Answer:
{x,y} = {89/37,-71/37}
Step-by-step explanation:
8x = 3y + 25
[2] x = 3y/8 + 25/8
Plug this in for variable x in equation [1]
[1] 6•(3y/8+25/8) + 7y = 1
[1] 37y/4 = -71/4
[1] 37y = -71
Solve equation [1] for the variable y
[1] 37y = - 71
[1] y = - 71/37
By now we know this much :
x = 3y/8+25/8
y = -71/37
Use the y value to solve for x
x = (3/8)(-71/37)+25/8 = 89/37
The answer for Group A. is 0.080 or 0.08
the answer for Group B. 1.07 or 1.070
Answer:
The answer is 8.3 bar notation.
Step-by-step explanation:
2 1/2 into improper is 5/2
3 1/3 into improper is 10/3
Now Multiply
5/2 x 10/3 is 300/36
Now Reduce.
25/3 = 8.3 bar notation
Mark Brainliest!

![\sf \left[\begin{array}{cc}\sf 4&\sf 6\\ \sf 5 &\sf 8 \\ \sf 3 &\sf -2\end{array}\right]-\left[\begin{array}{cc}\sf 2&\sf 3\\ \sf 1 &\sf 4 \\ \sf -5&\sf3\end{array}\right]](https://tex.z-dn.net/?f=%5Csf%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%204%26%5Csf%206%5C%5C%20%5Csf%205%20%26%5Csf%208%20%5C%5C%20%5Csf%203%20%26%5Csf%20-2%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%202%26%5Csf%203%5C%5C%20%5Csf%201%20%26%5Csf%204%20%5C%5C%20%5Csf%20-5%26%5Csf3%5Cend%7Barray%7D%5Cright%5D)
Just substract corresponding terms
![\\ \sf\longmapsto \left[\begin{array}{cc}\sf 2 &\sf 3\\ \sf 4&\sf4\\ \sf 8&\sf -5\end{array}\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%202%20%26%5Csf%203%5C%5C%20%5Csf%204%26%5Csf4%5C%5C%20%5Csf%208%26%5Csf%20-5%5Cend%7Barray%7D%5Cright%5D)
Option B
<h3>Answer:</h3>
(x, y) ≈ (1.49021612010, 1.22074408461)
<h3>Explanation:</h3>
This is best solved graphically or by some other machine method. The approximate solution (x=1.49, y=1.221) can be iterated by any of several approaches to refine the values to the ones given above. The values above were obtained using Newton's method iteration.
_____
Setting the y-values equal and squaring both sides of the equation gives ...
... √x = x² -1
... x = (x² -1)² = x⁴ -2x² +1 . . . . . square both sides
... x⁴ -2x² -x +1 = 0 . . . . . polynomial equation in standard form.
By Descarte's rule of signs, we know there are two positive real roots to this equation. From the graph, we know the other two roots are complex. The second positive real root is extraneous, corresponding to the negative branch of the square root function.