Function... Math problems. Help, how?

1.The minimum positive period is

Explanation: Split and merge.

y=cos（2x-π/60）+sin（2x+π/6）-1

cos2xcos（π/6）+sin2xsin（π/6）+sin2xcos（π/6）+sin（π/6）cos2x -1

3/2）cos2x+（1/2）sin2x+（√3/2）sin2x+（1/2）cos2x -1

√3/2)+(1/2)](cos2x+sin2x) -1

2[(√3/2)+(1/2)][2/2)cos2x+(√2/2)sin2x] -1

2[(√3/2)+(1/2)]cos[2x-(π/4)] 1

From the period formula t=(2 w): w=2 so t=k (k=n[integer]), so the original is the minimum positive period is

2."ln" is the logarithm of "e", "e" is a constant, and therefore "ln" is also a constant, ln

Just go straight into algebra.

Explanation: oa=(1,0) ob=(0,1).

moa=(m,0) （m-1）ob=(0,(m-1))

op=moa+(m-1)ob=(m,(m-1))=(x,y), so x=m,y=m-1, and then eliminate the element to obtain y=x-1

4.The minimum value is -1 and the set is x={(2k+1) 2}

From the meaning of the title: y=a*b=m(1+cos2x)+cos2x

Because the function image passes through the point ((4),2), substituting the original function gives m=2

Substituting m back to y=m(1+cos2x)+cos2x gives y=2(1+cos2x)+cos2x

Simplification yields y=3cos2x+2

So when "cos2x=-1", the minimum value of y is -1, and at this time 2x=(2k+1) so the set of x is x={(2k+1) 2}

Let's put point A first.

6k=b, because y decreases with the increase of x, so k<0 is b<0, and because the area of the buried car aob is liquid 12, so b=-4, k=-2 3

So y=-2 3x-4

1。 [a,0)

2.With parity, the domain is defined with respect to the origin symmetry 2a-3=a

So a=3

a150536819, hello:

There is a formula for changing the bottom of the log 4 (x)=1 2 log 2(x), and then substituting t=log 2(x) into it to get y=(t-2)(t 2-1 2)=1 2(t-2)(t-1)=1 2(t 2-3t+2).

Sell 1 share of the profit.

Return 1 winnings.

250 copies per month is a fixed profit, which can be sold in 30 days, and the profit is 250* yuan.

More than 250, each additional order, you can make a profit per month.

In order to maximize the profit, 400 shares should be made every day, and the profit should be 750 + 150 * yuan.

I.1[-Root number 3, root number 3]; 2.Classify a, when a>0, auspicious answer [a,1-a]; a<0, [-a,1+a].

II.11, f(-1)=0, f(-x)=f(x)+f(-1)=f(x), so it is an even blind banquet rent function.

3. Classification: A-2<0,4-A 2>2-A; a-2>0,4-a^2<2-a；4-a^2

!The three do which lease angle AOB is a right triangle, OC is the hypotenuse midline.

The hypotenuse midline is equal to half of the hypotenuse.

oc=1/2ab

So ab = 6 root number 5

OA squared. ob square = ab squared.

The length of the line segment oa,ob (oa less than ob) is the two real roots of the equation x square (2m+6)x+2m square 0 with respect to x.

So OA+OB=2M+6

oa*ob=2m squared.

OA + OB ) square - 2 * (OA * ob) = OA squared.

ob squared. Bring in the number to find m=6

Find the equation for pure trillion slow commotion and the two roots are 12

oa=6ob=12