Answer:
The image of ![\left[\begin{array}{c}4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D)
through T is ![\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
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 :
![T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%260%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%265%5Cend%7Barray%7D%5Cright%5D)
(Data of the problem)
![T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%261%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-2%267%5Cend%7Barray%7D%5Cright%5D)
(Data of the problem)
Writing the matrix A :
![A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now with the matrix A we can find the image of ![\left[\begin{array}{c}4&-4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D)
such as :
⇒
![T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
We found out that the image of through T is the vector
Answer:
33.33%
Step-by-step explanation:
We are told that the customer paid Rs. 2034 after getting 10% discount with 13% vat on marked price (m.p.)
hence:-
2034 = m.p. × 90/100 × 113/100
m.p = (2034 × 100 × 100)/(90 × 113)
m.p. = Rs.2000
Now, due to the fact that VAT (which in this question is given to be 13%) is not the profit of the retailer, thus the selling price (s.p.) of the bag would be given by;
s.p = m.p. × 90/100
s.p = 2000 × 90/100
s.p = Rs. 1800
We are told that the retailer made a profit of 20%
Thus:-
c.p. × 120/100 = s.p.
c.p.= s.p. × 100/120
c.p.= 1800 × 100/120
c.p. = Rs.1500
Therefore, the percentage with which he marked above the c.p is;
% mark up = (m.p - c.p)/c.p) × 100
Plugging in the relevant values, we have;
(2000 - 1500)/1500) × 100
(500/1500) × 100 = 33.33%
<h3>
Answer:</h3>
- using y = x, the error is about 0.1812
- using y = (x -π/4 +1)/√2, the error is about 0.02620
<h3>
Step-by-step explanation:</h3>
The actual value of sin(π/3) is (√3)/2 ≈ 0.86602540.
If the sine function is approximated by y=x (no error at x = 0), then the error at x=π/3 is ...
... x -sin(x) @ x=π/3
... π/3 -(√3)/2 ≈ 0.18117215 ≈ 0.1812
You know right away this is a bad approximation, because the approximate value is π/3 ≈ 1.04719755, a value greater than 1. The range of the sine function is [-1, 1] so there will be no values greater than 1.
___
If the sine function is approximated by y=(x+1-π/4)/√2 (no error at x=π/4), then the error at x=π/3 is ...
... (x+1-π/4)/√2 -sin(x) @ x=π/3
... (π/12 +1)/√2 -(√3)/2 ≈ 0.026201500 ≈ 0.02620
The average price of such a house is
(97.86/ft²)×(1400 ft²) = 97.86×1400 = 137,004
Aaa^3bxa^2b^3abx^4
=aaa^3a^2abb^3bxx^4
=a^(1+1+3+2+1)b^(1+3+1)x^(1+4)
=a^8^b^5x^5
