Written Assignement m my submission unit 5 PDF

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1. A retirement account is opened with an initial deposit of $8,500 and earns 8.12% interest compounded monthly. What will the account be worth in 20 years? What if the deposit was calculated using simple interest? Could you see the situation in a graph? From what point one is better than the other? In order to arrive at a solution, let us consider the following: A(t) = amount in savings after (t) year P = initial deposit a.k.a principal amount - $8500 t = number of time periods (years) = 20 n = number of times interest compounds within each time period = 12 r = interest rate = 0.0812 If a retirement account is opened with a principal value of $8500, monthly compounded interest of 8.12% and was allowed to run for 20 years. Then we must calculate as follows, considering A(t) = P (1 + r/n)nt A (20) = 8500 (1+0.0812/12)12*20 A (20) = 8500 (5.0456683766) A (20) = $42888.1812011 Where as if we were to calculate based on simple interest, we would use A(t) = P (1+rt), hence A (20) = 8500 (1 + 0.0812 * 20) A (20) = 8500 (2.624) A (20) = $22304 In terms of interest earned over the same period of 20 years, compounded interest is better since the return is 48% higher. Now let us prove this with the help of a graph, where our equation for compound interest is A(t) = 8,500(1+ 0.0812/12)12t and equation for simple interest is A(t) = 8,500(1+0.0812t). By using the aforementioned equations in the desmos graphing calculator, we can see that the red curve represents compound interest and the blue curve represents simple interest. As it can be seen, compound interest always results in higher invested amount if all other parameters are kept the same when t > 0. Only at the time period 0, would the two result in the same investment amount of 8,500. This can be seen when the red and blue lines cross at the y axis. After that pint, compound interest always yields better return that simple interest.

2. Graph the function f(x) = 5(0.5)-x and its reflection about the line y=x on the same axis, and give the x-intercept of the reflection. Prove that ax = exlna. [Suggestion: type y = 5(0.5)-x {- 7 < x < 2} {0 < y < 7} in desmos, and then type its inverse function.] y=x axis is the inverse reflection of the function f(x) = y = 5(0.5)-x inverse, switch x & y x = 5(0.5)-y ln(x) = ln(5(0.5)-y) ln(x) = ln (5) – yln (0.5) yln (0.5) = ln (5) – ln(x) y = 1n (5)/ln (0.5) – ln(x)/ (ln (0.5) If x is intercept, then y = 0 0 = ln (5)/ln (0.5) – ln(x)/ln (0.5) Therefore ln (5)/ln (0.5) = ln(x)/ (ln (0.5)

And x = 5. So (5,0) is the x-intercept of the reflection.

Proof: ax = exlna ln(ax) = ln(exln(a)) xln(a) = xln(a)ln(e) xln(a) = xln(a)(1) xln(a) = xlna Therefore, ax = exlna

3. How long will it take before twenty percent of our 1,000-gram sample of uranium-235 has decayed? [See Section 6.6 Example 13] The decay equation is A(t)=AoeKt, where t is the time for the decay, and K is the characteristic of the material. Suppose T is the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. Prove that K = ln0.5/T. What is K for the uranium-235? Show the steps of your reasoning. Prove K = ln (0.5)/T We know that half life (T) occurs when half of the material is decayed, A(T)/Ao = 0.5, which is the proportion of material that is left after half is decayed, 0.5. A(t)=AoeKt , substituting half-life (T) into the decay equation. A(T)/Ao = eKt and by dividing Ao on each side, we arrive at A(T)/Ao = 0.5.

By substituting A(T)/Ao = eKt, we know that eKt = 0.5and ln(eKt) = ln (0.5) by taking the ln of each side. By moving the exponent using logarithmic power rules KTln = ln (0.5) KT = ln (0.5), ln(e) = 1 By dividing each side by T, K = ln (0.5)/T T in this case is the half life of uranium 235, which is 703,800,000 years (Radioactivity.Eu.Com, n.d.). So how long does it take for 20% of our 1000 gms of uranium 235 to decay? A(t) = 1000 – 1000 * 0.20 = 800 Ao = 1000 K = ln (0.5)/T = ln (0.5)/703800000 A(t) = AoeKt 800 = 1000e (ln0.5/703800000) t Ln (0.8) = ln (0.5)/to3800000*t t = 703800000 * ln (0.8)/ln (0.5) = 226572993.182 years

References Abraham, J. (2017) Algebra and trigonometry. OpenStax, TX: Rice University. Retrieved from https://openstax.org/details/books/algebra-and-trigonometry Radioactivity.Eu.com. (n.d.). Uranium 238 and 235: A radioactive and strategic element. Retrieved from https://www.radioactivity.eu.com/site/pages/Uranium_238_235.htm...