Find the triangle, circle. The number represented by the square? Triangle + square = 63, triangle + circle = 46, circle + square = 91. Excuse me, circle =? Triangle =? Square =?

Find the triangle, circle. The number represented by the square? Triangle + square = 63, triangle + circle = 46, circle + square = 91. Excuse me, circle =? Triangle =? Square =?

method:
Because triangle + square = 63
Triangle + circle = 46
Circle + square = 91 ③
The three can be obtained by adding ① + ② + ③
2 * (triangle + square + circle) = 200
therefore
Triangle + square + circle = 100 ④
Substitute (1) for (4)
Get the circle = 37
Substitute (2) for (4)
Get block = 54
Substitute (3) for (4)
We get triangle = 9
I don't know. Do you understand?
Triangle 9 square 54 garden 37
Triangle 9 square 54 circle 37
37;9;54
Triangle + square = 63, triangle + circle = 46, circle + square = 91
○+▲=46
○+□=91
▲+□=63
Then □ = 54
▲=9
○=37
Circle = 37 square = 54 triangle = 9
Let f (x) be an odd function with period 3 defined on R, and f (- 1) = 2, then f (2012) + F (2011)=
If f (x) is a function with period 3, then f (x) = f (x + 3n) (n is an integer)
So: F (2012) + F (2011) = f (- 1 + 671 * 3) + F (1 + 670 * 3) = f (- 1) + F (1)
Because f (x) is an odd function, then: F (x) = f (- x)
So f (1) = - f (- 1) = - 2
That is: F (2012) + F (2011) = f (- 1) + F (1) = 0
2012=670×3+2，f(2102)=f(2)=f(-1)
2011=670×3+1, f(2011)=f(1)
So f (2012) + F (2011) = 0
Circle plus square = 91 triangle plus square = 63 circle plus triangle = 46 circle =? Triangle =? Square =?
The clever man helped me to calculate (thank you)
Isn't it to solve the ternary equation?
Circle + square = 91 triangle + square = 63 circle + triangle = 46
Circle + square + triangle = (91 + 63 + 46) / 2 = 100
Circle = circle + square + triangle - (triangle + square) 100-63 = 37
Square = circle + square + triangle - (circle + triangle) 100-46 = 54
Triangle = circle + square + triangle - (circle + square) 100-91 = 9
Circle + square + triangle = (91 + 63 + 46) / 2 = 100
Circle = 100-63 = 37
Square = 100-46 = 54
Triangle = 100-91 = 9
Circle = 37
Square = 54
Triangle = 9
Let f (x) be an odd function with a period of 3 and f (- 1) = - 1, then f (2012)=
The periodic function is x (3)
∵f(2012)=f(3*671-1)=f(-1)=-1
Because the period of F (x) is three, f (2012) = f (2009) = f (2006) =... = f (2) = f (- 1)
So f (2012) = - 1
Note: F (x + T) = f (x)
F (x) is a periodic function with a period of 3
∵f(2012)=f(3*671-1)=f(-1)=-1
Given that triangle divided by square equals 2, circle divided by triangle equals 4, circle minus square equals 56, then triangle () circle () square ()
Come on, I can't
Triangle (16), circle (64), square (8),
If for any x ∈ R, f (x) satisfies f (x + 2012) = - f (x + 2011), and f (2012) = - 2012, then f (- 1) = ()
A. 1B. -1C. 2012D. -2012
Let f = (2012 + F-2) = (2012 + F-1) = f (2012 + F-1) = f (2012 + F-1) = f (2012 + F-2) = F-2
Circle + triangle + square = 55, triangle + square + star = 70, circle + triangle + star = 60, circle + square + star = 55,
Square, star
Circle + triangle + square = 55,
Triangle + square + star = 70,
Circle + triangle + star = 60,
Circle + square + star = 55
The sum of the four formulas is as follows:
3 circle + 3 triangle + 3 square + 3 star = 55 + 70 + 60 + 55 = 240
Then, divide both sides by 3 to get: circle + triangle + square + star = 80
So, star = 80-55 = 25
Circle = 80-70 = 10
Square = 80-60 = 20
Triangle = 80-55 = 25
Circle + triangle + square = 55
Triangle + square + star = 70
Circle + triangle + star = 60
Circle + square + star = 55
Add all the four formulas to get
Circle + square + star + triangle = 80
Circle + square + star = 55
Circle + triangle + star = 60 (compared with circle + square + star + triangle = 80, square = 80-60 = 20)
Triangle + square + star = 70, so star = 70-25-20 = 2... Expand
Circle + triangle + square = 55
Triangle + square + star = 70
Circle + triangle + star = 60
Circle + square + star = 55
Add all the four formulas to get
Circle + square + star + triangle = 80
Circle + square + star = 55
Circle + triangle + star = 60 (compared with circle + square + star + triangle = 80, square = 80-60 = 20)
Triangle + square + star = 70, so star = 70-25-20 = 25
Circle + triangle + star = 60, so circle = 60-25-25 = 10
therefore
Circle = 10
Triangle = 25
Square = 20
Star = 25 ° up
A circle equals 10 and a square equals 20
Given that the function f (x) defined on R is an odd function, and the period of function f (2x + 1) is 2, f (1) = 5, then f (2011) + F (2012)=
Odd function → f (- 1) = - f (1) = - 5, f (0) = 0
F (2x + 1) period is 2 → f (x) period is 4
So f (2011) = f (4 × 503-1) = f (- 1) = - 5
f(2012)=f(4×503)=f(0)=0
f(2011)+f(2012) = -5
Ten
Two triangles = 3 squares, three squares = 4 circles, one triangle + one square + 2 circles = 400, then how many triangles are there? Squares
2 triangles = 3 squares
3 squares = 4 circles
Triangle + square + 2 circle = 400
Triangle = 1.5 square
Circle = 0.75 square
Triangle + square + 2 circle = 1.5 square + square + 2 * 0.75 square = 400
4 blocks = 400
Block = 100
Triangle = 1.5 square = 1.5 * 100 = 150
Circle = 0.75 square = 0.75 * 100 = 75
The known function f (x) = [2Sin (x-wu / 3) + 3sinx] × cosx + √ 3sinx ∧ 2 (x ∈ R)
Find the maximum and minimum value of function f (x) on [0, Wu / 2], (2) in the acute angle △ ABC, f (a) = √ 3, a = √ 7, find the area of triangle ABC^
F (x) = [2Sin (x-wu / 3) + 3sinx] × cosx ＋ 3sinx ∧ 2
=[sinx-√3cosx+3sinx]]cosx+√3sin²x
=4sinccosx-√3cos²x+√3cos²x
=2sin2x
∵ x ∈ [0, Wu / 2]
∴2x∈[0,π]
When 2x = π / 2, i.e. x = π / 4, f (x) reaches the maximum value of 2
When 2x = 0 or π, i.e. x = 0 or x = π / 2, the minimum value of F (x) is 0
Two
f(A)=2sin2A=√3
∴sin2A=√3/2
Ψ 2A = 2 π / 3 or π / 3
A = π / 3 or a = π / 6
∵a=√7
∴a²=b²+c²-2bccosA
∴7+2bccosA=b²+c²≥2bc
When a = π / 3, 7 + BC = B & # 178; + C & # 178; ≥ 2BC, BC ≤ 7
SΔ=1/2bcsinA≤7√3/4
The maximum area of triangle ABC is 7 √ 3 / 4
When a = π / 6, 7 + √ 3bC ≥ BC, BC has no maximum value and area has no maximum value