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|A| = (2*3) - (1*3) → 6 - 3 = 3

Mary(m), Ben(b) Jane(j) m = b +3 m = j - 5 where m = x b = x - 3 and j = x + 5 b+ j → x - 3 + x + 5 = 2x +2 Jane and Ben have $(2x+2)

if 20 students = 10000 : #500 per student #1,000,000 \div 500 = 2,000 cover students

M = \(\frac{nk}{p^2}\) k → \(\frac{mp^2}{n}\) = \(\frac{3x1^2}{2}\) k = \(\frac{3}{2}\) : m = \(\frac{3xn}{2p^2}\)

Three times a certain number (x) = 3x minus 2 = -2 is less than = < the number minus 6 = x -6 Mathematically; 3x - 2 < x - 6 collect like terms 3x - x < -6 + 2 2x < -4 x < -2 is the possible values of x

square both sides to remove the big square root → r\(^2\) = \(\frac{3v}{\pih}\) cross multiply 3v = r\(^2\) * \pih v = \(\frac{\pir^2h}{3}\)

square both sides to remove the square root k\(^2\) = m\(^2\) \(\frac{t-p}{r}\) \(\frac{k^2r}{m^2}\) = t - p t = \(\frac{k^2r}{m^2}\) + p t = \(\frac{k^2r + pm^2}{m^2}\)

If "x" is the same for both equations, then they are equivalent. (i.e., the equations have same roots), then the equations are equivalent. Then p = -5

6x\(^2\) = 5x - 1 → 6x\(^2\) - 5x + 1 = 0 Factorization: factors are -3x and -2x 6x\(^2\) - 3x -2x + 1 = 0 (6x\(^2\) - 3x) (-2x + 1) = 0 3x(2x-1) -1(2x-1) = 0 (3x-1) (2x-1) = 0 x = \(\frac{1}{2}\), \(\frac{1}{3}\)

\(\frac{5b+(a+b)^2}{(a-b)^2}\) = \(\frac{5 * -7 +(3+ -7)^2}{(3- -7)^2}\) = \(\frac{-35 + 16}{10^2}\) = \(\frac{-19}{100}\) or -0.19

Tn = a + (n-1)d 10th term = 32 32 = 3/2 + 9d 9d = 32 - 3/2 9d = 61/2 d = 61/18 : T4 = 3/2 + 3(61/18) T4 = 3/2 + 61/6 → 35/3

The common difference is 6. Just add 6 to get the next term 8, 14, 20 and 26

common difference d = -1 - [-6] → -1 + 6 d = 5; a = -6 T\(_n\) = a + (n-1)d : T\(_{17}\) = -6 + (17 - 1) 5 → -6 + (16)5 → -6 + 80 = 74

3 * 5 = 3 - 5 + (3 x 5)\(^2\) -2 + (15)\(^2\) -2 + 225 = 223

Volume of a rectangular pyramid = \(\frac{length * width * height}{3}\) or \(\frac{area * height}{3}\) = \(\frac{24 * 7.5}{3}\) → \(\frac{180}{3}\) Volume of the rectangular pyramid = 60cm\(^3\)