X+y/y−x when x=4 and y=3
1 answer:
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Notice that
13 - 9 = 4
17 - 13 = 4
so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :
149 = 9 + 4k
140 = 4k
k = 140/4
k = 35
This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.
Put another way, we have
but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.
Answer:
(A)Cost of Rental A, C= 15x
Cost of Rental B, C=5x+50
Cost of Rental C, C=9x+20
(B)
i. Rental C ii. Rental A iii. Rental B
Step-by-step explanation:
<u>Part 1</u>
Let x be the number of hours of the barbeque use by the club.
Rental A: $15/h
Cost of Rental A, C= 15hx
Rental B: $5/h + 50
Cost of Rental B, C=5x+50
Rental C: $9/h + 20
Cost of Rental C, C=9x+20
<u>Part 2</u>
The graph of the three models is attached below
<u>Part 3(11.05-4.30)</u>
If the barbecue's usage hour is from 11.05 to 4.30 when the football match ends.
Number of Hours between 11.05am and 4.30pm=4 hours 25 Minutes = 4.42 Hours
Cost of Rental A, C= 15x=15(4.42)=$66.30
Cost of Rental B, C=5x+50 =5(4.42)+50=$72.10
Cost of Rental C, C=9x+20=9(4.42)+20=$59.78
Rental C should be chosen as it offers the lowest cost.
<u>Part 4 (11.05-12.30)</u>
Number of Hours = 12.30 -11.05 =1 hour 25 Minutes = 1.42 Hours
- Cost of Rental A, C= 15x=15(1.42)=$21.30 Cost of Rental B, C=5x+50 =5(4.42)+50=$57.10 Cost of Rental C, C=9x+20=9(4.42)+20=$32.78
Rental A should be chosen since it offers the lowest cost.
<u>Part 5</u>
If the barbecue is returned the next day, say after 24 hours
- Cost of Rental A, C= 15x=15(24)=$360 Cost of Rental B, C=5x+50 =5(24)+50=$170 Cost of Rental C, C=9x+20=9(24)+20=$236
Rental B should be chosen as it offers the lowest cost.
