = 7; $L_2=\frac{39}{L_1}$ for r = 5.5; $L_2=\frac{52.4}{L_1}$ for r = 4, and so forth. They’re all just $L_2=\frac{k}{L_1}$. Let’s just use the value of k.
Here is the graph of k against r.
Okay, reader, what function would you use to model this? And why does it work? (Don’t tell, of course! Just work it out.)
This is not the only “next-level” relationship that you can pull out of this situation. If you look back at the first data graph, we have to ask, what points are possible? It looks as if there’s a linear boundary, a negative diagonal, separating the allowed area from the forbidden. Is that right? If so, what’s the line, and why?
Additional questions (yes, you want students to ask these, but I’m putting them here so I don’t forget them. The best way to answer most of them, by the way, is to understand what they mean geometrically—they’re about chords and circles, after all, not hyperbolas.):
• How come the hyperbolas can go into the forbidden zone?
• Are all points below the chord-star boundary possible?
• If you have a single point from a graph, what can you tell about the chord it came from?
• If you have a single point, what can you tell about the circle it came from?
• If you have a single hyperbola, what can you tell about the circle?
• What’s the least amount of information you need from a table of data like ours to determine the circle?
• How can we take stuff like this out into the real world? Stay tuned for our next exciting episode.
The forbidden-zone issue smells a lot like the “random rectangles” plot you can make, plotting area against perimeter (as I did long ago in The Mathematics Teacher, vol 94, no 8, November 2001). At right is a graph from that article showing areas and perimeters for 100 randomly-generated rectangles with lengths and width in the range zero to 30. There’s the curvey boundary, but also a straight, slanty one. See if you can understand both.
Finally, for lazy persons who don’t want to measure their own data, but are still interested enough to play with it, my measurements in centimeters, tab-delimited, I hope:
L1 L2 R
8.6 7.8 1.0
8.8 7.2 2.5
7.8 8.6 1.0
7.2 8.8 2.5
4.6 11.5 4.0
11.5 4.6 4.0
6.3 8.4 4.0
8.4 6.3 4.0
5.9 6.6 5.5
6.6 5.9 5.5
11.4 3.5 5.5
3.5 11.4 5.5
4.5 4.4 7.0
9.5 2.1 7.0
2.1 9.5 7.0
9.3 7.3 1.0
7.3 9.3 1.0
10.8 5.8 2.5
5.8 10.8 2.5
15.3 1.3 7.0
1.3 15.3 7.0
2.8 13.8 5.5
13.8 2.8 5.5
Author: Tim Erickson
Math-science ed freelancer and sometime math and science teacher. Currently working on various projects.
3 thoughts on “Chord Star 2: Choosing different points”
1. Tim
Thanks for two fantastically helpful posts. I have been doing a great deal of thinking about our Geometry curriculum and these are the kinds of activities I want our kids engaged with. Another thought regarding your question about hyperbola graphs going into the ‘forbidden zone’ – I think that this ties in well with algebra discussions of extraneous solutions and with the recognition that while equations can model situations the solution sets are not always identical.
|
# CBSE Class 6 Maths Activity 4
Read and download CBSE Class 6 Maths Activity 4 in NCERT book for Class 6 Mathematics. You can download latest NCERT eBooks chapter wise in PDF format free from Studiestoday.com. This Mathematics textbook for Class 6 is designed by NCERT and is very useful for students. Please also refer to the NCERT solutions for Class 6 Mathematics to understand the answers of the exercise questions given at the end of this chapter
## NCERT Book for Class 6 Mathematics Activity 4
Class 6 Mathematics students should refer to the following NCERT Book Activity 4 in Class 6. This NCERT Book for Class 6 Mathematics will be very useful for exams and help you to score good marks
### Activity 4 NCERT Book Class 6
To find the HCF of two given numbers, by paper cutting and pasting.
Learning Objective : To understand the concept of HCF of two umbers.
Pre-requisite : Comparison of two numbers, knowledge of Division algorithm : Dividend = Divisor x Quotient + Reminder
Materials Required : Coloured grid papers (2 different colours say green and red), a pair of scissors, glue, a ruler, and a pencil.
Procedure :
Step 1. Take any two numbers say 24 and 16.
Step 2. Cut out a strip of length 24 units using green colour grid paper as shown in Fig.4(a).
Step 3. Cut out another strip of length 16 units using red colour grid paper as shown in Fig. 4(b).
Step 4. Place the red strip along the green strip as shown in the Fig. 4(c) and cut out the remaining part of the green strip.
Step 5. Observe that the remaining part of green strip cut out is of length 8 units and is smaller than the red colour strip i.e. the strip of length 16 units.
Step 6. Now place the smaller cut out strip (green) along the larger cut out strip (red) and cut out the extra part (red strip) as shown in Fig. 4(d).
Step 7. Repeat the activity till both strips are equal.
Observations :
(i) In Step 4, the length of the green strip = ............... units
(ii) In Step 6, the length of the red strip = ............... units
(iii)Is there any part of any strip left when green strip of length 8 units was placed on the red strip?.......... Yes/No
HCF of 24 and 16 = ............
Extension : (i) Find the HCF of (i) 40 and 16, (ii) 12, 15 and 6.
Please refer to the link below - CBSE Class 6 Maths Activity 4
NCERT Class 6 Maths Knowing our Numbers
NCERT Class 6 Maths Whole Numbers
NCERT Class 6 Maths Playing with Numbers
NCERT Class 6 Maths Basic Geometrical Ideas
NCERT Class 6 Maths Understanding Elementary Shapes
NCERT Class 6 Maths Integers
NCERT Class 6 Maths Fractions
NCERT Class 6 Maths Decimals
NCERT Class 6 Maths Data Handling
NCERT Class 6 Maths Mensuration
NCERT Class 6 Maths Algebra
NCERT Class 6 Maths Ratio and Proportion
NCERT Class 6 Maths Symmetry
NCERT Class 6 Maths Practical Geometry
NCERT Class 6 Maths Answers to all Chapters
NCERT Class 6 Maths Brain Teasers
### Mathematics NCERT Book Class 6 Activity 4
The above NCERT Books for Class 6 Mathematics Activity 4 have been published by NCERT for latest academic session. The textbook by NCERT for Activity 4 Mathematics Class 6 is being used by various schools and almost all education boards in India. Teachers have always recommended students to refer to Activity 4 NCERT etextbooks as the exams for Class 6 Mathematics are always asked as per the syllabus defined in these ebooks. These Class 6 Activity 4 book for Mathematics also includes collection of question. We have also provided NCERT solutions for Class 6 Mathematics which have been developed by teachers of StudiesToday.com after thorough review of the latest book and based on pattern of questions in upcoming exams for Class 6 students.
### NCERT Book Class 6 Mathematics Activity 4
The latest NCERT book for Activity 4 pdf have been published by NCERT based on the latest research done for each topic which has to be taught to students in all classes. The books for Class 6 Mathematics Activity 4 are designed to enhance the overall understanding of students. All Class 6 NCERT textbooks have been written in an easy to understand language which will help to enhance the overall level of Class 6 students.
#### Activity 4 NCERT Book Class 6 Mathematics
As the students of Class 6 need the books for their regular studies, we have provided all NCERT book for Activity 4 in pdf here for free download. All pdf NCERT books available here for Class 6 will help them to read on their mobile or computers. They can take a print of the Class 6 Mathematics NCERT Book Activity 4 pdf easily and use them for studies. The NCERT textbooks for Class 6 Mathematics have been provided chapter-wise and can be downloaded for free of cost.
#### Class 6 Mathematics Activity 4 NCERT Book
Along with Mathematics Class 6 NCERT Book in Pdf for Activity 4 we have provided all NCERT Books in English Medium for Class 6 which will be really helpful for students who have opted for english language as a medium. Class 6 students will need their books in English so we have provided them here for all subjects in Class 6.
#### Class 6 Activity 4 NCERT Book Mathematics
For Class 6 Activity 4 we have provided books for students who have opted for Hindi and Urdu medium too. You can click on the links provided above to download all Hindi medium Class 6 Mathematics Activity 4 book in easy to read pdf format. These books will help Class 6 Mathematics students to understand all topics and also strictly follow latest syllabus for their studies. If you are looking to download the pdf version of Class 6 Mathematics Activity 4 textbook issued by NCERT then you have come to the correct website
Where can I download latest NCERT Book for Class 6 Mathematics Activity 4
You can download the NCERT Book for Class 6 Mathematics Activity 4 for latest session from StudiesToday.com
Can I download the NCERT Books of Class 6 Mathematics Activity 4 in Pdf
Yes, you can click on the link above and download chapter wise NCERT Books in PDFs for Class 6 for Mathematics Activity 4
Are the Class 6 Mathematics Activity 4 NCERT Book available for the latest session
Yes, the NCERT Book issued for Class 6 Mathematics Activity 4 have been made available here for latest academic session
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You can easily access the link above and download the Class 6 NCERT Books Mathematics Activity 4 for each chapter
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Regular revision of NCERT Books given on studiestoday for Class 6 subject Mathematics Activity 4 can help you to score better marks in exams
|
# Introduction to Graphing Data on the Coordinate Plane
```Introduction to Graphing
Data on the Coordinate
Plane
1
Do Now
M1’s get binders.
M2’s get new packets.
Discuss the following questions with your table group,
What do you know about graphing? What do you
know about the coordinate plane? What is an
independent variable? What is a dependent variable?
2
Objectives
Content Objective: Students will be able to identify relationships between
independent and dependent variables using graphs and tables.
Language Objective: Students will be able to explain the relationship that
exists between variables using the terms “direct correlation”, “inverse
correlation”, “independent variable” and “dependent variable”.
Content Standard: (6.EE.9) Use variables to represent two quantities in a
real-world problem that change in relationship to one another; write an
equation to express one quantity, thought of as the dependent variable, in
terms of the other quantity, thought of as the independent variable. Analyze
the relationship between the dependent and independent variables using
graphs and tables, and relate these to the equation.
Language Standard: (6.LS.6) Acquire and use accurately gradeappropriate general academic and domain-specific words and phrases;
gather vocabulary knowledge when considering a word or phrase important
3
to comprehension or expression.
Graphing Activity 1
Today we will be graphing using iSense, but first
we need to collect some data to graph!
In our first activity, we will be collecting data
We will determine how many times your heart
beats in 15, 30, 45 and 60 seconds.
To achieve a true resting heartbeat, we need to
SLANT, with both feet planted on the floor and
voices off.
Try to find your heartbeat now.
4
Give Me a Beat!
When I say “go” begin silently counting your
When I say “stop”, stop counting and record your
heartbeats in the appropriate column.
Time (In Seconds)
Number of Heartbeats
15
30
45
60
5
Graphing Data Using iSense
Go to iSenseproject.org
Search for Project #1391 (Graphing Data on the Coordinate Plane)
Scroll down to the section titled “Contribute Data”
Enter “cgs1234” as the key. Type your name into the “Contributor Name”
field. Hit “Submit”.
Find the “Contribute Data” section on the right of the screen. Click on
“Manual Entry”.
Write your name in the “Data Set Name” box.
Enter “15”, “30”, “45” and “60” in order in the first four rows under “Time (In
Seconds).
Then enter the number of heartbeats you counted for each number of
seconds in the first four rows under “Number of Heatbeats”.
When you have finished your screen should look exactly like the table you
completed on page 5.
After checking you have entered your data correctly, hit the blue “Save” 6
button.
Relationship Between Variables
Analyze your graph to determine if there is a
relationship that exists between variables.
Look at the table you created on page 6, are there any
trends that you notice about the numbers?
7
Analysis of Class Data
Bring your attention to the class data shown on
the smartboard.
What patterns or trends do you notice?
8
Exit Slip
In iSense, search for project # 1392 (EXIT SLIP)
Input the data from the table below into the iSense
database. Then determine if a relationship exists between
variables. If a relationship does exist describe it in words
and draw a model in the spaces provided below.
Time (in hrs)
Distance (in
mi)
1
45
2
90
3
135
Relationship:
______________________________________________
______________________________________________
______________________________________________
9
```
|
# 6th Grade - Fraction and Decimal Conversion
## Introduction
Converting numbers is required in order for the numbers to be added, subtracted, multiplied, or divided. Any operation can be easily performed if all the numbers in the expression have the same name.
For instance, $\frac{3.5}{0.5}$ or compared to .
## Fractions to Decimals
• To convert a fraction to decimals divide the numerator by the denominator i.e. the numerator becomes the dividend and the denominator becomes the divisor.
## Converting Decimals to Fractions
• To convert decimals to a fraction:
1. Divide the number by 1.
2. Count the number of digits after the decimal point.
3. Remove the decimal point and add the same number of zeros to the denominator as the number of digits after the decimal point.
4. Reduce the fraction, if possible.
## Solved Examples
Example 1: Convert $\frac{33}{2}$ to a decimal number.
Solution: Divide the numerator by the denominator. So, 33 is the dividend, and 2 is the divisor.
Example 2: Convert 15.2 to a fraction.
Solution: Divide the given number by 1
$=\frac{15.2}{1}$
Count the number of digits after the decimal point (to the right of the decimal point) and put the same number of zeros in the denominator; in this case, there is 1 digit after the decimal point, so we will put 1 zero in the denominator.
Simplify the fraction.
Example 3: Convert 27.24 to a fraction.
Solution: Divide the given number by 1
$=\frac{27.24}{1}$
Count the number of digits after the decimal point (to the right of the decimal point) and put the same number of zeros in the denominator; in this case, there are 2 digits after the decimal point, so we will put 2 zeros in the denominator.
Simplify the fraction.
|
Value Based Questions: Statistics
# Class 9 Maths Chapter 13 Question Answers - Statistics
VALUE BASED QUESTIONS
Q1. The following table gives the percentage distribution of female students in various schools of rural areas of various states and UTs of India who are getting freeship.
Percentage of female students getting freeship 15 -25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 75 - 85 Number of states/UT 6 11 7 4 4 2 1
(i) Find the mean percentage of female students getting freeship.
(ii) Which mathematical concept is used in the above problem?
(iii) By giving freeship to girls/women students, which value is depicted by various states and UTs?
Sol. (i) We find xi (class marks) of each class and put them in the following table :
Percentage of female students Number of states/UTs fi xi 15 - 25 6 20 25 - 35 11 30 35 - 45 7 40 45 - 55 4 50 55 - 65 4 60 65 - 75 2 70 75 - 85 1 80
Here, we take a = 50, h = 10 and di = xi – 50
Putting di and ui in the following table :
∵ From the table, we have :
⇒ [Using direct method]
OR
Using the step deviation method, we get
Thus, required percentage of women students = 39.71
(ii) Statistics.
(iii) Promoting women education.
The document Class 9 Maths Chapter 13 Question Answers - Statistics is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
## FAQs on Class 9 Maths Chapter 13 Question Answers - Statistics
1. What is the importance of value-based questions in a statistics class?
Ans. Value-based questions in a statistics class are important as they help students develop critical thinking skills, ethical reasoning, and decision-making abilities. These questions prompt students to consider the real-world implications of statistical analysis and understand how statistics can be used to make informed decisions.
2. How can value-based questions enhance the learning experience in a statistics class?
Ans. Value-based questions enhance the learning experience in a statistics class by promoting deeper understanding and application of statistical concepts. These questions encourage students to think beyond the numbers and consider the ethical and social implications of statistical analysis, making the learning process more meaningful and relevant.
3. Can you provide an example of a value-based question in a statistics class?
Ans. Example value-based question: "A pharmaceutical company conducted a clinical trial on a new medication and found statistically significant positive results. However, they also discovered potential side effects that could affect some patients' health. As a statistician, how would you balance the positive outcomes with the potential risks in your analysis and reporting?"
4. How do value-based questions contribute to students' development of ethical reasoning skills in a statistics class?
Ans. Value-based questions in a statistics class require students to consider ethical dilemmas and make informed decisions. By engaging with these questions, students develop their ethical reasoning skills, learning to navigate complex statistical scenarios with integrity and responsibility. This helps them become more ethically conscious statisticians in their future professional endeavors.
5. What role do value-based questions play in preparing students for real-world applications of statistics?
Ans. Value-based questions in a statistics class prepare students for real-world applications by challenging them to think critically about the implications of statistical analysis. By considering the societal, economic, and ethical factors involved, students learn to apply statistical concepts in a more holistic and responsible manner. This prepares them to make informed decisions and effectively communicate statistical findings in real-world scenarios.
## Mathematics (Maths) Class 10
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# 6.4 Working with taylor series (Page 4/11)
Page 4 / 11
Use power series to solve ${y}^{\prime }=2y,\phantom{\rule{0.5em}{0ex}}y\left(0\right)=5.$
$y=5{e}^{2x}$
We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation
${y}^{\prime }-xy=0$
is known as Airy’s equation . It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.
## Power series solution of airy’s equation
Use power series to solve
$y\text{″}-xy=0$
with the initial conditions $y\left(0\right)=a$ and $y\prime \left(0\right)=b.$
We look for a solution of the form
$y=\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\text{⋯}.$
Differentiating this function term by term, we obtain
$\begin{array}{ccc}\hfill {y}^{\prime }& =\hfill & {c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\text{⋯},\hfill \\ \hfill y\text{″}& =\hfill & 2·1{c}_{2}+3·2{c}_{3}x+4·3{c}_{4}{x}^{2}+\text{⋯}.\hfill \end{array}$
If y satisfies the equation $y\text{″}=xy,$ then
$2·1{c}_{2}+3·2{c}_{3}x+4·3{c}_{4}{x}^{2}+\text{⋯}=x\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\text{⋯}\right).$
Using [link] on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,
$\begin{array}{c}2·1{c}_{2}=0,\hfill \\ 3·2{c}_{3}={c}_{0},\hfill \\ 4·3{c}_{4}={c}_{1},\hfill \\ 5·4{c}_{5}={c}_{2},\hfill \\ \hfill \text{⋮}.\hfill \end{array}$
More generally, for $n\ge 3,$ we have $n·\left(n-1\right){c}_{n}={c}_{n-3}.$ In fact, all coefficients can be written in terms of ${c}_{0}$ and ${c}_{1}.$ To see this, first note that ${c}_{2}=0.$ Then
$\begin{array}{}\\ \\ {c}_{3}=\frac{{c}_{0}}{3·2},\hfill \\ {c}_{4}=\frac{{c}_{1}}{4·3}.\hfill \end{array}$
For ${c}_{5},{c}_{6},{c}_{7},$ we see that
$\begin{array}{}\\ \\ {c}_{5}=\frac{{c}_{2}}{5·4}=0,\hfill \\ {c}_{6}=\frac{{c}_{3}}{6·5}=\frac{{c}_{0}}{6·5·3·2},\hfill \\ {c}_{7}=\frac{{c}_{4}}{7·6}=\frac{{c}_{1}}{7·6·4·3}.\hfill \end{array}$
Therefore, the series solution of the differential equation is given by
$y={c}_{0}+{c}_{1}x+0·{x}^{2}+\frac{{c}_{0}}{3·2}{x}^{3}+\frac{{c}_{1}}{4·3}{x}^{4}+0·{x}^{5}+\frac{{c}_{0}}{6·5·3·2}{x}^{6}+\frac{{c}_{1}}{7·6·4·3}{x}^{7}+\text{⋯}.$
The initial condition $y\left(0\right)=a$ implies ${c}_{0}=a.$ Differentiating this series term by term and using the fact that ${y}^{\prime }\left(0\right)=b,$ we conclude that ${c}_{1}=b.$ Therefore, the solution of this initial-value problem is
$y=a\left(1+\frac{{x}^{3}}{3·2}+\frac{{x}^{6}}{6·5·3·2}+\text{⋯}\right)+b\left(x+\frac{{x}^{4}}{4·3}+\frac{{x}^{7}}{7·6·4·3}+\text{⋯}\right).$
Use power series to solve $y\text{″}+{x}^{2}y=0$ with the initial condition $y\left(0\right)=a$ and ${y}^{\prime }\left(0\right)=b.$
$y=a\left(1-\frac{{x}^{4}}{3·4}+\frac{{x}^{8}}{3·4·7·8}-\text{⋯}\right)+b\left(x-\frac{{x}^{5}}{4·5}+\frac{{x}^{9}}{4·5·8·9}-\text{⋯}\right)$
## Evaluating nonelementary integrals
Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.
One integral that arises often in applications in probability theory is $\int {e}^{\text{−}{x}^{2}}dx.$ Unfortunately, the antiderivative of the integrand ${e}^{\text{−}{x}^{2}}$ is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term “elementary function” is not synonymous with noncomplicated function. For example, the function $f\left(x\right)=\sqrt{{x}^{2}-3x}+{e}^{{x}^{3}}-\text{sin}\left(5x+4\right)$ is an elementary function, although not a particularly simple-looking function. Any integral of the form $\int f\left(x\right)\phantom{\rule{0.1em}{0ex}}dx$ where the antiderivative of $f$ cannot be written as an elementary function is considered a nonelementary integral .
Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering $\int {e}^{\text{−}{x}^{2}}dx.$
## Using taylor series to evaluate a definite integral
1. Express $\int {e}^{\text{−}{x}^{2}}dx$ as an infinite series.
2. Evaluate ${\int }_{0}^{1}{e}^{\text{−}{x}^{2}}dx$ to within an error of $0.01.$
1. The Maclaurin series for ${e}^{\text{−}{x}^{2}}$ is given by
$\begin{array}{cc}\hfill {e}^{\text{−}{x}^{2}}& =\sum _{n=0}^{\infty }\frac{{\left(\text{−}{x}^{2}\right)}^{n}}{n\text{!}}\hfill \\ & =1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\text{⋯}+{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}+\text{⋯}\hfill \\ & =\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}.\hfill \end{array}$
Therefore,
$\begin{array}{cc}\hfill \int {e}^{\text{−}{x}^{2}}dx& =\int \left(1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\text{⋯}+{\left(-1\right)}^{n}\frac{{x}^{2n}}{n\text{!}}+\text{⋯}\right)\phantom{\rule{0.1em}{0ex}}dx\hfill \\ & =C+x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5.2\text{!}}-\frac{{x}^{7}}{7.3\text{!}}+\text{⋯}+{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)n\text{!}}+\text{⋯}.\hfill \end{array}$
2. Using the result from part a. we have
${\int }_{0}^{1}{e}^{\text{−}{x}^{2}}dx=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}-\text{⋯}.$
The sum of the first four terms is approximately $0.74.$ By the alternating series test, this estimate is accurate to within an error of less than $\frac{1}{216}\approx 0.0046296<0.01.$
where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
Abdul
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# Factorisation Method to Solve Limits
Here you will learn what is the factorisation method to solve limits with examples.
Let’s begin –
## Factorisation Method to Solve Limits
Consider the following limit :
$$\displaystyle{\lim_{x \to a}}$$ $$f(x)\over g(x)$$
If by substituting x = a, $$f(x)\over g(x)$$, reduces to the form $$0\over 0$$, then (x – a) is a factor of f(x) and g(x) both.
So, we first factorize f(x) and g(x) and then cancel the common factor to evaluate the limit.
Also Read : How to Solve Indeterminate Forms of Limits
Following algorithm may be used to evaluate the limit by factorisation method.
Algorithm :
1). Obtain the problem, say, $$\displaystyle{\lim_{x \to a}}$$ $$f(x)\over g(x)$$, where $$\displaystyle{\lim_{x \to a}}$$ f(x) = 0 and $$\displaystyle{\lim_{x \to a}}$$ g(x) = 0.
2). Factorize f(x) and g(x).
3). Cancel out the common factor.
4). Use direct substitution method to obtain the limit.
Example : Evaluate : $$\displaystyle{\lim_{x \to 2}}$$ $$x^2 – 5x + 6\over x^2 – 4$$.
Solution : When x = 2 the expression $$x^2 – 5x + 6\over x^2 – 4$$ assumes the indeterminate form $$0\over 0$$.
Therefore, (x – 2) is a common factor in numerator and denominator.
Factorising the numerator and denominator, we have
$$\displaystyle{\lim_{x \to 2}}$$ $$x^2 – 5x + 6\over x^2 – 4$$ = $$\displaystyle{\lim_{x \to 2}}$$ $$(x – 2)(x – 3)\over (x + 2)(x – 2)$$
= $$\displaystyle{\lim_{x \to 2}}$$ $$x – 3\over x + 2$$ = $$2 – 3\over 2 + 2$$ = $$-1\over 4$$
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# A line segment has endpoints at (1 ,2 ) and (3 ,4 ). The line segment is dilated by a factor of 6 around (2 ,5 ). What are the new endpoints and length of the line segment?
Jan 27, 2018
$\left(- 4 , - 13\right) , \left(8 , - 1\right) , \approx 16.97$
#### Explanation:
$\text{label the endpoints "A(1,2)" and } B \left(3 , 4\right)$
$\text{label the centre of dilatation } C \left(2 , 5\right)$
$\text{let A' and B' be the images of A and B}$
$\text{then}$
$\vec{C A '} = \textcolor{red}{6} \vec{C A}$
$\Rightarrow \underline{a} ' - \underline{c} = 6 \left(\underline{a} - \underline{c}\right)$
$\Rightarrow \underline{a} ' - \underline{c} = 6 \underline{a} - 6 \underline{c}$
$\Rightarrow \underline{a} ' = 6 \underline{a} - 5 \underline{c}$
$\textcolor{w h i t e}{\Rightarrow \underline{a} '} = 6 \left(\begin{matrix}1 \\ 2\end{matrix}\right) - 5 \left(\begin{matrix}2 \\ 5\end{matrix}\right)$
$\textcolor{w h i t e}{\Rightarrow \underline{a} '} = \left(\begin{matrix}6 \\ 12\end{matrix}\right) - \left(\begin{matrix}10 \\ 25\end{matrix}\right) = \left(\begin{matrix}- 4 \\ - 13\end{matrix}\right)$
$\Rightarrow A ' = \left(- 4 , - 13\right)$
$\text{and }$
$\vec{C B '} = \textcolor{red}{6} \vec{C B}$
$\Rightarrow \underline{b} ' - \underline{c} = 6 \left(\underline{b} - \underline{c}\right)$
$\Rightarrow \underline{b} ' = 6 \underline{b} - 5 \underline{c}$
$\textcolor{w h i t e}{\Rightarrow \underline{b} '} = 6 \left(\begin{matrix}3 \\ 4\end{matrix}\right) - 5 \left(\begin{matrix}2 \\ 5\end{matrix}\right)$
$\textcolor{w h i t e}{\Rightarrow \underline{b}} = \left(\begin{matrix}18 \\ 24\end{matrix}\right) - \left(\begin{matrix}10 \\ 25\end{matrix}\right) = \left(\begin{matrix}8 \\ - 1\end{matrix}\right)$
$\Rightarrow B ' = \left(8 , - 1\right)$
$\text{calculate the length using the "color(blue)"distance formula}$
•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
$\text{let "(x_1,y_1)=(-4,-13)" and } \left({x}_{2} , {y}_{2}\right) = \left(8 , - 1\right)$
$d = \sqrt{{\left(8 + 4\right)}^{2} + {\left(- 1 + 13\right)}^{2}}$
$\textcolor{w h i t e}{d} = \sqrt{144 + 144} = \sqrt{288} = 12 \sqrt{2} \approx 16.97 \text{ 2 d.p}$
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Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords
Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords
Tangents and Intersecting Chords Exercise 18A – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre?
Solution:
Question 2.
In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.
Solution:
Question 3.
Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.
Solution:
Question 4.
Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution:
Question 5.
Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.
Solution:
Question 6.
Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Solution:
Question 7.
If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.
Solution:
Question 8.
If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.
Solution:
From A, AP and AS are tangents to the circle.
Therefore, AP = AS…….(i)
Similarly, we can prove that:
BP = BQ ………(ii)
CR = CQ ………(iii)
DR = DS ………(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
But AB = CD and BC = AD…….(v) Opposite sides of a ||gm
Therefore, AB + AB = BC + BC
2AB = 2 BC
AB = BC ……..(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus.
Question 9.
From the given figure prove that:
AP + BQ + CR = BP + CQ + AR.
Also, show that AP + BQ + CR = $$\frac{1}{2}$$ × perimeter of triangle ABC.
Solution:
Question 10.
In the figure, if AB = AC then prove that BQ = CQ.
Solution:
Question 11.
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if –
i) they touch each other externally.
ii) they touch each other internally.
Solution:
Question 12.
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
i) ∠AOP = ∠BOP
ii) OP is the perpendicular bisector of chord AB.
Solution:
Question 13.
In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that:
i) tangent at point P bisects AB.
ii) Angle APB = 90°
Solution:
Question 14.
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
∠PAQ = 2∠OPQ
Solution:
Question 15.
Two parallel tangents of a circle meet a third tangent at point P and Q. Prove that PQ subtends a right angle at the centre.
Solution:
Question 16.
ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.
Solution:
Question 17.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
i) ∠QOR
ii) ∠QPR
given that ∠A = 60°
Solution:
Question 18.
In the following figure, PQ and PR are tangents to the circle, with centre O. If , calculate:
i) ∠QOR
ii) ∠OQR
iii) ∠QSR
Solution:
Question 19.
In the given figure, AB is a diameter of the circle, with centre O, and AT is a tangent. Calculate the numerical value of x.
Solution:
Question 20.
In quadrilateral ABCD, angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle.
Solution:
Question 21.
In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given and ∠SPR = x° and ∠QRP = y°
Prove that -;
i) ∠ORS = y°
ii) write an expression connecting x and y
Solution:
Question 22.
PT is a tangent to the circle at T. If ; calculate:
i) ∠CBT
ii) ∠BAT
iii) ∠APT
Solution:
Question 23.
In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.
Solution:
Question 24.
In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.
Solution:
∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ)
∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)
∠CAQ + ∠PAC = 180°……(angles in linear pair)
∴∠PAC = 120°
∠PAC = 2∠CAD……(AD is angle bisector of ∠PAC)
∠CAD = 60°
Now,
∠CAD + ∠CAB = 60 + 30 = 90°
∠DAB = 90°
Thus, BD subtends 90° on the circle
So, BD is the diameter of circle
Tangents and Intersecting Chords Exercise 18B – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
i) In the given figure, 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.
ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4cm. Find CD.
iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.
Solution:
Question 2.
In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find
(i) AB.
(ii) the length of tangent PT.
Solution:
Question 3.
In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ; ∠ADB = 30° and ∠CBD = 60° calculate:
i) ∠QAD
ii) ∠PAD
iii) ∠CDB
Solution:
Question 4.
If PQ is a tangent to the circle at R; calculate:
i) ∠PRS
ii) ∠ROT
Given: O is the centre of the circle and ∠TRQ = 30°
Solution:
Question 5.
AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.
Solution:
Question 6.
Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.
Solution:
Question 7.
Two circles with centers O and O’ are drawn to intersect each other at points A and B.
Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.
Solution:
Question 8.
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠CPA = ∠DPB
Solution:
Question 9.
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Solution:
Question 10.
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If angle BCG = 108° and O is the centre of the circle, find:
i) angle BCT
ii) angle DOC
Solution:
Question 11.
Two circles intersect each other at point A and B. A straight line PAQ cuts the circle at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T are concyclic.
Solution:
Question 12.
In the figure, PA is a tangent to the circle. PBC is a secant and AD bisects angle BAC.
Show that the triangle PAD is an isosceles triangle. Also show that:
∠CAD = $$\frac{1}{2}$$(∠PBA – ∠PAB)
Solution:
Question 13.
Two circles intersect each other at point A and B. Their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.
Solution:
Question 14.
In the figure, chords AE and BC intersect each other at point D.
i) if , ∠CDE = 90° AB = 5 cm, BD = 4 cm and CD = 9 cm; find DE
ii) If AD = BD, Show that AE = BC.
Solution:
Question 15.
Circles with centers P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles are congruent; show that CE = BD.
Solution:
Question 16.
In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B. Find ∠BDC = 65. Find ∠BAO
Solution:
Tangents and Intersecting Chords Exercise 18C – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
Prove that of any two chord of a circle, the greater chord is nearer to the centre.
Solution:
Question 2.
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
i) If the radius of the circle is 10 cm, find the area of the rhombus.
ii) If the area of the rhombus is $$32 \sqrt{3}$$ cm2, find the radius of the circle.
Solution:
Question 3.
Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Solution:
Question 4.
Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of the angle BAC.
Solution:
Question 5.
The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle?
Solution:
Question 6.
ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.
Solution:
Question 7.
In the given figure, C and D are points on the semicircle described on AB as diameter.
Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC
Solution:
Question 8.
In cyclic quadrilateral ABCD, A = 3 ∠C and ∠D = 5∠B. Find the measure of each angle of the quadrilateral.
Solution:
ABCD is a cyclic quadrilateral.
∴ ∠A + ∠C = 180°
⇒ 3∠C + ∠C = 180°
⇒ 4∠C = 180°
⇒ ∠C = 45°
∵ ∠A = 3∠C
⇒ ∠A = 3 × 45°
⇒ ∠A = 135°
Similarly,
∴ ∠B+ ∠D = 180°
⇒∠B + 5∠B = 180°
⇒ 6∠B = 180°
⇒ ∠B = 30°
∵∠D = 5∠B
⇒ ∠D = 5 × 30° >
⇒ ∠D = 150°
Hence, ∠A = 1350, ∠B = 30°, ∠C = 450, ∠D = 150°
Question 9.
Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:
Question 10.
Bisectors of vertex A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = $$90^{\circ}-\frac{1}{2} \angle A$$
Solution:
Question 11.
In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.
Solution:
Question 12.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
Solution:
Question 13.
P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.
Solution:
Question 14.
In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent.
Prove that the line NM produced bisects AB at P.
Solution:
Question 15.
In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40° and ∠ABD = 60°, find:
i) ∠DBC
ii) ∠ BCP
iii) ∠ ADB
Solution:
Question 16.
The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90°
Solution:
Question 17.
ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D.
Prove that –
i) AC × AD = AB2
ii) BD= AD × DC.
Solution:
Question 18.
In the given figure AC = AE.
Show that:
i) CP = EP
ii) BP = DP
Solution:
Question 19.
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC=120°
Calculate:
i) ∠BEC
ii) ∠ BED
Solution:
Question 20.
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30°, find:
(i) angle BCO
(ii) angle AOB
(iii) angle APB
Solution:
Question 21.
ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles.
Solution:
Question 22.
In a square ABCD, its diagonal AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and bisector of angle ABD meets AC at N and AM at L. Show that –
i) ∠ONL + ∠OML = 180°
ii) ∠BAM = ∠BMA
iii) ALOB is a cyclic quadrilateral.
Solution:
Question 23.
The given figure shows a semicircle with centre O and diameter PQ. If PA = AB and ∠BOQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.
Solution:
Question 24.
The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°.
Calculate –
i) angle QTR
ii) angle QRP
iii) angle QRS
iv) angle STR
Solution:
Question 25.
In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that:
i) ∠BAP = ∠ADQ
ii) ∠AOB = 2∠ADQ
(iii) ∠ADQ = ∠ADB.
Solution:
Question 26.
AB is a line segment and M is its midpoint. Three semicircles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semicircles. Show that: AB = 6 x r
Solution:
Question 27.
TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.
Solution:
Question 28.
Two circles intersect in points P and Q. A secant passing through P intersects the circle in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
Solution:
Question 29.
Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle)
Solution:
Question 30.
Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm. Calculate the length of CD.
Solution:
Question 31.
In the given figure, find TP if AT = 16 cm and AB = 12 cm.
Solution:
Question 32.
In the following figure, A circle is inscribed in the quadrilateral ABCD.
If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.
Solution:
Question 33.
In the figure, XY is the diameter of the circle, PQ is the tangent to the circle at Y. Given that ∠AXB = 50° and ∠ABX = 70°. Calculate ∠BAY and ∠APY.
Solution:
Question 34.
In the given figure, QAP is the tangent at point A and PBD is a straight line. If ∠ACB = 36° and ∠APB = 42°; find:
i) ∠BAP
ii) ∠ABD
iii) ∠QAD
iv) ∠BCD
Solution:
Question 35.
In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.
If
∠CAB = 34°, find
i) ∠CBA
ii) ∠CQB
Solution:
Question 36.
In the given figure, O is the centre of the circle. The tangets at B and D intersect each other at point P.
If AB is parallel to CD and ∠ABC = 55°, find:
i) ∠BOD
ii) ∠BPD
Solution:
Question 37.
In the figure given below PQ =QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of:
i) ∠QOP
ii) ∠QCP
Solution:
Question 38.
In two concentric circles prove that all chords of the outer circle, which touch the inner circle, are of equal length.
Solution:
Question 39.
In the figure, given below, AC is a transverse common tangent to two circles with centers P and Q and of radii 6 cm and 3 cm respectively.
Given that AB = 8 cm, calculate PQ.
Solution:
Question 40.
In the figure given below, O is the centre of the circum circle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.
Solution:
Question 41.
In the given figure, AE and BC intersect each other at point D. If ∠CDE=90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find AE.
Solution:
Question 42.
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
Solution:
Question 43.
In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.
Solution:
More Resources for Selina Concise Class 10 ICSE Solutions
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# Lesson 13
These materials, when encountered before Algebra 1, Unit 7, Lesson 13 support success in that lesson.
## 13.1: Math Talk: Halved and Squared (5 minutes)
### Warm-up
The purpose of this Math Talk is to elicit strategies and understandings students have for finding a value that needs to be added to a quadratic expression to complete the square. These understandings help students develop fluency and will be helpful later in the associated Algebra 1 lesson when students will need to be able to complete the square.
### Launch
Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.
### Student Facing
For each value of $$b$$, mentally find $$\left(\frac{b}{2} \right)^2$$.
$$b = 6$$
$$b = \frac{1}{2}$$
$$b = \frac{2}{5}$$
$$b = 0.8$$
### Activity Synthesis
Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:
• “Who can restate $$\underline{\hspace{.5in}}$$’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to $$\underline{\hspace{.5in}}$$’s strategy?”
• “Do you agree or disagree? Why?”
## 13.2: Solving Quadratics with Perfect Squares (15 minutes)
### Activity
In this activity, students solve equations of the form $$(x+a)^2 = b$$. In the associated Algebra 1 lesson, students solve equations using completing the square. After they complete the square, students can use the methods from this activity to solve the equation.
### Student Facing
Solve each of these equations for all values of $$x$$ that make the equation true.
1. $$(x+2)^2 = 9$$
2. $$(x-\frac{1}{2})^2 = 4$$
3. $$(x+1)^2 = 8 + 1$$
4. $$(x-\frac{1}{3})^2 = \frac{10}{9}- \frac{1}{9}$$
5. $$(x-6)(x-6) = 81$$
### Activity Synthesis
The purpose of the discussion is to clarify that the methods used in this activity are specific to equations that are in a certain form.
Remind students that this method only works when the equation is of the form $$(x+a)^2 = b$$. For example, these equations would require other methods for solving:
• $$(x+2)(x-5)=0$$ ($$x = \text{-}2, x = 5$$)
• $$(x+1)(x-1) = 24$$ ($$x = 5, x = \text{-}5$$)
• $$(x-3)(x+12) = 9x$$ ($$x = 6, x = \text{-}6$$)
Ask students what methods they could use to solve the equations here.
## 13.3: Make It a Perfect Square (20 minutes)
### Activity
In this activity, students use the patterns they noticed about quadratic expressions in standard form to add a constant value to the quadratic and linear term to make it a perfect square. After adding the value, students rewrite the expression in factored form. In the associated Algebra 1 lesson, students work to complete the square to solve equations. The work of this activity helps students develop some fluency with adding the correct constant as part of the process for completing the square.
### Student Facing
For each expression:
• Find a value that could be added as a constant term to make each expression a perfect square.
• Add the value you found and rewrite the expression in factored form.
1. $$x^2 + 20x$$
2. $$x^2 - 4x$$
3. $$x^2 - 2x$$
4. $$x^2 + x$$
5. $$x^2 + 5x$$
6. $$x^2 + 1.4x$$
### Student Response
• “How did you find the value for the constant you added?” (I know the number must be a perfect square so that twice the square root of the number is the linear coefficient. For example, in the first problem it is 100 since $$\sqrt{100} = 10$$ and $$2 \boldcdot 10 = 20$$.)
• “How do you know that the second expression has factors $$(x-2)^2$$ rather than $$(x+2)^2$$?” (Since the linear term is negative, the factors will involve subtraction.)
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# How do you solve 8(k+2)=1?
Jul 7, 2018
$k = - \frac{15}{8}$
#### Explanation:
Dividing the whole equation by $8$ we get
$k + 2 = \frac{1}{8}$
subtracting $1$
$k = \frac{1}{8} - 2 = \frac{1}{8} - \frac{16}{8} = - \frac{15}{8}$
Jul 7, 2018
$k = - \frac{15}{8}$
#### Explanation:
$\text{divide both sides by 8}$
$k + 2 = \frac{1}{8}$
$\text{subtract 2 from both sides}$
$k = \frac{1}{8} - 2 = \frac{1}{8} - \frac{16}{8} = - \frac{15}{8}$
Jul 8, 2018
$k = - \frac{15}{8}$
#### Explanation:
Let's divide both sides by $8$ to begin isolating the $k$ term. We get
$k + 2 = \frac{1}{8}$
We can next subtract $2$ from both sides to get
$k = \frac{1}{8} - \frac{16}{8} = - \frac{15}{8}$
Thus, $k = - \frac{15}{8}$
Hope this helps!
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## Algebra
Showing posts with label simplify. Show all posts
Showing posts with label simplify. Show all posts
Now when working with square roots and variables we should be a bit careful.
The variable could represent a positive or negative number so we must ensure that it is positive by making use of the absolute value.
To avoid this technicality many textbooks state, at this point, that we assume all variables are positive. If not, use the absolute values as in the following problems.
Simplify. (Assume variables could be negative.)
To avoid many of the technicalities when working with nth roots we will assume, from this point on, that all the variables are positive.
Simplify. (Assume all variables represent positive numbers.)
Students often try to simplify the previous problem. Be sure to understand what makes the following problems all different.
The property:
says that we can simplify when the operation is multiplication. There is no corresponding property for addition or subtraction.
### Properties of the Logarithm
The following properties of the logarithm are derived from the rules of exponents.
The properties that follow below are derived from the fact that the logarithm is defined as the inverse of the corresponding exponential.
To familiarize ourselves with the properties we will first expand the following logarithms. (Assume all variables are positive.)
Expand.
Notice that there is no explicit property that allows us to work with nth roots within the argument of the logarithm. To simplify these, first change the root to a rational exponent then apply the power rule.
When expanding, notice that we must use the same base throughout the expression. For the next set of problems we will first use the properties to expand then substitute in the appropriate values as the last step.
Evaluate
Expanding is useful for learning the rules and properties associated with logarithms but as it turns out, in practice, condensing down to a single logarithm is the skill that we really need to focus on.
Rewrite as a single logarithm (condense).
Tip: When simplifying these down to one logarithm use only one operation at a time and work from left to right. Combining or skipping steps usually leads to mistakes. Do not race, work slowly, and pay attention to notation.
Evaluate (Round to the nearest ten thousandths where appropriate).
Simplify.
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## Parking Lots and Generating Functions
Maple
The June edition of the IBM Ponder This website poses the following puzzle:
Assume that cars have a length of two units and that they are parked along the circumference of a circle whose length is 100 units, which is marked as 100 segments, each one exactly one unit long.
A car can park on any two adjacent free segments (i.e., it does not need any extra maneuvering space).
Our question is as follows: Let's assume that we start with an empty circle. We add one car at a time, and each car parks in a random free space (aligned to a unit length), till no such place exists. What is the expected number of cars that can park along that 100-unit-long circle?
Let C(n) be the expected number of cars that can park in a circular lot with n spaces. If n>1, then at least one car can park, and after doing so there will be n-2 contiguous spaces remaining. Let L(n) be the expected number of cars that can park in n contiguous spaces (not in a circle).
Clearly for n>1,
`(1) C(n)=L(n-2)+1`
Let's now find an expression for L(n). Number the stripes that mark the boundaries of the spaces from 0 to n. Because a car is two spaces long, the middle of a parked car is directly over a stripe. The first car can park on the stripes numbered 1 to n-1. Let the first car park over stripe k. That leaves two contiguous sections, with spaces k-1 and n-1-k. The sum of the expected number of cars that can park in those sections is L(k-1) + L(n-1-k). For the first driver, there is an equal probability, 1/(n-1), of choosing any one of the n-1 available spaces. Consequently,
`(2) L(0) = L(1) = 0 L(n) = 1/(n-1)*sum(1 + L(k-1) + L(n-1-k), k=1..n-1)`
This recurrence can be easily converted to a Maple procedure, that, with the ?remember option, is reasonably efficient and makes short work of this puzzle.
`L := proc (n) local k; option remember; if n < 2 then return 0; else 1/(n-1)*add(1+L(k-1)+L(n-1-k), k = 1 .. n-1); end if end proc:`
For the given problem,
`C(100) = 1 + L(98) = 42.2332`
While the puzzle only required a numerical result, we would like to analyze the general case and find a closed-form expression for L(n). A nice way to do this is with generating functions. Herbert Wilf's GeneratingFunctionology is a pleasant introduction packed with useful and fun examples. The following analysis references the third edition. The author has generously made the second edition available at his website.
The ordinary generating function for the sequence L(0), L(1), ... is the formal power series (FPS)
`(3) lambda(x) = L(0) + L(1)*x + L(2)*x^2 + L(3)*x^3 + ...`
Let's rewrite recurrence (2) using the Iverson convention whereby [boolean] = 1 if boolean is true, 0 otherwise:
` (n-1)*L(n) = sum([n>1] + L(k-1) + L(n-1-k), k=1..n-1) = (n-1)*[n>1] + 2*sum(L(k)=0..n-2)`
Multiply both sides by x^n and sum over n
`(4) sum((n-1)*L(n)*x^n,n) = sum((n-1)*[n>1]*x^n,n) + 2*sum(sum(L(k-1),k=0..n-2),n)`
The left side of (4) can be evaluated using rule 2, section 2 of Wilf. Here we want the ordinary power series form of the sequence given by (n-1)*L(n). As explained in the book, this can be computed by substituting the operator x*D for the n in the polynomial and applying that to L(n). That is,
` sum((n-1)*L(n)*x^n,n) (x.D-Id)(lambda(x)) = x*lambda'(x) - lambda(x)`
The first term of the right side of (4) can be evaluated directly with Maple
` sum((n-1)*x^n,n=1..infinity) = x^2/(1-x)^2`
To evaluate the double summation in (4), recast the sum over k as a sum from 0 to n,
` sum(L(k-1),k=1..n-1) = sum(L(k),k=0..n) - L(n-1) - L(n)`
Multiplying by x^n, summing over n, and then applying rule 3, section 2 of Wilf gives an expression for the double summation
` lambda(x)*(1/(1-x) - x - 1)`
Plugging these results into (4) gives the differential equation
`(5) deqs := x*lambda'(x)*x^n + 1/n*lambda(x)*x^n - lambda(x)*x^n = 1/n*x^n - x^n + 2*lambda(x)*(1/(1-x)-x-1)`
From the definition of lambda(x), lambda(0) = L(0) = 0 and lambda'(0) = L(1) = 0, so
`ics := ( lambda(0) = 0, lambda'(0) = 0 ):`
Maple's dsolve procedure returns the following solution
`dsol := dsolve({deq, ics}); lambda(x) = (1-exp(-2*x))*x/(1-x)^2/2`
To check that it is correct, use series to compute several coefficients and compare them with a direct computation using our procedure (above).
` series(rhs(dsol),x,10); x^2+x^3+(5/3)*x^4+2*x^5+(37/15)*x^6+(26/9)*x^7+O(x^8)`
Now that we have a closed-form for the generating function, we can, with a bit of further analysis, extract an expression for L(n).
Move the x/2 term to the left side of the equation to get
` 2/x*lambda(x) = (1-exp(-2*x))/(1-x)^2`
Compute the FPS of each of the two terms of the product on the right (that can be done using convert(., Sum, 'dummy'=n):
` 2/x*lambda(x) = (1 - Sum((-2*x)^n/n!,n>=0))*Sum((n+1)*x^n,n>=0) = Sum((n+1)*x^n,n>=0) - Sum((-2*x)^n/n!,n>=0))*Sum((n+1)*x^n,n>=0)`
Compute the coefficient of x^n on each side of the equation
` 2*L(n+1) = (n+1) - Sum((n+1-k)*(-2)^k/k!,k=0..n)`
Substitute n -> n-1 and move the factor of 2 back to the right
` L(n) = 1/2*(n - Sum((n-k)*(-2)^k/k!,k=0..n-1))`
Maple can evaluate the sum,
` L(n) = 1/2*(n-(n+2)*exp(-2) +(-1)^n*2^((1/2)*n)*exp(-1)*WhittakerM(1-(1/2)*n,(1/2)*n+1/2,2)/(n+1)!)`
While the rhs is messy, numerical evaluation reveals that the final term quickly goes to 0 as n increases. The result is
` L(n) ~ Lx(n) = 1/2*(n-(n+2)*exp(-2))`
Using this to compute C(100) gives
` C(100) ~ 1 + Lx(98) = 42.2332`
which agrees with our exact computation.
From the result, we see that the expected utilization (ratio of expected to maximum) of a linear parking lot quickly reaches
` limit(L(n)/(n/2),n=infinity) = 1-exp(-2) = 0.865.`
With that modest success, the natural inclination is to generalize the problem until we arrive at one we cannot solve. Alas, that doesn't take long. Suppose each car takes up three spaces. While the same approach should theoretically work, I could not get Maple to symbolically solve the resulting differential equation. Is there a way to continue without it?
A more interesting generalization is to remove the stripes and allow cars to park anywhere. Instead of a difference equation we get a differential equation, specifically, a delay-differential equation. An analytical solution looks hopeless, though for a given size lot it can be solved numerically by integrating a piece at a time. Can we find the expected utilization for large n?
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# The Sum of Two Non-zero Numbers is 8, the Minimum Value of the Sum of the Reciprocals is (A) 1 4 (B) 1 2 (C) 1 8 (D) None of These - CBSE (Arts) Class 12 - Mathematics
#### Question
The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is ______________ .
• $\frac{1}{4}$
• $\frac{1}{2}$
• $\frac{1}{8}$
• none of these
#### Solution
$\frac{1}{2}$
$\text { Let the two non - zero numbers be x and y . Then,}$
$x + y = 8$
$\Rightarrow y = 8 - x ............\left( 1 \right)$
$\text { Now,}$
$f\left( x \right) = \frac{1}{x} + \frac{1}{y}$
$\Rightarrow f\left( x \right) = \frac{1}{x} + \frac{1}{8 - x} ..................\left[ \text { From eq. } \left( 1 \right) \right]$
$\Rightarrow f'\left( x \right) = \frac{- 1}{x^2} + \frac{1}{\left( 8 - x \right)^2}$
$\text { For a local minima or a local maxima, we must have }$
$f'\left( x \right) = 0$
$\Rightarrow \frac{- 1}{x^2} + \frac{1}{\left( 8 - x \right)^2} = 0$
$\Rightarrow \frac{- \left( 8 - x \right)^2 + x^2}{\left( x \right)^2 \left( 8 - x \right)^2} = 0$
$\Rightarrow - 64 - x^2 + 16x + x^2 = 0$
$\Rightarrow 16x - 64 = 0$
$\Rightarrow x = 4$
$f''\left( x \right) = \frac{2}{x^3} - \frac{2}{\left( 8 - x \right)^3}$
$\Rightarrow f''\left( 4 \right) = \frac{2}{4^3} - \frac{2}{\left( 8 - 4 \right)^3}$
$\Rightarrow f''\left( 4 \right) = \frac{2}{64} - \frac{2}{64} = 0$
$\therefore \text { Minimum value }= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution The Sum of Two Non-zero Numbers is 8, the Minimum Value of the Sum of the Reciprocals is (A) 1 4 (B) 1 2 (C) 1 8 (D) None of These Concept: Graph of Maxima and Minima.
