We know that<span>
<span>Figures can be proven similar if one, or more,
similarity transformations (reflections, translations, rotations, dilations)
can be found that map one figure onto another.
In this problem to prove circle 1 and circle 2 are similar, a
translation and a scale factor (from a dilation) will be found to map one
circle onto another.
we have that</span>
<span> Circle 1 is centered at (5,8) and has a
radius of 8 centimeters
Circle 2 is centered at (1,-2) and has a radius of 4 centimeters
</span>
step 1
<span>Move the center of the circle 1 onto the
center of the circle 2
the transformation has the following rule</span>
(x,y)--------> (x-4,y-10)
so
(5,8)------> (5-4,8-10)-----> (1,-2)
so
center circle 1 is now equal to center circle 2
<span>The circles are now concentric (they have the
same center)
</span>
step 2
<span>A dilation is needed to decrease the size of
circle 1 to coincide with circle 2
</span>
scale factor=radius circle 2/radius circle
1-----> 4/8----> 0.5
radius circle 1 will be=8*scale factor-----> 8*0.5-----> 4 cm
radius circle 1 is now equal
to radius circle 2
<span>A
translation, followed by a dilation will map one circle onto the other,
thus proving that the circles are similar
the answer is
</span></span>The circles are similar because you can translate Circle 1 using the transformation rule (x-4,y-10) and then dilate it using a scale factor of (0.5)
In a triangle when an angle between two equal length sides is 60 that means that the triangle is equilateral.
Answer:
y*(x+2)=k
so y= k/(x+2)
Step-by-step explanation:
The giving functions are from dofferent types. transforming a function does not change it's type so it is impossible to transfer f(x)=2 to f(x)=(x-3)^2-1.
The answer should be 112 degrees
