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Evgesh-ka [11]
4 years ago
13

Please help me with #57

Mathematics
1 answer:
omeli [17]4 years ago
6 0
To solve, you need to find the surface area of the box.  Since it is a rectangular prism, we know it has 6 sides.  The top and bottom will be the same area, the two long sides will be the same, and the two small sides will be the same.  So we find the area of one each of the top, long side, and small side then add them together and then multiply by two (to account for the other sides).

Top
A = L x W
A = 9 in x 7 in = 63 in^2

Long side
A = L x W
A = 9 in x 5 in = 45 in^2

Short side
A = L x W
A = 7 in x 5 in = 35 in^2

Add them together...

63 + 45 + 35 = 143 in^2  (that is the area of 3 sides so now we multiply by 2)

143 in^2 x 2 = 286 in^2

She needs

D. 286 in^2 of wrapping paper
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artcher [175]

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

  • Cost of Rental A, C= 15h

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

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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
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Rental B should be chosen as it offers the lowest cost.

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3 years ago
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\left(\dfrac{3a^{-3}b^2}{2a^{-1}b^0}\right)^2

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\left(\dfrac{3a^1b^2}{2a^3b^0}\right)^2=\left(\dfrac{3b^2}{2a^2}\right)^2=\dfrac{9b^4}{4a^4}

Then with a=-2 and b=-3, we have a^4=(-2)^4=16=4^2 and b^4=(-3)^4=81=9^2, so

\dfrac{9b^4}{4a^4}=\dfrac{9\cdot9^2}{4\cdot4^2}=\dfrac{9^3}{4^3}=\dfrac{729}{64}

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