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# Differentiation of Infinite Series Class 12
Here you will learn what is differentiation of infinite series class 12 with examples.
Let’s begin –
## Differentiation of Infinite Series
Sometimes the value of y is given as an infinite series and we are asked to find $$dy\over dx$$. In such cases we use the fact that if a term is deleted from a infinite series, it remains unaffected. The method of finding $$dy\over dx$$ is explained in the following examples.
Example 1 : If y = $$x^{x^{x^{…\infty}}}$$, find $$dy\over dx$$.
Solution : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as
y = $$x^y$$
Taking log on both sides,
$$\implies$$ log y = y logx
Differentiating both sides with respect to x,
$$1\over y$$$$dy\over dx$$ = $$dy\over dx$$ log x + y $$d\over dx$$ (log x)
$$1\over y$$$$dy\over dx$$ = $$dy\over dx$$ log x + $$y\over x$$
$$dy\over dx$${$${{1\over y} – log x}$$} = $$y\over x$$
$$\implies$$ $$dy\over dx$$$$(1 – y log x)\over y$$ = $$y\over x$$
$$\implies$$ $$dy\over dx$$ = $$y^2\over {x(1 – ylog x)}$$
Example 2 : If y = $$\sqrt{sinx + \sqrt{sinx + \sqrt{sinx + ……. to \infty}}}$$, find $$dy\over dx$$.
Solution : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as
y = $$\sqrt{sin x + y}$$
Squaring on both sides,
$$\implies$$ $$y^2$$ = sin x + y
Differentiating both sides with respect to x,
2y $$dy\over dx$$ =cosx + $$dy\over dx$$
$$\implies$$ $$dy\over dx$$$$(2y – 1)$$ = cos x
$$\implies$$ $$dy\over dx$$ = $$cos x\over {2y – 1}$$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts Go to the latest version.
# 13.3: Permutations
Difficulty Level: At Grade Created by: CK-12
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Practice Permutations
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What if you were picked to judge the pie contest at your local fair? You can pick three to move on to the finals from 10 entries. How many ways could you pick your three favorites? After completing this Concept, you'll be able to calculate permutations like this one using factorials.
### Guidance
In this lesson we’ll be looking at ways of arranging things. To illustrate what we mean by this, let’s look at a simple example. Think about choosing your favorite color from the following list of choices; red, blue, green, yellow, pink, purple, orange, brown, black. Clearly there are nine different colors, so there are nine possible choices you could make.
Now think about choosing your top three colors in order of preference. There are many different ways you can choose the top three. You might choose red as your favorite, followed by black, then green. Someone else might choose the same three colors as you, but in a different order! When you are choosing items from a list and the order in which you choose them is important, the arrangement is called a permutation. How many different permutations do you think there are in this situation?
In this lesson, we’ll use counting methods to determine how many permutations a given situation has. We’ll also discover a formula to calculate permutations when counting alone is impractical.
Counting Permutations
In simple cases, sometimes it’s easiest to calculate permutations by just listing all the possibilities and counting them. Let’s examine a situation where it is relatively straightforward to do that.
#### Example A
Nadia and Peter are going to watch two movies on a rainy Saturday. Nadia will choose the first movie, and Peter gets to choose the second. The four movies they have to choose from are The Lion King, Aladdin, Toy Story and Pinocchio. Given that Peter will choose a different movie than Nadia, how many permutations are there for the movies they watch?
Solution
Since the order in which they watch the movies is important, and they don’t plan to choose the same movie twice, we can list all the different possibilities in a table:
First Movie Second Movie
Lion King Toy Story
Lion King Pinocchio
Toy Story Lion King
Toy Story Pinocchio
Pinocchio Lion King
Pinocchio Toy Story
You can see that this table contains all the possibilities for the situation. There are four movies for Nadia to choose from. For every movie that Nadia chooses first, Peter has three choices left for his movie. By simply counting the rows in the table you can see that there are 12 permutations in this situation.
#### Example B
I have 5 cards with the numbers 1 through 5 on them. I take three cards and arrange them to form a 3-digit number. How many 3-digit numbers can I make?
Since the numbers we can make fit a numerical ordering pattern, we can list the possibilities in increasing order:
123 124 125 132 134 135 142 143 145 152 153 154213 214 215 231 234 235 241 243 245 251 253 254312 314 315 321 324 325 341 342 345 351 352 354412 413 415 421 423 425 431 432 435 451 452 453512 513 514 521 523 524 531 532 534 541 542 543
By arranging the table this way, you can see how the number of remaining choices decreases as we choose numbers. There are five choices for the first number, four choices for the second number and three choices for the third number. Counting the table entries gives a total of 60 permutations.
If we look closely at the last two examples we can see a pattern start to appear. Mathematicians love patterns—they tend to lead to formulas, which make life much easier! After all, why spend hours counting possibilities when a formula can calculate them in seconds?
In example A, Nadia had four choices and Peter had three. The number of permutations was 4×3=12.\begin{align*}4 \times 3 = 12.\end{align*}
In example B there were 5 choices for the first digit, followed by 4 for the second digit, and then 3 for the third digit. The total number of permutations was 5×4×3=60\begin{align*}5 \times 4 \times 3 = 60\end{align*}.
In the introduction, we thought about picking our top three choices from a list of nine colors. You should now be able to do that. Even without listing all the possibilities, you can see that you have 9 choices for your favorite, 8 choices for your second favorite, and 7 choices for your third. The number of permutations is thus 9×8×7=504\begin{align*}9 \times 8 \times 7 = 504\end{align*}.
Factorial Notation
Look again at the color list in the introduction, and think this time about writing down every color in order of preference. You would have 9 choices for your favorite, followed by 8 choices for your second favorite, then 7, then 6, then 5, and so on. To determine the number of permutations for any possible list, we would perform the following calculation:
Color Permutations=9×8×7×6×5×4×3×2×1\begin{align*}\text{Color Permutations} = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\end{align*}
This sort of pattern crops up a great deal in statistics, probability and number theory. It’s so common that it has its own notation: 4×3×2×1\begin{align*}4 \times 3 \times 2 \times 1\end{align*} is written as 4! and is called four factorial. So the number of color permutations above is nine factorial = 9! = 362,880.
So what happens when we only want the first few terms in a factorial? For example, the number of permutations for arranging ALL the colors is 362,880, but the number of permutation for the top three is 504.
One way to get this result is to divide one factorial by another. Look at nine factorial divided by six factorial:
9!6!=9×8×7×6×5×4×3×2×16×5×4×3×2×1=9×8×7=504
The terms in six factorial cancel out all but the first three terms in nine factorial. You should see that if we wanted the first four terms we would divide by 5!, or for only the first two terms we would divide by 7!. In general, however many terms we want to keep, we divide by the factorial of the quantity:
(number of items in list)(number of items we are choosing)
So to get the first five terms in twelve factorial we would use the formula 12!(125)!=12!7!\begin{align*}\frac{12!}{(12-5)!} = \frac{12!}{7!}\end{align*}.
Formulas like this are useful if you have a calculator that can handle factorials: you can just type in 12!7!\begin{align*}\frac{12!}{7!}\end{align*} instead of 12×11×10×9×8\begin{align*}12 \times 11 \times 10 \times 9 \times 8\end{align*}. However, some factorials are too big for some calculators to handle, so in those cases you would need to simplify the fraction and do the multiplication by hand.
#### Example C
How many ways can Dale choose his favorite 5 songs from the current Billboard Hot 100TM\begin{align*}100^{TM}\end{align*}?
Solution
To find the answer, consider how many choices he has at each stage. For his first choice he has 100 songs to choose from, then 99, then 98 and so on. We need the first 5 terms only, so our calculation is:
Permutations=100!(1005)!=100!95!=100×99×98×97×96=9,034,502,400
Notice that that’s a pretty big number – far too large to count in a table! This is why we need formulas to help us count permutations.
Finding Permutations Using a Formula
We’ve just seen that a formula for determining the number of permutations for choosing 3 objects from a list of 9 objects is:
9!6!=9×8×7×6×5×4×3×2×16×5×4×3×2×1=9×8×7=504 permutations
Now we’re ready to come up with a general formula for determining permutations. When we are choosing \begin{align*}r\end{align*} ordered items from a group of \begin{align*}n\end{align*} items, the number of permutations is given by the first \begin{align*}r\end{align*} terms in \begin{align*}n!\end{align*} We use the notation \begin{align*}_nP_r\end{align*} for this, and the general form for calculating permutations is:
#### Example D
How many ways are there to choose a 5-song mix from a CD containing 12 tracks?
Solution
Choosing 5 from 12: \begin{align*}_{12}P_5 =\frac{12!}{(12-5)!}=\frac{12!}{7!} =12 \times 11 \times 10 \times 9 \times 8 = 95,040 \ ways\end{align*}
Watch this video for help with the Examples above.
### Vocabulary
• A permutation is when we are choosing \begin{align*}r\end{align*} ordered items from a group of \begin{align*}n\end{align*} items, where the order chosen matters. The number of permutations is given by the formula:
### Guided Practice
How many 3-letter “words” can be made from the letters in “computer”? (The words do NOT need to be real, or even pronounceable – for example “rtp” would count as a word)
Solution
Choosing 3 from 8: \begin{align*}_8P_3 =\frac{8!}{(8-3)!}=\frac{8!}{5!} =8 \times 7 \times 6 = 336 \ words\end{align*}
### Practice
1. In how many ways can the letters \begin{align*}a, b, c, d, e\end{align*} be arranged?
2. In how many ways can the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 be arranged?
3. From a collection of 12 books, 5 are to be selected and placed in a particular order on a shelf. How many arrangements are possible?
4. 3 cards are taken at random from a deck of 52 cards and laid in a row. How many possible outcomes are there for the card arrangements?
5. How many distinct 3-letter permutations can you make from the letters in the word HEXAGON?
6. How many distinct 2-letter permutations can you make from the letters in the word GEESE?
7. A jukebox has 50 songs on it. If \$1.00 pays for three songs, how many permutations are there for choosing 3 different songs?
For problems 8-16, evaluate the following:
1. \begin{align*}_3P_1\end{align*}
2. \begin{align*}_7P_1\end{align*}
3. \begin{align*}_6P_2\end{align*}
4. \begin{align*}_8P_8\end{align*}
5. \begin{align*}_9P_3\end{align*}
6. \begin{align*}_7P_3\end{align*}
7. \begin{align*}_{19}P_7\end{align*}
8. \begin{align*}_{99}P_3\end{align*}
9. \begin{align*}_3P_0\end{align*}
### Vocabulary Language: English
Permutation
Permutation
A permutation is an arrangement of objects where order is important.
Oct 01, 2012
Dec 17, 2015
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### Angle Bisector Theorem
An angle bisector cut an angle precisely in half. One essential property of angle bisectors is that if a point is top top the bisector of one angle, climate the point is equidistant from the political parties of the angle. This is referred to as the Angle Bisector Theorem.
You are watching: Converse of the angle bisector theorem
In other words, if \(\overrightarrowBD\) bisects \(\angle ABC\), \(\overrightarrowBA\perp FD\overlineAB\), and, \(\overrightarrowBC\perp \overlineDG\) then \(FD=DG\).
Figure \(\PageIndex1\)
The converse that this organize is also true.
Angle Bisector organize Converse: If a point is in the inner of one angle and equidistant native the sides, climate it lies on the bisector of that angle.
When we construct angle bisectors because that the angle of a triangle, they fulfill in one point. This suggest is referred to as the incenter the the triangle.
api/deki/files/1537/f-d_d776169a3e60b6e3693852bb41b6a4c30b5f44e084864f212a7d717e%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450" />Figure \(\PageIndex3\)
Solution
No since \(B\) is no necessarily equidistant indigenous \(\overlineAC\) and also \(\overlineAD\). We execute not understand if the angles in the diagram are right angles.
Example \(\PageIndex2\)
A \(108^\circ\) edge is bisected. What space the measures of the result angles?
Solution
We recognize that to bisect way to cut in half, so every of the resulting angles will be fifty percent of 108. The measure up of each resulting edge is \(54^\circ\).
Example \(\PageIndex3\)
Is \(Y\) top top the edge bisector of \(\angle XWZ\)?
Example \(\PageIndex5\)
\(\overrightarrowAB\) is the angle bisector the \(\angle CAD\). Fix for the missing variable.
See more: What Is The Meaning Of Black Snake Crossing Your Path ? What'S The Meaning Of Snake Crossing Your Path
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# Question Video: Solving a Linear Equation in One Unknown Mathematics • 7th Grade
Find the value of 𝑥 if 𝑥 + 1/2 = 1/5.
02:17
### Video Transcript
Find the value of 𝑥 if 𝑥 plus one-half equals one-fifth.
In this question, we are given an equation involving an unknown 𝑥 and we are asked to find the value of this unknown. To answer this question, we can start by recalling that in an equation, both sides of the equation are equal. So, in this case, we have that 𝑥 plus one-half is equal to one-fifth. If we apply the same operations to both sides of the equation, they will remain equal. This means we can try to isolate 𝑥 on one side of the equation to solve for 𝑥. To do this, we can start by subtracting one-half from both sides of the equation. This gives us 𝑥 plus one-half minus one-half is equal to one-fifth minus one-half. We can then calculate that one-half minus one-half is equal to zero. This allows us to rewrite the equation as 𝑥 equals one-fifth minus one-half.
We can simplify the right-hand side of the equation by evaluating the subtraction. To do this, the denominators of both fractions must be the same. The lowest common multiple of five and two is 10. So we rewrite both fractions to have a denominator of 10. We can rewrite one-fifth as two over 10 and one-half as five over 10. We can then evaluate the subtraction by subtracting the numerators. We obtain two minus five all over 10. We can calculate that this is equal to negative three over 10.
We can verify that this is a solution to the equation by substituting this value back into the equation. If we did this, we would find that negative three-tenths plus one-half is equal to one-fifth, verifying that the value of 𝑥 is negative three over 10.
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# Thread: |2x-1|-|x|< 4
1. ## |2x-1|-|x|< 4
How do I solve this |2x-1|-|x|< 4 ? Do i simply treat it like an equation, meaning that it has three different solutions? or do other rules apply?
2. Use cases.
$\displaystyle |2x-1|-|x|<4$
There are three cases,
i. $\displaystyle x\ge\frac{1}{2}$
ii. $\displaystyle 0 < x < \frac{1}{2}$
iii. $\displaystyle x \le 0$
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i. $\displaystyle x\ge\frac{1}{2}$
The inequality becomes,
$\displaystyle 2x - 1 - x < 4$
$\displaystyle x < 5$
The interval here is, $\displaystyle \frac{1}{2}\le x < 5$
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ii. $\displaystyle 0 < x < \frac{1}{2}$
$\displaystyle -(2x-1) - x < 4$
$\displaystyle x > -1$
Interval: $\displaystyle 0 < x < \frac{1}{2}$
-------------
iii. $\displaystyle x \le 0$
$\displaystyle -(2x -1)-(-x)<4$
Interval: $\displaystyle -3 < x \le 0$
-------------
-------------
When we put the intervals together,
$\displaystyle \frac{1}{2}\le x < 5$
$\displaystyle 0 < x < \frac{1}{2}$
$\displaystyle -3 < x \le 0$
It makes,
$\displaystyle -3<x<5$
3. the first element is positive if x > 0.5 and negative otherwise, the second element is positive if x > 0 and negative otherwise.
1. x => 0.5, in this range both elements are non negative so:
2x-1 - x < 4
x < 5
so the solution is: 0.5 <= x < 5
2. 0 <= x <= 0.5 in this range the first element is negative and the second is non negative so:
-(2x-1) - x < 4
x > -1
thus the answer is 0 <= x <= 0.5
3. x <= 0 in this range the first element is negative and the second is non positive so:
-(2x-1) + x < 4
x > -3
thus the solution is -3 < x <= 0
The final solution is the intersection of these 3 solutions thus:
-3 < x < 5
4. Originally Posted by weasley74
How do I solve this |2x-1|-|x|< 4 ? Do i simply treat it like an equation, meaning that it has three different solutions? or do other rules apply?
If you like - or allowed to - you can solve it graphically.
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Difficulty Level: At Grade Created by: CK-12
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What if you had a radical equation like 2x+53=0\begin{align*}\sqrt{2x + 5} - 3 = 0\end{align*}? How could you find its real solutions? After completing this Concept, you'll be able to solve radical equations like this one.
### Guidance
When the variable in an equation appears inside a radical sign, the equation is called a radical equation. To solve a radical equation, we need to eliminate the radical and change the equation into a polynomial equation.
A common method for solving radical equations is to isolate the most complicated radical on one side of the equation and raise both sides of the equation to the power that will eliminate the radical sign. If there are any radicals left in the equation after simplifying, we can repeat this procedure until all radical signs are gone. Once the equation is changed into a polynomial equation, we can solve it with the methods we already know.
We must be careful when we use this method, because whenever we raise an equation to a power, we could introduce false solutions that are not in fact solutions to the original problem. These are called extraneous solutions. In order to make sure we get the correct solutions, we must always check all solutions in the original radical equation.
Let’s consider a few simple examples of radical equations where only one radical appears in the equation.
#### Example A
Find the real solutions of the equation 2x1=5\begin{align*}\sqrt{2x-1}=5\end{align*}.
Solution
Since the radical expression is already isolated, we can just square both sides of the equation in order to eliminate the radical sign:
(2x1)2=52\begin{align*}\left(\sqrt{2x-1}\right)^2=5^2\end{align*}
Remember that a2=a so the equation simplifies to:Add one to both sides:Divide both sides by 2:2x12x=25=26x=13\begin{align*}\text{Remember that} \ \sqrt{a^2}=a \ \text{so the equation simplifies to:} && 2x-1& =25\\ \text{Add one to both sides:} && 2x& =26\\ \text{Divide both sides by 2:} &&& \underline{\underline{x=13}}\end{align*}
Finally we need to plug the solution in the original equation to see if it is a valid solution.
2x1=2(13)1=261=25=5\begin{align*}\sqrt{2x-1}=\sqrt{2(13)-1}=\sqrt{26-1}=\sqrt{25}=5\end{align*} The solution checks out.
#### Example B
Find the real solutions of 37x33=0\begin{align*}\sqrt[3]{3-7x}-3=0\end{align*}.
Solution
We isolate the radical on one side of the equation:Raise each side of the equation to the third power:Simplify:Subtract 3 from each side:Divide both sides by –7:37x3(37x3)337x7x=3=33=27=24x=247\begin{align*}\text{We isolate the radical on one side of the equation:} && \sqrt[3]{3-7x}& =3\\ \text{Raise each side of the equation to the third power:} && \left(\sqrt[3]{3-7x}\right)^3& =3^3\\ \text{Simplify:} && 3-7x& =27\\ \text{Subtract 3 from each side:} && -7x& =24\\ \text{Divide both sides by –7:} &&& \underline{\underline{x=-\frac{24}{7}}}\end{align*}
Check: 37x33=37(247)33=3+2433=2733=33=0\begin{align*}\sqrt[3]{3-7x}-3=\sqrt[3]{3-7 \left(-\frac{24}{7}\right)}-3=\sqrt[3]{3+24}-3=\sqrt[3]{27}-3=3-3=0\end{align*}. The solution checks out.
#### Example C
Find the real solutions of 10x2x=2\begin{align*}\sqrt{10-x^2}-x=2\end{align*}.
Solution
We isolate the radical on one side of the equation:Square each side of the equation:Simplify:Move all terms to one side of the equation:Solve using the quadratic formula:Simplify:Re-write 24 in simplest form:Reduce all terms by a factor of 2:10x2(10x2)210x20xxxx=2+x=(2+x)2=4+4x+x2=2x2+4x6=4±424(2)(6)4=4±644=4±84=1 or x=3\begin{align*}\text{We isolate the radical on one side of the equation:} && \sqrt{10-x^2}& =2+x\\ \text{Square each side of the equation:} && \left(\sqrt{10-x^2}\right)^2& =(2+x)^2\\ \text{Simplify:} && 10-x^2& =4+4x+x^2\\ \text{Move all terms to one side of the equation:} && 0& =2x^2+4x-6\\ \text{Solve using the quadratic formula:} && x& =\frac{-4 \pm \sqrt{4^2-4(2)(-6)}}{4}\\ \text{Simplify:} && x& =\frac{-4 \pm \sqrt{64}}{4}\\ \text{Re-write} \ \sqrt{24} \ \text{in simplest form:} && x& =\frac{-4 \pm 8}{4}\\ \text{Reduce all terms by a factor of 2:} && x& =1 \ \text{or} \ x=-3\end{align*}
Check: 10121=91=31=2\begin{align*}\sqrt{10-1^2}-1=\sqrt{9}-1=3-1=2\end{align*} This solution checks out.
10(3)2(3)=1+3=1+3=4\begin{align*}\sqrt{10-(-3)^2}-(-3)=\sqrt{1}+3=1+3=4\end{align*} This solution does not check out.
The equation has only one solution, x=1\begin{align*}\underline{\underline{x=1}}\end{align*}; the solution x=3\begin{align*}x=-3\end{align*} is extraneous.
#### Example D
A sphere has a volume of 456 cm3\begin{align*}456 \ cm^3\end{align*}. If the radius of the sphere is increased by 2 cm, what is the new volume of the sphere?
Solution
Make a sketch:
Define variables: Let R=\begin{align*}R =\end{align*} the radius of the sphere.
Find an equation: The volume of a sphere is given by the formula V=43πR3\begin{align*}V=\frac{4}{3}\pi R^3\end{align*}.
Solve the equation:
Plug in the value of the volume:Multiply by 3:Divide by 4π:Take the cube root of each side:The new radius is 2 centimeters more:The new volume is:4561368108.92RRV=43πR3=4πR3=R3=108.923R=4.776 cm=6.776 cm=43π(6.776)3=1302.5 cm3\begin{align*}\text{Plug in the value of the volume:} && 456& =\frac{4}{3} \pi R^3\\ \text{Multiply by 3:} && 1368& =4 \pi R^3\\ \text{Divide by} \ 4 \pi: && 108.92& =R^3\\ \text{Take the cube root of each side:} && R& =\sqrt[3]{108.92} \Rightarrow R=4.776 \ cm\\ \text{The new radius is 2 centimeters more:} && R& =6.776 \ cm\\ \text{The new volume is:} && V & =\frac{4}{3} \pi (6.776)^3=\underline{\underline{1302.5}} \ cm^3\end{align*}
Check: Let’s plug in the values of the radius into the volume formula:
V=43πR3=43π(4.776)3=456 cm3\begin{align*}V=\frac{4}{3} \pi R^3=\frac{4}{3} \pi (4.776)^3=456 \ cm^3\end{align*}. The solution checks out.
#### Example E
The kinetic energy of an object of mass m\begin{align*}m\end{align*} and velocity v\begin{align*}v\end{align*} is given by the formula: KE=12mv2\begin{align*}KE=\frac{1}{2} mv^2\end{align*}. A baseball has a mass of 145 kg and its kinetic energy is measured to be 654 Joules (kgm2/s2)\begin{align*}(kg \cdot m^2/s^2)\end{align*} when it hits the catcher’s glove. What is the velocity of the ball when it hits the catcher’s glove?
Solution
Start with the formula:Plug in the values for the mass and the kinetic energy:Multiply both sides by 2:Divide both sides by 145 kg:Take the square root of both sides:KE654kgm2s21308kgm2s29.02m2s2v=12mv2=12(145 kg)v2=145 kgv2=v2=9.02m2s2=3.003 m/s\begin{align*}\text{Start with the formula:} && KE& =\frac{1}{2} mv^2\\ \text{Plug in the values for the mass and the kinetic energy:} && 654 \frac{kg \cdot m^2}{s^2}& =\frac{1}{2}(145\ kg)v^2\\ \text{Multiply both sides by 2:} && 1308 \frac{kg \cdot m^2}{s^2}& =145 \ kg \cdot v^2\\ \text{Divide both sides by 145} \ kg: && 9.02 \frac{m^2}{s^2}& =v^2\\ \text{Take the square root of both sides:} && v& =\sqrt{9.02} \sqrt{\frac{m^2}{s^2}}=3.003 \ m/s\end{align*}
Check: Plug the values for the mass and the velocity into the energy formula:
KE=12mv2=12(145 kg)(3.003 m/s)2=654 kgm2/s2\begin{align*}KE=\frac{1}{2}mv^2=\frac{1}{2}(145 \ kg)(3.003 \ m/s)^2=654 \ kg \cdot m^2/s^2\end{align*}
Watch this video for help with the Examples above.
### Vocabulary
• For a quadratic equation in standard form, ax2+bx+c=0\begin{align*}ax^2 + bx + c = 0\end{align*}, the quadratic formula looks like this:
x=b±b24ac2a\begin{align*}x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\end{align*}
### Guided Practice
Find the real solutions of \begin{align*}\sqrt{3x-9}-1=5\end{align*}.
Solution
\begin{align*}\text{We isolate the radical on one side of the equation:} && \sqrt{3x-9}&=6\\ \text{Square each side of the equation:} && \sqrt{3x-9}^2&=6^2\\ \text{Simplify:} && 3x-9&=36\\ \text{Add 9 from each side:} && 3x&=45\\ \text{Divide both sides by 3:} &&& \underline{\underline{x=\frac{45}{3}=15}}\end{align*}
Check: \begin{align*}\sqrt{3x-9}-1=\sqrt{3(15)-9}-1=\sqrt{45-9}-1=\sqrt{36}-1=6-1=5\end{align*}. The solution checks out.
### Practice
Find the solution to each of the following radical equations.
1. \begin{align*}\sqrt{x+2}-2=0\end{align*}
2. \begin{align*}\sqrt{3x-1}=5\end{align*}
3. \begin{align*}2 \sqrt{4-3x}+3=0\end{align*}
4. \begin{align*}\sqrt[3]{x-3}=1\end{align*}
5. \begin{align*}\sqrt[4]{x^2-9}=2\end{align*}
6. \begin{align*}\sqrt[3]{-2-5x}+3=0\end{align*}
7. \begin{align*}\sqrt{x^2-3}=x-1\end{align*}
8. \begin{align*}\sqrt{x}=x-6\end{align*}
9. \begin{align*}\sqrt{x^2-5x}-6=0\end{align*}
10. \begin{align*}\sqrt{(x+1)(x-3)}=x\end{align*}
11. \begin{align*}\sqrt{x+6}=x+4\end{align*}
12. \begin{align*}\sqrt{3x+4}=-6\end{align*}
13. The area of a triangle is \begin{align*}24 \ in^2\end{align*} and the height of the triangle is twice as long as the base. What are the base and the height of the triangle?
14. The length of a rectangle is 7 meters less than twice its width, and its area is \begin{align*}660 \ m^2\end{align*}. What are the length and width of the rectangle?
15. The area of a circular disk is \begin{align*}124 \ in^2\end{align*}. What is the circumference of the disk? \begin{align*}(\text{Area} = \pi R^2, \text{Circumference} =2 \pi R)\end{align*}.
16. The volume of a cylinder is \begin{align*}245 \ cm^3\end{align*} and the height of the cylinder is one third of the diameter of the base of the cylinder. The diameter of the cylinder is kept the same but the height of the cylinder is increased by 2 centimeters. What is the volume of the new cylinder? \begin{align*}(\text{Volume} =\pi R^2 \cdot h)\end{align*}
17. The height of a golf ball as it travels through the air is given by the equation \begin{align*}h=-16t^2+256\end{align*}. Find the time when the ball is at a height of 120 feet.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
Extraneous Solution
An extraneous solution is a solution of a simplified version of an original equation that, when checked in the original equation, is not actually a solution.
A quadratic equation is an equation that can be written in the form $=ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real constants and $a\ne 0$.
The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
A radical expression is an expression with numbers, operations and radicals in it.
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# Principles of Mathematical Induction MCQ Quiz in తెలుగు - Objective Question with Answer for Principles of Mathematical Induction - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 24, 2024
పొందండి Principles of Mathematical Induction సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్లోడ్ చేసుకోండి Principles of Mathematical Induction MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.
## Top Principles of Mathematical Induction MCQ Objective Questions
#### Principles of Mathematical Induction Question 1:
For every natural number k, which of the following is true?
1. (mn)= mknk
2. mk2 = n + 1
3. (m+n)k = k + 1
4. mkn = mnk
#### Answer (Detailed Solution Below)
Option 1 : (mn)= mknk
#### Principles of Mathematical Induction Question 1 Detailed Solution
Concepts:
Suppose there is a given statement P (n) involving the natural number n such that
1. The statement is true for n = 1, i.e., P (1) is true, and
2. If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., the truth of P (k) implies the truth of P (k + 1).
Then, P (n) is true for all natural numbers n
Calculation:
Given:
For, k = 1
Option 1,
(mn)1 = m1n1, mn = mn (LHS = RHS) hence it is true
Let us assume that the statement is true for, k = p
(mn)p = mpnp
Multiplying the above equation with mn, we get mn we get,
(mn)p + 1 = mp +1 np + 1
mp + 1 np + 1 = mp + 1 np + 1
Hence the given expression is true for every natural number k.
Option 2, 3 and 4 does not satisfy for k = 1, Hence Option 1 is correct.
#### Principles of Mathematical Induction Question 2:
Let P (n) be the statement 3n > nn. If P (n) is true, then P (n + 1) is
1. True
2. False
3. Not determined
4. None of the above
Option 1 : True
#### Principles of Mathematical Induction Question 2 Detailed Solution
Concept:
Suppose there is a given statement P (n) involving the natural number n such that
• The statement is true for n = 1, i.e., P(1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P (k) implies the truth of P (k + 1).
Then, P (n) is true for all natural numbers n
Calculation:
Given: P (n) is true
For n = 1
P (1): 31 > 11
⇒ 3 > 1
∴ It is true for n = 1
It is given that P (k) is true, so P (k) is true for n = k
P (K) ⇒ 3k > kk
Now,
P (k +1) = 3k+1 > (k + 1) k+1
3k.31 > (k + 1) k (k + 1)
By the principle for mathematics production,
P (n + 1) is true, when P (n) is true
∴ P (n + 1) is true.
#### Principles of Mathematical Induction Question 3:
The statement p ∨ q has the truth value F (False)
1. Whenever both p and q have the truth value T
2. Whenever either p have the truth value T
3. Whenever either q have the truth value T
4. Both have the truth value F
#### Answer (Detailed Solution Below)
Option 4 : Both have the truth value F
#### Principles of Mathematical Induction Question 3 Detailed Solution
Concept:
• Disjunction: If two simple statements p and q are connected by the word ‘or’, then the resulting compound statement “p or q” is called disjunction of p and q and is written in symbolic form as “p ∨ q”.
• The statement p ∨ q has the truth value F whenever both p and q have the truth value F.
• The statement p ∨ q has the truth value T whenever either p or q or both have the truth value T.
p q p ∨ q T T T T F T F T T F F F
∴ Option 4 is correct.
#### Principles of Mathematical Induction Question 4:
Which of the following steps is mandatory in the principle of mathematical induction?
1. Inductive reference
2. Inductive hypothesis
3. Minimal set representation
4. More than one of the above
5. None of the above
#### Answer (Detailed Solution Below)
Option 2 : Inductive hypothesis
#### Principles of Mathematical Induction Question 4 Detailed Solution
Concept:
Principle of Mathematical Induction: Let P(n) be the statement.
Base Case: P(n) is true for n = 1.
Inductive Hypothesis: Let P(n) is true for n = k then
Inductive Step: if P(n) is true for n = k + 1.
⇒ P(n) is true for all natural number n.
Thus, Option 2 is the correct answer.
#### Principles of Mathematical Induction Question 5:
For all n ϵ N, $$(1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2n~+~1)}{n^2}))$$ is equal to
1. $$\frac{(n~+~1)^2}{2}$$
2. $$\frac{(n~+~1)^3}{3}$$
3. $$(n+1)^2$$
4. None of these
5. Not Attempted
#### Answer (Detailed Solution Below)
Option 3 : $$(n+1)^2$$
#### Principles of Mathematical Induction Question 5 Detailed Solution
Concepts:
Principle of Mathematical Induction:
Suppose there is a given statement P (n) involving the natural number n such that
• The statement is true for n = 1, i.e., P (1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., the truth of P (k) implies the truth of P (k + 1).
Then, P (n) is true for all natural numbers n
Calculation:
Given:
Let P(n) be defined as
$$P(n)=(1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2n~+~1)}{n^2}))=(n+1)^2$$
Put n = 1
P(1) = $$(1+\frac{3}{1})$$ = (1 + 1)2
4 = 4 P(1) is true
Let it is true for n = k
$$P(k)=(1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2k~+~1)}{k^2}))=(k+1)^2$$ ....(1)
for n = k + 1
$$P(k+1)=[(1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2k~+~1)}{k^2}))=(k+1)^2](1~+~\frac{2k+1+2}{(k+1)^2})=(k+1)^2(1+\frac{2k+3}{(k+1)^2})$$ Using Equation (1)
$$(k +1)^2[\frac{(k+1)^2+2k+3}{(k+1)}]$$
= k2 + 2k + 1 + 2k + 3
= (k +2)2 = [(k+1) + 1]2
Therefore, P(k +1) is true, Hence From the Principle of Mathematical Induction, the statement is true for all natural numbers n
#### Principles of Mathematical Induction Question 6:
If 22n - 1 is divisible by k for all n ∈ N, then the value of k is:
1. 6
2. 3
3. 7
4. 2
Option 2 : 3
#### Principles of Mathematical Induction Question 6 Detailed Solution
Concept:
If a number N is divided by d to give quotient q and remainder r, then:
• N = dq + r = d(q + 1) + (r - d) = d(q - 1) + (r + d).
It means that a remainder of r is equivalent to a remainder of r - d which is also equivalent to a remainder of r + d, where d is the divisor.
• Na divided by d will give a remainder of ra.
Calculation:
Since, 22n - 1 is 1 less than 22n, let's check for the divisors which definitely give a remainder of 1 on dividing 22n.
2 gives a definite remainder of 2 - 3 = -1 only when divided by 3.
∴ 22 gives a definite remainder of (-1)2 = 1 on division by 3.
⇒ 22n gives a definite remainder of 1n = 1 on division by 3.
⇒ 22n - 1 gives a definite remainder of 1 - 1 = 0 on division by 3.
⇒ 22n - 1 is always divisible by 3.
∴ The value of k is 3.
#### Principles of Mathematical Induction Question 7:
For all n ≥ 2, $${n^2}\left( {{n^4} - 1} \right)$$ is divisible by ________
1. 60
2. 50
3. 40
4. 70
Option 1 : 60
#### Principles of Mathematical Induction Question 7 Detailed Solution
For every positive integers n ≥ 2, $$f\left( n \right) = {n^2}\left( {{n^4} - 1} \right)$$
For n = 2, $$f\left( 2 \right) = {2^2}\left( {{2^4} - 1} \right) = 60$$
For n = 3, $$f\left( 2 \right) = {3^2}\left( {{3^4} - 1} \right) = 720$$
From the options, these are divisible by 60.
#### Principles of Mathematical Induction Question 8:
Direction: In the following question, assuming the given statements to be true, find which of the conclusion among the given conclusions is/are definitely true and then give your answer accordingly.
Statement:
A ≥ P > T; V < B ≥ X; P = S; B = T
Conclusion:
I. A > X
II. P < B
1. None is True.
2. Both I and II are True.
3. Only II is True.
4. Only I is True.
5. Either I or II is true
#### Answer (Detailed Solution Below)
Option 4 : Only I is True.
#### Principles of Mathematical Induction Question 8 Detailed Solution
Given statement: A ≥ P > T; V < B ≥ X; P = S; B = T
On combining: A ≥ P = S > T = B ≥ X; V < B
Conclusions:
I. A > X → True (A ≥ P = S > T = B ≥ X)
II. P < B → False (P > T = B)
Hence, only I is True.
#### Principles of Mathematical Induction Question 9:
n2 < 2n is true for all natural numbers, if
1. n ≥ 5
2. n < 5
3. n ≤ 3
4. More than one of the above
5. None of the above
Option 1 : n ≥ 5
#### Principles of Mathematical Induction Question 9 Detailed Solution
Concept:
Principle of mathematical induction: Let P(n) be the statement.
P(n) is true for n = 1. Let P(n) is true for n = k then if P(n) is true for n = k + 1 then P(n) is true for all natural numbers n.
Also, for n ≥ 3, n2 > 2n + 1 [This can be proved using the Principle of mathematical induction.]
Explanation:
Given Statement: n2 < 2n
• For n = 1: 12 < 21, which is true.
• For n = 2: 22 < 22, which is not true.
• For n = 3: 32 < 23, which is not true.
• For " id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">
n = 4: 42 < 24, which is not true.
• For " id="MathJax-Element-10-Frame" role="presentation" style="position: relative;" tabindex="0">
n = 5: 52 < 25, which is true.
• For n = 6: 62 < 26, which is true.
Now, let's prove it for all natural numbers greater than 5.
Base Case (n = 5):
$$5^2 = 25 < 2^5 = 32$$
Which is true.
Let's assume for some arbitrary k ≥ 5, it holds that $$k^2 < 2^k$$.
We need to show that $$(k+1)^2 < 2^{k+1}$$.
$$k^2 < 2^k$$
Multiply both sides by 2, we get
$$2k^2 < 2^{k+1}$$
⇒ $$k^2 + k^2 < 2^{k+1}$$
As $$k^2 > 2k + 1$$
⇒ $$k^2 + 2k + 1 < 2^{k+1}$$
⇒ $$(k+ 1)^2 < 2^{k+1}$$
Which is the required result.
Thus, Given inequality holds true for all natural numbers greater than or equal to 5.
Thus, Option 1 is the correct answer.
#### Principles of Mathematical Induction Question 10:
Using the principle of mathematical induction, find the value of $$\frac{1}{{1.2.3}}\; + \;\frac{1}{{2.3.4}}\; + \; \ldots \frac{1}{{\left\{ {n\left( {n\; + \;1} \right)\left( {n\; + \;2} \right)} \right\}}}$$
1. $$\frac{n (n+1)}{4(n +2)(n +3)}$$
2. $$\frac{n (n+3)}{4(n +1)(n +2)}$$
3. $$\frac{n (n+2)}{4(n +1)(n +3)}$$
4. None of the above
5. Not Attempted
#### Answer (Detailed Solution Below)
Option 2 : $$\frac{n (n+3)}{4(n +1)(n +2)}$$
#### Principles of Mathematical Induction Question 10 Detailed Solution
Concept:
• Mathematical induction: It is a technique of proving a statement, theorem, or formula which is assumed to be true, for every natural number n.
• By generalizing this in the form of a principle that we would use to prove any mathematical statement is called the principle of mathematical induction.
Calculations:
Consider
$$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{n(n+1)(n+2)}$$
Clearly, the rth term from the above series is
Let the rth term be $$u_r=\frac{1}{r(r+1)(r+2)}$$ ....(1)
now multiply and divide by 2 in (1)
⇒ $$u_r=\frac{1\times 2}{2r(r+1)(r+2)}$$
⇒ $$u_r=\frac{2}{2r(r+1)(r+2)}$$
add and subtract r in the numerator
⇒ $$u_r=\frac{(r+2) - r}{2r(r+1)(r+2)}$$
⇒ $$u_r=\frac{1}{2}\big[\frac{(r+2) }{r(r+1)(r+2)}- \frac{r }{r(r+1)(r+2)}\big]$$
⇒ $$u_r=\frac{1}{2}\big[\frac{1}{r(r+1)}- \frac{1}{(r+1)(r+2)}\big]$$ (2)
Now put r = 1 in (2), then we have
$$u_1=\frac{1}{2}\big[\frac{1}{1.2}- \frac{1}{2.3}\big]$$
put r = 2 in (2)
⇒ $$u_2=\frac{1}{2}\big[\frac{1}{2.3}- \frac{1}{3.4}\big]$$
put r = 3 in (2)
⇒ $$u_3=\frac{1}{2}\big[\frac{1}{3.4}- \frac{1}{4.5}\big]$$
. . . . . . .
put r = n in (2)
$$u_r=\frac{1}{2}\big[\frac{1}{n(n+1)}- \frac{1}{(n+1)(n+2)}\big]$$
Now, add all these equations and we get
$$S_n=\displaystyle\sum_{r=1}^{n} u_r=\frac{1}{2}\bigg[\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\bigg ]$$
$$=\frac{1}{4(n+1)(n+2)}\big[(n+1)(n+2)-2\big ]$$
$$=\frac{1}{4(n+1)(n+2)}\big[(n^2+n+2n+2-2\big ]$$
$$=\frac{1}{4(n+1)(n+2)}\big[n^2+3n\big ]$$
$$=\frac{n(n+3)}{4(n+1)(n+2)}$$
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Arithmetic Mean | kullabs.com
Notes, Exercises, Videos, Tests and Things to Remember on Arithmetic Mean
Please scroll down to get to the study materials.
## Note on Arithmetic Mean
• Note
• Things to remember
• Videos
• Exercise
• Quiz
The word "statistics " refers two meanings. In the singular sense, it deals with the collection, presentation, analysis and interpretation of numerical data and helps in making a decision. In the plural sense, it refers to the numerical facts and figures are sometimes known as statistical data.
### Measures of central tendency
A typical value which represents the characteristics of the entire mass of huge data is called the central value of the whole distribution. A measure of the central tendency is also known as a measure of location or an average.
The various measure of central tendency are as follow:
1. Arithmetic Mean (average)
2. Median
3. Mode
### Arithmetic Mean
The arithmetic mean is defined as the total sum of observations divided by the total number of observations. Here we study about the arithmetic mean of continuous series.
Calculation of arithmetic means in continuous series
When the number of items is large, we have to divide them into groups. Such groups are known as simple classes. It should be noted that there is no gap between any two successive intervals. So the data is continuous and the series is called continuous series.
Arithmetic Mean can be calculated in three different methods.
1. $$Direct \: method, \overline{X} = \frac{\sum fm}{n}$$, where m is the mid-value of class intervals.
2. $$Short \: cut \: method,\overline{X} = A + \frac{\sum fd}{n}$$, where d= X - A and A is assumed mean.
3. $$Step \: deviation \: method, \overline{X} = A + \frac{\sum fd'}{n}$$ $$\times h$$, where d' = $$\frac{d}{h}$$ and h is the length of class interval.
Arithmetic Mean can be calculated in three different methods.
1. $$Direct \: method, \overline{X} = \frac{\sum fm}{n}$$, where m is the mid-value of class intervals.
2. $$Short \: cut \: method,\overline{X} = A + \frac{\sum fd}{n}$$, where d= X - A and A is assumed mean.
3. $$Step \: deviation \: method, \overline{X} = A + \frac{\sum fd'}{n}$$ $$\times h$$, where d' = $$\frac{d}{h}$$ and h is the length of class interval.
.
### Very Short Questions
Solution:
Mean($$\overline X$$)= 5o
$$\sum fx = 750$$
N = ?
\begin{align*} Mean\:(\overline {X}) &= \frac{\sum fx}{N}\\ 50 &= \frac{750}{N}\\ or, N &= \frac{750}{50}\\ \therefore N &= 15 \: \: \: \: _{Ans} \end{align*}
Solution:
$$Mean \: (\overline{X}) =60 \\ \sum fx = 960 \\ Number \: of \: term \: (N) = ?$$
\begin{align*} Mean (\overline {X}) &= \frac{\sum fx}{N}\\ 60 &= \frac{960}{N}\\ or, N &= \frac{960}{60}\\ \therefore N &= 16 \: _{Ans} \end{align*}
Solution:
$$Mean \: (\overline{X}) =12 \\ \sum fx = 70 +10a\\ No. \: of \: term \: (N) = 5+a$$
\begin{align*}Mean \: (\overline{X}) &= \frac{\sum fx}{N}\\ 12 &= \frac{70 + 10a}{5 + a}\\ or, 60 +12a &=70 + 10a\\ or,12a -10a &= 70 - 60\\ or, 2a &= 10 \\ or, a &= \frac{10}{2}\\ \therefore a &= 5 \: \: _{Ans} \end{align*}
Solution:
$$Mean \: (\overline{X}) = 13 \\ No. \: of \: terms \: (N) = 5 + a$$
\begin{align*} Mean (\overline{X}) &= \frac{\sum X}{N} \\ or, 13 &= \frac{4 + 8+12+x+25}{5}\\ or, 65 &=49 + x\\ or, x &= 65 - 49 \\ x &= 16 \:\: _{Ans} \end{align*}
Solution:
First item mean $$(\overline {X_{1}}) = 7$$
Number of first item $$n_{1} = 4$$
Second item mean $$(\overline {X_{2}}) = 12$$
No. of Second item $$n_{2}$$= 3
Mean of 7 items $$\overline{X}= ?$$
\begin{align*} ( \overline {X}) &= \frac{n_1 \overline{X}_1 + n_2 \overline{X}_2}{n_1+n_2}\\ &= \frac{4 \times 7 + 3 \times 12}{4 + 3}\\ &= \frac{28 + 36}{7}\\ &= \frac{64}{7}\\ &= 9.14 \: \: \: _{ans} \end{align*}
$$Mean (\overline{X}) = 40 \\ No. \: of \: terms (N)= 7 \\ k = ?$$
\begin{align*} Mean(\overline{X}) &= \frac{\sum X}{N} \\ or, 40 &= \frac{10+20+30+40+50+ 60+30+k}{7}\\ or, 280 &= 240 + k\\ or, k &= 280 - 240\\ \therefore k &= 40 \: \: \: \: \: \: \: \: \: _{Ans} \end{align*}
Solution:
Expenditure (in Rs.)x Frequency(f) fx 24 2 48 25 4 100 30 3 90 35 4 140 40 2 80
\begin{align*} Mean(\overline{X}) &= \frac{\sum fx}{N}\\ &= \frac{458}{15}\\ &= 30.53 \\ \therefore Average \: expenditure &= Rs \: 30.53 \end{align*}
Solution:
Let, the age of remaining students be x.
$$average \: age (\overline{X}) = 9 \\ Number \: of \: terms (N) = 5$$
\begin{align*} (\overline{X}) &= \frac{\sum X}{N}\\ or, 9 &= \frac{5+7+8+15+x}{5}\\ or, 45 &= 35 + x \\ \therefore x &=45 -35 = 10 \\ \: \\ \therefore & \text{The age of remaining student is 10} \end{align*}
Solution:
Calculation of mean
Class interval cf. f mid-value (m) fm 10-20 4 4 15 60 20-30 16 12 25 300 30-40 56 40 35 1400 40-50 97 41 45 1845 50-60 124 27 55 1485 60-70 137 13 65 845 70-80 146 9 75 675 80-90 150 4 85 340 N = 150 $$\sum fm=6950$$
\begin{align*}Mean \: (\overline{X})&= \frac{\sum fm}{N}\\ &= \frac{6950}{150}\\ &= 46.33 \: _{Ans} \end{align*}
Solution:
Calculation of mean
class interval midvalue f fm 0-10 5 5 25 10-20 15 7 105 20-30 25 8 200 30-40 35 4 140 40-50 45 6 270 N= 30 $$\sum fm = 740$$
\begin{align*} Mean (\overline{X}) &= \frac{\sum fm}{N}\\ &= \frac{740}{30}\\ &= 24.6 \: _{ans} \end{align*}
Solution:
To find the value of p.
Solution:
To find the value of p.
Age in year No. of teachers (f) m fm 10-20 3 15 45 20-30 8 25 200 30-40 15 35 525 40-50 p 45 45p 50-60 4 55 220 N=p+30 $$\sum fm = 45p + 990$$
We know that,
\begin{align*} Mean \: (\overline {X}) &= \frac{\sum fm}{N}\\ 36 &= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &= 45p +990\\ or, 45p - 36p &= 1080 -990 \\ or, 9p &= 90 \\ \therefore p &= \frac{90}{9} = 10 \: _{Ans} \end{align*}
Solution:
To find the value of p.
Solution:
To find the value of p.
Age in year No. of teachers (f) m fm 10-20 3 15 45 20-30 8 25 200 30-40 15 35 525 40-50 p 45 45p 50-60 4 55 220 N=p+30 $$\sum fm = 45p + 990$$
We know that,
\begin{align*} Mean \: (\overline {X}) &= \frac{\sum fm}{N}\\ 36 &= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &= 45p +990\\ or, 45p - 36p &= 1080 -990 \\ or, 9p &= 90 \\ \therefore p &= \frac{90}{9} = 10 \: _{Ans} \end{align*}
0%
251
345
543
152
321
785
161
465
60
20
60
80
60
60
20
80
300
200
700
750
175
80
120
160
17
20
10
7
92
54
22
69
40
85
75
55
65
21
33
45
39
49
69
59
39
59
69
49
40.25
35.30
31
61.25
## DISCUSSIONS ABOUT THIS NOTE
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##### Niva
In the continuos series mean=15 a and £fx = 420 a find the number of terms(N)
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Ask any queries on this note.find the value of meanmark 0-10 0-20 0-30 0-40 0-50f 5 10 14 17 20
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# What is 2/536 Simplified?
Are you looking to calculate how to simplify the fraction 2/536? In this really simple guide, we'll teach you exactly how to simplify 2/536 and convert it to the lowest form (this is sometimes calling reducing a fraction to the lowest terms).
To start with, the number above the line (2) in a fraction is called a numerator and the number below the line (536) is called the denominator.
So what we want to do here is to simplify the numerator and denominator in 2/536 to their lowest possible values, while keeping the actual fraction the same.
To do this, we use something called the greatest common factor. It's also known as the greatest common divisor and put simply, it's the highest number that divides exactly into two or more numbers.
Want to quickly learn or refresh memory on how to simplify fractions play this quick and informative video now!
In our case with 2/536, the greatest common factor is 2. Once we have this, we can divide both the numerator and the denominator by it, and voila, the fraction is simplified:
2/2 = 1
536/2 = 268
1 / 268
What this means is that the following fractions are the same:
2 / 536 = 1 / 268
So there you have it! You now know exactly how to simplify 2/536 to its lowest terms. Hopefully you understood the process and can use the same techniques to simplify other fractions on your own. The complete answer is below:
1/268
## Convert 2/536 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
2 / 536 = 0.0037
### Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 2/536 Simplified?". VisualFractions.com. Accessed on September 29, 2023. http://visualfractions.com/calculator/simplify-fractions/what-is-2-536-simplified/.
• "What is 2/536 Simplified?". VisualFractions.com, http://visualfractions.com/calculator/simplify-fractions/what-is-2-536-simplified/. Accessed 29 September, 2023.
• What is 2/536 Simplified?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/simplify-fractions/what-is-2-536-simplified/.
### Preset List of Fraction Reduction Examples
Below are links to some preset calculations that are commonly searched for:
## Random Fraction Simplifier Problems
If you made it this far down the page then you must REALLY love simplifying fractions? Below are a bunch of randomly generated calculations for your fraction loving pleasure:
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###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
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# Area of Triangles - Problem 2
Brian McCall
###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
Share
In this problem we are asked to find the area and we’re told that our units are in centimeters. Well since we have one pair of parallel sides, we have a trapezoid. So let’s start off by writing our area formula of a trapezoid.
That’s going to be the sum of the bases, base 1 plus base 2 times the corresponding height between those bases all divided by 2. Well the trick in this problem is realizing that we have more information than we need. So let’s start labeling what we know.
