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3241004551 [841]
3 years ago
15

Monroe is gift wrapping a box that has the shape of a right rectangular prism. The box has a length of 20 cm, a width of 7 cm, a

nd a height of 15 cm.
How many square centimeters of gift wrap will Monroe need to cover the entire outside of the box?
Mathematics
2 answers:
Ksenya-84 [330]3 years ago
5 0
Formula is 2(wl+hl+hw)
2[(7x20)+(15x20)+(15x7)]
=1090 cm^2
Nitella [24]3 years ago
4 0

Answer:

Amount of gift wrap required is 1090 cm².

Step-by-step explanation:

We are given the dimensions of the rectangular box as,

Length = 20 cm, Width = 7 cm and Height = 15 cm.

Since, we will cover the outside of the box. We have to find the surface area of the box to know the amount of gift wrap required.

As, Surface area of a rectangular prism = 2(l×w+h×l+h×w)

i.e. Area of the rectangular box = 2(20×7+15×20+15×7)

i.e. Area of the rectangular box = 2(140+300+105)

i.e. Area of the rectangular box = 2 × 545

i.e. Area of the rectangular box = 1090 cm²

Thus, the amount of gift wrap required is 1090 cm².

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We know that 12 equals one dozen. It is not asking for a dozen, but for 9 dozen. So, we need to multiply 9 by 12. 9*12 = 108. 

It says the rolls of first aid tape were divided equally into 4 boxes. We need to divide these 108 rolls into 4 equal groups, since that is what took place in the problem. So, we divide 108 by 4. 
108/4 = 27. There are 27 rolls in each box. 
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What is the difference x + 5/ x + 2 - x + 1 /x^2 + 2x
xxTIMURxx [149]

2x^3 + 2x^2 + 5x + 1/ x^2

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Find the area of the regular pentagon. 4.1 cm 6 cm Area = [?] cm? Enter your answer to the nearest tenth. ​
Assoli18 [71]

Answer:

Area of the given regular pentagon is 61.5 cm².

Step-by-step explanation:

Area of a regular polygon is given by,

Area = \frac{1}{2}aP

Here, a = Apothem of the polygon

P = Perimeter of the polygon

Apothem of the regular pentagon given as 4.1 cm.

Side of the pentagon = 6 cm

Perimeter of the pentagon = 5(6)

                                             = 30 cm

Substituting these values in the formula,

Area = \frac{1}{2}(4.1)(30)

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8 0
2 years ago
Two pools are being drained. To start, the first pool had 3850 liters of water and the second pool had 4370 liters of water. Wat
SashulF [63]

Answer:

(a)

Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes

Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes

(b)

3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes

Step-by-step explanation:

(a) With x being the minutes after which you want to calculate the amount of water in the pool, to calculate this amount of water you must subtract the water drained from the initial amount of water that the pool contains.

Knowing that, for example, the water from the first pool drains at a rate of 33 liters per minute, then after x minutes, the total water drained will be 33 liters per minute * x minutes. Then:

<u><em>Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes</em></u>

Reasoning in the same way:

<u><em>Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes</em></u>

(b)  Now want to know when the two pools would have the same amount of water.  If they have the same amount of water then you can express:

Amount of water in the first pool = Amount of water in the second pool

Replacing the expressions found in (a):

<u><em>3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes</em></u>

5 0
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Which value of a makes the equation 13=40-3a
weeeeeb [17]
The value of A is 9. This is because 40-13= 27
3x?=27 3x9=27
3 0
3 years ago
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