Solve the following math problems, solve the following math problems

Walking speed 3600 3600=1 meter second.

Set the train speed to v m seconds.

Then 15(v+1)=17(v-1), v=16 m-s.

Vehicle length = 15 (16 + 1) = 255 meters.

Kilometer-hour = 1-meter-second.

Suppose the speed of the train is x meters and seconds :

x+1）*15=（x-1）*17

x=16 So the train length: (16+1)*15=255 meters.

Let the speed of the vehicle be V1, the speed of the person be V2, and the length of the car be L.

v1*15+v2*15=l

v1*17-v2*17=l

v2 is a known quantity, which solves the upper system of equations.

i.e. results.

Kilometers = 1 meter second.

Train speed (17*1+15*1) (17-15)=16m/s.

So the length of the train is (16-1) * 17 = 255 meters.

Suppose: the train speed is x km/h; The length of the train is Y km.

Then:(15*(

x=32*y=(16*

The length of the train is 918 km.

x pulls obey n(10,p(x>11)=p[(x-10)>=1)]=p[(x-10) root》=1 root.

1-p[(x-10) root<1 root.

1- (Root.

1. 7/2 3√2－4

2. x≥5 x≠3/2

3.12 10 9 (any number other than 0 to the power of 0 is 1) a≠3

4.10 or.

5.(2,3)x is equal to y, and y is opposite.

6. a＝－11

7. （x-½）

8. 3y＋8x－3=0

Derived from the title. x=6-y 1zsquare=xy-9 2

1 substitution 2 to get z square = 6y-y square -9 to get z square = - (y-3) square.

Because the z-square cannot be negative, y=3

So x=y=3

x = 6 -y

z^2 = x*y - 9= 6y - y^2 -9 = -(y-3)^2

Z2 and (y-3)2 are both greater than or equal to zero.

So (y-3) 2 = 0

y = 3x = 3

Get x = y

x=6-y

z^2=xy-9=(6-y)y-9=-(-6y+y^2+9)=-(y-3)^

The same square will not have -, no solution.

Side area = surface area - two base areas.

288 (square centimeters).

Side area = bottom perimeter height.

Height = side area Bottom circumference.

8 (cm) volume = base area x height.

40x8320 (cubic centimeters).

Hope it helps

S side = S table - 2s bottom.

288(cm^2))

S side = C bottom h

H=s side c bottom.

8(cm)v=s bottom h

320(cm^3)

Set the bottom length to x, width to y, and height to z; 2(x+y)=36;xy=40；2（xy+yz+xz）=368

The solution is z=8, so v=xyz=320

It can be like this, A: This is a cubic trinomial", we know that this polynomial has ax 3;

B: The coefficient of the cubic term is 1", which shows that the trouser luck a=1;

c, the terms of this polynomial have a common factor" can be seen that this polynomial has bx 2 and cx;

D: The common factor should be used when factoring this polynomial" (note: this sentence seems to make little sense);

Therefore, this trinomial is:

x 3+bx 2+cx, where x is an unknown number, a, b, c are constants;

Factoring as:

x(x^2+bx+c)

A: This is a cubic trinomial", we know that this polynomial has ax 3;

B: The cubic term coefficient is 1", and we know that a=1;

c, the terms of this polynomial have a common factor" can be seen that this polynomial has bx 2 and cx;

D: The common factor should be used when factoring this polynomial" (note: this sentence seems to make little sense);

Therefore, this trinomial is:

x 3+bx 2+cx, where x is an unknown number, a, b, c are constants;

Factoring as:

x(x^2+bx+c)

A: This is a cubic trinomial", we know that this polynomial has ax 3;

B: The cubic term coefficient is 1", and we know that a=1;

C, the items of this friendly polynomial have common factors" shows that this polynomial has bx 2 and cx;

D: The common factor should be used when factoring this polynomial" (note: this sentence seems to make little sense);

Therefore, this triathlon is:

x 3+bx 2+cx, where x is an unknown number, a, b, c are constants;

Factoring as:

x(x^2+bx+c)

A 1The title is not clear.

2. y =(x+1)e^(-x),y' = e^(-x)-(x+1)e^(-x) = -xe^(-x).

Detached x = 0.

y'' = -e^(-x)+xe^(-x) = (x-1)e^(-x)

y''(0) = -1 < 0, then y has a maximum y(0) = 1

II 1 ∫cos√xdx/√x = 2∫cos√xd√x = 2sin√x + c

2.∫x[(sec(x^2)]^2dx = (1/2)∫[sec(x^2)]^2d(x^2)

1/2)tan(x^2) +c

3.∫cosx[3^(2sinx+1)]dx = (1/2)∫[3^(2sinx+1)]d(2sinx+1)

1/2)[3^(2sinx+1)]/ln3 + c = [1/(2ln3)]3^(2sinx+1) +c

1. The title is not clear.

2、 y=(x+1)^2e^(-x)

y'=2(x+1)e^(-x)+(x+1)^2[-e^(-x)]=(x+1)e^(-x)[2-(x+1)]=(x+1)e^(-x)(1-x)

y'>0

x+1)e^(-x)(1-x)>0

e^(-x)>0

x+1)(1-x)>0

x+1)(x-1)<0

10x<-1 or x>1

y'=0x=-1 or x=1

When x=1, y obtains a maximum: y=(1+1) 2e (-1)=4 ex=-1, y obtains a minima: y=(1-1) 2e (-1)=0 two, 1, 1 xcos xdx

2∫cos√xd√x

2sin√x+c

2、∫xsec^2(x^2)dx

1/2∫sec^2(x^2)d(x^2)=1/2tan(x^2)+c

3、∫cosx3^(2sinx+1)dx

3^(2sinx+1)dsinx

1/2∫3^(2sinx+1)d(2sinx+1)=1/2×3^(2sinx+1)/ln3+c=1/(2ln3)×3^(2sinx+1)+c

1. x 7+3 Solution: x 7+ x 7= x 7= x= x= Test: substitute x= into the equation on the left side of the equation Left = right, so x= is the solution of the equation ( Solution:

x= x= test: substitute x= into equation 3 on the left side of the equation ( left = right so x= is the solution of the equation (1)7 x= solution: 7 x=14 x=7 14 x= (2) ( solution:

x= x=