Question (PageIndex{1})

In order to refresh your matrix-vector multiply skills please calculate, by hand, the product (A^{T}GA) in the 3 compartment case and write out the 4 equations in the vector equation we arrived at in step(S4): (A^{T}GA extbf{x} = extbf{f})

**Feedback**

The second equation should read

[frac{-x_{1}+2x_{2}-x_{3}}{R_{i}}+frac{x_{2}}{R_{m}} = 0 label{1.2.1}]

Question (PageIndex{2})

We began our discussion with the 'hope' that a multicompartment model could indeed adequately capture the fiber's true potential and current profiles. In order to check this one should run fib1.m with increasing values of NN until one can no longer detect changes in the computed potentials.

- (a) Please run fib1.m with (N = 8, 16, 32), and 64. Plot all of the potentials on the
**same**(use`hold`

) graph, using different line types for each. (You may wish to alter`fib1.m`

so that it accepts NN as an argument).

Let us now interpret this convergence. The main observation is that the difference Equation ef{1.2.1}, approaches a differential equation. We can see this by noting that

[mathbb{d}(z) = frac{l}{N}]

acts as a spatial 'step' size and that (x_{k} mathbb{d}(z)) is approximately the value of the true potential at ((k-1) mathbb{d}(z)). In a slight abuse of notation, we denote the latter

[x((k-1) mathbb{d}(z))]

Applying these conventions to Equation ef{1.2.1} and recalling the definitions of (R_{i}) and (R_{m}) we see Equation ef{1.2.1} become

[frac{pi a^2}{ ho_{i}} frac{-x(0)+2x(mathbb{d}(z))-x(2 mathbb{d}(z))}{mathbb{d}(z)}+ frac{2 pi a mathbb{d}(z)}{ ho_{m}} x(d(z)) = 0]

or, after multiplying through by (frac{ ho_{m}}{pi a mathbb{d}(z)})

[frac{a ho_{m}}{ ho_{i}} frac{-x(0)+2x(mathbb{d}(z))-x(2mathbb{d}(z))}{mathbb{d}(z^2)}+2x(mathbb{d}(z)) = 0 label{1.2.3}]

We note that a similar equation holds at each node (save the ends) and that as (N ightarrow infty) and therefore (mathbb{d}(z) ightarrow 0) we arrive at

[frac{d^2}{dz^2} x(z)-frac{2 ho_{i}}{a ho_{m}} x(z) = 0 label{1.2.4}]

- (b) With (mu equiv frac{2 ho_{i}}{a ho_{m}}) show that

[x(z) = alpha sinh(sqrt{2 mu} z)+eta cosh(sqrt{2 mu}z) label{1.2.5}]

satisfies Equation ef{1.2.3} regardless of (alpha) and (eta)

We shall determine (alpha) and (eta) by paying attention to the ends of the fiber. At the near end we find

[frac{pi a^2}{ ho_{i}} frac{x(0)-x(mathbb{d}(z))}{mathbb{d}(z)} = i_{0}]

which, as (mathbb{d}(z) ightarrow 0) becomes

[frac{d}{dz} x(0) = -frac{ ho_{i} i_{0}}{pi a^{2}}]

At the far end, we interpret the condition that no axial current may leave the last node to mean

[frac{d}{dz} x(l) = 0]

- (c) Substitute Equation ef{1.2.4} into Equation ef{1.2.5} and solve for (alpha) and (eta) and write out the final (x(z)).
- (d) Substitute into (x) the (l, a,
ho_{i},
ho_{m}) values used in fib1.m, plot the resulting function (using, e.g.,
`ezplot`

) and compare this to the plot achieved in part (a).

## NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.2

**NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Exercise 1.2**

Ex 1.2 Class 8 Maths Question 1.

Represent these numbers on the number line

(i) (frac < 7 >< 4 >)

(ii) (frac < -5 >< 6 >)

Solution:

(i) (frac < 7 >< 4 >)

Here, point A represents (frac < 7 >< 4 >) on the number line.