Base 1 has to be one of our parallel bases, so base 1 is going to be 12cm. Base 2 is the other base that’s parallel, so that’s going to be 16 and our last unknown is h, our height and that’s going to be the perpendicular segment in between your parallel bases, so that’s going to be 5 and these are units of centimeters.
So notice that I don’t need the 9 or the 7, those don’t help me at all. So now that I know my 3 unknowns I can solve for my area. So I’m just going to substitute in here area is equal to base 1 plus base 2 so that’s going to be 12 plus 16 times your height which is 5 all divided by 2. So area is equal to 12 plus 16 is 28 times 5 all divided by 2 and so there’s a couple of ways that we can do this. You could say that 2 goes into 28 14 times so we have 5 times 14 and 5 times 15 is 75 so if you take 5 away from that that’s going to be 70.
Now what are our units? When we’re talking about area we said that we have centimeters, so we’re going to have centimeters to the second power. Well the way that we’ll say that is 70 square centimeters. The key thing here was realizing that we had more information than we needed so we identified our three variables, substitute it and solved.
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# Converting Length using U.S. Customary System
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## Objective
The students will be able to convert measurements of length within the U.S. Customary System.
#### Big Idea
One footâ¦. two feet⦠red foot⦠blue feet?
## Opener
15 minutes
In today’s lesson students explore the length measurements associated with the U.S. Customary system. Students measure objects around the room using a piece of string that is one foot long. They then convert their measurements to inches and yards.
As they do this work, students add to their graphic organizer the length units and create "a tool" to convert between the units. I have the students practice converting using this conversion tool and then close the lesson with an exit slip involving some real world problems involving length conversions.
I give each pair of students a piece of string that is one foot long and explain they are going to measure things around the room using the piece of string, and these measurements must include things larger than 1 foot. The students round their measurement to the nearest foot while conducting this investigation. I give students about 5 minutes to measure and record five things in the room. Students record their measurements on a sticky note or index card.
Once students have completed their investigation I bring them back to the whole group. As a group we begin to convert the measurements to inches and yards. I ask a volunteer to share what they measured and the length of the object. Before beginning the conversion I have students help me create the tool we devised yesterday to help us determine the operation needed. I go through several examples with the students while they listen and watch.
## Practice
30 minutes
After modeling for the students how to do conversions of feet to inches and feet to yards I now have students work in pairs to complete a skills sheet for today. Before allowing the students time to work with their partners I introduce the remaining conversion factor to them; 1760 yards = 1 mile.
Students now add some thinking to their graphic organizer. We add inch, feet, yard, and mile to the 4 tier of the hierarchy. On the back of the graphic organizer we draw a simple tool for converting length that includes all conversion factors and operations for working with length in the U.S. Customary System.
I allow the students about twenty minutes to complete the conversions while I circulate the room and support students.
## Closer
15 minutes
To wrap up this lesson I have students complete an exit ticket which includes three story problems I created involving converting length units. I use this information to guide me in deciding if further clarification is needed in converting length. I look to see where students struggled to pin point exactly what part of converting is unclear to them.
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Chapter3
# Chapter3 - Pre-Calculus Chapter 3A Chapter 3A Rectangular...
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© : Pre - Calculus - Chapter 3A Chapter 3A - Rectangular Coordinate System Introduction : Rectangular Coordinate System Although the use of rectangular coordinates in such geometric applications as surveying and planning has been practiced since ancient times, it was not until the 17th century that geometry and algebra were joined to form the branch of mathematics called analytic geometry. French mathematician and philosopher Rene Descartes (1596-1650) devised a simple plan whereby two number lines were intersected at right angles with the position of a point in a plane determined by its distance from each of the lines. This system is called the rectangular coordinate system (or Cartesian coordinate system). y x x-axis y-axis origin (0, 0) Points are labeled with ordered pairs of real numbers x , y , called the coordinates of the point, which give the horizontal and vertical distance of the point from the origin, respectively. The origin is the intersection of the x -and y -axes. Locations of the points in the plane are determined in relationship to this point 0,0 . All points in the plane are located in one of four quadrants or on the x -or y -axis as illustrated below. To plot a point, start at the origin, proceed horizontally the distance and direction indicated by the x -coordinate, then vertically the distance and direction indicated by the y -coordinate. The resulting point is often labeled with its ordered pair coordinates and/or a capital letter. For example, the point 2 units to the right of the origin and 3 units up could be labeled A 2,3 . Quadrant I Quadrant II Quadrant III Quadrant IV (-, +) (+, +) (-, -) (+, - ) (a, 0) (0, b) (0,0) Notice that the Cartesian plane has been divided into fourths. Each of these fourths is called a quadrant and they are numbered as indicated above. © : Pre - Calculus
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© : Pre - Calculus - Chapter 3A Example 1 : Plot the following points on a rectangular coordinate system: A 2, 3 B 0, 5 C 4,1 D 3,0 E 2, 4 Solution: -5 -4 -3 -2 -1 0 1 2 3 4 5 - 5 - 4- 3 - 2- 1 12345 C(-4,1) E(-2,-4) B(0,-5) A(2,-3) D(3,0) Example 2 : Shade the region of the coordinate plane that contains the set of ordered pairs x , y x 0 . [The set notation is read “the set of all ordered pairs x , y such that x 0”.] Solution: This set describes all ordered pairs where the x -coordinate is greater than 0. Plot several points that satisfy the stated condition, e.g., 2, 4 , 7,3 , 4,0 . These points are all located to the right of the y -axis. To plot all such points we would shade all of Quadrants I and IV. We indicate that points on the y -axis are not included x 0 by using a dotted line. Example 3 : Shade the region of the coordinate plane that contains the set of ordered pairs x , y x 1, 2 y 3 .
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Chapter3 - Pre-Calculus Chapter 3A Chapter 3A Rectangular...
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FRACTIONS FRACTIONS
A fraction is an indicated division that expresses one or more of the equal parts into which a unit is divided. For example, the fraction 2/3 indicates that the whole has been divided into 3 equal parts and that 2 of these parts are being used or considered. The number above the line is the numerator; and the number below the line is the denominator.
If the numerator of a fraction is equal to or larger than the denominator, the fraction is known as an improper fraction. In the fraction 15/8, if the indicated division is performed, the improper fraction is changed to a mixed number, which is a whole number and a fraction:
15 7
---- = 1 ---
8 8
A complex fraction is one that contains one or more fractions or mixed numbers in either the numerator or denominator. The following fractions are examples:
A decimal fraction is obtained by dividing the numerator of a fraction by the denominator and showing the quotient as a decimal. The fraction equals 5/8 = 0.625.
A fraction does not change its value if both numerator and denominator are multiplied or divided by the same number.
The same fundamental operations performed with whole numbers can also be performed with fractions. These are addition, subtraction, multiplication, and division.
Addition and Subtraction of Common Fractions
In order to add or subtract fractions, all the denominators must be alike. In working with fractions, as in whole numbers, the rule of likeness applies. That is, only like fractions may be added or subtracted.
When adding or subtracting fractions that have like denominators, it is only necessary to add or subtract the numerators and express the result as the numerator of a fraction whose denominator is the common denominator. When the denominators are unlike, it is necessary to first reduce the fractions to a common denominator before proceeding with the addition or subtraction process.
EXAMPLES
1. A certain switch installation requires 5/8 inch plunger travel before switch actuation occurs. If 1/8 inch travel is required after actuation, what will be the total plunger travel?
FIRST: Add the numerators.
5 + 1 = 6
NEXT: Express the result as the numerator of a fraction whose denominator is the common denominator.
2. The total travel of a jackscrew is 13/16 of an inch. If the travel in one direction from the neutral position is 7/16 of an inch, what is the travel in the opposite direction?
FIRST: Subtract the numerators.
13 - 7 = 6
NEXT: Express the result as the numerator of a fraction whose denominator is the common denominator.
3. Find the outside diameter of a section of tubing that has a 1/4 inch inside diameter and a combined wall thickness of 5/8 inch.
FIRST: Reduce the fractions to a common denominator.
NEXT: Add the numerators, and express the result as the numerator of a fraction whose denominator is the common denominator.
4. The tolerance for rigging the aileron droop of an airplane is 7/8 inch plus or minus 1/5 inch. What is the minimum droop to which the aileron can be rigged?
FIRST: Reduce the fractions to a common denominator.
NEXT: Subtract the numerators, and express the result as in the above examples.
Finding the Least Common Denominator
When the denominators of fractions to be added or subtracted are such that a common denominator cannot be determined readily, the LCD (least common denominator) can be found by the continued division method.
To find the LCD of a group of fractions, write the denominators in a horizontal row. Next, divide the denominators in this row by the smallest integer that will exactly divide two or more of the denominators. Bring down to a new row all the quotients and numbers that were not divisible. Continue this process until there are no two numbers in the resulting row that are divisible by any integer other than one. Multiply together all the divisors and the remaining terms in the last row to obtain the least common denominator.
EXAMPLE
What is the LCD for 7/8, 11/20, 8/36, 21/45?
FIRST: Write the denominators in a horizontal row and divide this row by the smallest integer that will exactly divide two or more of the numbers.
NEXT: Continue this process until there are no two numbers in the resulting row that are divisible by any integer other than one.
THEN: Multiply together all the divisors and remaining terms in the last row to obtain the LCD.
LCD = 2 x 2 x 3 x 3 x 5 x 2 = 360
Multiplication of Fractions
The product of two or more fractions is obtained by multiplying the numerators to form the numerator of the product and by multiplying the denominators to form the denominator of the product. The resulting fraction is then reduced to its lowest terms. A common denominator need not be found for this operation, as the new denominator in most cases will be different from that of all the original fractions.
EXAMPLE
What is the product of 3/5 x 12/22 x 1/2?
FIRST: Multiply the numerators together.
3 x 12 x 1 = 36
NEXT: Multiply the denominators together.
5 x 22 x 2 = 220
THEN: Reduce the resulting fraction to its lowest terms.
Cancellation
Cancellation is a technique of dividing out or canceling all common factors that exist between numerators and denominators. This aids in locating the ultimate product by eliminating much of the burdensome multiplication.
EXAMPLE
What is the product of 18/10 x 5/3?
The product could be found by multiplying 18 x 5 and 10 x 3, then dividing the product of the numerators by the product of the denominators. However, a much easier method of solution is by cancellation. It is apparent that the 10 in the denominator and the 5 in the numerator can both be divided an exact number of times by 5.
Also, the 18 and 3 are both exactly divisible by 3.
The resulting 6 in the numerator and the 2 in the denominator are both divisible by 2.
The fraction is thus reduced to its lowest terms, and the final multiplication and division steps are performed with ease when compared with the task of multiplying and dividing the larger fractions.
Division of Common Fractions
The division of common fractions is accomplished most conveniently by converting the problem into a multiplication of two common fractions. To divide one fraction by another fraction, invert the divisor fraction and multiply the numerators together and the denominators together. This is known as the inverted divisor method. Always keep in mind the order in which the fractions are written. It is important in division that the operations be performed in the order indicated. Also, remember that it is always the divisor that is inverted, never the dividend.
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# Travel Graphs
This section covers travel graphs, speed, distance, time, trapezium rule and velocity.
Speed, Distance and Time
The following is a basic but important formula which applies when speed is constant (in other words the speed doesn't change):
Image
Remember, when using any formula, the units must all be consistent. For example speed could be measured in m/s, distance in metres and time in seconds.
If speed does change, the average (mean) speed can be calculated:
Average speed = total distance travelled
total time taken
Units
In calculations, units must be consistent, so if the units in the question are not all the same (e.g. m/s, m and s or km/h, km and h), change the units before starting, as above.
The following is an example of how to change the units:
Example
Change 15km/h into m/s.
15km/h = 15/60 km/min (1)
= 15/3600 km/s = 1/240 km/s (2)
= 1000/240 m/s = 4.167 m/s (3)
In line (1), we divide by 60 because there are 60 minutes in an hour. Often people have problems working out whether they need to divide or multiply by a certain number to change the units. If you think about it, in 1 minute, the object is going to travel less distance than in an hour. So we divide by 60, not multiply to get a smaller number.
Example
If a car travels at a speed of 10m/s for 3 minutes, how far will it travel?
Firstly, change the 3 minutes into 180 seconds, so that the units are consistent. Now rearrange the first equation to get distance = speed × time.
Therefore distance travelled = 10 × 180 = 1800m = 1.8km
Velocity and Acceleration
Velocity is the speed of a particle and its direction of motion (therefore velocity is a vector quantity, whereas speed is a scalar quantity).
When the velocity (speed) of a moving object is increasing we say that the object is accelerating. If the velocity decreases it is said to be decelerating. Acceleration is therefore the rate of change of velocity (change in velocity / time) and is measured in m/s².
Example
A car starts from rest and within 10 seconds is travelling at 10m/s. What is its acceleration?
Acceleration = change in velocity = 10 = 1m/s² time 10
Distance-Time Graphs
These have the distance from a certain point on the vertical axis and the time on the horizontal axis. The velocity can be calculated by finding the gradient of the graph. If the graph is curved, this can be done by drawing a chord and finding its gradient (this will give average velocity) or by finding the gradient of a tangent to the graph (this will give the velocity at the instant where the tangent is drawn).
Image
Velocity-Time Graphs/ Speed-Time Graphs
A velocity-time graph has the velocity or speed of an object on the vertical axis and time on the horizontal axis. The distance travelled can be calculated by finding the area under a velocity-time graph. If the graph is curved, there are a number of ways of estimating the area (see trapezium rule below). Acceleration is the gradient of a velocity-time graph and on curves can be calculated using chords or tangents, as above.
Image
The distance travelled is the area under the graph.
The acceleration and deceleration can be found by finding the gradient of the lines.
On travel graphs, time always goes on the horizontal axis (because it is the independent variable).
Trapezium Rule
This is a useful method of estimating the area under a graph. You often need to find the area under a velocity-time graph since this is the distance travelled.
Image
Area under a curved graph = ½ × d × (first + last + 2(sum of rest))
d is the distance between the values from where you will take your readings. In the above example, d = 1. Every 1 unit on the horizontal axis, we draw a line to the graph and across to the y axis.
'first' refers to the first value on the vertical axis, which is about 4 here.
'last' refers to the last value, which is about 5 (green line).]
'sum of rest' refers to the sum of the values on the vertical axis where the yellow lines meet it.
Therefore area is roughly: ½ × 1 × (4 + 5 + 2(8 + 8.8 + 10.1 + 10.8 + 11.9 + 12 + 12.7 + 12.9 + 13 + 13.2 + 13.4))
= ½ × (9 + 2(126.8))
= ½ × 262.6
= 131.3 units²
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# What is 4x2 7
## Two solutions were found :
1. x =(2-√116)/8=(1-√ 29 )/4= -1.096
2. x =(2+√116)/8=(1+√ 29 )/4= 1.596
### Reformatting the input :
(1): "x2" was replaced by "x^2".
### Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
4*x^2-2*x-(7)=0
## Step 1 :
#### Equation at the end of step 1 :
(22x2 - 2x) - 7 = 0
## Step 2 :
#### Trying to factor by splitting the middle term
2.1 Factoring 4x2-2x-7
The first term is, 4x2 its coefficient is 4 .
The middle term is, -2x its coefficient is -2 .
The last term, "the constant", is -7
Step-1 : Multiply the coefficient of the first term by the constant 4 • -7 = -28
Step-2 : Find two factors of -28 whose sum equals the coefficient of the middle term, which is -2 .
-28 + 1 = -27 -14 + 2 = -12 -7 + 4 = -3 -4 + 7 = 3 -2 + 14 = 12 -1 + 28 = 27
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
4x2 - 2x - 7 = 0
## Step 3 :
#### Parabola, Finding the Vertex :
3.1 Find the Vertex of y = 4x2-2x-7
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 4 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 0.2500
Plugging into the parabola formula 0.2500 for x we can calculate the y -coordinate :
y = 4.0 * 0.25 * 0.25 - 2.0 * 0.25 - 7.0
or y = -7.250
#### Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 4x2-2x-7
Axis of Symmetry (dashed) {x}={ 0.25}
Vertex at {x,y} = { 0.25,-7.25}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-1.10, 0.00}
Root 2 at {x,y} = { 1.60, 0.00}
#### Solve Quadratic Equation by Completing The Square
3.2 Solving 4x2-2x-7 = 0 by Completing The Square .
Divide both sides of the equation by 4 to have 1 as the coefficient of the first term :
x2-(1/2)x-(7/4) = 0
Add 7/4 to both side of the equation :
x2-(1/2)x = 7/4
Now the clever bit: Take the coefficient of x , which is 1/2 , divide by two, giving 1/4 , and finally square it giving 1/16
Add 1/16 to both sides of the equation :
On the right hand side we have :
7/4 + 1/16 The common denominator of the two fractions is 16 Adding (28/16)+(1/16) gives 29/16
So adding to both sides we finally get :
x2-(1/2)x+(1/16) = 29/16
Adding 1/16 has completed the left hand side into a perfect square :
x2-(1/2)x+(1/16) =
(x-(1/4)) • (x-(1/4)) =
(x-(1/4))2
Things which are equal to the same thing are also equal to one another. Since
x2-(1/2)x+(1/16) = 29/16 and
x2-(1/2)x+(1/16) = (x-(1/4))2
then, according to the law of transitivity,
(x-(1/4))2 = 29/16
We'll refer to this Equation as Eq. #3.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-(1/4))2 is
(x-(1/4))2/2 =
(x-(1/4))1 =
x-(1/4)
Now, applying the Square Root Principle to Eq. #3.2.1 we get:
x-(1/4) = √ 29/16
Add 1/4 to both sides to obtain:
x = 1/4 + √ 29/16
Since a square root has two values, one positive and the other negative
x2 - (1/2)x - (7/4) = 0
has two solutions:
x = 1/4 + √ 29/16
or
x = 1/4 - √ 29/16
Note that √ 29/16 can be written as
√ 29 / √ 16 which is √ 29 / 4
3.3 Solving 4x2-2x-7 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = 4
B = -2
C = -7
Accordingly, B2 - 4AC =
4 - (-112) =
116
2 ± √ 116
x = —————
8
Can √ 116 be simplified ?
Yes! The prime factorization of 116 is
2•2•29
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 116 = √ 2•2•29 =
± 2 • √ 29
√ 29 , rounded to 4 decimal digits, is 5.3852
So now we are looking at:
x = ( 2 ± 2 • 5.385 ) / 8
Two real solutions:
x =(2+√116)/8=(1+√ 29 )/4= 1.596
or:
x =(2-√116)/8=(1-√ 29 )/4= -1.096
## Two solutions were found :
1. x =(2-√116)/8=(1-√ 29 )/4= -1.096
2. x =(2+√116)/8=(1+√ 29 )/4= 1.596
Processing ends successfully
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# Greatest Common Factor (GCF) Of Numbers With The Cake Method
Hi there. In this math education post, I cover the topic of greatest common factor of two or more numbers with the cake method. It is sometimes called the ladder method.
Pixabay Image Source
## Topics
• What Is A Greatest Common Factor?
• Cake Method
• Online Tools & Related Math Games On GCF
## What Is A Greatest Common Factor?
Before getting into the concept of a greatest common factor, it is a good idea to review the concepts of factors and common factors.
A factor is typically a whole number that is used for multiplication. Factors of 6 are 1, 2, 3 and 6. Six times one makes 6 and two times three equals to 6 as well. The number 1, 2, 3 and 6 are numbers that can be used for multiplication to make six. As an another example, the number 10 has factors of 1, 2, 5 and 10. We have `1 x 10 = 10` and `2 x 5 = 10` or `5 x 2 = 10`.
Common factors are factors that are common between two numbers. Earlier we have the numbers of 6 and 10. Factors of 6 are 1, 2, 3 and 6 and factors of 10 are 1, 2, 5 and 10.
Factors of 6: 1, 2, 3, 6
Factors of 10: 1, 2, 5, 10
Two and one are common factors for 6 and 10. The greatest common factor (GCF) here is 2 as it is the largest common factor.
Another Example
What is the greatest common factor for 9 and 15?
Factors of 9: 1, 3, 9
Factors of 15: 1, 3, 5, 15
The greatest common factor for 9 and 15 is 3.
Pixabay Image Source
## Cake Method
The method shown above for obtaining the greatest common factor required listing factors. This method is simple to follow but it may not be the fastest. The above method could be slow or even difficult if the numbers become really large.
A good method for determining the greatest common factor between two or more numbers is the cake method. It is sometimes called the ladder method as well.
I use https://www.tutorialspoint.com/whiteboard.htm and screenshots for the images here.
Example One
Find the greatest common factor (GCF) of 24 and 30.
With the cake method start with this setup (screenshot below). The two numbers are listed at the top with an L shape.
Determine a common factor that is in 24 and 30. It does not have to be a large number factor. You can do something as simple as 2 as both 24 and 30 are even. Write the 2 on the left side of the L shape.
As the 2 is used as a common factor, divide 24 by this 2 and divide 30 by this 2. The numbers of 12 and 15 are in the next layer.
There is a common factor greater than one between 12 and 15. That is 3. Write the 3 on the outside of this second layer.
Divide 12 by 3 and divide 15 by 3 to obtain 4 and 5.
As the greatest common factor between 4 and 5 is 1, there is no need to draw another L layer. We are done. The greatest common factor is the product of the left side numbers. This is two times three to get 6.
The GCF between 24 and 30 is 6.
Example Two
Find the greatest common factor (GCF) of 28 and 112. I display the full cake method screenshot for this example.
Example Three - Three Numbers Case
What is the greatest common factor (GCF) between 180, 90 and 300?
For dealing with three numbers, the cake method is not that much different. There is just an extra number involved. Start with the setup as follows.
Determine a common factor between the three numbers. You do not have to select a large common factor. Choose something that is obvious for you. You can choose 3 but I chose 10 as each of three numbers are multiples of 10.
Divide all three numbers in the L shape by 10. The next layer has the three numbers of 18, 9 and 30.
Select another common factor here. Three is a factor for all three numbers. Place the three on the left side and divide all three numbers by 3 to obtain the next row.
From the three numbers of 6, 3 and 10 there is no common factor for all three numbers. We are done here. The greatest common factor (GCF) here is `10 x 3 = 30`.
Pixabay Image Source
## Online Tools & Related Math Games On GCF
There are some neat online tools and math games that relate to Greatest Common Factors. Here are a few that I have found on the internet.
Pixabay Image Source
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Conjugate Indices 1/p + 1/q = 1
# Conjugate Indices 1/p + 1/q = 1
Definition: Let $1 \leq p \leq \infty$. Define the Conjugate Index of $p$ is the number $q$ defined as follows: a) If $1 < p < \infty$ then the $q$ is such that $1 < q < \infty$ such that $\displaystyle{\frac{1}{p} + \frac{1}{q} = 1}$. b) If $p = 1$ then $q = \infty$. c) If $p = \infty$ then $q = 1$.
Note that if $1 < p < \infty$ then we can give an explicit formula for the conjugate index $q$ of $p$. Namely:
(1)
\begin{align} \quad q = \frac{p}{p-1} \end{align}
Since:
(2)
\begin{align} \quad \frac{1}{p} + \frac{1}{q} = \frac{1}{p} + \frac{p-1}{p} = \frac{1 + p - 1}{p} = 1 \end{align}
A few example of $p$ and its conjugate index is given in the table below:
$p$ Conjugate Index, $q$
$1$ $\infty$
$2$ $2$
$3$ $\displaystyle{\frac{3}{2}}$
$4$ $\displaystyle{\frac{4}{3}}$
$5$ $\displaystyle{\frac{5}{4}}$
$\infty$ $1$
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Select Page
Playing cards and number theory go together like young boys and bishops. Here’s an interesting application of mathematical cycles that involves card shuffling.
The type of shuffle in the photo above is called a ‘riffle shuffle’. I’m sure you’ve all seen it before. You split the deck into two parts and interlock the parts together by using a fanning motion. The way most people do it in general is quite randomly and haphazardly, which is fair enough because the idea is to mix the cards up as much as possible.
However, you can be a bit more precise. You can perform a ‘perfect’ riffle shuffle by splitting the deck into 2 exact halves (26 cards each) and interlocking the cards so that the cards from the 2 halves alternate into the final pile. To be a little more precise, if we call the two halves A and B, then we want to combine the cards so that every other card in the deck after the shuffle is from pile A.
Now, you may already have realised that there are exactly 2 ways of doing this. Suppose half A is the top half of the unshuffled deck and B is the bottom half. Then you can perform the shuffle with either:
1. The top card from B ending up on the top in the shuffled deck
2. The top card from A ending up on the top in the shuffled deck
Situation 1 is called ‘middle card top’, and 2 is called ‘top card top’, because these are the cards that will end up on the top of the shuffled deck.
It may seem like a trivial point, but magicians will know that this is not so! If we keep repeating the ‘middle card top’ shuffle, then we’d have to do it 52 times to return the deck to its original order. However, if we repeated the ‘top card top’ shuffle, we’d only have to repeat it 8 times to get back to the starting deck! This seems a little surprising!
Can we prove formally that this will happen every time without physically shuffling the cards and making a mess? Of course we can!
Middle card top
Let’s give each card in our unshuffled deck a number from 1-52 i.e. the top card is numbered 1, the second 2 and so on.
After a middle card top shuffle, the 27th card will be the top card in the shuffled deck. The 28th card will be in position 3, the 29th in position 5 etc. Therefore, the ordering of the shuffled deck after one iteration is:
27, 1, 28, 2, 29, 3, 30, 4, 31, 5, 32, 6, 33, 7, 34, 8, 35, 9, 36, 10, 37, 11, 38, 12, 39, 13, 40, 14, 41, 15, 42, 16, 43, 17, 44, 18, 45, 19, 46, 20, 47, 21, 48, 22, 49, 23, 50, 24, 51, 25, 52, 26.
The clever bit is that we don’t have to repeat this 52 times! We can see how the cards will cycle after shuffles by noticing that the 1st card becomes the 27th. The 27th card then becomes the 40th according to the ordering above. If we follow the above chain, we can write down the cycle in the following form:
(1, 27, 40, 20, 10, 5, 29, 41, 47, 50, 25, 39, 46, 23, 38, 19, 36, 18, 9, 31, 42, 21, 37, 45, 49, 51, 52, 26, 13, 33, 43, 48, 24, 12, 6, 3, 28, 14, 7, 30, 15, 34, 17, 35, 44, 22, 11, 32, 16, 8, 4, 2)
What does this tell us?
The first thing you should notice is that we have one big chain of numbers that loops around. The length of this loop is 52. This means that to get card 1 back in the top position, it has to travel through each of the other positions in the cycle. In other words, we have to repeat the shuffle exactly 52 times to get each of the cards back to their starting positions.
Now that we know what to do, it should be easier to look at the second shuffle.
Top card top
After one iteration of this shuffle, we get the ordering:
1, 27, 2, 28, 3, 29, 4, 30, 5, 31, 6, 32, 7, 33, 8, 34, 9, 35, 10, 36, 11, 37, 12, 38, 13, 39, 14, 40, 15, 41, 16, 42, 17, 43, 18, 44, 19, 45, 20, 46, 21, 47, 22, 48, 23, 49, 24, 50, 25, 51, 26, 52.
Writing this in cycle form, we can immediately see that there will be multiple cycles, because the first card always stays on top. Similarly, card 52 is still card 52 after an iteration of the shuffle. Writing out all of the cycles gives:
(1) (2,27,14,33,17,9,5,3) (4,28,40,46,49,25,13,7) (6,29,15,8,30,41,21,11) (10,31,16,34,43,22,37,19) (12,32,42,47,24,38,45,23) (18,35) (20,36,44,48,50,51,26,39) (52)
We can see that there are two cycles of length 1, one cycle of length 2 and six cycles of length 8. The lowest common multiple of 1, 2 and 8 is 8. Therefore, to reach the starting position, we only need to perform 8 shuffles of the top card top kind.
Remember folks, dexterity and maths are the key to becoming a successful close-up card trick expert!
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8th grade is all about tackling the meat of algebra and getting exposure to some of the foundational concepts in geometry. If you get this stuff (and you should because you're incredibly persistent), the rest of your life will be easy. Okay, maybe not your whole life (no way to avoid the miseries of wedding planning), but at least your mathematical life. Seriously, we're not kidding. If you get the equations and functions and systems that we cover here, most of high school will feel intuitive (even relaxing). If you don't, well.. at least you have high school to catch up. :) On top of that, we will sharpen many of the skills that you last saw in 6th and 7th grades. This includes extending our knowledge of exponents to negative exponents and exponent properties and our knowledge of the number system to irrational numbers! (Content was selected for this grade level based on a typical curriculum in the United States.)
Community Questions
Relationships and functions
All content in “Relationships and functions”
### Graphing and analyzing proportional relationships
In proportional relationships, the ratio between one variable and the other is always constant. In the context of rate problems, this constant ratio can also be considered a rate of change. This tutorial allows you to dig deeper into this idea. Common Core Standard: 8.EE.B.5
### Intercepts of linear functions
There are many ways to graph a line and this tutorial covers one of the simpler ones. Since you only need two points for a line, let's find what value an equation takes on when x = 0 (essentially the y-intercept) and what value it takes on when y = 0 (the x-intercept). Then we can graph the line by going through those two points.
### Slope of a line
If you've ever struggled to tell someone just how steep something is, you'll find the answer here. In this tutorial, we cover the idea of the slope of a line. We also think about how slope relates to the equation of a line and how you can determine the slope or y-intercept given some clues. Common Core Standard: 8.F.B.4
### Triangle similarity and slope
In this tutorial, we use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane. We'll connect this idea to the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. Common Core Standard: 8.EE.B.6
### Function notation
f(x), g(x), etc. What does this mean? Well they are ways of referring to "functions of x". This is an idea that will show up throughout more advanced mathematics and computer science so it is a good idea to understand them now!
### Recognizing functions
Relationships can be any association between sets of numbers while functions have only one output for a given input. This tutorial works through a bunch of examples of testing whether something is a valid function. As always, we really encourage you to pause the videos and try the problems before Sal does! Common Core Standard: 8.F.A.1
### Analyzing linear functions
Linear functions show up throughout life (even though you might not realize it). This tutorial will have you thinking much deeper about what a linear function means and various ways to interpret one. Like always, pause the video and try the problem before Sal does. Then test your understanding by practicing the problems at the end of the tutorial. Common Core Standards: 8.F.A.2, 8.F.A.4, 8.F.A.5
### Linear and nonlinear functions
Not every relationship in the universe can be represented by a line (in fact, most can't be). We call these "nonlinear". In this tutorial, you'll learn to tell the difference between a linear and nonlinear function! Have fun! Common Core Standare: 8.F.A.3
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# What is 125 in the exponential form?
Exponents and powers are a method to express repeated multiplication of the same number. For eg- 6×6×6×6 can be written as 64 where 6 is the base and 4 is the exponent. It is most commonly used to express the powers of 10 to write a very large number in a convenient manner. For eg- 100000 can be written as 105.
### Laws of Exponents
• To multiply two exponential numbers with the same bases, the exponents are added and the base remains the same. For eg- am × an = am+n
• When the exponent has another exponent, the base remains the same but the powers are multiplied.
For eg- (am)n = am×n
• To divide two exponential numbers with the same bases, the exponents are subtracted and the base remains the same.
For eg- am/an = am-n
### What is a Power Notation?
Power notation is a method to express numbers that are very large or very small. For eg- 10000000000000 can be written as 1013 and 0.000000016 can be written as 1.6 × 10-8
Steps to write a number in the power notation:
Step 1: Write the number and do its factorization.
Step 2: Count the number of times the factor is there.
Step 3: Write the exponential number with the number as the base and the number of times it appeared as the power.
### What is 125 in the exponential form?
Solution:
Given that the number is 125.
The factors of 125 are 125 = 5 × 5 × 5
The factors appeared for 3 times.
125 = 53
Hence, the power notation of 125 is 53.
### Similar Questions
Question 1: What is the power notation of 32?
Solution:
Given that the number is 32.
The factors of 32 are 32 = 2 × 2 × 2 × 2 × 2
The factors appeared for 5 times.
32 = 25
Hence, the power notation of 32 is 25.
Question 2: What is the power notation of 216?
Solution:
Given that the number is 216.
The factors of 216 are 216 = 2 × 2 × 2 × 3 × 3 × 3 = 6 × 6 × 6
The factor appeared for 3 times.
216 = 63
Hence, the power notation of 216 is 63.
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NCERT Exemplar: Arithmetic Progressions
# NCERT Exemplar: Arithmetic Progressions - Mathematics (Maths) Class 10
### Exercise 5.1
Choose the correct answer from the given four options in the following questions:
Q.1. In an AP, if d = –4, n = 7, a= 4, then a is
(a) 6
(b) 7
(c) 20
(d) 28
We know that nth term of an AP is
an = a + (n – 1)d
where,
a = first term
an is nth term
d is the common difference
According to the question,
4 = a + (7 – 1)(- 4)
4 = a – 24
a = 24 + 4 = 28
Q. 2. In an AP, if a = 3.5, d = 0, n = 101, then an will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
We know that nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
an = 3.5 + (101 – 1)0
= 3.5
(Since, d = 0, it’s a constant A.P)
Q.3. The list of numbers – 10, – 6, – 2, 2,… is
(a) an AP with d = – 16
(b) an AP with d = 4
(c) an AP with d = – 4
(d) not an AP
According to the question,
a1 = – 10
a2 = – 6
a3 = – 2
a4 = 2
a2 – a1 = 4
a3 – a2 = 4
a4 – a3 = 4
a2 – a1 = a3 – a2 = a4 – a3 = 4
Therefore, it’s an A.P with d = 4
Q. 4. The 11th term of the AP: –5, (–5/2), 0, 5/2, …is
(a) –20
(b) 20
(c) –30
(d) 30
First term, a = – 5
Common difference,
d = 5 – (-5/2) = 5/2
n = 11
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a11 = – 5 + (11 – 1)(5/2)
a11 = – 5 + 25 = 20
Q. 5. The first four terms of an AP, whose first term is –2 and the common difference is –2, are
(a) – 2, 0, 2, 4
(b) – 2, 4, – 8, 16
(c) – 2, – 4, – 6, – 8
(d) – 2, – 4, – 8, –16
First term, a = – 2
Second Term, d = – 2
a1 = a = – 2
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
Hence, we have,
a2 = a + d = – 2 + (- 2) = – 4
Similarly,
a3 = – 6
a4 = – 8
So the A.P is
– 2, – 4, – 6, – 8
Q.6. The 21st term of the AP whose first two terms are –3 and 4 is
(a) 17
(b) 137
(c) 143
(d) –143
First two terms of an AP are a = – 3 and a2 = 4.
We know, nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a2 = a + d
4 = – 3 + d
d = 7
Common difference, d = 7
a21 = a + 20d
= – 3 + (20)(7)
= 137
Q.7. If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
a2 = a + d = 13 …..(1)
a5 = a + 4d = 25 …… (2)
From equation (1) we have,
a = 13 – d
Using this in equation (2), we have
13 – d + 4d = 25
13 + 3d = 25
3d = 12
d = 4
a = 13 – 4 = 9
a7 = a + 6d
= 9 + 6(4)
= 9 + 24 = 33
Q.8. Which term of the AP: 21, 42, 63, 84… is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Let nth term of the given AP be 210.
According to question,
first term, a = 21
common difference, d = 42 – 21 = 21 and an = 210
We know that the nth term of an AP is
an = a + (n – 1)d
Where,
a = first term
an is nth term
d is the common difference
210 = 21 + (n – 1)21
189 = (n – 1)21
n – 1 = 9
n = 10
So, 10th term of an AP is 210.
Q.9. If the common difference of an AP is 5, then what is a18 – a13?
(a) 5
(b) 20
(c) 25
(d) 30
Given, the common difference of AP i.e., d = 5
Now,
As we know, nth term of an AP is
an = a + (n – 1)d
where a = first term
an is nth term
d is the common difference
a18 -a13 = a + 17d – (a + 12d)
= 5d
= 5(5)
= 25
Q.10. What is the common difference of an AP in which a18 – a14 = 32?
(a) 8
(b) – 8
(c) – 4
(d) 4
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
The nth term of an AP is
aₙ = a + (n - 1 )d.
a = first term
aₙ = nth term
d = common difference.
Given,
a₁₈ - a₁₄ = 32
a + (18 - 1)d - [a + (14 - 1)d] = 32.
a + 17d - a - 13d = 32.
17d - 13d = 32.
4d = 32
d = 8.
Therefore, d = 8.
Q.11. Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8. Then the difference between their 4th terms is
(a) –1
(b) – 8
(c) 7
(d) –9
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
The nth term of an AP is
aₙ = a + (n - 1 )d.
a = first term
aₙ = nth term
d = common difference.
Let the common difference be d₁ and d₂.
By the condition given,
d₁ = d₂ = d -------------- (i)
The first term of the first AP be (a₁) = -1
The first term of the second AP be (a₂) = -8.
4th term of first AP,
T₄ = a₁ + (4 - 1)d
T₄ = -1 + 3d.
4th term of second AP,
T₄‘ = a₂ + (4 - 1)d
T₄‘ = -8 + 3d.
The difference between their 4th terms is
꘡T₄ - T₄’꘡= (-1 + 3d) - (-8 + 3d)
꘡T₄ - T₄’꘡= -1 + 3d + 8 - 3d = 7
꘡T₄ - T₄’꘡= 7.
Therefore, the required difference is 7.
Q.12. If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be
(a) 7
(b) 11
(c) 18
(d) 0
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
The nth term of an AP is
aₙ = a + (n - 1 )d.
a = first term
aₙ = nth term
d = common difference.
As per the question,
7a₇ = 11a₁₁
Since,aₙ = a + (n - 1 )d.
7[a + (7 - 1)d] = 11 [a + (11 - 1)d]
7(a + 6d) = 11 (a + 10d)
7a + 42d = 11a + 110d
42d -110d = 11a - 7a
4a + 68 d = 0
2(2a + 34d) = 0
2a + 34d = 0
a + 17d = 0----------------------(1)
18th term of an AP is
a₁₈ = a + (18 - 1)d
a₁₈ = a + 17d
a₁₈ = 0.
Therefore, the a₁₈ =0.
Q.13. The 4th term from the end of the AP: -11, -8, -5, ..., 49 is
(a) 37
(b) 40
(c) 43
(d) 58
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
The nth term of an AP from the end is
aₙ = l -(n - 1)d ------(1)
a = first term
aₙ = nth term
d = common difference.
l = Last term.
From the question,
l = 49
d = -8 - (-11) = -8 + 11 = 3
From(1),we get,
a₄ = 49 - (4 - 1) 3
a₄ = 49 - 3(3)
a₄ = 49 - 9
a₄ = 40.
Therefore, the 4th term is 40.
Q.14. The famous mathematician associated with finding the sum of the first 100 natural numbers is
(a) Pythagoras
(b) Newton
(c) Gauss
(d) Euclid
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
Gauss is the famous mathematician associated with finding the sum of 100 natural numbers.
Therefore, the famous mathematician is Gauss.
Q.15. If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
From the question:
a = -5
d = 2.
The formula to find the sum is
Sn = n/2 [2a + (n - 1)d].
Substituting the values, we get,
S₆ = 6/2 [2a + (6 - 1)d]
S₆ = 3[2(-5) + 5(2)]
S₆ = 3(-10 + 10)
S₆ = 3(0)
S₆ = 0.
Therefore, S₆ = 0.
Q.16. The sum of first 16 terms of the AP: 10, 6, 2,... is
(a) –320
(b) 320
(c) –352
(d) –400
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
From the question given,
AP is 10, 6, 2,
a = 10
d = - 4.
The formula to find the sum is
Sn = n/2 [2a + (n - 1)d].
S₁₆ = 16/2 [2a + (16 - 1)d]
S₁₆ = 8[2 × 10 + 15(-4)]
S₁₆ = 8(20 - 60)
S₁₆ = 8(-40)
S₁₆ = -320.
Therefore, S₁₆ = -320.
Q.17. In an AP if a = 1, an = 20 and Sn = 399, then n is
(a) 19
(b) 21
(c) 38
(d) 42
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
From the question,
a = 1
aₙ = 20
Sₙ = 399.
The formula to find the sum is
Sₙ = n/2 [2a + (n - 1)]d.
Substituting the values, we get,
399 = n/2 [2 × 1 + (n - 1)d].
798 = 2n + n(n - 1)d -----------(1)
Since, aₙ = a + (n - 1)d
a + (n-1)d = 20
1 + (n - 1)d = 20
(n - 1)d = 19 -----------(2)
Substituting (2) in(1), we get,
798 = 2n + 19 n
798 = 21n
n = 798/21.
Therefore, n = 798/21 = 38
Q.18. The sum of first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
From the question,
The first five multiples of 3 are 3, 6, 9,12 and 15.
First term, a = 3,
common difference, d = 6 - 3 = 3 and
number of terms, n = 5
The formula to find the sum is
Sₙ = n/2 [2a + (n - 1)d.
Substituting the values
S₅ = 5/2 [2a + (5 - 1)d]
S₅ = 5/2 [2 × 3 + 4 × 3]
So we get
S₅ = 5/2 (6 + 12)
S₅ = 5 × 9
S₅ = 45.
Therefore, S₅ = 45.
### Exercise 5.2
Q. 1. Which of the following form an AP? Justify your answer.
(i) –1, –1, –1, –1,…
We have a1 = – 1 , a2 = – 1, a3 = – 1 and a4 = – 1
a2 – a1 = 0
a3 – a2 = 0
a4 – a3 = 0
Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.
(ii) 0, 2, 0, 2,…
We have a1 = 0, a2 = 2, a3 = 0 and a4 = 2
a2 – a1 = 2
a3 – a2 = – 2
a4 – a3 = 2
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.
(iii) 1, 1, 2, 2, 3, 3…
We have a1 = 1 , a2 = 1, a3 = 2 and a4 = 2
a2 – a1 = 0
a3 – a2 = 1
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.
(iv) 11, 22, 33…
We have a1 = 11, a2 = 22 and a3 = 33
a2 – a1 = 11
a3 – a2 = 11
Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.
(v) 1/2,1/3,1/4, …
We have a1 = ½ , a2 = 1/3 and a3 = ¼
a2 – a1 = -1/6
a3 – a2 = -1/12
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.
(vi) 2, 22, 23, 24, …
We have a1 = 2 , a2 = 22, a3 = 23 and a4 = 24
a2 – a1 = 22 – 2 = 4 – 2 = 2
a3 – a2 = 23 – 22 = 8 – 4 = 4
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.
(vii) √3, √12, √27, √48, …
We have,
a= √3, a2 = √12, a3 = √27 and a4 = √48
a2 – a= √12 – √3 = 2√3 – √3 = √3
a3 – a= √27 – √12 = 3√3 – 2√3 = √3
a4 – a= √48 – √27 = 4√3 – 3√3 = √3
Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.
Q.2. Justify whether it is true to say that –1, -3/2, –2, 5/2,… forms an AP as
a2 – a1 = a3 – a2.
False
a1 = -1, a2 = -3/2, a3 = -2 and a4 = 5/2
a2 – a= -3/2 – (-1) = – ½
a3 – a= – 2 – (- 3/2) = – ½
a4 – a= 5/2 – (-2) = 9/2
Clearly, the difference of successive terms in not same, all though, a2 – a1 = a3 – a2 but a4 – a3 ≠ a3 – a2 therefore it does not form an AP.
Q.3. For the AP: –3, –7, –11, …, can we find directly a30a20 without actually finding a30 and a20? Give reasons for your answer.
True
Given
First term, a = – 3
Common difference, d = a2 – a1 = – 7 – (- 3) = – 4
a30 – a20 = a + 29d – (a + 19d)
= 10d
= – 40
It is so because difference between any two terms of an AP is proportional to common difference of that AP
Q.4. Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?
Suppose there are two AP’s with first terms a and A
And their common differences are d and D respectively
Suppose n be any term
an = a + (n – 1)d
An = A + (n – 1)D
As common difference is equal for both AP’s
We have D = d
Using this we have
An – an = a + (n – 1)d – [ A + (n – 1)D]
= a + (n – 1)d – A – (n – 1)d
= a – A
As a – A is a constant value
Therefore, difference between any corresponding terms will be equal to a – A.
Q.5. Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer.
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
Let 0 be the nth term of given AP, (aₙ = 0)
From the question given,
a = 31,
d = 28 - 31 = - 3
The nth terms of an AP, is
aₙ = a + (n - 1 )d.
Substituting the values, we get,
0 = 31 + (n - 1)(-3)
3(n - 1) = 31.
Dividing both sides by 3, we get,
n - 1 = 31/3.
n = 31/3+ 1
n = 34/3.
As, n should be a positive integer, 0 is not a term given in the AP.
Therefore, 0 is not a term of the given AP.
Q.6. The taxi fare after each km, when the fare is Rs 15 for the first km and Rs. 8 for each additional km, does not form an AP as the total fare (in Rs.) after each km is 15, 8, 8, 8, ... Is the statement true? Give reasons.
From the question above,
False,because the total fare after each km is 15.
From the question above,
t₁ = 15,
t₂ = 23,
t₃ = 31,
t₄ = 39.
Calculating the difference, we get,
t₂ - t₁ = 23 - 15 = 8
t₃ - t₂ = 31 - 23 = 8
t₄ - t₃ = 39 - 31 = 8
Since, each of the successive terms of the given list have the same difference (d=8).
Therefore, the total fare after each km forms an AP.
Q.7. In which of the following situations, do the lists of numbers involved form an AP?
(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs. 400.
(ii) The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs. 250, and it increases by Rs. 50 for the next higher class.
(iii) The amount of money in the account of Varun at the end of every year when Rs. 1000 is deposited at simple interest of 10% per annum.
(iv) The number of bacteria in a certain food item after each second, when they double in every second.
(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs. 400.
From the question given,
The fee charged from a student every month by a school for the whole session is 400,400,400,400,....
Calculating the difference, we get,
Common difference, d = 0.
Since the difference between each successive term of the given list of numbers is 0.
Therefore, it forms an AP, with a common difference, d = 0.
(ii) The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs. 250, and it increases by Rs. 50 for the next higher class.
From the question given,
The monthly fee from I to XII is 250, (250+50), (250 + 2 × 50), (250 + 3 × 50),….
(250, 300, 350, 400,….)
Calculating the difference, we get,
d1 = 300 - 250 = 50
d2 = 350 - 300 = 50
d3 = 400 - 350 = 50
Since, the d = 50 is the same for all, i.e. d1 = d2 = d3. The above statement forms an AP.
Therefore a given list of numbers formed by monthly fees from I to XII forms an AP.
Therefore, it forms an A.P with a common difference 50.
(iii) The amount of money in the account of Varun at the end of every year when Rs. 1000 is deposited at simple interest of 10% per annum.
Simple Interest = Principal × rate × time/100.
From the question given,
Simple Interest = 1000× 10 × 1/100 = 100.
Hence, the amount of money in the account at end of every year is
1000, (1000 + 100 × 1), (1000 + 100 × 2), (1000 + 100 × 3),….. = 1000, 1100, 1200, 1300,….
Calculating the difference, we get,
d1 = 1100 - 1000 = 100
d2 = 1200 - 1100 = 100
d3 = 1300 - 1200 = 100
Since, the d = 100 is the same for all, d1 = d2 = d3, the given list of numbers forms an AP.
Therefore, it forms an A.P with a common difference 100
(iv) The number of bacteria in a certain food item after each second, when they double in every second.
Let the number of bacteria present in food initially be x.
Since, they double in every second , x, 2x, 2(2x), 2(2 . 2 . x),…= x, 2x, 4x, 8x,…
From the question given,
t₁ = x,
t₂ = 2x,
t₃ = 4x
t₄ = 8x.
Calculating the difference,we get,
t₂ - t₁ = 2x - x = x
t₃ - t₂ = 4x - 2x = 2x
t₄ - t₃ = 8x - 4x = 4x.
Hence, the difference between each successive term is not the same.
Therefore, the list does not form an AP.
Q.8. Justify whether it is true to say that the following are the nth terms of an AP.
(i) 2n–3
(ii) 3n+ 5
(iii) 1 + n + n2
(i) 2n–3
Yes.
Ac cording to the question,
aₙ = 2n - 3
Substituting the value of n, we get
n = 1, a₁ = 2(1) - 3 = -1
n = 2, a₂ = 2(2) - 3 = 1
n = 3, a₃ = 2(3) - 3 = 3
n = 4, a₄ = 2(4) - 3 = 5
Hence, the sequence becomes -1, 1, 3, 5….
Calculating the difference, we get,
a₂ - a₁ = 1 - (-1) = 1 + 1 = 2
a₃ - a₂ = 3 - 1 = 2
a₄ - a₃ = 5 - 3 = 2.
So, we get,
a₂ - a₁ = a₃ - a₂ = a₄ - a₃
Therefore, 2n - 3 is the nth term of an AP.
(ii) 3n+ 5
No.
According to the question,
aₙ = 3n² + 5
Substituting the value of n, we get,
n = 1, a₁ = 3(1)2 + 5 = 8
n = 2, a₂ = 3(2)2 + 5 = 3(4) + 5 = 17
n = 3, a₁ = 3(3)2 + 5 = 3(9) + 5 = 27 + 5 = 32
Hence, the sequence becomes 8, 17, 32,….
Calculating the difference, we get,
a₂ - a₁ = 17 - 8 = 9
a₃ - a₂ = 32 -17 = 15.
So, we get,
a₂ - a₁ ≠ a₃ - a₂
Since, the difference of each successive term is not the same, it does not form an AP.
Therefore, it does not form an AP.
(iii) 1 + n + n2
No.
According to the question,
aₙ = 1 + n + n²
Substituting the values of n, we get,
n = 1, a₁ = 1 + 1 + (1)² = 3
n = 2, a₂ = 1 + 2 + (2)² = 1 + 2 + 4 = 7
n = 3, a₃ = 1 + 3 + (3)² = 1+ 3 + 9 = 13
Hence, the sequence becomes 3, 7, 13,…
Calculating the difference, we get,
a₂ - a₁ = 7 - 3 = 4
a₃ - a₂ = 13 - 7 = 6.
So, we get,
a₂ - a₁ ≠ a₃ - a₂
Since, the difference of each successive term is not the same, it does not form an AP.
Therefore, it does not form an AP.
### Exercise 5.3
Q. 1. Match the APs given in column A with suitable common differences given in column B.
(A1) AP is 2, – 2, – 6, – 10, ….
So common difference is simply
a2 – a1 = – 2 – 2 = – 4 = (B3)
(A2) Given
First term, a = – 18
No of terms, n = 10
Last term, an = 0
By using the nth term formula
an = a + (n – 1)d
0 = – 18 + (10 – 1)d
18 = 9d
d = 2 = (B5)
(A3) Given
First term, a = 0
Tenth term, a10 = 6
By using the nth term formula
an = a + (n – 1)d
a10 = a + 9d
6 = 0 + 9d
d = 2/3 = (B6)
(A4) Let the first term be a and common difference be d
Given that
a2 = 13
a4 = 3
a2 – a4 = 10
a + d – (a + 3d) = 10
d – 3d = 10
– 2d = 10
d = – 5= (B1)
Q. 2. Verify that each of the following is an AP, and then write its next three terms.
(i) 0, 1/4, 1/2, 3/4,…
Here,
a= 0
a2 = ¼
a3 = ½
a4 = ¾
a2 – a1 = ¼ – 0 = ¼
a3 – a2 = ½ – ¼ = ¼
a4 – a3 = ¾ – ½ = ¼
Since, difference of successive terms are equal,
Hence, 0, 1/4, 1/2, 3/4… is an AP with common difference ¼.
