-89x-x+41x+128= -58x-162x-29x+x= -5a+8(1+3a)= -9c-7(1-2c)= 3b-6(2-4b)= -6(3+x)+8(2-x)=. Created by pixelup228. matematika-ru

-89x-x+41x+128= -58x-162x-29x+x= -5a+8(1+3a)= -9c-7(1-2c)= 3b-6(2-4b)= -6(3+x)+8...

June 2022 2 5 Report

-89x-x+41x+128=
-58x-162x-29x+x=
-5a+8(1+3a)=
-9c-7(1-2c)=
3b-6(2-4b)=
-6(3+x)+8(2-x)=

koingrel

Verified answer

-89x-x+41x+128 = -49х + 128 = 128 - 49х
-58x-162x-29x+x = -249х + х = -248х
-5a+8(1+3a) = -5а + 8 + 24а = 8 + 19а
-9c-7(1-2c) = -9с - 7 + 14с = 5с - 7
3b-6(2-4b) = 3b - 12 + 24b = 27b - 12
-6(3+x)+8(2-x) = -18 - 6х + 16 - 8х = -2 - 14х

0 votes Thanks 1

bakovalada -89х-х+41х+128=-49х+128
-58х-162х-29х+х=-220-29х+х=-248х
-5а+8(1+3а)=-5а+8+24а=32а +8
-9с-7(1-2с)=-9с-7+14с=5с-7
3б-6(2-4б)=3б-12+24б=27б-12
-6(3+х)+8(2-х)=-18-6х+16-8х=-14х-2

0 votes Thanks 0

Answers & Comments

Verified answer

najdite znachenie vyrazheniyaa 34x 57h 66x 47x pri x 36 08 10b

10 3) -5y+7=42 4) 8y-(7y-142)=51 5) (6y+1)-(3-2y)=14...

opredelite massu serebryanoj lozhki obyomom 14sm otvet vyrazit v edinicah siste

imena geroev multika shkola horoshih maner nuzhno 20 ballov dayue08cacec966b4d22a0517558d213b854 34441

imena geroev multika shkola horoshih maner nuzhno 20 ballov dayu

рекомендуемые вопросы

Helpful Links

Helpful Social