Math Problem !! 100,000 urgent, math problems!! Pray to the Great God

32-20) (3 8 + 2 3-1) = 288.

Master Li plans to produce 288 parts.

Solution: Master Li plans to produce x parts.

3/8x+20+2/3x-32=x

3/8x+2/3x-x=12

1/24x=12

x=288A: Master Li plans to produce 288 parts.

Let the total number be x.

3 8 times x x + 20) + (2 3 times x x-32) = x solution to get x = 288

A: Master Li plans to produce 288 parts.

Let the number of units produced be x and the total number of units y

According to this column, x=eight-thirds y+20

y-x = two-thirds y-32

The two formulas stand in a row.

x=128, y=352

It is planned to produce 352 units.

Let's say x have been produced and y are planned to be produced.

then x-y 3 8=20

y×2/3-(y-x)=32

So the planned number of production is y=288.

Set up x parts.

3x/8+20+2x/3-32=x

The solution is x=288

Let the total number x, the number of production y, y=3 8x+20; x-y=2 3x-32 Add the two formulas to give 1 24x=12 so x=288 and y=128

Volume = (40-4 2) (30-4 2) 4 = 32 22 4 = 28416 (cm) =

Challenges

Water overflow:=36（dm³）=36（l）

61 (1+61)*61 2+61=1952, when the number of black dots increases to 61, there are a total of 1952 circles.

The white ones have been counted).

This is followed by 62 black circles.

No white circles appear.

So there are 61 white circles in total.

2,3,4,5,..Calculate the total number of items added to 2003. Then the answer is the total number of items minus 1

This is a problem to find patterns, you take each white circle as a breakpoint, you will find that the number of circles is 2, 3, 4, 5... This is a very simple series of equal differences. LZ simply does the math and the result is there.

Let the black + white between each be a rule.

a1=2 a2=3 ..an=n+1 so that each term has and only one sn=n(n+1+2) 2=2003

When n=61, there is sn=1952<2003

When n=62, there is sn=2015>2003

61 out of 61

Please point out any calculation errors.

1+2+3+4+……n+n≈2003

1+n)n 2+n and 2003

n(n+3) and 4006

n must be between 60 and 70.

n=60，n（n+3）=3780

n=61，n（n+3）=3904

n=62，n（n+3）=4030

So n=61

There are 3 in the first 3 groups, which correspond to exactly 3 black circles, and the black circles follow 1, 2, 3. Arrange. I put the black one aside and set it up n

Total number: 1+2+3+4....n+n=2003( nx(1+n))/2+n=2003

n = 88 and 2

That is, there are 88 + 1 = 89

1) The cost of using an energy-saving lamp is Q1, and the cost of using an incandescent lamp is Q2, according to the title, Q1 = * x * 49 = + 49

q2 = * x * 18 = + 18

2) According to the problem, Q1 = Q2, +49 = + 18, and x = 2000

Obtained from Q1 < Q2, x < 2000

Obtained from Q1 < Q2, x < 2000

When the lighting time is greater than 2000 hours, the cost of using incandescent lamps is low, and when the lighting time is less than 2000 hours, the cost of using energy-saving lamps is low.

3) If the incandescent lamp is used for x hours, then the energy-saving lamp (3000 - x) hours (200 <=x <=2800) is used, and the usage fee is q, according to the topic, q = * 3000 - x) *49 + x * 18

200 <=x <=2800

qmin = * 200 + =

To sum up, when using incandescent lamps for 200 hours and energy-saving lamps for 2800 hours, the cost is the lowest and is yuan.

Please open ** to see the detailed answer.

