V = (pi) * r^2 * h
h = 7
V = 252(pi)
252(pi) = r^2 * 7
252(pi) / 7 = r^2
36(pi) = r^2
sqrt 36(pi) = r
6 = r <==== radius is 6 cm
The answer would be (2,3)
Answer:
The range of T is a subspace of W.
Step-by-step explanation:
we have T:V→W
This is a linear transformation from V to W
we are required to prove that the range of T is a subspace of W
0 is a vector in range , u and v are two vectors in range T
T = T(V) = {T(v)║v∈V}
{w∈W≡v∈V such that T(w) = V}
T(0) = T(0ⁿ)
0 is Zero in V
0ⁿ is zero vector in W
T(V) is not an empty subset of W
w₁, w₂ ∈ T(v)
(v₁, v₂ ∈V)
from here we have that
T(v₁) = w₁
T(v₂) = w₂
t(v₁) + t(v₂) = w₁+w₂
v₁,v₂∈V
v₁+v₂∈V
with a scalar ∝
T(∝v) = ∝T(v)
such that
T(∝v) ∈T(v)
so we have that T(v) is a subspace of W. The range of T is a subspace of W.
perimeter =QR+RS+ST+TU+UQ =26.1
RQ = 5.39
RS = 4.47
ST = 3.61
TU = 4.12
UQ = 8.54
Answer:
Step-by-step explanation:
so you will move constant to the right side and change the sign. 72x=223-7
then you will subtract the numbers 72x=223-7 and that will give you 72=216
then you will divide both sides by 72 and that will give you
- x=3 .
