"8th grade online tutor" Introduction
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8th Grade Geometry
EX:3 the XYZ the angle Y is bisected and the bisector meets at XZ in W. If XYZ = 80 and YWZ = 95, find X and Z. From the FigureYWZ, 40 + 95 + Z = 180 Z = 180 135 = 45 Fro..
EX:3 the XYZ the angle Y is bisected and the bisector meets at XZ in W. If XYZ = 80 and YWZ = 95, find X and Z. From the FigureYWZ, 40 + 95 + Z = 180 Z = 180 135 = 45 Fro..8th grade experiments
8th grade experiments: Chemical Reactions Monitor closely experiments involving chemical reactions. Topics such as "Which road de-icer causes the most rust on steel?" or..
8th Grade Geometry
EX:1 How far away is a chord of length 10 cm from the mid point of the circle of radius 13 cm. XY is a chord of length 10 cm Z in the mid point of XY XZ = XY = 10 = 5 cm ; ..
EX:1 How far away is a chord of length 10 cm from the mid point of the circle of radius 13 cm. XY is a chord of length 10 cm Z in the mid point of XY XZ = XY = 10 = 5 cm ; ..8th grade writing
8th grade writing: Step 1 Determine if the sample is scorable or not. For the FCAT, a student's response is deemed "unscorable" if it meets a variety of criteria, includ..
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Question : I tutor math online math and statistics and would like to reach students 7-12 grade .
Any student with a computer can avail of tutoring services at their home at a time convenient to them and the content of the curriculum will be designed to fit individual needs.
I am looking to expand and do not know how.
Answer : Question interesting!! Good Luck!!
Answer : Question interesting!! Good Luck!!
Question : I need an online tutor,im really bad in math and social studies.plz help anyone
Answer : Hon, I wish I could help you, but asking strangers on the internet isn't the way to get help in school. (you could end up with a snotty seventh-grader making up stuff and giving you the wrong information/answers, for all you know). Tutoring in person, not online, is really the only way to ensure that both sides are communicating and that you're understanding the material. Have you tried asking your teachers for help? Most can help you outside of class. Or ask your parents, or check and see if you local public library and see if they have any sort of free homework help program. Any of those options would be better than asking anonymous internet users for help.
Answer : Hon, I wish I could help you, but asking strangers on the internet isn't the way to get help in school. (you could end up with a snotty seventh-grader making up stuff and giving you the wrong information/answers, for all you know). Tutoring in person, not online, is really the only way to ensure that both sides are communicating and that you're understanding the material. Have you tried asking your teachers for help? Most can help you outside of class. Or ask your parents, or check and see if you local public library and see if they have any sort of free homework help program. Any of those options would be better than asking anonymous internet users for help.
