CoolBird2006

An arithmetic sequence has a common difference between each term

Let the common difference be n

The first term = 2

The second term = 2+n

The third term = 2+2n

The fourth term = 2+3n

The fifth term = 2+4n

The sixth term = 2+5n

And so on...

The sum of the third and sixth terms is 25 so:

2+2n+2+5n = 25

7n+4 = 25

7n = 21

n = 3

Thus, the fourth term is 2+3n = 2+9 = 11

Each term is the previous term multiplied by 1/4

128 * 1/4 = 32 = x

32 * 1/4 = 8 = y

x + y = 32 + 8 = 40

The 21st of January would be Sunday.

y = a(x-1)(x-5)

7 = 1(-1)(-5)

7 = 5a

a = 7/5

Therefore, the quadratic formula is y = (7/5)(x-1)(x-5)

The answer is 1/2

@ElectricPavlov Thank you for working out this question!

10 is the answer.

1048576

The answer is 463 feet.

11 is the answer.

n = 20

(a).\(\sqrt{3} \over 4\)

(b).\(1 \over 2\)

(c).\(\sqrt{3} \over 4\)

\(ab + c = bc + a\\ ab - bc = a-c \\ b(a-c) = a-c \\ b(a-c) - (a-c) = 0\\ (a-c)(b-1)=0\\ a=c\\ b=1 \)

Therefore, a and c must equal to 6.5, though the question states that all of these numbers are integers.

\(6p^6 \times q^4 + 15p^4 \times q^2 + 3p^4 \times q^3 + 9p^5 \times q^2\)

\(j = {\pm \sqrt {32} - 3}\)