Pythagoras Theorem and Financial polynomials Essay Example for Free

Pythagoras Theorem and Financial polynomials EssayPythagoras Theorem and Financial polynomialsIntroduction Ahmed and Vanessa piss 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 Ahmeds half, the treasure is buried in the desert (2x + 6) paces form the Castle Rock while as per Vanessas half she has to passport (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 Vanessas case, I will let a=x, b =2x+4 and then c=2x+6. To follow, will be my efforts to arrange the measurements above into the real Pythagoras theorem equation as followsX2+ (2x+4)2=(2x+6)2 this is the equation formed out of the Pythagoras TheoremX2+42+16x+16 = 42+ 24x+36 are th e binomials squaredx2 42 on both sides can be subtracted out.X2+16x+16 = 24x +36 subtract 16x from both sidesX2+16 = 8x+36 now subtract 36 from both sidesX2-20 = 8xX2-8x-20=0 I will use to decide the function by factoring using the zero factor.(x-) (x+) the coefficient of x2Application 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 (x-10)(x+2)=0X-10=0 or x+2=0 creation of a Byzantine equationx=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 fragment A and then solve for p= $5670 + r = 3.5% for part B, without the parenthesis as followsP + P r + P r2/4 (the original polynomial) to reach this I followed the following go(1 + r/2)2 This is because it looks as if it is foilP(1 + r/2)P (1+r/2)(1+r/2) later the two equations I combine like terms. Because I am multiplying by 2 on r/2, it cancels out both 2s and I then get left with is r as followsP(1+ r/2 + r/2 + r2/4)P(1 + 2(r/2) + r2/4)I then write in descending dedicate (P + Pr + Pr2)To solve for P=2000 and r=10% the following followsP + Pr + Pr2/42000 + 2000 (0.10) +2000 0.10242000 + 200 + 5 = $2205P(1+ r/2)22000( 1 + .10)22000(1.05)22000( 1.1025) = $2205For part B I will solve for P=5670 and r= 3.5%P + Pr + P (r2/4)5670 + 5670 (0.035) + 5670 0.03525670 + 198.45 + 1.7364375 = 5870.1864375Th is is approximately ($5870.19)The problem 70 on page 311 has the following steps(-93 + 32 15x) (-3x)The Dividend is (-93 + 32 15x), and the Divisor is (-3x).The Dividend is (-93 + 32 15x), and the Divisor is (-3x).-93 + 32 15x-3xAfter I divide -9 by -3 which equals +3. The x on the bottom cancels the x from the top.-93 + 32 15x-3x -3x -3x-9* x*x* xI am now left with 32 for the first part of the polynomial.-3 * x-9*x *x * x-3 * xI first divide 3 by -3, which equals -1 and the x from the bottom cancels out one of the xs from the top.-93 + 32 15x-3x -3x -3x3 *x *xAt this point I am left with -1x, which simplifies to just x, as the second part of the polynomial. thus-3 *x3 *x * x-3 * xThen 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 whatsoever xs to carry onto the answer of the equation.-93 + 32 15x-3x -3x -3x-15 *xAt this point I am left with only 5 for the last part of the polynomial, and the answer is3 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 negatively charged sign, it changes the minus sign to a plus sign in the final answer, and the equation is in Descending order.ReferenceLarson, R., Hostetler, R. P. (2009). Elementary and intermediate algebra. Boston, Mass Houghton Mifflin showtime document