(1 pt) let an=n+1n+3. find the smallest number m such that: (a) |an−1|≤0.001 for n≥m m= 1997 (b) |an−1|≤0.1 for n≥m m= 23 (c) now use the limit definition to prove that limn→[infinity]an=1. that is, find the smallest value of m (in terms of t) such that |an−1|m. (note that we are using t instead of ϵ in the definition in order to allow you to enter your answer more easily). m= 2/t - 3t (enter your answer as a function of t)

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

translate k to n and reflect across the line containing jk.

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5