Jake is planning a trip to china. he has made a list of cities he would like to visit, as well as the approximate amount of money he plans to spend in each as a result of travel, lodging, shopping, and so on. all costs are listed in renminbi (¥).citycost (¥)tianjin557nanjing681zhengzhou595 beijing728wuhan449chengdu534unfortu nately, jake only has ¥2,920 available. what is the cheapest city that jake can remove from his travel plans and still stay under budget? a. tianjinb. wuhanc. nanjingd. beijing

The cheapest city that Jake can remove from his travel plans and still stay under his budget is C. Nanjing

Step-by-step explanation:

if we add all the totals he has for each city

449(Wuhan)+534(Chengdu)+557(Tianjin)+595(Zhengzhou)+681(Nanjing)+728(Beijing) = 3,544

It can't be A. Tianjin because 3,544-557= 2,987 which is still over his budget.

It can't be B. Wuhan because 3,544-449= 3,095 which is still over his budget.

It can't be D. Beijing because even though 3,544-728=2,816 it is under his budget, the question asks which one is the cheapest city that can be removed and Nanjing is cheaper than Beijing, therefore, the answer is C. Nanjing.

both equations and proportions are better.

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.