College Algebra MA 15200 Written Assignment Week 1 PDF

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College Algebra MA 15200 Written Assignment Week 1...

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PURDUE UNIVERSITY College Algebra MA 15200 Written Assignment Unit 1

Solution A (t) (account value) =? t (years) = 20 years p (initial amount) = $8500 r (rate) = 0.0812 n (period) = 12 years Formula: A(t) = p(1 + r/n) Hence; (A) 20 = 8500(1+8.12/100/12)12*20 8500(1+0.00677)240 8500(1.00677)240 Compounded interest= 42922.27 When deposit is compounded monthly with simple interest: A (t) = p(1+r/n(nt))

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P (1+rt) Therefore; 8500(1+0.0812x20) 8500(1.0812x20) 8500 x2.624 Simple interest= 22304 Compound interest quickens the growth of principal than the growth of simple interest as revealed in the both graphs. Compound interest: A (t) =8500(1.00677)12t Simple interest: A (t) = 8500(1+0.0812t)

Question 2 Solution 2

Proof: y = 5 x 0.5-x 5e –xln0.5 y/5 = e-xln0.5 ln(y/n) = -xln0.5 x = -ln(y/5)/ln0.5 y =- ln(x/5)/ln0.5

X-intercept of f-1(x) =( 5,0) Question 3 Solution T uranium = -203 is 703800000 A(t)=Aoe,(In(0.5)t/T)t 3

800 = 1000e ln0.5/703800000 t 800=1000e(ln(0.5)/703800000) after 20 years, 800 grams remaining ln0.8=ln0.5/703800000t 703800000 x ln0.8/ln0.5=t t=703800000 x 0.3321928 t=226576993 Solving for k in the equation A(t)=Aoekt Divide both sides by Aoe to get Ao/2 = Aoekt Multiply both sides by ½ 1/2ekt

ln0.5/T =k

Hence, k for uranium -235 is: Since k=ln0.5/T then, ln0.5/703,800,000 1/100,000,000 x ln0.5 1/100,000,000 x -0.099021026= -109 K= -109

which is true

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References Abramson, J. (2017). Algebra and trigonometry. OpenStax, TX: Rice University. https://openstax.org/details/books/algebra-and-trigonometry

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