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3 years ago

EXPERTS/ACE/TRUSTED HELPERS

Mathematics

1 answer:

LekaFEV [45]3 years ago

3 0

The sign of each Y coordinate changed

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3/8, -2 , 34% , -4/9 from least to greatest

Whitepunk [10]

-2, -4/9 , 34%, 3/8. This is your answer

6 0

3 years ago

2+4(x+1)-x<br>2+4x+-x<br>6+4x-x<br>6+3x

Nat2105 [25]

Answer:

What is your question

Step-by-step explanation:

5 0

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 $V=IR^{2}$

8 0

3 years ago

How would I solve this?

trasher [3.6K]

You had the right idea using the Pythagorean theorem to solve for b.

Problem is for that triangle to work, the 5 and the 2√2 would have to switch places. The length of a leg cannot be larger than the length of the hypotenuse for it to truly be a right triangle.

Pythagorean theorem only works for the right triangles. Only way to "solve this problem would be to bring in complex numbers.

5² + b² = (2√2)²

25 + b² = 2²(√2)²

25 + b² = 4(2)

25 + b² = 8

b² = 8 - 25

b² = - 17

b = √-17

b= (√17i)

Then the problem with THIS is a measurement/distance cannot be negative... which goes against exactly what that complex number i is.

8 0

3 years ago

Find the cubic equation that has -1 and 2i as roots

ryzh [129]

The cuic equation is -1 and 2i

6 0

3 years ago