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gulaghasi [49]
2 years ago
6

Determine if the described set is a subspace. Assume a, b, and c are real numbers. The subset of R3 consisting of vectors of the

form [a b c] , where at most one of a , b and c is non 0.
The set is a subspace.
The set is not a subspace.
If so, give a proof. If not, explain why not.
Mathematics
1 answer:
AveGali [126]2 years ago
7 0

Answer:

Not a subspace

Step-by-step explanation:

(4,0,0) and (0,4,0) are vectors in R3 with zero or one entries being nonzero, but their sum, (4,4,0) has two nonzero entries.

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AnnZ [28]
2y+4=3(y-1)
2y+4=3y-3
4+3=3y-2y
Y=7
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Y=-3x + 7?<br><br> what is slope of line y
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Answer:

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Step-by-step explanation:

coefficent of x enough to understand slope

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GrogVix [38]

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Step-by-step explanation:

7 0
2 years ago
Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati
Brrunno [24]
Remember that <span>an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.
 
Lets solve our equation to find out what is the extraneous solution:
</span>\sqrt{x-3} =x-5
(\sqrt{x-3})^2 =(x-5)^2
x-3=x^2-10x+25
x^2-11x+28=0
(x-4)(x-7)=0
x-4=0 and x-7=0
x=4 and x=7
<span>
So, the solutions of our equation are </span>x=4 and x=7. Lets replace each solution in our original equation to check if they are valid solutions:
- For x=7
\sqrt{x-3} =x-5
\sqrt{7-3} =7-5
\sqrt{4} =2
2=2
We can conclude that 7 is a valid solution of the equation.

- For x=4
\sqrt{x-3} =x-5
\sqrt{4-3} =4-5
\sqrt{1} =1
1 \neq 1
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: <span>D. the extraneous solution is x = 4</span>
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3 years ago
Please post an provide Simplify 5p^2\5p^3
Rom4ik [11]
( p - 3) or (2p + 1)
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