<span>If set X is made up of the possible ways five students, represented by A, B, C, D, and E, can be formed into groups of three, then the set X consists of such triples {ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE} (note that triple ABC is the same as triple ACB, or BCA, or BAC, or CAB, or CBA). The set X totally contains 10 elements (triples). The first statement is true.
</span>
<span>If set
Y is made up of the possible ways five students can be formed into
groups of three if student A must be in all possible groups, then </span><span>the set Y consists of such triples {ABC, ABD, ABE, ACD, ACE, ADE} and contains 6 elements. The second statement is also true.
</span>
<span>If person E must be in each group, then there can be only one group is false statement, because you can see from the set X that triples which contains E are 6.
</span>
<span>There are three ways to form a group if persons A and C must be in it. This statement is true and these groups are ABC, ACD, ACE.</span>
1. Angle-Side-Angle (ASA) Postulate.
2. corresponding parts of congruent triangles are congruent (CPCTC).
Your question is not clear make it a little clearer I like to help ypu
First you have to subtract 4/11 from the variable side of the equation and the right side as well. Then you'll have s=9/22-8/22, so your answer is 1/22
1-6x=x+14 so that means x=14/-5 i think