Now, consider the change problem in Binaryland. The input to this problem is an integer L. The output should be the minimum cardinality collection of coins required to make L cents of change (that is, you want to use as few coins as possible). In Binaryland, the coins are worth 1, 2, 22, 23, ..., 21000 cents. Assume that you have an unlimited number of coins of each type. Prove or disprove that the greedy algorithm (that takes as many coins of the highest value as possible) solves the change problem in Binaryland. HINT: The greedy algorithm is correct for one of the above two problems and is incorrect for the other.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.

20days louise can read300 pg

step-by-step explanation:

1 day 15 pg 300: 15= 20

x day 300 pg

the first option i believe

step-by-step explanation: