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AlladinOne [14]
3 years ago
14

The area of a rectangular mirror is 15 7 16 square feet. The width of the mirror is 3 1 4 feet. If there is a 5 foot tall space

on the wall to hang the mirror, will it fit?
Mathematics
1 answer:
Vlad1618 [11]3 years ago
8 0

Given:

Area of rectangle = 15\dfrac{7}{16} sq. feet.

Width of the mirror = 3\dfrac{1}{4} feet.

To find:

Whether it will fit on the wall whose length is 5 foot.

Solution:

We know that, area of a rectangle is

Area=length\times width

\dfrac{Area}{width}=length

Substituting the given values, we get

Length=\dfrac{15\dfrac{7}{16}}{3\dfrac{1}{4}}

Length=\dfrac{\dfrac{15\times 16+7}{16}}{\dfrac{3\times 4+1}{4}}

Length=\dfrac{\dfrac{240+7}{16}}{\dfrac{12+1}{4}}

Length=\dfrac{\dfrac{247}{16}}{\dfrac{13}{4}}

Length=\dfrac{247}{16}\times \dfrac{4}{13}

Length=4.75


So, length of mirror is 4.75 feet which is less than 5 feet.

Therefore, the mirror will fit on the wall.

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The sum of squares of lengths of the two sides of a rectangular field is 25m2.Their difference is 1m .what is the length of each
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Answer:

Perimeter of a rectangle = 2×(Length+Breadth)

(i) Length = 16.8 cm

Breadth = 6.2 cm

Perimeter = 2×(Length+Breadth)

= 2×(16.8+6.2) =46 cm

(ii) Length = 2 m 25 cm

=(200+25) cm (1 m = 100 cm )

= 225 cm

Breadth =1 m 50 cm

= (100+50) cm (1 m = 100 cm )

= 150 cm

Perimeter = 2×(Length+Breadth)

= 2×(225+150) =750 cm

(iii) Length = 8 m 5 dm

= (80+5) dm (1 m = 10 dm )

= 85 dm

Breadth = 6 m 8 dm

= (60+8) dm (1 m = 10 dm )

= 68 dm

Perimeter = 2×(Length+Breadth)

= 2×(85+68) =306

5 0
2 years ago
You have been invited to be part of the planning committee for the County Fair. In order to be an effective member you will need
kakasveta [241]
Firstly let's find the dimension of this large rectangle:(given)

Area of Rectangle = 660 x 66 =43,560 ft²

And we know that 1 acre = 43,560 ft², then each rectangle has an area of 1 acre & the 20 acres will correspond to 20 x 43560 = 871,200 ft²

We know that the 20 acres form a rectangle. We need to know what is their disposition:

1) We would like to know the layout of the rectangles since we have 4 possibilities FOR THE LAYOUTS

Note that W=66 & L=666 = 43,956 ft²/ unit )

lay out shape could be either:(in ft)
1 W by 20 L  (Final shape Linear 66 x 13320 = 879,120) or
2 W by 10 L  (Final shape Stacked  132 x 6660 = 879,120) or
4 W by   5 L  (Final shape Stacked  264 x 3330 = 879,120) or

2) We would like to know the number of participants so that to allocate equal space as well as the pedestrian lane, if possible, if not we will calculated the reserved space allocated for pedestrian/visitors)

3) Depending on the shape given we will calculate the visitor space & we will deduct it from the total space to distribute the remaining among the exhibitors.

4) (SUGGESTION) Assuming it's linear, we will reserve
20ft x 13320 ft = = 266,400 ft² and the remaining 612,720 ft² for exhibitors
5) Depending on the kind of the exhibition, we will divide the 612,720 ft² accordingly
6) How can we select the space allocated for each exhibitor:
 the 617,720 ft² could be written as a product of prime factors:
612720 = 2⁴ x 3² x 5 x 23 x 37
If you chose each space will be185 ft² , then we can accommodate up to 3,312 exhibitors.
Obviously you can choose any multiple of the prime factors to specify the area allocated & to calculate the number of exhibitors accordingly







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4 years ago
4b^2•2b^4=? step by step
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