Compute [p , q], where p(t)= 4+t , q(t)= 5- t^2

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

c

step-by-step explanation:

a) 40 more; b) old factory:   10%, new factory:   16%; c) week 4

explanation:

a) at week 0, we see in the graph that the new factory produces 190 specialty items.  

using the function, substituting 0 in for w,

this is a difference of 230-190 = 40.

b) the function for the old factory is written in the form f(x) = a(1+r)ˣ, where a is the original amount, x is the amount of time and r is the growth rate.   the equation for the old factory is ; 230 is in place of a and is the original amount; w is the amount of time; which means 1+r = 1.1.   this means that r must be 0.1 or 10%.

for the new factory, we find the percent of increase to find the growth rate; the formula for this is

between the first two points, the amount of change is 220-190 = 30; this gives us

30/190 = 0.158 ≈ 0.16 = 16%

c) at week 0, the old factory has 230 while the new factory has 190.

at week 1, the old factory has 230(1.1) = 253 while the new factory has 220.

at week 2, the old factory has 230(1.1²) = 278.3 while the new factory has 252.

at week 3, the old factory has 230(1.1³) = 306.13 while the new factory has 290.

at week 4, the old factory has 230(1.1⁴) = 336.743 while the new factory has 337.   this is the first week that the new factory surpasses the old.