5 X-9 x = 2.4 emergency

5 X-9 x = 2.4 emergency

Take 45 on both sides
9x-5x=2.4*45
4x=2.4*45
x=2.4*45÷4
x=27

There are two passenger and freight cars running from both sides at the same time. The passenger car runs 60km per hour and the freight car 72km per hour. When the passenger car runs 3 / 8 of the distance between Party A and Party B
The train is 6km away from the midpoint. How many kilometers is the whole journey

Freight car passing = 3 / 8 △ 60 × 72 = 9 / 20
Whole course = 6 ÷ (1 / 2-9 / 20) = 120 km

Calculation: 20032003 × 2003-20032002 × 2002-20032002=______ ．

20032003 × 2003-20032002 × 2002-20032002 = 20032003 × 2003-20032002 × (2002 + 1) = 20032003 × 2003-20032002 × 2003 = 2003 × (20032003-20032002) = 2003 × 1 = 2003

A and B trains run from AB to each other. A trains 120 km / h and B trains 90 km / h. when they meet, what is the distance ratio between a and B? What is the time ratio between a and B trains to complete the journey by themselves?

s=vt
The time is the same, so the distance ratio is 120:90 = 4:3
To complete the same journey, the formula shows that the inverse ratio of time and speed is 90:120 = 3:4

(6-2x)(8-2x)=24

(6-2x)(8-2x)=24
（3-x)(4-x)=6
12-3x-4x+x²=6
x²-7x+6=0
(x-1)(x-6)=0
x1=1,x2=6

The distance between the two places is 120 km. The two vehicles run from the two places for 2 / 3 hours at the same time. The speed ratio of the two vehicles is 4:5. How many kilometers do the two vehicles travel per hour?
It's better to have an explanation for the formula

Encounter problem: distance = speed and X time
Speed sum = distance △ time = 120 △ 2 / 3 = 180
Because the speed ratio of car a and car B is 4:5
So the speed of car a: 180 × 4 / (4 + 5) = 180 × 4 / 9 = 80 km / h
Speed of vehicle B: 180 × 5 / 9 = 100 km / h
A: the speed of car a is 80 km / h, and that of car B is 100 km / h

The difference between the number obtained by moving the decimal point one place to the right and the number obtained by moving the decimal point one place to the left is 11 &; 88, so what is the original number?

Move one bit to the right, 10 times the original,
Move one bit to the left to make one tenth of the original
 
So:
 
Original number
=11.88 (10-1 / 10)
=11.88÷9.9
=1.2
 

A passenger car and a freight car leave from a and B at the same time. They meet at a distance of 15 kilometers from the midpoint. The known speed ratio of passenger car and freight car is 7:5
A passenger car and a freight car leave from a and B at the same time. They meet at a distance of 15 kilometers from the midpoint. The speed ratio of the passenger car and the freight car is known to be 7:5. How many kilometers is the distance between a and B?

When the truck meets, it will travel 5 x km,
Because the speed ratio of passenger cars and freight cars is 7:5
So the bus runs 7x km
7x-5x=15*2
2x=30
x=15
7x + 5x = 12x = 12 * 15 = 180km
The distance between the two places is 180 kilometers

As shown in the figure, in ABC, ab = AC, BD ⊥ AC, prove that the square of BC = 2Ac times DC

If I understand your figure correctly, the problem-solving process is as follows:
First of all, BC ^ 2 is the square of BC
The topic is to prove that: BC ^ 2 = 2Ac * DC, because BDC is a right triangle, BD is perpendicular to DC, there is
BC ^ 2 = BD ^ 2 + DC ^ 2, so the equation to prove becomes
BD^2+DC^2=2AC*DC,
BD^2=2AC*DC-DC^2=DC*(2AC-DC)=DC*(AC+AC-DC)=DC*(AC+AD)=DC*AC+DC*AD
Because the triangle abd is a right triangle and BD is perpendicular to ad, there is a right triangle
BD^2=AB^2-AD^2
Since BD ^ 2 = DC * AC + DC * ad, we can see that
AB^2-AD^2=DC*AC+DC*AD
Since AB = AC, it can be concluded that
AC^2-AD^2=DC*AC+DC*AD
AC^2-DC*AC=AD^2+DC*AD
AC*(AC-DC)=AD*(AD+DC)
Ac * ad = ad * AC is an identity
The above is the idea of solving the problem. Writing this from the back to the front is the process of solving the problem

On a section of double track railway, two trains head-on. The speed of train a is 20 m / s, and that of train B is 25 m / s. if the total length of train a is 200 m, and that of train B is 160 m, the time for two trains to miss is 30 minutes______ Seconds

Suppose the time of two trains passing by is x seconds, then there is (25 + 20) x = 200 + 160, and the solution is x = 8