Can you post a clearer picture?
Answer:
1. 49a^6b^10
2. −81a^8b^2c^2
Step-by-step explanation:
N/A
Then that would be 3 and 9.
Is this from iready cuz I don’t I’ve had that question before
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.
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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.
