74Heaven Posted June 30, 2009 Posted June 30, 2009 we need help on several questions on this test = we have have been struggling and digging thru our books ofr days and hours and still can't get these and the college placement test is tomorrow. I would love some help.... Thist is the WA state math 'placement test' for several colleges - this is the practice test site - [website follows] we need help on problems on the advanced test for numbers 13,21,22,26,27,28, 29, 30, 31, 42 [we are esp frustrated on numbers 21,22,28] here is the website http://www.washington.edu/oea/services/testing_center/aptp/AdvancedSampleProblems.pdf THERE IS Another test that we are having problems with on the same site - that is the 'intermediate version of the test listed above on the same site - on the intermediate test, we are stumped on numbers 10,20, 37, 41 I know this is a huge request, but we really need to get these figured out and we are at the end of our expertise. We also have dial up so visiting all kinds of math helps sites is tedious and not very helpful. thank you for any help you can give. lisa j - ljdeerparkATaol.com Quote

kiana Posted June 30, 2009 Posted June 30, 2009 (edited) 21 - use similar triangles, x/3 = 5/10 and solve Edited: I meant (x+5)/10 22 - split it up, -3 < x+5 < 3, solve 28 - add 4x to both sides, subtract 5 from both sides, get 4x > -3, solve. Don't stress too much over the test -- you want to be placed in a passable course or even one that's slightly easy. Additionally, I found the cutoffs, it looks like 14/30 is precalc 1, 17/30 is precalc 2, 21/30 is calc 1. Edited June 30, 2009 by kiana I'm a doofus Quote

74Heaven Posted June 30, 2009 Author Posted June 30, 2009 for number 21, for the triangles x/3 = 5/10 = 3/2 and the answer key [on the last page of the test] says the answer to number 21 is 15/7th = x and I didn't think we could use similar triangles is the test did not tell us that the triangles were similar... I am not debating, I am just lost.. thanks... I think I understand number 22 and number 28 now... thanks... Can you tell me where you found the cut off scores of 14/30 - precalc 1; 17/30 precalc 2 and 21/30 is calculus.... lisaj Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 and I didn't think we could use similar triangles is the test did not tell us that the triangles were similar... They didn't explicitly tell you the triangles are similar, but you have enough information to see that they are similar: the acute angle in the lower left is common to the two triangles, and they both have a right angle. Therefore, all three angles are the same, which means they are similar. Off to print off the test and work thru the problems. I'll be back. Quote

kiana Posted June 30, 2009 Posted June 30, 2009 for number 21, for the triangles x/3 = 5/10 = 3/2 and the answer key [on the last page of the test] says the answer to number 21 is 15/7th = x and I didn't think we could use similar triangles is the test did not tell us that the triangles were similar... I am not debating, I am just lost.. thanks... I think I understand number 22 and number 28 now... thanks... Can you tell me where you found the cut off scores of 14/30 - precalc 1; 17/30 precalc 2 and 21/30 is calculus.... lisaj I lost the website now, I found it through google. They had a listing of cutoff scores. The triangles are similar because they have the same angles, the lower left corner is the same angle and they both have a right angle. I apologise for mistyping, it should have been set up as x/3 = (x+5)/10. Sorry for confusing you! Quote

74Heaven Posted June 30, 2009 Author Posted June 30, 2009 They didn't explicitly tell you the triangles are similar, but you have enough information to see that they are similar: the acute angle in the lower left is common to the two triangles, and they both have a right angle. Therefore, all three angles are the same, which means they are similar. Off to print off the test and work thru the problems. I'll be back. bless you... thanks - I see now that the triangles are similar - I still don't understand the answer being 15/7th.... and Kiana, I found the cut off scores are listed on each college's website link that is linked as part of the math test placement site webpage. lisaj, still working and printing problems as well - Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 Let's get rid of the denominator on the left by multiplying both sides of the equation by c, then: xc + bx = ac factor to get x(c + b) = ac divide both sides by (c +b) and remember that (b +c) is the same as (c +b), and we get the correct answer Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 21 - use similar triangles, x/3 = 5/10 and solve the correct proportion is: x/3 = (x +5)/10 which gives x = 15/7 Quote

