Desperate for a math problem! 50 points for answers!!!!!!!!!!!!!!!

Let the cone volume 2v bottom area 3s cylinder 3v base area 2s to meet the proposal.

Cone height = 3 * volume Base area = 2 * (v s) Cylinder height = volume Base area =

High ratio = 4:5

The high of the conic is 36 9*4=16

Cylindrical = 36 9 * 5 = 20

Let the height of the cone be x and the height of the cylinder be y

The base area of each may wish to be set to 3a and 2a, and the volume of 2b and 5b may be available. 1/3*3a*x=2b

2a*y=5b

For example, there is x y=4 5

There is x+y=36 again

Solvable. x=16

That is, the height of the cone is 16 cm.

If they were all cylinders, the volume ratio would be 6:5, 6:5 3:2=4:5,36 (4+5)*4=16 (cm), so the height of the cone would be 16 cm. That's it!!!

Because the height of the cone = 2 * 3 3 = 2 (parts).

Cylindrical height = 5 2 = 5 2 (parts).

So the ratio of high = 4:5

16 (cm).

Let the bottom area and height of the cone and the cylinder be S1, S2, H1, H2 respectively, according to the title.

s1/s2=3/2

s1h1/3s2h2=2/5

It can be obtained from the above 2 formulas.

h1/h2=4/5

again by h1 + h2 = 36

H1=16

Solution: Let the height of the cone be h, then the height of the cylinder is 36-h

Let the bottom area of the cone be s1 and the bottom area of the cylinder be s2, then s1:s2=3:2

There is a volume ratio of 2:5.

1 3*s1*h:s2*(36-h)=2:5 to obtain h=16

is 16s1 h1, which represents the base area and height of the cone, and s2 h2 represents the base area and height of the cylinder.

1/3)s1*h1)/(s2*h2)=(1/3)*(3/2)*(h1/h2)=2/5

So h1 h2=4 5;

h1=36*(4/9)=16;

Looking at the law of black circles, we can know that the first group has a black circle and a white garden, and the second group has 2 black circles and a white circle.

By analogy, there are n black circles and a white garden in the nth group.

So it's a series problem.

The sum of the first n groups of black circles is sn=(1+2+. n) = (1+n)n2 The sum of the first n groups of the white circle is n

The sum of all circles is tn=(1+n)n2+n, when n=62.

That is, in the 62nd group, the number of circles in front is 2015, which is greater than 2009, so the 2009 circles are not as white as the 62nd group.

Of the first 2009 circles, there are 61 hollow circles.

So if you don't understand, you can ask!!

Thanks for adopting!

1＋2＋3＋…n=(1 2)n(n 1)=2009 [Number of black circles] then: n=62, that is, there are 62 hollow circles.

61 is actually 2+3+4+. n close to 2009n (n + 2) 2 close to 2009

n=62 is.

A total of 1952 circles.

There are a total of 63 circles at n=2015.

So n=62

So the hollow circle is 61

Make a hollow circle in front.

1+n)n/2>=2009

n(n+1)>=4018

n>=20

There are (18) hollow circles.

Think of a solid circle and a hollow circle as a group.

Then this sequence is equivalent to 2+3+4+5+6.

Let the number of circles in which 2009 is located n

Then (2+n)*n2=2009

n should be an integer.

But this formula is solved so that n = 4019 —1 under the root number

Shows that the 2009th circle is not a hollow circle, that is, the last group by 2009, the nearest integer with the value of n is 62

That is, before 2009, the complete number of groups was 62.

Each group has a hollow circle.

It is said that there were a total of 62 hollow circles before 2009.

The law of black: 1+2+...n=1 2*n(n+1) There are n white ones.

Add the two: s=1 2*n(n+1)+n=1 2*(n2+3n)=1 2*n(n+3).

Mix up the numbers, n = 61, s = 1952, n = 62, s = 2015 but the 62nd does not appear white circles because the white one is behind the black one.

So there are 61 white ones.

Since there is only 1 hollow circle in each group, adding the hollow circle to each group of solid circles is regarded as a series of equal differences with the first term 2...It went down until the total was 2009.

According to the first n terms and formulas of the equal difference series: s(n)=n*a(1)+n*(n-1)*d 2 or s(n)=n*(a(1)+a(n)) 2. The first term is 2, the difference is 1, and the number of hollow circles can be obtained by finding n.

It is obtained: n*n+2n=2009*2

2009 is not divisible, indicating that there are some solid circles after the last hollow circle. With 3968, the test hollow circle is 64, 4018-3968=58There are also 58 solid circles at the back.

