How many math problems are solved? Solve these math problems

15.Solution: This kind of problem is generally transformed into a special angle, and it is easy to think of a method according to 60°=20°+40°.

Because tan60° = (tan20°+tan40°) (1-tan20°tan40°).

So tan20°+tan40°=tan60°*(1-tan20°tan40°).

So tan20°+tan40°+3tan20°tan40°

tan60°*(1-tan20°tan40°)+3tan20°tan40 °

1-2cosx>=0

2cosx<=1

cosx<=1/2

3 +2kπ<=x<=5π/3 +2kπ ，k∈z

Namely. π/3 +2kπ,5π/3 +2kπ ]k∈z

a - b cos3x

a - b = 3/2

a - b) = -1/2 => a + b = -1/2

Below - Upper: 2b = -2 => b = -1

a = 1/2

4a sin(3bx) = -4(1/2) sin(3 * 1 * x) = -2 * sin(3x) = 2sin(3x)

Cycle t = 2 (3) = 2 3

When 3x = -1, i.e. x = -1 3, the minimum value = -2

When 3x = 1, i.e. x = 1 3, the maximum value = 2

Let f(x) = 2sin(3x).

f(-x) = 2sin(3(-x))

2sin(-(3x))

2sin(3x)

f(x) f(x) is an odd function.

Then move the item and finally the root number 3.

It makes sense, it means that 1-2cosx o cosx 1 2 launches 2k +60 x 2k +300 bar,,, the idea is that this idea number is not necessarily right.

, the focal point is f(2,0), the alignment is x=-2, the distance from p to the y-axis is 4, then the distance from p to the alignment is 6, and thus the distance from p to the focus of the parabola is also 6

sin( sin( 3 5), so sin( 3 5), because is the angle of the second quadrant.

So tan = 3 4

Because tan2a=2*tana 1-(tana) 2 substitutes tan =3 4.

tan2a=-7/24

3x^2-2

f'(1)=1

The tangent equation is: y=x-1

4.Let the length of the major axis, the length of the minor axis, and the focal length be a, b, and c respectively, then b = a -c

Because a, b, and c are equal difference sequences, 2b = a+c, 4b = (a+c) so 4(a -c) = a+c).

4(a-c)(a+c)=(a+c)²

4(a-c)=(a+c)

3a=5ce=c/a=3/5

Monotonically increasing on (0, +infinity).

Therefore, if you don't understand it, you can ask directly

1.The distance from p to the y-axis is 4, x=+-4, and because y2>=0, x=4, y=4 root number 2, the distance from p to the parabolic focus is equal to the distance from p to the parabolic alignment (x=-2), so the distance is 6

2.sina = 3 5, and a is the second quadrant angle, cosa = -4 5, tana = -3 4, and tan2a = -24 7

3. y'=3x 2-2, slope = 1 at x = 1, and the curve passes through (1,0), and the tangent equation y=x-1

4.2b = a + c, a 2 = b 2 + c 2, e = c a, Lianli 5e 2 + 2 e-3 = 0, e = 3 5, choose b

5. c

1.A62Answer-24 7

The options are a bit incomprehensible, but you should be playing C.

If you have any questions, you can ask, please, thank you.

I'm only in the sixth grade, and I'm still taught.

The original profit of clothing A is 120-80=40, and the profit of clothing B is 90-60=30, and if clothing A is x, then clothing B is 100-x, and the total profit is y, then: y=(40-a)*x+(100-x)*30=40x-ax+3000-30x=(10-a)x+3000

As can be seen from the above formula, when a is 10, 10-a is a positive value, and the profit of clothing A is greater than that of clothing B, and the maximum value of x should be 75

At this point, the profit y=3750-75a

When a>10, 10-a is a negative value, and the profit of clothing A is less than that of clothing B, and the minimum value of X should be 65

At this time, the profit y=3650-65a

a+b+c=260 1 3a+1 4b=1 2c-22 by:c=2 3a+1 2b+44 will bring in: a+b+(2 3a+1 2b+44)=2605 3a+3 2b+44=2605 3a+3 2b=216b=(216-5 3a) 2 3b=144-10 9a will bring in to:

c = 2 3a + (144-10 9a) 1 2 + 44c = 11 9a + 116 substitute into a : a + 144-10 9