Pythagoras Theorem and Financial polynomials Essay
Pythagoras Theorem and Financial polynomials
Ahmed and Vanessa have interest in locating a treasure, which is buried. It is my responsibility to help the two locate it. First, I will help them locate it by the use of Pythagorean quadratic. As per Ahmed’s half, the treasure is buried in the desert (2x + 6) paces form the Castle Rock while as per Vanessa’s half she has to walk (x) paces to the north then walk (2x + 4) paces to the east. According to the Pythagorean theorem, every right angled triangle with length (a) and (b) as well as a hypotenuse (c), has a relationship of (a2 + b2 = c2) (Larson & Hostetler, 2009).
In Ahmed and Vanessa’s case, I will let a=x, b =2x+4 and then c=2x+6. To follow, will be my efforts to put the measurements above into the real Pythagoras theorem equation as follows:
X2+ (2x+4)2=(2x+6)2 this is the equation formed out of the Pythagoras Theorem
X2+4×2+16x+16 = 4×2+ 24x+36 are the binomials squared
x2 & 4×2 on both sides can be subtracted out.
X2+16x+16 = 24x +36 subtract 16x from both sides
X2+16 = 8x+36 now subtract 36 from both sides
X220 = 8x
X28x20=0 I will use to solve the function by factoring using the zero factor.
(x) (x+) the coefficient of x2
Application and selection from the following (2, 10: 10,2: 5,4; 4, 5)
In this case, it seems that I am going to use 10 and 2 is as per how the expression looks like this (x10)(x+2)=0
X10=0 or x+2=0 creation of a complex equation
x=10 or x=2 these are the two probable resolutions to this equation.
One of the two calculated solutions is an extraneous solutions, as it do not work with such sceneries. The remaining solution I only have is (X=10) as the number of paces Ahmed and Vanessa have to accomplish to find the lost treasure. As a result the treasure is 10 paces to the north 2x+4 connect the 10, now its 2(10)+4=24 paces to the east of Castle Rock, or 2x+6= 2(10)+6=26 paces from Castle Rock.
Financial polynomial
For the case of financial polynomials, I have first to write the polynomial without the parenthesis. Following the above, I have to solve for p= 2000 + r = 10% for part A and then solve for p= $5670 + r = 3.5% for part B, without the parenthesis as follows:
P + P r + P r2/4 (the original polynomial) to reach this I followed the following steps:
(1 + r/2)2 This is because it looks as if it is foil
P(1 + r/2)
P (1+r/2)(1+r/2) After the two equations I combine like terms. Because I am multiplying by 2 on r/2, it cancels out both 2’s and I then get left with is r as follows;
P(1+ r/2 + r/2 + r2/4)
P(1 + 2(r/2) + r2/4)
I then write in descending order (P + Pr + Pr2)
To solve for P=2000 and r=10% the following follows;
P + Pr + Pr2/4
2000 + 2000 ×(0.10) +2000× 0.1024
2000 + 200 + 5 = $2205
P(1+ r/2)2
2000×( 1 + .10)2
2000×(1.05)2
2000×( 1.1025) = $2205
For part B I will solve for P=5670 and r= 3.5%
P + Pr + P ×(r2/4)
5670 + 5670× (0.035) + 5670 × 0.0352
5670 + 198.45 + 1.7364375 = 5870.1864375
This is approximately ($5870.19)
The problem 70 on page 311 has the following steps;
(9×3 + 3×2 – 15x) ÷ (3x)
The Dividend is (9×3 + 3×2 – 15x), and the Divisor is (3x).
The Dividend is (9×3 + 3×2 – 15x), and the Divisor is (3x).
9×3 + 3×2 – 15x
3xAfter I divide 9 by 3 which equals +3. The x on the bottom cancels the x from the top.
9×3 + 3×2 – 15x
3x 3x 3x
9* x*x* x
I am now left with 3×2 for the first part of the polynomial.
3 * x
9*x *x * x
3 * x
I first divide 3 by 3, which equals 1 and the x from the bottom cancels out one of the x’s from the top.
9×3 + 3×2 – 15x
3x 3x 3x
3 *x *x
At this point I am left with 1x, which simplifies to just –x, as the second part of the polynomial.
Then
3 *x
3 *x * x
3 * x
Then I divide 15 by 3, which equals positive 5, and the x on the bottom cancels out the x on the top, so you do not have any x’s to carry onto the answer of the equation.
9×3 + 3×2 – 15x
3x 3x 3x
15 *x
At this point I am left with only 5 for the last part of the polynomial, and the answer is
3×2 – x + 5.
3 * x
15 * x
3 * x
The negative sign from the 3 x changes the plus sign in the equation to a minus sign, it changes the minus sign to a plus sign in the final answer, and the equation is in Descending order.
Reference
Larson, R., & Hostetler, R. P. (2009). Elementary and intermediate algebra. Boston, Mass: Houghton Mifflin
B

Subject: Other,

University/College: University of California

Type of paper: Thesis/Dissertation Chapter

Date: 6 December 2015

Words:

Pages:
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