How to solve the equation of 2x% * 15 1 / 2 = 11

How to solve the equation of 2x% * 15 1 / 2 = 11

2x/100*(31/2)=11
2200=62x
1100/31=x

How to solve this equation?

5 / 69 + 2x = 11 and 1 / 3
2X = 11 and 1 / 3-5 / 69 (with sum minus another addend)
2x=43/9
X = 43 / 18 (43 / 9 divided by 2 leads to 43 / 18)
Do you understand? If you don't understand, just ask!

If the line y = kx-2 intersects with the parabola y ^ 2 = 8x at two points a and B, if the abscissa of the midpoint of the line AB is 2, the length of the line AB is calculated

Y = kx-2 is substituted by Y & # 178; = 8x to get (kx-2) ^ 2 = 8x
k^2x^2-(4k+8)x+4=0
x1+x2=(4k+8)/k^2=2*2
K = 2 or - 1
The length of string | ab | is the root sign (1 + K ^ 2) * | x1-x2 | = 8 * root sign (K & # 178; + 1) * root sign (K + 1) / K & # 178;
When k = - 1, | ab | = 0, abandon
So k = 2, the length of string | ab | is 2, and the root sign is 15

An equation problem
The distance between a and B is 448km. A local train starts from place a and travels 60km per hour. One or two express trains start from place B and travels 80km per hour. Q: the two trains are facing each other. The local train starts at 28 minutes. How many hours after the express train starts, the two trains meet?

Suppose that the two cars meet after the express train leaves
60×7／15＋60χ＋80χ＝448
χ＝3
A: the two cars met three hours after the express train left

In Cartesian coordinates, the distance from point P (- 2,3) to the origin is
A. Root 5 B, root 13 C, root 11 d, root 2 D

Through the point P (- 2,3), make the vertical line of X axis,
Then the length of the vertical line segment is the ordinate of point P, which is 3,
And the distance from the origin to the perpendicular foot is the absolute value of the abscissa of point P, which is 2,
So the length of the hypotenuse of a right triangle is √ (2 ^ 2 + 3 ^ 20 = √ 13
That is, the distance from point P (- 2,3) to the origin is √ 13,
The answer is B

1. If x + y = 0, xy = - 6, then x & # 179; y-xy & # 179; - 5
2. Calculation: 101 & # 178; - 99 & # 178=
3.（-2）²º¹²+（-2）²º¹¹=
4. Factorization: (a-b) & # + 4 (a-b + 1)

X³Y-XY³-5
=xy(x^2-y^2)-5
=xy(x+y)(x-y)-5
=0-5
=-5
101^2-99^2
=(101-99)(101+99)
=2*200
=400
(-2)^2012+(-2)^2011
=2^2012-2^2011
=2*2^2011-2^2011
=(2-1)*2^2011
=2^2011
(a-b)^2+4(a-b+1)
=(a-b)^2+4(a-b)+4
=(a-b+2)^2

Data of land desertification area in China in 2010

In 2010, the desertified land in northern China was 375900 km2, of which 33.80% was mild desertified land, 22.84% was moderate desertified land, 22.16% was severe desertified land, and 21.21% was severe desertified land

Complete (it's better to have a formula),
The mass of iron and stone is 100g, the initial temperature is 10 ° C, they all absorb 2.76 * 1000j heat, and then put them together, try to determine whether they can have heat transfer through calculation

According to this formula: q = cm (t-to), the initial temperature to = 10 ° C, so the final temperature T = q / cm + to, the final temperature T of iron = 2760j / [460j / (kg * ~ C) * 0.1kg] + 10 ° C = 70 ° C, the final temperature T of stone = 2760j / [920j / (kg * ~ C) * 0.1kg] + 10 ° C = 40 ° C

In a certain month, if the date of three Sundays is even, what day is the 5th of this month?
It's Wednesday

When one Sunday is even, the next even Sunday is 14 days away
So three Sundays are even hours, with a total of 28 days between them
If the 2nd is Sunday, then the last Sunday is 2 + 28 = 30
If the 4th is Sunday, then the last Sunday is 4 + 28 = 32, which is obviously impossible
If the date later than the 4th is Sunday, it is impossible to have three even numbered Sundays this month
So there is only one situation, that is, the 2nd is Sunday
According to this, the 3rd is Monday, the 4th is Tuesday, and the 5th is Wednesday

To find the expressions of the chemical words learned in the first semester of the third grade of junior high school,

1. Sodium carbonate and a small amount of hydrochloric acid
Na2CO3+HCl=NaCl+NaHCO3
2. Sodium carbonate and excessive hydrochloric acid
Na2CO3+2HCl==CO2↑ +H2O+2NaCl
3. Add a small amount of sodium hydroxide to the sodium bicarbonate solution
NaOH+NaHCO3=Na2CO3+H2O
4. Add excess sodium hydroxide to the sodium bicarbonate solution
NaOH+NaHCO3=Na2CO3+H2O
5. Reaction of aluminum with hydrochloric acid
2Al+6HCl=2AlCl3+3H2 ↑
6. Alumina and hydrochloric acid
Al2O3 + 6HCl = 2AlCl3 + 3H2O
7. Alumina and sodium hydroxide
Al2O3 + 2NaOH = 2NaAlO2 + H2O
8. Aluminum hydroxide and hydrochloric acid
Al(OH)3+3HCl=AlCl3+3H2O
9. Aluminum hydroxide and sodium hydroxide solution
Al(OH)3+NaOH=NaAlO2+2H2O
10. Aluminum chloride solution and excess ammonia water
AlCl3+3NH3·H2O=Al(OH)3↓+3NH4Cl
11. Add excess sodium hydroxide to aluminum chloride solution
AlCl3+3NaOH=3NaCl+Al(OH)3
Al(OH)3+NaOH=NaAlO2+2H2O
12. Add excessive hydrochloric acid into sodium metaaluminate solution
NaAlO2+4HCL=NaCL+AlCl3+2H2O
13. Magnesium burns in air
2Mg+O2＝2MgO
3Mg+N2=Mg3N2
2mg + CO2 = = = ignition = = = 2MgO + C
14. Iron and hydrochloric acid
Fe+2HCl=FeCl2+H2↑
15. Iron oxide dissolves in hydrochloric acid
Fe2O3+6HCl=====2FeCl3+3H2O
16. Drop sodium hydroxide into ferric chloride solution
FeCl3 + 3NaOH =Fe(OH)3↓+ 3NaCl
17. Iron burns in oxygen
3Fe + 2O2 = (ignited) Fe3O4
18. Add chlorine water to the ferrous solution
Cl2+2FeCl2=2FeCl3
19. Add iron powder to ferric chloride solution
Fe + 2FeCl3 =3FeCl2
20. Reaction of ferric chloride solution with copper
Cu+2FeCl3=CuCl2+FeCl2
21. Production of limestone from carbon
C+O2=CO2
CO2+CaO=CaCO3
22. Industrial bleach
2Cl2＋2Ca(OH)2＝CaCl2＋Ca(ClO)2＋H2O
23. Reaction of bleach solution with carbon dioxide
Ca(ClO)2+CO2+H2O=CaCO3↓+ 2HClO