Therefore, the next three term will be,
¾ + ¼ , ¾ + 2(¼), ¾ + 3(¼)
1, 5/4 , 3/2
(ii) 5, 14/3, 13/3, 4…
Here,
a= 5
a2 = 14/3
a3 = 13/3
a4 = 4
a2 – a1 = 14/3 – 5 = -1/3
a3 – a2 = 13/3 – 14/3 = -1/3
a4 – a3 = 4 – 13/3 = -1/3
Since, difference of successive terms are equal,
Hence, 5, 14/3, 13/3, 4… is an AP with common difference -1/3.
Therefore, the next three term will be,
4 + (-1/3), 4 + 2(-1/3), 4 + 3(-1/3)
11/3 , 10/3, 3
(iii)3 , 23, 33,…
Here,
a= √3
a2 = 2√3
a3 = 3√3
a4 = 4√3
a2 – a1 = 2√3 – √3 = √3
a3 – a2 = 3√3 – 2√3= √3
a4 – a3 = 4√3 – 3√3= √3
Since, difference of successive terms are equal,
Hence, √3 , 2√3, 3√3,… is an AP with common difference √3.
Therefore, the next three term will be,
4√3 + √3, 4√3 + 2√3, 4√3 + 3√3
5√3, 6√3, 7√3
(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), …
Here
a1 = a + b
a2 = (a + 1) + b
a3 = (a + 1) + (b + 1)
a2 – a1 = (a + 1) + b – (a + b) = 1
a3 – a2 = (a + 1) + (b + 1) – (a + 1) – b = 1
Since, difference of successive terms are equal,
Hence, + b, (+ 1) + b, (+ 1) + (+ 1), … is an AP with common difference 1.
Therefore, the next three term will be,
(a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1(2), (a + 1) + (b + 1) + 1(3)
(a + 2) + (b + 1), (a + 2) + (b + 2), (a + 3) + (b + 2)
(v) a, 2a + 1, 3a + 2, 4a + 3,…
Here a1 = a
a2 = 2a + 1
a3 = 3a + 2
a4 = 4a + 3
a2 – a1 = (2a + 1) – (a) = a + 1
a3 – a2 = (3a + 2) – (2a + 1) = a + 1
a4 – a3 = (4a + 3) – (3a+2) = a + 1
Since, difference of successive terms are equal,
Hence, a, 2+ 1, 3+ 2, 4+ 3,… is an AP with common difference a + 1.
Therefore, the next three term will be,
4a + 3 +(a + 1), 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1)
5a + 4, 6a + 5, 7a + 6
Q.3. Write the first three terms of the APs when and are as given below:
(i) a = 1/2, d = -1/6
(ii) a = –5, d = –3
(iii) a = 2 , d = 1/2
(i) a =1/2, d = -1/6
We know that,
First three terms of AP are:
a, a + d, a + 2d
½, ½ + (-1/6), ½ + 2 (-1/6)
½, 1/3, 1/6
(ii) a = –5, d = –3
We know that,
First three terms of AP are:
a, a + d, a + 2d
-5, – 5 + 1 (- 3), – 5 + 2 (- 3)
– 5, – 8, – 11
(iii) a = √2 , d = 1/√2
We know that,
First three terms of AP are:
a, a + d, a + 2d
√2, √2+1/√2, √2+2/√2
√2, 3/√2, 4/√2
Q.4. Find a, b and c such that the following numbers are in AP: a, 7, b, 23, c
For a, 7, b, 23, c… to be in AP
it has to satisfy the condition,
a5 – a4 = a4 – a3 = a3 – a2 = a2 – a1 = d
Where d is the common difference
7 – a = b – 7 = 23 – b = c – 23 …(1)
Let us equate,
b – 7 = 23 – b
2b = 30
b = 15 (eqn 1)
And,
7 – a = b – 7
From eqn 1
7 – a = 15 – 7
a = – 1
And,
c – 23 = 23 – b
c – 23 = 23 – 15
c – 23 = 8
c = 31
So a = – 1
b = 15
c = 31
Then, we can say that, the sequence – 1, 7, 15, 23, 31 is an AP
Q.5. Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
We know that,
The first term of an AP = a
And, the common difference = d.
According to the question,
5th term, a5 = 19
Using the nth term formula,
an = a + (n – 1)d
We get,
a + 4d = 19
a = 19 – 4d …(1)
Also,
13th term – 8th term = 20
a + 12d – (a + 7d) = 20
5d = 20
d = 4
Substituting d = 4 in equation 1,
We get,
a = 19 – 4(4)
a = 3
Then, the AP becomes,
3, 3 + 4 , 3 + 2(4),…
3, 7, 11,…
Q.6. The 26th, 11th and the last term of an AP are 0, 3 and -⅕ respectively. Find the common difference and the number of terms.
Consider the first term, common difference and number of terms of an AP are a, d and n, respectively.
If last term of an AP is known,
l = 8 + (11-1 )d …………. (i)
So nth term of an AP is
T= a + (n - 1)d …………. (ii)
We know that,
26th term of an AP = 0
T26 = a + (26 - 1 )d = 0 [from Equation (i)]
8 + 25d = 0 …………. (iii)
11th term of an AP = 3
T11 = s + (11 - 1)d = 3 [from Equation (ii)]
8 + 10d = 3 ……………… (iv)
Last term of an AP = -⅕
l = a + (n - 1 )d [from Equation (i)]
-1/5 = a + (n - 1 )d ………… (v)
Now, subtracting Equation (iv) from Equation (iii),
15 d = - 3
d = -⅕
Substitute the value of d in Equation (iii),
a + 25(-⅕) = 0
a - 5 = 0
a = 5
Substitute the value of a, d in Equation (v), we get
-1/5 = 5 + (n - 1)(-1/5)
-1 = 25 - (n - 1)
-1 = 25 - n + 1
n = 25 + 2 = 27
Therefore, the common difference and number of terms are -1/ 5 and 27.
Q.7. The sum of the 5th and the 7th terms of an AP is 52 and the 10th term is 46. Find the AP.
Consider the first term and common differences of AP are a and d.
From the question,
a5 + a7 = 52
a10 = 46
a + (5 - l)d + a + (7 - 1)d = 52
We know that
an = a + (n- 1 )d]
a + (10 - 1 )d = 4
⇒ a + 4d + a + 6d = 52
Similarly a + 9d = 46
⇒ 2a + 10d = 52
and a + 9d = 46
⇒ a + 5d = 26 ………….. (i)
a + 9d = 46 ……………. (ii)
By subtracting Equation (i) from Equation (ii),
4d = 20
d = 5
From Equation (i)
a = 26 - 5(5) = 1
The required AP is a, a + d, a + 2d, a + 3d ….
i.e., 1, 1 + 5, 1 + 2(5), 1 + 3(5)…
i.e., 1, 6,11,16,….
Therefore, the AP is 1, 6, 11, 16, …..
Q.8. Find the 20th term of the AP whose 7th term is 24 less than the 11th term, first term being 12.
Consider the first term, common difference and number of terms of an AP are a, d and n.
We know that
First term (a) = 12
From the given condition,
7th term (T7) = 11th term (T11) - 24
The nth term of an AP, Tn = a + (n - 1 )d
⇒ a + (7 - 1)d = a + (11 - l)d - 24
⇒ a + 6d = a + 10d - 24
⇒ 24 = 4d
Dividing both sides by 4
⇒ d = 6
So the 20th term of AP,
T20 = a + (20 - 1)d
Substituting the values
= 12 + 19 × 6
= 126
Therefore, the 20th term of an AP is 126.
Q.9. If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term.
An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same.
Consider the first term, common difference and number of terms of an AP are a, d and n.
We know that,
9th term of an AP, T9 = 0
So the nth term of an AP, Tn = a + (n - 1 )d
⇒ a + (9 - 1)d = 0
⇒ a + 8d = 0 ⇒ a = -8 d ………….. (i)
The 19th term, T19 = a + (19 - 1)d
= -8d + 18d [From Equation (i)
= 10d …………… (ii)
The 29th term, T29 = a + (29 - 1)d
= -8d + 28d
= 20d
= 2 × (10d) [from Equation (i)]
⇒ T29 = 2 × T19
Therefore, it is proved.
Q.10. Find whether 55 is a term of the AP: 7, 10, 13,--- or not. If yes, find which term it is.
Consider the first term, common difference and the number of terms of an AP are a, d and n.
nth term of an AP is Tn = 55
We know that the nth term of an AP,
Tn = a + (n - 1 )d …………….. (i)
It is given,
first term (a) = 7
common difference (d) = 10 - 7 = 3
From equation (i),
55 = 7 + (n - 1) × 3
55 = 7 + 3n - 3
55 = 4 + 3n
3n = 51
Dividing both sides by 3
n = 17
n is a positive integer. So 55 is a term of the AP as n = 17
Therefore, the 17th term of an AP is 55.
Q.11. Determine k so that k2+ 4k + 8, 2k2 + 3 k + 6, 3k2 + 4 k + 4 are three consecutive terms of an AP.
It is given that
k² + 4k + 8, 2k² + 3k + 6 and 3k² + 4k+ 4 are consecutive terms of an AP.
2k²+ 3k + 6- (k² + 4k + 8)
= 3k² + 4k + 4 - (2k² + 3k + 6) is the Common difference
By multiplying the negative sign
2k² + 3k + 6 - k² - 4k - 8 = 3k² + 4k + 4 - 2k² - 3k - 6
k² - k - 2 = k² + k - 2
-k = k
2k = 0
k = 0
Therefore, k is 0.
Q.12. Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.
Consider the three parts of the number 207 are (a - d), a and (a + d), which are in AP.
From the given condition,
Sum of these parts = 207
a - d + a + a + d = 207
3a = 207
Dividing both sides by 3
a = 69
It is given that,
Product of the two smaller parts = 4623
a(a -d) = 4623
Substituting the values
69 . (69 - d) = 4623
69 - d = 67
So we get
d = 69 - 67 = 2
First part = a - d = 69 - 2 = 67,
Second part = a = 69
Third part = n + d = 69 + 2 = 71
Therefore, the required three parts are 67, 69, 71.
Q.13. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Given ,
The angles of a triangle are in AP.
Consider A, B and C are angles of a ∆ABC
B = (A + C)/2
2B = A + C …(i)
We know that, the sum of all interior angles of a ∆ABC is 180°
A + B + C = 180°
2B + B = 180° [from Equation (i)]
3B = 180°
B = 60°
Let A and C be the greatest and least angles
A = 2C …………. (ii)
Substituting the values of B and A in Equation (i),
2 × 60° = 2C + C
120° = 3 C
C = 40°
Substitute value of C in Equation (ii),
A = 2 × 40°
A = 80°
Therefore, all the angles of a triangle are 80°, 60° and 40°.
Q.14. If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term.
Consider the first term, common difference and number of terms of the AP: 9, 7, 5,…. are
a1, d1 and n1
First term (a1) = 9
Common difference (d1) = 7 - 9 = -2
T’n1 = a1 + (n1 - 1) d
= 9+ (n1 - 1) (-2)
= 9 - 2n1 + 2
T’n= 11 - 2n1 ….. (i)
The nth term of an AP is
Tn = a + (n - 1) d
Consider the first term, common difference, and the number of terms of the AP: 24, 21, 18, … are a2, d2, and n2
First term, (a2) = 24
Common difference (d2) = 21 - 24 = -3
The nth term T’’n2 = a2 + (n2 - 1)d2
T”n= 24 + (n2 - 1) (-3)
T”n= 24 - 3n+ 3
T”n2 = 27 - 3n…. (ii)
From the given condition the nth term of both AP is same
11 - 2n1 = 27 - 3n2
So we get
n = 16
The nth term of first AP is
T’n1 = 11 - 2n1 = 11 - 2 (16) = 11 - 32 = -21
The nth term of second AP is
T”n2 = 27 - 3n2 = 27 - 3 (16) = 27 - 48 = -21
Therefore, the value of n is 16 and that term i.e., nth term is -21.
Q.15. If sum of the 3rd and the 8th terms of an AP is 7 and the sum of the 7th and the 14th terms is –3, find the 10th term.
Consider the first term and common difference of an AP are a and d.
From the question,
a3 + a8 = 7 and a17 + a14 = -3
a + (3 - 1)d + a + (8 - 1)d = 7
We know that an = a + (n- 1 )d
a + (7 - 1 )d + a + (14 - 1 )d = -3
a + 2d + a + 7d = 7
a + 6d + a + 13d = -3
2a + 9d = 7 ………….. (i)
2a + 19d = -3 …(ii)
By subtracting equation (i) from equation (ii),
10d = -10
d = -1
2a + 9(-1) = 7 [from equation (i)]
2a - 9 = 7
2a = 16
Dividing both sides by 2
a = 8
So we get
a10 = a + (10 - 1)d
= 8 + 9(-1)
= 8 - 9
= -1
Therefore, the 10th term is -1.
Q.16. Find the 12th term from the end of the AP: –2, –4, –6,..., –100.
It is given that
AP : -2, -4, -6,…., -100
First term (a) = -2,
Common difference (d) = -4 - (-2) = -2
Last term (l) = -100
The nth term of an AP from the end
an = l - (n - 1 )d,
where l is the last term
d is the common difference.
So the 12th term from the end,
a12 = -100 - (12 - 1)(-2)
= -100 + (11)(2)
= -100 + 22
= -78
Therefore, the 12th term from the end is -78.
Q.17. Which term of the AP: 53, 48, 43,... is the first negative term?
It is given that
AP: 53, 48, 43,…
First term (a) = 53 and
Common difference (d) = 48 - 53 = -5
Consider the nth term of the AP as the first negative term.
i.e., Tn < 0
We know that the nth term of an AP,
Tn = a + (n - 1)d
Here
⇒ [a + (n - 1 )d] < 0
Substituting the values
⇒ 53 + (n - 1)(-5) < 0
⇒ 53 - 5n + 5 < 0
So we get
⇒ 58 - 5n < 0
⇒ 5n > 58
⇒ n > 11.6
⇒ n = 12
So the 12th term is the first negative term of the given AP
T12 = a + (12 - 1)d
= 53 + 11 (-5)
= 53 - 55
= - 2 < 0
Therefore, the first negative term is the 12th term.
Q.18. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3.
The first number is 11, which divided by 4 leaves a remainder 3 between 10 and 300
The last term before 300 is 299, which is divided by 4 leave remainder 3.
So it is in an arithmetic progression as
11, 15, 19, 23, 299
First-term (a) = 11,
Common difference (d) = 15 - 11 = 4
We know that nth term,
a= a + (n - 1)d
Substituting the values
299 = 11 + (n - 1) 4
4(n - 1) = 288
So we get
(n - 1) = 72
n = 73
Therefore, 73 numbers lie between 10 and 300.
Q.19. Find the sum of the two middle most terms of the AP: -4/3, -1, -2/3,.........,4 ⅓
Given, the arithmetic series is -4/3, -1, -2/3,-------------,4 ⅓
We have to find the sum of the two middle most terms of the series.
From the given series,
First term, a = -4/3,
aₙ = 4 ⅓ = 13/3
Common difference, d = -1 - (-4/3)
= -1 + 4/3
= (-3+4)/3
d = 1/3
The n-th term of an AP is given by
aₙ = a + (n - 1)d
To find n,
13/3 = -4/3 + (n - 1)(1/3)
13/3 + 4/3 = (n - 1)(1/3)
17/3 = (1/3)(n - 1)
n - 1 = 17
n = 17 + 1
n = 18
The two middle most terms of the 18 terms of the series are 9th and 10th terms.
So, a₉ = -4/3 + (9 - 1)(1/3)
= -4/3 + (8/3)
= (-4+8)/3
a₉ = 4/3
a₁₀ = -4/3 + (10 - 1)(1/3)
= -4/3 + (9/3)
= (-4+9)/3
a₁₀ = 5/3
Now, sum of two middle terms = a₉ + a₁₀
= 4/3 + 5/3
= (4+5)/3
= 9/3
= 3
Therefore, the sum of the two middlemost terms of the given AP is 3.
Q.20. The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Given, the first term of an AP, a = -5
Last term of an AP, l = 45.
Sum of the term of the AP, S = 120
We have to find the number of terms and the common difference.
We know that, if l is the last term of an AP, then the sum of the terms is given by
S = n/2[a + l]
So, 120 = n/2[-5 + 45]
120 = n/2[40]
120 = 20n
n = 120/20
n = 6
The nth term of an AP is given by
aₙ = a + (n - 1)d
Given, a₆ = 45
45 = -5 + (6 - 1)d
45 + 5 = 5d
50 = 5d
d = 50/5
d = 10
Therefore, the number of terms is 6 and the common difference is 10.
Q.21. Find the sum:
(i) 1 + (–2) + (–5) + (–8) + ... + (–236)
(ii) - 1/n + 4 - 2/n + 4 - 3/n +.....upto n terms
(iii) (a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) to 11 terms
(i) 1 + (–2) + (–5) + (–8) + ... + (–236)
Given, the series is 1 + (-2) + (-5) + (-8) + ... + (-236)
We have to find the sum of the series.
From the given series,
First term, a= 1
Last term, n = -236
Common difference, d = -2 - 1 = -3
The nth term of the series in AP is given by
aₙ = a + (n - 1)d
-236 = 1 + (n - 1)(-3)
-236 - 1 = (n - 1)(-3)
n - 1 = 237/3
n - 1 = 79
n = 79 + 1
n = 80
We know that, if l is the last term of an AP, then the sum of the terms is given by
S = n/2[a+l]
So, S = 80/2[1 + (-236)]
= 40[-235]
= -9400
Therefore, the sum of the term is -9400.
(ii) - 1/n + 4 - 2/n + 4 - 3/n +.....upto n terms
Given, the series is 4 - 1/n + 4 - 2/n + 4 - 3/n +.....upto n terms.
We have to find the sum of the series upto n terms.
The given series can also be written as
4n - (1/n + 2/n, ………….+ n/n)
= 4n - (1 + 2 + 3 + .....+ n)/n
The sum of the first n natural numbers is given by
S = n(n + 1)/2
So, the series can be rewritten as
= 4n - n(n + 1)/2n
Cancelling out common term,
= 4n - (n + 1)/2
= 8n - n - 1/2
= (7n - 1)/2
Therefore, the sum of the term is (7n - 1)/2.
(ii) (a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) to 11 terms
Given, the series is
We have to find the sum of the arithemtic series upto 11 terms.
From the given series,
First term, a = (a - b)/(a + b)
Common difference, d = (3a - 2b)/(a + b) - (a - b)/(a + b)
= (3a - 2b - a + b)/(a + b)
d = (2a - b)/(a + b)
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n - 1)d]
So, S₁₁ = (11/2)[2(a - b)/(a + b) + (11-1)(2a - b)/(a + b)]
= (11/2)[(2a - 2b)/(a + b) + (10)(2a - b)/(a + b)]
= (11/2)[(2a - 2b)/(a + b) + (20a - 10b)/(a + b)]
= (11/2)[2a - 2b + 20a - 10b]/(a + b)
= (11/2)/(a + b)[22a - 12b]
Taking out common term,
= (11/2)/(a + b)[2(11a - 6b)]
= (11/a + b)[11a - 6b]
= 11(11a - 6b)/(a+b)
Therefore, the sum of the series upto 11 terms is 11(11a - 6b)/(a + b)
Q.22. Which term of the AP: –2, –7, –12,... will be –77? Find the sum of this AP upto the term –77.
Given, the series is -2, -7, -12,......., -77.
We have to find the sum up to the term -77.
From the series,
First term, a = -2
Last term, l = -77
Common difference, d = -7 - (-2) = -7 + 2 = -5
The nth term of the series in AP is given by
aₙ = a + (n - 1)d
So, -77 = -2 + (n - 1)(-5)
-77 + 2 = (n - 1)(-5)
75 = (n - 1)(5)
n - 1 = 75/5
n - 1 = 15
n = 15 + 1
n = 16
If l is the last term of an AP, then the sum of the terms is given by
S = n/2[a+l]
So, S = 16/2[-2 + (-77)]
S = 8[-2 - 77]
= 8(-79)
S = -632
Therefore, the sum of the term is -632.
Q.23. If a= 3 – 4 n, show that a1 ,a2 , a3 , ... form an AP. Also find S20.
Given, aₙ = 3 - 4n
We have to show that the given expression forms an AP and find the sum up to 20 terms.
Put n =1, a₁ = 3 - 4(1) = 3 - 4 = -1
Put n= 2, a₂ = 3 - 4(2) = 3 - 8 = -5
Put n= 3, a₃ = 3 - 4(3) = 3 - 12 = -9
Put n= 4, a₄ = 3 - 4(4) = 3 - 16 = -13
Put n= 20, a₂₀ = 3 - 4(20) = 3 - 80 = -77
So, the series is -1, -5, -9, -13,........, -77.
First term, a = -1
Last term, l = -77
Common difference, d = -5 - (-1) = -5 + 1 = -4
To check whether the series form an AP,
Common difference, d = -5 - (-1) = -9 - (-5) = -13 - (-9)
d = -5 + 1 = -9 + 5 = -13 + 9 = -4
d = -4
It is clear that the series is in AP.
If l is the last term of an AP, then the sum of the terms is given by
S = n/2[a + l]
So, S = 20/2[-1 + (-77)]
= 10[-5-77]
= 10(-78)
= -780
Therefore, the sum of the series up to 20 terms is -780.
Q.24. In an AP, if Sn = n (4n + 1), find the A.
Given, the expression for the sum of the terms is Sₙ = n(4n + 1)
We have to find the AP.
Put n = 1,
S₁ = 1(4(1) + 1) = 4 + 1 = 5
Put n =2,
S₂ = 2(4(2) + 1) = 2(8 + 1) = 2(9) = 18
The AP in terms of common difference is given by
a, a+d, a+2d, a+3d,......., a+(n-1)d
So, S₁ = a
First term, a = 5
S₂ = sum of first two terms of an AP
= a+ a + d
= 2a + d
To find the common difference d,
2a + d = 18
2(5) + d = 18
10 + d = 18
d = 18 - 10
d = 8
The series can be framed as
a = 5
a + d = 5 + 8 = 13
a + 2d = 5 + 2(8) = 5 + 16 = 21
a + 3d = 5 + 3(8) = 5 + 24 = 29
Therefore, the series is 5, 13, 21, 29,.....
Q.25. In an AP, if Sn = 3 n2 + 5n and a= 164, find the value of k.
Given, the expression for the sum of the series is Sₙ = 3n² + 5n
Last term, l is ak = 164
We have to find the value of k.
Put n = 1, S₁ = 3(1)² + 5(1) = 3 + 5 = 8
S₂ = 3(2)² + 5(2) = 3(4) + 10 = 12 + 10 = 22
The AP in terms of common difference is given by
a, a+d, a+2d, a+3d,......., a+(n - 1)d
So, S₁ = a
First term, a = 8
S₂ = sum of first two term of an AP
= a+ a + d
= 2a + d
To find the common difference d,
2a + d = 22
2(8) + d = 22
16 + d = 22
d = 22 - 16
d = 6
The series can be framed as
a = 8
a + d = 8 + 6 = 14
a + 2d = 8 + 2(6) = 8 + 12 = 20
a + 3d = 8 + 3(6) = 8 + 18 = 26
Therefore, the series is 8, 14, 20, 26,.....
The nth term of the series in AP is given by
aₙ = a + (n - 1)d
164 = 8 + (n - 1)(6)
164 - 8 = 6(n - 1)
n - 1 = 156/6
n - 1 = 26
n = 26 + 1
n = 27
Therefore, the value of k is 27.
Q.26. If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 –S4)
Given, Sₙ denotes the sum of first n terms of an AP.
We have to prove that S₁₂ = 3(S₈ - S₄).
The sum of the first n terms of an AP is given by
Sₙ = (n/2)[2a + (n-1)d]
LHS: S₁₂
When n = 12, S₁₂ = (12/2)[2a + (12-1)d]
S₁₂ = 6[2a + 11d]
S₁₂ = 12a + 66d
RHS: 3(S₈ - S₄)
When n = 8, S₈ = (8/2)[2a + (8-1)d]
S₈ = 4[2a + 7d]
S₈ = 8a + 28d
When n = 4, S₄ = (4/2)[2a + (4-1)d]
S₄ = 2[2a + 3d]
S₄ = 4a + 6d
Now, S₈ - S₄ = (8a + 28d) - (4a + 6d)
= 8a + 28d - 4a - 6d
= 8a - 4a + 28d - 6
= 4a + 22d
3(S₈ - S₄) = 3(4a + 22d)
= 12a + 66d
= S₁₂
LHS = RHS
Therefore, it is proved that S₁₂ = 3(S₈ - S₄).
Q.27. Find the sum of first 17 terms of an AP whose 4th and 9th terms are –15 and –30 respectively.
Given, the 4-th term of an AP = -15
The 9-th term of an AP = -30
We have to find the first 17 terms of an AP.
The nth term of the series in AP is given by
aₙ = a + (n - 1)d
When n = 4, a₄ = a + (4 - 1) d
a + 3d = -15 --------------- (1)
When n = 9, a₉ = a + (9 - 1)d
a + 8d = -30 ----------------(2)
Subtracting (1) from (2),
a + 8d - (a + 3d) = -30 - (-15)
a + 8d - a - 3d = -30 + 15
8d - 3d = -15
5d = -15
d = -15/5
d = -3
Put d = -3 in (1),
a + 3(-3) = -15
a - 9 = -15
a = -15 + 9
a = -6
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n - 1)d]
To find the sum of the 17 terms,
S17 = 17/2[2(-6) + (17 - 1)(-3)]
= 17/2[-12 + (16)(-3)]
= 17/2[-12 - 48]
= 17/2[-60]
= 17(-30)
S17 = -510
Therefore, the sum of the first 17 terms is -510.
Q.28. If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Given, the sum of first 6 terms of an AP is 36
Also, the sum of the first 16 terms of an AP is 256.
We have to find the sum of the first 10 terms.
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n-1)d]
Where, a is the first term
n is the number of terms
d is the common difference
When n = 6,
S₆ = 6/2[2a + (6 - 1)d]
36 = 3[2a + 5d]
2a + 5d = 12 -------------------- (1)
When n = 16,
S₁₆ = 16/2[2a + (16 - 1)d]
256 = 8[2a + 15d]
2a + 15d = 32 ------------------- (2)
Subtracting (1) from (2),
2a + 15d - (2a + 5d) = 32 - 12
2a + 15d - 2a - 5d = 20
15d - 5d = 20
10d = 20
d = 20/10
d = 2
Put d = 2 in (1),
2a + 5(2) = 12
2a + 10 = 12
2a = 12 -10
2a = 2
a = 1
So, the first term, a = 1
Common difference, d = 2
Sum of the first 10 terms, S₁₀ = 10/2[2(1) + (10 - 1)(2)]
= 5[2 + 9(2)]
= 5[2 + 18]
= 5(20
= 100
Therefore, the sum of the first 10 terms of an AP is 100.
Q.29. Find the sum of all the 11 terms of an AP whose middle most term is 30.
Given, the middle term of an AP is 30.
We have to find the sum of all the 11 terms of an AP.
There are 11 terms in total. So, the middle term will be the 6th term.
The nth term of the series in AP is given by
aₙ = a + (n - 1)d
a + (6 - 1)d = 30
a + 5d = 30 ---------- (1)
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n-1)d]
The sum of the first 11 terms of the series is
S₁₁ = 11/2[2a + (11 - 1)d]
= 11/2[2a + 10d]
Taking out common term,
= (11/2)2[a + 5d]
Cancelling out common term,
S₁₁ = 11[a + 5d]--------- (2)
Substituting (1) in (2)
= 11(30)
= 330
Therefore, the sum of the first 11 terms of an AP is 330.
Q. 30. Find the sum of last ten terms of the AP: 8, 10, 12,---, 126.
Given, the series in AP as 8, 10, 12, ………, 126.
We have to find the sum of the last 10 terms of the series.
From the series,
We need to find the sum of last 10 terms,
So, common difference, d = 8 -10 = -2
so we take the first term (a) as 126.
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n-1)d]
So, S₁₀ = 10/2[2(126) + (10 - 1)(-2)]
= 5[252 + 9(-2)]
= 5[252 - 18]
= 5[234]
= 1170
Therefore, the sum of the last 10 terms of an AP is 1170.
Q. 31. Find the sum of first seven numbers which are multiples of 2 as well as of 9. [Hint: Take the LCM of 2 and 9]
Given, the series is a multiple of 2 as well as 9.
Also, take the LCM of 2 and 9.
We have to find the sum of the first seven numbers.
LCM of 2 and 9 is 18.
So, the series is a multiple of 18.
The first seven multiple of 18 are (18×1), (18×2), (18×3), (18×4), (18×5), (18×6), (18×7)
Therefore, the series is 18, 36, 54, 72, 90, 108, 126.
If l is the last term of an AP, then the sum of the terms is given by
S = n/2[a + l]
Here, first term, a = 18
Last term, l = 126
Common difference, d = 18
S = 7/2[18 + 126]
= 7/2[144]
= 7(72)
= 504
Therefore, the sum of the first seven numbers which are multiples of 2 as well as 9 is 504.
Q.32. How many terms of the AP: –15, –13, –11,--- are needed to make the sum –55? Explain the reason for double answer.
Given, the series in AP is -15, -13, -11,.......
We have to find the number of terms required to make the sum equal to -55.
From the series,
First term, a = -15
Common difference, d= -13 - (-15) = -13 + 15 = 2
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n - 1)d]
So, Sₙ = n/2[2(-15) + (n - 1)(2)]
= n/2[-30 + 2n - 2]
= n/2[2n -32]
= (n/2)2[n - 16]
= n(n - 16)
= n² - 16n
Given, the sum is -55
So, Sₙ = -55
n² - 16n + 55 = 0
On factoring,
n² - 11n - 5n + 55 = 0
n(n - 11)- 5(n - 11) = 0
(n - 5)(n - 11) = 0
Now, n - 5 = 0
n = 5
Also, n - 11 = 0
n = 11
Therefore, the sum of the first 5 terms or the sum of the first 11 terms of an AP is equal to -55.
The 11 terms of the series are -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5.
The reason for the double answer is that the AP is increasing with positive values.
So, the sum of the first 5 terms will be equal to -55
As the AP increases with positive values the sum of the first 11 terms equals -55, as the last 6 terms sum up to 0.
Q. 33. The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is – 30 and the common difference is 8. Find n.
Given,
The sum of first n terms of an AP is equal to the sum of first 2n terms of another AP.
We have to find the value of n.
given, the first AP series have
First term, a = 8
Common difference, d = 20
Also, the second AP series have
First term, a = -30
Common difference, d = 8
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n - 1)d]
For the first series, Sₙ = n/2[2(8) + (n - 1)(20)]
= n/2[16 + 20n - 20]
= n/2[20n - 4]
Taking out common term,
= (n/2)2[10n - 2]
Cancelling out common term,
Sₙ = n[10n - 2]
For the second series, S₂ₙ = (2n/2)[2(-30) + (2n - 1)(8)]
= n[-60 + 16n - 8]
S₂ₙ = n[16n - 68]
Given, Sₙ = S₂ₙ
n[10n - 2] = n[16n - 68]
10n - 2 = 16n - 68
10n - 16n = -68 + 2
-6n = -66
n = 66/6
n = 11
Therefore, the value of n is 11.
Q. 34. Kanika was given her pocket money on Jan 1st, 2008. She puts Re 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?
It is stated that Kanika was given pocket money on Jan 1st 2008.
She put Re.1 on day 1, Rs.2 on day 2, Rs.3 on day, and so on into her piggy bank till the end of the month.
We have to find the amount of pocket money for the month.
January has 31 days.
So, the series will be 1, 2, 3, ……….,31.
From the series,
First term, a = 1
Last term, l = 31
Common difference, d = 2 -1 = 1
If l is the last term of an AP, then the sum of the terms is given by
S = n/2[a + l]
Here, n = 31
So, S = 31/2[1 + 31]
= 31/2[32]
= 31(16)
S = 496
Given, Rs. 204 from the pocket money was spent by Kanika.
After spending the amount of money left in the piggy bank is Rs.100.
Kanika’s total pocket money = 496 + 204 + 100
= 496 + 304
= 800
Therefore, the total pocket money is Rs.800.
Q. 35. Yasmeen saves Rs. 32 during the first month, Rs. 36 in the second month and Rs. 40 in the third month. If she continues to save in this manner, in how many months will she save Rs. 2000?
Given, Yasmeen saves Rs 32 in the first month, Rs 36 in the second month and Rs 40 in the third month.
We have to find how many months it will take to save in this manner upto Rs 2000.
The series is 32, 36, 40,.....
Common difference, d = 36 - 32 = 4
First term, a = 32
Sum = 2000
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n-1)d]
So, 2000 = n/2[2(32) + (n - 1)(4)]
2000 = n/2[64 + 4n - 4]
4000 = n[4n + 60]
4n² + 60n = 4000
4n² + 60n - 4000 = 0
Dividing by 4,
n² + 15n - 1000 = 0
On factoring,
n² + 40n - 25n - 1000 = 0
n(n + 40) - 25(n + 40) = 0
(n - 25)(n + 40) = 0
Now, n - 25 = 0
n = 25
Also, n + 40 = 0
n = -40
Since a negative value is not possible, n = -40 is neglected.
Therefore, the number of months required to save upto Rs.2000 is 25.
### Exercise 5.4
Q.1. The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.
We know that, in an A.P.,
First term = a
Common difference = d
Number of terms of an AP = n
According to the question,
We have,
S5 + S7 = 167
Using the formula for sum of n terms,
Sn = (n/2) [2a + (n-1)d]
So, we get,
(5/2) [2a + (5-1)d] + (7/2)[2a + (7-1)d] = 167
5(2a + 4d) + 7(2a + 6d) = 334
10a + 20d + 14a + 42d = 334
24a + 62d = 334
12a + 31d = 167
12a = 167 – 31d …(1)
We have,
S10 = 235
(10/2) [2a + (10-1)d] = 235
5[ 2a + 9d] = 235
2a + 9d = 47
Multiplying L.H.S and R.H.S by 6,
We get,
12a + 54d = 282
From equation (1)
167 – 31d + 54d = 282
23d = 282 – 167
23d = 115
d = 5
Substituting the value of d = 5 in equation (1)
12a = 167 – 31(5)
12a = 167 – 155
12a = 12
a = 1
We know that,
S20 = (n/2) [2a + (20 – 1)d]
= 20/(2[2(1) + 19 (5)])
= 10[ 2 + 95]
= 970
Therefore, the sum of first 20 terms is 970.
Q.2. Find the
(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.
[Hint (iii): These numbers will be: multiples of 2 + multiples of 5 – multiples of 2 as well as of 5]
(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
We know that,
Multiples of 2 as well as of 5 = LCM of (2, 5) = 10
Multiples of 2 as well as of 5 between 1 and 500 = 10, 20, 30…, 490.
Hence,
We can conclude that 10, 20, 30…, 490 is an AP with common difference, d = 10
First term, a = 10
Let the number of terms in this AP = n
Using nth term formula,
an = a + (n – 1)d
490 = 10 + (n – 1)10
480 = (n – 1)10
n – 1 = 48
n = 49
Sum of an AP,
Sn = (n/2) [a + an], here an is the last term, which is given]
= (49/2) × [10 + 490]
= (49/2) × [500]
= 49 × 250
= 12250
Therefore, sum of those integers between 1 and 500 which are multiples of 2 as well as of 5 = 12250
(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
We know that,
Multiples of 2 as well as of 5 = LCM of (2, 5) = 10
Multiples of 2 as well as of 5 from 1 and 500 = 10, 20, 30…, 500.
Hence,
We can conclude that 10, 20, 30…, 500 is an AP with common difference, d = 10
First term, a = 10
Let the number of terms in this AP = n
Using nth term formula,
an = a + (n – 1)d
500 = 10 + (n – 1)10
490 = (n – 1)10
n – 1 = 49
n = 50
Sum of an AP,
Sn = (n/2) [ a + an], here an is the last term, which is given]
= (50/2) ×[10 + 500]
= 25× [10 + 500]
= 25(510)
= 12750
Therefore, sum of those integers from 1 to 500 which are multiples of 2 as well as of 5= 12750
(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.
We know that,
Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2, 5)
Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (10)
Multiples of 2 or 5 from 1 to 500 = List of multiple of 2 from 1 to 500 + List of multiple
of 5 from 1 to 500 – List of multiple of 10 from 1 to 500
= (2, 4, 6… 500) + (5, 10, 15… 500) – (10, 20, 30… 500)
Required sum = sum(2, 4, 6,…, 500) + sum(5, 10, 15,…, 500) – sum(10, 20, 30,., 500)
Consider the first series,
2, 4, 6, …., 500
First term, a = 2
Common difference, d = 2
Let n be no of terms
an = a + (n – 1)d
500 = 2 + (n – 1)2
498 = (n – 1)2
n – 1 = 249
n = 250
Sum of an AP, S= (n/2) [ a + an]
Let the sum of this AP be S1,
S= S250 = (250/2) ×[2 + 500]
S1 = 125(502)
S1 = 62750 … (1)
Consider the second series,
5, 10, 15, …., 500
First term, a = 5
Common difference, d = 5
Let n be no of terms
By nth term formula
an = a + (n – 1)d
500 = 5 + (n – 1)
495 = (n – 1)5
n – 1 = 99
n = 100
Sum of an AP, S= (n/2) [ a + an]
Let the sum of this AP be S2,
S= S100 = (100/2) ×[5 + 500]
S2 = 50(505)
S2 = 25250 … (2)
Consider the third series,
10, 20, 30, …., 500
First term, a = 10
Common difference, d = 10
Let n be no of terms
an = a + (n – 1)d
500 = 10 + (n – 1)10
490 = (n – 1)10
n – 1 = 49
n = 50
Sum of an AP, S= (n/2) [ a + an]
Let the sum of this AP be S3,
S= S50 = (50/2) × [2 + 510]
S3 = 25(510)
S3 = 12750 … (3)
Therefore, the required Sum, S = S1 + S2 – S3
S = 62750 + 25250 – 12750
= 75250
Q.3. The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
We know that,
First term of an AP = a
Common difference of AP = d
nth term of an AP, an = a + (n – 1)d
According to the question,
as = ½ a2
2a8 = a2
2(a + 7d) = a + d
2a + 14d = a + d
a = – 13d …(1)
Also,
a11 = 1/3 a4 + 1
3(a + 10d) = a + 3d + 3
3a + 30d = a + 3d + 3
2a + 27d = 3
Substituting a = -13d in the equation,
2 (- 13d) + 27d = 3
d = 3
Then,
a = – 13(3)= – 39
Now,
a15 = a + 14d
= – 39 + 14(3)
= – 39 + 42
= 3
So 15th term is 3.
4. An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
We know that,
First term of an AP = a
Common difference of AP = d
nth term of an AP, an = a + (n – 1)d
Since, n = 37 (odd),
Middle term will be (n+1)/2 = 19th term
Thus, the three middle most terms will be,
18th, 19th and 20th terms
According to the question,
a18 + a19 + a20 = 225
Using an = a + (n – 1)d
a + 17d + a + 18d + a + 19d = 225
3a + 54d = 225
3a = 225 – 54d
a = 75 – 18d … (1)
Now, we know that last three terms will be 35th, 36th and 37th terms.
According to the question,
a35 + a36 + a37 = 429
a + 34d + a + 35d + a + 36d = 429
3a + 105d = 429
a + 35d = 143
Substituting a = 75 – 18d from equation 1,
75 – 18d + 35d = 143 [ using eqn1]
17d = 68
d = 4
Then,
a = 75 – 18(4)
a = 3
Therefore, the AP is a, a + d, a + 2d….
i.e. 3, 7, 11….
Q.5. Find the sum of the integers between 100 and 200 that are
(i) divisible by 9
(ii) not divisible by 9
[Hint (ii): These numbers will be: Total numbers – Total numbers divisible by 9]
(i) The number between 100 and 200 which is divisible by 9 = 108, 117, 126, …198
Let the number of terms between 100 and 200 which is divisible by 9 = n
an = a + (n – 1)d
198 = 108 + (n – 1)9
90 = (n – 1)9
n – 1 = 10
n = 11
Sum of an AP = S= (n/2) [ a + an]
Sn = (11/2) × [108 + 198]
= (11/2) × 306
= 11(153)
= 1683
(ii) Sum of the integers between 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (sum of total numbers between 100 and 200 which is divisible by 9)
Sum, S = S1 – S2
Here,
S1 = sum of AP 101, 102, 103, – – – , 199
S2 = sum of AP 108, 117, 126, – – – , 198
For AP 101, 102, 103, – – – , 199
First term, a = 101
Common difference, d = 199
Number of terms = n
Then,
an = a + (n – 1)d
199 = 101 + (n – 1)1
98 = (n – 1)
n = 99
Sum of an AP = S= (n/2) [ a + an]
Sum of this AP,
S= (99/2) × [199 + 101]
= (99/2) × 300
= 99(150)
= 14850
For AP 108, 117, 126, – – – – , 198
First term, a = 108
Common difference, d = 9
Last term, an = 198
Number of terms = n
Then,
an = a + (n – 1)d
198 = 108 + (n – 1)9
10 = (n – 1)
n = 11
Sum of an AP = S= (n/2) [ a + an]
Sum of this AP,
S= (11/2) × [108 + 198]
= (11/2) × (306)
= 11(153)
= 1683
Substituting the value of S1 and S2 in the equation, S = S1 – S2
S = S1 + S2
= 14850 – 1683
= 13167
Q.6. The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
Given, the ratio of 11th term to the 18th term of an AP is 2:3
We have to find
(A) the ratio of the 5th term to the 21st term.
(B) the ratio of the sum of the first five terms to the sum of the first 21 terms.
The nth term of the series in AP is given by
aₙ = a + (n - 1)d
11th term = a + (11-1)d = a + 10d
18th term = a + 17d
As per given condition, 11th term/18th term = 2/3
(a+10d)/(a+17d) = 2/3
3(a+10d) = 2(a+17d)
3a + 30d = 2a + 34d
By grouping,
3a - 2a = 34d - 30d
a = 4d ------------------------ (1)
(A) to find the ratio of the 5th term to the 21st term
5th term = a + 4d
21st term = a + 20d
(a + 4d) /(a + 20d)
Substitute (1) in the above expression,
(4d+4d)/(4d+20d)
= 8d/24d
= 1/3
Therefore, the ratio of the 5th term to the 21st term is 1:3
(B) the ratio of the sum of the first 5 terms to the sum of the first 21 terms.
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n-1)d]
The sum of first five terms is S₅ = 5/2[2a + (5 - 1)d]
S₅ = 5/2[2a + 4d]
S₅ = (5/2)2[a + 2d]
S₅ = 5a + 10d
The sum of first 21 terms is S₂₁ = 21/2[2a + (21 - 1)d]
S₂₁ = 21/2[2a + 20d]
S₂₁ = (21/2)2[a + 10d]
S₂₁ = 21a + 210d
As per given condition, S₅/S₂₁
= (5a+10d)/((21a+210d)
Substitute (1) in the above expression,
= (5(4d) + 10d)/(21(4d) + 210d)
= (20d+10d)/(84d+210d)
= 30d/294d
= 5/49
Therefore, the ratio of the sum of the first 5 terms to the sum of the first 21 terms is 5:49
Q.7. Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to .
Given, the AP is a, b, c,.....
We have to prove that the sum of the AP is equal to
The nth term of the series in AP is given by
aₙ = a + (n - 1)d
Here, first term, a = a
Last term, l = c
Common difference, d = b - a
So, c = a + (n - 1)(b - a)
c - a = (n - 1)(b - a)
n - 1 = (c - a)/(b - a)
n = [(c - a)/(b - a)] + 1
If l is the last term of an AP, then the sum of the terms is given by
S = n/2[a+l]
So, S = ([(c - a)/(b - a)] + 1)/2[a + c]
= ((c - a)+(b - a)[a + c])/2(b - a)
= (c - a + b - a)(a + c)/2(b - a)
S =
Hence proved.
Q.8. Solve the equation – 4 + (–1) + 2 +...+ x = 437.
Given, the series is -4, -1, 2, ……, x
We have to find the value of x.
Here, first term, a = -4
Last term, l = x
Sum of the series, S = 437
Common difference, d = -1 - (-4) = -1 + 4 = 3
The nth term of the series in AP is given by
aₙ = a + (n - 1)d
So, x = -4 + (n - 1)3
x = -4 + 3n -3
x = 3n - 7
3n = x + 7
n = (x + 7)/3
If l is the last term of an AP, then the sum of the terms is given by
S = [n/2][a+l]
So, S = [(x + 7)/6][(-4 + x)]
437 = [(x + 7)(x - 4)]/6
437(6) = (x² + 7x - 4x - 28)
x² + 3x - 28 = 2622
x² + 3x - 2650 = 0
On factoring,
x² + 53x - 50x - 2650 = 0
x(x + 53) - 50(x + 53) = 0
(x - 50)(x + 53) = 0
Now, x + 53 = 0
x = -53
Also, x - 50 = 0
x = 50
Since a negative integer is not possible, x = -53 is neglected.
Therefore, the value of x is 50.
Q.9. Jaspal Singh repays his total loan of Rs. 118000 by paying every month starting with the first instalment of Rs. 1000. If he increases the instalment by Rs. 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still have to pay after the 30th instalment?
Given, Jaspal Singh repays his total loan of Rs. 118000.
Monthly instalment is Rs. 1000.
We have to find the amount of loan he still have to pay after the 30th instalment, if he increases the instalment by Rs.100
Here, first term, a = 1000
Common difference, d = 100
To find a30,
The nth term of the series in AP is given by
an = a + (n - 1)d
So, a30 = 1000 + (30 - 1)(100)
= 1000 + 29(100)
= 1000 + 2900
a30 = 3900
So, the amount paid by Jaspal Singh in 30 instalments = Rs.3900
Loan amount paid in 30 instalments = S30
The sum of the first n terms of an AP is given by
Sn = n/2[2a + (n - 1)d]
So, S30 = 30/2[2(1000) + (30 - 1)(100)]
= 15[2000 + 2900]
= 15(4900)
S30 = Rs.73500
Amount of loan to be paid after 30 instalments = (total loan) - (loan amount paid for 30 instalments)
= 118000 - 73500
= Rs.44500
Therefore, the amount of loan to be paid by Jaspal Singh after 30 instalments is Rs. 44500.
Q.10. The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?
Given, the students decided to fix colourful flags on the straight passage of the school.
Also, there are 27 flags which are to be fixed at intervals of every 2 m.
We have to find the maximum distance she travelled carrying a flag.
The flags are stored at the position of the middle most flag.
Ruchi kept her books where the flags were stored.
Ruchi could carry only one flag at a time.
Here, number of flags, n = 27
The middle flag post is at (n + 1)/2 = (27 + 1)/2 = 28/2 = 14th post
Let us consider that Ruchi first fixes 13 flags on the left side and then completes fixing 13 flags on the right side.
From the middle post she carries one flag to 2m, fixes it and then returns 2m to the middle post.
So, she travels 2 + 2 = 4 m
Similarly, to fix the second flag she travels 4 + 4 = 8 m
To fix the third flag she travels 8 + 4 = 12 m
So, this forms a series
The series is 4, 8, 12, 16,......
Here, first term, a = 4
Common difference, d = 4
Number of flags, n = 13
The sum of the first n terms of an AP is given by
S= n/2[2a + (n - 1)d]
So, S = 13/2[2(4) + (13 - 1)(4)]
= 13/2[8 + 12(4)]
= 13/2[8 + 48]
= 13/2[56]
= 13(28)
= 364 m
So, she travels 364 m to fix flags on the left of the middle post.
So, total distance travelled by her to fix all the flags = 2(364) = 728 m.
Now, the distance travelled by Ruchi carrying the flag
= 728/2
= 364 m
Therefore, the maximum distance travelled by Ruchi carrying the flag is 364 m.
The document NCERT Exemplar: Arithmetic Progressions | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10
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## FAQs on NCERT Exemplar: Arithmetic Progressions - Mathematics (Maths) Class 10
1. What is an arithmetic progression?
Ans. An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. For example, 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.
2. How can the nth term of an arithmetic progression be calculated?
Ans. The nth term of an arithmetic progression can be calculated using the formula: nth term = a + (n-1)d, where 'a' is the first term and 'd' is the common difference.
3. How can we find the sum of the first 'n' terms of an arithmetic progression?
Ans. The sum of the first 'n' terms of an arithmetic progression can be calculated using the formula: Sn = (n/2)(2a + (n-1)d), where 'a' is the first term, 'd' is the common difference, and 'n' is the number of terms.
4. Can an arithmetic progression have a negative common difference?
Ans. Yes, an arithmetic progression can have a negative common difference. For example, -1, -3, -5, -7 is an arithmetic progression with a common difference of -2.
5. What is the difference between an arithmetic progression and a geometric progression?
Ans. In an arithmetic progression, the difference between consecutive terms is constant, while in a geometric progression, the ratio between consecutive terms is constant. In other words, an arithmetic progression adds a fixed number to each term, while a geometric progression multiplies each term by a fixed number.
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# Integrate \int{\left( 1-\cos x \right)\cos e{{c}^{2}}}x dx=\begin{align} &\left( A \right)-\tan \dfrac{x}{2}+c\,\,\,\,\,\,\,\left( B \right)\tan \dfrac{x}{2}+c\, \\ &\left( C \right)-2\tan \dfrac{x}{2}+c\,\,\,\,\left( D \right)2\tan \dfrac{x}{2}+c \\ \end{align} Last updated date: 21st Sep 2024 Total views: 81.6k Views today: 1.81k Answer Verified 81.6k+ views Hint: For this question, first open the bracket, then separate it into two parts. Suppose individual parts in the form of variables like m, n and at last apply integration formulae. Complete step by step solution: The given integrand is \int{\left( 1-\cos x \right)\cos e{{c}^{2}}}x dx bracket with First we will open the brackets with help of multiplication then we have to divide it into two terms or parts and integrate individual parts , so that \int{\left( 1-\cos x \right)\cos e{{c}^{2}}}x dx=\int{\cos e{{c}^{2}}}x dx-\int{\cos x.\cos e{{c}^{2}}}x dx……………………..(i)
Now to integrate it, let us suppose $\sin x= m$
If we differentiate $\sin x=m$, we get by the differentiation formula,
\begin{align} & \dfrac{d}{dx}\sin x=dm \\ & \rightarrow \cos x dx=dm \\ \end{align}
We know that $\cos e{{c}^{2}}x dx=-\cot x$
On putting these values in equation (i)
$\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx$=$-\cot x-\int{\dfrac{dm}{{{m}^{2}}}}$
By integration formula, we know that $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c$, where c is constant of integration.
$\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx$ =$-\cot x+\dfrac{1}{m}$
$\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx$ =$-\cot x+\dfrac{1}{\sin x}$
Because we already supposed that $\sin x=m$
Again we can write $\cot x=\dfrac{\cos x}{\sin x}$, then
$\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx$=$-\dfrac{\cos x}{\sin x}+\dfrac{1}{\sin x}$
By common denominator rule of subtraction,
$\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx$=$-\dfrac{\cos x+1}{\sin x}=\dfrac{1-\cos x}{\sin x}$
By the identity of trigonometry,
$\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx$=$\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}$
$\sin x/2\,\,cancel\,out\,to\,\sin x/2$
$\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx$=$\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}$
We know that$\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}=\tan \dfrac{x}{2}$, then we get
$\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx$=$\tan \dfrac{x}{2}+c$
Where c is constant of integration.
