It is known that the domain of function f (x) is [1,9], and when 1 = 9, f (x) = x + 2, then function y = [f (x)] ^ 2 + F (x ^ 2) A[1,3] B[1,9] C[12,36] D[12,204] Many people say to choose D, but I think we should choose C, because f (x ^ 2) has a limit, 1 = 3 A function has only one domain, so they are all [1,3]

It is known that the domain of function f (x) is [1,9], and when 1 = 9, f (x) = x + 2, then function y = [f (x)] ^ 2 + F (x ^ 2) A[1,3] B[1,9] C[12,36] D[12,204] Many people say to choose D, but I think we should choose C, because f (x ^ 2) has a limit, 1 = 3 A function has only one domain, so they are all [1,3]

Given that the domain of function f (x) is [1,9], and when 1 = 9, f (x) = x + 2, then the range of function y = [f (x)] ^ 2 + F (x ^ 2) a [1,3] B [1,9] C [12,36] D [12204] analysis: ∵ the domain of function f (x) = x + 2 is [1,9] ∵ y = [f (x)] ^ 2 + F (x ^ 2), that is, function y is composed of two function relations [f (x)] ^ 2, f (x)] ^ 2

It is known that the domain of the function f (x) = x ^ 2-2x-3 is (1,9) to find the range of y = f (x ^ 2)
RT

The domain of y = f (x) is (1,9), then the domain of F (x ^ 2) is 1

The function f (x) = 2 + log3x (1) is known

F (x) is an increasing function and Y is also an increasing function
1=

The conclusion of the relationship between sliding friction and maximum static friction is explored by experiment

Sliding friction is equal to the friction when moving at constant speed, which is equal to the maximum static friction

When a is more or less, the square of equation (A-1) x + X-2 = 0 is a linear equation with one variable

That is, when a = 1, the original equation is a linear equation with one variable
Because the highest term of the original equation is a quadratic term, so as long as the two terms are equal to 0, the original equation is a linear equation with one variable
So a = 1

A and B start from ab at the same time and travel in opposite directions. A travels 45km per hour and B 55km per hour. If a increases 15km per hour, B increases 15km per hour
Plus 5km, the meeting time can be advanced by one fourth of an hour. What is the distance between AB and ab

It should be a quarter of an hour. It's a quarter of an hour
Let: the distance between two places be x km. X / (45 + 55) = x / (45 + 15 + 55 + 5) + 1 / 4
X / 100 = x / 120 + 1 / 4 multiply both sides by 600 to get 6x = 5x + 150
6x-5x = 150 x = 150 (km) a: the distance between the two places is 150 km

2-3 + 4-5 + 6-7 + 8-9 +... + 2000-2001 + 2002 =?
What is the final answer? Urgent

Grouping
Original formula = (2 + 4 + 6 +... + 2002) - (3 + 5 + 7 -... + 2001)
=[(2+2002)*1001/2] - [(3+2001)*1000/2] …… Summation formula of arithmetic sequence
=1003002 - 1002000
=1002
Or:
Original formula = (2-3) + (4-5) + (6-7) +... + (2000-2001) + 2002
= -1*1000+2002
= 1002

Simple operation 999 × 99 × 101 + 999

=999×[(100-1)×(100+1)+1]
=999×(100²-1+1)
=999×10000
=9990000

There are 63 candies in the three piles. The number of candies in the first pile is twice that in the second pile. The number of candies in the third pile is three times less than that in the second pile. How many candies are there in the second pile?

Given that 2B + 3 / 3A = 5, find the value of 6B + 9 / 9A + 3
/ it's a fraction

Because
　2b＋3／3a＝5
So
　6b＋9／9a＋3
　=3（2b+3)/ 3*(3a+1)
　=2b+3/3a+1
　=6/1