Velocity! Math Problems in Elementary School, Speed! Math problems!

Can I give points for what I can do?

Let barrel B be x kilograms, then the mass of bucket A is, you know.

If the kilogram is taken from bucket A and poured into bucket B, the weight of the two buckets is equal. This sentence tells us that before it was poured, bucket A was twice as heavy as bucket B, that is, kilograms.

So the column formula is: x

Solution: x This is the weight of barrel B.

So, a bucket is a kilogram.

It's all made clear, you have to give points.

Let A be x and B be y, and get x=, from the problem, x=8, y=

Solution: Let B have x kilograms, then A has kilograms;

B plus kilograms equals the mass of A minus kilograms, then there is.

x+ gets x=

So, B has kilograms, and A has 8 kilograms.

Let bucket B be x, bucket A is, because bucket A gives bucket B, from which we get:

Let B be x, then A is.

x = A: B is a kilogram, A is a kilogram.

Passing the point of the nuclear posture shed a as ah vertical book hungry in BC

In RT AHC.

Because the angle c is 60°

ac=10, so ah=5, root, number 3

hc=5 is changed in RT AHB.

There is the Pythagorean theorem.

bh=11, so bc=16

Directly set the formula Wang Zhou smiled.

c^2=a^2+b^2-2ab*cosc

The three corners of ABC correspond to the A, B, and C edges.

14 2 = a 2 + 10 2-2a * 10 * cos60 degrees.

196=a^2+100-20a*

a^2-10a-96=0

a-16)(a+6)=0

Solving the equation yields a=16 or a=-6

Negative roots are excluded, so a=16

So bc=16

Conical base area: *2) =4 (dm).

The sum of the volume of the two cones: 4 *4 3*2=32 3 (dm) Cylindrical base area: *4) =16 (dm).

Cylindrical height: 32 3 16 = 2 3 (dm) Answer: The water surface is 2 3 decimeters.

Because the cone is 1 3 of the volume of a cylinder of the same height of the same area

Now the ground is enlarged by 2*2 4 times, so there can be water 4d*2 4*1 3 (2 3)dm*10 (20 3)dm.

Set the water surface height to h

The volume of the cone is 1 3sh 1 3( r 2h)=1 3 *2 2*4=16 3

The volume of the cylinder sh r 2*h= *4 2*h=16 h2*the volume of the cone = the volume of the cylinder.

16/3∏*2=16∏h

h=2/3dm

The answer should be 2 3 decimeters, 2x1 3x x2 squared x4 = x4 squared xh

Set Zhang Liang has money x yuan. Then Huang Ming's money is 9x 5 columns.

9x/5-48=x-20

The solution is x= 35, so Huang Ming has 63 yuan.

If Huang Ming has 9x yuan, then Zhang Liang has 5x yuan.

9x-48=5x-20

x=7 Huang Ming has 63 yuan.

Let's say there is now $ x; Gotta :

x+48)/(x+20)=9/5

Launch x=15

So Huang Ming originally had 15 + 48 = 63 yuan.

Solution: Let Huang Ming have money x yuan and Zhang Liang have money y yuan, then there is x y=9 5, x-48=y-20, and the simultaneous equation is solved x=63 y=35

1.Lump sum deposit and withdrawal 1 year p.a. Lump sum deposit and withdrawal 2 years p.a. interest rate % 10540

2.days, there is still more than an hour to open a bottle of champagne to celebrate.

3.0 tons.

4.District 1: 35 people, District 2: 14 people.

Solution: From the graph, aob = aoc+ com+ mon+ nod+ dob

Because oc divides AOM equally, OD divides BOON

So aoc= com nod= dob, and because com+ mon+ nod= ,mon= so aob= aoc+ com+ mon+ nod+dob=2 com+ mon+2 nod

2∠aob=∠aoc+∠com+∠mon+∠nod+dob=2（∠com+∠mon+∠nod）-∠mon=2α -

com+ mon+ nod= and mon= so com+ nod= -

And OC divides AOM equally, OD divides BOON

So aom+ nob=2( com+ nod)=2( so aob= aom+ nob+ mon=2( =2 -

There are several situations that need to be divided: (you can draw a picture and calculate it yourself).

One is: d is between bn and c is between am.

aob=2α -

The second is: d, c between m and n.

aob=2α +