Ask for help for math problems in the first year of high school. High School Math Questions for help

12.According to tana = 1 3 0, so a is the first quadrant or third quadrant angle.

Let the coordinates of point p be (x,y), then.

tana=y/x=1/3

x²+y²=（3√10）²

Solution. x=9，y=3

or x=-9, y=-3

When a is the first quadrant angle, x=9 and y=3

sina=y/3√10=√10/10

cosa=x/3√10=3√10/10

cota=1/tana=3

When a is the third quadrant angle, x=-9 and y=-3

sina=y/3√10=-√10/10

cosa=x/3√10=-3√10/10

cota=1/tana=3

Draw a diagram, let the origin point be a, make a perpendicular line from the point p, the perpendicular foot is b, let pb be x, then ab is 3x, and x of the pythagorean theorem is 3

Then, sina = 10 10 under the root, cosa = 3 * 10 10 under the root, cota = 3

cccdc 17 because tana=1 3 so a is in one or three quadrants.

When a is in the first quadrant, sina = root, 10, 10, cosa= 3, root, 10, 10

When a is in the third quadrant is sina=-root, 10, 10, cosa=-3, root, 10, 10

The answer is (c), (commutation).

Solution: yf(x)=

4－cos²x－3sinx)／(2－sinx)（sin²x－3sinx＋3）／(2－sinx)【﹣sinx（2—sinx）＋（2－sinx）＋1】／(2－sinx)

sinx＋1＋1／(2－sinx)

2 sinx) 1 (2 sinx) 1 let t = 2 sinx, then 1 t 3

y=g（t）=t＋1／t－1

It is easy to know that the function y=g(t)=t 1 t is the "hook function", and the monotonicity is: g(t) is increasing on the interval [1].

g（t）min=g（1）=1，g（t）max=g（3）=7／31≤g（t）≤7／3

Namely: 1 y 7 3

The maximum value of y is 7 3

I don't know if you have learned the monotonicity of the hook function, you should have learned it, this is the knowledge of the higher function.

If you haven't learned, you can see here

For this "hook function" can be remembered as a conclusion. Its conclusions are as follows:

If f(x)=

ax＋b／x

a, b 0), then the monotonicity of f(x) is:

f(x) is a subtraction function over the interval [ ab), 0) and the interval (0, (ab)];

f(x) in the interval (ab)].

and intervals. (ab), which is the increasing function;

Proof: (You can use derivatives in the third year of high school, and you can use the definition method in the first year of high school, so I won't write the specifics, and leave it to myself to prove.) ）

Then, according to this conclusion, it is not difficult to understand the monotonicity of g(a).

Are you sure you got this question right? PA vector + PB vector + PC vector = 0 vector?

Normalization of the circle equation: (x+1) 2+(y-2) 2=2 drawing, it can be seen that there is only one tangent that meets the requirements, set to y=-x+b(b>0) because the tangent and the circle have and only one intersection point, so substitute the original equation, and obtain 2x 2+(6-2b)x+b 2-4b+3=0, and only one solution satisfies the equation.

Therefore, according to the formula of the solution of the unary quadratic equation, b 2-4ac=0 is obtained to b 2-2b-3 = 0, and b = 3 is obtained

Mistaken?

Obviously x -1= x -x +x -x+x-1=x (x-1)+x(x-1)+(x-1)=(x-1)(x +x+1).

You can also divide by x -1 by x-1 by x-1, as if it can't be broken down into (z-1) (z-x) (z-x).