kiana Posted June 30, 2009 Posted June 30, 2009 the correct proportion is: x/3 = (x +5)/10 which gives x = 15/7 Yeah, I fixed that in my next post but I'll go back and edit that post as well. Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 On absolute value / inequality problems, it's often easiest to test the possible answers to see which works. (It's really easy to get the < /> signs turned around.) So, here's how I would approach a problem like this: I see from the possible answers that -8 and -2 are the key points. I'd quickly sketch a number line, mark those points on it, and test the answers. I like to start with the middle answer, and go from there if it doesn't work. The first part of answer b says x<-8, so I choose -10 for x (because it's easy to work with and less than -8). If I put -10 in for x in the original inequality, I see that it holds true. I also check the x >-2 by choosing x=0. I see that this is also true, so I'm feeling really good about answer b. Answer a is only partly true, and answer c is never true. Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 Yeah, I fixed that in my next post but I'll go back and edit that post as well. sorry -- I didn't see that you'd done that. I think we were posting at the same time. Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 bless you... thanks - I see now that the triangles are similar - I still don't understand the answer being 15/7th.... from the proportion x/3 = (x +5)/10 we can cross multiply: 10 (x) = 3 (x + 5) distribute 10x = 3x + 15 subtract 3x from both sides 7x = 15 divide both sides by 7 x = 15/7 HTH Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 On graphs like this, pick a few numbers for x and see what it looks like. First, I'd choose x = 0: any number raised to the 0 power = 1, so that eliminates answer b. Now, we need to determine whether the value of y goes up rather quickly, as in answer c, or more slowly, as in answer a. If you plug in 1, 2, and 3 for x, you'll get 2, 4, 8, so answer c is best. HTH Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 Assuming you can use a calculator, pick a value for theta, say 30 degrees, and see which holds true. If you're picking a value for an angle, it's best to choose something less than 90, but not 45 degrees. Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 For this one, you need to remember the equations for the trig functions in a right triangle: use the mnemonic SOH CAH TOA (I'm going to use x rather than theta because it's easier to type) sin x = side opposite/hypotenuse cos x = side adjacent/hypotenuse tan x = side opposite/side adjacent In the figure, sin x = b/c and tan x = b/a so, sin x times tan x = (b/c) times (b/a) which = b^2/ca, which is answer c Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 for both of these, you want to be working with angles measured in radians rather than degrees #30: Pick a value for theta (in radians) and see which answer works #31: Here, you want to find what values of will cause y = 0. Since I've mostly been teaching Algebra 1 and Geometry the past several years, I don't remember these off the top of my head. But I would test each graph by plugging in the values of x where the graph crosses the x axis (and y=0). When you do that, you should see that a works. Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 You've got me -- I honestly don't remember seeing this nomenclature. I do see that the domain (possible values of x) can't be less than -2, but can go infinitely high, so I'm thinking that #3 is true. That eliminates answer a. The range is the possible values of f(x), and since that's a square root, it can't include -2, so #4 is out. That would leave us with answer b. Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 I actually found it fun to work thru these problems, LOL. Quote

MyThreeSons Posted June 30, 2009 Posted June 30, 2009 5 - 4x > 2 add 4x to both sides 5 > 2 + 4x subtract 2 from both sides 3 > 4x divide both sides by 4 3/4 > x which is the same as x < 3/4, which is answer a If you check their answer (x > 3/4) by testing with x =1, you get 5 - 4 > 2 or 1 > 2, which doesn't work on my planet, LOL. Quote

kiana Posted July 1, 2009 Posted July 1, 2009 5 - 4x > 2 add 4x to both sides 5 > 2 + 4x subtract 2 from both sides 3 > 4x divide both sides by 4 3/4 > x which is the same as x < 3/4, which is answer a If you check their answer (x > 3/4) by testing with x =1, you get 5 - 4 > 2 or 1 > 2, which doesn't work on my planet, LOL. Yep, they switched it. Quote

74Heaven Posted July 1, 2009 Author Posted July 1, 2009 from the proportion x/3 = (x +5)/10 we can cross multiply: 10 (x) = 3 (x + 5) distribute 10x = 3x + 15 subtract 3x from both sides 7x = 15 divide both sides by 7 x = 15/7 HTH I got it - and so did my dtr.... thanks so much really - you all have went over and above lisa Quote

74Heaven Posted July 1, 2009 Author Posted July 1, 2009 On the intermediate test - if anyone can help with these problems on the intermediate test - I would appreciate it - number 10 on the intermediate test must have the wrong answer... D is correct if it is 2.0 x 10 -4power 26 says if 9 to the 3x power = 3; then x = [and the choices are A 1; B 1/3 C 1/9 D 1/6 E 2 number 37 on the intermediate test number 41 on the intermediate test number 47 on the intermediate test [ I don't see how the answer could be B. it seems to me if chemical A is x; then the answer s/be answer A.] thanks again - this is great assistance - I am a math nerd wanna-be Quote

Martha in GA Posted July 1, 2009 Posted July 1, 2009 If 9^3x = 3: 9 is 3^2, so 9^3x = 3^6x; Setting 3^6x = 3^1 (3^1=3), setting your exponents equal to each other, 6x = 1, so x= 1/6. This is much harder to explain using the keyboard! Hope it is understandable. Martha Quote

74Heaven Posted July 1, 2009 Author Posted July 1, 2009 If 9^3x = 3: 9 is 3^2, so 9^3x = 3^6x; Setting 3^6x = 3^1 (3^1=3), setting your exponents equal to each other, 6x = 1, so x= 1/6. This is much harder to explain using the keyboard! Hope it is understandable. Martha thanks again, we are knocking these down one by one... lisaj [my dtr asks if this is a 'math help' site - ''well no, it is a homeschool mom help site....] Quote

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