Hi everyone. August 25th I took the CBEST Math exam and failed again for the fourth attempt. i got a scaled score of 31 but I have improved so much since the first three times, what I am a little unnerved by is that there were certain problems that were not discussed in workshops or even in any of the many copies of the books i studied out of. For example there are various hexagons in array of designs that asks to fine th perimeter. another problem was weight for example 58 blocks is equivalent to 27 lbs. 1 block is equivalent to X pounds but round to nearest hundredths... the questions are not very clear on those type of problems... theres a few other that i struggled with but i don't remember.... my brain is a little overwhelmed from yesterday... I am nonetheless going to be reviewing yet again so i can retake in a little over a month...

[QUOTE="cma211991 the CBEST Math exam and failed again for the fourth attempt. hexagons in array of designs that asks to fine the perimeter. 58 blocks is equivalent to 27 lbs. 1 block is equivalent to X pounds but round to nearest hundredths......[/QUOTE] Hi Cma: Have you tried TeachersTestPrep dot com? They were wonderful in helping me pass the CSET MS exam. They have online prep videos to help with passing CBEST. As for the hexagon problem...hmmm, if I remember correctly, count the number of sides, then multiply the side length by 6 ...(double-check that ). The 58 blocks question is a proportion. 58/27 = 1/X, find the answer, then round it off. Good luck

Perimeter is literally 'measuring around' the outside of a polygon. If the question specifies that a single polygon is regular, that means that all the sides are the same length and all the angles have the same measure, and then the strategy of multiplying the length of one side by the number of sides will work. A square is regular, so if one of its sides is 2 meters in length, the perimeter of the square is 2 meters • 4 sides = 8 meters. In the case described here, however, the regular polygons make up a bigger polygon - which won't necessarily be regular itself. Furthermore, there are sides of the regular polygons that aren't part of the greater polygon's perimeter, because they're not outside. To find the greater polygon's perimeter, it won't work to multiply the length of one side of a little polygon by the number of sides of that polygon and then by the number of polygons: that answer will be too big, because it includes sides that aren't actually part of the perimeter of the greater polygon. A good way to model this is to use toothpicks and think of Tetris. Build a square with toothpicks: if we decide each toothpick has a length of 1 unit (and we can: this is math, not reality), then the square's perimeter is 4 units. Now add three more squares onto this square to make one of these classic Tetris shapes: I-shape, L-shape, T-shape, four-square square. The first thing you'll notice is that you're not using FOUR toothpicks to build each added square: you're adding at most THREE, so it takes 13 toothpicks, not 16, to build the straight bar or the L-shape or the T-shape and just 12 to make the four-square square. Now remove any toothpick that isn't on the outside of the larger shape(s). The number of toothpicks left per larger shape is what you'd multiply by the length of one toothpick to get the perimeter of the Tetris shape. For the shapes that aren't square, that's 10 units; for the square, that's 8 units. The shapes on the test are rarely this simple, of course, and one won't have toothpicks. That being the case, the best bet is just to count the number of one-unit sides that make up the perimeter, then multiply by whatever length the problem specifies for the units. For instance, with the Tetris L-shape that's one unit at the top (north) side, three units down the left (west) side, two units across the bottom (south) side, one unit up on the right (east) side, one unit back on the short leg of the L, and two units up to the beginning: 1 + 3 + 2 + 1 + 1 + 2 = 10 units. If the unit is supposed to be 3 inches, the perimeter would thus be 10 units • 3 inches = 30 inches.

Thank you so much, this is soo helpful! are there any resources that might help me practice those type of problems?

The technical term for shapes that fit together the way squares or hexagons do without gaps or overlaps is "tesselation", though this can also be called "tiling". You could look up tesselation or tiling problems and then pick out the ones that look like what you want to work on. Alternatively, you could play with this concept on your own. You could look online for a blackline master (that is, a reproducible sheet) with simple regular shapes an inch or so in height: print the sheet on cardstock and cut out the equilateral triangle, the square, the hexagon, and the octagon. (Pentagons are cool but don't tesselate neatly; this, by the way, is why the outside of a soccer ball isn't totally made up of pentagons. Diamond shapes tesselate neatly, though they're not regular.) Choose one shape; put it in the middle of a piece of paper, trace around it, then shift it so one edge is against the shape you just traced and trace it again, and so on, till you've made a shape as complex as you feel like making. Then you can count all of the outside segments, marking them in a different color as you go so you make sure you get them all.