Moris T.

Moris T.

£30 - £32 /hr

Economics and Management (Bachelors) - Balliol College, Oxford University

5.0
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5 reviews

This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

13 completed lessons

About me

Moreover, as I have experienced the pathway to university in an academically challenging way, I am fully aware of the challenging pathway leading to great universities and I feel extremely confident in my abilities to incorporate many disciplines to provide the broad knowledge required in this journey.

 

Facing the demanding system in Oxford, I have learned to express my ideas at its best on a one-to-one basis. Also, providing my line of thought with the required detail is an area that I flourish.

 

Lastly, as I have participated in a tutoring program, which the local municipality conducted in my hometown, as a Student Leader, I have previous experience both on tutoring students going through a similar path that I went few years ago as well as leading over 15 classes and devising syllabuses for each different topic.

 

I am looking forward to meeting and having mutually fruitful lessons with all of my students!

Moreover, as I have experienced the pathway to university in an academically challenging way, I am fully aware of the challenging pathway leading to great universities and I feel extremely confident in my abilities to incorporate many disciplines to provide the broad knowledge required in this journey.

 

Facing the demanding system in Oxford, I have learned to express my ideas at its best on a one-to-one basis. Also, providing my line of thought with the required detail is an area that I flourish.

 

Lastly, as I have participated in a tutoring program, which the local municipality conducted in my hometown, as a Student Leader, I have previous experience both on tutoring students going through a similar path that I went few years ago as well as leading over 15 classes and devising syllabuses for each different topic.

 

I am looking forward to meeting and having mutually fruitful lessons with all of my students!

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About my sessions

I deeply value enabling my students to learn how to tackle the problems related to the topic by themselves. So, my Online Lessons would commence with the theoretical knowledge that is required to investigate the subject further with the questions. Shortly, I will walk through my students in example problems related to the topic. After, I would give time to my students to combine the knowledge they gained through the theoretical and practical part and tackle new problems by themselves.

 

As the students try to figure a way to tackle the problem they face, I will view their proposed solution and we will apply it with the student. Along this stage any incorrect parts of the proposed solution will be analyzed in detail in order to eliminate such problems. By the end, I will ensure that the students are capable of devising solutions to the questions they face in tests by themselves. This way I will be able to measure their progress and the students will be more than confident in the topic in the future.

I deeply value enabling my students to learn how to tackle the problems related to the topic by themselves. So, my Online Lessons would commence with the theoretical knowledge that is required to investigate the subject further with the questions. Shortly, I will walk through my students in example problems related to the topic. After, I would give time to my students to combine the knowledge they gained through the theoretical and practical part and tackle new problems by themselves.

 

As the students try to figure a way to tackle the problem they face, I will view their proposed solution and we will apply it with the student. Along this stage any incorrect parts of the proposed solution will be analyzed in detail in order to eliminate such problems. By the end, I will ensure that the students are capable of devising solutions to the questions they face in tests by themselves. This way I will be able to measure their progress and the students will be more than confident in the topic in the future.

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Ratings & Reviews

5
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5 customer reviews
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JL
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JO Parent from LONDON Lesson review 30 Nov, 16:30

30 Nov

I've been trying to do my personal statement for months now and that one hour session has been by far the most helpful thing and also the most progress I have made in these months.10/10, totally recommend.

VH
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Volodymyr Student Lesson review 26 Nov, 17:30

26 Nov

Really beneficial revision of key areas, especially before the exam

JL
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Jon Student Lesson review 28 Nov, 19:15

28 Nov

JL
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Jon Student Lesson review 26 Nov, 20:15

26 Nov

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Qualifications

SubjectQualificationGrade
MicroeconomicsUS AP Classes5/5
MacroeconomicsUS AP Classes5/5
Calculus BCUS AP Classes5/5
Calculus ABUS AP Classes5/5
Physics 1US AP Classes5/5
Physics 2US AP Classes4/5
English Literature and ComprehensionUS AP Classes4/5
English Language and ComprehensionUS AP Classes4/5
Math Level 2 SAT Subject TestUni admission test800/800
Math Level 1 SAT Subject TestUni admission test800/800
Physics SAT Subject TestUni admission test800/800
ACTUni admission test34,25/36
SATUni admission test1500/1600
IELTSUni admission test8,5/9
TOEFLUni admission test112/120
High School GPA (Science and Math Track)Diploma95.35

General Availability

MonTueWedThuFriSatSun
Pre 12pm
12 - 5pm
After 5pm

Pre 12pm

12 - 5pm

After 5pm
Mon
Tue
Wed
Thu
Fri
Sat
Sun

Subjects offered

SubjectQualificationPrices
EconomicsA Level£30 /hr
MathsA Level£30 /hr
EconomicsGCSE£30 /hr
MathsGCSE£30 /hr
EconomicsIB£30 /hr
MathsIB£30 /hr
Oxbridge PreparationMentoring£30 /hr
Personal StatementsMentoring£30 /hr
TSA OxfordUniversity£32 /hr

Questions Moris has answered

The function f has a local extreme at point (1,4). If f''(x)=3x^2+2x, then find f(0)?

A local extreme point would mean that f'(1)=0. Since integration is the inverse application of derivation, to obtain f'(x) function we integrate the f''(x) function. Bearing in mind the integration rule of increasing the power "n" to "n+1" and dividing the new number by "n+1", the integration of f''(x) would yield f'(x)=x^3+x^2+C (C is the constant generated by integrating and can be any constant value. (Remember, after derivation all constants are lost so we can not be certain whether such value of C is 0 or any other real number). As we know that f'(1)=0 (see first sentence), replacing x with 1 and solving the equation would lead to C=-2; thus, f'(x)= x^3+x^2-2.As we now have the f'(x) function, we can use similar integration methods to obtain f(x) function. Hence, f(x)=(x^4)/4+(x^3)/3-2x+C. The question itself provided that f(1)=4. Plugging in these values would give f(1)=1/4+1/3-2+C=4. Solving this, we obtain C=65/12, and f(x)=1/4+1/3-2+65/12. Finally, using this equation when x=0 , we get f(0)=65/12.A local extreme point would mean that f'(1)=0. Since integration is the inverse application of derivation, to obtain f'(x) function we integrate the f''(x) function. Bearing in mind the integration rule of increasing the power "n" to "n+1" and dividing the new number by "n+1", the integration of f''(x) would yield f'(x)=x^3+x^2+C (C is the constant generated by integrating and can be any constant value. (Remember, after derivation all constants are lost so we can not be certain whether such value of C is 0 or any other real number). As we know that f'(1)=0 (see first sentence), replacing x with 1 and solving the equation would lead to C=-2; thus, f'(x)= x^3+x^2-2.As we now have the f'(x) function, we can use similar integration methods to obtain f(x) function. Hence, f(x)=(x^4)/4+(x^3)/3-2x+C. The question itself provided that f(1)=4. Plugging in these values would give f(1)=1/4+1/3-2+C=4. Solving this, we obtain C=65/12, and f(x)=1/4+1/3-2+65/12. Finally, using this equation when x=0 , we get f(0)=65/12.

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2 months ago

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