Could you have a polynomial with solutions 2i and 5i? If not explain why not. If yes, then construct the polynomial.

If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi.  Imaginary solutions always come in pairs.

So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.

(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.

(x-2i)(x+2i)(x-5i)(x+5i)  = (x^2+4)(x^2+25)

                                   = x^4 + 29x^2 + 100

So the equation would be x^4 + 29x^2 + 100 = 0.

Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.

the correct answer is a) "catholic church".

the proposals that liberals in mexico made during 1855 were particularly focused on a person's individual rights, such as the freedom of speech, the freedom of conscience, freedom of assembly, and the right to possess weapons.

when the constitution was formally published 2 years later (1857). it also contained an article which permitted education free from any religious influence, the removal of the catholic church institutional privileges, and the sale of property belonging to the church. all of these provisions were clearly against the catholic church interests.  

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step-by-step explanation:

based on the question i am guessing that it is a typo for abc