UIL Math Questions

useless · Long-term Member

UIL Math Questions

bradp · Senior Member

7)

Right triangle a^2 + b^2 = c^2, acute triangle a^2 + b^2 > c^2, obtuse triangle a^2 + b^2 < c^2

44)

Harmonic mean of x and y is (G^2)/A where g is geometric mean and a is arithmetic mean. (g^2)/a simplifies to (2xy)/(x+y)

I'll answer some more when i get back from drivers ed

-- Edited by bradp at 07:33, 2006-06-08

bradp · Senior Member

8) Let point P be the centroid of triangle ABC. Let M be the midpt of BC. Find AP if MP is 14

medians intersect at the centroid. from http://www.andrews.edu/~calkins/math/webtexts/geom06.htm

"Medians always divide each other into a 1:2 ratio, with the larger portion (2/3) toward the vertex and the smaller portion (1/3) toward the opposite side"

so 14 is the 1/3 side, double it to get 28 for the 2/3 side, AP

29) take an equilateral triangle with sides 1 (perimeter of 3)

a = 1/2 b h

b = 1, h = sqrt(3) /2

a = sqrt(3) /4

perimeter / area = 3 / (sqrt(3) / 4) = 12 / sqrt(3) (simplify radical in denominator) = 4sqrt(3)

on the calculus problem, i don't know how to do that of the top off my head right now. i can only do more basic ones like the ones on NS tests and easier ones on mathematics tests(took algebra II this school year). you could probably figure it out without knowing how to actually do it by using your calculator

-- Edited by bradp at 10:23, 2006-06-08

fht · Long-term Member

33. I believe it was one of the Taylor's series. The function is sin(x)=x-x^3/3!+x^5/5!-x^7/7!..., so the answer is whatever sin(2) is.

34. Let f(x)=4x-x^2, and g(x)=x^2. Find the x-coord of the intersections, a and b, where a<b. And plug this in your calculator:

MATH -> fnInt: at this time, the screen should show fnInt( , then plug in the following parameters, fnInt(f-g, x, a, b), and enter, and the absolute value of the returned value is the area.

I will not discuss Calculus here, it's a matter of self-studying.

35. Because 24/5 is the sum of A and a fraction <1, so A is the greatest int <24/5, 4. If A=4, obtained remaining fraction's value =4/5. Flip the value to 5/4 to get rid of the 1/ over the part with B. B will be the greated int <5/4, 1. If B=1, 1/(C+1)=5/4-1=1/4, so C+1=4, C=3.

A+2B+3C=4+2(1)+3(3)=15

44. Harmonic means of n numbers is: n/(1/A1 + 1/A2 + 1/A3 +...+ 1/An)

8. The concurrencies of important lines in a triangles can be summarized as follows:

medians intersect at centroid; perpendicular bisectors at circumcenter; angular bisectors at incenter; altitudes at orthocenter.

useless · Long-term Member

Thank you both very much! Everything here makes perfect sense, even the method on how to obtain the area of two intersecting functions that have a bounded region between them. Thanks to you guys, and a bit of hard work on my part, I scored my first 200 on the UIL Math test, (On District I of 2003, easy test, I know...) and that is awesome for me especially since I got a 116 the pervious night on another test (7 reeeeeeeeeeeally dumb mistakes).

I still have a few questions, though.

6) Points A (3,1), B (6,-1), C(8,1), and D(x,y) form rectangle ABCD. Find x+y.

Is there a faster way than just taking the slope of BC and adding the change of y and x to point A's coordinates? I hate these because they take too long to do. I spent 4 fruitless minutes on it only to obtain the answer after the test was over.

9) A polyhedron has 9 faces and 9 verticies. How many edges does is have?

Formula?

13) One of the roots of ax^2 + bx + c = 0 is 3+5i. Find the discriminant when a =1.

I know the other root is 3-5i. How do we obtain the discriminant from there?

Strangely enough, numbers 14-41 were all easy.

Thanks again,

- Zack -

bradp · Senior Member

6) Points A (3,1), B (6,-1), C(8,1), and D(x,y) form rectangle ABCD. Find x+y.

think of A and C as opposite points... b is 2 below the diagonal AC, so, in order to equalize, d must be 2 above making D's y coord 3. now x

if b and d were on the same vertical line as each other, they'd be half way between a and c, 5.5. but since b is +.5 more, d is -.5 so its x coord is 5

5+3 = 8 (dunno if thats the right answer)

9) faces + vertices = edges + 2

fht · Long-term Member

13. The integer part is trash. What's important is the complex part because in quadratic formula, this is the part Sqrt(b^2-4ac)/(2a). a, in this case given, is 1, so Sqrt(b^2-4ac) in this case is 10i because of the 2 at the bottom. So b^2-4ac, the discriminant has to be -100.

Or:

since for quadratic root x1, x2 of x^2+bx+c=0, it suffices that x1+x2=-b, and x1x2=c, so (3-5i)+(3+5i) = -b = 6, and (3+5i)(3-5i) = c = 34. Thus b^2-4ac = (-6)^2-4(1)(34) = 36-136 = -100. Bingo.

Your test sounds like the Regional test this year. On the first page I skipped like 4, which scared the crap out of me.

useless · Long-term Member

Ok... You'll have to forgive me Sam, but I've forgotten how to generate quadratic formulas when given the roots. I remember some easier way than just plugging in (x-(3+5i))(x-(3-5i)).

I think it has to do with using the sum of the roots...

Maybe since the sum of the roots is 3+5i + 3-5i = 6, then -b/a is = 6.

-b/1 = 6. Ergo, b = -6. You can figure out c by using the product of the roots...

c/a = 34, c/1 = 34. c = 34.

36 - 4(1)(34) = -100.

YAY! However, I dont understand your faster way. Did you subtract 3+5i from 3-5i? How did you get to 10i?

ALSO, I've got one more question:

21) cot(theta + pi/2) = ?

The answer = -tan(theta).

How do you get to that answer? I tried using the angle addition formula for tangent, but I couldnt get that answer.

- Zack -

fht · Long-term Member

No, look at the quadratic formula: the only part that can get you the complex part of roots is +-Sqrt(b^2-4ac)/2a. And since a=1, so +-Sqrt(b^2-4ac)/2a = +-Sqrt(b^2-4ac)/2, which according to the complex part of the given root, is +-5i, so Sqrt(b^2-4ac)/2=5i, ergo Sqrt(b^2-4ac)=10i. And you square it to get the discriminant inside.

21. Try looking at it graphically.

Best Wishes