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dmitriy555 [2]
3 years ago
9

Suppose that {v1, v2, . . . , vn} spans V and T∈L(V, W). Prove that {T(v1), T(v2), . . . , T(vn)} spans range(T).

Mathematics
1 answer:
serious [3.7K]3 years ago
3 0

Answer:

Step-by-step explanation:

it’s 6

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I can't seem to answer this question may you guys help me
lina2011 [118]
Options A, B and E are the correct options.

Expenses are represented with negatives and income by positive.
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3 0
3 years ago
I don't know what to do
RUDIKE [14]
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6 0
3 years ago
1 -3,45<br> Rational or irrational
Naddik [55]

Answer: rational

Step-by-step explanation:

A rational number is a number that can be expressed as a fraction. Hence 1-3.45 = - 2.45 which is - 49/20 as a fraction

6 0
3 years ago
Read 2 more answers
Define fn : [0,1] --&gt; R by the
sasho [114]

Answer:

The sequence of functions \{x^{n}\}_{n\in \mathbb{N}} converges to the function

f(x)=\begin{cases}0&0\leq x.

Step-by-step explanation:

The limit \lim_{n\to \infty }c^{n} exists and converges to zero whenever \lvert c \rvert. But, if c=1 the sequence \{c^{n}\} is constant and all its terms are equal to 1, then converges to 1. Using this result, consider the sequence of functions \{f_{n}\} defined on the interval [0,1] by f_{n}(x)=x^{n}. Then, for all 0\leq x we have that \lim_{n\to \infty}x^{n}=0. Now, if x=1, then \lim_{n\to \infty }x^{n}=1. Therefore, the limit function of the sequence of functions is

f(x)=\begin{cases}0&0\leq x.

To show that the convergence is not uniform consider 0. For any n>1 choose x\in (0,1)  such that \varepsilon^{1/n}. Then

\varepsilon

This implies that the convergence is not uniform.

8 0
3 years ago
Express as a ratio in whole numbers 1/2 : 1/4 : 1/12
Gelneren [198K]

Answer:

6 : 3 : 1

Step-by-step explanation:

Given

\frac{1}{2} : \frac{1}{4} : \frac{1}{12} ( multiply each part by 12 , the LCM of 2, 3, 12 to clear the fractions )

= 6 : 3 : 1

7 0
3 years ago
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