Question

4x²-24y²-6x+96y-8=0 sacar centro, focos y vértices de la elipse ​

Answer 1

2(P + Q + R) = 54

P + Q + R = 27.(1)⇒ R = 15, P = 9 and Q = 3.P.Q.R = 405 (B)

10a + b.a + b = 4 + b –a + a = 4⇒ a = 2ab = 16b =8216a16==

= 100a + 10b + c.100a + 10b + c = 10(a + b + c) + 9⇒

90a = 9c +9c =

1a1099a90−=− As a = 1, c = 9 (9)

2x – 3, 2x – 1, 2x+12x +3 .(2x – 3)

2 + (2x – 1)2 + 64 = (2x + 1)2 + (2x + 3)2 4x2 – 12x + 9 + 4x

2 – 4x + 1 + 64= 4x2 + 4x + 1 + 4x2 + 12x9⇒ 32x = 64⇒

x = 2 1, 3, 5 and 738.

100a + 10b + c.b = 2c, a = 2b

⇒ a = 2 (2c) = 4c.100a + 10b + c – (100c + 10b + a) = 59499 (a – c) = 594

⇒ a – c = 6As a = 4c, 4c – c = 3c = 6⇒ c = 2b= 2(2) = 4a = 4(2) = 8∴(842)39.

10 a + b.10 a + b = 4 (a + b)10a – 4a = 4b – b

⇒ 6a = 3b⇒

b = 12, 24, 36, 48(D)

40. 100a + 10b + c.a = b + 2c = b – 2a + b + c = 3 b = 18⇒

b = 6so a = 8 and c = 864. : (864)

3n – d = 3→ (1)313dn=+ ⇒3n = d +

L, F = 3S + 5F + 15 = 2 (S + 15) F as 3S + 5 i 3S + 5 + 15 = 2S + 30S = 10)43.

.12a + 18b = 84

→ (1)I2a6(2a) + 16b = 8012a + 16b = 80→

(2)(2) from (1) 2b = 4⇒ b = 2 (B)44. 5x59x26x106x4++=−+(4x + 6) (5x + 5) = (2x + 9) (10x – 6)

⇒ 20x2 + 30x + 20x + 30= 20x2 + 90x – 12x– 54⇒ 28x = 84⇒ x

Respuesta:

R/

12x– 54⇒ 28x = 84⇒ x

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