Suppose a plane intersects a cube through three intersecting edges. How can you ensure that the triangular cross section is an equilateral triangle? a. the corner of the cube can be sliced off such that the three vertices of the triangle are at the same distance from the corner of the cube.

b. the corner of the cube can be sliced off such that the two vertices of the triangle are at the same distance from the corner of the cube.

c. the corner of the cube can be sliced off such that the plane intersects the base.

d. the corner of the cube can be sliced off such that the plane is perpendicular to the base.

a 4/6 = 9/12   not proportional

step-by-step explanation:

to find which are not proportional, we can use cross products

if it is true, they are proportional.   if it is not true, they are not proportional.

a 4/6 = 9/12

4* 12 = 6*9

48 = 54

not this is not proportional  

b 2/3 = 12/18  

2* 18 = 3 * 12

36 = 36

this is proportional

c 5/4 = 10/8  

5*8 = 4*10

40 = 40

this is proportional

d 7/5 = 35/25

7*25 = 5 * 35

175 = 175

is proportional

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation:

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: