Mr. garcia pays $834 for some model cars and model planes. each model car costs $52 and each model plane costs $14 less than each model car. mr. garcia buys 3 more model planes than cars. how many model planes did he buy?

Step-by-step explanation:

Given: Total Cost = $ 834

            Price of Car = $52

            Price of Plane = $14

Let, the number of model cars be x and model planes y.

⇒ 834 = 52x + 14y (1)

Also, The number of model planes are 3 more than that of the model cars.

⇒y = x+3 (2)

Substituting the value of y from (2) in (1)

    834 = 52x + 14(x+3)

⇒  834 = 52x + 14x +42

⇒834-42 = 66x

⇒66x = 792

⇒ x=12

Substituting the value of x in (2)

y = 12+3 = 15

Therefore, Mr. Garcia bought 15 model planes.

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see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2