n is the dividend right and 45 is the divisor then how do I find a divisor

Based on the length how do i get the coordinates

How do I apply the same for thw squares?

How do I calculate for A(200) after 101

Have you made *any* effort to search the questions on the internet?

Yes I did

1.For the first one I wanted to know whats incorrect in my understanding

2. I got this as the soln

*Assume Length of AB = x*

*so AB = CD = x and BC = 3x*

*so distance between AD = 5x=13β12=165x=13β12=16*

*or x=130x=130*

*now AC = 4x+13=430+13=1430or715*

3. For the third one I didnt follow the absolute and squaring relation

4. For the last one they did A(100) + (101)

but post 101 they did some simplification which I didnt follow

Just take n = 18. Only 9 and 6 are divisors of 18.

Now take (18 + 45) = 63. Only 9 is a divisor β thatβs the answer.

To solve it formally, notice that \frac{n - 18}{45} is equal to an integer. Call it k. Write n in terms of k and factorise.

Indeed, you have -1 \leq z \leq 1. The square of both negative and positive numbers is positive, which means that z^2 \leq 1.

B and C are easy to disprove by taking a negative value of z.

Probably did

A(200) = A(100) + (101 + 102 + β¦ + 200)

A(200) = A(100) + (100 + 1 + 100 + 2 + 100 + 3 + β¦ + 100 + 100)

A(200) = A(100) + 100 Γ 100 + (1 + 2 + 3 + β¦ + 100)

A(200) = 5050 + 10000 + 5050

A(200) = 20100

for 4

That 100 is from there right

Thanks for answering my remaining queries

Indeed - 100 is added 100 times.

Yes (that is, AC = AB + BC)