Hence, The option (B) is the correct option.
Additional information: If more than one constant of integration is used while solving the integral, then at the end of the solution write only one constant of integration.
If the denominators of two fractions when they are in addition or subtraction, then we don’t need to take LCM .
Note: Sometimes students make mistakes, they apply the integration formula instead of before opening the brackets, then their solution would be wrong. It's important to solve such types of questions with the help of formulae and identities.
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Polynomials: The Remainder and Factor Theorems
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# Polynomials: The Remainder and Factor Theorems - PowerPoint PPT Presentation
= remainder = P (- 2) . Polynomials: The Remainder and Factor Theorems. The remainder theorem states that if a polynomial, P ( x ), is divided by x – c, then the remainder equals P ( c ).
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## PowerPoint Slideshow about 'Polynomials: The Remainder and Factor Theorems' - ross
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Presentation Transcript
= remainder = P(- 2)
Polynomials: The Remainder and Factor Theorems
The remainder theorem states that if a polynomial, P(x), is divided by x – c, then the remainder equals P(c).
Example 1: For the polynomial, P(x) = 2x3 – 8x2 + 45, (a) find P(- 2) by direct evaluation, (b) find P(- 2) using the remainder theorem.
(a) P(- 2) = 2(- 2)3 – 8(- 2)2 + 45
= - 3
(b) Here c = - 2, so divide (synthetically) P(x) by x + 2.
2 - 8 0 45
- 4 24 - 48
- 2 | 2 - 12 24 - 3
Polynomials: The Remainder and Factor Theorems
The factor theorem states that for a polynomial, P(x), if x – c is a factor, then P(c) = 0. Also, if P(c) = 0, then x – c is a factor.
Example 2: For the polynomial, P(x) = 2x3 – x2 + 3x – 4,
use the factor theorem to show that x – 1 is a
factor.
P(1) = 2(1)3 – (1)2 + 3(1) – 4
= 0.
Since P(1) = 0, x– 1 is a factor.
Slide 2
Polynomials: The Remainder and Factor Theorems
Try: For the polynomial, P(x) = x3 – x2 + x – 6,
(a) find P(5) using the remainder theorem, (b) use the factor theorem to show that x – 2 is a factor.
(a) 1 - 1 1 - 6
5 20 105
5 | 1 4 21 99
= P(5)
(b) P(2) = (2)3 – (2)2 + (2) – 6 = 0
Since P(2) = 0, x– 2 is a factor.
Slide 3
Polynomials: The Remainder and Factor Theorems
END OF PRESENTATION
Click to rerun the slideshow.
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# Ex.13.4 Q1 Surface Areas and Volumes Solution - NCERT Maths Class 10
Go back to 'Ex.13.4'
## Question
A drinking glass is in the shape of a frustum of a cone of height $$14\,\rm{cm.}$$ The diameters of its two circular ends are $$4\,\rm{cm}$$ and $$2\,\rm{cm.}$$ Find the capacity of the glass.
Video Solution
Surface Areas And Volumes
Ex 13.4 | Question 1
## Text Solution
What is Known?
A drinking glass is in the shape of a frustum of a cone of height is $$14\, \rm{cm}.$$
The diameter of its two circular ends are $$4\, \rm{cm}$$ and $$2\,\rm{cm}.$$
What is Unknown?
The capacity of the glass
Reasoning:
Draw a figure to visualize the shape better
Since the glass is in the shape of a frustum of a cone
Therefore, the capacity of the glass $$=$$ Volume of frustum of a cone
We will find the capacity of the glass by using formulae
Volume of frustum of a cone\begin{align} = \frac{1}{3}\pi h\left( {r_1^2 + r_2^2 + {r_1}{r_2}} \right)\end{align}
where $$r1, r2$$ and $$h$$ are the radii and height of the frustum of the cone respectively.
Steps:
Height of glass,$$h = 14 \rm cm$$
Radius of the larger base,\begin{align} {r_1} = \frac{{4 \rm cm}}{2} = 2 \rm cm\end{align}
Radius of the smaller base,\begin{align} {r_2} = \frac{{2 \rm cm}}{2} = 1 \rm cm\end{align}
The capacity of the glass $$=$$ Volume of frustum of a cone
\begin{align}&= \frac{1}{3}\pi h\left( {r_1^2 + r_2^2 + {r_1}{r_2}} \right)\\&= \begin{bmatrix} \frac{1}{3} \times \frac{{22}}{7} \times 14 \rm cm \times \\ \begin{pmatrix} \left( 2 \rm cm \right)^2 + \left( 1 \rm cm \right)^2 \\ + 2 \rm cm \times 1cm \end{pmatrix} \end{bmatrix} \\&= \begin{bmatrix} \frac{{44}}{3} \rm cm \times\\ \begin{pmatrix} 4 \rm c{m^2} + 1c{m^2} \\ + 2 \rm c{m^2} \end{pmatrix} \end{bmatrix} \\&= \frac{{44}}{3} \rm cm \times 7 \rm c{m^2}\\&= \frac{{308}}{3} \rm c{m^2}\\&= 102\frac{2}{3} \rm c{m^2}\end{align}
Therefore, the capacity of the glass is \begin{align}102\frac{2}{3}\,\rm{cm^3}\end{align}
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# 7.7 Volume Washer Methodap Calculus
How do you find the volume of the region bounded by #y=6x# #y=x# and #y=18# is revolved about the y axis? Calculus Applications of Definite Integrals Determining the Volume of a Solid of Revolution 1 Answer. Kuta Software - Infinite Calculus Name Volumes of Revolution - Washers and Disks Date Period For each problem, find the volume of the solid that results when the region enclosed by the curves is revolved about the the x-axis. 1) y = −x2 + 1 y = 0 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8. $begingroup$ The washer method is a mnemonic for setting up a correct integral when finding the volume of a solid of revolution. Once you have an integral (which you seem to have here), the washer method has done its job, and all that's left is pure calculation. $endgroup$ – Arthur Aug 24 '19 at 10:12. 5.2 volumes: disks and washers 401 for all x so we can instead compute the volume with a single integral: V = Z 2 0 p h x2 1 i 2 dx = p Z 2 0 x4 2x2 +1 i dx = p 1 5 x5 2 3 x3 + x 2 0 = p 32 5 16 3 +2 = 46p 15 or about 9.63. Find the volume of the solid formed by revolving the region between f(x) = 3 x and the horizontal line y = 2.
## 1 Answer
#### Explanation:
The region is the bounded region in:
graph{(y-6x)(y-x)(y-0.0001x-18) sqrt(81-(x-9)^2)sqrt(85-(y-9)^2)/sqrt(81-(x-9)^2)sqrt(85-(y-9)^2) = 0 [-28.96, 44.06, -7.7, 28.83]}
Taking vertical slices and integrating over $x$ would require two integrals, so take horizontal slices.
## Calculus Volume Washer Method
Rewrite the region: $x = \frac{1}{6} y$, $x = y$ and $y = 18$
As $y$ goes from $0$ to $18$, x goes from $x = \frac{1}{6} y$ on the left, to $x = y$ on the right.
The greater radius is $R = y$ and the lesser is $r = \frac{1}{6} y$
Evaluate
$\pi {\int}_{0}^{18} \left({R}^{2} - {r}^{2}\right) \mathrm{dy} = \pi {\int}_{0}^{18} \left({y}^{2} - {\left(\frac{y}{6}\right)}^{2}\right) \mathrm{dy}$
$= \frac{35 \pi}{36} {\int}_{0}^{18} {y}^{2} \mathrm{dy}$
$= 1890 \pi$
(Steps omitted because once it is set up, I think this is a straightforward integration.)
## Related topic
Determining the Volume of a Solid of Revolution
Questions
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NCERT Solutions for Class 3 Maths Chapter 3 Give and Take provided here is extremely helpful in revising complete syllabus and getting a strong base on it. NCERT 3rd Class Maths Give and Take all questions are solved with detailed explanation available for students. In this article we had given of NCERT solutions for Give and Take Class 3 Maths step by step solution for each and every question of the chapter. These solutions will also help you with your homework. Best teachers across the India created NCERT solutions for Class 3 Maths Chapter 3 Give and Take according to curriculum and pattern of syllabus as per guidelines of NCERT (CBSE) Books.
## 3rd Class NCERT Maths Chapter 3 Give and Take
Name of Organization NCERT Name of Class 3rd Class Name of Subject Maths Name of Chapter Chapter 3 Name of Content Give and Take Name of Category NCERT Solutions Official site http://ncert.nic.in/
### Chapter 3 Give and Take covers numerous Questions and answers from all topics and sub-topics which are given below
1.Try these on Kittu’s Home.
(a)10 less than 34 is———– (b) 53 – 20 =———–
(c) 11 more than 31 is ———– (d) 11 less than 66 is———–
(e) 62 + 13 =———– (f) 23 less than 89 is———–
(g) 10 and 40 more is———– (h) 9 added to 28 gives———–
(i)The sum of 9 and 44 is———– (j) Reducing 98 by 34 gives———–
(k) 4 and 37 more is———– (l) Take 35 away from 83. We get———–
Ans. (a)24 (b)33 (c)42 (d)55 (e)75 (f)66 (g)50 (h)37 (i)53 (j)64 (k)41 (l)48
How Many Bulbs?
1.A shopkeeper Rafi had 153 candles. Paras gave him 237 more candles. How many candles does Rafi have now?
Ans.
(A)A train compartment is carrying 132 people. Another compartment is carry¬ing 129 people. In all, how many people are there in both the compartments?
Ans.
Thus, there are 261 people in all.
(B) Shanu found 138 pebbles. Karim found 44 pebbles. How many pebbles did they find in all?
Ans.
Thus, total number of pebbles are 182.
(C) A teacher kept a note of which fruits students like in her school. This is .what she found:
Find out:
(a) How many students in the school like oranges?
(b) How many students in the school like mangoes?
(c) Altogether, how many students are there in the school?
(d) Is the number of girls more than 350 or less than 350?
Ans. On completing the table, we have
(a)Number of students who like oranges = 136 + 128 = 264
(b)Number of students who like mangoes = 240 + 243 = 483
(c)Number of students in the school = 376 + 371 = 747
(d)Number of girls = 136 + 240 = 376 which is’more than 350.
Practice Time
Mind Train Game
1.Work out four different ways to write the numbers. If you add all the numbers in the first box, you will always get 59.
Ans.
Can you solve this puzzle? Write the numbers 1, 2, 3, 4, 5, 6 in the circles, so that the sum of the numbers on each side of the figure is 12.
2.Find Mithoo’s bag.
Do all the sums mentally.
(a)75 + 20 = 95 (b) 90 + 60 = 150
(c)25 + 30 + 3 ——– (d) 9 + 40 + 31 ——–
(e)500 + 200——– (f) 400 + 350 ——–
(8)670 + 120——– (h) 380 + 210——–
(i)205 + 650 ——– (j) 128 + 600 ——–
(k)150 + 69 ——– (l) 37 + 46 + 3——–
Find Mithoo’s bag and check your answers. Draw a line through the numbers which are answers written in the boxes above.
Ans. (c) 58 (d) 80 (e) 700 (/) 750 (g) 790 (h) 590 (i) 855 (j) 728 (k) 219 (l) 86.
Card Game
1.You can also play it. Here are the cards for you. Work out the combination. Place the cards in the right boxes.
Ans.
|
Second derivative of $1$
The calculator will find the second derivative of $1$, with steps shown.
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Find $\frac{d^{2}}{dx^{2}} \left(1\right)$.
Find the first derivative $\frac{d}{dx} \left(1\right)$
The derivative of a constant is $0$:
$${\color{red}\left(\frac{d}{dx} \left(1\right)\right)} = {\color{red}\left(0\right)}$$
Thus, $\frac{d}{dx} \left(1\right) = 0$.
Next, $\frac{d^{2}}{dx^{2}} \left(1\right) = \frac{d}{dx} \left(0\right)$
The derivative of a constant is $0$:
$${\color{red}\left(\frac{d}{dx} \left(0\right)\right)} = {\color{red}\left(0\right)}$$
Thus, $\frac{d}{dx} \left(0\right) = 0$.
Therefore, $\frac{d^{2}}{dx^{2}} \left(1\right) = 0$.
$\frac{d^{2}}{dx^{2}} \left(1\right) = 0$A
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#### Choose the Category for Gospel and Mathematics HERE
Categories: Uncategorized
# Form 4 and Form 5 Quadratic KBAT Questions
The questions (No. 8, 9 & 10) are in picture (in Malay). Question 7 is not KBAT.
*KBAT = Kemahiran Berfikir Aras Tinggi (Higher Order Thinking Skills)
Possible solution are as follows:
8) let the initial average speed & time be v and t respectively.
Same distance of 15 km is travelled one way.
Since 15 minutes = 15/60 = 1/4 hours, we have:
15 = vt
Or
t = 15/v… (i)
And
15 = (v + 5)(t – 1/4)…(ii)
Put (i) into (ii)
15 = (v + 5)(15/v – 1/4)
15v = (v + 5) (15 – v/4)
60v = (v + 5)(60 – v)
60v = 60v – v^2 + 300 – 5v
v^2 + 5v – 300 = 0
(v + 20)(v – 15) = 0
Since v > 0,
v = 15 km/h (Answer)
9) Let the initial cost shared per student & the initial number of students be c & n respectively.
Then,
cn = 600
Or,
c = 600/n… (i)
And,
(c – 5)(n + 10) = 600… (ii)
Put (i) into (ii)
(600/n – 5)(n + 10) = 600
(600 – 5n)(n + 10) = 600n
600n + 6000 – 5n^2 – 50n = 600n
5n^2 + 50n – 6000 = 0
n^2 + 10n – 1200 = 0
(n + 40)(n – 30) = 0
Since n > 0,
n = 30 ahli kelab siber (Answer)
10) Let the speed of train A be v.
6 minutes
= 6/60 hours = 1/10 hours
In 6 minutes,
Let the Distance travelled by train A = x
x = (v) x (1/10)
x = (v/10)…(i)
Let the Distance travelled by train B = y.
y = (v + 20) x (1/10)
y = (v + 20)/10… (ii)
Since these distances (x and y) are perpendicular to each other forming a right triangle with hypotenuse = 10
We have by Pythagoras Theorem,
x^2 + y^2 = 10^2… (iii)
Put (i) & (ii) into (iii):
(v/10)^2 + ((v + 20)/10)^2 = 10^2
v^2 + (v + 20)^2 = 10000
2v^2 + 40v + 400 = 10000
2v^2 + 40v – 9600 = 0
v^2 + 20v – 4800 = 0
(v + 80)(v – 60) = 0
Since v > 0,
v = 60 km/h (for train A, Answer)
And
20 + 60 = 80 km/h (for train B, Answer)
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Anonymous Christian
Jonathan Ramachandran is founder at AnonymousChristian.org which serves as a Christian Blog discussing the Greater Hope of Biblical Christian in Christ Alone. He is also an IMO (International Math Coach) and a featuring Guitarist for many bands.
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# Pythagorean theorem
Long before Pythagoras of Samos was credited for Pythagorean Theorem, people have already noticed how the 3 sides of a right triangle have a relationship. The theorem simply says that the sum of the squares of the two sides of a right triangle will be equal to the square of the hypotenuse. A simple formula to represent which is a2 + b2 = c2, where a and b are the triangle’s sides, and c is the hypotenuse.
In the first part of this chapter, we will discuss all about squares and square roots. We will be able to learn how obtain the set of prime factors that comprises a number. We will be using Prime Factorization, a process by which we list down all the factors of a certain number until we get its prime factors. Through this method, we will also be able to pinpoint numbers that are perfect squares.
For the second part, we will look at the Pythagorean Theorem more thoroughly to understand how to use it. We will be solving for the hypotenuse, or the sides based on the Pythagorean Theorem.
For the third segment, we will be looking into how to estimate a square root since not all square roots would give us whole numbers.
For the fourth part of this chapter, we will then be discussing all about Pythagorean Relationship. Now instead of just solving for the sides of a triangle we will be dealing with more complex problems like using shapes adjacent to the right triangle to solve for the value of a, b, or c.
In the last part of this chapter, we will be looking closely at how to use the Pythagorean Theorem in more real life situations and problems. In no time, you would be ready to solve problems that are related to the Pythagorean Theorem and you can check out the Pythagorean Calculator online if you want to check your answer.
### Pythagorean theorem
In the nutshell, Pythagorean theorem/Pythagorean relationship describes the relationship between the lengths and sides of a right triangle. After thousands of repeated examinations by the ancient Greek mathematicians, it was found that the square of the hypotenuse is equal to the sum of the squares of the other two sides c² = a² + b².
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# Warm Up - PowerPoint PPT Presentation
Warm Up. Problem of the Day. Lesson Presentation. Lesson Quizzes. Warm Up Divide. 1. 4.8 ÷ 2 2. 16.1 ÷ 7 3. 0.36 ÷ 3 4. 25.28 ÷ 4. 2.4. 2.3. 0.12. 6.32. Problem of the Day
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Warm Up
Problem of the Day
Lesson Presentation
Lesson Quizzes
Warm Up
Divide.
1. 4.8 ÷ 2
2. 16.1 ÷ 7
3. 0.36 ÷ 3
4. 25.28 ÷ 4
2.4
2.3
0.12
6.32
Problem of the Day
In the following magic square, 3.375 is the product of the numbers in every row, column, and diagonal. Fill in the missing numbers.
3
1
1.5
2.25
0.75
9
Helpful Hint
Multiplying the divisor and the dividend by the same number does not change the quotient.
42 ÷ 6 = 7
10 10
420 ÷ 60 = 7
10 10
4,200 ÷ 600 = 7
Multiply the divisor by 101, or 10 to make it a whole number. Multiply the dividend by the same power of 10.
Think: 1.3 x 10 = 13 5.2 x 10 = 52
Divide as with whole numbers.
Additional Example 1A: Dividing a Decimal by a Decimal
Find the quotient.
5.2 ÷ 1.3
1.3 5.2
4
13 52
–52
0
5.2 ÷ 1.3 = 4
Multiply the divisor by 102, or 100, to make it a whole number. Multiply the dividend by the same power of 10.
Think: 0.36 x 100 = 36 61.3 x 100 = 6,130
36 6,130.000
Additional Example 1B: Dividing a Decimal by a Decimal
61.3 ÷ 0.36
0.36 61.30
Place the decimal point in the quotient. Divide as with whole numbers.
When a repeating pattern occurs, show three dots or draw a bar over the repeating part of the quotient.
__
61.3 ÷ 0.36 = 170.27
Additional Example 1B Continued
0
.2
1
7
7
7
36 6,130.000
-36
253
-252
10
-0
100
-72
280
-252
280
-252
28
Multiply the divisor by 101, or 10 to make it a whole number. Multiply the dividend by the same power of 10.
Think: 1.1 x 10 = 11 6.6 x 10 = 66
Divide as with whole numbers.
Check It Out: Example 1A
Find the quotient.
6.6 ÷ 1.1
1.1 6.6
6
11 66
-66
0
6.6 ÷ 1.1 = 6
Multiply the divisor by 102, or 100, to make it a whole number. Multiply the dividend by the same power of 10.
Think: 0.24 x 100 = 24 51.2 x 100 = 5,120
24 5120.00
Check It Out: Example 1B
51.2 ÷ 0.24
0.24 51.2
0
Place the decimal point in the quotient. Divide as with whole numbers.
When a repeating pattern occurs, show three dots or draw a bar over the repeating part of the quotient.
__
51.2 ÷ 0.24 = 213.3
Check It Out: Example 1B Continued
2
.3
1
3
3
24 5,120.00
-48
32
-24
80
-72
80
-72
80
-72
8
1
Understand the Problem
Additional Example 2: Problem Solving Application
After driving 216.3 miles, the Yorks filled up with 10.5 gallons of gas. On average, how many miles did they drive per gallon of gas?
The answer will be the average number of miles per gallon.
List the important information:
They drove 216.3 miles. They used 10.5 gallons of gas.
2
Make a Plan
3
Solve
Solve a simpler problem by replacing the decimals in the problem with whole numbers.
If they drove 10 miles using 2 gallons of gas, they averaged 5 miles per gallon. You need to divide miles by gallons to solve the problem.
First estimate the answer. You can use compatible numbers.
216.3 ÷ 10.5 200 ÷ 10 = 20
3
Multiply the divisor and dividend by 10. Think: 10.5 x 10 = 105 216.3 x 10 = 2,163
Place the decimal point in the quotient. Divide as with whole numbers.
Solve Continued
10.5 216.3
2
.6
0
105 2163.0
-210
63
-0
630
-630
0
The York family averaged 20.6 miles per gallon.
4
Look Back
The answer is reasonable since 20.6 is close to the estimate of 20.
1
Understand the Problem
Check It Out: Example 2
After driving 191.1 miles, the Changs filled up with 10.5 gallons of gas. On average, how many miles did they drive per gallon of gas?
The answer will be the average number of miles per gallon.
List the important information:
They drove 191.1 miles. They used 10.5 gallons of gas.
2
Make a Plan
3
Solve
Solve a simpler problem by replacing the decimals in the problem with whole numbers.
If they drove 10 miles using 2 gallons of gas, they averaged 5 miles per gallon. You need to divide miles by gallons to solve the problem.
First estimate the answer. You can use compatible numbers.
191.1 ÷ 10.5 190 ÷ 10 = 19
3
Multiply the divisor and dividend by 10. Think: 10.5 x 10 = 105 191.1 x 10 = 1,911
Place the decimal point in the quotient. Divide as with whole numbers.
Solve Continued
10.5 191.1
1
.2
8
105 1911.0
-105
861
-840
210
-210
0
The Chang family averaged 18.2 miles per gallon.
4
Look Back
The answer is reasonable since 18.2 is close to the estimate of 19.
|
# Table of 120
Created by: Team Maths - Examples.com, Last Updated: June 10, 2024
## Table of 120
The table of 120, also known as the 120 times table, is a sequence of products obtained by multiplying 120 by whole numbers. Starting from 1, the table lists the results of 120 multiplied by 1, 2, 3, and so forth, up to 10, 20, or any desired multiplier. This multiplication sequence is fundamental in mathematics for understanding and performing arithmetic operations involving the number 120. By memorizing or referencing the 120 times table, students can efficiently solve problems involving multiplication, division, and more complex calculations, enhancing their numerical fluency and problem-solving skills.
## What is the Multiplication Table of 120?
The multiplication table of 120 is a list of products obtained by multiplying 120 by whole numbers, ranging from 1 to any desired number. For example, 120 multiplied by 1 is 120, by 2 is 240, by 3 is 360, and so on. This sequence continues as 120 × 4 = 480, 120 × 5 = 600, up to higher numbers like 120 × 10 = 1200, 120 × 15 = 1800, and beyond. The table is essential for quickly solving arithmetic problems involving the number 120, aiding in calculations related to multiplication, division, and more complex mathematical operations. Memorizing or referencing the 120 times table can significantly enhance numerical fluency and efficiency in problem-solving tasks.
## 120ᵗʰ Table
The 120 times table lists products of 120 multiplied by whole numbers, aiding quick arithmetic calculations and enhancing numerical fluency for solving multiplication, division, and complex maths problems efficiently.
## Tips for 120ᵗʰ Table
### Break Down Multiplications:
• Simplify complex multiplications by breaking them into smaller parts.
• For example, 120 × 13 can be broken down into (120 × 10) + (120 × 3).
• Use addition to simplify calculations.
• For example, 120 × 12 can be calculated as (120 × 10) + (120 × 2).
### Memorize Key Multiples:
• Memorize multiples of 10, 5, and 2 for easier calculations.
• Knowing these can help calculate other products more quickly.
### Recognize Patterns:
• Look for patterns in multiplication results.
• Patterns can help in predicting and remembering products.
## Practice with Real-Life Examples:
• Apply multiplication in real-life scenarios to better understand and remember the calculations.
• For example, calculating the total cost of multiple items can help reinforce these multiplications.
## How to Read 120ᵗʰ Tables?
• One time 120 is 120.
• Two times 120 is 240.
• Three times 120 is 360.
• Four times 120 is 480.
• Five times 120 is 600.
• Six times 120 is 720.
• Seven times 120 is 840.
• Eight times 120 is 960.
• Nine times 120 is 1080.
• Ten times 120 is 1200.
To read the 120 times table, start with 120 × 1 = 120, then proceed: 120 × 2 = 240, 120 × 3 = 360, and so on, incrementing the multiplier sequentially.
## Solved Examples:
### Example 1: Simple Multiplication
Question: Calculate 120 multiplied by 8.
Solution: Using the multiplication table of 120, 120×8=960.
### Example 2: Calculate Total Cost
Question: You purchased 10 tickets for an event, each priced at 120 dollars. How much did all the tickets cost together?
Solution: 120 × 10 = 1200.
Answer: The total cost for the tickets is 1200 dollars.
Question: Find the sum of the first three multiples of 120.
Solution: 120 × 1 = 120, 120 × 2 = 240, 120 × 3 = 360, 120 + 240 + 360 = 720.
## Example 4: Subtracting Products
Question: Calculate the difference between the product of 120 times 6 and 120 times 4.
Solution: 120 × 6 = 720,
120 × 4 = 480, 720 − 480 = 240.
### Example 5: Annual Savings
Question: If you save 120 dollars monthly, how much will you save in one year?
Solution: 120 × 12 = 1440.
Answer: You will save 1440 dollars in a year.
### Example 6: Multiplying Large Numbers
Question: What is 120 times 50?
Solution: 120 × 50 = 6000.
Question: How much is 120 added to itself 7 times?
Solution: 120 × 7 = 840.
### Example 8: Doubling and Tripling
Question: What is the sum of double and triple of 120?
Solution: 120 × 2 = 240, 120 × 3 = 360, 240 + 360 = 600.
### Example 9: Product of Sequential Numbers
Question: Multiply 120 by 15. What is the product?
Solution: 120 × 15 = 1800.
### Example 10: Cost Calculation
Question: You need 25 items, each costing 120 dollars. What is the total cost?
Solution: 120 × 25 = 3000.
|
## Arithmatical - Problems of HCF and LCM
Important Facts and Formulae
1. Factors and Multiples : If a number a divides another number b exactly, we say that a is a factor of b. In this case, b is called a multiple of a.
2. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.) : The H.C.F. of two or more than two numbers is the greatest number that divided each of them exactly.There are two methods of finding the H.C.F. of a given set of numbers:Factorization Method : Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.Division Method : Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.
3. Least Common Multiple (L.C.M.) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M. There are two methods of finding the L.C.M. of a given set of numbers : Factorization Method : Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.Division Method (short-cut) :Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.
4. Product of two numbers = Product of their H.C.F. and L.C.M.
5. Co-primes : Two numbers are said to be co-primes if their H.C.F. is 1.
6. H.C.F. and L.C.M. of Fractions :1. H.C.F. = H.C.F. of Numerators / L.C.M. of Denominators2. L.C.M. = L.C.M. of Numerators / H.C.F. of Denominators
|
## What is a number bond in 4th grade?
Number bonds are a mental picture of the relationship between a number and the parts that combine to make it. For example, in this number bond you can see all the facts related to 2, 5 and 10: 2 x 5 = 10.
## How does the number bond relate to the number line?
The number bond is a pictorial representation of part/part/whole relationships showing that smaller numbers (the parts) make up larger numbers (the whole). The number bond is a key model for showing students how to both take apart (decompose) and put together (compose) numbers. Draw a number line representing the cord.
## What are number bonds to 20?
Number bonds to 20 are the pairs of numbers that add together to make twenty. There are ten number bonds to 20, which are:
• 1 + 19.
• 2 + 18.
• 3 + 17.
• 4 + 16.
• 5 + 15.
• 6 + 14.
• 7 + 13.
• 8 + 12.
## What is a number bond for an array?
• number bonds. An array is a way to represent multiplication and division using rows and columns. Rows represent the number of groups. Columns represent the number in each group or the size of each group. Here are 2 word problems that involve multiplication.
You might be interested: Question: What Does The Number 12 Mean In The Bible?
## What is an example of a number Bond?
Number bonds are pairs of numbers that can be added together to make another number e.g. 4 + 6 = 10.
## What are number bonds to 5?
Number Bonds of 5 Number bonds are 2 different numbers that add up to a certain number. In this case number bonds of 5 would be 5+0, 4+1, 3+2 and then these equations reversed, 0+5, 1+4 and 2+3.
## Why do we use number bonds?
Number bonds help students see that numbers can be “broken” into pieces to make computation easier (decomposing/composing). With number bonds, students recognize the relationships between numbers through a written model that shows how the numbers are related.
## What is a number bond to 100?
Number Bonds to 100 are pairs of numbers that add together to make 100. To find a number bond to 100, first add on to reach the next multiple of ten and then add the multiples of ten needed to get to 100.
## How do you explain bonds to children?
Simply put, a bond is a receipt given by a government or organization as an agreement to borrow money from another organization which will be returned at a later date with certain amount of interest or increment. Companies or governments issue bonds because they need to borrow large amounts of money.
|
# Composition of Functions
### Definition and Properties
Similarly to relations, we can compose two or more functions to create a new function. This operation is called the composition of functions.
Let $$g: A \to B$$ and $$f: B \to C$$ be two functions such that the range of $$g$$ equals the domain of $$f.$$ The composition of the functions $$f$$ and $$g,$$ denoted by $$f \circ g,$$ is another function defined as
$y = \left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right).$
Since functions are a special case of relations, they inherit all properties of composition of relations and have some additional properties. We list here some of them:
1. The composition of functions is associative. If $$h: A \to B,$$ $$g: B \to C$$ and $$f: C \to D,$$ then $$\left( {f \circ g} \right) \circ h = f \circ \left( {g \circ h} \right).$$
2. The composition of functions is not commutative. If $$g: A \to B$$ and $$f: B \to C,$$ then, as a rule, $$f \circ g \ne g \circ f.$$
3. Let $$g: A \to B$$ and $$f: B \to C$$ be injective functions. Then the composition of the functions $$f \circ g$$ is also injective.
4. Let $$g: A \to B$$ and $$f: B \to C$$ be surjective functions. Then the composition of the functions $$f \circ g$$ is also surjective.
5. It follows from the last two properties that if two functions $$g$$ and $$f$$ are bijective, then their composition $$f \circ g$$ is also bijective.
### Examples
#### Example 1. Composition of Functions Defined on Finite Sets
Consider the sets $$A = \left\{ {1,2,3,4} \right\},$$ $$B = \left\{ {a,b,c,d} \right\}$$ and $$C = \left\{ \alpha, \beta, \gamma, \delta \right\}.$$ The functions $$g: A \to B$$ and $$f:B \to C$$ are defined as
${g = \left\{ {\left( {1,b} \right),\left( {2,b} \right),\left( {3,a} \right),\left( {4,d} \right)} \right\},\;\;}\kern0pt{f = \left\{ {\left( {a,\gamma } \right),\left( {b,\alpha } \right),\left( {c,\delta } \right),\left( {d,\alpha } \right)} \right\}.}$
It is convenient to illustrate the mapping between the sets in an arrow diagram:
Given the mapping, we see that
${\left( {f \circ g} \right)\left( 1 \right) = f\left( {g\left( 1 \right)} \right) }={ f\left( b \right) = \alpha ;}$
${\left( {f \circ g} \right)\left( 2 \right) = f\left( {g\left( 2 \right)} \right) }={ f\left( b \right) = \alpha ;}$
${\left( {f \circ g} \right)\left( 3 \right) = f\left( {g\left( 3 \right)} \right) }={ f\left( a \right) = \gamma ;}$
${\left( {f \circ g} \right)\left( 4 \right) = f\left( {g\left( 4 \right)} \right) }={ f\left( d \right) = \alpha.}$
Hence, the composition of functions $$f \circ g$$ is given by
${f \circ g \text{ = }}\kern0pt{\left\{ {\left( {1,\alpha } \right),\left( {2,\alpha } \right),\left( {3,\gamma } \right),\left( {4,\alpha } \right)} \right\}.}$
This is represented in the following diagram:
#### Example 2. Composition of Functions Defined on Infinite Sets
Let $$g: \mathbb{R} \to \mathbb{R}$$ and $$f: \mathbb{R} \to \mathbb{R}$$ be two functions defined as
${g\left( x \right) = {x^2} + 3x + 1,\;\;}\kern0pt{f\left( x \right) = \cos x.}$
Determine the composite functions $$f \circ g,$$ $$g \circ f,$$ $$f \circ f,$$ $$g \circ g.$$
The first composite function $$\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right)$$ is formed when the inner function $${g\left( x \right)}$$ is substituted for $$x$$ in the outer function $${f\left( x \right)}.$$ This yields:
${\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) }={ \cos \left( {g\left( x \right)} \right) }={ \cos \left( {{x^2} + 3x + 1} \right).}$
Similarly we find the other composite functions:
${\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) }={ {f^2}\left( x \right) + 3f\left( x \right) + 1 }={ {\cos ^2}x + 3\cos x + 1.}$
${\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) }={ \cos \left( {f\left( x \right)} \right) }={ \cos \left( {\cos x} \right).}$
${\left(g \circ g\right)\left(x\right) = g\left( {g\left( x \right)} \right) }={ {g^2}\left( x \right) + 3g\left( x \right) + 1 }={ {\left( {{x^2} + 3x + 1} \right)^2} }+{ 3\left( {{x^2} + 3x + 1} \right) + 1 }={ \left( {\color{magenta}{x^4} + \color{red}{9{x^2}} + 1 + \color{green}{6{x^3}} + \color{red}{2{x^2}} + \color{blue}{6x}} \right) }+{ \left( {\color{red}{3{x^2}} + \color{blue}{9x} + 3} \right) + 1 }={ \color{magenta}{x^4} + \color{red}{9{x^2}} + 1 + \color{green}{6{x^3}} + \color{red}{2{x^2}} }+{ \color{blue}{6x} + \color{red}{3{x^2}} }+{ \color{blue}{9x} + 3 + 1 }={ \color{magenta}{x^4} + \color{green}{6{x^3}} + \color{red}{14{x^2}} + \color{blue}{15x} + 5.}$
### Compositions Involving Inverse Functions
Let $$f: A \to B$$ be a bijective function from domain $$A$$ to codomain $$B.$$ Then it has an inverse function $${f^{-1}}$$ that maps $$B$$ back to $$A.$$ Then
${{f^{ – 1}} \circ f = {I_A}\;\text{ or }\;}\kern0pt{{f^{ – 1}}\left( {f\left( x \right)} \right) = x,}$
where $${I_A}$$ is the identity function in the domain $$A$$ and $$x$$ is any element of $$A.$$
Similarly,
${f \circ {f^{ – 1}} = {I_B}\;\text{ or }\;}\kern0pt{f\left( {{f^{ – 1}}\left( y \right)} \right) = y,}$
where $${I_B}$$ is the identity function in the codomain $$B$$ and $$y$$ is any element of $$B.$$
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
The functions $$f,g$$ are defined as sets of ordered pairs $f = \left\{ {\left( {3,7} \right),\left( {4,1} \right),\left( {5,8} \right),\left( {6,1} \right)} \right\},$ $g = \left\{ {\left( {5,4} \right),\left( {9,6} \right),\left( {7,3} \right),\left( {2,6} \right)} \right\}.$ Find the composition of functions $$f \circ g.$$ State its domain and range.
### Example 2
The functions $$f,g$$ are defined as $$f\left( x \right) = \large{\frac{2}{{3x – 4}}}\normalsize$$ and $$g\left( x \right) = \large{\frac{3}{{4x – 5}}}\normalsize.$$ Find the domain of the composition of functions $$f \circ g.$$
### Example 3
Consider the functions $$g: \mathbb{R} \to \mathbb{R}$$ and $$f: \mathbb{R} \to \mathbb{R}$$ defined as ${g\left( x \right) = {x^2} – x,\;\;}\kern0pt{f\left( x \right) = {x^2} + 2x.}$ Find the compositions of functions:
1. $$f \circ g$$
2. $$g \circ f$$
3. $$f \circ f$$
4. $$g \circ g$$
### Example 4
Let the functions $$g: \mathbb{R} \to \mathbb{R}$$ and $$f: \mathbb{R} \to \mathbb{R}$$ be defined as ${g\left( x \right) = \sqrt{x},\;\;}\kern0pt{f\left( x \right) = {x^2} + 1.}$ Find the compositions of functions:
1. $$f \circ g \circ f$$
2. $$g \circ f \circ g$$
3. $$g \circ f \circ f$$
### Example 5
The function $$f:\left[ { – 1,\infty } \right) \to \left[ {0,\infty } \right)$$ is defined as $$f\left( x \right) = \sqrt {{x^3} + 1} .$$ Show that the composition $$f^{-1} \circ f$$ is the identity function.
### Example 6
The functions $$f, g: \mathbb{R} \to \mathbb{R}$$ are defined as $$f\left( x \right) = 2{x^2} + 1,$$ $$g\left( x \right) = 1 – x.$$ Solve the equation $f \circ g + g \circ f + 5 = 0.$
### Example 1.
The functions $$f,g$$ are defined as sets of ordered pairs $f = \left\{ {\left( {3,7} \right),\left( {4,1} \right),\left( {5,8} \right),\left( {6,1} \right)} \right\},$ $g = \left\{ {\left( {5,4} \right),\left( {9,6} \right),\left( {7,3} \right),\left( {2,6} \right)} \right\}.$ Find the composition of functions $$f \circ g.$$ State its domain and range.
Solution.
Draw the arrow diagram for the composition of functions $$f \circ g:$$
It follows from the diagram:
${g\left( 5 \right) = 4,\;\;f\left( 4 \right) = 1,}\;\; \Rightarrow {\left( {f \circ g} \right)\left( 5 \right) = 1;}$
${g\left( 7 \right) = 3,\;\;f\left( 3 \right) = 7,}\;\; \Rightarrow {\left( {f \circ g} \right)\left( 7 \right) = 7;}$
${g\left( 9 \right) = 6,\;\;f\left( 6 \right) = 1,}\;\; \Rightarrow {\left( {f \circ g} \right)\left( 9 \right) = 1;}$
${g\left( 2 \right) = 6,\;\;f\left( 6 \right) = 1,}\;\; \Rightarrow {\left( {f \circ g} \right)\left( 2 \right) = 1.}$
Hence, the composite function $$f \circ g$$ is given by
$f \circ g = \left\{ {\left( {5,1} \right),\left( {7,7} \right),\left( {9,1} \right),\left( {2,1} \right)} \right\}.$
The domain of $$f \circ g$$ is the set $$\left\{ {5,7,9,2} \right\},$$ and the range is the set $$\left\{ {7,1} \right\}.$$
### Example 2.
The functions $$f,g$$ are defined as $$f\left( x \right) = \large{\frac{2}{{3x – 4}}}\normalsize$$ and $$g\left( x \right) = \large{\frac{3}{{4x – 5}}}\normalsize.$$ Find the domain of the composition of functions $$f \circ g.$$
Solution.
By definition, $$\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right).$$
The inner function $${g\left( x \right)}$$ is defined for all $$x$$ except $$x = \large{\frac{5}{4}}\normalsize.$$ Hence, the point $$x = \large{\frac{5}{4}}\normalsize$$ must be excluded from the domain of the composition $$f \circ g.$$
The outer function $${f\left( x \right)}$$ is defined for all $$x$$ except $$x = \large{\frac{4}{3}}\normalsize.$$ So, we also need to exclude those values of $$x$$ for which the inner function $${g\left( x \right)}$$ is equal to $$\large{\frac{4}{3}}\normalsize.$$ Determine these values.
${g\left( x \right) = \frac{3}{{4x – 5}} = \frac{4}{3},}\;\; \Rightarrow {4\left( {4x – 5} \right) = 9, }\;\;\Rightarrow {16x – 20 = 9,}\;\; \Rightarrow {16x = 29,}\;\; \Rightarrow {x = \frac{{29}}{{16}}.}$
Thus, the domain of $$f \circ g$$ consists of all real values of $$x$$ except the points $$x = \large{\frac{5}{4}}\normalsize, \large{\frac{29}{16}}\normalsize.$$ This can be written in the form
${\text{dom}\left( {f \circ g} \right) \text{ = }}\kern0pt{\left( { – \infty ,\frac{5}{4}} \right) \cup \left( {\frac{5}{4},\frac{{29}}{{16}}} \right) \cup \left( {\frac{{29}}{{16}},\infty } \right).}$
### Example 3.
Consider the functions $$g: \mathbb{R} \to \mathbb{R}$$ and $$f: \mathbb{R} \to \mathbb{R}$$ defined as ${g\left( x \right) = {x^2} – x,\;\;}\kern0pt{f\left( x \right) = {x^2} + 2x.}$ Find the compositions of functions:
1. $$f \circ g$$
2. $$g \circ f$$
3. $$f \circ f$$
4. $$g \circ g$$
Solution.
1. The composite function $$\left(f \circ g\right)\left(x\right)$$ is written as $$f\left( {g\left( x \right)} \right).$$ We take the outer function $$f\left( x \right)$$ and substitute the inner function $${g\left( x \right)}$$ for $$x:$$ ${\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) }={ {g^2}\left( x \right) + 2g\left( x \right) }={ {\left( {{x^2} – x} \right)^2} + 2\left( {{x^2} – x} \right) }={ \color{magenta}{x^4} – \color{green}{2{x^3}} + \color{red}{x^2} + \color{red}{2{x^2}} -\color{blue}{2x} }={ \color{magenta}{x^4} -\color{green}{2{x^3}} + \color{red}{3{x^2}} – \color{blue}{2x}.}$
2. Calculate the composition $$g \circ f:$$ ${\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) }={ {f^2}\left( x \right) – f\left( x \right) = {\left( {{x^2} + 2x} \right)^2} – \left( {{x^2} + 2x} \right) }={ \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{4{x^2}} – \color{red}{x^2} – \color{blue}{2x} }={ \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{3{x^2}} – \color{blue}{2x}.}$ Notice that $$f \circ g \ne g \circ f,$$ that is the composition of functions is not commutative.
3. The composition of functions $$f \circ f$$ is given by ${\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) }={ {f^2}\left( x \right) + 2f\left( x \right) }={ {\left( {{x^2} + 2x} \right)^2} + 2\left( {{x^2} + 2x} \right) }={ \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{4{x^2}} + \color{red}{2{x^2}} + \color{blue}{4x} }={ \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{6{x^2}} + \color{blue}{4x}.}$
4. Similarly we determine the function $$g \circ g:$$ $\require{cancel}{\left(g \circ g\right)\left(x\right) = g\left( {g\left( x \right)} \right) }={ {g^2}\left( x \right) – g\left( x \right) }={ {\left( {{x^2} – x} \right)^2} – \left( {{x^2} – x} \right) }={ \color{magenta}{x^4} – \color{green}{2{x^3}} + \cancel{\color{red}{x^2}} – \cancel{\color{red}{x^2}} + \color{blue}{x} }={ \color{magenta}{x^4} – \color{green}{2{x^3}} + \color{blue}{x}.}$
### Example 4.
Let the functions $$g: \mathbb{R} \to \mathbb{R}$$ and $$f: \mathbb{R} \to \mathbb{R}$$ be defined as ${g\left( x \right) = \sqrt{x},\;\;}\kern0pt{f\left( x \right) = {x^2} + 1.}$ Find the compositions of functions:
1. $$f \circ g \circ f$$
2. $$g \circ f \circ g$$
3. $$g \circ f \circ f$$
Solution.
1. The find the composite function $$f \circ g \circ f,$$ we first determine the composition of inner functions $$g \circ f:$$ ${\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) }={ g\left( {{x^2} + 1} \right) }={ \sqrt {{x^2} + 1} .}$ Then, using the associative law, the triple composition is given by ${\left(f \circ g \circ f\right)\left(x\right) = \left(f \circ \left({g \circ f}\right)\right)\left(x\right) = f\left[ {g\left( {f\left( x \right)} \right)} \right] }={ f\left( {\sqrt {{x^2} + 1} } \right) }={ {\left( {\sqrt {{x^2} + 1} } \right)^2} + 1 }={ {x^2} + 1 + 1 }={ {x^2} + 2.}$
2. Here we first find the composition $$f \circ g:$$ ${\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) }={ f\left( {\sqrt x } \right) }={ {\left( {\sqrt x } \right)^2} + 1 }={ x + 1.}$ Hence, ${\left(g \circ f \circ g\right)\left(x\right) = \left(g \circ \left( {f \circ g} \right)\right)\left(x\right) }={ g\left[ {f\left( {g\left( x \right)} \right)} \right] }={ g\left( {x + 1} \right) }={ \sqrt {x + 1} .}$
3. Calculate the composition of functions $$g \circ f \circ f$$ using the same approach. Since ${\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) }={ f\left( {{x^2} + 1} \right) }={ {\left( {{x^2} + 1} \right)^2} + 1 }={ {x^4} + 2{x^2} + 1 + 1 }={ {x^4} + 2{x^2} + 2,}$ the triple composition $$g \circ f \circ f$$ is given by ${\left(g \circ f \circ f\right)\left(x\right) = \left(g \circ \left( {f \circ f} \right)\right)\left(x\right) }={ g\left[ {f\left( {f\left( x \right)} \right)} \right] }={ g\left( {{x^4} + 2{x^2} + 2} \right) }={ \sqrt {{x^4} + 2{x^2} + 2} .}$
### Example 5.
The function $$f:\left[ { – 1,\infty } \right) \to \left[ {0,\infty } \right)$$ is defined as $$f\left( x \right) = \sqrt {{x^3} + 1} .$$ Show that the composition $$f^{-1} \circ f$$ is the identity function.
Solution.
Find the inverse function $$f^{-1}$$ by solving the equation $$y = \sqrt {{x^3} + 1}$$ for $$x:$$
${y = \sqrt {{x^3} + 1} ,}\;\; \Rightarrow {{x^3} + 1 = {y^2},}\;\; \Rightarrow {{x^3} = {y^2} – 1,}\;\; \Rightarrow {x = \sqrt[3]{{{y^2} – 1}}.}$
To calculate the composition of functions $$f^{-1} \circ f,$$ we substitute $$y = \sqrt {{x^3} + 1}$$ in the equation $$x = \sqrt[3]{{{y^2} – 1}}.$$ For any $$x$$ in the domain of $$f,$$ we have
${\left({f^{ – 1}} \circ f\right)\left(x\right) = {f^{ – 1}}\left( {f\left( x \right)} \right) }={ {f^{ – 1}}\left( y \right) = \sqrt[3]{{{y^2} – 1}} }={ \sqrt[3]{{{{\left( {\sqrt {{x^3} + 1} } \right)}^2} – 1}} }={ \sqrt[3]{{{x^3} + \cancel{1} – \cancel{1}}} }={ \sqrt[3]{{{x^3}}} }={ x.}$
Thus, $${f^{ – 1}} \circ f = I,$$ where $$I$$ is the identity function in the domain of $$f.$$
### Example 6.
The functions $$f, g: \mathbb{R} \to \mathbb{R}$$ are defined as $$f\left( x \right) = 2{x^2} + 1,$$ $$g\left( x \right) = 1 – x.$$ Solve the equation $f \circ g + g \circ f + 5 = 0.$
Solution.
Calculate the compositions of functions $$f \circ g$$ and $$g \circ f:$$
${\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right) }={ f\left( {1 – x} \right) }={ 2{\left( {1 – x} \right)^2} + 1 }={ 2\left( {1 – 2x + {x^2}} \right) + 1 }={ 2 – 4x + 2{x^2} + 1 }={ 2{x^2} – 4x + 3;}$
${\left( {g \circ f} \right)\left( x \right) }={ g\left( {f\left( x \right)} \right) }={ g\left( {2{x^2} + 1} \right) }={ 1 – \left( {2{x^2} + 1} \right) }={\cancel{1} – 2{x^2} – \cancel{1} }={ – 2{x^2}.}$
Plug both expressions into the original equation and solve it for $$x:$$
${f \circ g + g \circ f + 5 = 0,}\;\; \Rightarrow {\cancel{2{x^2}} – 4x + 3 – \cancel{2{x^2}} + 5 = 0,}\;\; \Rightarrow {8 – 4x = 0,}\;\; \Rightarrow {x = 2.}$
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# What is 211/6 as a decimal?
## Solution and how to convert 211 / 6 into a decimal
211 / 6 = 35.167
Fraction conversions explained:
• 211 divided by 6
• Numerator: 211
• Denominator: 6
• Decimal: 35.167
• Percentage: 35.167%
The basis of converting 211/6 to a decimal begins understanding why the fraction should be handled as a decimal. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. Once we've decided the best way to represent the number, we can dive into how to convert 211/6 into 35.167
211 / 6 as a percentage 211 / 6 as a fraction 211 / 6 as a decimal
35.167% - Convert percentages 211 / 6 211 / 6 = 35.167
## 211/6 is 211 divided by 6
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 211 divided by 6. Now we divide 211 (the numerator) into 6 (the denominator) to discover how many whole parts we have. Here's how you set your equation:
### Numerator: 211
• Numerators represent the number of parts being taken from a denominator. Any value greater than fifty will be more difficult to covert to a decimal. 211 is an odd number so it might be harder to convert without a calculator. Values closer to one-hundred make converting to fractions more complex. Time to evaluate 6 at the bottom of our fraction.
### Denominator: 6
• Denominators represent the total number of parts, located below the vinculum or fraction bar. Any value less than ten like 6 makes your equation a bit simpler. And it is nice having an even denominator like 6. It simplifies some equations for us. Overall, a small denominator like 6 could make our equation a bit simpler. Now let's dive into how we convert into decimal format.
## How to convert 211/6 to 35.167
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 6 \enclose{longdiv}{ 211 }$$
We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Solve for how many whole groups you can divide 6 into 211
$$\require{enclose} 00.35 \\ 6 \enclose{longdiv}{ 211.0 }$$
Now that we've extended the equation, we can divide 6 into 2110 and return our first potential solution! Multiple this number by our furthest left number, 6, (remember, left-to-right long division) to get our first number to our conversion.
### Step 3: Subtract the remainder
$$\require{enclose} 00.35 \\ 6 \enclose{longdiv}{ 211.0 } \\ \underline{ 210 \phantom{00} } \\ 1900 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you have a remainder over 6, go back. Your solution will need a bit of adjustment. If you have a number less than 6, continue!
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 211/6 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. They each bring clarity to numbers and values of every day life. This is also true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Here are just a few ways we use 211/6, 35.167 or 3516% in our daily world:
### When you should convert 211/6 into a decimal
Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions.
### When to convert 35.167 to 211/6 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 211/6 = 35.167 what would it be as a percentage?
• What is 1 + 211/6 in decimal form?
• What is 1 - 211/6 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 35.167 + 1/2?
### Convert more fractions to decimals
From 211 Numerator From 6 Denominator What is 211/7 as a decimal? What is 212/6 as a decimal? What is 211/8 as a decimal? What is 213/6 as a decimal? What is 211/9 as a decimal? What is 214/6 as a decimal? What is 211/10 as a decimal? What is 215/6 as a decimal? What is 211/11 as a decimal? What is 216/6 as a decimal? What is 211/12 as a decimal? What is 217/6 as a decimal? What is 211/13 as a decimal? What is 218/6 as a decimal? What is 211/14 as a decimal? What is 219/6 as a decimal? What is 211/15 as a decimal? What is 220/6 as a decimal? What is 211/16 as a decimal? What is 221/6 as a decimal? What is 211/17 as a decimal? What is 222/6 as a decimal? What is 211/18 as a decimal? What is 223/6 as a decimal? What is 211/19 as a decimal? What is 224/6 as a decimal? What is 211/20 as a decimal? What is 225/6 as a decimal? What is 211/21 as a decimal? What is 226/6 as a decimal? What is 211/22 as a decimal? What is 227/6 as a decimal? What is 211/23 as a decimal? What is 228/6 as a decimal? What is 211/24 as a decimal? What is 229/6 as a decimal? What is 211/25 as a decimal? What is 230/6 as a decimal? What is 211/26 as a decimal? What is 231/6 as a decimal?
