Triangle minus circle = 40, triangle = circle + circle + circle + circle + circle?

Triangle minus circle = 40, triangle = circle + circle + circle + circle + circle?

Triangle = 5 circles
Triangle rounding = 40
4 circle = 40
Circle = 10
Triangle = 50

Triangle circle when there are several triangles, there are 18 circles

18 △ 3 * 2 = 12

Circle + circle + triangle - Triangle = 28 circle + circle + triangle + triangle = 37 triangle = () circle = ()

Circle + circle + triangle - Triangle = 28
Circle + circle + triangle + triangle + triangle = 37
Triangle = (3)
Circle = (14)
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If three circles + one triangle = 3.6, three triangles + one circle = 4.4, then triangle =? Circle =?

Add two
4 circles + 4 triangles = 3.6 + 4.4 = 8
1 circle + 1 triangle = 2
2 circles = 3.6-2 = 1.8
1 circle = 0.9
1 triangle = 2-0.9 = 1.1

Triangle circle = 3 4 * (triangle + circle) = 100 triangle =? Circle =?
Find triangle and circle
Be sure to give me the answer before Qingming Festival in 2011!

Triangle = 14
Circle = 11

Square + triangle + triangle = 23 square + triangle + triangle + triangle = 29 triangle = () square = ()

Triangle = (3) square = (17)

Triangle + triangle + triangle = circle + circle; circle + circle + circle = box + box

Triangle + triangle + triangle = circle + circle
Triangle: circle = 2:3
Circle + circle + circle + circle = box + box
Circle: box = 1:2
Triangle: Circle: box = 2:3:6

Triangle plus square equals 240, triangle equals square plus square plus square, how much is triangle and square?

Triangle + square = 240 -- 1 triangle = 3 square ----- 2 from 1: Triangle = 240 square into 2: 240 square = 3 square 4 square = 240 square = 60, then triangle = 240-60 = 180

The sum of the first n terms of an is Sn, and the common ratio is Q. if S3 is the median term of S1 and S2, A1-A3 = 3, find the sum of Q and S5

a1-a1q^2=3,a1(1-q^2)=3 (1)
2*s3=s1+s2,2* [ a1(1-q^3)/(1-q)]=a1+a1(1-q^2/(1-q)) ,(2)
From (2), 2q ^ 2-q-1 = 0 so q = - 1 / 2 or q = - 1 (rounding)
Substituting (1) gives A1 = 4
s5=a1(1-q^5)/(1-q)=11/4

1. Given that the equation x & sup2; + 2mx + 2m-2 = 0 has a negative root greater than - 2 and a positive root less than 2, the value range of M is obtained
2. If the root of the equation (m-2x) = - X has two unequal real roots, then the value range of real number m is (- 1 < m ≤ 0). I want to ask, why should m be less than or equal to 0?
3. Solving the equations {X & sup2; - 2xy-y & sup2; = 1,2x & sup2; - 5xy-3y & sup2; = 0} (no mistakes in the title)

1. Let f (x) = x2 + 2mx + 2m-2, and the opening of the function be upward. Because the negative root is greater than - 2, and a positive root is less than 2, f (2) > 0, f (- 2) 0, and then solve three inequalities~
2 because the root sign (m-2x) = - x, so - x > = 0, then x