Answer:
the answer is 56.25
Step-by-step explanation:
<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>
Answer:
x^2+y^2=r^2 is quation of circle whose centre is (0,0)
Answer:
x = 2
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
m(x) = 4x - 10
m(x) = -2
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute: -2 = 4x - 10
- Isolate <em>x</em> term: 8 = 4x
- Isolate <em>x</em>: 2 = x
- Rewrite: x = 2