Try this solution:
1. Note, that 100 is divisible by 4, and 999 is not divisible by it, only 996. This is an arithmetic sequence.
2. a1;a2;a3;a4;...a(n) the sequence, where a1=100; a2=104; a3=108; a4=112; ... etc., and a(n)=996. n=?
3. using a formula for n-term of the sequence: a(n)=a1+d(n-1), where a(n)=996; a1=100 and d=4 (according to the condition ' is divisible by 4'). Then 100+4(n-1)=996; ⇒ 4n=900; ⇒ n=225 (including 100).
answer: 225
(8.6 x 108) x (3.2 x103
place them in order
(8.6 x 3.2) x (108 x 103)
27.52 x (108 x 103)
27.52 x 11124
That is your answer to this question.
(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
Answer:
choice 1) <u>-7 + 4 = -3</u>
Step-by-step explanation:
Since -3--7 = 4, -3 = -7 + 4, -7 = -3-4
4 greater than -7 is 3.
Answer:
Step-by-step explanation:
.65 x 12. Take answer and add 12
