A pyramid 8cm high stands on a rectangular base 6cm by 4cm. Calculate the volume of the pyramid

Answer 1

Volume of all "pointy" things = (1/3)(area of the base)(height)
you have the data.
let me know what you get so I can check it

Answer 2

Volume=1/3*(6cm*4cm)*8cm
=64cm^3

Answer 3

I can't say

Answer 4

Like it

Answer 5

V=1/3 times6cm^2 times4cm times8cm=64cm^3

from Prthogoras' theorem
DB^2=(6cm)^2+(4cm)^2
=36^2+16^2
oB^2=52cm^2
Find the square root of 52cm^2=
DB=7.21cm
XB=7.21/2=36.1cm

Answer 6

There seems to be an error in your calculation. Here's the correct solution:
Using the formula V = (1/3) * base area * height, where base area = length * width:
V = (1/3) * 6cm * 4cm * 8cm
V = 64 cm^3

Therefore, the volume of the pyramid is 64 cm^3.

For the second part of the problem:
Using the Pythagorean theorem, we have:
DB = sqrt((6cm)^2 + (4cm)^2)
DB = sqrt(36cm^2 + 16cm^2)
DB = sqrt(52cm^2)
DB = 7.2111cm (rounded to 4 decimal places)

Since D is the midpoint of XB, we have:
XB = 2 * DB
XB = 2 * 7.2111cm
XB = 14.4222cm (rounded to 4 decimal places)

Therefore, XB is approximately 14.4222cm.

Answer 7

16c+20
C=20-16
=4

Answer 8

That is correct.

If you have an expression 16c + 20 and you need to evaluate it when c = 4, you would substitute 4 in place of c to get:

16(4) + 20 = 64 + 20 = 84