let's recall that a year has 12 months, so 15 months is really 15/12 years.
From the graph given
For f(x), where x = 3, on the graph f(3) is
For g(x), where x = 3, on the graph g(3) is
Hence,
The answer is the second option
The correct answer is 20 because if you solve A=hbb 1=5 8 2=20 Hope this helps! ;D
(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
x^2 -12 x = -28
(-12/2) ^ 2 =36
x^2 -12 x + 36 = -28 + 36
x^2 -12x + 36 = 8
(x-6)^2 = 8
take the square root of each side
x-6 = sqrt (8) x-6 = -sqrt (8)
you get a positive and a negative when taking the square root
x= 6 + sqrt (8)
x = 6 - sqrt (8)
sqrt (8) = 2 sqrt (2)
x= 6 + 2 sqrt (2)
x = 6 - 2 sqrt (2)
Answer:
x= 6 + 2 sqrt (2)
x = 6 - 2 sqrt (2)
