Topic : Polygon
Question : List the names of Ploygons
Solution :
Sides ------ Name
n ----------N-gon
3 ----------Triangle
4 ----------Quadrilateral
5 ----------Pentagon
6 ----------Hexagon
7 ----------Heptagon
8 ----------Octagon
10 ---------Decagon
12 ---------Dodecagon
Thursday, April 9, 2009
Monday, April 6, 2009
Question to Prove a Theorem
Topic : Theorem
Theorem : Prove that if gcd(a,p²)=p and gcd(b,p²)=p² then gcd(ab,p^4)=p³. where a and b are integers and p is a prime number.
Solution :
GCD (a, p²) = p
implies that a ia a multiple of p or p is a divisor of a.
So let a = kp
where k is a constant
Similarly GCD(b, p²)=p²
implies that b is the multiple of p² or p² is a divisor of b
So let b = mp²
where m is a constant
So a = kp and b = mp²
ab = kp . mp²
ab = kmp³
implies that ab is a multiple of p³ and km is the constant
So greatest common dividor of kmp³ and p^4 is p³
Hence GCD (ab, p^4) = p³
Hence proved.
Theorem : Prove that if gcd(a,p²)=p and gcd(b,p²)=p² then gcd(ab,p^4)=p³. where a and b are integers and p is a prime number.
Solution :
GCD (a, p²) = p
implies that a ia a multiple of p or p is a divisor of a.
So let a = kp
where k is a constant
Similarly GCD(b, p²)=p²
implies that b is the multiple of p² or p² is a divisor of b
So let b = mp²
where m is a constant
So a = kp and b = mp²
ab = kp . mp²
ab = kmp³
implies that ab is a multiple of p³ and km is the constant
So greatest common dividor of kmp³ and p^4 is p³
Hence GCD (ab, p^4) = p³
Hence proved.
Tuesday, March 31, 2009
Wednesday, March 25, 2009
Simple Linear Equation to Find the Value of a Variable
Topic : Linear Equation
Problem : Solve 0.7n - 1.5 + 7.3n = 14.5
Solution :
0.7n - 1.5 + 7.3n = 14.5
8.0n - 1.5 = 14.5
8.0n = 14.5 + 1.5
8.0n = 16.0
n = 16.0/8.0
n = 2.0
= n = 2
Problem : Solve 0.7n - 1.5 + 7.3n = 14.5
Solution :
0.7n - 1.5 + 7.3n = 14.5
8.0n - 1.5 = 14.5
8.0n = 14.5 + 1.5
8.0n = 16.0
n = 16.0/8.0
n = 2.0
= n = 2
Friday, March 20, 2009
Integration by Tabular Method
Topic : Integration by Tabular Method
Question : Find ∫x³ Cos 2x dx
Hints :
∫x^n Cos ax dx
Take u = x^n
dv = Cos ax dx
For patterns like
∫x^n Sin ax dx , ∫x^n Cos ax dx ,
∫x^n e^(ax) dx we use tabular method
Solution :
∫x³ Cos 2x dx
u = x³ and dv = Cos 2x
+x³(Sin 2x)/2 - 3x²(-Cos 2x)/4 + 6x(-Sin 2x)/8 - 6(Cos 2x)/16
=(x³Sin 2x)/2 + 3/4 *(x²Cos 2x) - 6/8 *(xSin 2x) - 6/16 *(Cos 2x) + c
Question : Find ∫x³ Cos 2x dx
Hints :
∫x^n Cos ax dx
Take u = x^n
dv = Cos ax dx
For patterns like
∫x^n Sin ax dx , ∫x^n Cos ax dx ,
∫x^n e^(ax) dx we use tabular method
Solution :
∫x³ Cos 2x dx
u = x³ and dv = Cos 2x
+x³(Sin 2x)/2 - 3x²(-Cos 2x)/4 + 6x(-Sin 2x)/8 - 6(Cos 2x)/16
=(x³Sin 2x)/2 + 3/4 *(x²Cos 2x) - 6/8 *(xSin 2x) - 6/16 *(Cos 2x) + c
Tuesday, March 17, 2009
Problem on Factorization
Topic : Factorization
Problem : Solve a^4 +3a^3+27a+81
Solution :
a^4 +3a^3+27a+81
First let's group the terms
(a^4 +3a^3)+(27a+81)
Now we take common factors
from each parenthesis.
a^3(a+3)+ 27(a+3) as 81/27 = 3
Again we have a common factor
(a+3)
So, (a+3)(a^3+27)
or (a+3)(a^3+3^3)
Applying the formula x^3+y^3 = (x+y)(x^2-xy+y^2) for a^3+b^3
we get (a+3)(a+3)(a^2-3a+3^2)
(a+3)(a+3)(a^2-3a+9)
or
(a+3)^2(a^2-3a+9)
Problem : Solve a^4 +3a^3+27a+81
Solution :
a^4 +3a^3+27a+81
First let's group the terms
(a^4 +3a^3)+(27a+81)
Now we take common factors
from each parenthesis.
a^3(a+3)+ 27(a+3) as 81/27 = 3
Again we have a common factor
(a+3)
So, (a+3)(a^3+27)
or (a+3)(a^3+3^3)
Applying the formula x^3+y^3 = (x+y)(x^2-xy+y^2) for a^3+b^3
we get (a+3)(a+3)(a^2-3a+3^2)
(a+3)(a+3)(a^2-3a+9)
or
(a+3)^2(a^2-3a+9)
Thursday, March 12, 2009
Word Problem on Finance
Topic : Finance
Question : A car dealer offers you a choice of 0% financing for 60 months or $2500 cash back on a new vehicle. You have a pre-approved 60-month loan you can use from your credit union at a 4% interest rate. If the monthly payments at 0% are $16.67 per $1000 financed, and the monthly payments at 4% are $18.41 per $1000 financed, what is the range of new car prices for which the cash back option will cost you less? For what range of car prices should you take the 0% financing?
Solution :
Let the price at which both the options will give same result be x thousand (dollor)
Total payment for cash purchase,
= 16.67 x.60
and total payment for credit purchase
=18.41x.60 - 2500
Since these are supposed to be equal,
18.41x.60-2500=16.67x.60
1104.6x-2500=1000.2x
-1000.2x = -1000.2x
------------------------
104.4x-2500=0
+2500=+2500
-----------------
104.4x=2500
x=2500/104.4
x= 23.946
So below 23.946thousand = $23.946, the cash back option will cost less.
Above $23.946 you should take 0% financing
Question : A car dealer offers you a choice of 0% financing for 60 months or $2500 cash back on a new vehicle. You have a pre-approved 60-month loan you can use from your credit union at a 4% interest rate. If the monthly payments at 0% are $16.67 per $1000 financed, and the monthly payments at 4% are $18.41 per $1000 financed, what is the range of new car prices for which the cash back option will cost you less? For what range of car prices should you take the 0% financing?
Solution :
Let the price at which both the options will give same result be x thousand (dollor)
Total payment for cash purchase,
= 16.67 x.60
and total payment for credit purchase
=18.41x.60 - 2500
Since these are supposed to be equal,
18.41x.60-2500=16.67x.60
1104.6x-2500=1000.2x
-1000.2x = -1000.2x
------------------------
104.4x-2500=0
+2500=+2500
-----------------
104.4x=2500
x=2500/104.4
x= 23.946
So below 23.946thousand = $23.946, the cash back option will cost less.
Above $23.946 you should take 0% financing
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