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MariettaO [177]
2 years ago
7

Need HELP ASAPPPPP !!!! 100 points !!!!

Mathematics
1 answer:
Leokris [45]2 years ago
6 0

Answer:

C

Step-by-step explanation:

Using the trigonometric ratios, C would be correct.

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3.perpendicular

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Step-by-step explanation:

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3 years ago
Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?
sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have V=IR^{2} and let H be the subset of V of all points on the line

4x+3y=12

We need to find if H is a subspace of the vector space V.

In IR^{2} all the possibilities for own subspace of the vector space IR^{2} are :

  • IR^{2} itself.
  • The vector 0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]
  • All lines in IR^{2} that passes through the origin  (  0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]  )

We know that H is the subset of IR^{2} of all points on the line 4x+3y=12

If we look at the equation, the point \left[\begin{array}{c}0&0\end{array}\right] doesn't verify it because :

4x+3y=12\\4(0)+3(0)=12\\0=12

Which is an absurd. Therefore, H doesn't contain the origin (and H is a line in IR^{2}). Finally, it can't be a vector space of V=IR^{2}

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