Please solve this math problem, thanks

km) 84 2 = 168 (km).

Because it's 5-4 1 anyway, it has to be divided by a fifth, and the current solution is half the distance, so multiply by 2. The final result was 168 km.

The distance from place B to place A is x kilometers.

Because distance = speed * time (time is equal) (here the time is the time after the departure of the bus), the speed ratio is equal to the time ratio).

x=168

The distance from place B to place A is x kilometers.

Because distance = speed * time (time is equal).

So the velocity ratio is equal to the time ratio).

x=168

I'm different from everyone else. If the truck is a kilometer away from ground A, assuming that the bus and the truck depart at the same time, then when the truck travels a kilometer, the passenger car should travel a kilometer. If so, the bus will travel five out of the ninth of the way when they meet.

But in fact, the bus traveled half of the way, and the difference was 21 kilometers. Use another five-ninths minus one-half = one-eighteenth. That's the corresponding score of 21.

Another 21 one-eighteenth = 378 km.

The time when the passenger car travels the whole distance 1 2 The truck should travel the whole distance of (1 2) * (4 5) = 2 5

So the whole journey is kilometers.

3 bags of rice and 3 bags of soybeans weighing 75 kg in total, 9 bags of rice and 9 bags of soybeans weighing 225 kg in total, 5 bags of rice and 9 bags of soybeans weighing 185 kg in total.

Rice: (225-185) (9-5) = 10 (kg) Soybeans: 75 3-10 = 15 (kg).

Set each bag of rice x kg? Y kg of soybeans per bag, 3x+3y=75, 5x+9y=185, so 3x+3y=75 multiplied by 3 and then subtracted by 5x+9y=185 is 9x+9y-5x-9y=75 multiplied by 3-185, and the solution is x=10, y=15

10 kg of rice, 15 kg of soybeans.

I can't draw it, you can draw it on scratch paper.

The perpendicular line of the AC crossed by Q and the extension of the AC at F

It is easy to know that aep cfq(ap=cq, two right angles, pae= qcf) so ae=cf, pe=qf

and dep dqf(pe=qf, the corners correspond to equal) so de=df

So ac=ae+de+dc=cf+de+dc=de+df=2de, so de is a fixed value, equal to half the length of the sides of the equilateral triangle.

The perpendicular line of the AC crossed by Q and the extension of the AC at F

It is easy to know that aep cfq(ap=cq, two right angles, pae= qcf) so ae=cf, pe=qf

and dep dqf(pe=qf, the corners correspond to equal) so de=df

So ac=ae+de+dc=cf+de+dc=de+df=2de, so de is a fixed value, equal to half the length of the sides of the equilateral triangle.

Please use the quadratic formula to solve, please explain the process.

Question 1: If.

Square. Target.

Edge length. For a, his area is equal to the area of a rectangle with a length of 96 cm and a width of 12 cm.

a^2=96*12=1152

a = root 1152 = 24 root 2

Question 2: When x is the value of the nucleus, the following numbers are in.

Real number. Makes sense in scope?

1) 3-x under the root number

3-x>=0

That is: x<=3

2) 2x-1/1 under the root number

2x-1>0

i.e.: x>1 2

Question 3: Simplification: (1) 500 under the Gen's alarm

Root (5 * 100) = 10 root 5

2) 12x under the root number = root (4 * 3x) = 2 roots (3x), (x is greater than or equal to 0) 3) 4 and 2/3 under the root number = root (14 3) = root (14 * 3) (3 * 3) = 1 3 roots 42

4) 2 squares of 3a under the root number = root(2 3a 2) = root(6 9a 2) = |1/3a|Root 6

1) a>0, above equation = 1 3a root 6

2) a<0, above formula = -1 3a root 6

Solution: See the figure below. As can be seen from the diagram, B starts up 1 hour after A departs.

A number of degrees after 1 hour v1 = (240-60) (7-1) = 30km hr; B's velocity v2=240 (5-1)=60km hr;

1) Let B meet A at t hours of departure, then: 60+30t=60t; t=60/(60-30)=2(hr)。

2) Set a departure for 1 hour, and the distance between the two cars is 25km;

60*1+30*(t1-1)+/-25=60t1;

i) t1 = (60 + 30 + 25) (60-30) = 3 + 50 60 = 3 hours 50 minutes;

ii) t1 = (60 + 30-25) (60-30) = 3-50 60 = 2 hours and 10 minutes. Both of these answers are solutions to this problem.

Answer: Slightly. Finished.

This question should be more careful, and the last question has four moments when the two cars are 25km apart

The polyline equation for calculating A is y1=60x,0 x 1,=30x+30,1 x 7

Calculate the linear equation of B y2=60x-60, 1 x 5, 60x-60=30x+30, get x=3, at this time y=3*60-60=120, that is, B starts to meet in 3 hours, and the distance b is 240-120=120 kilometers when car B does not move, A y=60x=25, get x=25 60=5 12 hours = 25 minutes, y1-y2-60 absolute value = 30x-30 =25, get x=11 6, or x=1 6, round.

That is, 1 hour and 50 minutes after the departure of A, the distance between the two vehicles is 25km

It's not that simple.

Substitute the correct solution x=8 and y=10 into the equation:

8a+10b=62 (1)

8c-200=-224

Solution: c=-3

Substitute the solution x=11,y=6 into the first equation:

11a+6b=62 (2)

1) (2) Forming a system of equations, solving a=4, b=3

So a+b+c=4

The first set is brought into the equations.

8a+10b=62

c = -3 The second set is brought into the equation.

11a+6b=62

then a= can be found from the equation of 2 abouts

then a+b+c=4

48-12-12=24 (sheets).

24-4=20 (sheets).

Answer: Xiaolan and her class originally had 20 movie tickets.

Xiaolanban's original movie tickets: 48-12-4=32 (tickets).

(1) Solution:

Analysis) In the triangle Bai shape, the vertex angle of this du can be determined to be the angle of A, and the downward movement of the straight line EF can be set to the intersection of the high line of the triangle through point A as d, and there is S capacity abc=24So there is ad=24 10*From the meaning of the title, it can be seen that when y=0, x=0, the speed of linear motion v=2 units of seconds; y=24 5, x=12 5;

So y = 2*x; 0<=x<=(2) There can be two methods: direct method and indirect method.

Direct: s=1 2(10*(24 5-2*x)*5 24+10)(24 5-2*x).

Indirect: s=24-1 2*(2*x810*5 24)*2*x=24-25 6*x.2 (25/6 times x squared).

Analysis, (1) ca*cb=a*b*cosc

c = a +b -2ab*cosc=b -a is du, zhib = c +a

Therefore, b=90 and sinb=1

Again, asina*sinb+b*cos a= 2aa=sina*2r, b=sinb*2r

Therefore, sin a+cos a= 2sina

sina = 2 2, ie.

The DAO is a=45

Then c=45

2) c = 2 2, three within.

The angular ABC is an isosceles right-angled accommodating triangle, and B is a right-angled triangle.

c=a=2√2

s=1/2*(2√2)²=4。