Matemáticas, pregunta formulada por Ariadna1628, hace 1 año

Por favor ayúdenme con esto!!

Adjuntos:

Respuestas a la pregunta

Contestado por Akenaton
1
Expresion: h²j - 5j²∛(h) - 3

Para h = -1: j = 2

h²j = (-1)²(2) = (1)(2) = 2

5j²∛(h) = 5(2)²∛(-1) = 5(4)(-1) = 20(-1) = -20

h²j - 5j²∛(h) - 3 = 2 - 20 - 3 = -21

b) h = 1; j = -2

h²j - 5j²∛(h) - 3

h²j = (1)²(-2) = (1)(-2) = -2

5j²∛(h) = 5(-2)²∛(1) = 5(4)(1) = 20(1) = 20

h²j - 5j²∛(h) - 3 = - 2 + 20 - 3 = 15

2) (1/3)x²y - (2/5)xy + 2y²

a) Para x = 1/2; y = -1/3

(1/3)x²y = (1/3)((1/2)²)(-1/3) = (-1/9)(1/4) = -1/36

- (2/5)xy = -(2/5)(1/2)(-1/3) = [(-2x1x-1)/(5x2x3)] = 2/30 = 1/15

2y² = 2(-1/3)² = 2(1/9) = 2/9

(1/3)x²y - (2/5)xy + 2y² = -1/36 + 1/15 + 2/9

m.c.m de (36,15,9)

36     15     9 l 2
18     15     9 l 2
  9     15     9 l 3
  3       5     3 l 3
  1       5     1 l 5
           1

m.c.m de (36,15,9) = 2x2x3x3x5 = 4x9x5 = 180

-1/36 + 1/15 + 2/9 = [(-1x5) + (1x12) + (2x20)]/180 = (-5 + 12 + 40)/180

-1/36 + 1/15 + 2/9 = 47/180

(1/3)x²y - (2/5)xy + 2y²  = 47/180

b) x = -1; y = -1

(1/3)x²y - (2/5)xy + 2y²

(1/3)x²y = (1/3)(-1)²(-1) = (1/3)(1)(-1) = -1/3

- (2/5)xy = -(2/5)(-1)(-1) = -(2/5)(1) = -2/5

2y² = 2(-1)² = 2(1) = 2

(1/3)x²y - (2/5)xy + 2y² = -1/3 - 2/5 + 2

-1/3 - 2/5 + 2/1

m.c.m de (3 , 5) = 3 x 5 = 15

-1/3 - 2/5 + 2/1 = [(5x-1) - (2x3) + 2(15)]/15 = [-5 - 6 + 30]/15 = 19/15

(1/3)x²y - (2/5)xy + 2y² = 19/15

3)

a) P = 3q² - q + 8: Para q = 1.3

3q² = 3(1.3)² = 3(1.69) = 5.07

-q = -(1.3) = -1.3

P = 3q² - q + 8 = 5.07 - 1.3 + 8 = 11.77

P = 2q³ - q² + 5q - 3: Para q = -1.5

2q³ = 2(-1.5)³ =  2(-3.375) = -6.75

-q² = -(-1.5)² = -(2.25) = -2.25

5q = 5(-1.5) = -7.5

P = 2q³ - q² + 5q - 3 = -6.75 - 2.25 - 7.5 - 3 = -19.5

 

 




 



    



 









 





















Otras preguntas