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2 years ago

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Calculate the distance between the points K= (-8, 0) and Q= (-3, -3) in the coordinate plane round your answer to the nearest hu

ndredth

1 answer:

Drupady [299]2 years ago

3 0

Answer:

5.83

Step-by-step explanation:

Make a third point such that line KQ is the hypotenuse of a right triangle. See attachment.

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Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac

Degger [83]

Answer:

Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set $w_1,\dots w_n$ is linearly independent if and only if the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$ implies that

$\lambda_1 = \cdots = \lambda_n$ .

Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$

8 0

3 years ago

If x = 11 units and h = 7 units, then what is the area of the triangle shown?

m_a_m_a [10]

The answer is A because of base times height

3 0

3 years ago

Read 2 more answers

In this geometric sequence, what is the common ratio?

mrs_skeptik [129]

Answer:

-2

Step-by-step explanation:

Each value is divided by -2 to get to the next term

4 0

4 years ago

Read 2 more answers

Canada has a population that is about 1/10 as large as the United States. If Canada's population is about 32,000,000, about how

Nezavi [6.7K]

Canada's populations times 10 and you will get the answer.

8 0

4 years ago

Please please please help me I'm really confused and I don't understand it

maxonik [38]

Answer : p=2

2p-2-5x=-2(7-2x)

First, work on the parentheses. So the -2 on the outside as to be multiplied by the 7 , which gets -14 . Then you multiply the -2 by the 2x, which is 4x.

It will go from -2(7-2x) to -14+4x
But if the -14+4x looks too confusing, you can just switch them around to look “normal”. So it would be 4x-14 .

Now that we’re finished with one side of the problem, we’ll go to the other side.

On the left side , there are two variables, or letters. Because they are the same letter (p) all you have to do is add them together.

3p
-5p
+—-
-2p

And then just carry the -2 that was originally there , so it’s -2p-2

Now the equation is
-2p-2 = 4p-14

So we’re going to add 2p to both sides.

The -2p on the left cancels out, and the 4p on the right side becomes 6p

-2 = 6p-14

Next, you are going to want to isolate the variable. (Leave the letter by itself) so you add 14 to the -14 , so it can cancel out. (AND DO THE SAME ON THE OTHER SIDE)
So, you add 14 to the -2 on the left of the equation. Which comes out to 12.

12 = 6p

Lastly, divide the 6p , so you can get the letter on it’s own. And do the same on the other side.

12 divided by 6 is 2

2 = p

OR

p = 2

4 0

4 years ago