Q 37, Somebody borrows $30000 with 15% interest Per Anuum. He has to pay it all off after 20 years in equal monthly repayments. He also gets the first 6 months interest free. Find
a) the monthly fee
b) how much he pays altogether
Oops looks like I was too slow. And adgala, there's a mistake in your second answer. Anyway, make my last question on Exponential Decay 37(b) =D
Answer to 37(a):
Q 37, Somebody borrows $30000 with 15% interest Per Anuum. He has to pay it all off after 20 years in equal monthly repayments. He also gets the first 6 months interest free. Find
a) the monthly fee
b) how much he pays altogether
P = 30,000
r = 15% per annum
= (15/12)% per month
n = 12 yrs = 240 months
Let A(n) be the amount owing after n months
Let M be the equal monthly repayment
Let R = 1 + 15/1200
The first 6 months are interest free, therefore:
A(1) = 30000 - M
A(2) = A(1) - M = 30000 - 2M
...
A(6) = 30000-6M
interest starts being charged at the 7th month:
A(7) = A(6) * R - M
= (30000-6M)R - M
A(8) = A(7) * R - M
= (30000-6M)R2 - MR - M
= (30000-6M)R2 - M (R + 1)
...
A(n) = (30000-6M)Rn - 6 - M (1+ R + R2 + ... + Rn-7)
I didn't see a question 38?
Ah well, answer to Q38:
LHS=(cosx/sinx+1/sinx)2
=[(cosx+1)/sinx]2
=(cosx+1)2/sin2x
=(1+cosx)(1+cosx)/(1+cosx)(1-cosx)
=(1+cosx)/(1-cosx)
=RHS
Hence,
(1+cosx)/(1-cosx)=1/cosx
cosx+cos2x=1-cosx
cos2x+2cosx-1=0
using quadratic formula
cos x= -1+sqrt2 or cosx=-1-sqrt2 for 0<2x<4pi
2x=65o32', 294o28', 425o32', 654o28'
.: x=32o46', 147o14', 212o46', 327o14'
Q39: Given that the 2 roots of a quadratic are α and ß, show that the sum of the roots of the quadratic equation ax2+bx+c=0 is given by -b/a and the product of the roots is given by c/a.
Q39: Given that the 2 roots of a quadratic are α and ß, show that the sum of the roots of the quadratic equation ax2+bx+c=0 is given by -b/a and the product of the roots is given by c/a.
Q41:
Jeremy goes to the casino and plays a new game involving two unbiased dice in which the sum of numbers on the uppermost faces is recorded after each throw. If the sum is 7 or 11 after the first throw, he wins at once. If the sum is 2, 3, or 12 on the first throw he loses at once. If the sum is any other number (apart from 2, 3, 7, 11, or 12) on the first throw, this number is noted and becomes his "score". Jeremy then wins by throwing his "score" again or loses by throwing a sum of 7. Show that in the long run the odds for winning the game are in the casino's favour.
Q41:
Jeremy goes to the casino and plays a new game involving two unbiased dice in which the sum of numbers on the uppermost faces is recorded after each throw. If the sum is 7 or 11 after the first throw, he wins at once. If the sum is 2, 3, or 12 on the first throw he loses at once. If the sum is any other number (apart from 2, 3, 7, 11, or 12) on the first throw, this number is noted and becomes his "score". Jeremy then wins by throwing his "score" again or loses by throwing a sum of 7. Show that in the long run the odds for winning the game are in the casino's favour.
1st throw
So the probability of winning on the 1st throw is 2/9, while the probability of losing is 1/9
2nd throw
P(lose on 2nd throw) = P(sum of 7) = 1/6 NB: Does not depend on the previous event (i.e. 1 * 1/6)
P(sum of 4) * P(sum of 4) = P(sum of 10) * P(sum of 10) = (3/36)^2 = (1/12)^2 or
P(sum of 5) * P(sum of 5) = P(sum of 9) * P(sum of 9) = (4/36)^2= (1/9)^2 or
P(sum of 6) * P(sum of 6) = P(sum of 8) * P(sum of 8) = (5/36)^2
Thus, P(win on 2nd throw):
2 * (1/12)^2 = 1/72 or
2 * (1/9)^2 = 2/81 or
2 * (5/36)^2 = 25/648
Thus, you are nearly twice as likely to lose than to win in the long run for each subsequent throw after the first throw. Thus the odds are favourable for the casino.
I'm not to sure how to format this answer, I'll do my best.
Let I stand for the integration from 0 to 12,
A = I (12-x)^1/2.(12+x)1/2
A = I [(12-x)(12+x)]1/2
A = I (144-x2)1/2
F(A) = {(144-x2)3/2/-3x}
Sub integral limits,
and you get a division by 0 and I cry in shame not being
able to see where my mistake lies.
Question 44
Prove that the sum Sn of an arithmetic series is 1/2 n (a+l) where n is the number of terms in the series, a is the first term and l is the last term.
evaluating this integral is equivalent to the area of the first quadrant of a circle radius 12 units
so, Sqrt(12-x) * Sqrt(12+x) from 0 to 12 = 144pi/4 = 36pi
Question 43
Find the values of k such that k(k + 3) + (k + 3)x - x2 is a negative definite.
The standard form I (ax+b)^n = (ax+b)^(n+1) / a(n+1) + C is only for linear equations of x (i.e. ax + b). In the question, a quadratic is being raised to a power, so you cannot use that standard form. Please see "word."'s solution to the problem below, which is the only way to do the integral using 2U level mathematics. Note that in 3U maths, the integral can be solved using trigonometric substitution of x = 12sin(u).
MATHS IS GAAAAAAY
I mean sure there woz a time i wantd 2 be a algebra manager & perhaps evn a top class bracket expander
BUT COMEOn PPL do we really need %90 of this shit they teach us i got all that shit loaded into my brain then wen it cums 2 da algebra in physics frgrt
Answer to question 44:
Prove that the sum Sn of an arithmetic series is 1/2 n (a+l) where n is the number of terms in the series, a is the first term and l is the last term.
Sn = a+(a+d)+(a+2d)+...+l
Sn = l+(l-d)+(l-2d)+...+a
2Sn = (a+l)+(a+l)+(a+l)+...+(a+l)
2Sn = n(a+l)
.:. Sn = (1/2).n(a+l)