if f(x)=3/x+2-sqrtx-3 complete the following statement. The domain for f(x) is all real numbers___ than or equal to 3

1 answer:

6 0

The domain of the function is the set of all the values of x that would allow the function to have real values. Since the radicand is x - 3, the value of x cannot be less than 3 because that would make the value of sqrt (x - 3) an imaginary number. Thus, the domain should be all numbers GREATER than or equal to 3.

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Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

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Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$