(ii) (frac < -5 >< 6 >)

Here, point B represents (frac < -5 >< 6 >) on the number line.

Ex 1.2 Class 8 Maths Question 2.

Represent (frac < -2 >< 11 >), (frac < -5 >< 11 >) , (frac < -9 >< 11 >) on a number line.

Solution:

We have (frac < -2 >< 11 >), (frac < -5 >< 11 >) and (frac < -9 >< 11 >)

Here, point A represents (frac < -2 >< 11 >) , point B represents (frac < -5 >< 11 >), point C represents (frac < -9 >< 11 >)

Ex 1.2 Class 8 Maths Question 3.

Write five rational numbers which are smaller than 2.

Solution:

Required five rational numbers smaller than 2 are

1, 0, (frac < 1 >< 2 >), (frac < 1 >< 3 >) and (frac < 1 >< 4 >)

Ex 1.2 Class 8 Maths Question 4.

Find ten rational numbers between (frac < -2 >< 5 >) and (frac < 1 >< 2 >).

Solution:

Ex 1.2 Class 8 Maths Question 5.

Find five rational numbers between

(i) (frac < 2 >< 3 >) and (frac < 4 >< 5 >)

(ii) (frac < -3 >< 2 >) and (frac < 5 >< 3 >)

(iii) (frac < 1 >< 4 >) and (frac < 1 >< 2 >)

Solution:

Ex 1.2 Class 8 Maths Question 6.

Write five rational numbers greater than -2.

Solution:

Required rational numbers greater than -2 are -1, 0, (frac < 1 >< 2 >), (frac < 3 >< 4 >) , 1.

Ex 1.2 Class 8 Maths Question 7.

Find ten rational numbers between (frac < 3 >< 5 >) and (frac < 3 >< 4 >).

Solution:

Given rational numbers are (frac < 3 >< 5 >) and (frac < 3 >< 4 >).

If you need solutions in Hindi, Click for Hindi Medium solutions of 10 Maths Exercise 1.2

Exercise 1.1

#### 10 Maths Chapter 1 Exercise 1.2 in Hindi

To get the solutions in English, Click for English Medium solutions.

##### Important Questions of Real Numbers Exercise 1.2 for Practice

- Sate whether 7 × 11 × 13 + 7 is a composite number or a prime number. (Answer: Composite)
- Find the value of m if HCF of 65 and 117 is expressible in the form 65m – 117. (Answer: 2)
- Find the least number which is divisible by all numbers from 1 to 10 (both inclusive).
- The number 525 and 3000 are divisible by 3, 5, 15, 25 and 75. What is the HCF of 525 and 3000. (Answer: 2520)
- Can two numbers have 18 as their HCF and 380 as their LCM? Give reason. (Answer: No, HCF is not a factor of LCM)
- What is the digit at unit’s place of 9^n? (Answer: Even power = 1, Odd power = 9)
- Show that 12^n cannot end with the digit 0 or 5 for any natural number n.
- State fundamental theorem of Arithmetic and hence find the unique factorization of 120. (Answer: 2 × 2 × 2 × 3 × 5)
- Find HCF and LCM of 56 and 112 by prime factorization method. (Answer: HCF = 56 and LCM = 112)
- Three friends Salman, Hrithik and John were very good friends. They used to go for morning walk together. Once, on a morning walk, they step off together and their steps measure 40 cm, 42 cm and 45 cm respectively.

(a) What is the minimum distance each should walk so that each can cover the same distance in complete steps? (Answer: 25.2 m)

(b) What have you learnt (values/lesson) from above activity of three friends. (Answer: Morning walk is good for health, religion doesn’t matter in friendship)

- Aakriti decided to distribute milk in an orphanage on her birthday. The supplier brought two milk containers which contain 398 l and 436 l of milk. The milk is to be transferred to another containers so 7 l and 11 l of milk is left in both the containers respectively.

