Answer:
Step-by-step explanation:
These come directly from my textbook, so I'm not sure if your teacher will accept this kind of work.
1. Angle construction:
Given an angle. construct an angle congruent to the given angle.
Given: Angle ABC
Construct: An angle congruent to angle ABC
Procedure:
1. Draw a ray. Label it ray RY.
2. Using B as center and any radius, draw an arc that intersects ray BA and ray BC. Label the points of intersection D and E, respectively.
3. Using R as center and the same radius as in Step 2, draw an arc intersecting ray RY. Label the arc XS, with S being the point where the arc intersects ray RY.
4. Using S as center and a radius equal to DE, draw an arc that intersects arc XS at a point Q.
5. Draw ray RQ.
Justification (for congruence): If you draw line segment DE and line segment QS, triangle DBE is congruent to triangle QRS (SSS postulate) Then angle QRS is congruent to angle ABC.
You can probably also Google videos if it's hard to imagine this. Sorry, construction is super hard to describe.
5:65h
6:110x
7:12x
I hope this helps
0.78 converted to a percent would be 78%
Answer:
Step-by-step explanation:
Since the box contains a cylinder, the height of the box must be gt; = cylinder height
And the area at the bottom of the box is definitely larger than the area at the bottom of the cylinder.
The area at the bottom of the cylinder is PI r squared,
If the bottom of the box is a square, then the area of the bottom is 4r^2, whereas the bottom tends to be polygon, which is equivalent to cutting the circle into equal parts by circle cutting.
So the area at the bottom of the box is 
，
It follows that the volume of the box is greater than the volume of the cylinder
