Equation of straight line (149:35:1) It is known that the straight line passes through the point P (3,2) and intersects with the positive half axis of x-axis and y-axis respectively, and points a (a, 0), B (0, b) and o are the origin of coordinates (1) when the area of △ AOB is the smallest, the equation of line L is obtained; (2) when a + B is the minimum, the equation of line L is obtained

Let the equation of line l be Y-2 = K (x-3) (K0) if and only if - 9K = - 4 / K, i.e. k = - 2 / 3, the area of triangle ABO is the minimum, mins (ABO) = 12, then the equation of line L is Y-2 = (- 2 / 3) (x-3), i.e. y = - 2x / 3 + 4 (2) a + B = (3-2 / k) + (2-3K) = 5-2 / k-3k, because K0 - 3K >

If the positive integers a and B make the equation a + (a + b) (a + B-1) / 2 = 2009 hold, find the values of a and B
Today is the day
sorry,

Let a ≥ B ≥ 0, x = a + B, y = A-B, then x ≥ y ≥ 0. The original equation is equivalent to x ^ 2 + y = 4018, then x ^ 2 + X ≥ 4018 ≥ x ^ 2, 62 * 63 = 3906, 64 ^ 2 = 4096, so x = 63, y = 49. The non negative integer solution is a = 56, B = 7; or a = 7, B = 56
(Note: This is a parabola with infinite lattice points (points whose abscissa and ordinate are integers)

10x x = 5500?

10x x = 5500
11x=5500
x=5500÷11
x=500
complete!

Evaluation: (1). (1 + cot 75 °) / (1-cot 75 °) (2). Sin70 ° sin65 ° - sin20 ° sin25 °
(3) . Tan (α + 5 π) = 1 / 2, then (COS α - 1 / 2Sin α) / (COS α + sin α) is solved in detail

(1+cot75°)/(1-cot75°)
=(1+1/tan75°)/(1-1/tan75°)
=[(tan75°+1)/tan75°]/[(tan75°-1)/tan75°]
=(tan75°+1)/(tan75°-1)
=-(tan75°+1)/(1-tan75°)
=-(tan75°+tan45°)/(1-tan75°tan45°)
=-tan(75°+45°)
=-tan120°
=-tan(180°-60°)
=tan60°
=√3
sin70°sin65°-sin20°sin25°
=sin70°sin(90°-25°)-sin(90°-70°)sin25°
=sin70°cos25°-cos70°sin25°
=sin(70°-25°)
=sin45°
=√2/2
tan(α+5π)=1/2,
tan(α+4π+π)=1/2,
tan(α+π)=1/2,
tanα=1/2,
The denominator of (COS α - 1 / 2Sin α) / (COS α + sin α) is also divided by cos α
=(cosα/cosα-1/2sinα/cosα)/(cosα/cosα+sinα/cosα)
=(1-1/2tanα)/(1+tanα)
=(1-1/2*1/2)/(1+1/2)
=(1-1/4)/(3/2)
=(3/4)/(3/2)
=3/4*2/3
=1/2

Who can give me five grade 100 equations? Decimal

If we know Tan alpha = 3 and Tan β = - 1 / 2, then what is Tan (a - β) =?

tan(a-β)
=(tana-tanβ)/(1+tanatanβ)
=(3+1/2)/(1-3/2)
=-7

In the plane rectangular coordinate system, the straight line L = ax + by + C = 0 is translated to the right along the x-axis by 2 units, and then down by 3 units to calculate the slope
In the plane rectangular coordinate system, the straight line L = ax + by + C = 0 is translated 2 units to the right along the x-axis, and then 3 units to the down. If the obtained line coincides with the original line, the slope of the straight line l will decrease
A 2/3 B -3/2 C 3/2 D -2/3
How to find the slope? I don't have any idea. Mathematics is always bad
Please give me more details about the idea and process

Write it as a function:
Y = - ax / B-C / b translates 2 units to the right along the X axis, and then translates 3 units down
It becomes: y = - A (X-2) / B-C / B-3
Compared with the original function, the result is as follows
2a/b-c/b-3=-c/b
So a / b = 3 / 2, but the slope has a minus sign
So choose B

Nine out of four equals () hours () minutes

2:15

If the sum of the intercept of the line 3x-4y + k = 0 on the coordinate axis is 1, then K=

3x-4y=-k
3x/(-k)+4y/k=1
that
-k/3+k/4=1
-k/12=1
k=-12
perhaps
Let x = 0, y = K / 4
Let y = 0, x = - K / 3
k/4-k/3=1
k=-12

Solving the equation SiNx / 2 + cosx = 1

sinx/2+cosx=1
sinx/2=1-cosx=2sin²x/2
2sin²x/2-sinx/2=0
sinx/2（2sinx/2-1）=0
Solution
sinx/2=0 x/2=kπ x=2kπ
2sinx/2-1=0 sinx/2=1/2
X / 2 = 2K π + π / 6 or X / 2 = 2K π + 5 π / 6
X = 4K π + π / 3 or x = 4K π + 5 π / 3
So the solution of the equation is:
x1=2kπ
x2=4kπ+π/3
x3=4kπ+5π/3

If the positive integers a and B make the equation a + (a + b) (a + B-1) / 2 = 2009 hold, find the values of a and B Today is the day sorry,