(a) What will be the maximum capacity of the drum? (Answer: 17)

(b) What qualities / values were shown by Aakriti. (Answer: Charity, Concern for others, etc.)

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## Real Numbers : Exercise 1.2 (Mathematics NCERT Class 10th)

**Q.1 Express each number as product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (vi) 7429**

**Sol.****(i)**We use the division method as shown below :

Therefore, 140 = 2 × 2 × 5 × 7

**(ii)** We use the division method as shown below:

Therefore, 156 = 2 × 2 × 3 × 13

**(iii)** We use the division method as shown below :

Therefore, 3825 = 3 × 3 × 5 × 5 × 17

**(iv)** We use the division method as shown below :

Therefore, 5005 = 5 × 7 × 11 × 13

**(v)** We use the division method as shown below :

Therefore, 7429 = 17 × 19 × 23

**Q.2 Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. **

**(i) 26 and 91**

**(ii) 510 and 92**

**(iii) 336 and 54**

**Sol.****(i)**26 and 91

26 = 2 × 13 and 91 = 7 × 13

Therefore, LCM of 26 and 91 = 2 × 7 × 13 = 182

and HCF of 26 and 91 = 13

Now, 182 × 13 = 2366 and 26 × 91 = 2366

Since, 182 × 13 = 26 × 91

Hence verified.

**(ii)** 510 and 92

510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23

Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460

and HCF of 510 and 92 = 2

Now, 23460 × 2 = 46920 and 510 × 92 = 46920

Since 23460 × 2 = 510 × 92

Hence verified.

**(iii)** 336 and 54

336 = 2 × 2 × 2 × 2 × 3 × 7

and 54 = 2 × 3 × 3 × 3

Therefore, LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024

and HCF of 336 and 54 = 2 × 3 = 6

Now, 3024 × 6 = 18144

and 336 × 54 = 18144

Since, 3024 × 6 = 336 × 54

Hence verified.

**Q.3 Find the LCM and HCF of the following integers by applying the prime factorisation method **

**(i) 12,15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25**

**Sol.****(i)**First we write the prime factorisation of each of the given numbers.

12 = 2 × 2 × 3 = × 3, 15 = 3 × 5 and 21 = 3 × 7

Therefore, LCM = × 3 × 5 × 7 = 420

and, HCF = 3

**(ii)** First we write the prime factorisation of each of the given numbers.

17 = 17, 23 = 23 and 29 = 29

Therefore, LCM = 17 × 23 × 29 = 11339

and HCF = 1

**(iii)** First we write the prime factorisation of each of the given numbers**.** 8 = 2 × 2 × 2 , 25 = 5 × 5 =

Therefore, LCM

and HCF = 1

**Q.4 Given that HCF (306, 657) = 9, find LCM (306, 657). **

*We know that the product of the HCF and the LCM of two numbers is equal to the product of the given numbers.*

**Sol.**Therefor, HCF (306,657) × LCM (306,657) = 306 × 657

9 × LCM (306 × 657) = 306 × 657

LCM (306,657) = 22338

**Q.5. Check whether can end with the digit 0 for any natureal number n. **

**If the number , for any n ends with the digit zero, then it is divisible by 5. That is, the prime factorisation of contains the prime 5. That is, not possible as the only prime in the factorisation of is 2 and 3 and the uniqueness of the Fundamental Theorem of Arihmetic guarantees that there are no other primes in the factorisation of . So, there is no for which ends with the digit zero.**

*Sol.* **Q.6 Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. **

*Since, 7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1)*

**Sol.**= 13 × (77 + 1)

= 13 × 78

It is a composite number.

Again, 7 × 6 × 5 × 4 × 3 × 1 × 1 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 1 × 1 + 1)

It is a composite number.

**Q.7 There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the path, while Ravi takes 12** **minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how** **many minutes will they again at the starting point ? **

*To find the LCM of 18 and 12, we have*

**Sol.**

18 = 2 × 3 × 3 and 12 = 2 × 2 × 3

LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36

So, Sonia and Ravi will meet again at the starting point after 36 minutes.

