*Official* November 2014 Math 2 Thread

<p>do u remember what the exact problem for 49 was? i dont think i got that</p>

<p>wait I got like xz = wy for that one;</p>

<p>here’s an example:</p>

<p>a = 3, x = 2, b = 9, y = 1
w = 4, z = 2</p>

<p>3^2 = 9^1
3^4 = 9^2</p>

<p>We have xz = 2<em>2 = 4 = wy = 4</em>1.
x + y = 3 ≠ 6 = w + z.</p>

<p>@happiface‌ same</p>

<p>can someone check these? i only can think of these problems. also if you remember more (even the super easy ones), can u post them

  1. 2^(3/5) = 1.516
    from f(g(x), f(x) = 2/sqrt(x)
    sqrt(16-x2)/2
    sin/cos 22.5 -> ii only (sin2 + cos2 = 1)
    standard deviation 15 -> 1.2
    parametric = -y2-1
    vector 130 = 11
    series sum 2x+1 = 71
    xz = wy
    distance btwn 2 3d points (forgot answer, only remember question)
    there is an x-intercept
    period of 8 -> 15
    y/4 = 1/(2-2x) (something like that)
    29 degrees for 10/10/5 triangle
    2 distinct real roots -> (x2-1)(x2+1)
    355 chair -> 25
    symmetry -> (0.2)
    2sin2x area = pi/2
    inverse the same -> i and iii only (1/x and 1/x-1 + 1)
    compounded interest, months = 105
    -didn’t get the temp/month one or the last triangle one</p>

<p>For the compound interest question, the question wanted the time(in months) it takes for the initial amount (1000) to double using an 8 percent annual interest rate. So, the answer would be 108 months because, 2000 = 1000(1+0.08)^x. So 2 = (1.08)^x and by using logs, X would come out to be 9. However this value is in years, so by multiplying 9 by 12, you get 108 months. </p>

<p>108 Months</p>

<p>i left like 6 blank andi got three wrong that i know of. Maybe a 770?</p>

<p>no but it was compounded monthly so it would be 2 = (1 + 0.08/12)^x</p>

<p>I rushed the last question so I don’t remember it to clearly, but if the question was just asking for the side length then you could have used law of sines to find the angles required and used law cosine to find the side length.</p>

<p>Anyway as stated above, discussing concepts on the test or used to solve the problem is find but we should avoid using the actual questions/answers.</p>

<p>did you guys use law of cosines for the vector sum?</p>

<p>for the vector sum i used Law of cosines and i got 6.2 or something…probably not the right approach or…? i just verrified with my calculator…and some people are getting 11 or somethin…idek</p>

<p>If it was compounded monthy, then doesnt it have to be 2 = (1+0.08/12)^12(x)</p>

<p>Vector sums are tip to tail which would, in a sense, make it a triangle. Thus if you know two sides and the angle of the 3rd side Law of Cosines is the ideal route to go.</p>

<p>@emotionalboyz‌ yeah i think so, but you have to multiply x by 12 anyway since x in that equation = number of years</p>

<p>For the annual interest question;</p>

<p>2 = (1+0.08/12)^12t;</p>

<p>Time is measured in years (though the interest is compounded monthly, the rate is still annual). Therefore, we would solve for 12t actually.</p>

<p>log(2) = 12t[log(1+1.08/12)]</p>

<p>12t = [log(1+1.08/12)]/log(2)</p>

<p>12t ~ 105</p>

<p>For the second last question (ship and two radars)</p>

<p>I did this with the “law of sines”. In the larger (given) triangle we know that angle C is 32, side BC = 10.6, AC = 6. We can then solve for angle A to get 69.42. This way we can get the angles for the second triangle formed by drawing a line from point B (the first sonar to a point A’ (representing ships new location) on line AC. In this new smaller triangle on the right, we have angle B’ and the internal angle A’. We also have isoscles triangle AA’B on the left from which we can calculate angle A’ and therefore angle B’ [A’ = 110.58 and B’ = 37.42]. Doing a second round of “law of sines”, we find distance from the second sonar which is A’C = 10.6 x sin(37.42) / sin (110.58) = 6.88</p>

<p>For the area of the triangle under the sine graph:</p>

<p>Area = 1/2 x base x height
Height = Amplitude = 2
Base = 1/2 period = pi/2 [f(x) = 2sin(2x) so the period of the function is pi. The base is equal to only half the period since we have only half a wave).
Area = pi/2</p>

<p>For the x,y,w,z question:</p>

<p>a^x = b^y
log(b)/log(a) = x/y</p>

<p>a^w = b^z
log(b)/log(a) = w/z</p>

<p>Therefore: x/y = w/z
So: xz = wy </p>

<p>I might have mixed up the order of the exponents but the concept remains the same.</p>

<p>Hey, quick question. From what I can gather, I got 3 wrong and left 3 blank. What score do you guys think that is with this test’s curve?</p>

<p>Depending on the curve, 780-790
It is possible that the curve is at 5 total points lost = 800 (3 wrong = 4 points)
I don’t think this test was hard enough to call for a curve of 7 = 800.</p>

<p>What did you guys get for the inverse question?</p>

<p>Never mind. Figured it out.</p>

<p>i thought this was hard lol</p>

<p>what was the answer to hte one with the -y|-y| < -y|y|? was it all real numbers?</p>

<p>I think it was none, because |-y|=y and |y|=y so the values will always be equal</p>