Which lengths can be used, directly or indirectly, to calculate the volume of the hexagonal right pyramid? select three options.
xy and st
vu and tw
xs and xw
tx and wx
vu and yz

See below.

Step-by-step explanation:

The volume is 1/3 * height * area of the base so we need the height and this area to find the volume. Note: if we join the point S to each of the vertices of the base we get 6 equilateral triangles so knowing the length of a side of the base we can find the area of the base.

The following lengths can be used:

XY and ST  We can get the area of the base knowing the length of XY and ST is the height.

VU and TW. From VU we can get the area of the base and the length of SW and from the length of TW and SW we can get the height.

TX and WX . From WX we can get the area of the base and the length of SX and from TX and SX we can find the height.

understand that i don't know the answers

will someone answer his question

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.