Consider the following two statements:I. Any | vedclass

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(B) Statement $I$ is false. $A$ pair of consistent linear equations can have either a unique solution (intersecting lines) or infinitely many solution

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both $I$ and $II$ are false

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(B) Statement $I$ is false. $A$ pair of consistent linear equations can have either a unique solution (intersecting lines) or infinitely many solutions (coincident lines).Statement $II$ is false. Let the two consecutive integers be $x$ and $x+1$. According to the problem,$x^2 + (x+1)^2 = 365$.$x^2 + x^2 + 2x + 1 = 365$$2x^2 + 2x - 364 = 0$$x^2 + x - 182 = 0$$(x + 14)(x - 13) = 0$So,$x = 13$ or $x = -14$.If $x = 13$,the integers are $13$ and $14$. Check: $13^2 + 14^2 = 169 + 196 = 365$.Since such integers exist,statement $II$ is false.Therefore,both $I$ and $II$ are false.

Consider the following two statements:
$I$. Any pair of consistent linear equations in two variables must have a unique solution.
$II$. There do not exist two consecutive integers,the sum of whose squares is $365$.
Then,

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Consider the following two statements:$I$. Any pair of consistent linear equations in two variables must have a unique solution.$II$. There do not exist two consecutive integers,the sum of whose squares is $365$.Then,