The area of triangle ABC is calculated when the image of the first-order function y = 2 / 3x + m and y = - 2 / 1X + n passes through point a (- 2,0) and intersects two points B and C with y axis respectively

The area of triangle ABC is calculated when the image of the first-order function y = 2 / 3x + m and y = - 2 / 1X + n passes through point a (- 2,0) and intersects two points B and C with y axis respectively

From the problem: m - (4 / 3) = 0n + 1 = 0, the solution is: M = 4 / 3; n = - 1, then the first-order function: y = (2 / 3) x + (4 / 3); y = (- 1 / 2) X-1, because the function and Y axis intersect B and C respectively, so: B (0,4 / 3), C (0, - 1) | BC | = 1 + (4 / 3) = 7 / 3 | OA | = 2S △ ABC = (& # 189;) × | OA | × | BC | = (& # 189;) × 2 × (7 / 3) = 7 / 3
Take a (- 2, 0) into y = 2 / 3x + 4 / 3, y = - 1 / 2x-1, so it intersects with y axis B (0, 4 / 3) C (0, - 1), so s triangle ABC = 2 * 1 / 2 * 7 / 3 = 7 / 3, that's the way of thinking, but I have problems in calculation!!
Calculate the value of (2x ^ 3 + 3x ^ 2y-9xy ^ 2) - 0.5 (6x ^ 2Y + 4x ^ 3), where x = 1 / 3, y = - 2. Student a regards y = - 2 as y = 2, but the calculation result is still correct. What do you think of this?
(2x^3+3x^2y-9xy^2)-0.5(6x^2y+4x^3)
=2x^3+3x^2y-9xy^2-3x^2y-2x^3
=-9xy^2
When y = - 2 and y = 2, y ^ 2 is 4
So when y = 2 and y = - 2, the result is the same
So it's still true
If the images of linear functions y = 2 / 3x + 3 and y = - 1 / 2x + 9 pass through point a (m, 0), and intersect with y axis at points B and C respectively, try to find the area of triangle ABC
Let x = 0, Y1 = 3, y2 = 9
Area = | y1-y2 | * m / 2 = 3M
I'm sorry. I forgot to beg
Two equations can be directly substituted into (m, 0) to get M
Calculation: (1) 3x ^ 2Y ^ 4 / 8Z ^ 3 * 10z62 / - 6x ^ 2Y ^ 2 (2) 4x ^ 2-y ^ 2 / 3x ^ 2Y △ 2x-y / XY
（1）3x^2y^4/(8z^3)*10z^2/(-6x^2y^2)=-(3/8*10/6)*x^2y^4/z^3*z^2/(x^2y^2)=-5/8*y²/z=-(5y²)/(8z) (2)(4x^2-y^2)/(3x^2y)÷(2x-y)/(xy)=(2x+y)(2x-y)/(3x²y)×xy/(2x-y)=(2x+y)/(3x)
As shown in the figure, if the image of the first-order function y = ax + 2 and y = KX + B intersects at point a (2,1), then the solution of the system of first-order equations with respect to x, y two variables (1) ax-y = - 2 is, related
② kx－y= -b
The solution set of inequality ax + 2 < KX + B in X is
Ax-y = - 2 can be changed into y = ax + 2, kx-y = - B can be changed into y = KX + B, which is the expression of two first-order functions, so the solution of binary first-order equation is the intersection of two functions (2,1)
In the second problem, when ax + 21 / 2, the solution set is (2, positive infinity)
A ^ 6x = 3, a ^ 2Y = 2, find the value of a ^ 3x, a ^ 2x + y
a^6x=(a^3x)²=3
∴a^3x=±√3
a^2x+y
=a^2x*a^y
=³√a^6x*²√a^2y
=³√3*²√2
∵a^6x=3，a^2y=2，
∴a^3x=±√3
a^y=±√2
a^(2x+y)
=a^2x*a^y
=³√3(±√2)
=±³√3√2
Let P (- 2,2) be the intersection of a positive scale function and a linear function
And the ordinate of the intersection Q of the Y-axis of the graph of the first-order function is 4
1. Find out the analytic expressions of the two functions;
2. Find the area of triangle pqo
If y = KX. 2 = k * (- 2), then k = - 1
The linear function (- x) / x + 2 = (4-y) / (Y-2) is sorted out as y = x + 4
The area of triangle pqo is 2 * 4 / 2 = 4
Solve the following inequality: ① 10-4 (x-4) is less than or equal to 2 (x-1) ② x-3 of two
1）10-4(x-4)≤2(x-1)
Remove the bracket, get: 10-4x + 16 ≤ 2x-2
Transfer term: - 4x-2x ≤ - 2-10-16
By combining the similar terms: - 6x ≤ - 28 is obtained
When the coefficient is changed to 1, X ≥ 14 / 3 is obtained
2）(x-3)/2
10-4(x-4)≤2(x-1)
10-4x+16≤2x-2
4x+2x≥10+16+2
6x≥28
x≥14/3
(x-3)/2
Help me summarize the image properties of linear function and positive proportion function
The image of the first-order function f (x) = KX + B is a straight line, while the image of the positive proportion function f (x) = KX is a special first-order function, so the image of the first-order function is a special straight line passing through the origin;
(1) X-2 ＞ 1; (2) 2x-1 ＜ 1; (3) 3x ≥ 6; (4) - 2x ≥ 2
[find the solution set of the following inequality]
（1）x-2＞1；（2）2x-1＜1；（3）3x≥6；（4）-2x≥2；
(5) X ≤ 0; (6) x ＞ - 2.5; (7) x ＜ 2 / 3; (8) x ＞ 4
Solution set of inequality
（1）x-2＞1 →{x|x＞3}
（2）2x-1＜1；→{x|x＜1}
（3）3x≥6； → {x|x≥2}
（4）-2x≥2；→{x|x≥－1}
（5）x≤0；→{x|x≤0}
（6）x＞-2.5；→{x|x＞－2.5}
(7) X ＜ 2 / 3; → {x | x ＜ 2 / 3}
（8）x＞4 → {x|x＞4}