## 1.2: Chapter 1 Exercises - Mathematics

- Mondays 6:00 p.m. - 7:50 p.m. online (Zoom meeting ID: 700-529-221)
- The office hour on Monday, May 25 has been rescheduled to Tuesday, May 26, 10:00 a.m. - 11:50 a.m. because of Memorial Day.
- Final exam week office hour: Monday, June 8, 8:00 p.m. - 10:50 p.m. at the same Zoom meeting ID

- by Yehuda Pinchover and Jacob Rubinstein by Yehuda Pinchover and Jacob Rubinstein by Yehuda Pinchover and Jacob Rubinstein

Update (May 19): For some of you, I had to assess a 5 point penalty to your Homework 10 scores for copying the textbook solutions verbatim. I do *not* tolerate plagiarism. Remember to write your homework solutions in your own words!

Here are all the whiteboard slides from our Section 011 discussion meetings on Zoom that I have saved. All theorems, examples, and exercises referenced in these slides correspond to those of the Pinchover and Rubinstein textbook.

Please also see the whiteboard slides from our Section 012 discussion webpage, where it contains whiteboard slides that may differ from the ones in Section 011.

- Week 1
- Week 2
- Week 3
- Week 4
- Week 5
- Week 6
- Week 7
- Week 8
- Week 9
- Week 10

- Week 1
- Homework 1
- Exercise 1.2(a) of Pinchover and Rubinstein
- My solution to Exercise 1.2(a) can be found in my suggested homework solutions for Chapter 1 (linked below)

- Exercise 1.4(a) of Pinchover and Rubinstein
- My solution to Exercise 1.4(a) can be found in my suggested homework solutions for Chapter 1 (linked below)

- by Zilong Song

- My Homework 3 solutions
- Exercise 2.6 of Pinchover and Rubinstein
- My solution to Exercise 2.6 can be found in my suggested homework solutions for Chapter 2 (linked below)
- My Homework 5 solutions
- Exercise 5.6(a) of Pinchover and Rubinstein by Zilong Song
- My solution to Exercise 5.6(a) can be found in my suggested homework solutions for Chapter 5 (linked below)
- My Homework 7 solutions
- My Homework 8 solutions
- My Homework 9 solutions
- Exercise 7.15 of Pinchover and Rubinstein
- My solution to Exercise 7.15 can be found in my suggested homework solutions for Chapter 7 (linked below)
- My Homework 11 solutions
- My Homework 12 solutions
- Exercise 8.8 of Pinchover and Rubinstein by Zilong Song by Zilong Song
- My solution to Exercise 8.8 can be found in my suggested homework solutions for Chapter 8 (linked below)
- Exercise 8.5 of Pinchover and Rubinstein
- My solution to Exercise 8.5 can be found in my suggested homework solutions for Chapter 8 (linked below)
- My Homework 15 solutions
- Exercise 9.5 of Pinchover and Rubinstein
- My solution to Exercise 9.5 can be found in my suggested homework solutions for Chapter 9 (linked below)
- Chapter 1
- Exercises 1, 2, 4, 7 of Pinchover and Rubinstein
- My suggested homework solutions for Chapter 1

- Exercises 1, 2, 3, 4, 6, 7, 11, 12, 16, 21 of Pinchover and Rubinstein
- My suggested homework solutions for Chapter 2

- Exercises 1, 3(a), 4(a), 5, 6, 7, 8, 9, 10(a), 15 of Pinchover and Rubinstein
- My suggested homework solutions for Chapter 5

- Exercises 2, 3, 4, 7, 8, 11, 14(a), 15, 20, 22 of Pinchover and Rubinstein
- My suggested homework solutions for Chapter 7

- Exercises 5, 6, 7, 8, 11 of Pinchover and Rubinstein
- My suggested homework solutions for Chapter 8

- Exercises 3, 4, 5, 6, 7 of Pinchover and Rubinstein
- My suggested homework solutions for Chapter 9

- and midterm exam solutions by Zilong Song and final exam solutions by Zilong Song - The professor made a typo in Question 8 of his solution: The 1/2 term in the expression of u(r,t) should actually be a 1, which means u(2,1)=1 and u(2,2)=5/4.

