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.
Answer:
3 x^2 +12x-9
Step-by-step explanation:
5x^2 - (2x-3)^2
We need to FOIL the second term
(2x-3)^2 = (2x-3) (2x-3) = 4x^2 -6x-6x+9 = 4x^2 -12x+9
Replacing it in to the original equation
5x^2 - (4x^2 -12x+9)
Distribute the minus sign
5x^2 - 4x^2 +12x-9
Combine like terms
x^2 +12x-9
<h3>
Answer: A. 7</h3>
Work Shown:
Replace every 'a' with 2, and replace every b with 3
a^2+b = 2^2 + 3 = 4 + 3 = 7
i believe your answers would be
1/2c+7 and 2H-9
Parallelogram because a rhombus is basicly a staanding up straight parallelogram.