Let {an} be an arithmetic sequence with a tolerance of - 2 if a1 + A4 + A7 + +A97 = 50, then A3 + A6 + A9 + +A99 equals () A. 82B. -82C. 132D. -132

Let {an} be an arithmetic sequence with a tolerance of - 2 if a1 + A4 + A7 + +A97 = 50, then A3 + A6 + A9 + +A99 equals () A. 82B. -82C. 132D. -132

Because {an} is an arithmetic sequence with tolerance of - 2, A3 + A6 + A9 + + A99 = (a1 + 2D) + (A4 + 2D) + (A7 + 2D) + +(A97 + 2D) = a1 + A4 + A7 + + A97 + 33 × 2D = 50-132 = - 82

The {an} is an arithmetic sequence, d = - 2, a1 + A4 + A7 + +A7 = 50, then A3 + A6 + A9 + a99=

a1+a4+a7+…… +A97 = 50a3 + A6 + A9 +. + A99 is different from each known term. 2da3 = a1 + 2da6 = A4 + 2da9 = A7 + 2D... A99 = A97 + 2D ｜ = a1 + 2D + A4 + 2D + A7 + 2D +. A97 + 2D = (a1 + A4 + A7 +... + A97) + n * 2DN is the number of terms in this column. Let's see a1 + A4 + A7 +. A97 is equivariant

It is known that the image of a function of degree passes through the point m (- 3,2) and is parallel to the straight line y = 4x-1. (1) find the analytic expression of the function; (2) find the function graph
The area of a triangle enclosed by an image

(1) Because the image of a function is parallel to the straight line y = 4x-1, we can set the analytic expression of a function as y = 4x + B. substituting x = - 3 and y = 2 into the analytic expression, we can get 2 = - 12 + B and B = 14. So the analytic expression of this function is y = 4x + 14 (2). This problem should be the area of the triangle formed by the image of a function and the coordinate axis

The image of a given function is parallel to the straight line y = - 3 / 4x + 2 and passes through the point (- 1,1)
(1) Find the analytic expression of this function
(2) Find the area of the triangle formed by the function image and two coordinate axes

(1) If the image of a linear function is parallel to the straight line y = - 3 / 4x + 2 and the analytic expression of the linear function y = - 3 / 4x + B passes through (- 1,1), then: 1 = - 3 / 4 * (- 1) + B B = 1 / 4 the analytic expression of the linear function y = - 3 / 4x + 1 / 4 (2) the intersection point of the linear function y = - 3 / 4x + 1 / 4 on the X axis is (1 / 3,0)

The image of a linear function passes through - 3,2 and intersects with the line y = 4x + 6 at a point on the x-axis,
This kind of problem tells us how to calculate the analytic formula when a function passes through a point? What about the two points?

y=4x+6
Let y = 0
x=-3/2
Then the straight line passes through (- 3 / 2,0) and (- 3,2)
Let x = KY-3 / 2 be substituted into (- 3,2)
-3=2k-3/2
2k=-3/2
k=-3/4
x=-3/4y-3/2
That is y = - 4 / 3x-2

It is known that the image of a function of degree is parallel to the image of y = - 1 / 2x and intersects with the Y axis (0,3)

y=-1/2x + 3
Because the two image lines have the same slope, k = - 1 / 2x
And because the intercept of the image on the Y axis is 3, B = 3
Therefore, the analytic expression of the function is y = - 1 / 2x + 3

Given that the graph of a function of degree passes through two points a (2, - 1) and B, where B is the intersection of y = - 1 / 2x + 3 and Y axis, the analytic expression of the function of degree is obtained,

Because point B is the intersection of the function y = - 1 / 2x + 3 and the Y axis, let x = 0 and y = 3, that is, B (0,3)
Let the analytic expression of this function be y = KX + B. the image of this function is known to pass through two points a (2, - 1) and B (0,3)
2k+b=-1,0+b=3
The solution is k = - 2, B = 3
So the analytic expression of this linear function is y = - 2x + 3

The image of a function of degree passes through the intersection of y = 2x-3 and y = - x + 3, and the coordinate of the intersection with X axis is - 2,0. When x takes what value, Y > 0

First order function over (2,1) (- 2,0)
y=x/4+1/2>0
x>-2

The intersection points of the image and x-axis of the first-order function are a and B respectively. The abscissa of the intersection point of the image of the first-order function y = 2x + 1 is 2, and the ordinate of the intersection point with the function y = - x + 2 is 1. Find the analytic expression of the first-order function and the area of ⊿ AOB

y=4x-3,9/8

The image of a function of degree is known to pass through points (2,1) and (- 1, - 3). (1) find the analytic expression of the function of degree; (2) find the image of the function of degree and two points

y=kx+b
Then 1 = 2K + B
-3=-k+b
subtract
3k=4
k=4/3
b=k-3=-5/3
So y = 4x / 3-5 / 3
'
x=0,y=-5/3
y=0,x=5/4
So the area of the triangle enclosed by the coordinate axis = | - 5 / 3 × 5 / 4 | - 2 = 25 / 48