The number of cars (c) in a parking lot increases when the parking fee (f) decreases. Write the correct equation for this scenario, and solve for the number of cars when the fee is $6. Fee Cars 20 15 30 10 c = − 300 over f c = −50 c = 300 over f c = 50 c = 3f c = 18 c = 3f; c = 2

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step-by-step explanation:

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

3,750 deer

step-by-step explanation:

1,500 is 15 times larger than 100

lets make 250 15 times larger.

3750 deer

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%