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3 years ago

The dimensions of a standard tennis court are 36 ft. 78 ft. with a net that is 3 ft. high in the center. The

Mathematics

1 answer:

Andreas93 [3]3 years ago

7 0

Answer:

The dilation factor of the width is 3/4

The dilation factor of the length is 10/13

The standard court and the modified court are not similar

Step-by-step explanation:

The given dimension of a tennis courts are given as follows;

The width of the standard tennis court = 36 ft.

The length of the standard tennis court = 78 ft.

The height of the net of the standard tennis court = 3 ft.

The width of the modified tennis court = 27 ft.

The length of the modified tennis court = 60 ft.

The height of the net of the standard tennis court = 3 ft.

The dilation factor of the width = (Modified court width)/(Standard court width)

∴ The dilation factor of the width = (27 ft.)/(36 ft.) = 3/4

Therefore, the width of the modified court = 3/4 × The width of the standard court

∴ The dilation factor of the length = (60 ft.)/(78 ft.) = 10/13

Therefore, the length of the modified court = 10/13 × The length of the standard court

The dilation factor for the width is different from the dilation factor for the length and the net of both courts are of equal dimension, therefore, based on similarity theorem, both courts are not similar.

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Step-by-step explanation:

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2 years ago

Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form

Alex787 [66]

Answer:

$y=-\frac{2}{5}x-\frac{1}{5}$

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

m is the slope

b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

$m = -\frac{2}{5}$

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

$y=-\frac{2}{5}x + b$

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,1). When x of the line is -3, y of the line must be 1.
(2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: $y=-\frac{2}{5}x + b$ . $b$ is what we want, the - $-\frac{2}{5}$ is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

(-3,1). y = mx + b or $1=-\frac{2}{5} * -3 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(-3)$ . $b = -\frac{1}{5}$ .
(2,-1). y = mx + b or $-1=-\frac{2}{5} * 2 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .