Answer:
length, width , and perimeter
Step-by-step explanation:
27 8
Answer:
Answer:
They both have q+3/2p, so that means that 2PQ=CB and that means they are parallel to each other
Step-by-step explanation:
PQ=PA+QA
PQ=1/2(2q-p)+2/5*5p=q-1/2p+2p=q+3/2p
CB=2q+3p=2(q+3/2p)
Other explanation: It should be written like this PQ=q+3/2p and CB=2q+3p=2(q+3/2p) they are parallel bcs CB=2*PQ.
Answer:
A. 4
Step-by-step explanation:
Hope this helps!

- Determine the surface area of the right square pyramid.

The formula for finding the surface area of a right square pyramid is ⇨ b² + 2bl, where
- b = base of the right square pyramid
- l = slant height of the right square pyramid.
In the given figure,
- base (b) = 4 ft.
- slant height (l) = 8 ft.
Now, let's substitute the values of b & l in the formula & solve it :-

So, the surface area of the right square pyramid is <u>8</u><u>0</u><u> </u><u>ft²</u><u>.</u>