<h3>
Answer:</h3>
(x, y) = (7, -5)
<h3>
Step-by-step explanation:</h3>
It generally works well to follow directions.
The matrix of coefficients is ...
![\left[\begin{array}{cc}2&4\\-5&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%264%5C%5C-5%263%5Cend%7Barray%7D%5Cright%5D)
Its inverse is the transpose of the cofactor matrix, divided by the determinant. That is ...
![\dfrac{1}{26}\left[\begin{array}{ccc}3&-4\\5&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B26%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-4%5C%5C5%262%5Cend%7Barray%7D%5Cright%5D)
So the solution is the product of this and the vector of constants [-6, -50]. That product is ...
... x = (3·(-6) +(-4)(-50))/26 = 7
... y = (5·(-6) +2·(-50))/26 = -5
The solution using inverse matrices is ...
... (x, y) = (7, -5)
3x−x+4=4(2x−1)
Step 1: Simplify both sides of the equation.<span><span><span><span>3x</span>−x</span>+4</span>=<span>4<span>(<span><span>2x</span>−1</span>)</span></span></span><span>Simplify: (Show steps)</span><span><span><span>2x</span>+4</span>=<span><span>8x</span>−4</span></span>Step 2: Subtract 8x from both sides.<span><span><span><span>2x</span>+4</span>−<span>8x</span></span>=<span><span><span>8x</span>−4</span>−<span>8x</span></span></span><span><span><span>−<span>6x</span></span>+4</span>=<span>−4</span></span>Step 3: Subtract 4 from both sides.<span><span><span><span>−<span>6x</span></span>+4</span>−4</span>=<span><span>−4</span>−4</span></span><span><span>−<span>6x</span></span>=<span>−8</span></span>Step 4: Divide both sides by -6.<span><span><span>−<span>6x/</span></span><span>−6</span></span>=<span><span>−8/</span><span>−6</span></span></span><span>x=<span>4/3</span></span>Answer:<span>x=<span>4/<span>3</span></span></span>
This involves quite a lot of arithmetic to do manually.
The first thing you do is to make the first number in row 2 = to 0.
This is done by R2 = -3/2 R1 + R2
so the matrix becomes
( 2 1 1) ( -3 )
( 0 -13/2 3/2) (1/2 )
(5 -1 2) (-2)
Next step is to make the 5 in row 5 = 0
then the -1 must become zero
You aim for the form
( 1 0 0) (x)
(0 1 0) (y)
(0 0 1) ( z)
x , y and z will be the required solutions.
General equation of a circle with centre (h, k) is given by:
![(x - h)^{2} + (y - k)^{2} = r^{2}](https://tex.z-dn.net/?f=%28x%20-%20h%29%5E%7B2%7D%20%2B%20%28y%20-%20k%29%5E%7B2%7D%20%3D%20r%5E%7B2%7D)
Now, the origin is the centre and radius is 20, so substituting these points in yields:
![x^{2} + y^{2} = 20^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%20y%5E%7B2%7D%20%3D%2020%5E%7B2%7D)
6t-10is the awsner. If s is the number of nautical miles separating the ships, express s in terms of t(the number of hours after 12 noon). I formed the equation s squared to the second power equals (12t) squard to the second power + (10 - 16) squared to the second power