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Pani-rosa [81]
3 years ago
7

Order the decimals in each from least to greatest .178 .084 .009 .2

Mathematics
2 answers:
xxTIMURxx [149]3 years ago
5 0

Answer:

.009, .084, .178, .2

Step-by-step explanation:

nignag [31]3 years ago
3 0
.009 , .084, .178, .2
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Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec
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Answer:

(2,6,6) \not \in \text{Span}(u,v)

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Step-by-step explanation:

Let b=(b_1,b_2,b_3) \in \mathbb{R}^3. We have that b\in \text{Span}\{u,v\} if and only if we can find scalars \alpha,\beta \in \mathbb{R} such that \alpha u + \beta v = b. This can be translated to the following equations:

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2.2\alpha+4 \beta = b_2

3. 3 \alpha +2 \beta = b_3

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for \alpha,\beta and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

-\alpha + 3 \beta = 2

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whose unique solutions are \alpha =1 = \beta, but note that for this values, the third equation doesn't hold (3+2 = 5 \neq 6). So this vector is not in the generated space of u and v.

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Using equations 1 and 2 we get

-\alpha + 3 \beta = -9

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Prove ABE congruent ACD​
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