Can somebody help me with this problem

Mathematics

2 answers:

lesya [120]3 years ago

7 0

3/5 If I'm not mistaken.

igomit [66]3 years ago

7 0

It would be 5/10, or 1/2 for the simplest form.

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The vertices of a parallelogram are A(x1, y1), B(x2, y2), C(x3, y3), and D(x4, y4). Which of the following must be true if paral

d1i1m1o1n [39]

Answer:

The following points are not arranged in a parallelogram or rectangle order.

Step-by-step explanation:

Well first we need to graph the following.

A(1,1) B(2,2) C(3,3) D(4,4)

By looking at the image below we can tell it is not any shape, it’s not a parallelogram or a rectangle.

It is a line with a slope of 1 or x.

8 0

3 years ago

If T: (x, y) → (x + 6, y + 4), then T-1: (x,y) → _____.

alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

$\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]$

The inverse of the matrix is

$\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]$

So, the Inverse Transformation is given by

$T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$