Given the vertices of M(1, 5) A(-1, 2), T(1, -1) H(3, 2), prove that MATH is a rhombus.
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About 102.5625... I think
We need to start from the innermost parenthesis and work our way out.
The first parenthesis is . These are not like terms because one involves a variable, while the other is a constant term. Two terms are summable if they involve the same power(s) of the same variable(s).
So, we can take one step outwards, and we arrive to the square brackets. We have
and 2z and -13z are like terms, so we can sum them:
Finally we arrive to the whole expression, which is
Because, again, 5z and 11z were like terms.