Mathematics, 23.06.2019 07:40 maevemboucher78
Sal is trying to determine which cell phone and service plan to buy for his mother. the first phone costs $100 and $55 per month for unlimited usage. the second phone costs $150 and $51 per month for unlimited usage. how many months will it take for the second phone to be less expensive than the first phone? the inequality that will determine the number of months, x, that are required for the second phone to be less expensive is 100 + 55x > 150 + 51x100 + 55x < 150 + 51x100x+ 55 > 150x+ 51100x+ 55 < 150x+ 51.the solution to the inequality is x > 2.4x < 2.4x < 12.5x > 12.5.sal’s mother would have to keep the second cell phone plan for at least 231213 months in order for it to be less expensive
Answers: 1
Mathematics, 21.06.2019 18:30, Angelanova69134
Someone answer this asap rn for ! a discount store’s prices are 25% lower than department store prices. the function c(x) = 0.75x can be used to determine the cost c, in dollars, of an item, where x is the department store price, in dollars. if the item has not sold in one month, the discount store takes an additional 20% off the discounted price and an additional $5 off the total purchase. the function d(y) = 0.80y - 5 can be used to find d, the cost, in dollars, of an item that has not been sold for a month, where y is the discount store price, in dollars. create a function d(c(x)) that represents the final price of an item when a costumer buys an item that has been in the discount store for a month. d(c(x)) =
Answers: 1
Mathematics, 22.06.2019 00:30, savthespice
Bo is buying a board game that usually costs bb dollars. the game is on sale, and the price has been reduced by 18\%18%. which of the following expressions could represent how much bo pays for the game? choose 2 answers: choose 2 answers: (choice a) a 0.82b0.82b (choice b) b 1.18b1.18b (choice c) c b-0.18b−0.18 (choice d) d b-18b−18 (choice e) e b-0.18bb−0.18b
Answers: 2
Sal is trying to determine which cell phone and service plan to buy for his mother. the first phone...
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