4 and 5 are coprime so that the greatest common factor is 1 and the least common multiple is 4*5=20

6 and 16 both have a common factor of 2, so their greatest common factor is 2, and the least common multiple is 6*16 2=48

15 and 25 both have a common factor of 5, so their greatest common factor is 5, and the least common multiple is 15*25 5=75

21 and 63 They both have a common factor 21, so their greatest common factor is 21, which is one of the two numbers, then the least common multiple is the other number 63

1. Set x kilometers. （1/7）x+(1/7)x+16=94 x=273km

2. Set x kilometers. （1/7）x+(1/7)x+16=x-94 x=154km

3. Set the total length of x meters. (1/3)x-12+(1/4)x+15+102=x x=252m

B repaired: (1 4) * 252 + 15 = 78m

4、r=,s=

5. Let the radius of the large circle r, then the radius of the small circle (3 5)r, (r 2-(r 2=100 s=

6. Set the radius to r. Then the length of the rectangle is: 2*under the root number (r 2-(1 4)r 2).

2*Under the root number (r 2-(1 4)r 2)+r)*2=solution r, find s = r 2 = 65cm 2

7. If the master wants to do x, the apprentice will do 140-x.

Solve x = 80, 60 apprentices.

8、c=6+3*

s=(1/2)*

1.Let the whole journey be x km, then there is 1 7x+1 7x+16=94, and x=273 km can be obtained;

2.Let the whole journey be x km, then there is x-1 7x-(1 7x+16)=94, then x=154 km can be obtained;

I'm off work, and I'm going to ...... to be continued

Set the maximum number to be reached on day t, and divide the 31 days of May into two phases:

Ascending phase (1 day ---t day):

Day 1: 1000

Day 2: 1000+100...

Day t: 1000+100(t-1)=900+100t

Decreasing phase ((t+1) days---31 days):

Day T+1: 1000+100(T-1)-100=800+100T

Day t+2: 1000+100(t-1)-100*2=700+100t...

Day 31: 1000+100(t-1)-100*(31-t)=-2200+200t

These two phases are in a series of equal differences, which are summed respectively by the formula sn=(a1+an)*n 2, and then added to get the equation:

1000+（900+100t）]t/2+[(800+100t)+(2200+200t)](32-t)/2

The solution is t=18

After answering, give me extra points, hehe.

In this problem, we need to use the formulas of the equal difference series: sum = (first and last) terms 2 last terms = first term tolerance (number of terms - 1).

Assuming that 100 is incremented every day, the total amount in a month = (1000 1000 100 30) 31 2 = 77500 tons.

77,500-59,300 = 18,200 tons more than the actual maximum amount due to a decrease of 100 per day after actually reaching the maximum amount

The daily amount is 100 2 = 200 tons and the increase is n+1 days.

200＋200＋200n )(n+1)/2 = 18200(n+1)(n+2)=182

Decomposition 182 = 13 14

i.e. n+1=13

So the date for the maximum quantity is 31-13=18 and the maximum quantity is 1000 100 17=2700 tons.

This is obviously a number series problem.

Well, a month is counted as 31 days.

First calculate how much it transports when it does not grow: 31 1000 = 31000 tons, and 59300-31000 = 28300 tons are transported in the process of growth, and the maximum amount is x tons.

Yes: 1 2 x 31 = 28300

This gives us x = 1825 tons divided by 100 as the number of days i.e. 18 days.

There are pictures here and detailed answers.

Let the maximum quantity be x, and the day with the largest is day a.

1000+（a-1)*100=x

1/2(1000+x)*a+1/2[x-100+x-(31-a)*100](31-a)=59300

Can you see it.

Solution: Assuming that the height of the tank is x cm, then the equation is 5*5*x=200

The equation for the solution x=80 (cm).

Volume = base area Height = 200 liters = 200,000cm Base area = 50 50 = 2500cm Height = 200,000 2,500 = 80cm

A square of 50 cm area = 2500 square centimeters = square decimeters.

200 dm=8m

200 liters 200,000 cubic centimeters.

Water tank height 200000 (50, 50) 80 (cm).

200 liters = 200 cubic decimeters = 200,000 cubic centimeters, so the height is.

200000 (50 50) = 80 (cm) A, the height of the tank is 80 cm.

200 liters 200,000 cubic centimeters.

The base area is 50 50 = 2500 square centimeters.

200000 2500 = 80 cm.

A: The height of the water tank is 80 cm.

200 liters = 200 cubic decimeters = 200,000 cubic centimeters.

The height of the tank = 200000 (50 50) = 80 cm.