My brother asks you a few math problems! 40

I got it. Wait.

1 meter = 10 decimeters.

Each piece to be painted is a quarter of the surface area of the cylinder plus two rectangles, the height of the rectangle is 10 decimeters, and the width is the radius = 1 decimeter.

The area of the rectangle = 10 * 1 = 10 cubic decimeters.

Cylindrical surface area = two bottom surfaces + one side = 2*

Each piece to be painted area = 1 4 * cubic decimeters.

5 cm 15 cm = 1 3, that is, the volume of the empty part is 1:3 compared to the volume of the drink part, and the volume of the drink part is liters.

Cone volume = 1 3sh

1/3sh=5*5*5

1/3*15*h=125h=25

Cone volume = 1 3sh

The empty part of the top is also a cone, and the radius of the bottom surface is half of the radius of the entire bottom surface of the cone according to the similarity of the triangles.

Whole volume = 1 3sh = 1 3*

Empty part volume = 1 3*

1/3sh-1/12sh=

1/4sh=

1/12sh=

Empty partial volume =

That is, it can also be loaded with liters.

Volume = 1 4 * 1 3sh

Cubic metre. Mass = kg 640 kg.

1m = 10dm 2 * square decimeter square * 2 = approximately.

15 + 5 = 20cm liters = 1600 ml 1600 20 = 80 square centimeters 80 * 15 = 1200 ml = liters.

Keep going.

5*5*5=125 cubic centimeters 125*3 15=25 centimeters.

It's an inverted cone!

The part that holds the water is a small cone, and the container is a large cone.

The height of the small cone: The height of the large cone = 1:2

Rule. Bottom radius of the small cone: Bottom radius of the large cone = 1:2 The base area of the small cone: The base area of the large cone = 1:4

The volume of the small cone: the volume of the large cone = (1*1 :(4*2:2 so. The volume of the conical container = 3 * 2 = 6 liters.

This container can also hold liters.

1 3* square * meters.

640 kg.

I'm very poor, I'm in 5th grade, and I think the easiest thing is 1+1=2, how about it, it's very simple

The 4 balls are arbitrarily numbered, A, B, C, D

Called a, b: the first case: if a = b, then a and b are equally heavy;

2 is called c and d, c cannot be equal to d; If C > D, then C is heavier, D is lighter, if CC, 3 is called A and C, if A=C, then B is heavier, A, C is the same, D is lighter, if AC, then B is heavier, C is lighter, A, D is the same heavy;

The third case: if a>b, that is, b can see this method, in most cases it can be judged by weighing twice, and in a few cases, weighing three times;

Hello big brother. The question can be written like this: Let the weight of the four apples be a, b, c, d, 1, put a, b on the left side of the balance and put it on the right side Because there are 2 apples of equal mass, there are two cases: a+b c+d or a+b c+d remove ac, if the balance is balanced, then put ac into the left and right ends, if a>c, c is heavy, a is clear, and vice versa, a is heavy, and b is light.

Label the apples with ABCD separately

One on both sides of the scale, until the balance is balanced, then the two are the same weight, and the remaining two are placed on both sides of the scale, and the weight can be seen at a glance, then the four apples are distinguished.

Two in a group, divided into two groups.

If two are the same weight in a group.

The first time the two groups are weighed, the two groups can be identified.

If two are the same weight separate in both groups.

The second time the two heavier of the two groups are weighed, the heavier is the heaviest, the lightest one is the lightest in the same group, and the other two are equally heavy.

1 Take out two arbitrarily, put one on each side of the scale, if it is balanced, it means that it is the two that weigh the same weight, then the remaining two are weighed again, one must be heavy and one light.

2 Take out two at random, and if the balance is uneven, replace the light one with the remaining two, so that you can find the heaviest one, and in the same way, you can find the lightest one.

This is only a text explanation, not a list of formulas.

Suppose the lightest is A, the same weight is B, and the heaviest is C.

Take two apples to weigh.

In the first case, if two apples are the same weight, then they are both b. Then weigh the other two, so heavy is C, and light is A.

The second case: two apples, one light and one heavy. Also weigh the other two.

situation, the remaining two apples are the same weight, then they are both B, then the weight of the four apples also comes out, the previous ones were A and C.

case, the remaining apples are one light and one heavy, so use two light ones to weigh. These two are light, and the one that is lighter in it is the lightest A, and the one that is slightly more important in it is the two B that are equally heavy. Then the one who was heavier than B was C, and the one who was heavier than A was B.

1.30% completion for the first time, 80% completion for the second and first attempts.

Then the second time it was 50% done and 45 trees were planted, so it was 45 50% = 90 trees 2The ratio of bamboo pole to shadow length is 3:2, so the ratio of flagpole to shadow length is also 3:2

The shadow is 10 meters long, so the flagpole is 10 2 * 3 = 15 meters long.

3.The first time 20 * 25% = 5 meters, the second time 20 * 1 2 = 10 meters, a common 25% + 50% (that is, one-half) = 75%.

1.Solution: There are a total of x trees.

x=902.Solution: Set the length of the flagpole to be x meters.

3/2=x/10

x=153.Solution: {(20-20*.)

This proof is wrong.

To prove that tanx x increases monotonically, we can find the minimum value of the given interval of [xsec 2(x) -tan(x)].

xsec^2(x) -tan(x)]‘= sec^2(x) +2xsec^2(x)tan(x) -sec^2(x) = 2xsec^2(x)tan(x) >0

When x = 0, [xsec 2(x) -tan(x)] = 0, so [xsec 2(x) -tan(x)] 0 on (0, proof complete.

Let the original speed of the car be xkm h, then the speed after the failure is (x-9) km h

108 x+24 60=108 (x-9) gives x1=54 and x2=-45

The test shows that x=54 is the solution of the original equation.

If you solve it with an equation, you will definitely be able to solve it!

reduce f(x) to f(x)=a+(a 2-a-1) (2 x+1);

The denominator of the latter term is the increasing function, and the fall is the decreasing function, and only the numerator needs to be greater than 0

i.e. a 2-a-1>0

Find out.

Monotonically decreasing.

then f'(x)<=0 to get the range of a.