What is sin65 ° * cos20 ° - sin65 ° * sin20 ° equal to?

What is sin65 ° * cos20 ° - sin65 ° * sin20 ° equal to?

=Sin (65-20) = sin45 = (radical 2) / 2
What is 130 degrees
sin130º
=sin（180º-50º)
=sin50º
=≈0.766
This is my conclusion after meditation,
If not, please ask, I will try my best to help you solve it~
If you are not satisfied, please understand~
Given a ∈ {x log (2) x + x = 0, then the increasing interval of the logarithm of F (x) = x * 2-2x-3 with a as the base is
A ∈ {x log (2) x + x = 0}
x>3,x
It is known that the intersection of the image of the first-order function y = - 2 + 4 and the y-axis is B, the intersection of the image of y = 3x + 1 and the y-axis is C, and the intersection of the two function images is a
(1) Finding the coordinates of a, B and C (2) finding the area of triangle ABC
(1) If y = - 2x + 4 intersects with y axis at point B, then the coordinates of B are (0,4);
If y = 3x + 1 and Y axis intersect at point C, then the coordinate of point C is (0,1);
If the intersection of y = - 2x + 4 and y = 3x + 1 is a, then the coordinates of point a are (3 / 5,14 / 5);
(2) Area of triangle ABC = 1 / 2 * BC * 3 / 5 = 9 / 10
∵ and y-axis intersection coordinate is (0, b) ∵ B point coordinate is (0, 6) ∵ C point coordinate is (0, 1), then - 2x + 6 = 3x + 1, x = 1 ∵ y = 4 ∵ a (1,4) ∵ s △ ABC = (6-1) × 1 △ 2
1. Because y = - 2 + 4 intersects the Y axis at point B
The coordinates of B are (0,4);
Because y = 3x + 1 intersects the Y axis at point C
So the coordinates of point C are (0,1)
From the meaning of the title;
y=-2+4
y=3x+1
The solution is obtained;
X=1/3
Y=2
therefore
The coordinates of a are (1 / 3, 2)
2、
Area of triangle ABC = 1 / 2x3x1 / 3 = 1 / 2
There is something wrong with this question
1. Because y = - 2 + 4 intersects the Y axis at point B
The coordinates of B are (0,4);
Because y = 3x + 1 intersects the Y axis at point C
So the coordinates of point C are (0,1)
From the meaning of the title;
y=-2+4
y=3x+1
The solution is obtained;
X=1/3
Y=2
therefore
The coordinates of a are (1 / 3, 2)
2、
Area of triangle ABC = 1 / 2x3x1 / 3 = 1 / 2
There is something wrong with this question
How can y = - 2 + 4 be a linear function
We know that the logarithm of F (x) = log with a as the base (3-ax) is monotonically decreasing on [0,2], so we can find the value range of A
Given that the logarithm of F (x) = log with a as the base (3-ax) is monotonically decreasing on [0,2], the value range of a is obtained,
When 00,3-2a > 0, a
When 00,3-2a > 0, a
Given the quadratic function y = 3x ^ - 8x + 4, how many intersections do the graph of this function have with the X axis
Y = 3X² - 8X + 4
△ = 8² - 4×3×4 = 16 〉0
So there are two intersections between the function image and the X axis
(- 8) ^ 2-4 * 3 * 4 = 16 > 0, so there are two intersections with the X axis
Did not understand your function, 3x ^ - 8x
Given the significance of log (x-1) (x ^ 2-5x + 6), find the value range of X
(1,2) Union (3, positive infinity)
Given the function y = 3x-2. Find the coordinate intersection of function image and x-axis, y-axis
Let y = 0, then x = 2 / 3
Let x = 0, then y = - 2
So the intersection points of function image and x-axis and y-axis are (2 / 3,0) (0, - 2) respectively
If this problem really can't, LZ should have a good look at the book
The characteristic of x-axis is that the value of Y is 0
The characteristic of y-axis is that the value of X is 0
So if y = 0, we can calculate that the intersection of x = 2 / 3 and X axis is (2 / 3, 0)
So if x = 0, we can calculate that the intersection of y = - 2 and X axis is (0, - 2)
X:(2/3,0);Y(0,-2)
The logarithm of solving inequality log with base 2 (x + 4) is more than 4
㏒2（X+4）＞4
X+4＞2^4
X+4＞16
X＞12
If we know that 2y-3 is in direct proportion to 3x + 1, and x = 2, y = 5. If point (a, 2) is on this function image, we can find a
2y-3 is proportional to 3x + 1,
Let 2y-3 = K (3x + 1)
And when x = 2, y = 5
Then 10-3 = 7K
K=1
Then 2y-3 = 3x + 1
Point (a, 2) is on this function image
Then 4-3 = 3A + 1
Then 3A = 0
A=0