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4 years ago

10

Please help me on problem 2 of Solving Unit-Rate Problems

1 answer:

nlexa [21]4 years ago

8 0

The answer is A. c and d don't have the numbers and B has them backwards.

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3 years ago

The size of a television is the length of the diagonal of its screen in inches. The aspect ratio of the screens of older televis

Shalnov [3]

Answer:

The width and height of the old 35-inch television are 28 inches and 21 inches, respectively.

Step-by-step explanation:

35-inch television is a television whose screen has an hypotenuse ( $l$ ) of 35 inches and the aspect ratio of 4 : 3 means that 4 inches width per each 3 inches height. And by Pythagorean Theorem:

$r_{l} =\sqrt{r_{w}^{2}+r_{h}^{2}}$

Where:

$r_{l}$ - Hypotenuse rate, dimensionless.

$r_{w}$ - Width rate, dimensionless.

$r_{h}$ - Height rate, dimensionless.

If $r_{w} = 4\,in$ and $r_{h} = 3\,in$ , the hypotenuse rate is:

$r_{l} = \sqrt{4^{2}+3^{2}}$

$r_{l} = 5$

The width and height of the old television can be found with the help of trigonometric functions:

Width of 35-inch old television ( $w$ )

$w = l\cdot \cos \theta$

$w = l\times\left(\frac{r_{w}}{r_{l}} \right)$

Height of 35-inch old television ( $h$ )

$h = l\cdot \sin \theta$

$h = l\times\left(\frac{r_{h}}{r_{l}} \right)$

Where $\theta$ is the direction of the hypotenuse with respect to width, measured in sexagesimal degrees.

If $r_{w} = 4$ , $r_{h} = 3$ and $r_{l} = 5$ and $l = 35\,in$ , the width and height of the old 35-inch television:

$w = (35\,in)\times \left(\frac{4}{5} \right)$

$w = 28\,in$

$h =(35\,in)\times \left(\frac{3}{5} \right)$

$h = 21\,in$

The width and height of the old 35-inch television are 28 inches and 21 inches, respectively.

3 0

3 years ago