Ok so first, write down the equation. We know that x=3 and y=4. You just basically plug in the terms into the variables. So it would be 3(3)+1/4(4)^2. Simplify that equation and you should get your answer :)
Y=55x+200, you multiply the number of weeks by the amount of money, and add 200 because they started with that.
Answer:
Option C
Step-by-step explanation:
We are given a coefficient matrix along and not the solution matrix
Since solution matrix is not given we cannot check for infinity solutions.
But we can check whether coefficient matrix is 0 or not
If coefficient matrix is zero, the system is inconsistent and hence no solution.
Option A)
|A|=![\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%266%5C%5C2%261%263%5C%5C-2%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%3D0)
since II row is a multiple of I row
Hence no solution or infinite
OPtion B
|B|=![\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%26-2%5C%5C-7%261%265%5C%5C4%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2810%29-2%2810%29%3D0)
Hence no solution or infinite
Option C
![\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%26-2%5C%5C-2%260%266%5C%5C1%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2836-2%29%3D68)
Hence there will be a unique solution
Option D
=0
(since I row is -5 times III row)
Hence there will be no or infinite solution
Option C is the correct answer
Answer:
The Symmetric Property states that for all real numbers x and y , if x=y , then y=x . Transitive Property. The Transitive Property states that for all real numbers x ,y, and z, if x=y and y=z , then x=z .
Step-by-step explanation: you couldnt have looked it up on ggle?...haha -_-'
Answer:
9·x² - 36·x = 4·y² + 24·y + 36 in standard form is;
(x - 2)²/2² - (y + 3)²/3² = 1
Step-by-step explanation:
The standard form of a hyperbola is given as follows;
(x - h)²/a² - (y - k)²/b² = 1 or (y - k)²/b² - (x - h)²/a² = 1
The given equation is presented as follows;
9·x² - 36·x = 4·y² + 24·y + 36
By completing the square, we get;
(3·x - 6)·(3·x - 6) - 36 = (2·y + 6)·(2·y + 6)
(3·x - 6)² - 36 = (2·y + 6)²
(3·x - 6)² - (2·y + 6)² = 36
(3·x - 6)²/36 - (2·y + 6)²/36 = 36/36 = 1
(3·x - 6)²/6² - (2·y + 6)²/6² = 1
3²·(x - 2)²/6² - 2²·(y + 3)²/6² = 1
(x - 2)²/2² - (y + 3)²/3² = 1
The equation of the hyperbola is (x - 2)²/2² - (y + 3)²/3² = 1.