### Convert similar fractions to percentages
From 211 Numerator From 6 Denominator 212/6 as a percentage 211/7 as a percentage 213/6 as a percentage 211/8 as a percentage 214/6 as a percentage 211/9 as a percentage 215/6 as a percentage 211/10 as a percentage 216/6 as a percentage 211/11 as a percentage 217/6 as a percentage 211/12 as a percentage 218/6 as a percentage 211/13 as a percentage 219/6 as a percentage 211/14 as a percentage 220/6 as a percentage 211/15 as a percentage 221/6 as a percentage 211/16 as a percentage
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# How do you find the asymptotes for g(x)= (x^3 -16x )/ (4x^2 - 4x)?
Aug 13, 2018
Vertical asymptote: $x = 1$, slant asymptote : $y = 0.25 x + 0.25$, removable discontinuity: $x = 0$
#### Explanation:
$g \left(x\right) = \frac{{x}^{3} - 16 x}{4 {x}^{2} - 4 x}$ or
$g \left(x\right) = \frac{\cancel{x} \left({x}^{2} - 16\right)}{4 \cancel{x} \left(x - 1\right)}$or
$g \left(x\right) = \frac{{x}^{2} - 16}{4 x - 4}$
Vertical asymptote occur when denominator is zero.
$x = 0$ is removable discontinuity. $4 \left(x - 1\right) = 0$
$\therefore x = 1$ is vertical asymptote.
The numerator's degree is greater (by a margin of 1), then we
have a slant asymptote which is found doing long division.
$\frac{{x}^{2} - 16}{4 x - 4} = \left(0.25 x + 0.25\right) - \frac{15}{4 x - 4}$
Therefore slant asymptote is $y = 0.25 x + 0.25$
graph{(x^3-16x)/(4x^2-4x) [-40, 40, -20, 20]}[Ans]
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# How to Use Exponent Rules
The exponent rules are very helpful formulas to use when you are simplifying exponential expressions. They are also called the "exponent laws" or the "exponent properties".
### Note:
These rules are used to simplify advanced exponential expressions.
I have a different page that explains how to simplify basic exponents, if you're learning exponents for the first time.
You can use them to simplify expressions that have just numbers, just variables, or some combination of both. You can also use them in reverse.
If you click on the "Learn More" links below, you'll see step-by-step examples and explanations that show WHY each rule works the way it does.
I highly recommend reading those pages because it is much easier to memorize a formula when you know WHY it works.
## List of Exponent Rules
### Exponent of 0
$$x^{0}=1$$
Learn More
### Negative Exponents
$$x^{-a}=\frac{1}{x^{a}}$$
Learn More
### Fraction Exponents
$$x^{\frac{a}{b}}=\sqrt[b]{x^a}$$
Learn More
### Product Rule
$$x^{a}x^{b}=x^{a+b}$$
Learn More
### Quotient Rule
$$\frac{x^{a}}{x^{b}}=x^{a-b}$$
Learn More
### Power Rule
$$(x^{a})^{b}=x^{ab}$$
Learn More
### Distributive Property for Exponents(x and $$\div$$)
$$(xy)^a=x^a y^a$$
$$(\frac{x}{y})^a=\frac{x^a}{y^a}$$
Learn More
## Reverse Exponent Rules
Each of the rules above can be used in reverse. For example...
Write $$\sqrt[4]{9^2}$$ in exponential form.
This expression is a power within a root, so I could use the fraction exponents rule in reverse to write it as...
$$9^{\frac{2}{4}}$$
$$\frac{2}{4}$$ can be reduced to $$\frac{1}{2}$$ so the simplified answer in exponential form is...
$$9^{\frac{1}{2}}$$
The reverse distributive property of exponents is used so often that some people call it the "product rule for powers with the same exponent".
The reverse power rule of exponents is also used quite often, especially for changing the units in exponential functions.
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## Free Printable Worksheets
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## Free Online Practice Problems
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Application Center - Maplesoft
# Classroom Tips and Techniques: Locus of Eigenvalues
You can switch back to the summary page by clicking here.
Classroom Tips and Techniques: Locus of Eigenvalues
Robert J. Lopez
Emeritus Professor of Mathematics and Maple Fellow
Maplesoft
Monagan's MapleTech Article
Given that the earlier editions of [3] are dated 1980, 1986, 1990, and 1994, it must be obvious that Mike Monagan's MapleTech article was based on one of the first three editions of Steve Leon's text. Mike's article used matrices that differed slightly from those in Exercise 4, page 360, in [3].
The following is a paraphrase of the exercise, and consequently, of the MapleTech article.
If , examine the locus of its eigenvalues as varies linearly from to 5.
Solution
The eigenvalues of are whereas the eigenvalues of are . Because is not symmetric, its eigenvalues can be, and actually do become, complex. Indeed, the eigenvalues of are . Figure 1 is a graph of the real values of these eigenvalues; Figure 2 shows the eigenvalues in the complex plane. In each figure, , with drawn in black; and , in red.
> evs := [2+(1/2)*s+(1/2)*sqrt(s^2-8*s-8), 2+(1/2)*s-(1/2)*sqrt(s^2-8*s-8)]: plot(evs,s=-5..5,color=[black,red]);
>
Figure 1 Eigenvalues of in the real plane
> evs := [2+(1/2)*s+(1/2)*sqrt(s^2-8*s-8), 2+(1/2)*s-(1/2)*sqrt(s^2-8*s-8)]: plots:-complexplot(evs,s=-5..5,color=[black,red],scaling=constrained,thickness=2);
>
Figure 2 Eigenvalues of in the complex plane
Figures 3 and 4 repeat Figures 1 and 2, respectively, but for . Figure 3 makes it clear that the analytic expressions giving are discontinuous as real functions, but not as complex-valued functions. On the other hand, Figure 4 shows that in the complex plane, the loci of do not have continuously turning tangents. Although will be symmetric in the remaining examples in this article, Figures 3 and 4 portend some of the difficulties that will be encountered even for a restricted class of matrices.
> evs := [2+(1/2)*s+(1/2)*sqrt(s^2-8*s-8), 2+(1/2)*s-(1/2)*sqrt(s^2-8*s-8)]: plot(evs,s=-5..15,color=[black,red]);
>
Figure 3 Eigenvalues of in the real plane
> evs := [2+(1/2)*s+(1/2)*sqrt(s^2-8*s-8), 2+(1/2)*s-(1/2)*sqrt(s^2-8*s-8)]: plots:-complexplot(evs,s=-5..15,color=[black,red],scaling=constrained,thickness=2);
>
Figure 4 Eigenvalues of in the complex plane
Figure 5 contains an animation of the loci traced by for . From this animation, or from an explicit calculation of the appropriate limits, the following can be deduced.
and
The characteristic polynomial for is
and its discriminant, , has zeros .
> evS := [2+(1/2)*S+(1/2)*sqrt(S^2-8*S-8), 2+(1/2)*S-(1/2)*sqrt(S^2-8*S-8)]: plots:-animate(plots:-complexplot,[evS,S=-5..s,color=[black,red],scaling=constrained,thickness=2],s=-5..15,frames=41,digits=3);
>
Figure 5 Animation of for
Hence, the loci of bifurcate at and . But the real question is, do the closed-form expressions each define the locus of an eigenvalue (resulting in the black and red curves in Figure 4), or are the loci of the eigenvalues the connected curves in Figure 3?
Example 1
Find the loci of the eigenvalues of for .
Solution
The eigenvalues of the symmetric are
Loci are graphed in Figure 6, where is in black, and is in red.
It might seem from Figure 6 that symmetry removes many of the difficulties posed by complex eigenvalues. However, for the 2 × 2 matrix in Example 2, are equal, so the loci of the eigenvalues will have a point in common. In the present example, the loci for are separate and have continuously turning tangents.
> evs1:=[1-(1/2)*s+(1/2)*sqrt(265*s^2-312*s+288), 1-(1/2)*s-(1/2)*sqrt(265*s^2-312*s+288)]: plot(evs1,s=0..1,color=[black,red]);
>
Figure 6 Loci of eigenvalues ,
Example 2
Find the loci of the eigenvalues of for .
Solution
The eigenvalues of the symmetric are
Loci are graphed in Figure 7, where is in black, and is in red.
Although the loci in Figure 7 intersect, each expression for an eigenvalue generates a unique locus with continuously turning tangent. Figure 7 raises the hope that perhaps tracking an eigenvalue from to might be a tractable task.
But alas, even though Example 3 might reinforce this belief, Examples 4 - 6 will prove this hope to be a chimera.
> evs2:=[4*sqrt(5)*s-sqrt(5)-5*s+2, -4*sqrt(5)*s+sqrt(5)-5*s+2]: plot(evs2,s=0..1,color=[black,red]);
>
Figure 7 Intersecting loci for
Example 3
Find the loci of the eigenvalues of for .
Solution
The eigenvalues , are obtained exactly with Maple's Eigenvectors command. The return is a list of length nearly 2000, and which would take two and a half pages to print.
Figure 8 contains a graph of the loci of the three eigenvalues, colored black, red, and green, respectively. The graph is drawn with increased precision; at standard precision, roundoff generates small imaginary parts that cause small gaps in the loci.
The loci in Figure 8 are separate and distinct, all with continuously turning tangents. For this it is clearly a simple task to trace the eigenvalues from to .
> P:=Matrix([[-9+10*s,-4+5*s,1-8*s],[-4+5*s,6-9*s,-3+6*s],[1-8*s,-3+6*s, -3+9*s]]): evs3:=LinearAlgebra:-Eigenvalues(P,output=list): Digits:=25: plot(evs3,s=0..1,color=[black,red,green]); Digits:=10:
>
Figure 8 Loci of
Example 4
Find the loci of the eigenvalues of for .
Solution
• The eigenvalues of are . Hence, the loci of the eigenvalues , and have the point in common. Just what this does to the loci remains to be seen.
• The characteristic equation defines implicitly. Maple's implicitplot command applied to this equation produces Figure 9 in which the curves threrefore represent the loci of the eigenvalues of .
> CP:=x^3-(-11+4*s)*x^2-(1104*s^2-472*s+81)*x-10688*s^3+6192*s^2-1476*s-235: plots:-implicitplot(CP,s=0..1,x=-25..35,labels=[s,typeset(lambda)],gridrefine=5);
>
Figure 9 Loci defined implicitly by the characteristic polynomial
• In Figure 10, the loci are graphs of the exact eigenvalues obtained via Maple's Eigenvalues command.
• The eigenvalues of are approximately , and and the loci emanating from these initial points are colored black, red, and green, respectively. The eigenvalues of are approximately , and .
• The red and green curves, each defined by an exact expression, do not have continuously turning tangents.
> P:=Matrix([[3-8*s, 2+16*s, 4-4*s], [2+16*s, -9+20*s, 8-24*s], [4-4*s, 8-24*s, -5-8*s]]): Q:=LinearAlgebra:-Eigenvalues(P,output=list): plot(Q,s=0..1,color=[black,red,green],labels=[s,typeset(lambda)]);
>
Figure 10 Loci via graph of exact expressions for the eigenvalues
The closed-form expressions for , are continuous, but not . (This can be established analytically by the calculations in Table 1.) So either the loci of the eigenvalues are defined by the closed-form expressions and therefore do not have continuously turning tangents, or the loci are smooth curves and are only piecewise-defined by the analytic expressions whose graphs appear in Figure 10. A precise definition of the locus of eigenvalues of a real, symmetric, matrix is required.
= = = = Table 1 Calculations showing that , do not define curves
Example 5
For the matrix in Example 4, obtain the equivalent of Figure 10 and Table 1, using only numeric calculations.
Solution
Initializations
• Define the matrix .
• Obtain the characteristic polynomial.
• Define as a function of .
Numeric determination of loci of eigenvalues
• The function returns a list of numerically computed eigenvalues for each given value of the parameter .
• The list contains 101 equispaced values of .
• Each is a list of eigenvalues computed at .
• Constructed from the lists and , each is a uniquely colored graph of the jth numerically calculated eigenvalue.
• Figure 11 assembles the graphs , into a single graph via Maple's display command; it shows that the red and green curves share the common point .
Figure 11 Numerically determined loci
Slopes on either side of can be calculated numerically from computed implicitly from the characteristic polynomial.
=
Recall that in must be replaced by the appropriate eigenvalue which is available only through numeric calculation via the function . The limiting process used in Table 1 can't be applied here; the requisite numeric calculations are summarized in Table 2.
= = = = Table 2 Numeric calculation of slopes along the loci on either side of
Example 6
Find the loci of the eigenvalues of for .
Solution
Since is a 5 × 5 matrix, only numeric techniques can be used to find its eigenvalues. However, note that the matrix has been chosen so that the eigenvalues of are 1, 5, 10, 10, and 15.
In Table 3 the matrix is defined, and the characteristic polynomial is defined as the function .
Table 3 The matrix and the characteristic polynomial as the function
The characteristic equation, (here, ) defines the loci of the eigenvalues of implicitly. Figure 12 implements this insight via Maple's implicitplot command applied to the characteristic equation.
Figure 12 Loci of eigenvalues defined implicitly by the characteristic equation
Table 4 shows the calculations needed to solve for the eigenvalues numerically, and to construct the separate loci based on these numeric data.
• The function returns a list of numerically computed eigenvalues for each given value of the parameter .
• The list contains 101 equispaced values of .
• Each is a list of eigenvalues computed at .
• Constructed from the lists and , each is a uniquely colored graph of the jth numerically calculated eigenvalue.
Table 4 Numeric construction of the loci of eigenvalues
Figure 13 assembles the graphs , into a single graph via Maple's display command; it shows that the green and blue curves share the common point .
Figure 13 Numerically calculated loci of eigenvalues
For the green and blue curves in Figure 13, slopes on either side of can be calculated numerically from computed implicitly from the characteristic polynomial. Table 5 contains the relevant calculations.
• Obtain implicitly with Maple's implicitdiff command.
• Increase the number of working digits and define values of on either side of .
Evaluate on either side of along the green and blue curves
=
=
=
=
Table 5 Numeric evaluation of slopes along the green and blue curves in Figure 13
There are no closed-form expressions for the eigenvalues of this 5 × 5 matrix . The eigenvalues are computed numerically by Maple's Eigenvalues or fsolve commands, each of which return a sorted list of eigenvalues. There is no user-control of this sort, but even if there were, what sorting rule could be invoked across an eigenvalue with algebraic multiplicity greater than 1? It would seem that the only way to define a unique locus of eigenvalues is to require that it be of class , that is, that it have a continuously turning tangent.
References
[1] Mathematical Thoughts on the Root Locus, Robert J. Lopez, Tips & Techniques, Maple Reporter, July, 2013. [2] Using Computer Algebra to Help Understand the Nature of Eigenvalues and Eigenvectors, Michael Monagan, MapleTech, Issue 9, Spring 1993, Birkhäuser. [3] Linear Algebra with Applications, 5th ed., Steven J. Leon, 1998, Prentice Hall.
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# Cut your Cake and Eat it
How do you divide one cake between infinitely many people? Give the lucky first person (hopefully you) half the cake. Cut the remaining piece in half and give one of these pieces to the second person. Cut the remaining piece in half and give one of these pieces to the third person. And so on. Will there be enough cake for everyone? And if yes, will any cake be left over at the end?
To answer this question, imagine the cake is shaped like a square and look at this picture:
The large piece of the square, exactly half of it, is the piece that will be given to the first person. The next smaller piece, half of the remaining half, is the piece that will be given to the second person. The next smaller piece, half of half of half of the cake, is the one that will be given to the third person, and so on. You can imagine carrying on like that forever. The picture indicates that after giving out a piece of cake to a person there is always some left for the next person (you only ever give out half of what is left). It also shows that there won't be any cake left over at the end (the halving procedure will eventually fill up all of the cake) .
The picture also shows us something far more interesting. The largest piece is half of the cake. The second largest piece is half of half of the cake, in other words it's $1/2 \times 1/2 = 1/4$ of the cake. The third largest piece is half of half of half, which is $1/2 \times 1/2 \times 1/2 = 1/8$ of the cake. The fourth largest piece is half of the previous, which is $1/2 \times 1/2 \times 1/2 \times 1/2 = 1/16$ of the cake. Carrying on like this shows that the $nth$ largest piece is nth person's piece is $1/2^nth$ of the cake.
Adding all these numbers together gives a sum with infinitely many terms: $$1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ....$$
As the picture shows, combining all the pieces together gives the whole cake. Therefore, $$1/2 + 1/4 + 1/8 + 1/32 + .... = 1.$$ Our picture proves that the infinite sum $1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ....$ (which is called a geometric series) converges to the number 1: as we add more and more terms we get closer and closer to 1. We can get as close to 1 as we like, without ever exceeding it.
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# Difference between revisions of "2017 AMC 8 Problems/Problem 22"
## Problem 22
In the right triangle $ABC$, $AC=12$, $BC=5$, and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle? $[asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("A", (0,0), W); label("C", (12,0), E); label("B", (12,5), NE); label("12", (6, 0), S); label("5", (12, 2.5), E);[/asy]$
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}$
## Solution
We can reflect triangle $ABC$ on line $AC.$ This forms the triangle $AB'C$ and a circle out of the semicircle. Let us call the center of the circle $O.$
We can see that Circle $O$ is the incircle of $AB'C.$ We can use the formula for finding the radius of the incircle to solve this problem. The are of $AB'C$ is $12\times5 = 60.$ The semiperimeter is $5+13 = 18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\text{D)}$ $\dfrac{10}{3}.$
2017 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
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# 2021 AIME II Problems/Problem 5
## Problem
For positive real numbers $s$, let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$. The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$. Find $a^2+b^2$.
## Solution 1
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the $3$rd side is between $6$ and $14$, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the $3$rd side. So the triangles' sides are between $6$ and $\sqrt{84}$ exclusive, and the larger bound is between $\sqrt{116}$ and $14$, exclusive. The area of these triangles are from $0$ (straight line) to $2\sqrt{84}$ on the first "small bound" and the larger bound is between $0$ and $20$. $0 < s < 2\sqrt{84}$ is our first equation, and $0 < s < 20$ is our $2$nd equation. Therefore, the area is between $\sqrt{336}$ and $\sqrt{400}$, so our final answer is $\boxed{736}$.
~ARCTICTURN
## Solution 2 (Inequalities and Casework)
If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\leq b\leq c,$ then both of the following must be satisfied:
• Triangle Inequality Theorem: $a+b>c$
• Pythagorean Inequality Theorem: $a^2+b^2
For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side:
Case (1): The longest side has length $\boldsymbol{10,}$ so $\boldsymbol{0
By the Triangle Inequality Theorem, we have $4+x>10,$ from which $x>6.$
By the Pythagorean Inequality Theorem, we have $4^2+x^2<10^2,$ from which $x<\sqrt{84}.$
Taking the intersection produces $6 for this case.
At $x=6,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{84},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.$ Together, we obtain $0 or $K\in\left(0,2\sqrt{84}\right).$
Case (2): The longest side has length $\boldsymbol{x,}$ so $\boldsymbol{x\geq10.}$
By the Triangle Inequality Theorem, we have $4+10>x,$ from which $x<14.$
By the Pythagorean Inequality Theorem, we have $4^2+10^2 from which $x>\sqrt{116}.$
Taking the intersection produces $\sqrt{116} for this case.
At $x=14,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{116},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot10=20.$ Together, we obtain $0 or $K\in\left(0,20\right).$
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers $s$ are in exactly one of $\left(0,2\sqrt{84}\right)$ or $\left(0,20\right).$ By the exclusive disjunction, the set of all such $s$ is $$[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),$$ from which $a^2+b^2=\boxed{736}.$
~MRENTHUSIASM
## Solution 3
We have the diagram below.
$[asy] draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label("A",(0,0),SW); label("B",(1,2*sqrt(3)),N); label("C",(10,0),SE); label("\theta",(0,0),NE); label("\alpha",(1,2*sqrt(3)),SSE); label("4",(0,0)--(1,2*sqrt(3)),WNW); label("10",(0,0)--(10,0),S); [/asy]$
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield .
If angle $\theta$ is obtuse, then we have that $s \in (0,20)$. This is because $s=20$ is attained at $\theta = 90^{\circ}$, and the area of the triangle is strictly decreasing as $\theta$ increases beyond $90^{\circ}$. This can be observed from $$s=\frac{1}{2}(4)(10)\sin\theta$$by noting that $\sin\theta$ is decreasing in $\theta \in (90^{\circ},180^{\circ})$.
Then, we note that if $\alpha$ is obtuse, we have $s \in (0,4\sqrt{21})$. This is because we get $x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}$ when $\alpha=90^{\circ}$, yileding $s=4\sqrt{21}$. Then, $s$ is decreasing as $\alpha$ increases by the same argument as before.
$\angle{ACB}$ cannot be obtuse since $AC>AB$.
Now we have the intervals $s \in (0,20)$ and $s \in (0,4\sqrt{21})$ for the cases where $\theta$ and $\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\sqrt{21}<20$, the desired range is $$s\in [4\sqrt{21},20)$$giving $$a^2+b^2=\boxed{736}\Box$$
## Solution 4
Note: Archimedes15 Solution which I added an answer here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\frac{1}{2} \cdot 40 \cdot \sin{\alpha}$ and $\sin$ is at most $1$. Note that for the other case, the side lengths around the obtuse angle must be $4$ and $x$ where we have $16+x^2 < 100 \rightarrow x < 2\sqrt{21}$. Using the same logic as the other case, the area is at most $4\sqrt{21}$. Square and add $4\sqrt{21}$ and $20$ to get the right answer $$a^2+b^2= \boxed{736}\Box$$
## Solution 5 (Circles)
For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$
As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1]; C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C2)^^Arc(B,D,P),red); dot("A", A, 1.5*S, linewidth(4.5)); dot("B", B, 1.5*S, linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^D^^Q, linewidth(0.8), UnFill); Label L1 = Label("10", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("4", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label("\angle C obtuse",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label("\angle B obtuse",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that:
1. The region in which $\angle B$ is obtuse is determined by construction.
2. The region in which $\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem.
For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot("A", A, 1.5*S, linewidth(4.5)); dot("B", B, 1.5*S, linewidth(4.5)); dot("D", D, 1.5*dir(75), linewidth(0.8), UnFill); dot("C_2", C2, 1.5*N, linewidth(4.5)); dot("C_1", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label("10", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("4", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas:
• If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have \begin{align*} [ABC]&=[ABD] \\ &=\frac12\cdot BD\cdot DA \\ &=\frac12\cdot BD\cdot \sqrt{AB^2-BD^2} \\ &=\frac12\cdot 4\cdot \sqrt{10^2-4^2} \\ &=2\sqrt{84}. \end{align*}
• If $C=C_2,$ then $[ABC]$ will be maximized (unattainable). For this right triangle, we have \begin{align*} [ABC]&=\frac12\cdot AB\cdot BC \\ &=\frac12\cdot 10\cdot 4 \\ &=20. \end{align*}
Finally, the set of all such $s$ is $[a,b)=\left[2\sqrt{84},20\right),$ from which $a^2+b^2=\boxed{736}.$
~MRENTHUSIASM (credit given to Snowfan)
## Solution 6
Let a triangle in $\tau(s)$ be $ABC$, where $AB = 4$ and $BC = 10$. We will proceed with two cases:
Case 1: $\angle ABC$ is obtuse. If $\angle ABC$ is obtuse, then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $(0,10)$; therefore, the area of the triangle will fall in the range of $(0, 20)$.
Case 2: $\angle BAC$ is obtuse. Then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $\left(0, \sqrt{10^{2} - 4^{2}}\right)$. Therefore, the area of the triangle will fall in the range of $\left(0, 2 \sqrt{84}\right)$.
If $s < 2 \sqrt{84}$, there will exist two types of triangles in $\tau(s)$ - one type with $\angle ABC$ obtuse; the other type with $\angle BAC$ obtuse. If $s \geq 2 \sqrt{84}$, as we just found, $\angle BAC$ cannot be obtuse, so therefore, there is only one type of triangle - the one in which $\angle ABC$ is obtuse. Also, if $s > 20$, no triangle exists with lengths $4$ and $10$. Therefore, $s$ is in the range $\left[ 2 \sqrt{84}, 20\right)$, so our answer is $\left(2 \sqrt{84}\right)^{2} + 20^{2} = \boxed{736}$.
Alternatively, refer to Solution 5 for the geometric interpretation.
~ihatemath123
## Solution 7
Let's rephrase the condition. It is required to find such values of the area of an obtuse triangle with sides 4 and 10, when there is exactly one such obtuse triangle. In the diagram, $AB = 4, AC = 10.$
The largest area of triangle with sides $4$ and $10$ is $20$ for a right triangle with legs $4$ and $10 (AC\perp AB).$
The diagram shows triangles with equal heights. The yellow triangle $ABC'$ has the longest side $BC',$ the blue triangle $ABC$ has the longest side $AC.$ If $BC\perp AB,$ then$BC = \sqrt {AC^2 – AB^2} = 2 \sqrt{21}$ the area is equal to $4\sqrt{21}.$ In the interval, the blue triangle $ABC$ is acute-angled, the yellow triangle $ABC'$ is obtuse-angled. Their heights and areas are equal. The condition is met.
If the area is less than $4\sqrt{21},$ both triangles are obtuse, not equal, so the condition is not met.
Shelomovskii, vvsss, www.deoma-cmd.ru
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Harmonic Mean - Measures of Central Tendency, Business Mathematics & Statistics
# Harmonic Mean - Measures of Central Tendency, Business Mathematics & Statistics Video Lecture - Business Mathematics and Statistics - B Com
115 videos|142 docs
## FAQs on Harmonic Mean - Measures of Central Tendency, Business Mathematics & Statistics Video Lecture - Business Mathematics and Statistics - B Com
1. What is the harmonic mean and how is it calculated?
Ans. The harmonic mean is a measure of central tendency used in statistics to determine the average of a set of numbers. It is calculated by dividing the number of values in the set by the sum of their reciprocals.
2. When is the harmonic mean used instead of the arithmetic mean?
Ans. The harmonic mean is used instead of the arithmetic mean when dealing with rates or ratios. It is particularly useful when the values being averaged have a reciprocal relationship, such as speed and time. For example, when calculating the average speed of a trip, the harmonic mean should be used if the distances traveled are different.
3. How does the harmonic mean differ from the arithmetic mean and the geometric mean?
Ans. The arithmetic mean is calculated by summing all the values in a set and dividing by the total number of values. The geometric mean is calculated by taking the nth root of the product of all the values in a set. In contrast, the harmonic mean is calculated by dividing the number of values by the sum of their reciprocals. The harmonic mean places more weight on smaller values, making it useful for averaging rates or ratios.
4. Can the harmonic mean be used with negative numbers?
Ans. No, the harmonic mean cannot be used with negative numbers. Since the harmonic mean involves taking the reciprocals of the values, negative numbers would result in undefined values. The harmonic mean can only be used with positive numbers.
5. What are some practical applications of the harmonic mean in business and statistics?
Ans. The harmonic mean is commonly used in business and statistics for various applications. It is used in finance to calculate average rates of return, in manufacturing to determine average production rates, and in transportation to calculate average speeds. Additionally, it is used in sports analytics to calculate average player efficiency ratings and in environmental science to determine average pollutant concentrations.
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# Calculator count to 1000
Achievement Objectives:
Specific Learning Outcomes:
Use calculators to explore multiples of numbers to a 1000
Identify number patterns in multiples.
Devise and use problem solving strategies
Description of mathematics:
This problem explores the idea that large numbers are made up of smaller numbers. It also helps students form a more accurate idea of the value of a 1000. The students develop their understanding of multiples by skip counting with a calculator. Skip counting with a calculator is easy and intriguing. (You can use the repeat function on most calculators to skip count. For example skip count by sevens by pressing: 7 + = = = etc) The calculator allows the students to deal with a greater range of large number problems than they could approach if they had to perform all the computation with pen and paper. In this problem the calculator allows them to look for patterns in the multiples of numbers.
Required Resource Materials:
Calculators
Copymaster of the problem (English)
Copymaster of the problem (Māori)
Activity:
### The Problem
Select a number less than 20. Skip count with it to 1000. How many numbers can you find that will hit no hundreds through to 1000.
### Teaching sequence
1. Introduce the problem by using the calculators to skip count in 8's.
2. Record the multiples of 8 as they are found up to 200.
3. Ask the students to predict which other hundreds they think would be "hit" (multiples of 8). Give them time to find the hits (200, 400, 600, 800)
4. Discuss the pattern found.
5. Read the problem for the students to work on in pairs.
6. Visit with the pairs as they work asking:
What are the next hundreds you will hit exactly? How do you know?
Which number will you pick next? Will that number hit a hundreds number? Which one?
Can you pick a number that will hit every hundred through to 1000? Are there others?
7. Encourage the students to organise their findings into a list and use the list to look for patterns.
8. Share lists and patterns.
#### Extension to the problem
What numbers will hit every hundred through 1000?
### Solution
11, 13, 17, 19
Solution to the Extension
1, 2, 4, 5, 10
AttachmentSize
CalculatorCount.pdf96.37 KB
CalculatorCountMaori.pdf109.61 KB
## The Escape
Find strategies for investigating number problems;
Find a common multiple of a group numbers.
## Counting Cubes
skip count in 2s, 3s, and 5s
devise and use problem solving strategies (make an organised list)
## Well,Well!
Skip count in twos
Investigate and find patterns in addition and subtraction
Devise and use problem solving strategies (act it out, draw a picture)
## Mrs Parore's Laundry
Use simple problem solving strategies;
Solve a simple problem involving number.
## The Three Cold Kittens
Count groups of two to 6;
Use equipment to model and add sets of twos.
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# Writing linear equations
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• 1. Writing Linear Equations
• 2. Slope and Y-Intercept5/2/13• To write the equation of a line:▫ Use y = mx + b▫ Put the given slope and y-intercept in theequation▫ Remember: m is slopeb is y-interceptAnswer needs both x and y in it!!
• 3. Examples:• Write an equation of the line with:• slope = -2y-intercept = 4• slope =y-intercept = -1
• 4. • Slope = 1y-intercept =• Slope = -3• y-intercept = 0
• 5. You Try!• Write an equation of the line with:• slope = 4y-intercept = -5
• 6. Point-Slope Form5/3/13• Another form of a linear equation ispoint-slope form.• It is:• m is still slope!• (x1, y1) is a point on the line• Answer still needs x and y in it!
• 7. Examples:• Write an equation in point-slope formof the line:• Slope = 2• Through (1, 3)• Slope = -4• Through (-2, -6)
• 8. • Slope = 1• Through (0, 5)• Slope = 1/3• Through (-4, 0)
• 9. You Try!• Slope = -2• Through (3, -1)
• 10. Point-Slope toSlope-Intercept Form 5/6/13• Given a point and the slope, we canuse point-slope form.• Remember: y – y1 = m(x – x1)• Then, to get in slope-intercept form:▫ Solve for y▫ Distribute first!▫ Add/subtract to get y alone.
• 11. Examples:• Write an equation of the line with thegiven slope that passes through thegiven point.• Slope = -2• Through (1, 4)
• 12. • Slope = 1/3• Through (-2, 5)• Slope = 1• Through (0, 8)
• 13. You Try!• Write an equation of the line withslope = -1, and through the point (4, -2).
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Bilinear Interpolation Calculator
A Bilinear interpolation calculator is a mathematical calculator configured to determine the location of a point provided by two coordinates. This calculator is helpful in practical situations when two coordinates and a point to bridge the gap must be known.
Corner points coordinates
Corner points values
Interpolated point
What is bilinear interpolation?
Bilinear interpolation is a common technique for two-dimensional interpolation on a rectangle. Values are estimated between known values using Bilinear interpolation. It is a method for figuring out where one point on a curve is in relation to the locations of the other two points.
Bilinear Interpolation Formula
The following actions must be taken to find point P(x, y) using bilinear interpolation:
One linear interpolation along the y-axis, finding point P, and two linear interpolations along the x-axis, discovering intermediate points $$R_1$$ and $$R_2$$.
You can see that bilinear interpolation consists of one linear interpolation along the y-axis and two linear interpolations along the x-axis.
The definition of Point $$R_1(x, y)$$ is
$$R_1(x, y) = \dfrac{Q_{11} (x_2 – x)}{(x_2 – x_1)} + \dfrac{Q_{21} (x – x_1)}{(x_2 – x_1)}$$
Point $$R_2(x, y)$$ is given as follows:
$$R_2(x, y) = \dfrac{Q_{12} (x_2 – x)}{(x_2 – x_1)} + \dfrac{Q_{22} (x – x_1)}{(x_2 – x_1)}$$
We define the interpolated point P(x, y) as follows:
$$P(x, y) = \dfrac{R_1 (y_2 – y)}{(y_2 – y_1)} + \dfrac{R_2 (y – y_1)}{(y_2 – y_1)}$$
Why Do We Use Bilinear Interpolation?
The most often utilized applications of bilinear interpolation in engineering and science are as follows:
• Basic image processing and resampling methods are part of computer vision
• As a method to extract values from a specified set of data, embedded control systems
FAQs
What is the Linear interpolation formula?
The linear interpolation formula is used to fit curves with linear polynomials. In essence, the interpolation approach generates new values for any function using the existing values.
The Linear Interpolation Formula is
$$y = y_1 + (\dfrac{x – x_1}{x_2 – x_1})(y_2 – y_1)$$
Where x is the known and y is the unknown value, x1 and y1 are the coordinates below the x value, and x2 and y2 are the coordinates above the x value.
What is the use of bilinear interpolation in MATLAB?
We can do image zooming using Bilinear Interpolation in MATLAB. Other software has an option of zooming an image too. For example in photoshop, you can zoom your image quite easily by clicking a button. We can do the same thing with MATLAB Bilinear Interpolation using code.
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# Parameters of Discrete Random Variables
## (How to Find the Mean, Median, Mode, Variance & Standard Deviation of a Discrete Random Variable)
In this section we learn how to find the , mean, median, mode, variance and standard deviation of a discrete random variable.
We define each of these parameters:
• mode
• mean (expected value)
• variance & standard deviation
• median
in each case the definition is given and we illustrate how to calculate its value with a tutorial, worked examples as well as some exercises all of which are solved in short video tutorials.
It is worth spending a bit of time on this section as all that is taught here applies to all discrete random variable probability distributions, such as the Binomial Distribution as well as the Poisson Distribution.
## Mode
Given a discrete random variable $$X$$, its mode is the value of $$X$$ that is most likely to occur.
Consequently, the mode is equal to the value of $$x$$ at which the probability distribution function, $$P\begin{pmatrix}X = x \end{pmatrix}$$, reaches a maximum.
## Example
A discrete random variable $$X$$can take-on the values: $x = \left \{ 2, \ 3, \ 4, \ 5, \ 6 \right \}$ and has probability distribution function: $P\begin{pmatrix} X = x \end{pmatrix} = \frac{x^2}{90}$
1. Construct the probability distribution table for $$X$$.
2. Find the mode of the discrete random variable.
#### Solution
1. To construct the probability distribution table we calculate each of the probabilities $$P\begin{pmatrix}X = 2\end{pmatrix}$$, $$P\begin{pmatrix}X = 3\end{pmatrix}$$, ... , $$P\begin{pmatrix}X = 6\end{pmatrix}$$ and summarize the results in a table like the one shown here:
2. The discrete random variable's mode is the value that $$X$$ is most likely to take-on.
Looking at this table we can see that the greatest probability is $$P\begin{pmatrix}X = 6 \end{pmatrix} = \frac{36}{90}$$.
So the mode is $$6$$.
## Mean $$\mu$$ (or Expected Value $$E\begin{pmatrix}X\end{pmatrix}$$)
The expected value of a discrete random variable $$X$$ is the mean value (or average value) we could expect $$X$$ to take if we were to repeat the experiment a large number of times. It is calculated with: $E(X) = \sum x.P \begin{pmatrix} X = x \end{pmatrix}$ The expected value is also known as the mean $$\mu$$ of the random variable, in which case we write: $\mu = \sum x.P \begin{pmatrix} X = x \end{pmatrix}$ Note: it doesn't matter whether we refer to $$E(X)$$ or $$\mu$$, but it is important to know that they both refer to the same thing.
## Example
Consider the simple experiment of rolling a single unbiased dice once.
Define the discrete random variable $$X$$ as: $X:\text{the value obtained when rolling the dice}$ Find the mean value of this discrete random variable.
#### Solution
The discrete random variable $$X$$ can take-on any of the $$\left \{ 1, \ 2, \ 3, \ 4, \ 5, \ 6 \right \}$$.
We can illustrate this probability distribution in a table:
We now calculate the mean value $$\mu$$ of $$X$$: \begin{aligned} \mu & = \sum_{x=1}^6 x.P \begin{pmatrix} X = x \end{pmatrix} \\ & = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + 5 \times \frac{1}{6} + 6 \times \frac{1}{6} \\ & = \frac{21}{6} \\ \mu & = 3.5 \end{aligned} This mean tells us that if we were to roll a dice a large number of times and were to calculate the average of all the values we obtained the result would be close to $$3.5$$. In other words the mean value we can expect to obtain when rolling a dice is $$3.5$$.
## Tutorial 1
In the following tutorial we show how to find the mode and the mean of a discrete random variable, using the rules we just read (above).
We do this for a discrete random variable $$X$$ that has the following probability distribution table:
## Exercise 1
1. A discrete random variable $$X$$ has the following probability distribution table:
1. State the mode.
2. Calculate this discrete random variable's mean value.
2. A discrete random variable $$X$$ can take on the values: $x = \left \{ 2, \ 4, \ 6 \right \}$ and has probability distribution function defined as: $P\begin{pmatrix} X = x \end{pmatrix} = \frac{8x - x^2}{40}$
1. Construct a probability distribution table for $$X$$.
2. State the mode of $$X$$.
3. Calculate this discrete random variable's mean value.
3. A discrete random variable $$X$$ has probability distribution table:
1. Find the value of $$k$$.
2. State the mode of $$X$$.
3. Calculate the expected value $$E\begin{pmatrix}X\end{pmatrix}$$.
Note: this exercise can be downloaded as a worksheet to practice with:
### Variance & Standard Deviation
#### Variance, $$Var\begin{pmatrix}X\end{pmatrix}$$ or $$\sigma^2$$
Given a discrete random variable $$X$$, we calculate its Variance, written $$Var\begin{pmatrix}X \end{pmatrix}$$ or $$\sigma^2$$, using one of the following two formula:
##### Formula 1
$Var\begin{pmatrix}X \end{pmatrix} = \sum \begin{pmatrix}x - \mu \end{pmatrix}^2 . P\begin{pmatrix} X = x \end{pmatrix}$
##### Formula 2
$Var\begin{pmatrix} X \end{pmatrix} = E\begin{pmatrix}X^2 \end{pmatrix} - \mu^2$ Where: $E\begin{pmatrix}X^2 \end{pmatrix} = \sum x^2.P\begin{pmatrix} X = x \end{pmatrix}$
#### Standard Deviation, $$\sigma$$
The standard deviation, $$\sigma$$, of a discrete random variable $$X$$ tells us how far away from the mean $$\mu$$ we can expect the value of $$X$$ to be.
We calculate $$\sigma$$ using the formula: $\sigma = \sqrt{ Var \begin{pmatrix} X \end{pmatrix}}$ Note: it is important to realize and keep in mind that the value of $$\sigma$$ is an average and could be expected to be observed after a sufficiently large number of trials.
## Example
A discrete random variable $$X$$ can take the values $$x = \left \{ 3, \ 4, \ 6, \ 7 \right \}$$ and has a probability distribution function $$P\begin{pmatrix} X = x \end{pmatrix} = \frac{x}{20}$$.
1. Calculate the mean value of $$X$$.
2. Calculate the variance and the standard deviation.
### Solution
1. To calculate the mean value we use the formula: $\mu = \sum x.P\begin{pmatrix}X = x \end{pmatrix}$ Given $$X$$ can take-on either of the values $$3$$, $$4$$, $$6$$ and $$7$$, this formula leads to: \begin{aligned} \mu & = 3 \times P \begin{pmatrix} X = 3 \end{pmatrix} + 4\times \begin{pmatrix} X = 4 \end{pmatrix} + 6 \times \begin{pmatrix} X = 6 \end{pmatrix} + 7 \times \begin{pmatrix} X = 7 \end{pmatrix} \\ & = 3 \times \frac{3}{20} + 4 \times \frac{4}{20} + 6 \times \frac{6}{20} + 7 \times \frac{7}{20} \\ & = \frac{9}{20} + \frac{16}{20} + \frac{36}{20} + \frac{49}{20} \\ & = \frac{110}{20} \\ \mu & = 5.5 \end{aligned} Finally, we can state: the mean value of this discrete random variable is $$\mu = 5.5$$.
2. To calculate the variance we use the formula: $Var\begin{pmatrix} X \end{pmatrix} = E\begin{pmatrix}X^2 \end{pmatrix} - \mu^2$ Where: $E\begin{pmatrix} X^2 \end{pmatrix} = \sum x^2.P\begin{pmatrix} X = x \end{pmatrix}$ We start by calculating $$E\begin{pmatrix}X^2\end{pmatrix}$$.
Using the values $$x$$ that the discrete random variable can take the formula leads to: \begin{aligned} E\begin{pmatrix} X^2 \end{pmatrix} & = 3^2\times P\begin{pmatrix}X = 3 \end{pmatrix} + 4^2 \times P\begin{pmatrix}X = 3 \end{pmatrix} + 6^2 \times P\begin{pmatrix}X = 6 \end{pmatrix} + 7^2 \times P\begin{pmatrix}X = 7 \end{pmatrix} \\ & = 3^2 \times \frac{3}{20} + 4^2 \times \frac{4}{20} + 6^2\times \frac{6}{20} + 7^2 \times \frac{7}{20} \\ & = 9 \times \frac{3}{20} + 16 \times \frac{4}{20} + 36 \times \frac{6}{20} + 49 \times \frac{7}{20} \\ & = \frac{27}{20} + \frac{64}{20} + \frac{216}{20} + \frac{343}{20} \\ & = \frac{650}{20} \\ E\begin{pmatrix} X^2 \end{pmatrix} & = 32.5 \end{aligned} Now that we know the value of $$E\begin{pmatrix} X^2 \end{pmatrix} = 32.5$$, we can calculate the variance: \begin{aligned} Var\begin{pmatrix} X \end{pmatrix} = E\begin{pmatrix}X^2\end{pmatrix} - \mu^2 \\ & = 32.5 - 5.5^2 \\ & = 32.5 - 30.25 \\ Var\begin{pmatrix} X \end{pmatrix} & = 2.25 \end{aligned} The variance is $$2.25$$.
Now that we know the variance, we can calculate this discrete random variable's standard deviation: \begin{aligned} \sigma & = \sqrt{Var\begin{pmatrix} X \end{pmatrix}} \\ & = \sqrt{2.25}\\ \sigma & = 1.5 \end{aligned} Finally, we can state: the standard deviation is $$\sigma = 1.5$$.
## Tutorial 2
In tutorial 2 we learn how to calculate the variance and standard deviation of a discrete random variable.
We do this for the following example:
A discrete random variable $$X$$ can take the values $$x = \left \{ 1, \ 2, \ 3, \ 4 \right \}$$.
It has probability distribution function $$P\begin{pmatrix}X = x \end{pmatrix} = \frac{x}{10}$$.
Find: the variance and the standard deviation of $$X$$.
## Exercise 2
1. A discrete random variable $$X$$ has probability distribution function defined as: $P\begin{pmatrix} X = x \end{pmatrix} = \frac{x^2}{120}$ Where $$X$$ can take-on the values $$1$$, $$3$$, $$5$$, $$6$$ and $$7$$.
1. Calculate this discrete random variable's mean value.
2. Calculate the standard deviation.
2. A discrete random variable $$X$$ has the following probability distribution table:
1. Calculate the mean value of $$X$$.
2. Calculate the variance of $$X$$.
3. Calculate the standard deviation of $$X$$
3. A discrete random variable $$X$$ can take-on the values: $x = \left \{ 1, \ 2, \ 3, \ 4 \right \}$ and has probability distribution function: $P\begin{pmatrix} X = x \end{pmatrix} = \frac{x}{10}$
1. Construct a probability distribution table for this discrete random variable.
2. State its mode.
3. Calculate the mean value of $$X$$.
4. Calculate the variance of $$X$$, hence calculate its standard deviation.
Note: this exercise can be downloaded as a worksheet to practice with:
## Median
Given a discrete random variable $$X$$ and its cumulative distribution function $$P\begin{pmatrix} X \leq x \end{pmatrix} = F(x)$$, the median of the discrete random variable $$X$$ is the value $$M$$ defined as: $M = \frac{x_1+x_2}{2}$ Where:
• $$x_1$$ is the greatest value $$X$$ can take such that: $$P\begin{pmatrix}X \leq x_1 \end{pmatrix} \leq 0.5$$.
• $$x_2$$ is the smallest value $$X$$ can take such that: $$P\begin{pmatrix}X \leq x_2 \end{pmatrix} \geq 0.5$$.
The median $$M$$ of $$X$$ is the middle value.
The probability that $$X$$ takes-on a value less than $$M$$ is $$0.5$$. Similarly, the probability that $$X$$ takes-on a value greater than $$M$$ is $$0.5$$.
In other words there is an equal chance that $$X$$ be greater or less than $$M$$.
## Example
A discrete random variable $$X$$ has the following cumulative probability distribution table:
Find the median value of $$X$$.
### Solution
We need to find $$x_1$$ and $$x_2$$.
Remember:
• $$x_1$$ is the greatest value $$X$$ can take such that: $$P\begin{pmatrix}X \leq x_1 \end{pmatrix} \leq 0.5$$.
• $$x_2$$ is the smallest value $$X$$ can take such that: $$P\begin{pmatrix}X \leq x_2 \end{pmatrix} \geq 0.5$$.
Looking at the cumulative distribution table we see that:
• the greatest value of $$P\begin{pmatrix} X \leq x \end{pmatrix}$$ that is less than, or equal to, $$0.5$$ is $$P\begin{pmatrix} X \leq x \end{pmatrix} = 0.3$$, which corresponds to $$X = 1$$.
• the smallest value of $$P\begin{pmatrix} X \leq x \end{pmatrix}$$ that is greater than, or equal to, $$0.5$$ is $$P\begin{pmatrix} X \leq x \end{pmatrix} = 0.6$$, which corresponds to $$X = 2$$.
So:
$$x_1 = 1$$ and $$x_2 = 2$$
The median value of $$X$$ is therefore: \begin{aligned} M &= \frac{x_1 + x_2}{2} \\ & = \frac{1+2}{2} \\ & = \frac{3}{2} \\ M & = 1.5 \end{aligned} The median value of $$X$$ is $$1.5$$, which tells us there is a $$50\%$$ chance that $$X$$ be less than $$1.5$$ and a $$50\%$$ chance it be greater than $$1.5$$.
## Tutorial 3
In the following tutorial we learn how to find the median of a discrete random variable.
We start by reminding ourselves how to construct a cumulative probability distribution table and then learn how to use it to find the median value.
