Prove that if a ≡ b modulo m and c ≡ d modulo m, then (a + c) ≡ (b + d) modulo m. Conclude that there is a well-defined addition of equivalence classes modulo m given by [a]+[b] = [a+b].

answer is c

step-by-step explanation:

81/12=324/x

x=48

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629

step-by-step explanation:

let pq = qr = rs = x, and xz = y and zu = z.

if the horizontal lines are parallel, then corresponding segments are in proportion:

substitute:

therefore 3 = y = z.

xu = y + z → xu = 3 + 3 = 6