25) If log(base 9) x = klog(base 3)y . If y = x^a, then a = ______________
I did this: log(base 9) x = 2 log(base 3) x = ka log(base 3) x. You then get 2 = ka and 2/k =a.
Thats wrong. The answer is 1/2k. I was wondering what I did wrong.
42) The slant asymptote of y = (12x^2 - 9x +1) / (4x-3) is: ______________
No clue.
46) The line 2x + by = c is tangent to the parabola y = x^2 + x + k at the point (2,-1). Find the value of c.
Calc?
49) The total surface area of a right circular cone is 100 sq uints. If the slant side forms an angle of 20 degrees with the base, find the volume to the nearest cubic unit.
How can you do this one without the slant hight?
50) The graph of r = 4sin(theta) is a _________
(answer - circle) I used my graphing calculator and got this wrong. How do you do this one?
52) A triangle has sides of 8 and 13, with an included angle of 48 degrees. To the nearest hundredth unit, find the radius of its inscribed circle.
No clue.
60) Let the R value of a solid be the numerical ratio of its surface area to its volume. Find the R value for a sphere if the sphere is inscribed in a cube whose R value is 2.
I found that the side of the cube must be 4. But upon trying to find the R value of the sphere I got 1.5, which isnt the right answer. ug.
25) kept trying to do this one and i screwed up everytime... not sure here
"I did this: log(base 9) x = 2 log(base 3) x"
that's not correct.. should be 1/2 log(base3) x = log(base 9) x, therefore ka = 1/2, a = 1/(2k) (you meant to put parentheses around 2k right?)
i might have done this wrong; i think y is supposed to be involved in the solution..
42) Slant Asymptotes: A rational function has a slant asymptote if the numerator is one degree higher than the denominator. The equation of this asymptote is found by dividing (long or synthetic) the numerator by the denominator. Take the remainder off of this result and that is the equation of the asymptote (it's linear)
49) try using sin or cos or tan of 20 find the proportions of the radius vs. slant height vs. height. all cones w/ 20 deg angle between frustum and base must be similar.. so the trick is finding the one with a SA of 100. that should work
60) side of cube must be 3: SA = 6(s^2), V = s^3 R = 2. Through some math skills- s = 3 because SA is double the volume so 6 must be twice as much as s (compare 6(s^2) to s*(s^2) )
46. Given (2,-1) on y = x^2 + x + k, k = -7. And by differentiating y = x^2 + x + k, obtain that on (2,-1), slope = 4. Then if the tangent has a slope of 4, b = -1/2, and since the line has (2,-1), c can be known also.
Here is how you find the slope of f(x) @ a:
MATH -> 8 -> nDeriv( -> nDeriv(f, x, a) -> enter
50. MODE -> "fun(ction)" to "pol(ar)" -> enter -> y= -> r = 4sin(theta)
On 46, I took the derivative of the first function, but never found the connection between the slope of the line and the slope at point (2,-1). But now that it isnt 2:00 in the morning it makes perfect sense:
First, like you did, solve for what k is by using the point (2,-1)
y = x^2 + x + k ==> -1 = 4 + 2 + k ==> k = -7.
Next, take the derivative of the quadrantic function, so that we can find the slope of the line at the point (2,-1).
Der(x^2+x-7) = 2x + 1. The slope is: 2(2) + 1 which equals 5.
Now that we know the slope, we need to find what the b value of the linear function:
2x + by = c. ==> y = 2x/b - c/b, (This is the part that threw me) 2(-1)/b = 5 ==> b = -2/5 2(2) + (-2/5)(-1) = c
c = 4.4. Yay!!!!!!!!
Anyways...
I was wondering if anyone could explain Brad's approach to the cone problem. And nobody has explained 52 yet. I've been playing with the numbers, but nothing has really come of it yet.
I'm starting to be a consistant 200+ scorer now on the Invite and District tests from 2003-2005 tests, which is strange considering I sucked this year at the meets. I guess its because I'm getting to around the 45th problem which is some 15 better than what I did at the meets. I think its because of the Number Sense. That, and I've found that many of the questions are repeated, in UIL that is. In my opinion, I think that the TMSCA tests are a bit more challenging. They seem to have several one-step problems that require knowing obscure things, and I like that. Anyway, I'm really starting to get into the UIL mathematics contests, and I hope that come next May, I can give you guys a run for your money (Not really, just making it to state would be awesome).
"49) The total surface area of a right circular cone is 100 sq uints. If the slant side forms an angle of 20 degrees with the base, find the volume to the nearest cubic unit."
i'll give a more in depth explanation, say l is the slant height
SA of cone = 100 = pi * r * l + pi * r * r
now, cos(20) = r/l - getting the value of cos20 will give us something like .4l = r (im just making up .4 cuz i dont have a calc with me.) now just put in .4l for all of the radii and solve for l... you now have the slant height and can find radius and height using sin20 cos20 etc. then take those variables and solve for the volume
and with 200s.. you're not only giving me a run for my money but beating me.. all three of my scores (inv a, inv b, distr) last year were right around 160. But I will improve.
However, I'm not going to post my methods yet, but I will post my answer. And Zack, if it's right, tell me and I'll post the method.
r = 2.519
Is it right?
One more comment on that derivative problem. Even if you haven't taken Cal yet, it's important for you to know that the derivative of a function at a given point is the slope of the function ergo (I like that word from "cogito ergo sum") the tangent at that point. It's the rate. Thus if f is about distance, f' is speed. If f is speed, f' is acceleration.
I'll look at the other perplexing problem tomorrow. I'M CONSUMED IN SOCCER NOW. Matches tomorrow have France AND Brazil --- VERY EXCITING.
You were really close on 49, Brad. Your slant length and radius were right on. The height you obtained was a little off. I got 1.429, which gives you a volume of 23, which is the answer. Ka ching! I can't believe I didnt think of that... I've been out of physics for too long.
And you are 100% correct on #52 Sam.
Thanks for all your help so far, you guys. Y'all rock.
Ah... That makes sense because of Hero's formula. The derivation looks kind of daunting just based on the intial algebra. Thanks for the formula. Its awesome!
, if my answer is correct, then I declare with confidence that the radius of inscribed circle in a triangle is (drumroll):
