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Mathematics 11 Online
OpenStudy (anonymous):

Refer to the following scenario to answer the next four questions: Brandon makes and sells car speakers. The fixed cost for his work space is $2000. The cost of materials and labor to build the speakers is $98 per set. Brandon sells the speakers for $150 per set. State two functions to represent Brandon’s income i(x) and his total costs c(x). a.) Income: i(x) = 98x, Cost: c(x) = 150x b.) Income: i(x) = 150x, Cost: c(x) = 98x c.) Income: i(x) = 150x, Cost: c(x) = 98x + 2000 d.) Income: i(x) = 98x, Cost: c(x) = 150x + 2000

OpenStudy (anonymous):

@danish071996

OpenStudy (anonymous):

the answer is c i think

OpenStudy (anonymous):

can you help me on two more they have to do with this question?

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

31) Brandon’s profit is the difference between his income and his total costs. State a function to represent Brandon’s profit, p(x). a.) Profit: p(x) = 150x – 98x b.) Profit: p(x) = 150 x + 98x c.) Profit: p(x) = 150x – 98x + 2000 d.) Profit: p(x) = 150x – (98x + 2000)

OpenStudy (anonymous):

A

OpenStudy (anonymous):

32) Use the profit function determined in the previous problem to calculate Brandon’s profit for selling 75 speakers sets? 33) Brandon “breaks even” when his profit is $0. He must sell at least how many speaker sets to break even? Give the answer as a whole number.? those are the last two

OpenStudy (anonymous):

um oh its about i think 17 or 20 and/or 33

OpenStudy (anonymous):

im not so sure about that part

OpenStudy (anonymous):

for which one 32 or 33?

OpenStudy (anonymous):

33

OpenStudy (anonymous):

32 is 140 i believe

OpenStudy (anonymous):

thank you so much!

OpenStudy (anonymous):

u r sooooooooooo welcome really

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