Aused car dealer sold two cars for $1980 each. on the first car he made a 10 percent profit. on the second car he took a loss of 10 percent. taken together, did he profit or lose on the deals?

he neither made profit nor loss on both deals

Step-by-step explanation:

The cost of the first car sold by the car dealer is $1980. On the first car he made a 10 percent profit. This means that the profit made on the first car would be

10/100 × 1980 = 0.1 × 1980 = $198

Amount for which he sold the first car would be

1980 + 198 = $2178

The cost of the second car sold by the car dealer is $1980. On the second car he made a 10 percent loss. This means that the loss made on the second car would be

10/100 × 1980 = 0.1 × 1980 = $198

Amount for which he sold the second car would be

1980 - 198 = $1782

Total cost of both cars = 1980 + 1980 = $3960

Total selling price = 2178 + 1782 = $3960

Profit or loss = selling price - cost price. Therefore

Profit or loss = 3960 - 3960 = $0

Step-by-step explanation:

Neither.  He came out even.  The profit made on the first sale was exactly wiped out by the loss on the second sale.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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