1. 1 (3x-2)^2 Use mathway.com.. get 9x^2 – 12x + 4
1.2 (2x^3 – x + 1)(x^2 – 2x -2) gets: 2x^5 – 4x^4 – 5x^3 + 3x^2 – 2
1.3 (x^3)(y^-2)/((x^-3)(y^5)) gets: x^6/y^7
1.4 2^(-3/4) gets: about.. .5946
1.5 (x^(-2/3))^(3/4) gets: x^(-1/2) or 1/(x^(1/2)
3. Table starts.. (1, 10.3) (2, 10.61) (3, 10.927)…..(6, 11.94)
4. You get the same value.. You are always multiplying by the ratio of something around 1.03, or the growth rate
6. A = P(1 + r/n)^nt Where r = .04, n=12 and P = 10,000
This gives you A + 10000(1+.04/12)^12t The normal growth equation is: A= P b^t, or A = 10000b^t. We want to find b. b = (1 + .04/12)^12
Calculate this and get: 1.0407, or a growth rate of 4.07 percent
NOTE: this is a little higher than the anual rate of 4 percent, because since you get interest every month, you get intrest also on your interest, thus adding this small fraction. It can make a big difference over time, however
7. The puddle starts at 100 square feet. It is 10 percent less after an hour, or 90% as big.. multiply the 100 initial size by .9 to get the area after an hour of (90 square feet) The next hour, the 90 square foot puddle is only 90 percent as big, so multiply 90 by .9 and get its new size after 2 hours of (81 square feet). Continue this process until you get a puddle of 50 square feet, or half its original size. You will note that this will take about 6 total hours.
8. Note that in your graph, the linear function (the first one, y=2x) always increases the same amount when you go over the same distance, say from x=3 to x=4, In each case, you increase by 2 every time you go over 1.. this is the constant slope. In y=x^2, note that you do not have the same increase. from 1 to 2 you increase by 3, from 2 to 3 you increase 5, from 3 to 4 you increase by 7 (9 to 16..). Your increase is increasing by 2 each time, or you are adding 2 to your change each time. In the last equation, notice your increase is being MULTIPLIED by 2 each time. In a similar fashion, you can look at any table and figure out if it represents a linear, polynomial, or exponential function
8. You invest 5,000 at 12 percent for 10 years, compounded continuously. How much do you have after 10 years? How much less would you have if the interest was compounded quarterly instead of continuously after 10 years?
P=5000, r=.12 t=10 and you are compounding continuously, so you use this function A = Pe^(rt) or A = 5000(2.72)^((.12)(10)) $16,613