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3 years ago

6

A sand dune stands 5 feet above sea level. The hill is eroding at a rate of 1 foot per 20 years. Let y represent the height of t

he sand dune after x years. Which equation represents the situation?

2 answers:

Soloha48 [4]3 years ago

7 0

Answer:

$y=-\frac{1}{20}x+5$

Step-by-step explanation:

Here we are given that the present height of the sand dune is 5 ft and it is eroding 1 ft in every 20 years. If x represent the number of years and y represents the height of the sand dune. Hence we can consider the coordinates be (0,5) (20,4) (40,3) (60,2) (80,1) (100,0)

Let us evaluate the equation of the line plotted on these coordinates

Here we can see the y intercept is 5

Let us find the slope

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{5-4}{0-20}\\m=\frac{1}{-20}\\m=-\frac{1}{20}$

The equation in slope intercept form is given as

y=mx+c

where c is y intercept i.e. 5

$y=-\frac{1}{20}x+5$

Hence this is our required equation

avanturin [10]3 years ago

4 0

y = -1/2x + 5 should be your equation because it is 5 ft above sea level and its eroding, or going down, 1 ft per 20 years. So every year, the sand dune erodes 1/20 ft more.

Hope this helps!

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Radius: $r =\frac{\sqrt {21}}{6}$

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Step-by-step explanation:

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$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

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$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

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$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

3 years ago