Same for me it won’t let me pick a certain topic
Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be 2 - 5i and:

Answer B.
If a graph is proportional then the line will go through the origin at point (0, 0). If the equation is proportional then it will be in the form of y=kx with no other operations after. The constant of proportionality is another way to say the slope and in your specific equation the slope would be 1/5.
Answer is <span>16y + 7z
</span><span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>−<span>4z
</span></span><span>=<span><span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>+</span>−<span>4z
</span></span></span>Combine Like Terms:
<span>=<span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>+<span>−<span>4z
</span></span></span></span><span>=<span><span>(<span><span>9y</span>+<span>7y</span></span>)</span>+<span>(<span><span>11z</span>+<span>−<span>4z</span></span></span>)
</span></span></span><span>=<span><span>16y</span>+<span>7z</span></span></span><span>
</span>
Answer:
The given coordinates of the points A and C of the right triangle ABC are (2, 5) and (-2, 3) respectively
The possible pairs of the coordinates of the point B are found as follows;
1) Draw a horizontal line from the point C to the right and a vertical line down from the point A to get the point B at (2, 3) which is the point of the perpendicular intersection of the two constructed lines above
2) Draw a vertical line from the point C upwards and a horizontal line to the left from the point A to get the point B at (-2, 5) which is the point of the perpendicular intersection of the two newly constructed lines here
Step-by-step explanation: