9th grade geometry practice test
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Answers:You understand the material but your test scores are really low? You need to study the material more, and practice doing problems before tests. Also do your homework all the time, I promise it helps.
Answers:there are probably books at the bookstore, i dont no for sure but i think there would probably be one of those geometry books for dummies or sumthin like that. u might be shy, but if your grades really matter to you and you want to do better, than you can't let your teacher's disagreeable attitude affect your actions. go in and ask for help, and if you start to feel bad because of something he says, just tell yourself that it's on him. he's the teacher and he's supposed to help you when you don't get something. if u still don't go in for help, ask your friends to help you, or if there is a math club try going to some tutoring sessions where other students tutor you.
Answers:Sorry, I havent taken it. ALG 2 ISNT ON THE TEST!! ITS 10th GRADE.. I got it confirmed from the makers. ITS ONLY ALL OF GEO(which has basic Trig). Tips...keep practicing with the time you have, a good 1 month or so. Redo all of geo. Barrons SHSAT is good and challenging. Im trying the same exact thing, brooklyn tech. GL, its in the 530s, EZ lj4lyfe6~AIM, firstname.lastname@example.org, email@example.com
Answers:Rather than a graphical technique, I prefer an analytical technique. Assuming the standard x-y coordinate system, then you can use the Pythagorean Theorem to calculate the length of each side of the original triangle. Simply take the difference in x-components squared plus the difference in y-components squared and square-root. For instance, side AB has a length of: AB = [(-5 - 7)^2 + (2 - 4)^2}^(1/2) = 12.166 units Similarly, the lengths of BC and CA are both 13.601 units. So one immediately determines that this is an isosceles triangle. The point D is midway between A and B (the midpoint between -5 and 7 is 1, the midpoint between 2 and 4 is 3) so the line CD is a bisector of the line AB. Because this is an isosceles triangle and the line is bisecting the unequal leg, then it is a perpendicular bisector of the line AB (it is splitting the triangle into two congruent triangles with mirror-symmetry). It is also bisecting the angle ACB (I suppose I could calculate the angles, but that would be a waste of time since it is obvious). Of course, the definitions for altitude (in a triangle, the altitude of a triangle is the perpendicular acute segment from a vertex to a line that contains the opposite side) and median (a median of a triangle is a line segment joining a vertex to the midpoint of the opposing side) also apply to the line CD. So the answer is that line segment CD is all four.