I don t know how to solve the math problem in the second year of junior high school, my sisters and

1 Set the original speed x km/h. Column Equations:

60 x + 1 2 = 30 x + 30 * 5 4x solution 240 + 2x = 120 + 150

x=1515*4 5=12 The speed of the second half of the journey is 12km h2, and the number of parts is set to be B x parts for days, and A3 is 2x days.

Column: 8000 3x=3000x-1

Solution 3x=1000

x=1000 3 3 2*1000 3=5003(1000 3+500)=2500 can be made for 2500.

Set the speed of the second half to x km h

30 Solution. x=12

Let the efficiency of B be x days, then A is a day.

Solution. x=1000/3

A and B cooperate for three days: (1000 3 + 1500 3) * 3 = 2500 (pcs).

Original speed x, rear speed (1-1 5) x

According to the second half of the journey, the difference in time is half an hour.

30/(1-1/5)x]-30/x=30/60x=15.So the second half of the speed is 12km h

To require the total workload of A and B, and the time is known, it is necessary to find their respective work efficiency first. Let the ergonomics of B be x. Then A is.

Total amount of work = ergonomics * time The equation can be found according to the time difference between the two people.

3000 x-4000 is multiplied on both sides of the equation.

x=1000/3

Then the cooperation of A and B is the total amount of A + the total amount of B.

In the previous question, the speed of the second half of the course is a, and the formula is obtained.

60/2a+60/2(

Find a = 11 km hours.

In the latter question, if B can make a part in a day, then A can make a part in a day to get the formula.

3000/a-4000/

Find a=1000 3.

The two worked together for three days.

3 ( pcs.

1 Set the original speed x km/h.

60/x+1/2=30/x+30*5/4x240+2x=120+150

Solution x=15

A: The second half speed is 12kmh

2 set B x days, A 3 2x days.

8000/3x=3000/x-1

3x=1000

Solution x=1000 3

A: 2500 pieces can be made.

1. Set the speed to x

60/x+1=60/(1/5x)

A: The speed of the halfway is 1-1 5x

Let A be x4000 x+1=3000

Find out how much x is equal to.

x+ A.

1 square.

c perimeter s area a side length.

Circumference Side length 4

c = 4a area = side length side length.

s=a a2 cube.

v: volume a: edge length.

Surface area = edge length Ridge length 6

S table = a a 6

Volume = edge length edge length edge length.

v=a×a×a

3 rectangles.

c perimeter s area a side length.

Circumference = (length + width) 2

c=2(a+b)

Area = length and width.

s=ab 4 cuboid.

V: Volume S: Area A: Length B: Width H: Height (1) Surface area (length and width + height + width and height) 2s=2(ab+ah+bh).

2) Volume = length, width and height.

v=abh5 triangle.

S area A bottom H height.

Area = base height 2

s=ah÷2

Triangle height = area 2 bases.

Triangular base = area 2 high.

6 Parallelogram.

S area A bottom H height.

Area = base height.

s=ah 7 trapezoidal.

S area: A, upper bottom, b, lower bottom, H high.

Area = (top bottom + bottom bottom) height 2

s=(a+b)× h÷2

8 Circle s area c perimeter d = diameter r = radius (1) perimeter = diameter = 2 radius.

c=∏d=2∏r

2) Area = Radius Radius

9 Cylinders.

V: Volume h: High S; Base area r: bottom surface radius c: bottom perimeter (1) side area = bottom perimeter height.

2) Surface area = side area + bottom area 2

3) Volume = base area high.

4) Volume side area 2 radius.

10 cones.

V: Volume h: High S; Base area r: bottom radius volume = bottom area 1/3 of the height of the view.

Because ocb=60°, oc=2ec, ec=three-thirds of the root number three*oe=2 3*root number three, bc=4 3 root number three = ad, ab = root number three bc=4, so the circumference = (8 3 root number three + 8) cm is expected to be adopted.

Since the angle ACB = 60 so the angle coe=30, let ec=x then oc=2x then x 2 +4=(2x) 2 so x = 2/3 times the root number 3, so bc=2ec=4/3 times the root number 3, ab =2oe=4, so the perimeter = 2(ab+bc)=.

By tan60 degrees = root number 3, ec is 2 3 root number 3, then bc = 4 3 root number 3, then by tan30 degrees = root number 3 3, then bc ab = tan30 degrees, then ab = 4, then ad=bc, ab = dc then the perimeter is 8 3 root number 3 + 8

Solution: Let the relative atomic mass of the m atom be x

12 to the minus 26 power = x to the minus 26 power.

After solving it, you get x 40 answers.

The relative atomic mass of the m atom = the mass of the m atom (mass of the carbon atom (C12)).

MN under the root number

Multiply by the power of the same base.

Question added: Get rid of the detailed process. Thank you x=1 (3 4)2x=1-7 16 So, 3 4=2x root number 9 16 i.e.: 2x=2, get x=1Solution: 3 4 to the power of 2x = 3 4 to the power of 2

Divide the formula to the power of 5 by the formula to the power of 3 to get x 2=n m

The square of both sides gives x 4=(n m) 2

I'm also in my second year of junior high school, so I don't seem to have done anything like this, so I'll give it a try.

According to the topic. x^3=m

x^5=nn/m=x^5/x^3=x^2

x^2*x^2=x^4

So n m*n m=n 2 m 2=x 4

If the walking speed is x km/h, the speed of a bicycle is x+8 km/h, and the speed of a car is x+16,000 knanam-h.

2 x+10 (x+16)=12 (x+8) multiplied by x(x+16) (x+8) on both sides at the same time

Get. 2 (x+16) (x+8) + 10x (x+8) = 12x (x+16) 2x potato sis +48x + 256 + 10x +80x = 12x +192x48x + 80x - 192x = -256

64x=256

x = 4 walking speed = 4 kilometers and hours of moderation.

Bicycle speed = 4 + 8 = 12 km/h.

Car speed = 4 + 16 = 20 km h.

Because the midline on the hypotenuse is long.

So the hypotenuse is 13cm long

Let the two right-angled sides be a and b

ab=60a²+b²=169

a=12b=5 or a=5, b=12

Here the solution of the equation can be solved as a + b + 2ab = 289, and a + b = 17 is obtained, which is still full of simplicity).

The right-angled sides are 12 and 5

Let a b then 2ab = 120 hypotenuse length 13 a +b =13 =169 add (a + b) = 289 = 17

Subtraction of the two formulas (a-b) = 49 = 7

a=12b=5 or a=5, b=12

The midline of the rt hypotenuse is equal to half of the hypotenuse, hypotenuse = ab 2 = 30

a²+b²=13²

Find a b

Because the middle line on the hypotenuse is equal to half of the hypotenuse.

So hypotenuse=2

Let the two right-angled sides be x, y,

We get the system of equations xy divided by 2=30

x +y = root number 13

The solution is x=5

y=12

The middle line on the hypotenuse of a right triangle is half of the hypotenuse, so the length of the hypotenuse is 13, let the two right-angled sides be a and b, use the Pythagorean theorem to list an equation, and then use the area formula to list an equation.