1.Употребите глаголы во второй форме (Past SimpleMeet go draw have help ask like be visit2.Выбери правильную форму глагола was или were.The weather___stormy and cloudy.There___two chairs in the bedroom.His eyes big and kindMary slim three days ago.3.Напишите отрицательные предложенияWe visited Moscow last summer.Mary asked Dan about this story. The girl looked sad and angry. 4. Напишите сравнительную и превосходную степени прилагательных:Strong Big Good NiceHappy 5. Раскрой скобки. Поставь прилагательные в сравнительную или превосходную степень:The ocean is (large) than the sea.The sea is (big) than the river.The houses in town are (tall) than the houses in country.​

1. met, went, drew, helped, asked, liked, was/were, visited

2. was; were; were; was

3.

We didn't visit Moscow last summer

Mary didn't ask Dan about this story

The girl didn't look sad and angry

4.

Strong-stronger-the strongest

Big-bigger-the biggest

Good-better-the best

Nice-nicer-the nicest

Happy-happier-the happiest

5. larger; bigger; taller

а) предложения, в которых ing-форма является подлежащим

The driving wheel of the machine is broken.

These happenings are remarkable.

Travelling is a pleasant way of improving one’s education.

Asking him for help is useless.

в) предложения, в которых ing-форма является обстоятельством

Every trust arranges for the marketing of its products.

It is no use crying over spilt milk.

We have every chance of passing our examinations well.

You don’t know what you miss, not having the desire to listen to good music.

с) предложения, в которых ing-форма является частью сказуемого.

Happily we escaped being delayed on our way.

Having been knocked down by a passing car, the poor man was at once taken to hospital.

I was told of a great friendship existing between two captains.

Driving in a motor car, we passed many villages.

Объяснение:

Сначала формулы . затем формулы двойного угла sin2α=2sinαcosα cos2α=cos²α-sin²α=cos²α-(1-cos²α)=2cos²α-1⇒  2cos²α=1+cos2α cos2α=cos²α-sin²α=1-sin²α-sin²α=1-2sin²α⇒  2sin²α=1-cos2α а) сos75°cos105°=cos(90°-15°)·cos(90°+15°)== sin15°(-sin15°)=-sin²15°=-(1-cos30°)/2=(cos30°-1)/2= ((√3/2)-1)/2=0,25√3-0,5б) sin75°sin15°=°sin(90°-15°)sin15°=cos15° sin15°=sin30°/2=1/4=0,25в) sin105°cos15°=sin(180°-75°)cos15°=sin75°cos15°=sin(90°-15°)cos15°=cos15°cos15°=(1+cos30°)/2=(1+(√3/2))/2=0,5 +0,25√3 2 способ формулы cosα·cosβ=0,5cos(α-β)+0,5cos(α+β) sinα·sinβ=0,5cos(α-β)-0,5cos(α+β) sinα·cosβ=0,5sin(α+β)+0,5sin(α-β) а) сos75°cos105°=0,5cos(75°-105°)+0,5cos(75°+105°)=0,5cos(-30°)+0,5 cos180°==0,5· √3/2+0,5·(-1)=0,25√3-0,5б) sin75°sin15°=0,5cos(75°-15°)-0,5cos(75°+15° )=0,5cos60°-0,5 cos90°=0,5·0,5=0,25в) sin105°cos15°=0,5sin(105°+15°)+0,5sin(105°-15°)= =0,5sin120°+0,5sin90°==0,5 sin(180°-60°)+0,5·1=0,5 sin 60°+0,5=0,25√3+0,5