Answer:
i think it will be 20 unit diatance .
Answer:
x = y = 22
Step-by-step explanation:
It would help to know your math course. Do you know any calculus? I'll assume not.
Equations
x + y = 44
Max = xy
Solution
y = 44 - x
Max = x (44 - x) Remove the brackets
Max = 44x - x^2 Use the distributive property to take out - 1 on the right.
Max = - (x^2 - 44x ) Complete the square inside the brackets.
Max = - (x^2 - 44x + (44/2)^2 ) + (44 / 2)^2 . You have to understand this step. What you have done is taken 1/2 the x term and squared it. You are trying to complete the square. You must compensate by adding that amount on the end of the equation. You add because of that minus sign outside the brackets. The number inside will be minus when the brackets are removed.
Max = -(x - 22)^2 + 484
The maximum occurs when x = 22. That's because - (x - 22) becomes 0.
If it is not zero it will be minus and that will subtract from 484
x + y = 44
xy = 484
When you solve this, you find that x = y = 22 If you need more detail, let me know.
Sorry people send you links instead of actually helping you
Answer:
a) 
b) 
Step-by-step explanation:
a) 
1. Distribute the second power (2) outside the first pair of parenthesis:
= 
2. Distribute the third power (3) outside the second pair of parenthesis:
= 
3. Combine like terms:
--------------------------------------------
b) 
1. Factor the number 6 (= 2 · 3):
2. Cancel the common factor (2):
3. Cancel out 
in the numerator an denominator:
hope this helps!
Answer:
We have the equation
![c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=c_1%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_2%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_3%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_4%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Then, the augmented matrix of the system is
![\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D0%260%260%268%5C%5C0%260%264%264%5C%5C0%263%263%263%5C%5C1%261%261%261%5Cend%7Barray%7D%5Cright%5D)
We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:
![\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C0%263%263%263%5C%5C0%260%264%264%5C%5C0%260%260%268%5Cend%7Barray%7D%5Cright%5D)
This matrix is in echelon form. Then, now we apply backward substitution:
1.
2.
3.
4.
Then the system has unique solution that is 
and this imply that the vectors 
are linear independent.
