Решите уравнения первым первый уравнение 8*(x+2)=32+4*(x+2) 15*(a-4)=42+8*(a-4) 8x+8*2=32+4x+4*2 (y+1)*18=36+12*(y+1) - id249951520220307 от Andrey7745 07.03.2022 19:58

Question

Математика: Решите уравнения первым первый уравнение 8*(x+2)=32+4*(x+2) 15*(a-4)=42+8*(a-4) 8x+8*2=32+4x+4*2 (y+1)*18=36+12*(y+1) 8x+16=32+4x+8 (k+5)*7-54=(k+5)*4 8x+16=40+4x 17*(c-2)-36=19+(c-2)*12 8x+16-4x=40+4x-4x 4x+16=40

Andrey7745 · Accepted Answer

Пусть m - это средняя линия трапеции.
1) AD^2 = (1 - (-1))^2 + (6 - 2)^2 = 2^2 + 4^2 = 4 + 16 = 20
AD = 2*sqrt(5)
BC^2 = (2 - 1)^2 + (0 - (-2))^2 = 1^2 + 2^2 = 1 + 4 = 5
BC = sqrt(5)
m = (AD + BC)/2 = (2sqrt(5) + sqrt(5))/2 = 3sqrt(5)/2

2) AD^2 = (4 - 4)^2 + (6 - 0)^2 = 0^2 + 6^2 = 36
AD = 6
BC^2 = (-1 - (-1))^2 + (3 - 0)^2 = 0^2 + 3^2 = 9
BC = 3
m = (AD + BC)/2 = (6 + 3)/2 = 9/2

3) AD^2 = (-3-(-2))^2 + (1 - (-2))^2 = (-1)^2 + 3^2 = 1 + 9 = 10
AD = sqrt(10)
BC^2 = (2,5 - 3)^2 + (2,5 - 1)^2 = 0,5^2 + 1,5^2 = 0,25 + 2,25 = 2,5 = 10/4
BC = sqrt(10) / 2
m = (AD + BC)/2 = (sqrt(10) + sqrt(10)/2) /2 = 3sqrt(10)/4

4) AD^2 = (7 - (-3))^2 + (7 - 1)^2 = 10^2 + 6^2 = 136
AD = 2sqrt(34)
BC^2 = (3-(-2))^2 + (1 - (-2))^2 = 5^2 + 3^2 = 25 + 9 = 34
BC = sqrt(34)
m = (AD + BC)/2 = 3sqrt(34)/2

sqrt - это корень.

Решите уравнения первым первый уравнение 8(x+2)=32+4(x+2) 15(a-4)=42+8(a-4) 8x+82=32+4x+42 (y+1)18=36+12(y+1) 8x+16=32+4x+8 (k+5)7-54=(k+5)4 8x+16=40+4x 17(c-2)-36=19+(c-2)12 8x+16-4x=40+4x-4x 4x+16=40

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