Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps e2 into y2. Find the images of 4 −4 and

The image of through T is

Step-by-step explanation:

We know that  → is a linear transformation that maps into ⇒

And also maps into  ⇒

We need to find the image of the vector

We know that exists a matrix A from (because of how T was defined) such that :

for all x ∈

We can find the matrix A by applying T to a base of the domain ().

Notice that we have that data :

{}

Being the cannonic base of

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

  (Data of the problem)

  (Data of the problem)

Writing the matrix A :

Now with the matrix A we can find the image of such as :

⇒

We found out that the image of through T is the vector

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

191 students play sports and ninety-seven 10th graders were surveyed.

step-by-step explanation: