High School Math Problem Finding Seconds Senior 1 Math Finding Seconds

7+ immovable fixed input.

Variable input for. n is a natural number greater than or equal to 1.

Then by the nth year his expenditure is.

7+ then his average payout is:

7+ just asks for his minimum value.

i.e. (the minimum value of 7+.)

In fact, it becomes a matter of the minimum value of the 1 n+n type.

A+B》2 root number under ab

That is, the minimum value of 144 n+n.

When n=12 is the minimum value obtained.

So it's 12 years.

Set to be used for x years.

The maintenance cost for year x is (1+x) thousand dollars.

The average usage fee is s=[70+2+(2+3+4+..)1+x)]/x[71+（3+x)(1+x)/2]/x

71/x+x/2+2+3/2x

145/2x+x/2+2

x>0 uses important inequalities.

So s = 145+2

If and only if 145 2x=x 2

i.e. x = 145 12

was established. So it is best to use it for 12 years.

The minimum is 145+2

The best useful life is the year when the average annual cost is the smallest.

Assuming the optimal period of x years, the average annual cost is y

then y=70 x+2x+(2+(2+(x-1)*1))*x 2y=70 x+(x2+7x) 2

I forgot not to return the ball later.

1 Total solution: f(x) is an odd function, f(1) = -f(-1) = 1 f(x) is an increasing function at [-1,1].

The maximum value of f(x) at [-1,1] is f(1) = 1f(x) =t 2-2at+1, which is equivalent to t 2-2at + 1>=1 for any real number a [-1,1].

That is, t 2-2at>= 0 for any real number a [-1,1] holds 1When t=0, then 0>=0 is constant.

2.When t>0, then a<=t 2, is constant. ∴t/2>=1 t>=2;

3.When t<0, a>=t2, is constant. ∴t/2<= -1,t<=-2;

In summary, the value range of t is (-2] [2,+

Solution: Odd function f(x), then there is f(1)=-f(-1)=1;

f(x) is an increasing function at [-1,1].

So 1 == f(x)max =1

That is, t 2-2at>= 0 holds for any real number a [-1,1], and when t> 0, a<=t 2, always holds.

So t 2>=1, we get t>=2;

When t<0, a>=t2, is constant.

So t 2>=1, we get t>=2;

In summary, t can be valued in the range of (negative infinity, -2] or [2, positive infinity) or 0

According to the title: f(x)=f(x-1)-f(x-2)=f(x-2)-f(x-3)-f(x-2)=-f(x-3).

i.e. f(x) = -f(x-3) = f(x-6).

So f(2009)=f(335*6-1)=f(-1) and log2(1-x) 0 x, so f(-1)=1

So f(2009)=1

f(x+4)=-1/f(x+2)=-1/(-1/f(x))=f(x)

And because it is an even function, there is f(-x)=f(x)f(

Solution: According to the condition, there are 6 line segments, which are AD, AC, AB, DC, DB and CB.

ad=1/2ac

ac=acab=2ac

dc=1/2ac

db=dc+cb=1/2ac+ac=3/2accb=ac

ad+ac+ab+dc+db+cb=1 2ac+ac+2ac+1 2ac+3 2ac+ac=13 2ac=23, and the solution is ac=46 13

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Solution: Knowing from the meaning of the question.

Because a-1 a=5

So (a-1 a) 2=25

again a 2-(1 a) 2=(a-1 a) 2-2, so the original formula = 25-2 = 23

Idea (a-b) 2+2ab=a 2+b 2

tana=1

a=pi 4 or 5pi 4 is so obvious.

Cosa sina is not the root number 2 2 or - the root number 2 2 2 special horn can't be remembered?