Theinvestment isdecreasing atthe rate of 23%every year.x =216No, in Step 2, Haileyshould have alsomultipliedthe exponent of thecoefficient by 2 to get32.8,579yearsf (x) =4(1.05)^nf (x) =975 (1 .03)^x1.4yearsf(x) =268(0.86)^xlog3243 = xx =log2(3)f( x )=2^x/4subtracttheexponentsAllison'sby$3.95The functionrepresentsexponentialgrowth.$0.54perweekf(x) =200(0.92)^x3.29log ba = xIn10/4f(x) is less thang(x) for thesame values ofx asx approachesnegative infinity.x^22/3f(x) =(0.005)(1.005)xk(x) > f(x)when x >0Theinvestment isdecreasing atthe rate of 23%every year.x =216No, in Step 2, Haileyshould have alsomultipliedthe exponent of thecoefficient by 2 to get32.8,579yearsf (x) =4(1.05)^nf (x) =975 (1 .03)^x1.4yearsf(x) =268(0.86)^xlog3243 = xx =log2(3)f( x )=2^x/4subtracttheexponentsAllison'sby$3.95The functionrepresentsexponentialgrowth.$0.54perweekf(x) =200(0.92)^x3.29log ba = xIn10/4f(x) is less thang(x) for thesame values ofx asx approachesnegative infinity.x^22/3f(x) =(0.005)(1.005)xk(x) > f(x)when x >0

Unit 5 - Call List

(Print) Use this randomly generated list as your call list when playing the game. There is no need to say the BINGO column name. Place some kind of mark (like an X, a checkmark, a dot, tally mark, etc) on each cell as you announce it, to keep track. You can also cut out each item, place them in a bag and pull words from the bag.


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  1. The investment is decreasing at the rate of 23% every year.
  2. x = 216
  3. No, in Step 2, Hailey should have also multiplied the exponent of the coefficient by 2 to get 3 2 .
  4. 8,579 years
  5. f (x) = 4(1.05)^n
  6. f (x) = 975 (1 . 03)^x
  7. 1.4 years
  8. f(x) = 268(0.86)^x
  9. log3 243 = x
  10. x = log2 (3)
  11. f( x )= 2^x/4
  12. subtract the exponents
  13. Allison's by $3.95
  14. The function represents exponential growth.
  15. $0.54 per week
  16. f(x) = 200(0.92)^x
  17. 3.29
  18. log b a = x
  19. In10/4
  20. f(x) is less than g(x) for the same values of x as x approaches negative infinity.
  21. x^2
  22. 2/3
  23. f(x) = (0.005)(1.005)x
  24. k(x) > f(x) when x > 0