The fifth grade students carry out dancing and singing competitions. The number of students participating in the competitions accounts for 80% of the whole grade. The number of students participating in dancing competitions accounts for 30% of the total number of students participating in the competitions. The number of students participating in singing competitions accounts for 80% of the total number of students participating in the competitions. There are 24 students participating in both competitions. How many students are there in the fifth grade

All participants = 30% + 80% - 100% = 10%
There are 24 of these 10%
So the total number of participants is 24 divided by 10% = 240
240 people account for 80% of the total number. 240 divided by 80% = 300 people

Girls account for 48% of the students in the school, 80 less than boys. How many students are there in this school?

There are x girls and Y boys
X/(X+Y)=0.48
Y-X=80
X=960
X+Y=2000
2000 people in total

In a school, the ratio of grade one, grade two and grade three is 8:12:21, the number of grade one is 80 less than that of grade two, and the number of students in each of the three grades

80/(12-8)=20
Grade one 20 * 8 = 160
Grade 2 20 * 12 = 240
Grade 3 20 * 21 = 420

A development exercise of grade 4 in primary school (solving simple equation)
Dongfang primary school bought 4 football and 5 basketball, which cost 283 yuan. It is known that each basketball is 8.9 yuan more expensive than football. How much is each football and basketball
Format football:
Basketball:
The solution of the demand equation

Let each basketball be X Yuan and each football be y yuan
X-Y =8.9
5X+4Y=283
(solving equation)
X=8.9+Y
5(8.9+Y)+4Y=283
Y=26.5
X=35.4
Answer: each basketball is 35.4 yuan, each football is 26.5 yuan

Given that the function f (x + 1) is equal to x-2x-3, find the range of F (x)

F (x + 1) = x ^ 2-2x-3, right?
Let t = x + 1, x = T-1
f(t)=(t-1)^2-2(t-1)-3=t^2-2t+1-2t+2-3=t^2-4t
That is, f (x) = x ^ 2-4x = (X-2) ^ 2-4 > = - 4
So the range is [- 4, + infinity)

If f (x) = x − 1 x, then the zero of G (x) = f (4x) - x is zero______ ．

∵ f (x) = x − 1X, ∵ f (4x) = 4x − 14x, let g (x) = f (4x) - x = 0, that is 4x − 14x − x = 0, the solution is x = 12, so the answer is 12

Solving the equation of the fifth grade disjunctive calculation problem (simple operation)

408-12 × 24 (46 + 28) × 60 42 × 50-1715 △ 532 + 105 △ 5 (108 + 47) × 52 420 × (327-238) (4121 + 2389) △ 7 671 × 15-974 469 × 12 + 1492 405 × (3213-3189) 5000-56 × 23 125 × (97-81) 6942 + 480 △ 3 304 × 32-1

1. Solve the linear equation of the point P (1, e) on the curve y = e ^ X and perpendicular to the tangent of the curve at this point 2. Find the tangent and straight line of the point m on the curve y = 1 / 5x ^ 5
2. The tangent at a point m on the curve y = 1 / 5x ^ 5 is perpendicular to the straight line y = - x-3, and the tangent equation is obtained
There is a lot of money

The first problem is to find the derivative on both sides of y = e ^ x, and get: y '= e ^ x, the tangent slope of y = e ^ X of the curve passing through point P (1, e) = e, the slope of the required straight line = - 1 / E, and the equation of the required straight line is y-e = - (1 / E) (x-1), that is, x + ey-1-e ^ 2 = 0

It is known that ABC belongs to Z, and ab + BC + AC = 0. It is proved that there is at least one even number in ABC by using the counter argument

Suppose that a, B, C are not even numbers, and they are all integers, then only they are odd numbers. According to "the product of odd numbers and odd numbers is still odd numbers", AB is odd. Similarly, BC and AC are all odd numbers, and the sum of three odd numbers must be odd, that is, AB + BC + AC is odd, which is contrary to the known AB + BC + AC = 0, so the hypothesis is not tenable. Therefore, there is at least one even number in a, B, C

What is the formula for finding the final term of the arithmetic sequence

Arithmetic sequence
Sum = (first term + last term) × number of terms △ 2
Number of items = (last first item) △ tolerance + 1
First term = 2 and △ number of terms - last term
Last term = 2 and △ number of terms - first term
Last item = first item + (number of items - 1) × tolerance