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Nutka1998 [239]
3 years ago
8

For questions 8 – 14, use the following functions:

Mathematics
1 answer:
pogonyaev3 years ago
8 0

we are given

f(x)=5

g(x)=\sqrt{x+2}

(8)

(f+g)(x)=f(x)+g(x)

we can plug it

(f+g)(x)=5+\sqrt{x+2}

(9)

(f-g)(x)=f(x)-g(x)

we can plug it

(f-g)(x)=5-\sqrt{x+2}

(10)

(f*g)(x)=f(x)*g(x)

we can plug it

(f*g)(x)=5\sqrt{x+2}

(11)

(\frac{f}{g} )(x)=\frac{f(x)}{g(x)}

we can plug it

(\frac{f}{g} )(x)=\frac{5}{\sqrt{x+2}}

(12)

(\frac{g}{f} )(x)=\frac{g(x)}{f(x)}

we can plug it

(\frac{g}{f} )(x)=\frac{\sqrt{x+2}}{5}

(13)

(fog)(x)=f(g(x))

f(x)=5

we can replace g(x)

we get

(fog)(x)=5

(14)

(gof)(x)=g(f(x))

f(x)=5

we can replace f(x)

(gof)(x)=\sqrt{f(x)+2}

we get

(gof)(x)=\sqrt{5+2}

(gof)(x)=\sqrt{7}


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Find the coordinates of the circumcenter for ∆DEF with coordinates D(1,3) E (8,3) and F(1,-5).
NemiM [27]

Let the coordinates of the circumcentre O of the triangle DEF be (x, y). Circumcentre of a triangle is equidistant from each of the vertices.

1. Distance OD=distance OE, then:

(x-1)^2+(y-3)^2=(x-8)^2+(y-3)^2.

2. Distance OD=distance OE, then:

(x-1)^2+(y-3)^2=(x-1)^2+(y+5)^2.

Solve the system:

\left\{\begin{array}{l}          (x-1)^2+(y-3)^2=(x-8)^2+(y-3)^2 \\          (x-1)^2+(y-3)^2=(x-1)^2+(y+5)^2         \end{array}\right.,

\left\{\begin{array}{l}          x^2-2x+1+y^2-6y+9=x^2-16x+64+y^2-6y+9 \\          x^2-2x+1+y^2-6y+9=x^2-2x+1+y^2+10y+25         \end{array}\right.,

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Then

x=\dfrac{63}{14} =\dfrac{9}{2} =4.5,\\  \\ y=-1.

Answer: the coordinates of the circumcenter for ∆DEF are x=4.5, y=-1.


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