Advanced Maths Tutors for STEM Activities in Agartala
If you are looking for Advanced Maths tutors for STEM Activities in Agartala, LearnPick can help. We have a large number of experienced teachers and instructors, who can help you taking classes and lessons for Advanced Maths tutors for STEM Activities. Both one to one and group class options are available near your place or online. View the profiles of the teachers and trainers detailing their qualification, expertise, teaching techniques, hourly rates and availability. Post your requirement for free and find an excellent tutor.
I got a degree in teaching english. i´ve been teaching for almost 15 years. it´s pleasure to teach young kids and teenagers. right now i´m looking for extra resources to help my students learn more about tech.
I got a degree in teaching English. I´ve been teaching for almost 15 years. It´s pleasure to teach young kids and teenagers. Right now I´m looking for extra resources to help my students learn more about tech.
According to Pythagoras theorem,the area of a square drawn on the hypotenuse of a right angled triangle equals the sum of the areas of the squares drawn on the other two sides.The sides of right triangled triangles bear an integral ratio and are known as Pythagorean triples.Therfore any square ,the length of whose side(element of N) is the hypotenuse of a right angled triangle can be dissected into smaller squares (no two of which are of same size).
The Pythagorean triples can be generated by the relation (m^2 -n^2), 2mn, (m^2+n^2),where m&n are elements of N. Thus any square whose side length can be expressed as the sum of two square numbers or simple multiples of such side lengths can be dissected into smaller squares, of which no two squares are of same size.
Starting with 5 we can see that 5 is the sum of 2 square numbers 1&4(5= 1+4) and so 5^2 = 3^2+4^2. The next possible number is 10 which is 1+9 and so we can express 10^2 = 6^2+8^2. The next number is 13 which is 4+9 and so 13^2 =5^2+12^2.The next number in sequence is 15.15 cannot be expressed as the sum of two square numbers but it is a multiple of 5,which is sum of two square numbers.Therefore 15^2 = 9^2 +12^2. Thus you can generate side lengths of all squares which can be dissected into smaller squares of which no two squares are of the same size. Following are the side lengths of squares between N= 1 to 60 that can be dissected into smaller squares (of which no two of them are of the same size) : 5, 10,13,15,17,20,25,26,29,30,34,35,37,40,41,45,50,51,52,53,55,58,60
A cube can also be dissected into a finite number of smaller cubes out of which no two cubes are of the same size.For example
6^3 = 3^3+4^3 + 5^3.
9^3 = 1^3+6^3 +8^3
13^3 = 1^3 + 5^3 + 7^3 +12^3
14^3 = 2^3 +3^3 + 8^3 + 13^3
However, unlike squares, no formula is available to generate all such cubes. They have to be found by trial & error only.
Further all squares having a Square number (N) as their lengths can be dissected into N number of smaller identical sized squares.For example a square of side length 4 units can be dissected into 4 squares of side length of 2 units. That is 4^2 = 4*(2^2) = 16 . Similarly a square of side length 9 units can be dissected into 9 squares of side length 3 units. That is 9^2 = 9*(3^2) =81 and so on
Similarly all cubes having a Cube number(N) as their side length can be dissected into N smaller identical sized cubes. For example a cube of size 8 units can be dissected into 8 smaller cubes of side length 4 units. That is 8^3 = 8*(4^3) = 512.Similarly a cube of side length 27 units can be dissected into 27 cubes of side length 9 units. That is 27^3 = 27*(9^3) = 19683 and so on.
Total selling price,SP= 504+504= Rs 1008
Gain after selling one toy= 12%
Selling Price, SP of the toy=(1+.12) Cost Price,CP
or, CP=SP/1.12=504/1.12=450 Rs
Now loss after selling another toy=4%
So,total cost price,CP= 450+525= 975 Rs
So, total profit %= (1008-975)/975*100=33/975 *100=44/13= 3 5/13 (ans)
⭐ How can students improve their knowledge in Advanced Maths ?
Students can improve their Advanced Maths knowledge and skills in a number of ways like:
Practicing solutions regularly.
Understand the underlying concepts/formulas clearly.
Emphasize conceptual understanding over procedure.
Solving additional exercises.
Sharing a positive attitude about Advanced Maths.
⭐ How can Advanced Maths tutors help students improve their score and skills in Advanced Maths ?
There are many ways students can improve their skills in Advanced Maths . But experienced Advanced Maths tutors in Agartala can help to:
Build confidence in the student.
Encourage questioning and make space for curiosity.
Emphasize conceptual understanding over procedural learning.
Provide authentic problems that increase students’ drive to engage with Advanced Maths.
Share a positive attitude about Advanced Maths.
⭐ How many Advanced Maths tutors are available in Agartala to teach Advanced Maths ?
We have a massive database of 1 verified and experienced STEM Activities tutors in Agartala to teach Advanced Maths . You can view their profiles with their qualification, expertise, teaching techniques, hourly rates and availability. Post your requirement for free to find the best STEM Activities tutors for Advanced Maths in Agartala.
⭐ What is the tuition fee charged by STEM Activities tutors in Agartala to teach Advanced Maths ?
Tuition fees of STEM Activities tutors in Agartala depend on a number of factors like tutoring hours, experience and qualifications. You can find out the STEM Activities tutor from our list as per your estimated fee with your preferred location.
⭐ Do STEM Activities tutors in Agartala provide training for competitive examinations in Advanced Maths?
Yes, most of them do. However, we would request you to discuss the same with the STEM Activities tutor of your choice for clarification of any extra hours, fees, etc.
⭐ What is the normal duration of tuition classes hosted by STEM Activities tutors in Agartala for Advanced Maths?
Usually, STEM Activities tutors on LearnPick conduct a session for 1 to 2 hours a day for Advanced Maths. But it can vary depending on the arrangements made between the student and the tutor at the time of hiring.
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