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denis23 [38]
3 years ago
15

What is v in 55 -7v 6?

Mathematics
1 answer:
Oxana [17]3 years ago
7 0
The v is the variable
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I need help don’t understand
nadya68 [22]
If f(x)=x+1 and then x became 2, you would have the function f(x)=3. So basically for that function you would be going up three over 1. That function is already g(x)=4x. If X became 2, you would have g(x)=8x. The rate of up 8 and then over one. Because of that, g(x) would be higher
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3 years ago
Read 2 more answers
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}

8 0
3 years ago
Which pair of angles are vertical angles?
Vitek1552 [10]
Answer is C. HMG and LMK
see attached picture to explain
Vertical angles are congruent and opposite of each other where two lines cross

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2 years ago
Find the slope of the line that passes through (4, 6) and (2, 3).
Verizon [17]
The slope is 3/2

To find the slope, you have to use the slope formula

y2-y1/x2-x1

6-3/4-2

3/2

So the slope is 3/2
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3 years ago
Help please! This is for geometry! Question below!!
labwork [276]
<span>1.) Circumscribed angle 2.) Minor arc 3.) Central Angle 4.) Inscribed Angle 5.)Major Arc</span>
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3 years ago
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