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Rider
- Rate ₹4,776
- Response 1h
₹4,776/hr
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- Maths
- Algebra
- Trigonometry
- Statistics
- Algorithms
Mathematics Bachelors Degree from Harding University, teaches high school and college level math (algebra, trigonometry, precalculus, calculus, statistics), holistic and individualized approach.
- Maths
- Algebra
- Trigonometry
- Statistics
- Algorithms
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About Rider
I am a well-rounded tutor with experience teaching math and other subjects. I have been recognized for being compassionate and understanding towards students who are struggling, but also have rigorous high standards. I am willing to work with anyone no matter how far behind they may feel. I meet students where they are and take a step-by-step approach to achieving their goals.
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About the class
- Classes 1 to 5
- Classes 6 to 8
- Class 10
- +9
levels :
Classes 1 to 5
Classes 6 to 8
Class 10
Classes 11 & 12
Class 12
BTech
Class 11
Adult Literacy
Masters
PhD
MBA
Kindergarten (LKG, UKG)
- English
All languages in which the class is available :
English
I like to spend time discussing with the student to identify any specific challenges they may be facing with the subject at hand. This allows me to come up with practice work that sits in the zone of proximal development, that is, "the space between what a learner is capable of doing unsupported and what the learner cannot do even with support. It is the range where the learner is able to perform, but only with support from a teacher or a peer with more knowledge or expertise". I believe this is the most efficient and enjoyable path to progress in learning. By targeting work in the zone of proximal development, the student exercises their brain by doing new and challenging work, but also remains motivated by being able to successfully understand and solve the problems at hand, rather than being lost and confused.
I also believe that practice tests are one of the best tools available. I try to encourage students to see tests as part of the learning process rather than as an evaluation of their skills to help reduce test anxiety. A good practice test will have questions very similar to those that appear on "real" tests and will effectively exercise the students ability to recall the material. It also provides a valuable diagnostic for identifying strengths and weaknesses, areas of understanding and areas for improvement.
I am happy to work with learners in high school and college who want to improve their grades, their test taking ability, or their understanding of the material.
Sample Lesson Plan (this is just an outline, the full lesson would include pictures/graphs, demonstrations, and interactive discussions/explanations):
GRAPHING SYSTEMS OF INEQUALITIES WITH ABSOLUTE VALUES
An inequality is an equation like y > x or y < x + 3, etc.
A point (x, y) satisfies an inequality if the statement is true when you plug in the values.
Example: (1, 1) satisfies y < x + 3 but not y > x (because 1 > 1 is false).
When you graph an inequality you can graph the line for the equation that replaces the inequality sign (less than, greater than, less than or equal to, greater than or equal to) with an equals sign (=). If the inequality is “inclusive” (“or equal to”), graph the line with a solid line, if it is a strict inequality (not “or equal to”), it will be a dotted or dashed line. Then you must shade the side of the line that includes the points (x, y) that satisfy the inequality.
A system of inequalities just means you have multiple inequalities and you want to graph the points that solve all the inequalities in the system. So you will graph the lines and shaded areas as above, and the overlap of the shaded areas is the solution (all the points (x, y) that satisfy each of the inequalities).
Remember that when you graph an equation that has an absolute value in it, you will get a right angled corner in your graph.
When you graph an absolute value you should break the function up into its “cases” (the case where the contents inside the absolute value are greater than or equal to zero, and the case where they are less than zero). Each case will give you an equation (or an inequality). Plot both of the lines and see where they intersect, note that the x-value of the intersection point will be the point where the cases “switch” (so if your absolute value gave you cases x < 1 and x ≥ 1, the intersection point will have an x-value of 1. Keep the sides of the lines that correspond to the case on that side of the intersection point and erase the other ends of the lines.
To figure out which sides of lines to shade on, pick any point (I usually pick (1,1)) and see if it satisfies the inequality corresponding to each line, if it satisfies the inequality, shade on the side of the line that the point (1,1) is on, if it does not satisfy the inequality shade on the opposite side of the line.
EXAMPLE WITH STEPS:
Graph this system of inequalities:
y ≥ |x + 3| - 2
y > |x - 1|
Pick one to start with, I will start with y ≥ |x + 3| - 2
Break the absolute value down into its cases
|x+3| = x+3 if x+3 ≥ 0 which is equivalent to x ≥ -3
|x+3| = -(x+3) = -x - 3 if x+3 < 0 which is equivalent to x < -3
Break the inequality down into its cases based on the absolute value cases
If x ≥ -3 then the inequality is y ≥ x + 3 - 2 = x + 1
If x < -3 then the inequality is y ≥ -x - 3 - 2 = -x - 5
Plot the line for each case (remember whether it should be a solid or dotted/dashed line)
y = x + 1
y = -x - 5
Determine which line to keep on either side of the intersection point by looking at the “if condition” (in this example, we have x + 1 “if x ≥ -3”, so keep the y = x + 1 line on the right side of x = -3, and keep the other line on the other side of x = -3)
Shade the area that satisfies the inequality (test points like (1,1) to determine which area to shade)
Repeat steps 2-6 with the other inequality
Now the final solution will be the overlap of the two shaded areas.
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- ₹4,776
Pack prices
- 5h: ₹20
- 10h: ₹40
online
- ₹4,776/hr
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