For this we suppose we're given a discrete random variable $$X$$ with the following probability distribution table:
## Exercise 3
1. A discrete random variable $$X$$ has the following cumulative distribution table:
1. Find $$P\begin{pmatrix}X = 4\end{pmatrix}$$
2. Find the median value of $$X$$.
2. A discrete random variable $$X$$ has probability distribution table defined as:
1. Construct this random variable's cumulative distribution table.
2. Find the median value of $$X$$.
3. Given the discrete random variable $$X$$, with probability distribution function: $P\begin{pmatrix}X = x \end{pmatrix} = \frac{x^2}{90}$ Where: $x = \left \{ 2, \ 3, \ 4, \ 5, \ 6 \right \}$
1. Construct a probability distribution table for $$X$$.
2. State the mode of $$X$$.
3. Calculate the mean value of $$X$$.
4. Calculate this discrete ranom variable's standard deviation.
5. Construct a cumulative distribution table for $$X$$.
6. Find the median value of $$X$$.
Note: this exercise can be downloaded as a worksheet to practice with:
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## using coordinates to find length of sides
There are two ways. How to draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. How to find the angle of a right triangle. Find area of triangle if two vectors of two adjacent sides are given. Example: (0, 0), (5, 3), (5, 7), (0, 4). (Using A,D will produce the same result). Step By Step. However, you can see that other numbers can be used instead of 3, 4 and 5. However, if only two sides of a triangle are given, finding the angles of a right triangle requires applying some basic trigonometric functions: Set up an equation using a sohcahtoa ratio.Since we know the hypotenuse and want to find the side opposite of the 53° angle, we are dealing with sine $$sin(53) = \frac{ opposite}{hypotenuse} \\ sin(53) = \frac{ \red x }{ 12 }$$ Let us look into some examples to understand the above concept. Calculating the length of another side of a triangle If you know the length of the hypotenuse and one of the other sides, you can use Pythagoras’ theorem to find the length of the third side. 2. But still i haven't been able to find the coordinates. It turns out that in a 30-60-90 triangle, you can find the measure of any of the three sides, simply by knowing the measure of at least one side in the triangle. Scalene-- No two sides are congruent (equal in length) Here’s a formula to use, based on the counterclockwise entry of the coordinates of the vertices of the triangle ( x 1 , … The hypotenuse is equal to twice the length of the shorter leg, which is the side across from the 30 degree angle. In this lesson students will recognize that by determining the distance and slopes of various lines of a quadrilateral, they can classify the shape as a parallelogram, rectangle, rhombus, square or none of these. if triangle PQR has the co-ordinates (-8,5), (2,1), (7,9), what are the lengths of each individual side. Multiply both sides by θ, so for an angle θ radians. Find the length of the sides in an Isosceles triangle, given its area and perimeter? Where a and b are the parallel sides of the trapezium and h is its height. Using the formula for the distance between two points, this is Things to try. Thanks In this case, we can explore the nature of polygons. Using the formula for the distance between two points, this is The length of a diagonals is the distance between B and D. (Using A,C will produce the same result). Carefully find the midpoints of two of the sides, and then draw the two medians to those midpoints. You can work out the length of an arc by calculating what fraction the angle is of the 360 degrees for a full circle. I'd like to calculate both positions. We usually use the distance formula for finding the length of sides of polygons if we know coordinates of their vertices. I just wasn't sure how to use those to get the coordinates of that specific point. $\begingroup$ Actually, I was already aware of how to get the lengths of the sides and internal angles of the triangle. The length of the other two sides are given. Find its centroid. The attached pictures can help to visualize the problem. STEP 4: So, to find x, we substitute a with x,b with x+8, h with 7 and A with 91. How to find the length of a line using coordinates, Microsoft online services sign in assistant, The length of this line can be found using the distance formula: \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2). Explanation: . A triangle's height is the length of a perpendicular line segment originating on a side and intersecting the opposite angle.. First, calculate the length of all the sides. It will be instructive to learn both methods. To find the coordinates of a point in the polar coordinate system, consider Figure $$\PageIndex{1}$$. A triangle gets its name from its three interior angles. Active 2 years, 11 months ago. Suppose, we have a as shown in the diagram and we want to find its area.. Let the coordinates of vertices are (x1, y1), (x2, y2) and (x3, y3). Step 3: Calculating the semi perimeter using the formula (a+b+c)/2. Big Ideas: Coordinate geometry can be used to determine existing properties of a quadrilateral and classify it. By their sides, you can break them down like this: Sides. Real World Problems Using Length between Two Points. Although we can write semi perimeter … Then apply above formula to get all angles in radian. How to Find the Length of an Arc. I believe there can only be two possible solutions for the coordinates of C; the one drawn above, and one with C reflected about the line c, approximately at C'. But there’s an even better choice, based on the determinant of a matrix. If you know one angle apart from the right angle, calculation of the third one is a piece of cake: Givenβ: α = 90 - β. Givenα: β = 90 - α. Every triangle has three heights, or altitudes, because every triangle has three sides. A = bh It's easiest to show by actually doing an example. (You can draw in the third median if you like, but you don’t need it to find the centroid.) Q.1: Find the lengths of the sides of the triangle with vertices A(-4,0), B(3,4) and C(4,1).show that the triangle ABC is isosceles? How to find the length of a median of a triangle with vertices : Here we are going to see how to find the length of a median of a triangle with vertices. Using Pythagoras to calculate the Length of an Unknown Side–Study Guide 9 Page 2 The 3 – 4 – 5 Method is a popular term for this rule, as it is an easy way to show how the rule works. Take the coordinates of two points you want. The area of a rectangle ( A ) is related to the length ( L ) and width ( W ) of its sides by the following relationship: A = L × W. If you know the width, it's easy to find the length by rearranging this equation to get. Slope formula is given by-----If the product of any two slopes is -1. then the angle between them is 90 degrees and It is a right triangle.What is slope? This problem is then solved using the above mentioned theorem. 1. Area of Triangle = Now, we can easily derive this formula using a small diagram shown below. In the figure at the top of the page, click on "hide details" . Given the area and perimeter of a triangle, find its coordinates. Ask Question Asked 2 years, 11 months ago. Defining Polar Coordinates. First calculate the difference between the coordinates of corresponding points on the x-axis and the y-axis. Find the length and equation of its sides. The distance formula is-----Using the slope formula, you can find whether the triangle is right triangle or not. A = bh use distance formula to find b = base; use perpendicular distance from a line to a point formula to find h = height Given: coordinates of a parallelogram. Isosceles triangle and scalene triangle. How do you find the length of sides to a triangle using co-ordinates? The points (0, 0), (5,3) represent the base. The point $$P$$ has Cartesian coordinates $$(x,y)$$. You can classify triangles either by their sides or their angles. The line segment connecting the origin to the point $$P$$ measures the distance from the origin to $$P$$ and has length … These are the four steps to follow: Step 1 Find the names of the two sides we are using, one we are trying to find and one we already know, out of Opposite, Adjacent and Hypotenuse. This gives: STEP 5: Now, we simplify the equation: STEP 6: Notice that, to find x, we need to remove 7. Using the sides length information i can find out the three angles inside. Example 1 : Find the length of the medians of the triangle whose vertices are (1 , … In order to find , we must first find .The formula for the area of a parallelogram is: We are given as the area and as the base. To find the altitude, we first need to know what kind of triangle we are dealing with. Step 1: In this C program, User will enter the three sides of the triangle a, b, c. Step 2: Calculating the Perimeter of the Triangle using the formula P = a+b+c. Take the coordinates of the points of one side and caluclate the length of the line using the formula. Finding the side length of a rectangle given its perimeter or area - In this lesson, we solve problems where we find one missing side length while one side length … We calculate the perimeter by adding the lengths of the sides. Click hereto get an answer to your question ️ If the coordinates of the mid - points of the sides of a triangle are (1, 1), (2, - 3) and (3, 4). solving real-world and mathematical problems, Common Core Grade 6, 6.g.3, length of sides, examples and step by step solutions We draw perpendiculars AP, BQ and CR to x-axis. How to Find the Height of a Triangle. 24, Feb 17. Click hereto get an answer to your question ️ The co - ordinates of the midpoint of the sides of a triangle ABC are D(2,1),E(5,3),f(3,7) . Now, we can use trigonometry to solve for .With respect to , we know the opposite side of the right triangle and we are looking for … Find area. Use the formula to calculate the cut length, if your task explicitly set the coordinates of the vertices of the triangle.For this follow some simple steps. This is 3 separate problems that can be solved using the same equation. The centroid is where these medians cross. So to convert radians to degrees, multiply by 180/π. I know the coordinates of A and B, the lengths of a and c, and that the angle C will always be a right angle. ... Find coordinates of the triangle given midpoint of each side. Then convert angles from radian into degrees. ; Step 2 Use SOHCAHTOA to decide which one of Sine, Cosine or Tangent to use in this question. If we compare the lengths of two or more line segments, we use the formula for the distance between two points. The results … Below is implementation of above steps. You can always use the distance formula, find the lengths of the three sides, and then apply Heron’s formula. 09, Oct 18. θ radians = 360/(2π) x θ = (180/π)θ degrees. As expected, a bit of basic algebric manipulation (linear equation solution) is required to obtain the length … Geometry Quizlet DBA You can find the length of the three sides by using the distance formula. L = \frac{A}{W} If you know the length and want the width, rearrange to get. , ( 5,3 ) represent the base the base take the coordinates the parallel sides of the triangle Actually an... Draw the two medians to those midpoints be solved using the formula for finding using coordinates to find length of sides length of line. Of that specific point name from its three interior angles side across from the 30 degree.., given its area and perimeter of a perpendicular line segment originating a. We first need to know what kind of triangle we are dealing.. Properties of a quadrilateral and classify it a side and intersecting the angle... This is Things to try between the coordinates of their vertices their angles the hypotenuse is equal twice! Take the coordinates of their vertices use the distance formula for the distance between two points, is! Was already aware of how to find the midpoints of two or more line segments we... } { W } if you know the length of sides to a triangle using co-ordinates you find the of... Triangle 's height is the side across from the 30 degree angle by adding the of! Of their vertices -- No two sides are congruent ( equal in )... Be solved using the sides two of the trapezium and h is height... Name from its three interior angles system, consider Figure \ ( (,. Be solved using the formula for the distance formula semi using coordinates to find length of sides using above. Polar coordinate system, consider Figure \ ( \PageIndex { 1 } \ ) those midpoints 360/ 2π... Able to find the altitude, we can explore the nature of polygons coordinate geometry can be solved the... Width, rearrange to get the lengths of the triangle given midpoint of each.! Or more line segments, we use the distance formula for the distance formula --. Still i have n't been able to find the altitude, we can explore the nature of if! 3: calculating the semi perimeter using the distance between two points, this is Things try! Using the formula for the distance formula for finding the length of three! The polar coordinate system, consider Figure \ ( ( x, y ) \.! Of a point in the polar coordinate system, consider Figure \ ( P\ ) has coordinates! The line using the distance between two points, this is Things to try, you can whether... To use those to get the coordinates of their vertices in an Isosceles,! And internal angles of the sides, and then draw the two medians to those midpoints all in. And 5 centroid. 5,3 ) represent the base perimeter of a quadrilateral and classify it and... ( 180/π ) θ degrees of how to use in this case, we use the formula for distance... Which one of Sine, Cosine or Tangent to use in this question to midpoints. By calculating what fraction the angle is of the sides length information i can find out length... To degrees, multiply by 180/π for the distance formula still i have n't been able to the! Actually, i was already aware of how to find the midpoints of two adjacent sides are congruent ( in! We can explore the nature of polygons if we know coordinates of a point in the Figure at top. Do you find the coordinates of that specific point need to know what kind of we. Vectors of two of the line using the formula for the distance formula by 180/π by 180/π ) represent base. Degrees for a full circle ; Step 2 use SOHCAHTOA to decide which one of Sine, Cosine Tangent. Third median if you know the length of the three angles inside all angles in radian -- the! And want the width, rearrange to get all angles in radian of triangle we are dealing with top..., rearrange to get the coordinates of the sides rearrange to get lengths... Polygons if we compare the lengths of the three angles inside you like, but you don t. The above mentioned theorem the centroid. 's easiest to show by Actually doing an example we usually the!, we first need to know what kind of triangle if two vectors of two of the.... Can break them down like this: sides the length of the sides internal. We first need to know what kind of triangle if two vectors of two adjacent are... Coordinate geometry can be used to determine existing properties of a point in polar. The angle of a quadrilateral and classify it angle is of the sides length information i can find angle..., based on the x-axis and the y-axis you like, but you don ’ t need it find! Do you find the angle of a perpendicular line segment originating on a side and intersecting the angle... Dealing with point \ ( \PageIndex { 1 } \ ) polygons if we coordinates. It to find the angle is of the triangle us look into some examples to understand above. -Using the slope formula, you can break them down like this: sides parallel of... Show by Actually doing an example is the side across from the 30 degree angle radians to degrees multiply! To get all angles in radian show by Actually doing an example are. You like, but you don ’ t need it to find the length of the,! } if you know the length of a quadrilateral and classify it length of triangle... Other two sides are given in the third median if you know the length of an by! Like, but you don ’ t need it to find the length of sides to triangle! Based on the x-axis and the y-axis sides or their angles a point in the Figure at the top the. The three sides by using the distance formula is -- -- -Using the slope formula, you can whether! \Begingroup \$ Actually, i was already aware of how to use in this,... Of polygons sides, you can find out the length of a and! Can help to visualize the problem help to visualize the problem we the! Of one side and intersecting the opposite angle twice the length of sides. The page, click on hide details '' or more line segments, first! Sure how to get the lengths of the sides length information i can find whether the.... Can draw in the Figure at the top of the points of one side intersecting. Above mentioned theorem degree angle formula is -- -- -Using the slope formula, can. Line using the above mentioned theorem kind of triangle we are dealing with using coordinates to find length of sides segments we... Even better choice, based on the determinant of a perpendicular line segment originating on a side intersecting! Right triangle or not we usually use the distance formula at the top of the sides see. Sides by using the above concept gets its name from its three interior angles is the length of sides a... Or not in the third median if you know the length and want the width, to., multiply by 180/π adding the lengths of two or more line segments, we use the formula ( )...
0000
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#### Mixed Math for Easter
Have fun together with your lower elementary homeschooler doing these Easter themed math rpoblems and activities!
Word Problems
1. Trent and Jenny colored eggs with their Sunday School class. If they each colored 5 eggs, how many eggs did they color all together?
2. Peter Rabbit hid 15 eggs around his neighborhood for the children to find. How many eggs were left if the children only found ten of them?
Jelly Bean Problems
*Jelly Beans Needed for #'s 3-7! There are alternate types of jelly beans that can be fond at health food and specialty stores such as gluten free and sugar free if needed for health purposes.
3. Count out 25 jelly beans. Eat five of them. How many are left?
4. Group the jelly beans by color. Which color has the most number jellybeans? Which color has the least amount of jellybeans?
5. Count 50 jelly beans out by two's and five's.
6. Create a pattern using three colors of jelly beans.
7. Read Arthur's Jelly Beans together and complete the corresponding book activities throughout the book.
Real Eggs
8. Egg Math- Using a hard cooked egg, show your child the various fractions of one third, one half, and one fourth via slices. You may need several eggs for this!
Estimate and Place Value
9. Create an Estimation Jar- Fill a jar with jelly beans and have your homeschoolers esimate how many jelly beans are in the jar. Discuss the idea of an estimation, or a guess. After all family members have guessed you can count out the real amount together!
10. Make a tens and ones place value worksheet, with two wide columns, one for tens and one for ones. Next, place fifteen jelly beans for your child to count out and group into a tens and ones column. Discuss how to separate the groups and write the number on an additional sheet of paper, too. Then start again with another tens/ones combination number, like eighteen, eleven, etc. You can move into the twenties, and so forth if your child is ready, and have 2 groups of ten, etc.
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# How to Make a Sign Diagram
## What is a Sign Diagram?
A sign diagram shows you where a function has positive or negative values.
How to make a sign diagram: General Steps [1]
1. Find the critical points. These will be anywhere the function has zeros(roots) or vertical asymptotes.
2. Figure out if your points in Step 1 are zeros, vertical asymptotes or holes.
3. Test a point in each interval to see if it is positive or negative in that interval.
4. Draw a number line with the critical points. Label each point underneath with its x-value.
5. Ad a ± in each interval, indicating where the function is positive or negative (or note that it is undefined).
## How to Make a Sign Diagram: Example
Example question: Make a sign diagram of the following function:
Step 1: Find the critical points. In order to find the critical points for this particular function, we need to factor it:
The function has three critical values: one which will make the function equal zero (x = 2) and two which make the function undefined (x = -3).
Step 2: Figure out what’s happening at the critical points. You can do this by plugging in some values into the equation for the function:
• x = -3 is a vertical asymptote. You can check for this by testing a few values around x = -3; They will start to deviate rapidly.
• x = 2 is a zero(root). This is where the function crosses the x-axis. Plug x = 2 into the function and y = 0.
• x = 4 is a hole. If you plug x = 4 into the function, it will be undefined.
The point at x = 4 is a hole and does not change the sign of the function, so we can omit it from our sign diagram.
Step 3: Draw a number line with the points from Step 3 (usual practice is to place the labels underneath the line):
Step 4: Test a random point in each interval. For this sign diagram, we have three intervals:
• (-∞, -3)
• (-3, 2)
• (2, ∞}
At this point, you could refer to the graph. But if you aren’t able to graph the function, test a point by plugging it into the function. For example, in the middle interval you might choose 0:
This is a negative value, so the sign on the center interval will be negative.
Step 5: Complete the sign diagram by placing a “-” or “+” in each interval, based on your results from Step 4.
## References
[1] How to Construct a Sign Diagram: Retrieved July 13, 2021 from: https://people.highline.edu/dwilson/Previous_Classes/2006.4_Fall/2006_fall_math_115/sign_diagrams/how_to.htm
[2] Smith, K. (2013). Elementary Functions. Retrieved July 13, 2021 from: https://www.shsu.edu/kws006/precalculus/2.6_Rational_Functions_files/2.6%20Rational%20Functions%20(slides,%204%20to%201).pdf
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HOW TO DO ALGEBRA - SPECIAL PRODUCTS, EXAMPLES, EXERCISE, USING FORMULAS
Algebra Formulas before How To Do Algebra
if you have not already done so.
There we have listed out all the
Algebra Formulas that need to be
remembered.
Proofs of the first six Formulas
Proofs of the Last six Formulas
if you have not already done so.
Here we present Solved Examples
and Exercise problems
on application of those Formulas.
Example 1 of How To Do Algebra
Find the quotient (27x3 - 64)÷(9x2 + 12x + 16) without actual division.
Solution to Example 1 of How To Do Algebra :
Let P = (27x3 - 64)÷(9x2 + 12x + 16)
We know 27 = 3 x 3 x 3 = 33 ⇒27x3 = 33x3 = (3x)3
64 = 4 x 4 x 4 = 43
∴ Numerator of P = (27x3 - 64) = (3x)3 - 43
This looks like (a3 - b3) with (3x) in place of a and (4) in place of b
Denominator of P = (9x2 + 12x + 16) = (3x)2 + (3x)(4) + 42
This looks like (a2 + ab + b2) with (3x) in place of a and (4) in place of b
We have (a3 - b3) = (a - b)(a2 + ab + b2) (See Formula 5)
⇒ (a3 - b3)÷(a2 + ab + b2) = (a - b)
P is like the L.H.S. of this with (3x) in place of a and (4) in place of b
∴ P = (3x - 4) Ans.
Example 2 of How To Do Algebra
If 2x + 3y - 4z = 10 and 3xy - 6yz - 4zx = 15, find 4x2 + 9y2 + 16z2.
Solution to Example 2 of How To Do Algebra :
Let P = 4x2 + 9y2 + 16z2 = (2x)2 + (3y)2 + (4z)2
This looks like a2 + b2 + c2
with (2x) in place of a, (3y) in place of b and (4z) in place of c
Then the given data looks like a + b - c = 10 and (1⁄2 )(ab - bc - ca) = 15
We have (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (See Formula 8)
Putting (-c) in place of c, we get
(a + b - c)2 = a2 + b2 + (-c)2 + 2ab + 2b(-c) + 2(-c)a (See Formula 8)
= a2 + b2 + c2 + 2ab - 2bc - 2ca
a2 + b2 + c2= (a + b - c)2 - 2(ab - bc - ca)
Applying this to P, we get
P = (2x)2 + (3y)2 + (4z)2
= (2x + 3y - 4z)2 - 2{(2x)(3y) - (3y)(4z) - (4z)(2x)}
= (2x + 3y - 4z)2 - 2 x 2{(x)(3y) - (3y)(2z) - (4z)(x)}
= (2x + 3y - 4z)2 - 4{(3xy) - (6yz) - (4zx)}
By data, 2x + 3y - 4z = 10 and 3xy - 6yz - 4zx = 15
Using these in P, we get
P = (10)2 - 4(15) = 100 - 60 = 40. Ans.
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Example 3 of How To Do Algebra
Solved Example 3 on How To Do Algebra :
Find the quotient
(x3 + 27y3 + 8z3 - 18xyz)÷(x + 3y + 2z)
without actual division.
Solution to Example 3 of How To Do Algebra :
We have a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) [See Formula 9]
⇒ (a3 + b3 + c3 - 3abc)÷(a + b + c) = (a2 + b2 + c2 - ab - bc - ca).....(i)
Let P = (x3 + 27y3 + 8z3 - 18xyz)÷(x + 3y + 2z)
= {x3 + (3y)3 + (2z)3 - 3(x)(3y)(2z)}÷(x + 3y + 2z)
This looks like Equation (i)
with x in place of a, 3y in place of b and 2z in place of c
Applying Equation (i) here, we get
P = {x3 + (3y)3 + (2z)3 - 3(x)(3y)(2z)}÷(x + 3y + 2z)
= {(x)2 + (3y)2 + (2z)2 - (x)(3y) - (3y)(2z) - (2z)(x)}
= {x2 + 9y2 + 4z2 - 3xy - 6yz - 2zx} Ans.
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Example 4 of How To Do Algebra
Solved Example 4 on How To Do Algebra :
Find the products
(i) (x + 1)(x - 3) (ii) (2x + 5)(5x - 3) (iii) (x + 2)(x + 4)(x + 5)
using the Algebra Formulas.
Solution to Example 4 of How To Do Algebra :
(i) Let P = (x + 1)(x - 3)
We have (x + a)(x + b) = x2 + x(a + b) + ab [See Formula 10]
Comparing P with this Formula, a = 1 and b = -3
∴ P = (x + 1)(x - 3) = x2 + x{1 + (-3)} + {(1)(-3)}
= x2 + x{-2} + {(-3)} = x2 - 2x - 3 Ans.
(ii) Let P = (2x + 5)(5x - 3)
We have(ax + b)(cx + d) = acx2 + x(ad + bc) + bd
Comparing P with this Formula, a = 2, b = 5, c = 5 and d = -3
∴ P = (2x + 5)(5x - 3) = (2)(5)x2 + x{(2)(-3) + (5)(5)} + (5)(-3) = 10x2 +19x - 15. Ans.
Example 5 of How To Do Algebra
Solved Example 5 on How To Do Algebra :
If l + m + n = 0, prove that l3 + m3 + n3 = 3lmn
Solution to Example 5 of How To Do Algebra :
l + m + n = 0 ⇒ l + m = -n .........(i)
Cubing both sides, we get
(l + m)3 = (-n)3
l3 + m3 + 3lm(l + m) = -n3 [See Formula 6]
Using the value of (l + m) from (i), we get
l3 + m3 + 3lm(-n) = -n3
l3 + m3 - 3lmn = -n3
l3 + m3 + n3 = 3lmn (Proved.)
Exercise : How To Do Algebra
Solve the following problems on How To Do Algebra
1. Find the quotient (8x3 - 343)÷(2x - 7) without actual division.
2. If x + 2y + 3z = 12 and x2 + 4y2 + 9z2 = 44, find 2xy + 6yz + 3zx
3. Find the quotient
(l3 + 27m3 + 343n3 - 63lmn)÷(l2 + 9m2 + 49n2 - 3lm - 21mn - 7nl)
without actual division.
4. Find the products
(i) (x - 9)(x - 7) (ii) (5x + 2)(7x + 5) (iii) (x - 3)(x - 8)(x + 6)
using the Algebra Formulas.
5. If p + q = r, prove that p3 + q3 + 3pqr = r3
For Answers See at the bottom of the page.
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Answers to Exercise : How To Do Algebra
Answers to Problems on How To Do Algebra :
1. 4x2 + 14x + 49
2. 50
3. l + 3m + 7n
4. (i) x2 -16x + 63 (ii) 35x2 + 39x + 10 (iii) x3 - 5x2 - 42x + 144
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# Triangles
Here we will learn about triangles, including what a triangle is and how to solve problems involving their sides and their angles.
There are also triangles worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What are triangles?
Triangles are 2D shapes with three straight sides and three vertices. These are the only two key properties of triangles.
To solve problems involving triangles, we need to be able to gather information about its side lengths and/or its interior angles and apply these to the problem.
The sum of interior angles of a triangle is \bf{180°}.
For example,
29^{\circ}+68^{\circ}+83^{\circ}=180^{\circ}.
If a triangle’s angles are all acute angles (less than 90^{\circ} ), it is sometimes known as an acute triangle.
If a given triangle has one obtuse angle it is sometimes known as an obtuse triangle.
Step-by-step guide: Angles in a triangle
### Types of triangles
There are different types of triangles that have different properties.
Being able to recognise different types of triangles is known as classifying triangles.
Step-by-step guide: Types of triangles
### Equilateral triangles
An equilateral triangle has the following properties,
• All sides are equal length.
• All angles are equal size (60^{\circ}) .
An equilateral triangle has 3 lines of symmetry. The lines go from one vertex to the middle of the opposite side.
The lines of symmetry are angle bisectors of the vertices and are also perpendicular bisectors of the sides of the equilateral triangle.
Step-by-step guide: Equilateral triangles
### Isosceles triangles
An isosceles triangle has the following properties,
• Two sides are equal length.
• Two equal angles.
The two equal angles are called base angles and they are opposite the sides of equal length. The unequal side is known as the base and is opposite the unequal angle.
Step-by-step guide: Isosceles triangles
### Scalene triangles
A scalene triangle has the following properties,
• No equal sides.
• No equal angles.
Step-by-step guide: Scalene triangles
### Right-angled triangles
A right-angled triangle has the following property,
• One right angle.
The longest side of a right-angled triangle is opposite the right-angle. It is called the hypotenuse.
The Pythagorean theorem is used to determine missing side lengths of a right-angled triangle. If a question involves both the sides of the triangle and one of the angles, we could use trigonometry.
A special case of a right-angled triangle is a right-angled triangle which also has two equal sides. It can be known as an isosceles right triangle or an isosceles right-angled triangle. The two equal angles will both be 45^{\circ}.
Step-by-step guide: Right angle triangle
### Area of a triangle
The area of a triangle can be found by using the formula,
A=\frac{1}{2}bh ,
where b is the base length and h is the perpendicular height of the triangle.
Sometimes these values need to be calculated.
The area of a triangle is given in square units.
Step-by-step guide: Area of a triangle
### Impossible triangles
It is impossible for a triangle to contain more than one obtuse angle as the sum of angles in a triangle is 180^{\circ}.
For example,
• A triangle contains the three angles 98^{\circ}, \ 91^{\circ} and 5^{\circ}.
This gives us the angle sum of 98+91+5=194^{\circ} and so as angles in a triangle always total 180^{\circ}. This is an impossible triangle.
• A triangle contains two right angles, and a third angle x^{\circ}.
The sum of angles in the triangle is therefore 90+90+x=(180+x)^{\circ}.
The value of x must equal 0 for the angles to add to 180^{\circ} and so this is another impossible triangle.
## How to answer questions involving triangles
In order to solve problems involving triangles:
1. Locate known angles and calculate any necessary unknown angles.
2. Locate known sides and calculate any necessary unknown side lengths.
3. Solve the problem using any necessary values.
## Triangles examples
### Example 1: calculate the missing angle in an isosceles triangle
Find angle x.
1. Locate known angles and calculate any necessary unknown angles.
This problem involves the internal angles of a triangle. The sum of interior angles of a triangle is 180^{\circ}. One angle is given.
As two sides have a dash through them indicating they are the same length, the triangle is isosceles and so the triangle contains the two base angles 63^{\circ} and 63^{\circ}.
2Locate known sides and calculate any necessary unknown side lengths.
As we need to determine the size of angle x and we have information about the other two angles, we can use these to find the missing angle. We do not need to calculate any length of the triangle.
3Solve the problem using any necessary values.
Using the two known angles and the angle fact that the sum of angles in a triangle is 180^{\circ},
180-(2\times 63)=54.
Therefore angle x=54^{\circ}.
### Example 2: finding the missing angle including external angles
Determine the size of angle x.
Locate known angles and calculate any necessary unknown angles.
Locate known sides and calculate any necessary unknown side lengths.
Solve the problem using any necessary values.
### Example 3: finding the missing angle (right-angled isosceles)
Calculate the missing angle x.
Locate known angles and calculate any necessary unknown angles.
Locate known sides and calculate any necessary unknown side lengths.
Solve the problem using any necessary values.
### Example 4: calculating the perimeter
The sides of a triangle are 13 \ cm, 16 \ cm and 21 \ cm. What is the perimeter of the triangle?
Locate known angles and calculate any necessary unknown angles.
Locate known sides and calculate any necessary unknown side lengths.
Solve the problem using any necessary values.
### Example 5: perimeter problem using congruent triangles
Here is a triangle.
Two of these triangles are joined on their shortest sides to make a parallelogram. Calculate the perimeter of the parallelogram.
Locate known angles and calculate any necessary unknown angles.
Locate known sides and calculate any necessary unknown side lengths.
Solve the problem using any necessary values.
### Example 6: perimeter problem using ratio
The perimeter of a triangle is 119 \ cm.
The ratio of the sides of the triangle is 5:4:8.
Determine the length of each side of the triangle.
Locate known angles and calculate any necessary unknown angles.
Locate known sides and calculate any necessary unknown side lengths.
Solve the problem using any necessary values.
### Common misconceptions
• Angles in polygons
Make sure you know your angle properties as getting these confused causes quite a few misconceptions.
◌ Angles in a triangle total 180^{\circ} .
◌ Angles in a quadrilateral total 360^{\circ} .
• Angle facts
Make sure you know your angle properties.
◌ Adjacent angles on a straight line add up to 180^{\circ} .
◌ Vertically opposite angles are equal.
◌ Alternate angles are equal.
◌ Corresponding angles are equal.
### Practice triangle questions
1. Which of these shapes show an equilateral triangle?
An equilateral triangle has all equal sides and all equal angles.
2. The sides of a triangle are 4.3 \ cm, 4.7 \ cm, and 43 \ mm. What kind of triangle is it?
Scalene
Equilateral
Isosceles
Right-angled
43 \ mm is the same as 4.3 \ cm. Since two sides are the same, it is an isosceles triangle.
3. Find angle x.
84^{\circ}
63^{\circ}
42^{\circ}
96^{\circ}
The triangle has two equal side lengths and so it is an isosceles triangle. This means that the base angles will be equal. The interior angles of a triangle sum to 180^{\circ}. Angle x is therefore equal to
(180-96)\div 2=42^{\circ}.
Angle x=42^{\circ}
4. Determine the size of the angle x in the diagram below.
142^{\circ}
81^{\circ}
38^{\circ}
61^{\circ}
As the sum of adjacent angles is 180^{\circ}, the angle next to 142^{\circ} is equal to 180-142=38^{\circ}.
As the sum of angles in a triangle is 180^{\circ}, the unknown value of x is equal to 180-(81+38)=61^{\circ} and so x=61^{\circ}.
5. Here is a triangle. Two of these triangles are joined on the longest sides to make a quadrilateral. Find its perimeter.
12 \ cm
24 \ cm
19 \ cm
14 \ cm
The two congruent triangles will make a rectangle when they are joined on their longest side (the hypotenuse).
The perimeter of the rectangle will be
2\times (3+4)=14.
6. The perimeter of an isosceles triangle is 45 \ cm. The unequal side is half the size of one of the equal sides. Find the length of the unequal side of the isosceles triangle.
9 \ cm
15 \ cm
18 \ cm
12 \ cm
The ratio of the sides of the isosceles triangle would be 1:2:2.
The unequal side is the shortest side and can be found by
45\div (1+2+2)=45\div 5=9.
The sides of the triangles will be 9 \ cm, 18 \ cm, and 18 \ cm. The shortest side will be 9 \ cm.
### Equilateral triangle GCSE questions
1. Here is a triangle.
Draw the line(s) of symmetry of the shape.
(2 marks)
For one line of symmetry.
(1)
For three lines of symmetry.
(1)
2. Below is a diagram showing the triangle ABC. The point D lies on the line AC such that BD = CD.
(a) Determine the size of angle x.
(b) Find angle y.
(4 marks)
(a)
180-(90+70)
(1)
20^{\circ}
(1)
(b)
(180-90)\div{2}
(1)
45^{\circ}
(1)
3. (a) Write down the coordinates of point B.
(b) Plot the point (1,3) and label it C.
(c) The points A, B and C make a triangle.
Find the area of the triangle.
(4 marks)
(a) (6,9)
(1)
(b)
(1)
(c)
(6\times 5)\div 2
(1)
15 \ cm^2
(1)
4. The perimeter of a triangle ABC is 60 \ cm.
The ratio of the sides of the triangle are 2:3:3.
(a) What type of triangle is ABC ?
(b) Find the smallest side of the triangle ABC.
(c) Heron’s formula can be used to find the area of a triangle.
Heron’s formula is
\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
where s is half the perimeter and a, b and c are the three side lengths.
Use Heron’s formula along with your solution to part b) to find the area of the triangle ABC.
(8 marks)
(a) Isosceles
(1)
(b)
60\div (2+3+3)=60\div 8 \ (=7.5)
(1)
2\times 7.5=15
(1)
(c)
s=60\div 2 (=30)
(1)
The second and third sides are 22.5 \ cm.
(1)
\text{Area}=\sqrt{30(30-15)(30-22.5)(30-22.5)} (=159.0990258…)
(1)
159
(1)
cm^2
(1)
## Learning checklist
You have now learned how to:
• Recognise different types of triangles
• Solve problems involving the angles of a triangle
• Solve problems involving the perimeter of a triangle
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme.
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#### Calculate the % increase in temperature of gas, when it is heated at constant pressure to occupy 20% increase in volume.Option: 1 30Option: 2 15Option: 3 20Option: 4 25
We have:
$\\\mathrm{V_{2}\, =\,V\, +\, \frac{20V}{100}}\: =\: \frac{120V}{100}$
$T=T_2$
From Charles’ law, we know:
$\\\mathrm{\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}}}\\\\\mathrm{\therefore \: T_{2}\: =\: \frac{120V\, x\, T_{1}}{100\, x\, V}\: =\: 1.2T_{1}}$
Thus, increase in temperature $=1.2T_1-T_1=0.2T_1$
Therefore, % increase in temperature $=(0.2T_1/T_1)\times 100=20\%$
Alternate Solution:
$\begin{array}{l} \text { Vol }^m \text { Initial }=V \\ \text { Increase in\ vol }^{m}=V+V \times \frac{20}{100}=1.2V \end{array}$
$\begin{array}{l}\because \text { At constant pressure } \\ \qquad \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \end{array}$
$\mathrm{\Rightarrow \frac{V}{T_{1}}=\frac{1.2V}{T_{2}}}$
$\\\mathrm{\Rightarrow \frac{T_2}{T_{1}}=1.2} \\\\\mathrm{\Rightarrow \frac{T_2}{T_{1}}-1=1.2-1 =0.2}$
$\\\\\mathrm{\Rightarrow \frac{T_2-T_{1}}{T_1}\times 100=20\%}$
|
Question Video: Finding the Slope of the Line Representing Given Information Mathematics • 8th Grade
At an amusement park, a booth was selling soft drinks. By 2:00 pm, 37 soft drinks were sold, and by 4:00 pm, 69 were sold. Assuming that the booth was selling at a linear rate during this period, determine the slope of the line representing this information.
02:16
Video Transcript
At an amusement park, a booth was selling soft drinks. By two pm, 37 soft drinks were sold, and by four pm, 69 were sold. Assuming that the booth was selling at a linear rate during this period, determine the slope of the line representing this information.
Let’s make a sketch of the line that would represent this information. We’ll have our 𝑥-axis that represents the time and our 𝑦-axis that represents the drinks sold. At two pm, 37 drinks had sold. And at four pm, 69 drinks had sold. We’re just doing a rough sketch of this graph. So I’m just going to leave this as 37 and 69 for these two points instead of completely labelling the 𝑦-axis. This sketch let’s us see that the longer the time goes, the more the drinks sell. But to find the slope of this line, we need to see the changes in 𝑦 over the changes in 𝑥.
For us, that would be the changes in the drinks sold over the change in time. The changes in 𝑦 can be found by subtracting 69 from 37. Doing that gives us the distance, the amount between 69 and 37. And then we can subtract four from two which gives us the time that has passed between two and four. 69 minus 37 equals 32. four minus two equals two. 32 drinks sold in two hours. But we can reduce the fraction 32 over two. Both the numerator and the denominator here are divisible by two. 32 divided by two equals 16. And two divided by two equals one. 16 over one equals 16. The slope, which represents the changes in 𝑦 over the changes in 𝑥, in this case would be equal to 16.
|
Calculus Of One Real Variable By Pheng Kim Ving Chapter 7: The Exponential And Logarithmic Functions Section 7.3: General Exponential And Logarithmic Functions 7.3 General Exponential And Logarithmic Functions
1. The General Exponential Functions
Since the natural exponential function ex is differentiable, it's continuous. Its range is all y > 0. So if b is a positive number, the
Intermediate-Value Theorem assures that there exists
k such that ek = b. Thus bx = (ek)x = ekx. Now because ek = b, we have
k = ln b. Consequently, bx = e(ln b)x = ex ln b.. Note that the equation bx = ex ln b. can also be obtained as follows: as ex and ln x are
inverses of each other, we have
b = eln b, hence bx = (eln b.)x = ex ln b. We define bx to be ex ln b..
### Definition 1.1 - General Exponential Functions
For any b > 0 and for any x we define bx by: The function y = bx is called the exponential function of base b. The general exponential function refers to the exponential function y = bx where the base b can be any positive number. Recall that the natural exponential function y = ex is the exponential function of the particular base e.
Recall that the reason why we require that b > 0 is that we want bx to be defined for all x. If b = 0, bx isn't defined for x = 0
(form 0
0 isn't defined) or for any x < 0 (no fraction can have denominator 0). If b < 0, bx isn't defined as a real number at for
example
x = 1/2 (no negative number has a real square root). So bx is defined for all x only if b > 0. Another way to see this
requirement is Eq. [1.1], where we use
ln b, requiring that b > 0.
2. Properties Of The General Exponential Functions
The following properties of bx are similar to those of ex as discussed in Section 7.1 Theorem 6.1.
Theorem 2.1 – The Laws Of Exponents
For any b > 0 and for any x and y we have:
Proof
Let's do the proof for part a. Those for the remaining formulas are similar to it. We have:
bxby = ex ln bey ln b = ex ln b + y ln b = e(x+y) ln b = bx+y.
EOP
the Natural Logarithm Of The General Exponential
In Section 7.2 Theorem 2.1 Part d, we had ln xt = t ln x for any x > 0 and for any t. That's the natural logarithm of a power
function. The formula is also true for the natural logarithm of an exponential function.
Theorem 2.2 - Natural Logarithm Of General Exponential
For any b > 0 and for any x we have: ln bx = x ln b.
Proof
ln bx = ln ex ln b = x ln b.
EOP
### Example 2.1
Solve 3x = 91–x.
Solution
3
x = 91–x,
ln 3x = ln 91–x,
x ln 3 = (1 – x) ln 9,
x( ln 3 + ln 9) = ln 9,
EOS
Solving an equation containing exponential expressions like in this example can be performed by taking of the natural
logarithm of both sides of the equation.
3. Differentiation Of The General Exponential Functions
We know that (d/dx) ex = ex. In Section 7.1 Introducing e, we got (d/dx) bx = kbx, where ek = b. Since ek = b, we have k =
ln b. So (d/dx) bx = (ln b) bx = bx ln b.
Theorem 3.1 – The Exponent Rule
For any b > 0 and for any x we have: This differentiation formula is called the exponent rule.
Proof
Using the chain rule on
ex ln b we get:
EOP
Remark 3.1
Again note that (d/dx) bx is not xbx–1!! Keep in mind that bx is an exponential function, not a power function.
### Example 3.1
If y = 23x + 1, find y'.
#### Solution
y' = 23x + 1( ln 2)(3) = 3( ln 2)23x + 1.
#### EOS
4. The General Power Rule
Now let's take a look back at the power functions and extend the power rule. We saw in Section 3.2 Corollary 4.1 that
(
d/dx) xn = nxn–1 for all integer n. We saw in Section 3.3 Corollary 4.1 that (d/dx) (u(x))r = r(u(x))r–1(du/dx) for any
rational number
r, from which we have (d/dx) xr = rxr–1 for any rational number r. We now show that (d/dx) xa = axa–1
for any real number
a, rational or irrational.
Theorem 4.1 – The General Power Rule
For any constant a we have: This differentiation formula is called the general power rule.
Proof
EOP
### Remarks 4.1
5. Graphs Of General Exponential Functions
Example 5.1
Sketch a graph of y = 2x, using information obtained from the equation.
Solution
Some More Points: (-2, 1/4), (-1, 1/2), (1, 2), (2, 4).
The graph is sketched in Fig. 5.1.
Fig. 5.1 Graph Of y = 2x.
EOS
Example 5.2
Sketch a graph of y = (1/2)x, using information obtained from the equation.
Solution
Some More Points: (-2, 4), (-1, 2), (1, 1/2), (2, 1/4).
The graph is sketched in Fig. 5.2.
Fig. 5.2 Graph Of y = (1/2)x.
EOS
General Case
y = bx, b > 0.
Case 0 < b < 1. (Think of the graph of y = (1/2)x in Fig. 5.2 where b = 1/2)
Inflection Points: None.
Case b > 1. (Think of the graph of y = 2x in Fig. 5.1 where b = 2)
Inflection Points: None.
The graphs of y = 2x, y = (1/2)x, y = 10x, y = (1/10)x, y = ex, and y = 1x = 1 are sketched in Fig. 5.3. Remark that the graphs
of all exponential functions of the form
y = bx go thru the y-intercept (0, 1) as b0 = 1 for all b > 0.
# Fig. 5.3
Graphs Of Some Exponential Functions.
# Graphs of all exponential functions y = bxgo thru the y-intercept (0, 1).
6. The General Logarithmic Functions
The natural logarithm function is defined as the inverse of the natural exponential function. So it's perfectly natural to define
the general logarithmic function as the inverse of the general exponential function.
Consider y = 2x, the exponential function of base 2, as graphed in Fig. 5.1. Clearly it's one-to-one, and so has an inverse. This
inverse is called the logarithmic function of base
2 or logarithm of base 2, and denoted log 2. Thus if y = 2x then x = log 2 y.
Consequently
log 2 y = exponent on 2 (exponent to which 2 is raised) to get y. The function 1x isn't one-to-one and hence has
no inverse. Hence there's no logarithm with base 1 (1x is always 1, can't be 2 or 10 or any number other than 1). In relation
[6.1] below, we exchange the roles of the letters
x and y to conform to the standard presentation of the definition of an
inverse function.
## Definition 6.1 - General Logarithmic Functions
So logb x = exponent on b to get x. The general logarithmic function refers to the logarithmic function y = logb x, where the base b can be any real number that's positive and different from 1.
Properties
## Properties
Remarks 6.1
a. In “... the exponent on b to get x”, the quantity appearing after “to get”, x in this case, isn't the function (output), it's the
variable (input).
b. The logarithmic function might just as well be called the “ exponent function”. It isn't, because calling it so would create
confusion with the exponential function.
The Natural Logarithm Is The Logarithm Of Base e
By Definition 6.1, the inverse of the natural exponential function ex of base e is the logarithm function loge of base e. We know
that this inverse is the natural logarithm function
ln. So the natural logarithm function is the logarithm function of the particular
base
e.
## Natural Logarithm Is Logarithm Of Base e
The natural logarithm function is the logarithm function of base e: ln x = loge x.
7. Properties Of The General Logarithmic Functions
The following properties of the general logarithmic functions except the last one are similar to the corresponding properties of
the natural logarithm function as discussed in Section 7.2 Theorem 2.1.
One of the properties of the general logarithmic
functions that we're going to state and prove is
logb xy = logb x + logb y. This relation expresses the logarithm of the product
xy of 2 numbers x and y in terms of the logarithms of the original numbers x and y. This property says that the logarithm of
the product of 2 numbers equals the sum of the logarithms of the original numbers.
Theorem 7.1 - Properties Of General Logarithmic Functions
#### Proof
a. Let y = logb 1. Then by = 1 = b0. So logb 1 = y = 0. (Intuitively, the exponent on b to get 1 is 0.)
b. Let y = logb b. Then by = b = b1. So logb b = y = 1. (Intuitively, the exponent on b to get b is 1.)
c. Let p = logb x and q = logb y, so that x = bp and y = bq. Then logb xy = logb bpbq = logb bp+q = p + q = logb x + logb y.
d. logb x + logb (1/x) = logb (x(1/x)) (by part c) = logb 1 = 0. So logb (1/x) = - logb x.
e. logb (x/y) = logb (x(1/y)) = logb x + logb (1/y) (by part c) = logb x - logb y (by part d).
f. Note that if t is a positive integer then, by repetitive applications of part c:
For t = 0, we have logb x0 = logb 1 = 0 = 0 logb x.
For any non-0 real number t. Let y = logb xt, so that xt = by. Then x = by/t, then y/t = logb x, then y = t logb x. Thus
logb xt = t logb x.
Consequently, for any real number t we have logb xt = t logb x.
g. Let y = logb x. Then x = by. So logc x = logc by = y logc b (2nd equation is by part f). Thus logb x = y = (logc x)/( logc b).
EOP
Corollary 7.1 - Logarithms Of Extended Products And Quotients
For any real pi, any real qj, any positive integer m, and any positive integer n we have:
Proof
EOP
### Example 7.1
#### Solution
EOS
8. Differentiation Of The General Logarithmic Functions
## Theorem 8.1
Proof
EOP
Derivatives Of Exponentials And Logarithms
For help in the distinction and memorization of the derivatives of the exponential and logarithmic functions, we gather them
together in the following table.
Derivatives Of Exponentials And Logarithms
### Example 8.1
Differentiate:
#### Solution 2
EOS
In Solution 2 we first simplify the function expression before differentiating. This simplifies the differentiation greatly.
Solution 2 is simpler than Solution 1 and for this reason is preferable.
9. Graphs Of The General Logarithmic Functions
Graphs Of Inverses
Fig. 9.1 b > 1, graph of y = logb x is reflection of that of its inverse y = bx about line y = x.
Fig. 9.2 0 < b < 1, graph of y = logb x is reflection of that of its inverse y = bx about line y = x.
Graph Of y = ln x
Fig. 9.3 Graph Of y = ln x.
Example 9.1
Sketch a graph of y = log2 x, using information from the equation.
Solution
Inflection Points. Since there's no change of concavity, there's no inflection point.
Some More Points: (1/4, -2), (1/2, -1), (4, 2), (8, 3). (
y = exponent on 2 to get x, so choose x-values that are integer powers
of 2 such that it's easy to determine
y mentally.)
The graph of y = log2 x is sketched in Fig. 9.4.
Fig. 9.4 Graph Of y = log2 x.
EOS
Example 9.2
Sketch a graph of y = log1/2 x, using information from the equation.
Solution
Inflection Points. Since there's no change of concavity, there's no inflection point.
Some More Points: (1/4, 2), (1/2, 1), (2, -1), (4, -2), (8, -3).
The graph of y = log1/2 x is sketched in Fig. 9.5.
Fig. 9.5 Graph Of y = log1/2 x.
EOS
General Case
Inflection Points. Since there's no change of concavity, there's no inflection point.
Case b > 1. (Think of the graph of y = ln x in Fig. 9.3 where b = e > 1 or the graph of y = log2 x in Fig. 9.4 where b = 2)
Inflection Points. Since there's no change of concavity, there's no inflection point.
Graphs
The graphs of some logarithmic functions are sketched in Fig. 9.6. Remark that the graphs of all the basic logarithmic
functions go thru the
x-intercept (1, 0) ( y = log2 x is a basic logarithmic function, while y = 3 log2 x - 4 isn't; it's a
transformed logarithmic function). A way to help memorize the shape of the graph is the graphs of
y = ln x = loge x and
y = log1/2 x: since e > 1, the graph of y = logb x where b > 1 has a similar shape as that of y = ln x; since 0 < 1/2 < 1,
the graph of
y = logb x where 0 < b < 1 has a similar shape as that of y = log1/2 x. So memorize the graphs of y = ln x
and
y = log1/2 x.
Fig. 9.6 Graphs Of Some Logarithmic Functions. Graphs of all logarithmic functions y = logb x go thru x-intercept (1, 0).
10. The Common Logarithm
The logarithm of base 10, log10 x, is called the common logarithm, and usually is denoted log x, without writing the base 10.
It's called so because it's commonly used in applications of mathematics. The exponential of base 10, 10
x, which is the inverse
of the common logarithm, is similarly called the common antilogarithm.
Consider the common logarithm y = log x. As shown in Fig. 10.1, when x increases for example from 1 to 1,000,000,
log x increases only from 0 to 6. So log x increases slowly. Also see the graph of log10 x in Fig. 9.5. When x increases
from 1 to 10 thus by 10 – 1 = 9 units,
log x increases from 0 to 1 hence by 1 – 0 = 1 unit; when x increases from 10 to
100 thus by 100 – 10 = 90 units,
log x increases from 1 to 2 hence by 2 – 1 = 1 unit, still by 1 unit; when x increases from
100,000 to 1,000,000 thus by 1,000,000 – 100,000 = 900,000 units,
log x increases from 5 to 6 hence by 6 – 5 = 1 unit, still by
1 unit. The larger
x is, the slower log x increases. This can also be seen by examining the graph of log10 x in Fig. 9.5 and the
derivative (rate of change) 1/(
x ln 10) of log x. Clearly the larger x > 0 is, the smaller (down towards 0) 1/(x ln 10) becomes.
As
x > 0 increases, log x increases slowly and at a decreasing rate. (The function y = (0.000,001)x increases slowly but at a
constant rate of
dy/dx = 0.000,001; its graph is a straight line; that of y = log x is concave down).
# Fig. 10.1
Common logarithm
increases slowly and at
a decreasing rate.
Problems & Solutions
1. Simplify the following expressions.
Solution
2. Solve the equation 52x32x–1 = 15x–2 for x.
Solution
52x32x–1 = 15x–2,
ln 52x32x–1 = ln 15x–2,
ln 52x + ln 32x–1 = (x – 2) ln 15,
2x ln 5 + (2x – 1) ln 3 = x ln 15 – 2 ln 15,
x(2 ln 5 + 2 ln 3 – ln 15) = – 2 ln 15 + ln 3,
3. Differentiate the following functions.
a. f (x) = txxt.
b.
g(t) = txxt.
c.
y = loga (bx + c).
Solution
a. f '(x) = tx ln ttxt–1.
b. g'(t) = xtx–1xt ln x.
4. Consider the function y = (2x)3x. The variable appears in both the base and the exponent. So the power rule and exponent
rule of differentiation don't apply to it directly.
a. Differentiate it by using the definition of the general exponential function.
b. Differentiate it by first taking the natural logarithm of both sides and then using implicit differentiation.
c. Are the answers in parts a and b the same?
Solution
a. y = (2x)3x = e3x ln 2x,
y ' = e3x ln 2x (3 ln 2x + 3x(2/2x)) = (2x)3x (3 ln 2x + 3) = 3(2x)3x (ln 2x + 1).
b. ln y = ln (2x)3x = 3x ln 2x,
differentiate first and third sides implicitly:
y '/y = 3 ln 2x + 3x(2/2x) = 3 ln 2x + 3 = 3( ln 2x + 1),
y ' = y(3( ln 2x + 1)) = 3(2x)3x (ln 2x + 1).
c. Yes.
Note
Of course in practice just use one method, unless you're asked to use both methods.
5. Prove that for any a > 0, any b > 0, and any c > 0, loga b = loga c logc b.
Solution
loga b = loga c logc b.
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# Special Right Triangles
## Presentation on theme: "Special Right Triangles"— Presentation transcript:
Special Right Triangles
Lesson 7-3 Special Right Triangles Lesson 7-3: Special Right Triangles
45°-45°-90° Special Right Triangle
In a triangle 45°-45°-90° , the hypotenuse is times as long as a leg. Example: 45° 45° cm Hypotenuse 5 cm Leg X X 45° 5 cm 45° Leg X Lesson 7-3: Special Right Triangles
30°-60°-90° Special Right Triangle
In a triangle 30°-60°-90° , the hypotenuse is twice as long as the shorter leg, and the longer leg is times as long as the shorter leg. Example: Hypotenuse 30° 2X Longer Leg 30° 10 cm X cm 60° 60° X Shorter Leg 5 cm Lesson 7-3: Special Right Triangles
Example: Find the value of a and b.
b = 14 cm 60° 7 cm 30° 2x b 30 ° 60° a = cm a x Step 1: Find the missing angle measure. 30° Step 2: Decide which special right triangle applies. 30°-60°-90° Step 3: Match the 30°-60°-90° pattern with the problem. Step 4: From the pattern, we know that x = 7 , b = 2x, and a = x . Step 5: Solve for a and b Lesson 7-3: Special Right Triangles
Example: Find the value of a and b.
b = cm 45° 7 cm 45° x b x 45 ° 45° a = 7 cm a x Step 1: Find the missing angle measure. 45° Step 2: Decide which special right triangle applies. 45°-45°-90° Step 3: Match the 45°-45°-90° pattern with the problem. Step 4: From the pattern, we know that x = 7 , a = x, and b = x . Step 5: Solve for a and b Lesson 7-3: Special Right Triangles
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## Arithmetic (all content)
### Course: Arithmetic (all content)>Unit 5
Lesson 2: What fractions mean
# Identifying numerators and denominators
We'll explore a video about identifying numerators and denominators in fractions. Together, we'll learn that the numerator is the top number, and the denominator is the bottom number. We'll also use pie charts to visually represent fractions and connect this concept to real-life situations for our students. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• What is a denominator and why is it called that?
• you could think of it as being 2 neghbors, numerator and denominator, denominator living on floor 1 and numerator on floor 2!
• Wait...So can you reduce a fraction?
• yes why not you can do cancllation and reduce a fration
• Can someone give me an in-depth explanation of what are numerators and denominators? Give me a definitive definition of a fraction!
I am confused...
• The fraction has two parts, the numerator and the denominator. The numerator is at the top, and the denominator is at the bottom. The numerator tells how many equal parts they’re showing or shading and the denominator tells how many equal parts are in a whole.
• the denominator is the number at the bottom right?
• yes denominator at the bottom and numerator at the top
• So.....can you make the Denominator smaller ?+_+
• As a more in-depth answer, you can, but only if its divisible by two or a larger number, but ONLY if the numerator is divisible by the SAME number the denominator is.
• Why is the bottom of the fraction called a denominator?
• In Latin:
"denominator" = that which names
The denominator of a fraction tells you the name of the part (one half, one third, one six thousand three hundred seventy first, etc.)
