Try 122.
<h3 />
Step-by-step explanation:
Since <em>x</em> = 1, x^6 = 1.
Since <em>y</em> = 11, 11^2 = 121.
1 + 121 = 122.
<h3 /><h3>
So, the value of <em>
x</em>
^ 6 + <em>
y </em>
^ 2 = 122.</h3><h3 />
That would a set of 5 numbers in ascending order with 4 being the middle one and the other 4 being the highest possible So it is
2, 3, 4, 9, 10
Mean is 28/5 = 5.6 answer
( I am assuming that each number is different)
Of duplicates is allowed then its 4 4 4 10 10 whose mean is 6.4
For the first one is definitely 0.12
The second one I was having trouble with, but I know it is not .25 or .75. Looking at the leftover answers I would say .36 because I already used .12. Also .52 and .25 are the givens.
Answer:
The completed proof is presented as follows;
The two column proof is presented as follows;
Statements 
Reason
1. 
║ 
, J is the midpoint of 
1. Given
2. ∠IHJ ≅ ∠JLK 2. Alternate angles are congruent
3. ∠IJH ≅ ∠KJL 3. Vertically opposite angles
4. 
≅ 
4. Definition of midpoint
5. ΔHIJ ≅ ΔLKJ 5. By ASA rule of congruency
Step-by-step explanation:
Alternate angles formed by the crossing of the two parallel lines and , by the transversal are equal
Vertically opposite angles formed by the crossing of two straight lines 
and are always equal
A midpoint divides a line into two equal halves
Angle-Side-Angle, ASA rule of congruency states that two triangles ΔHIJ and ΔLKJ, that have two congruent angles, ∠IHJ in ΔHIJ ≅ ∠JLK in ΔLKJ and ∠IJH in ΔHIJ ≅ ∠KJL in ΔLKJ, and that the included sides between the two congruent angles is also congruent ≅ , then the two triangles are congruent, ΔHIJ ≅ ΔLKJ.
(p + q)⁵
(p + q)(p + q)(p + q)(p + q)(p + q)
{[p(p + q) + q(p + q)][p(p + q) + q(p + q)](p + q)}
{[p(p) + p(q) + q(p) + q(q)][p(p) + p(q) + q(p) + q(q)](p + q)}
(p² + pq + pq + q²)(p² + pq + pq + q²)(p + q)
(p² + 2pq + q²)(p² + 2pq + q²)(p + q)
{[p²(p² + 2pq + q²) + 2pq(p² + 2pq + q²) + q²(p² + 2pq + q²)](p + q)}
{[p²(p²) + p²(2pq) + p²(q²) + 2pq(p²) + 2pq(2pq) + 2pq(q²) + q²(p²) + q²(2pq) + q²(q²)](p + q)}
(p⁴ + 2p³q + p²q² + 2p³q + 4p²q² + 2pq³ + p²q² + 2pq³ + q⁴)(p + q)
(p⁴ + 2p³q + 2p³q + p²q² + 4p²q² + p²q² + 2pq³ + 2pq³ + q⁴)(p + q)
(p⁴ + 4p³q + 6p²q² + 4pq³ + q⁴)(p + q)
p⁴(p + q) + 4p³q(p + q) + 6p²q²(p + q) + 4pq³(p + q) + q⁴(p + q)
p⁴(p)+ p⁴(q) + 4p³q(p) + 4p³q(q) + 6p²q²(p) + 6p²q²(q) + 4pq³(p) + 4pq³(q) + q⁴(p) + q⁴(q)
p⁵ + p⁴q + 4p⁴q + 4p³q² + 6p³q² + 6p²q³ + 4p²q³ + 4pq⁴ + pq⁴ + q⁵
p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵
