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kvv77 [185]
3 years ago
10

An architect makes a model of a new house. The model shows a tile patio in the backyard. In the​ model, each tile has length 1/4

in. and width 1/6 in. The actual tiles have length 1/6 ft and width 1/9 ft.
What is the ratio of the length of a tile in the model to the length of an actual​ tile? What is the ratio of the area of a tile in the model to the area of an actual​ tile
Mathematics
1 answer:
Semmy [17]3 years ago
6 0

Answer:

The answer is below

Step-by-step explanation:

The model is a tile with length 1/4 in. and width 1/6 in while the actual tile has length 1/6 ft. and width 1/9 ft.

Firstly we have to convert the model length and width from inches to feet.

1 feet = 12 inches

model length =  1/4 in. = 1/4 in * (1/12) ft./in = 1/48 ft.

model width =  1/6 in. = 1/6 in * (1/12) ft./in = 1/72 ft.

Therefore:

Ratio of model length to actual length of tile = (1/48 ft.) / (1/6 ft.) = 1 / 8

Ratio of model length to actual length of tile = (1/72 ft.) / (1/9 ft.) = 1 / 8

Area of model tile = length * width = 1/48 ft. * 1/72 ft. = 1/3456 ft²

Area of actual tile = length * width = 1/6 ft. * 1/9 ft. = 1/54 ft²

Ratio of model area to actual area of tile = (1/3456 ft²) / (1/54 ft²) = 1 / 64

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I'm blanking on how to do this, I learned it so long ago, any help would be greatly appreciated. More interested on how to do it
anyanavicka [17]

Answer:

\dfrac{16 y^{22}}{x^{10}z^{10}}

Step-by-step explanation:

Given expression is ,

\sf\longrightarrow \bigg(\dfrac{2x^3y^{-5}z^8}{8x^{-2}y^6z^3}\bigg)^{-2}

This would be simplified using the law of exponents , some of which I will use here are ,

  • (an)^m = a^m n^m
  • \dfrac{a^m}{a^n}=a^{m-n}

  • a^m a^n = a^{m+n}
  • a^{-n} = \dfrac{1}{a^n}

Using the above laws ,

\sf \longrightarrow \bigg[ \dfrac{2}{8} \bigg(\dfrac{x^3}{x^{-2}}\bigg)\bigg(\dfrac{y^{-5}}{y^6}\bigg)\bigg(\dfrac{z^8}{z^3}\bigg)  \bigg]^{-2}

Using the second law mentioned above , we have,

\sf \longrightarrow \bigg[ \dfrac{1}{4}(x^{3+2})(y^{-5-6})(z^{8-3})\bigg]^{-2}

Simplify ,

\sf \longrightarrow \bigg[\dfrac{1}{4} x^5y^{-11}z^5\bigg]^{-2}

Using the first law mentioned above , we have,

\sf \longrightarrow \bigg[ \dfrac{1}{4^{-2}} x^{5(-2)} y^{-11(-2)} z^{5(-2)}\bigg]

Simplify,

\sf \longrightarrow 4^2x^{-10}y^{22} z^{-10}

Finally using the fourth law mentioned above , we have ,

\sf \longrightarrow \boxed{\bf \dfrac{16 y^{22}}{x^{10}z^{10}}}

<h3>Option K is the correct answer.</h3>
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2 years ago
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