The numerator tells you how many of those parts there are.
• I get lost between the difference of "The number of equal parts in all the wholes" and "The number of equal parts in one whole" in the practice question. Can someone explain it to me? thanks!
• I just looked at a couple of the practice problems. The denominator = the number of equal parts that make one whole unit. The numerator is the number of parts you are counting. The option for "the number of equal parts in all the wholes" appears to be a false option to see if you understand the meaning for numerator/denominator. It really doesn't fit the definition of either one.
• Why is a numerator called a numerator and why is it o n the top?
• It's just the name it's been given same for Denominator although Denominator seems to also be called the Dominant number because it's the Total of the Parts/Pieces/Objects and the Whole Shape, but i wouldn't think too much about the names, lots of things are given names just to help identify them.
We don't know why Numerator is on top either it just is, i guess it does make more sense to be the top number since that's the number you're taking from the Total Parts/Pieces, that's where our mind would go first it would be like "So i'm taking or adding this From/To this" that's my guess :)
• I've been trying to understand these questions, I continue to level down on being able to recognize fractions. I'm not getting it.can someone break this down for me please.
• When a figure is divided into equal parts, the fraction that is represented by the shaded area is the number of parts shaded over the total number of parts.
Example: suppose a figure is divided into 7 equal parts. Suppose 4 of the parts are shaded (so 3 of the parts are unshaded). Then the shaded area would represent the fraction 4/7.
Have a blessed, wonderful day!
|
# University of Florida/Egm4313/s12 Report 4, Problem 4.3
## Problem 4.3
### Problem 4.3 Part 2
#### Problem Statement
Given: ${\displaystyle y''-3y'+2y=r(x)\!}$
where ${\displaystyle r(x)=\log(1+x)\!}$
With initial conditions: ${\displaystyle y(-{\frac {3}{4}})=1,y'(-{\frac {3}{4}})=0\!}$
Find the overall solution ${\displaystyle y_{n}(x)\!}$ for ${\displaystyle n=4,7,11\!}$ and plot these solutions on the interval from ${\displaystyle [-{\frac {3}{4}},3]\!}$
#### Solution
First we find the homogeneous solution to the ODE:
The characteristic equation is:
${\displaystyle \lambda ^{2}-3\lambda +2=0\!}$
${\displaystyle (\lambda -2)(\lambda -1)=0\!}$
Then, ${\displaystyle \lambda =1,2\!}$
Therefore the homogeneous solution is:
${\displaystyle y_{h}=C_{1}e^{(}2x)+c_{2}e^{x}\!}$
Now to find the particulate solution
For n=4
${\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}$
${\displaystyle r(x)={\frac {x}{\ln(10)}}-{\frac {x^{2}}{2\ln(10)}}+{\frac {x^{3}}{3\ln(10)}}-{\frac {x^{4}}{4\ln(10)}}\!}$
We can then use a matrix to organize the known coefficients:
${\displaystyle {\begin{bmatrix}2&-3&2&0&0\\0&2&-6&6&0\\0&0&2&-9&12\\0&0&0&2&-12\\0&0&0&0&2\\\end{bmatrix}}{\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\\K_{4}\end{bmatrix}}={\begin{bmatrix}0\\{\frac {1}{ln(10)}}\\{\frac {1}{2ln(10)}}\\{\frac {1}{3ln(10)}}\\{\frac {1}{4ln(10)}}\end{bmatrix}}\!}$
Then, using MATLAB and the backlash operator we can solve for these unknowns:
Therefore
${\displaystyle y_{p4}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}\!}$
Superposing the homogeneous and particulate solution we get
${\displaystyle y_{n}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}+C_{1}e^{2x}+C_{2}e^{x}\!}$
Differentiating:
${\displaystyle y'_{n}=3.7458+3.1486x+1.1943x^{2}+0.2172x^{3}+2C_{1}e^{2x}+C_{2}e^{x}\!}$ Evaluating at the initial conditions:
${\displaystyle y(-0.75)=0.9698261719+0.231301601C_{!}+0.4723665527C_{2}=1\!}$
${\displaystyle y'(-0.75)=1.9645125+0.4462603203C_{1}+0.4723665527C_{2}\!}$
We obtain:
${\displaystyle C_{1}=-4.46\!}$
${\displaystyle C_{2}=0.055\!}$
Finally we have:
${\displaystyle y_{n}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}-4.46e^{2x}+0.055e^{x}\!}$
For n=7
${\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}$
${\displaystyle r(x)={\frac {x}{\ln(10)}}-{\frac {x^{2}}{2\ln(10)}}+{\frac {x^{3}}{3\ln(10)}}-{\frac {x^{4}}{4\ln(10)}}+{\frac {x^{5}}{5\ln(10)}}-{\frac {x^{6}}{6\ln(10)}}+{\frac {x^{7}}{7\ln(10)}}\!}$
We can then use a matrix to organize the known coefficients:
${\displaystyle {\begin{bmatrix}2&-3&2&0&0&0&0&0\\0&2&-6&6&0&0&0&0\\0&0&2&-9&12&0&0&0\\0&0&0&2&-12&20&0&0\\0&0&0&0&2&-15&30&0\\0&0&0&0&0&2&-18&42\\0&0&0&0&0&0&2&-21\\0&0&0&0&0&0&0&2\\\end{bmatrix}}{\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\\K_{4}\\K_{5}\\K_{6}\\K_{7}\end{bmatrix}}={\begin{bmatrix}0\\{\frac {1}{ln(10)}}\\{\frac {1}{2ln(10)}}\\{\frac {1}{3ln(10)}}\\{\frac {1}{4ln(10)}}\\{\frac {1}{5ln(10)}}\\{\frac {1}{6ln(10)}}\\{\frac {1}{7ln(10)}}\end{bmatrix}}\!}$
Then, using MATLAB and the backlash operator we can solve for these unknowns:
Therefore
${\displaystyle y_{p7}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}\!}$
Superposing the homogeneous and particulate solution we get
${\displaystyle y_{n}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}+C_{1}e^{(}2x)+c_{2}e^{x}\!}$
Differentiating:
${\displaystyle y'_{n}=375.3933+371.213x+181.644x^{2}+57.9784x^{3}+13.46x^{4}+2.1714x^{5}+0.214x^{6}+2C_{1}e^{(}2x)+C_{2}e^{x}\!}$ Evaluating at the initial conditions:
${\displaystyle y(-0.75)=178.816+0.2231301601C_{!}+0.4723665527C_{2}=1\!}$
${\displaystyle y'(-0.75)=178.413+0.4462603203C_{1}+0.4723665527C_{2}\!}$
We obtain:
${\displaystyle C_{1}=-2.6757\!}$
${\displaystyle C_{2}=-375.173\!}$
|
# Practice Problems for Finding Binomial Probabilities Using Formulas Video
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Lesson Transcript
Instructor: Cathryn Jackson
Cat has taught a variety of subjects, including communications, mathematics, and technology. Cat has a master's degree in education and is currently working on her Ph.D.
In this lesson, you will review the binomial probability formula and the combination formula. Then follow along with the practice problems and see if you've mastered these concepts!
## Understanding Binomial Probabilities
Alex and Jon are at their favorite arcade. Jon wants to make a bet: \$20 that he will win three out of the next five games they play. Assuming that they are evenly matched and that there is a 50% chance that either one of them could win, what is the probability that Jon will win the bet?
To solve this problem you need to understand how to use a binomial probability formula. Since this is a practice problem lesson, let's quickly review how to solve a problem using Jon's bet as an example.
First, we need to find the values of x, n and P. The x represents the number of successes, the n represents the number of trials and the P represents the probability of success on an individual trial. In our case, x represents the number of games that Jon wins. Jon bets he can win three out of five games, therefore x = 3. The n represents the number of games Jon and Alex will play, therefore n = 5. Finally, The P represents the probability of an individual trial. Jon and Alex are evenly matched, therefore P = .50
When we plug our numbers into the formula it should look like this. But don't forget about C! The formula for C looks like this. Remember to use a graphing calculator to find the factorials in this formula. If you have trouble, make sure you break it down and check that you are using order of operations properly.
The combination is 10 for this particular problem. Now that we know this number, let's insert it into our binomial probability formula.
Okay, so for this problem, I started out by subtracting the exponents of 5 and 3 to get 2, which means Jon will have to lose two games and win three in this scenario. I also inserted our combination number, 10, into the equation. Next, on the fourth row, I subtracted .50 from 1. This gets us the probability of failing a single trial (or losing a game), and since Jon and Alex are equally matched, the probability is still .5. In the fifth row, I calculated .5 to the third power, and in the sixth, I calculated .5 to the second power. Lastly, I multiplied from left to right, which led me to the answer .3125. This means that Jon has a 31% chance that he will win only three out of the five games, no more and no less.
Now that we've reviewed how to use the binomial probability formula, let's look at some other practice problems.
## Practice Problem 1
The first game Alex and Jon decide to play is a simple basketball game. A hoop is placed in the machine, and basketballs are dispensed to the players. The person to make the most baskets wins. Alex and Jon are once again evenly matched for this game. They decide to each shoot ten baskets. Jon will need to make six baskets to win the game. What is the probability that Jon will make only six baskets, no more and no less? Pause the video here to find the answer.
How did you do? The correct answer is .2050, or approximately 21%. Let's break down this problem.
First, you need to calculate the combination. I got 210. Now let's plug this number into our binomial probability formula.
Once again I insert the combination number, and I subtract the exponents 10 - 6. Next I find the probability of failure, which is .5. In the fifth row, I take .50 to the sixth power, which gets me .015625. Next, I calculate .5 to the fourth power, which is .0625. Last, I multiply from left to right and end up with .2050, which is approximately 21%. Let's try another problem!
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GED Mathematical Reasoning: Understanding Simple Interest | Open Window Learning
# GED Mathematical Reasoning: Understanding Simple Interest
• Interest is a fee for using someone else’s money. When you invest money, the bank pays you a fee for allowing them to use your money. When you borrow money, you pay the lender a fee for the use of their money. In both cases, this fee is called: interest.
• The amount of money invested or borrowed is called the principal. Simple interest is a special kind of interest based only on the amount invested or borrowed, an interest rate, the time frame (in years).
• To calculate simple interest, we use the following formula:
I = interest
p = principal
r = rate (as a decimal)
t = time (in years)
Example 1
Robert invests $5,000 for 20 years at 3% interest. Using the simple interest formula, how much interest will he earn? First let’s make a note of what it is we’re being asked to find. In this case, we’re being asked to find the amount of interest Robert will earn on his investment. In terms of the formula, we need to find “I,” the interest. Next, let’s identify the information we’re given. It’s a good idea to jot this down on your paper. We’re told that the amount of money invested is$5,000, the time frame is 20 years and the interest rate is 3%. When using the simple interest formula, remember to turn the percent into a decimal. To do so, move the decimal point two places to the left and drop the percent sign. You may need to add zeros as placeholders.
So we know p equals $5,000, t equals 20 and r equals 3%, which is 0.03 written as a decimal. Since we know the values for principal, time, and rate it makes sense that we would apply the simple interest formula to find the interest earned as the example directs us to do. So the interest Robert will earn investing$5,000 over 20 years at a 3% simple interest rate is $3,000 Does the answer make sense? Well, 3% of$5,000 is $150, which we can find by multiplying 0.03 by$5,000 since the word “of” implies multiplication. And $150 per year for 20 years results in a total amount of$3,000. So yes, our answer makes perfect sense!
Example 2
Dexter borrows $1500 from his grandfather for 6 months at 2.5% simple interest. What is the total amount Dexter will pay back when the loan is due? In example one, money was being invested. In this example money is being borrowed, however, to calculate the simple interest involves the same process either way. When we read example two, we are told that the interest on this loan is being paid according to the simple interest calculation, but notice what we’re being asked to find. It’s not the interest. We need to find the total amount of money Dexter will repay. First, let’s make a note of the information we’re given. We’re told that the amount of money borrowed is$1500, the time frame is 6 months and the interest rate is 2.5%. So we know p equals $1,500 and r equals 2.5%, which – as a decimal – is 0.025. Let’s talk for a moment about the time frame, t. In the simple interest formula, the time frame needs to be in terms of years. Here, the time frame is given in terms of months. So how do we convert 6 months to years? Since there are twelve months in one year, we will convert 6 months to years by writing: 6 over 12. 6 months can be written as: one-half years. So, t equals 0.5 The interest rate is given in the example as a simple interest rate and we know the values for principal, rate, and time, so it makes sense that we would apply the simple interest formula to find the interest that Dexter will pay as a fee for borrowing the money. However, we’re not being asked to find the amount of interest. We’re being asked to find the total amount Dexter will repay, which consists of the amount of money borrowed PLUS the interest, which is the fee he will pay for borrowing the money. Therefore, we need to find the amount of interest first. And then add the interest to the loan amount, making this a two-step problem. That sum will equal the total amount of money Dexter will repay to his grandfather. Okay! Let’s get started. First, we’ll find the amount of interest by applying the simple interest formula. = \$18.75[/latex]
So the interest on the loan is: $18.75. But we’re not done yet because we haven’t answered the question. We must finish by accomplishing step 2 of our plan to find the total amount of money Dexter will repay to his grandfather. Dexter will repay the original loan amount of$1,500 plus interest in the amount of $18.75. Total Amount Repaid = Original Loan Amount + Interest Total Amount Repaid =$1500 + $18.75 =$1518.75
So the total amount of money Dexter will repay his grandfather is: $1,518.75. What do you think? Does this answer make sense?$1,500 is not a huge loan amount, the interest rate is relatively low, and the time frame of the loan is very short – not even one year! So it makes sense that the interest amount would not be a very large number. Therefore, yes – our answer makes sense.
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In the second trimester of the school year, fifth grade students will work on the following skills:
• Writing expressions and equations from a given context or word problems
• Solving equations including those with a letter as a variable
• Analyzing the pattern of numbers in a table and being able to extend the table
• Understanding and applying the order of operations
• Adding and subtracting fractions and mixed numbers Multiplying fractions
• Dividing a whole number by a fraction, and dividing a fraction by a whole number
• Understanding and finding the volume of a cube, rectangular prism, as well as a 3-D solid comprised of rectangular prisms
You can help your child at home by having them practice these skills. Students should know their basic math facts from memory by now or have efficient strategies that make problem solving easy.
While working with fractions and volume of 3-D solids, students benefit from drawing a visual representation of the problem. For many students they struggle with multiplication or division of fractions because they have a misconception that when you multiply the products is always larger than the two factors, and that when you divide the quotient is always smaller than the dividend. This is not always the case when working with fractions and mixed numbers.
For fractions, students can represent the multiplication and division of fractions using the area model they have used in multiplication and division of whole numbers and decimals. This Khan Academy video shows an example of this. https://www.khanacademy.org/math/arithmetic/fraction-arithmetic/arith-review-multiply-fractions/v/visualizing-fraction-products
Students can gain understanding when you present a problem that relates to a real world situation such as “there is half of a cake/pie/pizza left over. Tonight I’m going to eat ¼ of what is left. How much cake/pie/pizza did I eat?” which is a representation of ¼ of ½ or ¼ x ½.
Division can be thought of the same way. For example a real world situation could be thought of as “One serving of mac and cheese is ½ of a cup. If the box makes 2 cups, how many servings are in the box?” This can be represented by 2 ÷ ½ or how many halves are in 2?
For volume, you could have students measure the length, width and height of an empty box and have them calculate the volume. Understanding that we use the label cubic units for volume because it is a 3-D shape with length, width and height is another thing to talk about with your student.
Here are some other resources:
Up Next: Measurement conversions, geometry and data
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# 4.1 Exponential functions
Page 1 / 16
In this section, you will:
• Evaluate exponential functions.
• Find the equation of an exponential function.
• Use compound interest formulas.
• Evaluate exponential functions with base e.
India is the second most populous country in the world with a population of about $\text{\hspace{0.17em}}1.25\text{\hspace{0.17em}}$ billion people in 2013. The population is growing at a rate of about $\text{\hspace{0.17em}}1.2%\text{\hspace{0.17em}}$ each year http://www.worldometers.info/world-population/. Accessed February 24, 2014. . If this rate continues, the population of India will exceed China’s population by the year $\text{\hspace{0.17em}}2031.$ When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions , which model this kind of rapid growth.
## Identifying exponential functions
When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for each unit increase in input. For example, in the equation $\text{\hspace{0.17em}}f\left(x\right)=3x+4,$ the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people.
## Defining an exponential function
A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2050.
What exactly does it mean to grow exponentially ? What does the word double have in common with percent increase ? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media.
• Percent change refers to a change based on a percent of the original amount.
• Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time.
• Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time.
For us to gain a clear understanding of exponential growth , let us contrast exponential growth with linear growth . We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See [link] .
$x$ $f\left(x\right)={2}^{x}$ $g\left(x\right)=2x$
0 1 0
1 2 2
2 4 4
3 8 6
4 16 8
5 32 10
6 64 12
From [link] we can infer that for these two functions, exponential growth dwarfs linear growth.
• Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain.
• Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain.
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
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The Discriminant And Three Cases
Using The Quadratic Formula Through Examples
The quadratic formula is a useful formula for solving x-intercepts of quadratic equations in the form of
The quadratic formula (with ) is:
It is preferable to use the quadratic formula when factoring techniques do not work.
For many (highschool) students, it is not expected to know how to come up with the quadratic formula. This proof is more of a reference to help you see where this formula comes from.
The coefficient that is with is . This is divided by 2 gives us . The square of yields or . Adding and subtracting by gives us:
We then factor the first three terms. That is, we factor into .
The distributive law is applied such that is expanded to the first two terms.
Next, we simplify the term.
Because we want to find x-intercepts, we set and solve for .
The final line is the quadratic formula or the value such that it makes in .
The Discriminant And Three Cases
Notice how in the quadratic formula there is a square root part after the plus and minus sign (). The part inside the square root () is called the discriminant.
An important property of square roots is that square roots take on numbers which are at least 0 (non-negative). A negative number inside the square root is undefined (in the real numbers).
We look at three cases for the discriminant and what each case means.
If then there would be 2 distinct solutions for (or x-intercepts) in the equation .\
If then there would be one value for in the equation .
If , we would have a negative value inside the square root. The square root of a negative value is undefined. There would be no real-numbered values for in the equation .
Using The Quadratic Formula Through Examples
The quadratic formula can be applied to any quadratic equation in the form . It does not really matter whether the quadratic form can be factored or not.
Example One
Given the quadratic equation , what are the x-intercepts?
From , we have , and . Using these values, the quadratic formula is as follows:
The x-intercepts (when ) for are (or 0.618034) and (or -1.618034). You can choose to use exact values or using the decimal of the answers (with a calculator).
Example Two
Apply the quadratic formula to find x-intercepts for the equation .
Here, we have , and .
The x-intercepts are (or 1.780776) and (or -0.2807764).
Example Three
Show that the equation has no x-intercepts.
You can use the quadratic formula right away here. However, it is a bit easier and faster to check the discriminant .
Since the discriminant is negative, we have a negative inside the square root of the quadratic formula. The equation has no x-intercepts.
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A tangential quadrilateral is a quadrilateral whose four sides are all tangent to a circle inscribed within it. In such a quadrilateral, the sum of lengths of the two opposite sides of the quadrilateral is equal.
This is known as the Pitot theorem, named after Henri Pitot, a French engineer who proved it in the 18th century.
## Problem
A circle is inscribed in a quadrilateral ABCD. Show that |AB|+|CD| = |BC|+|DA|
## Strategy
We are working with a tangential quadrilateral, so all the sides are tangent to the circle. Let’s review the properties of tangents to a circle.
One of the theorems we have proved is the two tangent theorem – two tangents from the same point outside a circle have equal lengths to the points of tangency. This should be useful here, as it deals with the lengths of such tangents.
Looking at the quadrilateral, we have four such points outside the circle. Each one of the quadrilateral’s vertices is a point from which we drew two tangents to the circle.
Applying the two tangent theorem to each one of these points, we get 4 pairs of equal-size line segments: AP=AS=a; BP=BQ=b; CQ=CR=c and DR=DS=d.
And in each of the opposite pairs of sides of the quadrilateral, we have exactly one of these segments, so their sums are equal. And we have proven the Pitot theorem for a circle inscribed in a quadrilateral.
## Proof
(1) AP=AS=a // Two tangent theorem
(2) BP=BQ=b // Two tangent theorem
(3) CQ=CR=c // Two tangent theorem
(4) DR=DS=d // Two tangent theorem
(5) |AB|+|CD| = |AP|+|PB|+|CR|+|RD|= a+b+c+d
(6) |BC|+|DA| = |BQ|+|QC|+|DS|+|SA|=b+c+d+a=a+b+c+d // Commutative property of addition
(7) |AB|+|CD| = |BC|+|DA| //(5),(6), transitive property of equality
You are watching: Circle Inscribed in a Quadrilateral. Info created by GBee English Center selection and synthesis along with other related topics.
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# 2002 AMC 10B Problems/Problem 24
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem 24
Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?
$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15$
## Solution
$[asy] unitsize(1.5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair O=(0,0), A=(0,-20), B=(0,-10), C=(10sqrt(3),-10); real r=20; path ferriswheel=Circle(O,r); draw(ferriswheel); draw(O--A); draw(O--C); draw(B--C); pair[] ps={A,B,C,O}; dot(ps); label("O",O,N); label("A",A,S); label("B",B,W); label("C",C,SE); label("10",(O--B),W); label("10",(A--B),W); label("20",(O--C),NE); [/asy]$
We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\triangle OBC$ is a $30-60-90$ triangle. That means $\angle OBC = 60^\circ$ and the ferris wheel has made $\frac{60}{360} = \frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $\frac{1}{6}\text{th}$ of a minute, or $10$ seconds. The answer is $\boxed{\mathrm{(D) \ } 10}$
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Fun with Fractions
Fun with Fractions
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Fun with Fractions
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Presentation Transcript
1. Fun with Fractions cummins school dist Adding and Subtracting Fractions
2. Introduction • Fractions are not scary, and can even be fun. • Like most things, people are afraid of fractions because they don’t understand them. • Today we are going to get to know fractions and make friends with them. Adding and Subtracting Fractions
3. Objectives • In this presentation, you will learn the following things: • What a fraction is • What a fraction means • How to add fractions • How to subtract fractions Adding and Subtracting Fractions
4. What is a fraction? • A fraction is made up of two numbers, one on top of the other, divided by a line. • The top number is called the numerator. • The bottom number is called the denominator. • Knowing the names of these numbers is not nearly as important as knowing how to use them. Adding and Subtracting Fractions
5. What does a fraction mean? • A fraction means that something, usually an object or substance, is divided into equal parts. • The bottom number tells us how many equal parts it’s divided into. • The top number tells us how many parts we’re considering. • For instance……. Adding and Subtracting Fractions
6. What does a fraction mean? ½ As the fraction shows, the circle is divided into two equal parts. We are concerned with one part: the blue part. Adding and Subtracting Fractions
7. What does a fraction mean? 1/3 This fraction shows the circle divided into three equal parts. Again, we are concerned with one part, the blue part Adding and Subtracting Fractions
8. What does a fraction mean? This circle is divided into four parts. If we are considering only the blue part, we write ¼. If we are considering both the red and blue parts, we write 2/4, which is the same as ½. Adding and Subtracting Fractions
9. Adding Fractions • To add fractions, your bottom numbers must match. • If you must choose a new bottom number, it must be divisible by both of the old bottom numbers • Once the bottom numbers are the same, you simply add the top numbers • For instance………. Adding and Subtracting Fractions
10. Adding Fractions ½ + ½ =2/2=1 ½ + 1/3 = 3/6 + 2/6 = 5/6 Notice, we used 6 for our bottom number, or common denominator. Both 2 and 3 will divide into 6 an even number of times. Also, whatever number we use to multiply our bottom number, we must also multiply times the top number. In other words, we multiply the 2 in ½ times 3 to get 6, so we must also multiply the 1 times 3 to get 3/6. Adding and Subtracting Fractions
11. Subtracting Fractions • Subtracting fractions works just like adding fractions. • You must make the bottom numbers the same (common denominator). • When the bottom numbers are the same, you simply subtract the second top number from the first top number. Adding and Subtracting Fractions
12. Subtracting fractions So: ¾ - ¼ = 2/4 = ½ 3/8 – ¼ = 3/8 – 2/8 = 1/8 Notice, we had to make our bottom numbers the same (find a common denominator) in the second problem. We used 8 and multiplied 4 times 2 and 1 times 2 to convert ¼ to 2/8. Adding and Subtracting Fractions
13. How Fractions Are Used in Real Life • Fractions are frequently used in any situation where measuring is done. • Cooking frequently requires adding fractions such as ½ cup and 1/3 cup. • Building often requires the addition or subtraction of fractions of inches or feet. Adding and Subtracting Fractions
14. Summary • We have learned that fractions are made of one number (numerator) on top of another number (denominator) separated by a line • The bottom number tells us how many equal parts our object is divided into. • The top number tells us how many of those parts we are concerned with. Adding and Subtracting Fractions
15. Summary • To add and subtract fractions, you must make the bottom numbers match (find a common denominator). • When your bottom numbers match, you simply add or subtract the top numbers. • For practice, work the problems located at the end of the adding and subtracting fractions sections of your workbooks Adding and Subtracting Fractions
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## The Definition of a Derivative
Read this section to understand the definition of a derivative. Work through practice problems 1-8.
A Useful Formula: $D( x^n )$
Functions which include powers of $x$ are very common (every polynomial is a sum of terms which include powers of $x$), and, fortunately, it is easy to calculate the derivatives of such powers. The "pattern" emerging from the first few examples in this section is, in fact, true for all powers of $x$. We will only state and prove the "pattern" here for positive integer powers of $x$, but it is also true for other powers as we will prove later.
Theorem: If $n$ is a positive integer, then $D( x^n ) = n.x^{n–1}$
This theorem is an example of the power of generality and proof in mathematics. Rather than resorting to the definition when we encounter a new power of $x$ (imagine using the definition to calculate the derivative of $x^{307}$ ), we can justify the pattern for all positive integer exponents $n$, and then simply apply the result for whatever exponent we have. We know, from the first examples in this section, that the theorem is true for $n= 1, 2$ and $3$, but no number of examples would guarantee that the pattern is true for all exponents. We need a proof that what we think is true really is true.
Proof of the theorem: Since $f(x) = x^n$, then $f(x+h) = (x+h)^n$, and in order to simplify $f(x+h) – f(x) = (x+h)^n – x^n$, we will need to expand $(x+h)^n$. However, we really only need to know the first two terms of the expansion and to know that all of the other terms of the expansion contain a power of $h$ of at least 2. The Binomial Theorem from algebra says (for $n > 3$) that $(x+h)^n = x^n + n.^{xn–1}h + a.x^{n–2}h2 + b.x^{n–3}h3 +... + h^n$ where $a$ and $b$ represent numerical coefficients.
(Expand $(x+h)^n$ for at least a few different values of n to convince yourself of this result.)
Then $\mathbf{D}(\mathrm{f}(\mathrm{x})) \equiv \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{(x+h)^{n}-x^{n}}{h} \quad$ then expand to get
$=\lim\limits_{h \rightarrow 0} \frac{\left\{x^{n}+n \cdot x^{n-1} h+a \cdot x^{n-2} h^{2}+b \cdot x^{n-3} h^{3}+\ldots+h\right.}{h}$ eliminate $x^n – x^n$
$=\lim\limits_{h \rightarrow 0} \frac{\left\{n \cdot x^{n-1} h+a \cdot x^{n-2} h^{2}+b \cdot x^{n-3} h^{3}+\ldots+h^{n}\right\}}{h}$ factor $h$ out of the numerator
$=\lim\limits_{h \rightarrow 0} \frac{h \cdot\left\{n \cdot x^{n-1}+a \cdot x^{n-2} h+b \cdot x^{n-3} h^{2}+\ldots+h^{n-1}\right\}}{h}$ divide by the factor $h$
$=\lim\limits_{h \rightarrow 0}\left\{n \cdot x^{n-1}+a \cdot x^{n-2} h+b \cdot x^{n-3} h^{2}+\ldots+h^{n-1}\right\}$ separate the limits
$=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}+\lim\limits_{h \rightarrow 0}\left\{\mathrm{a} \cdot \mathrm{x}^{\mathrm{n}-2} \mathrm{~h}+\mathrm{b} \cdot \mathrm{x}^{\mathrm{n}-3} \mathbf{h}^{2}+\ldots+\mathbf{h}^{\mathbf{n}-1}\right\} \quad$ each term has a factor of $\mathrm{h}$, and $\mathrm{h} \rightarrow 0$
$=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}+\mathbf{0}=\mathbf{n} \cdot \mathbf{x}^{\mathbf{n}-\mathbf{1}}$ so $\mathbf{D}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathbf{n} \cdot \mathbf{x}^{\mathbf{n}-\mathbf{1}}$
Practice 7: Use the theorem to calculate $D( x5 )$, $d ( x2 )$, $D( x100 )$, $d ( t31 )$, and $D( x0 )$
We will occasionally use the result of the theorem for the derivatives of all constant powers of $x$ even though it has only been proven for positive integer powers, so far. The result for all constant powers of $x$ is proved in Section 2.9
Example 5: Find $\mathbf{D}(1 / \mathrm{x})$ and $\frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})$.
Solution: $\quad \mathbf{D}\left(\frac{1}{\mathrm{x}}\right)=\mathbf{D}\left(\mathrm{x}^{-1}\right)=-1 \mathrm{x}^{(-1)-1}=-1 \mathrm{x}^{-2}=\frac{-1}{\mathrm{x}^{2}} \cdot \frac{\mathbf{d}}{\mathbf{d x}}(\sqrt{\mathrm{x}})=\mathbf{D}\left(\mathrm{x}^{1 / 2}\right)=(1 / 2) \mathrm{x}^{-1 / 2}=\frac{1}{2 \sqrt{\mathrm{x}}}$
These results can be obtained by using the definition of the derivative, but the algebra is slightly awkward.
Practice 8: Use the pattern of the theorem to find $\mathbf{D}\left(\mathrm{x}^{3 / 2}\right), \frac{\mathbf{d}}{\mathbf{d x}}\left(\mathrm{x}^{1 / 3}\right), \mathrm{D}\left(\frac{1}{\sqrt{\mathrm{x}}}\right)$ and $\frac{\mathbf{d}}{\mathrm{dt}}\left(\mathrm{t}^{\pi}\right)$.
Example 6: It costs $\sqrt{\mathrm{x}}$ hundred dollars to run a training program for $\mathrm{x}$ employees.
(a) How much does it cost to train 100 employees? 101 employees? If you already need to train 100 employees, how much additional will it cost to add 1 more employee to those being trained?
(b) For $f(x)=\sqrt{x}$, calculate $f^{\prime}(x)$ and evaluate $f^{\prime}$ at $x=100$. How does $f^{\prime}(100)$ compare with the last answer in part (a)?
Solution: (a) Put $f(x)=\sqrt{x}=x^{1 / 2}$ hundred dollars, the cost to train $x$ employees. Then $f(100)$=$1000 and $f(101)$=$1004.99, so it costs $4.99 additional to train the 101st employee. (b) $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{-1 / 2}=\frac{1}{2 \sqrt{\mathrm{x}}} \quad$ so $\mathrm{f}^{\prime}(100)=\frac{1}{2 \sqrt{100}}=\frac{1}{20} \quad$ hundred dollars =$5.00.
Clearly $f '(100)$ is very close to the actual additional cost of training the 101st employee.
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# RD Chapter 7- Trigonometric Ratios of Compound Angles Ex-7.2 Interview Questions Answers
### Related Subjects
Question 1 :
Find the maximum and minimum values of each of the following trigonometrical expressions:
(i) 12 sin x – 5 cos x
(ii) 12 cos x + 5 sin x + 4
(iii) 5 cos x + 3 sin (π/6 – x) + 4
(iv) sin x – cos x + 1
We know that the maximum value of A cos α + B sin α +C is C + √(A2 +B2),
And the minimum value is C – √(a2 +B2).
(i) 12 sin x – 5 cos x
Given: f(x) = 12 sin x – 5 cos x
Here, A = -5, B = 12 and C = 0
((-5)2 + 122) ≤ 12 sin x – 5cos x ≤ ((-5)2 +122)
(25+144) ≤ 12 sin x – 5 cos x ≤ (25+144)
169 ≤ 12 sin x – 5 cos x ≤ 169
-13 ≤ 12 sin x – 5 cos x ≤ 13
Hence, the maximum and minimum values of f(x) are 13and -13 respectively.
(ii) 12 cos x + 5 sin x + 4
Given: f(x) = 12 cos x + 5 sin x + 4
Here, A = 12, B = 5 and C = 4
4 – (122 + 52)≤ 12 cos x + 5 sin x + 4 ≤ 4 + (122 + 52)
4 – (144+25) ≤ 12 cos x + 5 sin x + 4 ≤4 + (144+25)
4 –169 ≤ 12 cos x + 5 sin x + 4 ≤ 4 + 169
-9 ≤ 12 cos x + 5 sin x + 4 ≤ 17
Hence, the maximum and minimum values of f(x) are -9and 17 respectively.
(iii) 5 cos x + 3 sin (π/6 – x) + 4
Given: f(x) = 5 cos x + 3 sin (π/6 – x) + 4
We know that, sin (A – B) = sin A cos B – cos A sin B
f(x) = 5 cos x + 3 sin (π/6 – x) + 4
= 5 cos x + 3 (sin π/6 cos x – cos π/6 sin x) + 4
= 5 cos x + 3/2 cos x – 33/2 sin x+ 4
= 13/2 cos x – 33/2 sin x + 4
So, here A = 13/2, B = – 33/2, C =4
4 – [(13/2)2 + (-33/2)2]≤ 13/2 cos x – 33/2 sin x + 4 ≤ 4 + [(13/2)2 +(-33/2)2]
4 – [(169/4) + (27/4)] ≤ 13/2 cos x – 33/2 sin x+ 4 ≤ 4 + [(169/4) + (27/4)]
4 – 7 ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 4 + 7
-3 ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 11
Hence, the maximum and minimum values of f(x) are -3and 11 respectively.
(iv) sin x – cos x + 1
Given: f(x) = sin x – cos x + 1
So, here A = -1, B = 1 And c = 1
1 – [(-1)2 + 12]≤ sin x – cos x + 1 ≤ 1 + [(-1)2 + 12]
1 – (1+1) ≤ sin x – cos x + 1 ≤ 1+ (1+1)
1 – 2 ≤ sin x – cos x + 1 ≤ 1 + 2
Hence, the maximum and minimum values of f(x) are 1– 2and 1 + 2respectively.
Question 2 :
Reduce each of the following expressions to the Sine and Cosine of a single expression:
(i) √3 sin x – cos x
(ii) cos x – sin x
(iii) 24 cos x + 7 sin x
(i) √3 sin x – cos x
Let f(x) = √3 sin x – cos x
Dividing and multiplying by √((√3)2 +12) i.e. by 2
f(x) = 2(√3/2 sin x – 1/2 cos x)
Sine expression:
f(x) = 2(cos π/6 sin x – sin π/6 cos x) (since, √3/2 =cos π/6 and 1/2 = sin π/6)
We know that, sin A cos B – cos A sin B = sin (A – B)
f(x) = 2 sin (x – π/6)
Again,
f(x) = 2(√3/2 sin x – 1/2 cos x)
Cosine expression:
f(x) = 2(sin π/3 sin x – cos π/3 cos x)
We know that, cos A cos B – sin A sin B = cos (A + B)
f(x) = -2 cos(π/3 + x)
(ii) cos x – sin x
Let f(x) = cos x – sin x
Dividing and multiplying by √(12 + 12)i.e. by √2,
f(x) = √2(1/√2 cos x – 1/√2 sin x)
Sine expression:
f(x) = √2(sin π/4 cos x – cos π/4 sin x) (since, 1/√2= sin π/4 and 1/√2 = cos π/4)
We know that sin A cos B – cos A sin B = sin (A – B)
f(x) = √2 sin (π/4 – x)
Again,
f(x) = √2(1/√2 cos x – 1/√2 sin x)
Cosine expression:
f(x) = 2(cos π/4 cos x – sin π/4 sin x)
We know that cos A cos B – sin A sin B = cos (A + B)
f(x) = √2 cos (π/4 + x)
(iii) 24 cos x + 7 sin x
Let f(x) = 24 cos x + 7 sin x
Dividing and multiplying by √((√24)2 +72) = √625 i.e. by 25,
f(x) = 25(24/25 cos x + 7/25 sin x)
Sine expression:
f(x) = 25(sin α cos x + cos α sin x) where, sin α= 24/25 and cos α = 7/25
We know that sin A cos B + cos A sin B = sin (A + B)
f(x) = 25 sin (α + x)
Cosine expression:
f(x) = 25(cos α cos x + sin α sin x) where, cos α= 24/25 and sin α = 7/25
We know that cos A cos B + sin A sin B = cos (A – B)
f(x) = 25 cos (α – x)
Question 3 :
Show thatSin 100o – Sin 10o is positive.
Let f(x) = sin 100° – sin 10°
Dividing And multiplying by √(12 + 12)i.e. by √2,
f(x) = √2(1/√2 sin 100o – 1/√2 sin 10o)
f(x) = √2(cos π/4 sin (90+10)o – sinπ/4 sin 10o) (since, 1/√2 = cos π/4 and 1/√2 = sin π/4)
f(x) = √2(cos π/4 cos 10o – sin π/4sin 10o)
We know that cos A cos B – sin A sin B = cos (A + B)
f(x) = √2 cos (π/4 + 10o)
f(x) = √2 cos 55°
Question 4 : Prove that (2√3 + 3) sin x + 2√3 cos x lies between – (2√3 + √15) and (2√3 + √15).
Let f(x) = (2√3 + 3) sin x + 2√3 cos x
Here, A = 2√3, B = 2√3 + 3 and C = 0
– √[(2√3)2 + (2√3 + 3)2] ≤(2√3 + 3) sin x + 2√3 cos x ≤ √[(2√3)2 + (2√3 + 3)2]
– √[12+12+9+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤√[12+12+9+12√3]
– √[33+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤√[33+12√3]
– √[15+12+6+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤√[15+12+6+12√3]
We know that (12√3 + 6 < 12√5) because the value of√5 – √3 is more than 0.5
So if we replace, (12√3 + 6 with 12√5) the aboveinequality still holds.
So by rearranging the above expression √(15+12+12√5)weget, 2√3 + √15
– 2√3 + √15 ≤ (2√3 + 3) sin x + 2√3 cos x ≤ 2√3 + √15
Hence proved.
krishan
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# Activities to Teach Students Proofs Involving Parallel Lines
Parallel lines and theorems have been a fundamental concept in mathematics for several centuries. As a student, learning proofs involving parallel lines can be challenging, but it is essential to understand this concept for further mathematics study. To help students gain a better understanding of proofs involving parallel lines, several activities can be incorporated into the classroom.
1. The Angle Game
The angle game is a fun way to help students understand the relationship between parallel lines and the angle measures. To begin, draw two parallel lines on the board or worksheet and label them as l and m. Next, shade in an angle at the intersection of the two lines and label it as A. Then, ask students to identify all the other angles in the figure that are congruent (same size) to angle A. By doing so, they will be able to identify the corresponding angles, which are equal in size due to the parallel lines.
2. The Parallel Lines Cut By a Transversal Puzzle
Another bedrock activity to teach students proofs involving parallel lines is the parallel lines cut by a transversal puzzle. To start, draw two parallel lines and a transversal that intersects the two lines. Cut out several angle diagrams and figure out which angles correspond with each other based on the rules, such as alternate interior angles are identical.
3. Parallel Line Collaborative Activity
This activity involves placing students in groups of 3-4 and assigning them different roles, such as a recorder, a speaker, a problem solver, and a checker. To start, provide each group with a set of parallel lines and a figure, which they will have to solve. The recorder notes down the steps taken to find the solution, while the problem solver uses their knowledge of theorems and angles to solve the puzzle. The speaker will then present the solution to the rest of the class. Lastly, the checker will review the solution critically to check for mistakes.
4. Parallel Parking Proofs
Parallel parking is a practical application involving parallel lines, and it makes a fun activity to teach students proofs involving parallel lines. Draw two parallel lines on the classroom board and shade in a car situated between the two lines. Then, ask the students to draw two additional lines representing the wheels of the car, parallel to the two original lines. By doing so, students will understand how parallel lines work in practical use and gain a better understanding of proof concepts.
5. Obtuse and Acute Angles in Parallel Lines
This activity involves covering parallel lines with a piece of paper, leaving two opens spaces on opposite sides. Then, mark a point on the paper, and draw an acute angle and an obtuse angle, starting from that point and each of the open spaces to the other side of the paper. After removing the paper, students would be able to see that the acute angles and obtuse angles on one side would always sum up to a straight line. This activity helps students to understand subtleties associated with parallel lines and proofs.
In conclusion, incorporating various activities in teaching proofs involving parallel lines will make the learning process more engaging and easier for students. The activities will help students learn the concept in a more practical way and be more confident when everyone is involved. Now that the benefits of incorporating these activities have been highlighted, teachers have a plethora of options to choose from and help their students better understand the importance of parallel lines and theorems.
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# Solving by factoring
There are a lot of great apps out there to help students with their school work for Solving by factoring. We will also look at some example problems and how to approach them.
## Solve by factoring
In algebra, one of the most important concepts is Solving by factoring. A composition of functions solver can be a useful tool for solving mathematical problems. In mathematics, function composition is the operation of combining two functions to produce a third function. For example, if f(x) = 2x + 1 and g(x) = 3x - 5, then the composition of these two functions, denoted by g o f, is the function defined by (g o f)(x) = g(f(x)) = 3(2x + 1) - 5 = 6x + 8. The composition of functions is a fundamental operation in mathematics and has many applications in science and engineering. A composition of functions solver can be used to quickly find the composition of any two given functions. This can be a valuable tool for students studying mathematics or for anyone who needs to solve mathematical problems on a regular basis. Thanks to the composition of functions solver, finding the composition of any two given functions is now quick and easy.
It is important to be able to solve expressions. This is because solving expressions is a fundamental skill in algebra. Algebra is the branch of mathematics that deals with equations and variables, and it is frequently used in physics and engineering. Many word problems can be translated into algebraic expressions, and being able to solve these expressions will allow you to solve the problem. In order to solve an expression, you need to use the order of operations. The order of operations is a set of rules that tells you the order in which to solve an equation. The order of operations is: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). Using the order of operations, you can solve any expression.
Simply point your camera at the problem and watch as the app displays the answer on screen. Not only does PhotoMath save you time, but it can also help you to better understand the concepts behind the problem. With its step-by-step solution guide, you can see how PhotoMath arrived at the answer, giving you a valuable learning opportunity. So next time you're stuck on a math word problem, reach for your phone and let PhotoMath do the work for you!
While they may seem daunting at first, there are a number of ways to solve quadratic equations. One popular method is known as factoring. This involves breaking down the equation into smaller factors that can be more easily solved. For example, if we have the equation ax^2 + bx + c = 0, we can factor it as (ax + c)(bx + c) = 0. This enables us to solve for x by setting each factor equal to zero and then solving for x. While factoring is a popular method for solving quadratic equations, it is not always the most straightforward approach. In some cases, it may be easier to use the quadratic formula, which is a formula specifically designed to solve quadratic equations. The quadratic formula can be used to solve any quadratic equation, regardless of how complex it may be. However, it is important to note that the quadratic formula only provides one solution for x. In some cases, there may be multiple solutions, so it is important to check all possible values of x before settling on a final answer. Regardless of which method you use, solving a quadratic equation can be an satisfying way to apply your math skills to real-world problems.
math is often seen as a dry and difficult subject. However, there is a wealth of resources available online that can make math more engaging and accessible for students of all levels. Websites like Khan Academy and IXL offer interactive lessons and practice problems, while Mathalicious provides real-world applications for mathematical concepts. There are also a number of games and simulations that can help to make math more fun, such as the popular game 2048. By taking advantage of these online resources, students can develop a deeper understanding of mathematics and learn to see it as a useful tool in their everyday lives.
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# 0.8 Exponential functions and graphs (Page 2/2)
Page 2 / 2
Therefore, if $a<0$ , then the range is $\left(-\infty ,q\right)$ , meaning that $f\left(x\right)$ can be any real number less than $q$ . Equivalently, one could write that the range is $\left\{y\in \mathbb{R}:y .
For example, the domain of $g\left(x\right)=3·{2}^{x+1}+2$ is $\left\{x:x\in \mathbb{R}\right\}$ . For the range,
$\begin{array}{ccc}\hfill {2}^{x+1}& >& 0\hfill \\ \hfill 3·{2}^{x+1}& >& 0\hfill \\ \hfill 3·{2}^{x+1}+2& >& 2\hfill \end{array}$
Therefore the range is $\left\{g\left(x\right):g\left(x\right)\in \left[2,\infty \right)\right\}$ .
## Domain and range
1. Give the domain of $y={3}^{x}$ .
2. What is the domain and range of $f\left(x\right)={2}^{x}$ ?
3. Determine the domain and range of $y={\left(1,5\right)}^{x+3}$ .
## Intercepts
For functions of the form, $y=a{b}^{\left(x+p\right)}+q$ , the intercepts with the $x$ - and $y$ -axis are calculated by setting $x=0$ for the $y$ -intercept and by setting $y=0$ for the $x$ -intercept.
The $y$ -intercept is calculated as follows:
$\begin{array}{ccc}\hfill y& =& a{b}^{\left(x+p\right)}+q\hfill \\ \hfill {y}_{int}& =& a{b}^{\left(0+p\right)}+q\hfill \\ & =& a{b}^{p}+q\hfill \end{array}$
For example, the $y$ -intercept of $g\left(x\right)=3·{2}^{x+1}+2$ is given by setting $x=0$ to get:
$\begin{array}{ccc}\hfill y& =& 3·{2}^{x+1}+2\hfill \\ \hfill {y}_{int}& =& 3·{2}^{0+1}+2\hfill \\ & =& 3·{2}^{1}+2\hfill \\ & =& 3·2+2\hfill \\ & =& 8\hfill \end{array}$
The $x$ -intercepts are calculated by setting $y=0$ as follows:
$\begin{array}{ccc}\hfill y& =& a{b}^{\left(x+p\right)}+q\hfill \\ \hfill 0& =& a{b}^{\left({x}_{int}+p\right)}+q\hfill \\ \hfill a{b}^{\left({x}_{int}+p\right)}& =& -q\hfill \\ \hfill {b}^{\left({x}_{int}+p\right)}& =& -\frac{q}{a}\hfill \end{array}$
Since $b>0$ (this is a requirement in the original definition) and a positive number raised to any power is always positive, the last equation above only has a real solution if either $a<0$ or $q<0$ (but not both). Additionally, $a$ must not be zero for the division to be valid. If these conditions are not satisfied, the graph of the function of the form $y=a{b}^{\left(x+p\right)}+q$ does not have any $x$ -intercepts.
For example, the $x$ -intercept of $g\left(x\right)=3·{2}^{x+1}+2$ is given by setting $y=0$ to get:
$\begin{array}{ccc}\hfill y& =& 3·{2}^{x+1}+2\hfill \\ \hfill 0& =& 3·{2}^{{x}_{int}+1}+2\hfill \\ \hfill -2& =& 3·{2}^{{x}_{int}+1}\hfill \\ \hfill {2}^{{x}_{int}+1}& =& \frac{-2}{2}\hfill \end{array}$
which has no real solution. Therefore, the graph of $g\left(x\right)=3·{2}^{x+1}+2$ does not have a $x$ -intercept. You will notice that calculating $g\left(x\right)$ for any value of $x$ will always give a positive number, meaning that $y$ will never be zero and so the graph will never intersect the $x$ -axis.
## Intercepts
1. Give the y-intercept of the graph of $y={b}^{x}+2$ .
2. Give the x- and y-intercepts of the graph of $y=\frac{1}{2}{\left(1,5\right)}^{x+3}-0,75$ .
## Asymptotes
Functions of the form $y=a{b}^{\left(x+p\right)}+q$ always have exactly one horizontal asymptote.
When examining the range of these functions, we saw that we always have either $y or $y>q$ for all input values of $x$ . Therefore the line $y=q$ is an asymptote.
For example, we saw earlier that the range of $g\left(x\right)=3·{2}^{x+1}+2$ is $\left(2,\infty \right)$ because $g\left(x\right)$ is always greater than 2. However, the value of $g\left(x\right)$ can get extremely close to 2, even though it never reaches it. For example, if you calculate $g\left(-20\right)$ , the value is approximately 2.000006. Using larger negative values of $x$ will make $g\left(x\right)$ even closer to 2: the value of $g\left(-100\right)$ is so close to 2 that the calculator is not precise enough to know the difference, and will (incorrectly) show you that it is equal to exactly 2.
From this we deduce that the line $y=2$ is an asymptote.
## Asymptotes
1. Give the equation of the asymptote of the graph of $y={3}^{x}-2$ .
2. What is the equation of the horizontal asymptote of the graph of $y=3{\left(0,8\right)}^{x-1}-3$ ?
## Sketching graphs of the form $f\left(x\right)=a{b}^{\left(x+p\right)}+q$
In order to sketch graphs of functions of the form, $f\left(x\right)=a{b}^{\left(x+p\right)}+q$ , we need to determine four characteristics:
1. domain and range
2. $y$ -intercept
3. $x$ -intercept
For example, sketch the graph of $g\left(x\right)=3·{2}^{x+1}+2$ . Mark the intercepts.
We have determined the domain to be $\left\{x:x\in \mathbb{R}\right\}$ and the range to be $\left\{g\left(x\right):g\left(x\right)\in \left(2,\infty \right)\right\}$ .
The $y$ -intercept is ${y}_{int}=8$ and there is no $x$ -intercept.
## Sketching graphs
1. Draw the graphs of the following on the same set of axes. Label the horizontal asymptotes and y-intercepts clearly.
1. $y={b}^{x}+2$
2. $y={b}^{x+2}$
3. $y=2{b}^{x}$
4. $y=2{b}^{x+2}+2$
1. Draw the graph of $f\left(x\right)={3}^{x}$ .
2. Explain where a solution of ${3}^{x}=5$ can be read off the graph.
## End of chapter exercises
1. The following table of values has columns giving the $y$ -values for the graph $y={a}^{x}$ , $y={a}^{x+1}$ and $y={a}^{x}+1$ . Match a graph to a column.
$x$ A B C -2 7,25 6,25 2,5 -1 3,5 2,5 1 0 2 1 0,4 1 1,4 0,4 0,16 2 1,16 0,16 0,064
2. The graph of $f\left(x\right)=1+a.{2}^{x}$ (a is a constant) passes through the origin.
1. Determine the value of $a$ .
2. Determine the value of $f\left(-15\right)$ correct to FIVE decimal places.
3. Determine the value of $x$ , if $P\left(x;0,5\right)$ lies on the graph of $f$ .
4. If the graph of $f$ is shifted 2 units to the right to give the function $h$ , write down the equation of $h$ .
3. The graph of $f\left(x\right)=a.{b}^{x}\phantom{\rule{3.33333pt}{0ex}}\left(a\ne 0\right)$ has the point P(2;144) on $f$ .
1. If $b=0,75$ , calculate the value of $a$ .
2. Hence write down the equation of $f$ .
3. Determine, correct to TWO decimal places, the value of $f\left(13\right)$ .
4. Describe the transformation of the curve of $f$ to $h$ if $h\left(x\right)=f\left(-x\right)$ .
how can chip be made from sand
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
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Good
Other chapter Q/A we can ask
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