5(3)+2
15+2
17 monies
Just plug the amount of games in for n
Use pythagoras
d = √(Δx² + Δy²)
d = √(12-2)² + (9-4)²
d = √(10² + 5²)
d = √(100 + 25)
d = √125
d = 5√5
d = 5 × 2.236
d = 11.18
The nearest tenth is 11.2, so the distance is near to 11.2
Answer:
Step-by-step explanation:
all you have to do is * 7 by 5 and divide by 2
Answer:
11.9
Step-by-step explanation:
71.40 divided by 6
To answer the question, you need to determine the amount Mr. Traeger has left to spend, then find the maximum number of outfits that will cost less than that remaining amount.
Spent so far:
... 273.98 + 3×7.23 +42.36 = 338.03
Remaining available funds:
... 500.00 -338.03 = 161.97
The cycling outfits are about $80 (slightly less), and this amount is about $160 (slightly more), which is 2 × $80.
Mr. Traeger can buy two (2) cycling outfits with the remaining money.
_____
The remaining money is 161.97/78.12 = 2.0733 times the cost of a cycling outfit. We're sure he has no interest in purchasing a fraction of an outfit, so he can afford to buy 2 outfits.
