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5 mod 317. 317 = 5 * 63 + 2 5 = 2 * 2 + 1 2 = 1 * 2 + 0 GCD(317, 5) = 1. 1 = 5 - 2 * 2. How would you finish this via substitution? For example, 2 = 317 - 5 * 63. Correction: 1 = 5 - 2 * 2 1 = 5 * 1 + 2 * - 2 1 = 5 * 1 + (317 - 5 * 63) * - 2 1 = 5 (1 + 126) + 317 (-2) 127 = inverse.

5 mod 317. 317 = 5 * 63 + 2 5 = 2 * 2 + 1 2 = 1 * 2 + 0 GCD(317, 5) = 1. 1 = 5 - 2 * 2. How would you finish this via substitution? For example, 2 = 317 - 5 * 63. Correction: 1 = 5 - 2 * 2 1 = 5 * 1 + 2 * - 2 1 = 5 * 1 + (317 - 5 * 63) * - 2 1 = 5 (1 + 126) + 317 (-2) 127 = inverse.

A conical fuel tank is 2.5m deep and has a top diameter of 2m. Fuel is withdrawn from the tank at a rate of 0.25 m³ /min. At what rate is the level of fuel falling at the instant when the depth of fuel is 1.5m?

A conical fuel tank is 2.5m deep and has a top diameter of 2m. Fuel is withdrawn from the tank at a rate of 0.25 m³ /min. At what rate is the level of fuel falling at the instant when the depth of fuel is 1.5m?

y = { 0 when x <= 0 { x when 0 < = x

(3, 2, 0) and is parallel to (1, 2, 3). Write the equation of L in parametric form. Please explain the steps.

The Q. is to sketch the region formed by this expression --> arg(z + 2 - 3i) = π / 6. Let z = x + iy...

If w1 = -√3 + i is a cube root of z. Find z. Is there any quick method to getting the original angle (arg(z))? I don’t understand that part, in particular.

... look like if k was allowed to vary over all possible complex no.s? Note, z = 3+9i. k = every possible complex no. (& obv. all the purely reals too).

Simplify (1 + i / 1 - √3i)9. Many thanks!

Use De Moivre's theorem to find all z where (i) z^2 = i

Find the modulus of the following complex no. without multiplying into cartesian form: (a) 2 + 8i / 2 - 3i

Factorise (a) 3(a2 + 2b2) – (a + b)(a2 + 2b2) (b) (a + b)4 + (a - b)2(a + b)2 (c) (x + y)x2 – (x + y)y2

Solve (x2+6) + xi = 3(2+3i)(1-i) I ended up with (x2+6) + xi = 15 + 3i & solved xi = 3i to get x = 3. I'm wondering how else people do this?

Find z1z2 and z12 if z1 = cis(π / 6) & z2 = 3cis(-5π/6) Answer: z1z2 = 3e-2iπ / 3 = 3/2 (-1 - √3i) z12 = eiπ / 3 = ½(1 + √3i) Can someone please show working to reach this? π = pi.

Prove logb(x) = y is equivalent to b^y = x. If logb(x) = 5, then logb(x^2) is equal to? TBH, I'm more interested to see how people prove the log law (second Q. is just bringing down exponent).

1. 4x / (2x²+3) 2. x - 1 / x²

Given the point (2, -8) is a turning point of the curve y = f(x), f''(x) = 12x + 18.

Prove LHS equals the RHS: 1. (sec²Ѳ - 1)cos²Ѳ = sin²Ѳ 2. 1 / cot²Ѳ + 1 = 1 / cos²Ѳ

y = x^3-2x^2-4x+8 0 = x^3-2x^2-4x+8 I got x = + / - 2, but I'd really like to see how someone else derives their answer. Thanks.