Answer:
No for all of the 4 values
Step-by-step explanation:
The sum of the internal angles of a quadrilateral is always 360°, so we have that:
m∠A + m∠B + m∠C + m∠D = 360
11x - 8 + 3x2 + 1 + 15x + 32 + 2x2 - 1 = 360
5x2 + 26x - 320 = 0
Solving the quadratic function using Bhaskara's formula, we have:
Delta = 26^2 + 4*5*320 = 7076
sqrt(Delta) = 84.12
x1 = (-26 + 84.12)/10 = 5.81
x2 = (-26 - 84.12)/10 = -11.01 (this value will generate negative angles, so it is not valid).
Now, finding the angles, we have:
m∠A = 11*5.81 - 8 = 55.91°
m∠B = 3*(5.81)^2 + 1 = 102.27°
m∠C = 15*5.81 + 32 = 119.15°
m∠D = 2*(5.81)^2 - 1 = 66.51°
As all these values are different from the values shown, the answer is No for all of them.
on the matrix gives![\left[\begin{array}{cc|c}5&5&5\\1&-\dfrac23&5\end{array}\right]\to\left[\begin{array}{cc|c}1&-\dfrac23&5\\5&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%265%265%5C%5C1%26-%5Cdfrac23%265%5Cend%7Barray%7D%5Cright%5D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%26-%5Cdfrac23%265%5C%5C5%265%265%5Cend%7Barray%7D%5Cright%5D)