Solution: (1) The area of the triangle can be subtracted from the area of three small triangles to find the area of the rectangle xy=80 100 2-y(100-x) 2-(80-y)x 2xy=4000-y(100-x) 2-x(80-y) 22xy=8000-100y-80x+2xy, both sides of which are subtracted by 2xy

0=8000-100y-80x

100y=8000-80x

y=（8000-80x）÷100

y=（400-4x）÷5

2) The defg is a square.

i.e. x=y, so x=(400-4x) 5

Solution x=400 9

x²=1600/81

The profit of each piece does not exceed 40%, that is, the unit price does not exceed 50 + 50 * 40% = 70, so the party range of x is (50, 70].

Let the functional relationship between y and x be y=ax+c, then there is 400=60a+c and 300=70a+c, i.e., a= -10, c=1000, so the functional relationship is y= -10x+1000, and 50 is w=(x-50)*y=(x-50)*(10x+1000)= -10x2+1500x-50000

When x = -b 2a = -1500 2*(-10)=75, the maximum value of w is 6250

The range of x's gatherings is (50, 70].

Let the functional relationship between y and x be y=ax+c, then there is 400=60a+c and 300=70a+c, i.e., a= -10, c=1000, so the functional relationship is y= -10x+1000, and 50 is w=(x-50)*y=(x-50)*(10x+1000)= -10x2+1500x-50000

When x = -b 2a = -1500 2*(-10)=75, the maximum value of w is 6250

Okay, that's all.

Do everything possible to form groups of three or five.

n=2a+1 type, a total of 50.

n=6a+2 type, a total of 16.

n=10a+4 type, a total of 6.

n=14a+6 type, a total of 4.

There are 76 in total.

1. Hundreds of groups of three or five.

2. The total stroke of the puppy is the speed of the puppy multiplied by the time taken by the puppy, and the time taken by the puppy is the time when the two meet.

10*[km.]

The total distance traveled by the puppy is km3

Hundreds, thousands, groups of threes and fives.

Divide the elements in the original set into 3 groups:

3Obviously, the numbers in each group must be selected at the same time, and the set a is non-empty, so all three groups of numbers cannot be selected at all.

This can be seen as a problem of the number of non-empty subsets of a set of 3 elements (each of which refers to the set of numbers mentioned above), so the set that satisfies the condition is 2 3-1 = 7.

The numbers in the set a must add up to 6, and if there is a number a, there must be 6-a, which are {15}{24}{3}{135}{234}{12345}{1245} respectively. The idea of 2*4-1=7 is as follows: first, put a pair of pairs of numbers that satisfy a {a,6-a} in the set, there are three kinds of {a,6-a}, and then put four other pairs besides {a,6-a} (and one that doesn't choose anything) Anyone who has learned probability knows that one of them is repeated.

∵a∈a→(6-a)∈a

There are only two possibilities for an element in a:

1) a a, b a and a + b = 6

2) a a, and 2a = 6

Therefore, the elements of a can be divided into 3 groups:

The set that satisfies the conditions of the question can only be selected in groups among these 3 sets of numbers, so that the total number of methods selected is .

c(3,1)+c(3,2)+c(3,3)=2 3-1=7 means that the number of sets that meet the requirements of the problem is 7.

Note: c(n,m) denotes the number of all combinations of m(m n) elements selected from n different elements.

There are 2 3 subsets in total, and then subtract what is a true subset in itself, 2 3-1 = 7

Because 6-a belongs to a, there are three pairs of 1,5. 2,4 3,3 So the total total set is 2 to the 3rd power. Subtracting 1 more empty set is the answer.

, then a=

If there are two elements, then a=, if there are three elements, then a=, and if there are four elements, then a=

There are five elements, then a=

So there are 7 of them.

The non-empty set a satisfies a and if a belongs to a, then 6-a belongs to a, and the number of sets that satisfy the above conditions ( ).

The answer is 7.

Solution: First consider the number of elements of a as few as possible. is one of them, a total of 3 kinds;

Then, two of them are taken to get a union, and there are 3 kinds; Take all of them, so there are 3 + 3 + 1 = 7.

Subtract 6 from each possibility of a, count it again, and you're good to go.