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)
Answer:
60°
Step-by-step explanation:
Total angle measurement for a triangle is 180°
Equilateral triangle measures 60 on all angles. (60 + 60 + 60 = 180°)
This triangle is equilateral so it's angle is 60°.
Why is it equilateral? Because all sides of the triangle are equal (10 on all sides).
Hope this helps
Answer:
Ty :)
Step-by-step explanation:
I lost all my points bc a moderator (LukeG1) >:(
If you are talking about the binomial being expanded then it would be:
8x^3 + 12x^2y + 6xy^2 + y^3
The y in the second term is not part of the exponent
And since you are raising the binomial to the third, you would be using the third row of Pascal's triangle.
Hope this helped!
