The GCF of 6x^2y−24x^3+18xy^4 is

SU and VT are chords that intersect at point R.

vovikov84 [41]

Answer:

<h2>The length of the line segment VT is 13 units.</h2>

Step-by-step explanation:

We know that SU and VT are chords. If the intersect at point R, we can define the following proportion

$\frac{RS}{RT} =\frac{RV}{RU}$

Where

$RS=SR=x+6\\RT=x+4\\VR=RV=x+1\\RU=x$

Replacing all these expressions, we have

$\frac{x+6}{x+4} =\frac{x+1}{x}$

Solving for $x$ , we have

$x(x+6)=(x+4)(x+1)\\x^{2} +6x=x^{2} +x+4x+4\\6x-5x=4\\x=4$

Now, notice that chord VT is form by the sum of RT and RV, so

$VT=VR+RT\\VT=x+1+x+4\\VT=2x+5$

Replacing the value of the variable

$VT=2(4)+5\\VT=8+5\\VT=13$

Therefore, the length of the line segment VT is 13 units.

5 0

3 years ago

Read 2 more answers

Help I need the answers in 20 minutes

Gelneren [198K]

2. 5 hours

3. $200

4. 9.5 hours. If it has to be whole numbers, 10 hours

Hope that helps!

6 0

3 years ago

Read 2 more answers

Rezolvati 1) -2x+1= -7 2) 6+3x= -54+x 3) 2x+9= 7 +x 4) 3-2x= -x -6 5) -3(x+1) = -6 6) -4 (x -2)=12 7) -7(x-3)= -14

Natalija [7]

<span>Răspunsurile sunt

1. x= 4
2. x = -30
3. x = -2
4. x = 9
5. x = 1
6. x = -1
7. x = 5
</span>

3 0

3 years ago

PLEASE HELP!

sesenic [268]

Answer:

$36\sqrt{3}$

Step-by-step explanation:

Hi!

Since an equilateral triangle means that every side is equal, our triangle will have 12 on all sides.

To find the height of an equilateral triangle we use $\frac{a\sqrt{3}}{2}$ .

$\frac{12\sqrt{3} }{2} = 6\sqrt{3}$ .

So the height is $6\sqrt{3}$ .

Now we have to solve 12 * $6\sqrt{3}$ ÷ 2.

$12 \cdot 6\sqrt{3} = 72\sqrt{3}$

$72\sqrt{3}\: \div\: 2 = 36\sqrt{3}$ .

Thus, the area of the triangle is $\boxed{36\sqrt{3}}$ .

Hope this helps!

4 0

3 years ago

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