Math help. Do me a favor, math help, help?

20% = 480 (hectares).

6 (1+ (apple) grapes).

8 (1-20%) = 10 (hours).

2 [42% 5 7-(1-42%-42% 5 7)] = 100 (Ben).

1) Set the original plan to afforestate x hectares, according to the meaning of the topic, the actual afforestation x (1 + 20%) is actually 96 hectares more than the original plan, and the equation can be listed: x (1 + 20%) - x = 96 to obtain x = 480 The original plan to afforestate 480 hectares.

2) Let the original cost price of each product be y yuan, (

Find y=44 The original cost price of each product is 44 yuan.

3) The area of the apple orchard is regarded as unit one, then the area of the vineyard is (1+8%), and the area of the apple orchard is s, then the area of the vineyard is (1+8%)ss+(1+8%)s=

Find s = 120 hectares of vineyards and 120 hectares of apple orchards.

Question: If the speed of 1 car is V, then the speed of 2 cars V (1-20%) 1 car takes 8 hours to complete the journey, then the whole journey is 8V

2. The time required for the complete journey of the car t=s v(1-20%=8vv(1-20%)=10 (hours)).

There are a total of x books in this batch of x, so the books assigned to class 1 are 42%x, the books assigned to class 2 are 5 7*42%x, and the books assigned to class 3 are 5 7*42%x-2, and this batch of books is just finished.

42%x+5 7*42%x+5 7*42%x-2=xObtain x=100 There are 100 x books in this batch.

1.The increase of 20% is the additional 96 hectares, the original plan = 96 2Now it is 85% of the original cost, then the original =

3.Let 2 area be y, then 1 area is obtained by y=120, i.e., 1 area is 120 hectares, and 2 area hectares.

Assuming that the speed of 1 car is V, then the time required for 2 cars to speed 2 cars is hours.

Suppose there is m in the book, 1 class, 2 classes 5 7 * (, 3 classes 5 7 * (, the solution is m=100

1. If the original plan of afforestation x hectares, then (1 + 20%) x = x + 96, then x = 480, A...

2. Let the original cost price of each product be x yuan, then, then x=44, A...

3. Let (2) be x hectares, then (1+8%) x+x=, then x=120, then (1) is, A...

Question 1: Let the speed of 1 be the unit 1, and the time for 2 to complete the line is x hours, then 8=(1-20%)x, that is, x=10, answer...

Question 2: 42%x+42%x*5 7+42%*5 7-2=x, then x=100, answer.

Hello, try the cosine theorem.

cos75 degrees quarters (root number six minus root number two).

If there are x people in addition to the 88 students, then there are 74x+88=76*(x+1) to get x=6

So these 6 people did a total of 74 * 6 = 444, each person did a minimum of 72, 6 * 72 = 432, so there are at most 12 more to be assigned.

So the most of these 6 people did 72 + 12 = 84 <88Therefore, the fastest one was 88.

There are x people.

76*x=74*(x-1)+88

x=6 A total of 6 people did 76*4=456.

1 makes 88.

Because each student does at least 72.

72*5=360.

Even if the extra 8 were made by a classmate.

The fastest student did 88.

There are a total of x people in the first place.

76x-88=74*(x-1)

x=7 (person).

If there are 6 students who do 88 students, suppose one of them does the fastest, and the other 5 people are 72, see if he does more than 88:

So the 88 students who do it are the fastest!

88 columns of equations, set x people.

76x=74(x-1)+8

Calculate x=7, making a total of 532.

Then, if the remaining 6 people only do 72, then there are 532-6*72-88=12, which means that even if all 12 are done by one person, there are still only 84, so the fastest is 88.

Solution: Let the three numbers be b1, b2, b3, the common ratio is q, and let the tolerance of the number series be db1b2b3=b2 =8

b2=2b1=2/q，b3=2q

b1+2, b2+2, b3+1 into an equal difference series, then 2(b2+2)=(b1+2)+(b3+1)b2=2 b1=2 q b3=2q substitution, sorting, 2q -5q+2=0

q-2)(2q-1)=0

q=2 or q=1 2, the three numbers are arranged from smallest to largest, q=1 2 rounded off b1=b2 q=2 2=1 b3=qb2=2 2=4a3=b1+2=1+2=3 a4=b2+2=2+2=4 a5=b3+1=4+1=5

d=a5-a4=a4-a3=5-4=1

The general formula for the series an=a1+(n-1)d=a3+(n-3)d=3+1 (n-3)=n is an=n

bn=a(n+1)/an +an/a(n+1)=(n+1)/n +n/(n+1)

1+ 1/n +(n+1-1)/(n+1)=1+1/n +1 -1/(n+1)

2+ 1/n -1/(n+1)

tn=b1+b2+..bn

2n +[1/1-1/2+1/2-1/3+..1/n-1/(n+1)]

2n +1 -1/(n+1)

2n + n/(n+1)

From the title: 2a+3+b-2=10

2a+b=9

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This problem belongs to the regular problem in the number series.