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1- For the points (-2,7) and (-4,5),(a) Find the exact distance between the points.(b) Find the midpoint of the line segment whose endpoints are the given points.----------------------------------------------------------------------------------------------------------------------------------2- Graph the equation y= -3x+2 on the viewing window defined by [-10,10,1} by [-10,10,1].---------------------------------------------------------------------------------------------------------------------------3-The endpoints of a diameter of a circle are (-7,2) and(-13,10) .(a) Write an equation of the circle in standard form.(b) Graph the circle.---------------------------------------------------------------------------------------------------------------------------4- Refer to the function . f= {(6,10), (-5,8), (4,6), (2,7)}For what value of X for which f(x)=10 is {.......} ?-------------------------------------------------------------------------------------------------------------------------5-Write the domain in interval notation.(a) w(x)= lx+1l+4(b) y(x)= X ➗ lx+1l+4(c) Z(x)= X ➗ lx+1l-4----------------------------------------------------------------------------------------------------------------------------6- Graph the equation and identify the x- and y-intercepts.3x=2y

Sagot :

1.

(a) square root of 8

[tex]\sqrt{8}[/tex]

(b) midpoint: (-3,8)

Explanation:

Data:

Point 1 : (-2,7)

Point 2: (-4,5)

Formula:

. Distance

[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

. Midpoint

[tex]M_{pt}=(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})[/tex]

Solution:

(a)

[tex]\begin{gathered} D=\sqrt{(-4-(-2))^2+(5-7)^2} \\ D=\sqrt{(-2)^2+(-2)^2} \\ D=\sqrt{4+4} \\ D=\sqrt{8} \end{gathered}[/tex]

(b)

[tex]\begin{gathered} M_{pt}=(\frac{-4+(-2)}{2},\frac{5+7}{2}) \\ M_{pt}=(\frac{-6}{2},\frac{12}{2}) \\ M_{pt}=(-3,6) \end{gathered}[/tex]

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