Boyları 120 cm, 135 cm ve 180 cm olan üç demir çubuk eşit büyüklükte olmak üzere her biri ayrı ayrı parçalara ayrılacaktır. Bu iş için en az kaç kesim yapılması gerektiğini bulunuz.

1)1.7-14=-7                                                                                                                                                         2.-28:-7=4                                                                                                                                                            3.4-10=-6                                                                                                                                                            задача а: 36*2/9=8 лет                                                                                                                                    задача б: 5:4*100=125%                                                                                                                                задача в 40%=20 слов                           40%=0,4                                                                                         20:0,4=50 слов                                                                                                                                                  10(1-0,2у)-(1-у)=17                                                                                                                                            10-2у-1+у=17                                                                                                                                                      9-у=17                                                                                                                                                                   у=9-17=-8                                                                                                                                                            щас остальные ответы напишу                                                                                                               задача:60 + 8=68км/час - второй поезд                                                                                               24*1/3=8 часов-ехали поезда                                                                                                                 60*8+68*8=1004 км                                                                                                                                   задача:72-12=60м-стала длина забора                                                                                              26-7=19м-стала ширина забора                                                                                                            60*19=1140 м2

Пусть Х---скорость первого байкера

тогда Х-16---скорость второго байкера на первой половине пути

путь они проехали один и тот же S

время потратили одинаковое:

время первого байкера S/X

время второго байкера S/(2*(X-16)) + S/(2*120)

получим уравнение

S/X = S/(2*(X-16)) + S/(2*120)

1/X = 1/(2*(X-16)) + 1/(2*120)

2/X - 1/(X-16) = 1/120

(2*(X-16) - Х) / (Х*(X-16)) = 1/120

120*(X-32) = Х*(X-16)

120*X - 120*32 - Х*X + 16*Х = 0

Х*X - 136*X + 120*32= 0

D = 136*136 - 4*120*32 = 8*17*8*17 - 4*2*60*8*4 = 8*8*(17*17 - 60*4) = 8*8*(289-240) = 8*8*49

X1 = (136 + 8*7)/2 = 68 + 4*7 = 96

X2 = (136 - 8*7)/2 = 68 - 4*7 = 40---не удовлетворяет условию

ответ: скорость первого байкера 96 км/час

ПРОВЕРКА:

время в пути первого байкера S/96

время в пути второго байкера S/2 : (96-16) + S/2 : 120 = S/(2*80) + S/(2*120) = S/160 + S/240 = (3S+2S)/480 = 5S/480 = S/96

т.е. они доехали одновременно