This photo is the problem I have to solve for x help please
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A diagram of parallelogram MNOP is attached below
We have side MN || side OP and side MP || NO
Using the rule of angles in parallel lines, ∠M and ∠P are supplementary as well as ∠M and ∠N.
Since ∠M+∠P = 180° and ∠M+∠N=180°, we can conclude that ∠P and ∠N are of equal size.
∠N and ∠O are supplementary by the rules of angles in parallel lines
∠O and ∠P are supplementary by the rules of angles in parallel lines
∠N+∠O=180° and ∠O+∠P=180°
∠N and ∠P are of equal size
we deduce further that ∠M and ∠O are of equal size
Hence, the correct statement to complete the proof is
<span>∠M ≅ ∠O; ∠N ≅ ∠P
</span>
We have side MN || side OP and side MP || NO
Using the rule of angles in parallel lines, ∠M and ∠P are supplementary as well as ∠M and ∠N.
Since ∠M+∠P = 180° and ∠M+∠N=180°, we can conclude that ∠P and ∠N are of equal size.
∠N and ∠O are supplementary by the rules of angles in parallel lines
∠O and ∠P are supplementary by the rules of angles in parallel lines
∠N+∠O=180° and ∠O+∠P=180°
∠N and ∠P are of equal size
we deduce further that ∠M and ∠O are of equal size
Hence, the correct statement to complete the proof is
<span>∠M ≅ ∠O; ∠N ≅ ∠P
</span>
The fundamental theorem of algebra states that a polynomial with degree n has at most n solutions. The "at most" depends on the fact that the solutions might not all be real number.
In fact, if you use complex number, then a polynomial with degree n has exactly n roots.
So, in particular, a third-degree polynomial can have at most 3 roots.
In fact, in general, if the polynomial has solutions
, then you can factor it as
So, a third-degree polynomial can't have 4 (or more) solutions, because otherwise you could write it as
But this is a fourth-degree polynomial.