George purchases a sack of apples, a bunch of bananas, a cantaloupe, and a carton of dates for $20. If a carton of dates costs twice as much as a sack of apples and the price of a cantaloupe is equal to the price of a sack of apples minus a bunch of bananas, how much would it cost George to purchase a bunch of bananas and a cantaloupe?

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  $5

Step-by-step explanation:

Let the variables A, B, C, D represent a sack of apples, a bunch of bananas, a cantaloupe, and a carton of dates, respectively. Then the problem statement gives rise to 3 equations:

  A +B +C +D = 20

  D = 2A

  C = A -B

Substituting the last two equations into the first gives ...

  A +B + (A -B) +2A = 20

  4A = 20 . . . simplify

  A = 5 . . . . . divide by 4

Substituting this into the last equation, we have ...

  C = 5 -B

  B + C = 5 . . . . add B

The cost to purchase a bunch of bananas and a cantaloupe is $5.

step-by-step explanation:

its 512 because parenthesis and to the 2nd power is 512 so it will be 512

-3(b-8)-5=9(b+2)+1

distribute

-3b+24-5=9b+18+1

add/ subtract

-3b+20=9b+19

20-19= 1

-3b+1=9b

9b+3b= 12b

1=12b

1 divided by 12 = 1/12

answer: 1/12 = b