(tan40°+sin20°)°= The result is root sign

(tan40°+sin20°)°= The result is root sign

1/sin 20°-1/tan 40 °
= (2cos20°)/(2sin20°cos20°)-cos40°/sin40°
=(2cos20°-cos40°)/sin40°
=[2cos(60°-40°)-cos40°]/sin40°
=(2cos60°cos40°+2sin60°sin40°-cos40°)/sin40°
=... unfold
1/sin 20°-1/tan 40 °
= (2cos20°)/(2sin20°cos20°)-cos40°/sin40°
=(2cos20°-cos40°)/sin40°
=[2cos(60°-40°)-cos40°]/sin40°
=(2cos60°cos40°+2sin60°sin40°-cos40°)/sin40°
=2sin60°sin40°/sin40°
=√3
Is that ok? Follow up: 1 / sin 20 ° - 1 / Tan 40 ° and
(tan40°+sin20°)°=
Is it the same? I'm a junior high school teacher
Tan40 = how many / how many
zero point eight three nine one
Tan40 = sin40 / cos40
If - 1
Inequality can be reduced to loga (1 / a)
No, not before!
If the image of a function is known to pass through a (0,3) and B (2, - 3), the analytic expression of the function is obtained, and whether C (- 2,8) is on the image of the function is judged
Let the analytic expression of the first-order function be y = KX + B (k is not equal to 0) --- this is the general form of all first-order functions. By substituting a (0,3) and B (2, - 3), we get that the meaning of a (0,3) on the function image is that for this first-order function, when x = 0, y = 3, and B (2, - 3) on the function image is that
(1) Y = KX + B brings two points AB in
B=3
2K + B = - 3, the solution k = 0, B = 3, the analytic formula is y = 0x + 3
(2) Point C is not on the image of a linear function
Let the analytic expression of a function be y = KX + B
Substituting a (0,3) and B (2, - 3), we get
3=k*0+b
-3 =2k+b
The solution is k = - 3, B = 3, and the analytic formula is y = - 3x + 3
When x = - 2, y = - 3 * (- 2) + 3 = 9 is not equal to 8
So C is not on the image
1. Solution. Let y = KX + B 2. When x = - 2, y = (- 3) × (- 2) + 3 = 9
In ∵ C (- 2,8), y = 8
... unfold
1. Solution. Let y = KX + B 2. When x = - 2, y = (- 3) × (- 2) + 3 = 9
In ∵ C (- 2,8), y = 8
9 ≠ 8
Ψ not on the image of the function
(1)b=3
(2）-3=2k+b
The solution is k = - 3
Substituting B = 3 into y = KX + B, we get y = - 3x + 3
1. Solution. Let y = KX + B 2. When x = - 2, y = (- 3) × (- 2) + 3 = 9
In ∵ C (- 2,8), y = 8
... unfold
1. Solution. Let y = KX + B 2. When x = - 2, y = (- 3) × (- 2) + 3 = 9
In ∵ C (- 2,8), y = 8
9 ≠ 8
Ψ not on the image of the function
(1)b=3
(2）-3=2k+b
The solution is k = - 3
If B = 3 is substituted by y = KX + B, then y = - 3x + 3 agrees with 0 | put it away
If 2 / 3 of loga is greater than 1, then the value range of a is?
Please, brother!
I want to see if my answer is right
loga 2/3>1
Equal to log a 2 / 3 > log a
Then we discuss the scope of A,
If a > 1, it is an increasing function,
At this time, 2 / 3 ＞ a ＞ 1;
Obviously wrong
If a is less than 1, it is a decreasing function,
At this time, 2 / 3 < A, a and 0 < a < 1,
SO 2 / 3 < a < 1
If there is a problem, it should be corrected immediately. Even if there are some problems in the reference books, it is better to ask the teacher immediately if there is any difference, so as to know who is right. It is more important for students
2 / 3 of loga is a decreasing function in the domain of definition
When a = 2 / 3, it is equal to 1
So 0
It is known that the linear function y = (2-m) x-2m square + 8 1. When the value is, the image passes (4,18)?
Substitute the values of X and Y into the square of 18 = (2-m) * 4-2m2 + 8 (M + 1) = 0 m = - 1
Substituting points (4,18) into the matrix, we can get the following results
(2-m)×4-2m^2+8=18
The solution is: M = - 1
If (loga 2 / 3) ∧ 2 ＜ 1, then the value range of a is?
RT, why my classmates and I have all kinds of answers!
The questions are: (loga 2 / 3) ∧ 2 < 1
Then: - 13 / 2
If 0
When 0
When we know the value of a linear function y = (m-2) x-quarter m square + 1. What is the value of M, the function image is parallel to the straight line y = 2x? (important process)
All property
The two lines are parallel
Then m-2 = 2
The solution is m = 4
Function image parallel and straight line y = 2x
It shows that m-2 = 2
M = 4: process
Given that the function f (x) = loga [(1a-2) x + 1] is always positive in the interval [1,2], the value range of real number a is obtained
① When a ＞ 1, to make f (x) constant positive, we only need the real number (1a − 2) x + 1, when x ∈ [1,2] is constant greater than 1, let y = (1a − 2) x + 1, the function is monotone on [1,2], so we only need (1a − 2) × 1 + 1 ＞ 1 (1a − 2) × 2 + 1 ＞ 1, no solution; when 0 ＜ a ＜ 1, to make f (x) constant positive, we only need true
Given that the image of the first-order function and the image of the inverse scale function y = - 2 / X intersect at the point (1, m) and pass through the point (0, - 3), find the first-order function
Let the analytic expression of the function be y = KX + B
Because y = KX + B and y = - 2 / X intersect at point (1, m)
So x = 1
Then in y = - 2 / x, - 2 / 1 = y = M
So m = - 2
Because y = KX + B passes through points (1, - 2), (0, - 3)
So K + B = - 2
b=－3
So k = 1
b=－3
So the analytic expression of the first-order function is y = x-3