Вычислите объем и площадь поверхности куба с ребром 4 дм.

V = a^3 = 4*4*4 = 64дм^3 - объём
S = 6*a^2 = 6*4*4 = 96дм^2 - площадь

1)0,6- 1,6*(х-4)= 3*(7-0,4х)

0,6-1,6х+6,4 = 21-1,2х

-1,6х+1,2х = -0,6-6,4+21

-0,4х = 14

х = 14:(-0,4)

х =-35

0,6-1,6*(-35-4) = 3*(7-0,4*(-35))

0,6+62,4 = 3*21 = 63

ответ: -35.

 

2) 3/4*(1/2х+2/5)=3х+2 1/14

 3/8х+3/10 = 3х+2 1/14

3/8х-3х = 2 1/14-3/10

3/8х -2 8/8х = 29/14-3/10

-2 5/8х = 290/140-42/140

-21/8х = 248/140 

-21/8х = 62/35

х = 62/35:(-21/8)

х = 62/35*(-8/21)

х = - 496/735

3/4*(1/2*(-496/735)+2/5) = 3*(-496/735)+2 1/14

3/4*(-248/735+294/735) = -496/245+29/14

3/4*46/735 = -992/490+1015/490 = 23/490

ответ: -496/735. 

 

1)

sin(x)*sin(3x)

так как

sin (3x)= sin(2x + x) = sin(2x) cos(x) + sin(x)cos(2x), то

sin(x)*sin(3x)=sin(x)*[ sin(2x) cos(x) + sin(x)cos(2x)]=

=sin(x)*[2sin(x)cos(x)*cos(x)+sin(x)*(2cos^2(x)-1)]=

=sin^2(x)*[2cos^2(x)+2cos^2(x)-1]=sin^2(x)*[4cos^2(x)-1]=

=4sin^2(x)cos^2(x)-sin^2(x)

 

  a.  int(4sin^2(x)cos^2(x))dx=int(2sin(x)cos(x))^2dx=int(sin(2x)^2dx=

=int((1/2)*(1-cos(2*2x)))dx=(1/2)*(x-(1/4)*sin(4x))+c

 

б.  int(sin^2(x))dx=(-1/2)int(1-cos(2x))dx=(-1/2)*[x-(1/2)sin(2x))]+c

 

итого

int sin(x)*sin(3x)dx=(1/2)*[x-(1/4)*sin(4x)]+c1+(-1/2)*[x-(1/2)sin(2x)]+c2=

=(1/2)*[(1/2)sin(2x)-(1/4)sin(4x)]+c