We know that
<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.
</span>we have that
<span>Circle 1 is centered at (4,3) and has a radius of 5 centimeters
</span><span> Circle 2 is centered at (6,-2) and has a radius of 15 centimeters
</span>
step 1
<span>Move the center of the circle 1 onto the center of the circle 2
</span>the transformation has the following rule
(x,y)--------> (x+2,y-5)
so
(4,3)------> (4+2,3-5)-----> (6,-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
A dilation is needed to increase the size of circle 1<span> to coincide with circle 2
</span>
scale factor=radius circle 2/radius circle 1-----> 15/5----> 3
radius circle 1 will be=5*scale factor-----> 5*3-----> 15 cm
radius circle 1 is now equal to radius circle 2
A translation, followed by a dilation<span> will map one circle onto the other, thus proving that the circles are similar</span>
Answer:
36 milliliters of rain.
Step-by-step explanation:
The rate at which rain accumluated in a bucket is given by the function:

Where r(t) is measured in milliliters per minute.
We want to find the total accumulation of rain from <em>t</em> = 0 to <em>t</em> = 3.
We can use the Net Change Theorem. So, we will integrate function <em>r</em> from <em>t</em> = 0 to <em>t</em> = 3:

Substitute:

Integrate:

Evaluate:

36 milliliters of rain accumulated in the bucket from time <em>t</em> = 0 to <em>t</em> = 3.
Answer:
9
Step-by-step explanation:
Answer:
The volume of the prism is 1/4 of the volume of the rectangular box.
Step-by-step explanation:
The figure alluding to the exercise is required, according to the description I will attach the one that must be to be able to solve the exercise.
The first thing is that the cross sections of the prism are triangles and in addition those triangles are congruent to each other with areas equal to the area of the base triangle.
By congruence we can say that the triangle has 1/4 of the area of the base rectangle. We can affirm that the height of the prism is equal to the height of the rectangular box.
Now, the Cavalieri principle states that if two solids have the same height and their cross-sectional areas taken parallel and at equal distances from their bases are always equal, then they have the same volume.
now in this case the cross-sectional areas (parallel to the base) of the prism and the cross-sectional areas (parallel to the base) of the cuboid with a height equal to that of the rectangular box and the length, width of half of the sizes of the rectangular box are always the same.
Which means that the volume of the parallelepiped is 1/4 the volume of the rectangular box and thanks to this we can say that the volume of the prism is 1/4 of the volume of the rectangular box.