Answer:
(A)Cost of Rental A, C= 15h
Cost of Rental B, C=5h+50
Cost of Rental C, C=9h+20
(B)
i. Rental C
ii. Rental A
iii. Rental B
Step-by-step explanation:
Let h be the number of hours for which the barbeque will be rented.
Rental A: $15/h
Rental B: $5/h + 50
- Cost of Rental B, C=5h+50
Rental C: $9/h + 20
- Cost of Rental C, C=9h+20
The graph of the three models is attached below
(b)11.05-4.30
When you keep the barbecue from 11.05 to 4.30 when the football match ends.
Number of Hours = 4.30 -11.05 =4 hours 25 Minutes = 4.42 Hours
-
Cost of Rental A, C= 15h=15(4.42)=$66.30
- Cost of Rental B, C=5h+50 =5(4.42)+50=$72.10
- Cost of Rental C, C=9h+20=9(4.42)+20=$59.78
Rental C should be chosen as it offers the lowest cost.
(c)11.05-12.30
Number of Hours = 12.30 -11.05 =1 hour 25 Minutes = 1.42 Hours
- Cost of Rental A, C= 15h=15(1.42)=$21.30
- Cost of Rental B, C=5h+50 =5(4.42)+50=$57.10
- Cost of Rental C, C=9h+20=9(4.42)+20=$32.78
Rental A should be chosen as it offers the lowest cost.
(d)If the barbecue is returned the next day, say after 24 hours
- Cost of Rental A, C= 15h=15(24)=$360
- Cost of Rental B, C=5h+50 =5(24)+50=$170
- Cost of Rental C, C=9h+20=9(24)+20=$236
Rental B should be chosen as it offers the lowest cost.
(-4n^(2))* (5n^(7)):-20n to the power of 9
(2x)^(3)*(5x^(4 ))^(2): 200x to the power of 11
To find the answer, it is best to use formulas. Area is mentioned, so lets use the formula for area, A=lw. Plugging in, you get

. While this might look confusing, it is easier than you think: divide both sides by (x-70) for

. From here, try factoring the top: there is a -90 difference between -70 and -160 and luckily -70 times -90 equals 6300. So, your equation should look like

. You can eliminate (x-70) from the top and bottom, so you should get l=x-90 . Since there is no way to find x, x-90 is your length.
Answer:
1) $6,374.72
2) $3,386.40
Step-by-step explanation:
1,200.32
+985.43
+1,200.65
+1,987.34
+1,000.98= 6,374.72
1,200.32
+985.43
+1,200.65
= 3,386.4