Which pair of angles shares ray At as a common side ?

If AGHJ - ALMK, with a scale factor of 5:6, find the perimeter of AGHJ.

Sergeeva-Olga [200]

Answer:

I am pretty sure the answer is 10: 12 have a great day.

Step-by-step explanation:

7 0

4 years ago

The midpoint of AB is M(-5,1). If the coordinates of A are (-4,-5), what are the coordinates of B?

Elenna [48]

<u>ANSWER:</u>

The midpoint of AB is M(-5,1). The coordinates of B are (-6, 7)

<u>SOLUTION: </u>

Given, the midpoint of AB is M(-5,1).

The coordinates of A are (-4,-5),

We need to find the coordinates of B.

We know that, mid-point formula for two points A $(x_{1}, y_{1})$ and B $(x_{1}, y_{2})$ is given by

$M\left(x_{3}, y_{3}\right)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Here, in our problem, $\mathrm{x}_{3}=-5, \mathrm{y}_{3}=1, \mathrm{x}_{1}=-4 \text { and } \mathrm{y}_{1}=-5$

Now, on substituting values in midpoint formula, we get

$(-5,1)=\left(\frac{-4+x_{2}}{2}, \frac{-5+y_{2}}{2}\right)$

On comparing, with the formula,

$\frac{-4+x_{2}}{2}=-5 \text { and } \frac{-5+y_{2}}{2}=1$

$-4+\mathrm{x}_{2}=-10 \text { and }-5+\mathrm{y}_{2}=2$

$\mathrm{x}_{2}=-6 \text { and } \mathrm{y}_{2}=7$

Hence, the coordinates of b are (-6, 7).

5 0

4 years ago

PLEASE LET ME ADD PICS BEFORE ANSWERING!!!!!!

Helen [10]

1. inconsistant. point in quadrent 2.
2. equivilant. point in quadrent 1.
3. consistent. point on neg x-axis

Im pretty sure this is right but if not tell me

3 0

4 years ago

Read 2 more answers

Carlos uses a 3-D printer that melts plastic to produce custom action figures. The growth shows the amount of plastic it melts O

IceJOKER [234]

Answer and Step-by-step explanation:

It be more than 2 ounces and less than 1.5 ounces

The amount of plastic that is melted in 5.7 minutes is 1.71

To find this, you would use the point on the graph. Divide both of those numbers by 4.

(4, 1.2)

4/4 = 1.

1.2/4 = 0.3.

So, for every minute, 0.3 of an ounce of plastic is melted.

We then multiply 0.3 by 5.7 to get 1.71.

#teamtrees #PAW (Plant And Water)

8 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