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.
Answer:
this (｡ŏ_ŏ) statement is correct.
J) 180° - m<7 = m<1
Answer: D = 70
A = 30
Explanation: d is parallel to 70 degrees, 40 is parallel to e, and the whole set of angles is 180 so 180 - 2(70) + 2(40) = 30 so D = 70 and A = 30
The growth of the plant last year was 25 inches if the normal growth was ten inches more than twice the amount of last year.
<h3>What is linear equation?</h3>
It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.
If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.
The normal yearly growth of a plant is 60 inches.
Let's suppose the growth of the plant last year was x
The normal growth was ten inches more than twice the amount of last year.
From the above statement:
10 + 2x = 60
2x = 50
x = 25 inches
Thus, the growth of the plant last year was 25 inches if the normal growth was ten inches more than twice the amount of last year.
Learn more about the linear equation here:
brainly.com/question/11897796
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