A) 1. we expect to hear — him — the end — the month when the ship has reached kamchatka. 2. have you heard — the latest discovery made — the physics laboratory? 3. why take his words seriously? i am sure he said it all — fun. 4. if you are not sure — the spelling, look the word — — the dictionary.
5. i am — — him, i've had enough of his empty promises. 6. the trouble—that fellow is that he is interested — many things but good — nothing. 7. i don't advise you to mention the news — her, it will be all — the place — once. 8. why remind him — something he is trying so hard to forget? 9. they
hadn't been warned — the change and missed the first report. 10. he had missed a lot of lectures — the beginning — term and was working hard now to catch — — the rest—the group. 11. ann didn't stay — the end but rose to go long before the picture was —. she just couldn't stand it any more: — her
right sat a man who had been smoking all the time, — her left was a woman who had slept — the greater part — the film. 12. i am not sure i'll ever get used — his strange ways, but working-— him is a pleasure. 13. we'll be looking you — one of these days. 14. she didn't see how greatly disappointed
we were — her work. 15. it's impossible to keep all these facts — the head, i must make some notes. 16. as — grace, her parents didn't want her to take part — their long trip because that would mean missing a semester — college and be — the other students when she returned. 17. — first i called him
mr blake, but very soon he said: "leave — the 'mister', just call me plain blake." 18. as — the price he mentioned, no good businessman could make such an offer.
Плоскость АВ₁С₁ - это плоскость АВ₁С₁D
По теореме Пифагора DC₁²=6²+8²=100
DC₁=10
РК- средняя линия треугольника DCC₁
PK=5
PT|| AD и PT || ВС
РТ=4
AD⊥CD ⇒ РТ⊥СD
AD⊥DD₁ ⇒ РТ⊥ DD₁
РТ перпендикулярна двум пересекающимся прямым плоскости DD₁C₁C, значит перпендикулярна любой прямой лежащей в этой плоскости, в том числе прямой РК
РТ⊥ РК
Аналогично, МТ ⊥МК
Сечение представляет собой прямоугольник
Р(cечения)=Р( прямоугольника ТМКР)=2·(4+5)=18
ММ₁К₁К - трапеция
СС₁- средняя линия трапеции
СС₁=(ММ₁+КК₁)/2=(16+6)/2=11
2) Точка M имеет абсциссу х=√(12) =2√3 ординату у=0
Точка К имеет асбциссу х=-2 ордината у находится из уравнения
у²=12-4
у=√8
у=2√2
точка O (0;0)
ОМ имеет длину 2√3
ОМ- радиус вектор
ОМ=2√3
ОМ=ОК=2√3
tg∠КОМ=-√2 ( так как тангенс смежного с ним угла α равен √2 tg α=2√2/2=√2)
cos²∠КОМ= 1/(1+tg²∠KOM)=1/3
sin²∠КОМ=1-cos²∠KOM=1-(1/3)=2/3
sin ∠KOM=√(2/3)
S=ОК·ОМ· sin ∠KOM/2= (2√3)²·(√(2/3))/2=2√6 кв. ед