Update (June 7): Your final exam on June 9 will be a take-home exam. Please complete the exam by yourself, and do not consult your classmates. The professor has announced that collaboration with classmates is prohibited and will result in an automatic course grade of D for all parties involved. If I detect any evidence of collaboration while I am grading the final exams, I will report it to the professor.

*Please do not cheat on the final exam.*Also, if you decide to use my solutions to help you with an exam question, please write your exam solution in your own words, just in case another student may have also copied from my solutions. You do not want the professor to think you collaborated with another student because the two of you independently copied the wording of my solutions.

## NCERT Solutions for Class 8 Maths Rational Numbers

###### Class –VIII Mathematics (Ex. 1.2)

###### 1. Represent these numbers on the number line:

NCERT Solutions for Class 8 Maths Exercise 1.2

###### 2. Represent on the number line.

**Ans 2.**Here, B = C = and D =NCERT Solutions for Class 8 Maths Exercise 1.2

###### 3. Write five rational numbers which are smaller than 2.

**Ans 3.**and so on.###### 4. Find ten rational numbers between and

Here, L.C.M. of 5 and 2 is 10.

Ten rational number between and are .

NCERT Solutions for Class 8 Maths Exercise 1.2

###### 5. Find five rational numbers between:

**Ans 5.**(i) andFive rational numbers between and are .

Five rational numbers between and are .

Five rational numbers between and are

NCERT Solutions for Class 8 Maths Exercise 1.2

###### 6. Write 5 rational numbers greater than

**Ans 6.**Five rational numbers greater than are:**[Other rational numbers may also be possible]**NCERT Solutions for Class 8 Maths Exercise 1.2

###### 7. Find ten rational numbers between and

Five rational numbers between and are:

## Mathematics Form 4 Chapter 1 Exercise With Answer

**Mathematics form 4 chapter 1 exercise with answer**. Now is the time to redefine your true self using slader s algebra 1. Is easier to understand by students when newly brought the topic this way it would save time to speed up to finish the chapter as the students do not have to copy down the questions into their books and more effecients compared to. Integration content page concept map 2 4 1 integration of algebraic functions 3 4 exercise a 5 4 2 the equation of a curve from functions of gradients. Coordinate geometry chapter 7.Additional mathematics form 4 click to download your exercises here chapter 1. A common core curriculum textbook solutions reorient your old paradigms. Indices and logarithms chapter 6. 1 2 law of indices 6 chapter standard form 30 2 1 significant figures 32 2 2 standard form 37 chapter consumer mathematics.

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Simultaneous equations chapter 5. Quadratic equations chapter 3. Form 4 and form 5 additional mathematics note module exercise answer. Ncert solutions for class 10 maths chapter 1 exercise 1 4 in english medium and hindi medium are given below to use online or download in pdf form.

The download link is given on the page or visit class 10 maths chapter 1 solutions main page to download. Quadratic equations modul 4 2 1 concept map 2 2 2 forming a quadratic equations from given roots and 3 2 2 1 form a quadratic equation with the roots 4 exercises 1 4 2 2 2 determine the sum of the roots and product of the roots of the 5 following quadratic equations. Functions quadratic equations. Module add maths form 4.

A common core curriculum answers. Quadratic functions chapter 4. The answer can leave with degree and minute or in. Exercises 2 5 2 3.

## Watch the video: ΜΑΘΗΜΑΤΙΚΑ Γ ΛΥΚΕΙΟΥ: Συναρτήσεις Θεωρία part I (September 2021).

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