Solution:
a * b = a + b + ab
b * a = b + a + ba
⇒ a * b = b * a
So * is commutative.
Now,
(a * b) * c
= (a + b + ab) * c
= a + b + ab + c + ca + cb + abc
a * (b * c)
= a * (b + c + bc)
= a + b + c + bc + ab + ac + abc
⇒ (a * b) * c = a * (b * c)
So * is associative.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
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($1$) Let $p_i$ be the probability that a randomly chosen point has $i$ many friends, $i=0,1,2,3,4$. Let $X$ be a random variable such that for $i=0,1,2,3,4$, the probability $P(X=i)=p_i$. Then the value of $7 E(X)$ is
($2$) Two distinct points are chosen randomly out of the points $A_1, A_2, \ldots, A_{4 g}$. Let $p$ be the probability that they are friends. Then the value of $7 p$ is
$z^5=1$ then value of $\left| {\begin{array}{*{20}{c}}
{{e^\alpha }}&{{e^{2\alpha }}}&{{e^{3\alpha + 1}}}&{ - {e^{ - \delta }}} \\
{{e^\beta }}&{{e^{2\beta }}}&{{e^{3\beta + 1}}}&{ - {e^{ - \delta }}} \\
{{e^\gamma }}&{{e^{2\gamma }}}&{{e^{3\gamma + 1}}}&{ - {e^{ - \delta }}}
\end{array}} \right|$
$\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
$\vec{\text{r}}.(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=7$
$\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})=7$
$\text{None of these}$